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https://www.physicsforums.com/threads/solving-the-schroedinger-equation-for-free-electrons.935906/ | Tags:
1. Jan 2, 2018
### SeM
Dear all, sorry I made a new post similar to the previous post "Initial conditions..", however, a critical point was missed in the previous discussion:
The initial conditions y(0)=1 and y'(0)=0 are fine and help in solving the Schrödinger equation, however, studying free electrons, the equation cannot be solved neither for the particle in the box , or for a finite linear motion harmonic oscillator as far as I can see from the literature.
When I solve the eqn, HY=EY , I can set E= p/2m or E= ground state energy. In either cases, I get a result where some unknown constant appears.
This constant can be solved using the Born Sommerfeld condition, but because the solution is not square integrable, it gives complex values.
How can I solve the unknown constant for a solution of the Schrödinger eqn. where the above given initial conditions are used?
Thanks!
2. Jan 2, 2018
### vanhees71
I'm a bit insecure what you mean with your "initial conditions" and most of the rest of your posting, but let's briefly discuss free non-relativistic electrons. The Hamiltonian reads
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m}.$$
Working in the position representation ("wave mechanics") we have
$$\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}.$$
For simplicity also let's neglect the spin of the electron. Then to get a complete set of energy eigenvectors it's clear that we can as well look for a complete set of common eigenvectors of $\hat{\vec{p}}$ (note that $[\hat{p}_j,\hat{p}_k]=0$, and that thus there's a set of common eigenvectors of all three momentum components). To find these eigenvectors we have to solve the eigenvalue equation
$$\hat{\vec{p}}u_{\vec{p}}(\vec{x})=-\mathrm{i} \hbar \vec{\nabla} u_{\vec{p}}(\vec{x}) =\vec{p} u_{\vec{p}}(\vec{x}).$$
It's easy to see that the solution is
$$u_{\vec{p}}(\vec{x})=N(\vec{p}) \exp \left (\frac{\mathrm{i} \vec{p} \cdot \vec{x}}{\hbar} \right), \quad \vec{p} \in \mathbb{R}^3.$$
It's convenient to normalize the eigenvectors "to a $\delta$ distribution":
$$\langle u_{\vec{p}} |u_{\vec{p}'} \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} N^*(\vec{p}) N(\vec{p}') \exp \left [\frac{\mathrm{i} \vec{x} \cdot (\vec{p}'-\vec{p})}{\hbar} \right] =|N(\vec{p})|^2 (2 \pi)^3 \hbar^3 \delta(\vec{p}-\vec{p}') \; \Rightarrow \; N(\vec{p})=\frac{1}{(2 \pi \hbar)^{3/2}}.$$
So finally we have
$$u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi \hbar)^{3/2}} \exp \left (\frac{\mathrm{i} \vec{p} \cdot \vec{x}}{\hbar} \right).$$
Of course, each $u_{\vec{p}}$ is an energy eigenstate,
$$\hat{H} u_{\vec{p}}(\vec{x})=\frac{\vec{p}^2}{2m} u_{\vec{p}}(\vec{x})=E_{\vec{p}} u_{\vec{p}}(\vec{x}).$$
The initial value is the wave function at time $t_0$ (not some strange other "initial values" you quote in the OP). The general solution of the initial value problem is now easily given in terms of the just found energy eigenfunctions. Defining
$$\tilde{\psi}_0(\vec{p})=\langle u_{\vec{p}}|\psi_0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \langle u_{\vec{p}}|\vec{x} \rangle \langle \vec{x}|\psi_0 \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}}^*(\vec{x}) \psi_0(\vec{x}),$$
you get
$$\psi(t,\vec{x})=\langle \vec{x} | \exp[-\mathrm{i} \hat{H}(t-t_0)/\hbar]|\psi_0 \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \langle \vec{x}|\exp[-\mathrm{i} \hat{H}(t-t_0)/\hbar] u_{\vec{p}} \rangle \langle u_{\vec{p}}|\psi_0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} u_{\vec{p}}(\vec{x}) \exp[-\mathrm{i} E_{\vec{p}}(t-t_0)/\hbar] \tilde{\psi}_0(\vec{p}).$$
3. Jan 2, 2018 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.976318359375, "perplexity": 443.3190017365884}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583513760.4/warc/CC-MAIN-20181021052235-20181021073735-00461.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/119426-prove-s-group-print.html | # Prove S is a Group
• December 8th 2009, 05:59 PM
mesmo
Prove S is a Group
Hi again.
Well, I have another question.
Suppose A is a ring with unity denoted by 1. Let S be the set of elements in A that have multiplicative inverses in A.
I want to prove S is a group under mult. I know the four cases that must be proved. However, I'm having a little difficulty as to how I should prove that these particular things are in the set of elements S.
For example, the last case is to prove that for every element in S there is a multiplicative inverse. However, I am having a hard time thinking as to how I should prove the the inverses reside in S.
If this is vague, I am more than happy to explain what I mean.
• December 9th 2009, 01:00 AM
Swlabr
Quote:
Originally Posted by mesmo
Hi again.
Well, I have another question.
Suppose A is a ring with unity denoted by 1. Let S be the set of elements in A that have multiplicative inverses in A.
I want to prove S is a group under mult. I know the four cases that must be proved. However, I'm having a little difficulty as to how I should prove that these particular things are in the set of elements S.
For example, the last case is to prove that for every element in S there is a multiplicative inverse. However, I am having a hard time thinking as to how I should prove the the inverses reside in S.
If this is vague, I am more than happy to explain what I mean.
You have 4 things to show: associativity, identity, closure, and inverses.
Associativity holds as your multiplication is just the multiplication in the ring, which is associative.
Identity holds as $1 \in S$.
So closure and inverses remain to be proven. However, if we prove closure we have proven inverses (why?).
So, $x \in S$, $y \in S$. This means we can find $x^{\prime}$, $y^{\prime}$ such that $xx^{\prime}=1$ and $yy^{\prime}=1$. Then can you think up an inverse for $xy$?
• December 9th 2009, 04:48 AM
mesmo
Thanks Swlabr....i think i am understanding.
Okay, because this is a ring with unity, it automatically follows that this unity has to be an element of S? (same thing with associative multiplication?) I was thinking along those lines but I kept seeing S as something detached from the ring.
xx' = 1 and yy' = 1. xx' = yy'....? Is this going the write way? Since they are both equal to one is it correct to equate them to each other? however, this is as far as I've gotten. Is closure the same way? Thanks for the help so far.
• December 9th 2009, 04:56 AM
Swlabr
Quote:
Originally Posted by mesmo
Thanks Swlabr....i think i am understanding.
Okay, because this is a ring with unity, it automatically follows that this unity has to be an element of S? (same thing with associative multiplication?) I was thinking along those lines but I kept seeing S as something detached from the ring.
xx' = 1 and yy' = 1. xx' = yy'....? Is this going the write way? Since they are both equal to one is it correct to equate them to each other? however, this is as far as I've gotten. Is closure the same way? Thanks for the help so far.
You want an element $r \in R$ such that $(xy)r=1$. We know there is an element which gives us $xyr_1 = x$. What is this element $r_1$? Can you think of an element $r_1 \in R$ such that $xyr_1r_2=xr_1=1$? Set $r=r_1r_2$ and you are done!
I tend to think of this set " $S$" as everything close to the identity. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9298868775367737, "perplexity": 267.1211284976545}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368701910820/warc/CC-MAIN-20130516105830-00054-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.clutchprep.com/chemistry/practice-problems/65481/please-refer-to-the-charts-a-what-is-the-value-of-the-activation-energy-of-the-u-1 | # Problem: Please refer to the charts.a. What is the value of the activation energy of the uncatalyzed reaction?b. What is the value of the enthalpy change of the uncatalyzed reaction?c. What is the value of the activation energy of the uncatalyzed reaction in reverse?d. What is the value of the enthalpy change of the uncatalyzed reaction in reverse?
###### FREE Expert Solution Show answer
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###### Problem Details
a. What is the value of the activation energy of the uncatalyzed reaction?
b. What is the value of the enthalpy change of the uncatalyzed reaction?
c. What is the value of the activation energy of the uncatalyzed reaction in reverse?
d. What is the value of the enthalpy change of the uncatalyzed reaction in reverse?
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Energy Diagrams concept. You can view video lessons to learn Energy Diagrams Or if you need more Energy Diagrams practice, you can also practice Energy Diagrams practice problems .
What is the difficulty of this problem?
Our tutors rated the difficulty of Please refer to the charts.a. What is the value of the activ... as medium difficulty.
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Our expert Chemistry tutor, Dasha took 6 minutes to solve this problem. You can follow their steps in the video explanation above.
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Based on our data, we think this problem is relevant for Professor Freschl's class at UWM. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8237505555152893, "perplexity": 1021.4218004461713}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00111.warc.gz"} |
https://www.cuemath.com/maths/4-digits-numbers-and-place-values/ | # Numbers up to 4-Digits
Go back to 'Numbers and Place values'
## Introduction to Numbers up to $$4$$-Digits
When we multiply a unit with ten, we get a two digit number. For example, $$2$$ multiplied by $$10$$ gives $$20.$$ For $$3$$ digit numbers it’s \begin{align} 10 \times 10 = 100. \end{align}
What happens when we multiply a unit digit with $$1000?$$ In such a number, the following place values are covered:
Each of the above place values would have a number, and so you would get a number with $$4$$ digits. After $$999,$$ the first $$1$$ digit number begins with $$1000.$$ The next $$9000$$ numbers are all $$4$$ digit numbers, for example $$3469, 9210, 2583$$ etc.
## The Big Idea: Numbers up to $$4$$-Digits
### A simple idea: The Place Value of Numbers
When we covered $$3$$-digit numbers, we saw that the hundreds place value was the units place multiplied by a hundred. Like you’ve already read, for four digit numbers a new place value is introduced which is the units place multiplied by $$1000.$$
It’s not that hard to visualise actually, take any $$3$$ digit and multiply it with $$10,$$ and there you go! A four digit number!
### How To Decompose $$4$$-digit numbers
We already saw that a $$4$$ digit number has the following place values – thousands, hundreds, tens and units. That is why the multiplier to be used for each of the digits of a $$4$$-digit number are $$1000, 100, 10$$ and $$1$$ respectively.
Let us examine the number $$1729.$$ We can decompose the number as
\begin{align} { 1729 = ( 1 \times 1000 ) + ( 7 \times 100 ) + ( 2 \times 10 ) + ( 9 \times 1 ) = 1000 + 700 + 20 + 9 = } { 1729 } \end{align}
Consider the number $$1792, 1279$$ and $$2179$$ as well. And notice the values of the same digit in each of these numbers.
\begin{align} &{ 1729 - 2 \text { has ten's place and a place value of } 20 } \\ &{ 1792 - 2 \text { has unit's place and a place value of } 2 } \\ &{ 1279 - 2 \text { has hundred's place and a place value of } 200 } \\& { 2179 - 2 \text { has thousand's place and a place value of } 2000 } \end{align}
What this tells us is that the value of a digit is not the only thing that determines its place value in a $$4$$-digit number. In $$1729,$$ for instance, the number $$1$$ has a lower value than the number $$2,$$ but in terms of place value, $$1$$ has a place value of $$1000$$ (being the $$4^{\text{th}}$$ digit from right) and $$2$$ has a place value of $$10$$ (being the second digit from right).
You can try some more $$4$$-digit numbers here to see how they can be decomposed into their place values.
### Commas in $$4$$-digit numbers
In every $$4$$-digit number, the $$4^{\text{th}}$$ digit on the extreme left represents the thousand’s place. It is a common convention to apply a comma between the $$4^{\text{th}}$$ and $$3^{\text{rd}}$$ digit (thousand’s place and hundred’s place). So, while the $$3$$-digit number $$374$$ is written simply as $$374,$$ the $$4$$-digit number $$4539$$ is written as $$4,539.$$
There is no mathematical reason for doing this. The $$4$$-digit number is longer than a $$3$$-digit or $$2$$-digit number, and the comma only helps to make the number more readable.
## How is it important?
### The Amazing Kaprekar’s Routine
As per this amazing law, there is a magic number $$6174.$$ Let us see how we get it. These are the steps to be followed.
1. Arrange any $$4$$-digit number in ascending order of digit and descending order of digits.
2. This was you get two distinct $$4$$-digit numbers.
3. Find out the difference between the two numbers.
4. Repeat the above $$3$$ steps on this number.
5. You will reach $$6174$$ in at most $$7$$ iterations.
Doesn’t it sound amazing enough for us to try it out? Ok, so let’s say I choose the number $$2873.$$ The digits arranged in ascending order give us $$2378,$$ and $$8732.$$
What is $$8732 - 2378?$$ The answer is $$6354.$$
Again this gives two numbers $$3456$$ and $$6543.$$
What is $$6543 - 3456?$$ The answer is $$3087.$$
After that, $$8730 - 0378 = 8325.$$
Finally $$8532 - 2358 = 6174!!$$
Whatever $$4$$ digit number you choose, you will reach $$6174$$ within at most $$7$$ iterations.
Now what happens after $$6174$$ is reached? It gives the two numbers $$7641$$ and $$1467,$$ and the difference between these two numbers is $$6174$$ again, so this sequence keeps getting repeated ad infinitum. This interesting phenomenon was first observed and formalized by the Indian mathematician Kaprekar. That is why this sequence is known as the Kaprekar routine.
### An Interesting Story to End
Ramanujam was one the greatest Indian mathematicians. One day he fell ill, and his friend Hardy came to visit him. Hardy decided to tease his mathematician friend by saying that the cab I came in had a very uninteresting number plate – $$1729.$$ Even though he was not keeping well, Ramanujam’s sharp brain was still working, and he immediately retorted that $$1729$$ was actually a very interesting number, because it was the smallest number which could be expressed as the sum of two cubes in two different and unique ways!! If this was how sharp he was when unwell, imagine what his brain was capable of when functioning fully. In honor of this incident, the number $$1729$$ has been referred to as the Hardy – Ramanujam number. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 6, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8739849328994751, "perplexity": 397.8217727046093}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655878639.9/warc/CC-MAIN-20200702080623-20200702110623-00156.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/52486-proofs-invertibility.html | 1. ## Proofs of Invertibility
I'm supposed to prove an invertible, but I'm not sure how, as the prof didn't talk about it in class, and the book is rather vague...
The question is this:
Let u and v be column vectors in R^n, and let A= I + u(v^T). Show that if (u^T)v =/= -1, then A is invertible and
A^-1 = I - (1/(1+((u^T)v)) (u(v^T))
Sorry about the last equation being so confusing, but that's it. I'm not sure how to prove this, since I don't even know how it works. I'm terrible at this stupid inverse matrix stuff...
2. Originally Posted by Hellreaver
Let u and v be column vectors in R^n, and let A= I + u(v^T). Show that if (u^T)v =/= -1, then A is invertible and A^-1 = I - (1/(1+((u^T)v)) (u(v^T))
this is a nice problem! first we need a little point:
$\boxed{1}$ if $u,v$ are two vectors in $\mathbb{R}^n,$ then $uv^Tuv^T=uv^Tu^Tv.$
Proof. note that $u^Tv$ and $v^Tu$ are both scalars and obviously $u^Tv=v^Tu.$ also a scalar commutes with a vector. thus: $uv^Tu v^T=uu^Tvv^T=uv^Tu^Tv. \ \ \ \Box$
back to your problem: we have $A^2=(I+uv^T)^2=I + 2uv^T+uv^Tuv^T=I+2uv^T+uv^Tu^Tv. \ \ \ \ \ \ \ \text{by} \ \boxed{1}$
hence: $A^2=I+uv^T + uv^T + uv^Tu^Tv=A+uv^T(1+u^Tv)=A+(A-I)(1+u^Tv),$ which gives us: $A(A-I-(1+u^Tv)I)=-(1+u^Tv)I. \ \ \ \ \ \ (*)$
now $1+u^Tv \neq 0.$ so we can divide both sides of $(*)$ by $-(1+u^Tv).$ also we have $A-I=uv^T.$ thus $(*)$ gives us: $A \left(I - \frac{1}{1+u^Tv}uv^T \right)=I.$
3. Friggen complicated enough?
I so don't understand this stuff. I was pretty good at trig proofs, but that was back when I actually had a teacher. This linear algebra stuff just doesn't make sense...
4. Originally Posted by Hellreaver
Friggen complicated enough?
I so don't understand this stuff. I was pretty good at trig proofs, but that was back when I actually had a teacher. This linear algebra stuff just doesn't make sense...
his response is very easy to follow, grab a pen and paper and write it carefully down. You should be able to understand it.
5. I understand his response perfectly enough (thanks, NonCommAlg). I just don't understand the majority of the concept behind it. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9429147839546204, "perplexity": 515.5698491171253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934808972.93/warc/CC-MAIN-20171124214510-20171124234510-00070.warc.gz"} |
https://openstax.org/books/chemistry-atoms-first/pages/13-2-equilibrium-constants | Chemistry: Atoms First
# 13.2Equilibrium Constants
Chemistry: Atoms First13.2 Equilibrium Constants
### Learning Objectives
By the end of this section, you will be able to:
• Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
• Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
• Relate the magnitude of an equilibrium constant to properties of the chemical system
Now that we have a symbol $(⇌)(⇌)$ to designate reversible reactions, we will need a way to express mathematically how the amounts of reactants and products affect the equilibrium of the system. A general equation for a reversible reaction may be written as follows:
$mA+nB+⇌xC+yDmA+nB+⇌xC+yD$
We can write the reaction quotient (Q) for this equation. When evaluated using concentrations, it is called Qc. We use brackets to indicate molar concentrations of reactants and products.
$Qc=[C]x[D]y[A]m[B]nQc=[C]x[D]y[A]m[B]n$
The reaction quotient is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation. For example, the reaction quotient for the reversible reaction $2NO2(g)⇌N2O4(g)2NO2(g)⇌N2O4(g)$ is given by this expression:
$Qc=[N2O4][NO2]2Qc=[N2O4][NO2]2$
### Example 13.1
#### Writing Reaction Quotient Expressions
Write the expression for the reaction quotient for each of the following reactions:
(a) $3O2(g)⇌2O3(g)3O2(g)⇌2O3(g)$
(b) $N2(g)+3H2(g)⇌2NH3(g)N2(g)+3H2(g)⇌2NH3(g)$
(c) $4NH3(g)+7O2(g)⇌4NO2(g)+6H2O(g)4NH3(g)+7O2(g)⇌4NO2(g)+6H2O(g)$
#### Solution
(a) $Qc=[O3]2[O2]3Qc=[O3]2[O2]3$
(b) $Qc=[NH3]2[N2][H2]3Qc=[NH3]2[N2][H2]3$
(c) $Qc=[NO2]4[H2O]6[NH3]4[O2]7Qc=[NO2]4[H2O]6[NH3]4[O2]7$
Write the expression for the reaction quotient for each of the following reactions:
(a) $2SO2(g)+O2(g)⇌2SO3(g)2SO2(g)+O2(g)⇌2SO3(g)$
(b) $C4H8(g)⇌2C2H4(g)C4H8(g)⇌2C2H4(g)$
(c) $2C4H10(g)+13O2(g)⇌8CO2(g)+10H2O(g)2C4H10(g)+13O2(g)⇌8CO2(g)+10H2O(g)$
(a) $Qc=[SO3]2[SO2]2[O2];Qc=[SO3]2[SO2]2[O2];$ (b) $Qc=[C2H4]2[C4H8];Qc=[C2H4]2[C4H8];$ (c) $Qc=[CO2]8[H2O]10[C4H10]2[O2]13Qc=[CO2]8[H2O]10[C4H10]2[O2]13$
The numeric value of Qc for a given reaction varies; it depends on the concentrations of products and reactants present at the time when Qc is determined. When pure reactants are mixed, Qc is initially zero because there are no products present at that point. As the reaction proceeds, the value of Qc increases as the concentrations of the products increase and the concentrations of the reactants simultaneously decrease (Figure 13.6). When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change.
Figure 13.6 (a) The change in the concentrations of reactants and products is depicted as the $2 SO 2 ( g ) + O 2 ( g ) ⇌ 2 SO 3 ( g ) 2 SO 2 ( g ) + O 2 ( g ) ⇌ 2 SO 3 ( g )$ reaction approaches equilibrium. (b) The change in concentrations of reactants and products is depicted as the reaction $2 SO 3 ( g ) ⇌ 2 SO 2 ( g ) + O 2 ( g ) 2 SO 3 ( g ) ⇌ 2 SO 2 ( g ) + O 2 ( g )$ approaches equilibrium. (c) The graph shows the change in the value of the reaction quotient as the reaction approaches equilibrium.
When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the equilibrium constant (K) of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as Kc.
That a reaction quotient always assumes the same value at equilibrium can be expressed as:
$Qcat equilibrium=Kc=[C]x[D]y...[A]m[B]n...Qcat equilibrium=Kc=[C]x[D]y...[A]m[B]n...$
This equation is a mathematical statement of the law of mass action: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.
### Example 13.2
#### Evaluating a Reaction Quotient
Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:
$2NO2(g)⇌N2O4(g)2NO2(g)⇌N2O4(g)$
When 0.10 mol NO2 is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO2] = 0.016 M and [N2O4] = 0.042 M.
(a) What is the value of the reaction quotient before any reaction occurs?
(b) What is the value of the equilibrium constant for the reaction?
#### Solution
(a) Before any product is formed, $[NO2]=0.10mol1.0L=0.10M,[NO2]=0.10mol1.0L=0.10M,$ and [N2O4] = 0 M. Thus,
$Qc=[N2O4][NO2]2=00.102=0Qc=[N2O4][NO2]2=00.102=0$
(b) At equilibrium, the value of the equilibrium constant is equal to the value of the reaction quotient. At equilibrium, $Kc=Qc=[N2O4][NO2]2=0.0420.0162=1.6×102.Kc=Qc=[N2O4][NO2]2=0.0420.0162=1.6×102.$ The equilibrium constant is 1.6 $××$ 102.
Note that dimensional analysis would suggest the unit for this Kc value should be M−1. However, it is common practice to omit units for Kc values computed as described here, since it is the magnitude of an equilibrium constant that relays useful information. As will be discussed later in this module, the rigorous approach to computing equilibrium constants uses dimensionless quantities derived from concentrations instead of actual concentrations, and so Kc values are truly unitless.
For the reaction $2SO2(g)+O2(g)⇌2SO3(g),2SO2(g)+O2(g)⇌2SO3(g),$ the concentrations at equilibrium are [SO2] = 0.90 M, [O2] = 0.35 M, and [SO3] = 1.1 M. What is the value of the equilibrium constant, Kc?
Kc = 4.3
The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A large value for Kc indicates that equilibrium is attained only after the reactants have been largely converted into products. A small value of Kc—much less than 1—indicates that equilibrium is attained when only a small proportion of the reactants have been converted into products.
Once a value of Kc is known for a reaction, it can be used to predict directional shifts when compared to the value of Qc. A system that is not at equilibrium will proceed in the direction that establishes equilibrium. The data in Figure 13.7 illustrate this. When heated to a consistent temperature, 800 °C, different starting mixtures of CO, H2O, CO2, and H2 react to reach compositions adhering to the same equilibrium constant (the value of Qc changes until it equals the value of Kc). This value is 0.640, the equilibrium constant for the reaction under these conditions.
$CO(g)+H2O(g)⇌CO2(g)+H2(g)Kc=0.640T=800°CCO(g)+H2O(g)⇌CO2(g)+H2(g)Kc=0.640T=800°C$
It is important to recognize that an equilibrium can be established starting either from reactants or from products, or from a mixture of both. For example, equilibrium was established from Mixture 2 in Figure 13.7 when the products of the reaction were heated in a closed container. In fact, one technique used to determine whether a reaction is truly at equilibrium is to approach equilibrium starting with reactants in one experiment and starting with products in another. If the same value of the reaction quotient is observed when the concentrations stop changing in both experiments, then we may be certain that the system has reached equilibrium.
Figure 13.7 Concentrations of three mixtures are shown before and after reaching equilibrium at 800 °C for the so-called water gas shift reaction: $CO ( g ) + H 2 O ( g ) ⇌ CO 2 ( g ) + H 2 ( g ) . CO ( g ) + H 2 O ( g ) ⇌ CO 2 ( g ) + H 2 ( g ) .$
### Example 13.3
#### Predicting the Direction of Reaction
Given here are the starting concentrations of reactants and products for three experiments involving this reaction:
$CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g)+H2O(g)⇌CO2(g)+H2(g)$
$Kc=0.64Kc=0.64$
Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.
Reactants/Products Experiment 1 Experiment 2 Experiment 3
[CO]i 0.0203 M 0.011 M 0.0094 M
[H2O]i 0.0203 M 0.0011 M 0.0025 M
[CO2]i 0.0040 M 0.037 M 0.0015 M
[H2]i 0.0040 M 0.046 M 0.0076 M
#### Solution
Experiment 1:
$Qc=[CO2][H2][CO][H2O]=(0.0040)(0.0040)(0.0203)(0.0203)=0.039.Qc=[CO2][H2][CO][H2O]=(0.0040)(0.0040)(0.0203)(0.0203)=0.039.$
Qc < Kc (0.039 < 0.64)
The reaction will shift to the right.
Experiment 2:
$Qc=[CO2][H2][CO][H2O]=(0.037)(0.046)(0.011)(0.0011)=1.4×102Qc=[CO2][H2][CO][H2O]=(0.037)(0.046)(0.011)(0.0011)=1.4×102$
Qc > Kc (140 > 0.64)
The reaction will shift to the left.
Experiment 3:
$Qc=[CO2][H2][CO][H2O]=(0.0015)(0.0076)(0.0094)(0.0025)=0.48Qc=[CO2][H2][CO][H2O]=(0.0015)(0.0076)(0.0094)(0.0025)=0.48$
Qc < Kc (0.48 < 0.64)
The reaction will shift to the right.
Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.
(a) A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl:
$2NO(g)+Cl2(g)⇌2NOCl(g)Kc=4.6×1042NO(g)+Cl2(g)⇌2NOCl(g)Kc=4.6×104$
(b) A 5.0-L flask containing 17 g of NH3, 14 g of N2, and 12 g of H2:
$N2(g)+3H2(g)⇌2NH3(g)Kc=0.060N2(g)+3H2(g)⇌2NH3(g)Kc=0.060$
(c) A 2.00-L flask containing 230 g of SO3(g):
$2SO3(g)⇌2SO2(g)+O2(g)Kc=0.2302SO3(g)⇌2SO2(g)+O2(g)Kc=0.230$
(a) Qc = 6.45 $××$ 103, shifts right. (b) Qc = 0.23, shifts left. (c) Qc = 0, shifts right
In Example 13.2, it was mentioned that the common practice is to omit units when evaluating reaction quotients and equilibrium constants. It should be pointed out that using concentrations in these computations is a convenient but simplified approach that sometimes leads to results that seemingly conflict with the law of mass action. For example, equilibria involving aqueous ions often exhibit equilibrium constants that vary quite significantly (are not constant) at high solution concentrations. This may be avoided by computing Kc values using the activities of the reactants and products in the equilibrium system instead of their concentrations. The activity of a substance is a measure of its effective concentration under specified conditions. While a detailed discussion of this important quantity is beyond the scope of an introductory text, it is necessary to be aware of a few important aspects:
• Activities are dimensionless (unitless) quantities and are in essence “adjusted” concentrations.
• For relatively dilute solutions, a substance's activity and its molar concentration are roughly equal.
• Activities for pure condensed phases (solids and liquids) are equal to 1.
As a consequence of this last consideration, Qc and Kc expressions do not contain terms for solids or liquids (being numerically equal to 1, these terms have no effect on the expression's value). Several examples of equilibria yielding such expressions will be encountered in this section.
### Homogeneous Equilibria
A homogeneous equilibrium is one in which all of the reactants and products are present in a single solution (by definition, a homogeneous mixture). In this chapter, we will concentrate on the two most common types of homogeneous equilibria: those occurring in liquid-phase solutions and those involving exclusively gaseous species. Reactions between solutes in liquid solutions belong to one type of homogeneous equilibria. The chemical species involved can be molecules, ions, or a mixture of both. Several examples are provided here.
$C2H2(aq)+2Br2(aq)⇌C2H2Br4(aq)Kc=[C2H2Br4][C2H2][Br2]2C2H2(aq)+2Br2(aq)⇌C2H2Br4(aq)Kc=[C2H2Br4][C2H2][Br2]2$
$I2(aq)+I−(aq)⇌I3−(aq)Kc=[I3−][I2][I−]I2(aq)+I−(aq)⇌I3−(aq)Kc=[I3−][I2][I−]$
$Hg22+(aq)+NO3−(aq)+3H3O+(aq)⇌2Hg2+(aq)+HNO2(aq)+4H2O(l)Hg22+(aq)+NO3−(aq)+3H3O+(aq)⇌2Hg2+(aq)+HNO2(aq)+4H2O(l)$
$Kc=[Hg2+]2[HNO2][Hg22+][NO3−][H3O+]3Kc=[Hg2+]2[HNO2][Hg22+][NO3−][H3O+]3$
$HF(aq)+H2O(l)⇌H3O+(aq)+F−(aq)Kc=[H3O+][F−][HF]HF(aq)+H2O(l)⇌H3O+(aq)+F−(aq)Kc=[H3O+][F−][HF]$
$NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Kc=[NH4+][OH−][NH3]NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)Kc=[NH4+][OH−][NH3]$
In each of these examples, the equilibrium system is an aqueous solution, as denoted by the aq annotations on the solute formulas. Since H2O(l) is the solvent for these solutions, its concentration does not appear as a term in the Kc expression, as discussed earlier, even though it may also appear as a reactant or product in the chemical equation.
Reactions in which all reactants and products are gases represent a second class of homogeneous equilibria. We use molar concentrations in the following examples, but we will see shortly that partial pressures of the gases may be used as well.
$C2H6(g)⇌C2H4(g)+H2(g)Kc=[C2H4][H2][C2H6]C2H6(g)⇌C2H4(g)+H2(g)Kc=[C2H4][H2][C2H6]$
$3O2(g)⇌2O3(g)Kc=[O3]2[O2]33O2(g)⇌2O3(g)Kc=[O3]2[O2]3$
$N2(g)+3H2(g)⇌2NH3(g)Kc=[NH3]2[N2][H2]3N2(g)+3H2(g)⇌2NH3(g)Kc=[NH3]2[N2][H2]3$
$C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)Kc=[CO2]3[H2O]4[C3H8][O2]5C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)Kc=[CO2]3[H2O]4[C3H8][O2]5$
Note that the concentration of H2O(g) has been included in the last example because water is not the solvent in this gas-phase reaction and its concentration (and activity) changes.
Whenever gases are involved in a reaction, the partial pressure of each gas can be used instead of its concentration in the equation for the reaction quotient because the partial pressure of a gas is directly proportional to its concentration at constant temperature. This relationship can be derived from the ideal gas equation, where M is the molar concentration of gas, $nV.nV.$
$PV=nRTPV=nRT$
$P=(nV)RTP=(nV)RT$
$=MRT=MRT$
Thus, at constant temperature, the pressure of a gas is directly proportional to its concentration.
Using the partial pressures of the gases, we can write the reaction quotient for the system $C2H6(g)⇌C2H4(g)+H2(g)C2H6(g)⇌C2H4(g)+H2(g)$ by following the same guidelines for deriving concentration-based expressions:
$QP=PC2H4PH2PC2H6QP=PC2H4PH2PC2H6$
In this equation we use QP to indicate a reaction quotient written with partial pressures: $PC2H6PC2H6$ is the partial pressure of C2H6; $PH2,PH2,$ the partial pressure of H2; and $PC2H6,PC2H6,$ the partial pressure of C2H4. At equilibrium:
$KP=QP=PC2H4PH2PC2H6KP=QP=PC2H4PH2PC2H6$
The subscript P in the symbol KP designates an equilibrium constant derived using partial pressures instead of concentrations. The equilibrium constant, KP, is still a constant, but its numeric value may differ from the equilibrium constant found for the same reaction by using concentrations.
Conversion between a value for Kc, an equilibrium constant expressed in terms of concentrations, and a value for KP, an equilibrium constant expressed in terms of pressures, is straightforward (a K or Q without a subscript could be either concentration or pressure).
The equation relating Kc and KP is derived as follows. For the gas-phase reaction $mA+nB⇌xC+yD:mA+nB⇌xC+yD:$
$KP=(PC)x(PD)y(PA)m(PB)nKP=(PC)x(PD)y(PA)m(PB)n$
$=([C]×RT)x([D]×RT)y([A]×RT)m([B]×RT)n=([C]×RT)x([D]×RT)y([A]×RT)m([B]×RT)n$
$=[C]x[D]y[A]m[B]n×(RT)x+y(RT)m+n=[C]x[D]y[A]m[B]n×(RT)x+y(RT)m+n$
$=Kc(RT)(x+y)−(m+n)=Kc(RT)(x+y)−(m+n)$
$=Kc(RT)Δn=Kc(RT)Δn$
The relationship between Kc and KP is
$KP=Kc(RT)ΔnKP=Kc(RT)Δn$
In this equation, Δn is the difference between the sum of the coefficients of the gaseous products and the sum of the coefficients of the gaseous reactants in the reaction (the change in moles of gas between the reactants and the products). For the gas-phase reaction $mA+nB⇌xC+yD,mA+nB⇌xC+yD,$ we have
$Δn=(x+y)−(m+n)Δn=(x+y)−(m+n)$
### Example 13.4
#### Calculation of KP
Write the equations for the conversion of Kc to KP for each of the following reactions:
(a) $C2H6(g)⇌C2H4(g)+H2(g)C2H6(g)⇌C2H4(g)+H2(g)$
(b) $CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g)+H2O(g)⇌CO2(g)+H2(g)$
(c) $N2(g)+3H2(g)⇌2NH3(g)N2(g)+3H2(g)⇌2NH3(g)$
(d) Kc is equal to 0.28 for the following reaction at 900 °C:
$CS2(g)+4H2(g)⇌CH4(g)+2H2S(g)CS2(g)+4H2(g)⇌CH4(g)+2H2S(g)$
What is KP at this temperature?
#### Solution
(a) Δn = (2) − (1) = 1
KP = Kc (RT)Δn = Kc (RT)1 = Kc (RT)
(b) Δn = (2) − (2) = 0
KP = Kc (RT)Δn = Kc (RT)0 = Kc
(c) Δn = (2) − (1 + 3) = −2
KP = Kc (RT)Δn = Kc (RT)−2 = $Kc(RT)2Kc(RT)2$
(d) KP = Kc (RT)Δn = (0.28)[(0.0821)(1173)]−2 = 3.0 $××$ 10−5
Write the equations for the conversion of Kc to KP for each of the following reactions, which occur in the gas phase:
(a) $2SO2(g)+O2(g)⇌2SO3(g)2SO2(g)+O2(g)⇌2SO3(g)$
(b) $N2O4(g)⇌2NO2(g)N2O4(g)⇌2NO2(g)$
(c) $C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)C3H8(g)+5O2(g)⇌3CO2(g)+4H2O(g)$
(d) At 227 °C, the following reaction has Kc = 0.0952:
$CH3OH(g)⇌CO(g)+2H2(g)CH3OH(g)⇌CO(g)+2H2(g)$
What would be the value of KP at this temperature?
(a) KP = Kc (RT)−1; (b) KP = Kc (RT); (c) KP = Kc (RT); (d) 160 or 1.6 $××$ 102
### Heterogeneous Equilibria
A heterogeneous equilibrium is a system in which reactants and products are found in two or more phases. The phases may be any combination of solid, liquid, or gas phases, and solutions. When dealing with these equilibria, remember that solids and pure liquids do not appear in equilibrium constant expressions (the activities of pure solids, pure liquids, and solvents are 1).
Some heterogeneous equilibria involve chemical changes; for example:
$PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)Kc=[Pb2+][Cl−]2PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)Kc=[Pb2+][Cl−]2$
$CaO(s)+CO2(g)⇌CaCO3(s)Kc=1[CO2]CaO(s)+CO2(g)⇌CaCO3(s)Kc=1[CO2]$
$C(s)+2S(g)⇌CS2(g)Kc=[CS2][S]2C(s)+2S(g)⇌CS2(g)Kc=[CS2][S]2$
Other heterogeneous equilibria involve phase changes, for example, the evaporation of liquid bromine, as shown in the following equation:
$Br2(l)⇌Br2(g)Kc=[Br2]Br2(l)⇌Br2(g)Kc=[Br2]$
We can write equations for reaction quotients of heterogeneous equilibria that involve gases, using partial pressures instead of concentrations. Two examples are:
$CaO(s)+CO2(g)⇌CaCO3(s)KP=1PCO2CaO(s)+CO2(g)⇌CaCO3(s)KP=1PCO2$
$C(s)+2S(g)⇌CS2(g)KP=PCS2(PS)2C(s)+2S(g)⇌CS2(g)KP=PCS2(PS)2$
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http://www.physicsforums.com/showthread.php?s=1fba1c6a9ad157df997a9e363598630d&p=4722412 | # SuSy Algebra
by ChrisVer
Tags: algebra, susy
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P: 1,015 So I'm trying to show that one choice of representation for the SuSy generators fulfills the SuSy algebra.... (one of which is $\left\{ Q_{a},\bar{Q_{\dot{b}}} \right\}= 2 \sigma^{\mu}_{a\dot{b}} p_{\mu}$)... For $Q_{a}= \partial_{a} - i σ^{μ}_{a\dot{β}} \bar{θ^{\dot{β}}} \partial_{\mu}$ $\bar{Q_{\dot{b}}}= -\bar{\partial_{\dot{b}}} + i θ^{a}σ^{μ}_{a\dot{b}} \partial_{\mu}$ I am not sure how I could go on with computing this anticommutation.... $Q_{a}\bar{Q_{\dot{b}}}= (\partial_{a} - i (σ^{μ}\bar{θ})_{a} \partial_{\mu})(-\bar{\partial_{\dot{b}}} + i (θσ^{μ})_{\dot{b}} \partial_{\mu})$ $Q_{a}\bar{Q_{\dot{b}}}= -\partial_{a}\bar{\partial_{\dot{b}}}+\partial_{a} (θσ^{μ})_{\dot{b}} p_{\mu}+(σ^{μ}\bar{θ})_{a}\bar{\partial_{\dot{b}}} p_{\mu}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}} ∂_{μ}∂_{ν}$ In a similar way I can write: $\bar{Q_{\dot{b}}}Q_{a}= -\bar{\partial_{\dot{b}}}\partial_{a}+\bar{\partial_{\dot{b}}}(σ^{μ}\bar {θ})_{a} p_{\mu}+(θσ^{μ})_{\dot{b}}\partial_{a}p_{\mu}+ (θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a} ∂_{μ}∂_{ν}$ Now the first terms after addition cancel because the grassmann derivatives are like spinor fields, so they anticommute: $\left\{ \partial_{a},\bar{\partial_{\dot{b}}} \right\}=0$ Is that a correct statement? (I think it is, but I am also asking) The last terms give: $[(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}}] ∂_{μ}∂_{ν}=0$ I am saying it's equal to zero, because I'm again seeing the parenthesis (...) as spinors, so they anticomutte... and because the partial derivatives are symmetric under the interchange μ to ν, and the [...] is antisymmetric it's going to give zero: $[(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}}] ∂_{μ}∂_{ν}=[(θσ^{μ})_{\dot{b}}(σ^{ν}\bar{θ})_{a}+(σ^{ν}\bar{θ})_{a} (θσ^{μ})_{\dot{b}}] ∂_{ν}∂_{μ}$ $=[-(σ^{ν}\bar{θ})_{a}(θσ^{μ})_{\dot{b}}- (θσ^{μ})_{\dot{b}}(σ^{ν}\bar{θ})_{a}] ∂_{ν}∂_{μ}$ renaming again: $[(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}}] ∂_{μ}∂_{ν}=-[(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}+(σ^{μ}\bar{θ})_{a} (θσ^{ν})_{\dot{b}}] ∂_{μ}∂_{ν} =0$ So far I am certain I'm on a correct way (because I don't want as a result the double derivatives-is it in spinor or lorentz spaces)... However I'm not certain about my reasonings.... For example, in the last one, I also thought of writing: $(θσ^{ν})_{\dot{b}}(σ^{μ}\bar{θ})_{a}=(θσ^{ν}\bar{1})(1σ^{μ}\bar{θ}) \propto n^{\mu\nu} (θ1)(\bar{1}\bar{θ})$ But it wouldn't work out I think...I would also lose the spinoriac indices... Then I have the middle terms: $[\partial_{a} (θσ^{μ})_{\dot{b}} +(σ^{μ}\bar{θ})_{a}\bar{\partial_{\dot{b}}}+ \bar{\partial_{\dot{b}}} (σ^{μ} \bar{θ})_{a} +(θσ^{μ})_{\dot{b}}\partial_{a}]p_{\mu}$ And here I'd like to ask, if you have any suggestion of how this can work out.... I'd really appreciate it.....
P: 1,015 Does for example, in the 1st term in the last expression, the derivative act on θ? (to give a delta Kroenicker)..... But then I'm not sure about the derivatives appearing on the left... Also I could try acting with that thing on some spinors (for example a $θ\bar{θ}$ and use a chain rule derivative???)...But I am not sure if the dotted derivatives "see" the undotted spinors and the opposite...
P: 1,020 You are right about the first and last term, they vanish as per requirement of anticommutation and symmetry of partial derivative respectively. But you are making mistakes with middle terms, you should write them in the anticommutator form like one of the term is, ##i\left\{∂_a,θ^cσ^v_{cḃ}∂_v\right\}=i(∂_aθ^c)σ^v_{cḃ}∂_v-(iθ^cσ^v_{cḃ}∂_v)∂_a+(iθ^cσ^v_{cḃ}∂_v)∂_a=i(∂_aθ^c)σ^v_{cḃ}∂_v=iδ^c_aσ^ v_{cḃ}∂_v##, because as you pass the derivative w.r.t. grassmann variable through a grassmann variable, there is a sign change and hence the second term. Similar term arises from differentiation w.r.t. ##{\dot{b}}##,which combined with above give desired result.
P: 1,015 SuSy Algebra So in fact, you did a chain rule in the 1st two terms- first the derivative acting on the spinor $\theta$ and then acting on "something" that will appear next... correct? I understand that since $\partial_{a} (\theta^{b} \theta^{c})= \delta_{a}^{b} \theta^{c} - \theta^{b} \delta_{a}^{c}$ there also appears the minus
P: 1,020 Yes.
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[121] Marina Ratner. Distribution rigidity for unipotent actions on homogeneous spaces. Bull. Amer. Math. Soc. 24 (1991) 321-325. MR 1069988. Abstract, references, and article information View Article: PDF [122] James W. Lewis. Infinitesimal rigidity for the action of ${\rm SL}(n,{\bf Z})$ on ${\bf T}\sp n$ . Trans. Amer. Math. Soc. 324 (1991) 421-445. MR 1058434. Abstract, references, and article information View Article: PDF This article is available free of charge [123] Christopher Bishop and Tim Steger. Three rigidity criteria for ${\text{PSL}}\left( {2,{\mathbf{R}}} \right)$. Bull. Amer. Math. Soc. 24 (1991) 117-123. MR 1065010. Abstract, references, and article information View Article: PDF [124] Colin Adams, Martin Hildebrand and Jeffrey Weeks. Hyperbolic invariants of knots and links . Trans. Amer. Math. Soc. 326 (1991) 1-56. MR 994161. Abstract, references, and article information View Article: PDF This article is available free of charge [125] Luis Hernández. Maximal representations of surface groups in bounded symmetric domains . Trans. Amer. Math. Soc. 324 (1991) 405-420. MR 1033234. Abstract, references, and article information View Article: PDF This article is available free of charge [126] Garrett Stuck. On the characteristic classes of actions of lattices in higher rank Lie groups . Trans. Amer. Math. Soc. 324 (1991) 181-200. MR 986031. Abstract, references, and article information View Article: PDF This article is available free of charge [127] D. J. Benson and F. R. Cohen. Mapping class groups of low genus and their cohomology. Memoirs of the AMS 90 (1991) MR 1052554. Book volume table of contents [128] Steven Hurder. Deformation rigidity for subgroups of $SL\left( {n,{\mathbf{Z}}} \right)$ acting on the $n$-torus. Bull. Amer. Math. Soc. 23 (1990) 107-113. MR 1027900. Abstract, references, and article information View Article: PDF [129] Richard K. Skora. Splittings of surfaces. Bull. Amer. Math. Soc. 23 (1990) 85-90. MR 1031584. Abstract, references, and article information View Article: PDF [130] Alessandra Iozzi. A nonlinear extension of the Borel density theorem: Applications to invariance of geometric structures and to smooth orbit equivalence. Bull. Amer. Math. Soc. 23 (1990) 115-120. MR 1021792. Abstract, references, and article information View Article: PDF [131] Lowell Edwin Jones and F Thomas Farrell. Classical Aspherical Manifolds. CBMS Regional Conference Series in Mathematics 75 (1990) MR MR1056079. Book volume table of contents [132] F. T. Farrell and L. E. Jones. Rigidity and other topological aspects of compact nonpositively curved manifolds. Bull. Amer. Math. Soc. 22 (1990) 59-64. MR 1001606. Abstract, references, and article information View Article: PDF [133] Dave Witte. Topological equivalence of foliations of homogeneous spaces . Trans. Amer. Math. Soc. 317 (1990) 143-166. MR 942428. Abstract, references, and article information View Article: PDF This article is available free of charge [134] Alexander Lubotzky. Lattices of minimal covolume in ${\rm SL}\sb 2$: a non-Archimedean analogue of Siegel's theorem $\mu\geq\pi/21$ . J. Amer. Math. Soc. 3 (1990) 961-975. MR 1070003. Abstract, references, and article information View Article: PDF This article is available free of charge [135] Mark Pollicott. Kleinian groups, Laplacian on forms and currents at infinity . Proc. Amer. Math. Soc. 110 (1990) 269-279. MR 1012936. Abstract, references, and article information View Article: PDF This article is available free of charge [136] Guofang Wei. On the fundamental groups of manifolds with almost-nonnegative Ricci curvature . Proc. Amer. Math. Soc. 110 (1990) 197-199. MR 1021214. Abstract, references, and article information View Article: PDF This article is available free of charge [137] Chal Benson and Carolyn S. Gordon. K\"ahler structures on compact solvmanifolds . Proc. Amer. Math. Soc. 108 (1990) 971-980. MR 993739. Abstract, references, and article information View Article: PDF This article is available free of charge [138] Hyman Bass and Ravi Kulkarni. Uniform tree lattices . J. Amer. Math. Soc. 3 (1990) 843-902. MR 1065928. Abstract, references, and article information View Article: PDF This article is available free of charge [139] Alexander Lubotzky. Trees and discrete subgroups of Lie groups over local fields. Bull. Amer. Math. Soc. 20 (1989) 27-30. MR 945301. Abstract, references, and article information View Article: PDF [140] Mason S. Osborne. Lefschetz formulas on nonelliptic complexes. Math. Surveys Monogr. 31 (1989) 171-222. Book volume table of contents View Article: PDF [141] D. I. Wallace. Terms in the Selberg trace formula for ${\rm SL}(3,{\bf Z})\backslash {\rm SL}(3,{\bf R})/{\rm SO}(3,{\bf R})$ associated to Eisenstein series coming from a maximal parabolic subgroup . Proc. Amer. Math. Soc. 106 (1989) 875-883. MR 963577. Abstract, references, and article information View Article: PDF This article is available free of charge [142] Robert J. Zimmer. Representations of fundamental groups of manifolds with a semisimple transformation group . J. Amer. Math. Soc. 2 (1989) 201-213. MR 973308. Abstract, references, and article information View Article: PDF This article is available free of charge [143] F. T. Farrell and L. E. Jones. A topological analogue of Mostow's rigidity theorem . J. Amer. Math. Soc. 2 (1989) 257-370. MR 973309. Abstract, references, and article information View Article: PDF This article is available free of charge [144] F. T. Farrell and L. E. Jones. Topological rigidity for hyperbolic manifolds. Bull. Amer. Math. Soc. 19 (1988) 277-282. MR 940487. Abstract, references, and article information View Article: PDF [145] G. D. Mostow. On discontinuous action of monodromy groups on the complex $n$-ball . J. Amer. Math. Soc. 1 (1988) 555-586. MR 932662. Abstract, references, and article information View Article: PDF This article is available free of charge [146] Robert J. Zimmer. Arithmeticity of holonomy groups of Lie foliations . J. Amer. Math. Soc. 1 (1988) 35-58. MR 924701. Abstract, references, and article information View Article: PDF This article is available free of charge [147] Diego Benardete. Topological equivalence of flows on homogeneous spaces, and divergence of one-parameter subgroups of Lie groups . Trans. Amer. Math. Soc. 306 (1988) 499-527. MR 933304. Abstract, references, and article information View Article: PDF This article is available free of charge [148] Y. L. Tong and S. P. Wang. Construction of cohomology of discrete groups . Trans. Amer. Math. Soc. 306 (1988) 735-763. MR 933315. Abstract, references, and article information View Article: PDF This article is available free of charge [149] Benjamin Fine and Morris Newman. The normal subgroup structure of the Picard group . Trans. Amer. Math. Soc. 302 (1987) 769-786. MR 891646. Abstract, references, and article information View Article: PDF This article is available free of charge [150] Ted Chinburg. A small arithmetic hyperbolic three-manifold . Proc. Amer. Math. Soc. 100 (1987) 140-144. MR 883417. Abstract, references, and article information View Article: PDF This article is available free of charge
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http://againstthegraindesigns.com/6xa09y3/x9dhjm.php?12904d=how-hard-is-partial-differential-equations | If you're seeing this message, it means we're having trouble loading external resources on our website. Having a good textbook helps too (the calculus early transcendentals book was a much easier read than Zill and Wright's differential equations textbook in my experience). A variable is used to represent the unknown function which depends on x. We plan to offer the first part starting in January 2021 and … First, differentiating ƒ with respect to x … Vedantu The differential equations class I took was just about memorizing a bunch of methods. There are Different Types of Partial Differential Equations: Now, consider dds (x + uy) = 1y dds(x + u) − x + uy, The general solution of an inhomogeneous ODE has the general form: u(t) = u. This defines a family of solutions of the PDE; so, we can choose φ(x, y, u) = x + uy, Example 2. This is the book I used for a course called Applied Boundary Value Problems 1. thats why first courses focus on the only easy cases, exact equations, especially first order, and linear constant coefficient case. As a general rule solving PDEs can be very hard and we often have to resort to numerical methods. Analysis - Analysis - Partial differential equations: From the 18th century onward, huge strides were made in the application of mathematical ideas to problems arising in the physical sciences: heat, sound, light, fluid dynamics, elasticity, electricity, and magnetism. Such a method is very convenient if the Euler equation is of elliptic type. A central theme is a thorough treatment of distribution theory. Log In Sign Up. The definition of Partial Differential Equations (PDE) is a differential equation that has many unknown functions along with their partial derivatives. We also just briefly noted how partial differential equations could be solved numerically by converting into discrete form in both space and time. Ordinary and Partial Differential Equations. We first look for the general solution of the PDE before applying the initial conditions. 2 An equation involving the partial derivatives of a function of more than one variable is called PED. Hence the derivatives are partial derivatives with respect to the various variables. They are a very natural way to describe many things in the universe. Even though Calculus III was more difficult, it was a much better class--in that class you learn about functions from R^m --> R^n and … Differential Equations 2 : Partial Differential Equations amd Equations of Mathematical Physics (Theory and solved Problems), University Book, Sarajevo, 2001, pp. A Partial Differential Equation commonly denoted as PDE is a differential equation containing partial derivatives of the dependent variable (one or more) with more than one independent variable. While I'm no expert on partial differential equations the only advice I can offer is the following: * Be curious but to an extent. The derivation of partial differential equations from physical laws usually brings about simplifying assumptions that are difficult to justify completely. Partial differential equations can describe everything from planetary motion to plate tectonics, but they’re notoriously hard to solve. So the partial differential equation becomes a system of independent equations for the coefficients of : These equations are no more difficult to solve than for the case of ordinary differential equations. A differential equationis an equation which contains one or more terms which involve the derivatives of one variable (i.e., dependent variable) with respect to the other variable (i.e., independent variable) dy/dx = f(x) Here “x” is an independent variable and “y” is a dependent variable For example, dy/dx = 5x A differential equation that contains derivatives which are either partial derivatives or ordinary derivatives. The complicated interplay between the mathematics and its applications led to many new discoveries in both. In the above six examples eqn 6.1.6 is non-homogeneous where as the first five equations … So in geometry, the purpose of equations is not to get solutions but to study the properties of the shapes. In this book, which is basically self-contained, we concentrate on partial differential equations in mathematical physics and on operator semigroups with their generators. Since we can find a formula of Differential Equations, it allows us to do many things with the solutions like devise graphs of solutions and calculate the exact value of a solution at any point. The first definition that we should cover should be that of differential equation.A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. Read this book using Google Play Books app on your PC, android, iOS devices. Viewed 1k times 0 $\begingroup$ My question is why it is difficult to find analytical solutions for these equations. A partial differential equation requires, d) an equal number of dependent and independent variables. Using linear dispersionless water theory, the height u (x, t) of a free surface wave above the undisturbed water level in a one-dimensional canal of varying depth h (x) is the solution of the following partial differential equation. differential equations in general are extremely difficult to solve. Two C1-functions u(x,y) and v(x,y) are said to be functionally dependent if det µ ux uy vx vy ¶ = 0, which is a linear partial differential equation of first order for u if v is a given … What To Do With Them? The RLC circuit equation (and pendulum equation) is an ordinary differential equation, or ode, and the diffusion equation is a partial differential equation, or pde. For example, dy/dx = 9x. In algebra, mostly two types of equations are studied from the family of equations. Nonlinear differential equations are difficult to solve, therefore, close study is required to obtain a correct solution. There are many ways to choose these n solutions, but we are certain that there cannot be more than n of them. • Ordinary Differential Equation: Function has 1 independent variable. Get to Understand How to Separate Variables in Differential Equations The precise idea to study partial differential equations is to interpret physical phenomenon occurring in nature. Get to Understand How to Separate Variables in Differential Equations As indicated in the introduction, Separation of Variables in Differential Equations can only be applicable when all the y terms, including dy, can be moved to one side of the equation. Ordinary and partial differential equations: Euler, Runge Kutta, Bulirsch-Stoer, stiff equation solvers, leap-frog and symplectic integrators, Partial differential equations: boundary value and initial value problems. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. In case of partial differential equations, most of the equations have no general solution. pdex1pde defines the differential equation (i) Equations of First Order/ Linear Partial Differential Equations, (ii) Linear Equations of Second Order Partial Differential Equations. Log In Sign Up. (y + u) ∂u ∂x + y ∂u∂y = x − y in y > 0, −∞ < x < ∞. Now isSolutions Manual for Linear Partial Differential Equations . Differential equations are the key to making predictions and to finding out what is predictable, from the motion of galaxies to the weather, to human behavior. In addition to this distinction they can be further distinguished by their order. An equation is a statement in which the values of the mathematical expressions are equal. Sorry!, This page is not available for now to bookmark. A partial differential equation has two or more unconstrained variables. But first: why? Would it be a bad idea to take this without having taken ordinary differential equations? A differential equation having the above form is known as the first-order linear differential equation where P and Q are either constants or … (diffusion equation) These are second-order differential equations, categorized according to the highest order derivative. Furthermore, the classification of Partial Differential Equations of Second Order can be done into parabolic, hyperbolic, and elliptic equations. Most often the systems encountered, fails to admit explicit solutions but fortunately qualitative methods were discovered which does provide ample information about the … Included are partial derivations for the Heat Equation and Wave Equation. Algebra also uses Diophantine Equations where solutions and coefficients are integers. These are used for processing model that includes the rates of change of the variable and are used in subjects like physics, chemistry, economics, and biology. For eg. I find it hard to think of anything that’s more relevant for understanding how the world works than differential equations. For partial differential equations (PDEs), neural operators directly learn the mapping from any functional parametric dependence to the solution. The reason for both is the same. Differential equations involve the differential of a quantity: how rapidly that quantity changes with respect to change in another. We will show most of the details but leave the description of the solution process out. Using differential equations Radioactive decay is calculated. It is used to represent many types of phenomenons like sound, heat, diffusion, electrostatics, electrodynamics, … The definition of Partial Differential Equations (PDE) is a differential equation that has many unknown functions along with their partial derivatives. This example problem uses the functions pdex1pde, pdex1ic, and pdex1bc. We solve it when we discover the function y(or set of functions y). Polynomial equations are generally in the form P(x)=0 and linear equations are expressed ax+b=0 form where a and b represents the parameter. PETSc for Partial Differential Equations: Numerical Solutions in C and Python - Ebook written by Ed Bueler. I'm taking both Calc 3 and differential equations next semester and I'm curious where the difficulties in them are or any general advice about taking these subjects? Thus, they learn an entire family of PDEs, in contrast to classical methods which solve one instance of the equation. The partial differential equation takes the form. Do you know what an equation is? Press J to jump to the feed. The following is the Partial Differential Equations formula: We will do this by taking a Partial Differential Equations example. All best, Mirjana Press question mark to learn the rest of the keyboard shortcuts. Section 1-1 : Definitions Differential Equation. The derivatives re… User account menu • Partial differential equations? Sometimes we can get a formula for solutions of Differential Equations. Learn differential equations for free—differential equations, separable equations, exact equations, integrating factors, and homogeneous equations, and more. An ode is an equation for a function of The general solution of an inhomogeneous ODE has the general form: u(t) = uh(t) + up(t). There are many other ways to express ODE. And different varieties of DEs can be solved using different methods. All best, Mirjana Differential equations (DEs) come in many varieties. Differential Equations • A differential equation is an equation for an unknown function of one or several variables that relates the values of the function itself and of its derivatives of various orders. Differential equations (DEs) come in many varieties. . Even more basic questions such as the existence and uniqueness of solutions for nonlinear partial differential equations are hard problems and the resolution of existence and uniqueness for the Navier-Stokes equations in three spacial dimensions in particular is … How hard is this class? Equations are considered to have infinite solutions. For multiple essential Differential Equations, it is impossible to get a formula for a solution, for some functions, they do not have a formula for an anti-derivative. In the above six examples eqn 6.1.6 is non-homogeneous where as the first five equations … . RE: how hard are Multivariable calculus (calculus III) and differential equations? . Even though we don’t have a formula for a solution, we can still Get an approx graph of solutions or Calculate approximate values of solutions at various points. Compared to Calculus 1 and 2. No one method can be used to solve all of them, and only a small percentage have been solved. In this eBook, award-winning educator Dr Chris Tisdell demystifies these advanced equations. How to Solve Linear Differential Equation? The number $k$ and the number $l$ of coefficients $a _ {ii} ^ {*} ( \xi )$ in equation (2) which are, respectively, positive and negative at the point $\xi _ {0}$ depend only on the coefficients $a _ {ij} ( x)$ of equation (1). The simple PDE is given by; ∂u/∂x (x,y) = 0 The above relation implies that the function u(x,y) is independent of x which is the reduced form of partial differential equation formulastate… Method of Lines Discretizations of Partial Differential Equations The one-dimensional heat equation. to explain a circle there is a general equation: (x – h). This is a linear differential equation and it isn’t too difficult to solve (hopefully). A PDE for a function u(x1,……xn) is an equation of the form The PDE is said to be linear if f is a linear function of u and its derivatives. The definition of Partial Differential Equations (PDE) is a differential equation that has many unknown functions along with their partial derivatives. 258. Partial differential equations form tools for modelling, predicting and understanding our world. Partial Differential Equation helps in describing various things such as the following: In subjects like physics for various forms of motions, or oscillations. L u = ∑ ν = 1 n A ν ∂ u ∂ x ν + B = 0 , {\displaystyle Lu=\sum _ {\nu =1}^ {n}A_ {\nu } {\frac {\partial u} {\partial x_ {\nu }}}+B=0,} where the coefficient matrices Aν and the vector B may depend upon x and u. Here are some examples: Solving a differential equation means finding the value of the dependent […] 258. (See [2].) In general, partial differential equations are difficult to solve, but techniques have been developed for simpler classes of equations called linear, and for classes known loosely as “almost” linear, in which all derivatives of an order higher than one occur to the first power and their coefficients involve only the independent variables. Example 1: If ƒ ( x, y) = 3 x 2 y + 5 x − 2 y 2 + 1, find ƒ x, ƒ y, ƒ xx, ƒ yy, ƒ xy 1, and ƒ yx. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. See Differential equation, partial, complex-variable methods. Here are some examples: Solving a differential equation means finding the value of the dependent […] It was not too difficult, but it was kind of dull. To apply the separation of variables in solving differential equations, you must move each variable to the equation's other side. Would it be a bad idea to take this without having taken ordinary differential equations? The degree of a partial differential equation is the degree of the highest order derivative which occurs in it after the equation has been rationalized, i.e made free from radicals and fractions so for as derivatives are concerned. You can classify DEs as ordinary and partial Des. And we said that this is a reaction-diffusion equation and what I promised you is that these appear in, in other contexts. Today we’ll be discussing Partial Differential Equations. Differential equations are the equations which have one or more functions and their derivatives. Partial Differential Equations: Analytical Methods and Applications covers all the basic topics of a Partial Differential Equations (PDE) course for undergraduate students or a beginners’ course for graduate students. Calculus 2 and 3 were easier for me than differential equations. Well, equations are used in 3 fields of mathematics and they are: Equations are used in geometry to describe geometric shapes. If a hypersurface S is given in the implicit form. Likewise, a differential equation is called a partial differential equation, abbreviated by pde, if it has partial derivatives in it. In the equation, X is the independent variable. In general, partial differential equations are difficult to solve, but techniques have been developed for simpler classes of equations called linear, and for classes known loosely as “almost” linear, in which all derivatives of an order higher than one occur to the first power and their coefficients involve only the independent variables. That's point number two down here. In this chapter we introduce Separation of Variables one of the basic solution techniques for solving partial differential equations. endstream endobj 1993 0 obj <>stream pdepe solves partial differential equations in one space variable and time. Introduction to Differential Equations with Bob Pego. Therefore, each equation has to be treated independently. Most of the time they are merely plausibility arguments. So, to fully understand the concept let’s break it down to smaller pieces and discuss them in detail. In addition to this distinction they can be further distinguished by their order. This is intended to be a first course on the subject Partial Differential Equations, which generally requires 40 lecture hours (One semester course). Don’t let the name fool you, this was actually a graduate-level course I took during Fall 2018, my last semester of undergraduate study at Carnegie Mellon University.This was a one-semester course that spent most of the semester on partial differential equations (alongside about three weeks’ worth of ordinary differential equation theory). There are two types of differential equations: Ordinary Differential Equations or ODE are equations which have a function of an independent variable and their derivatives. Some courses are made more difficult than at other schools because the lecturers are being anal about it. Differential equations have a derivative in them. Press J to jump to the feed. to explain a circle there is a general equation: (x – h)2 + (y – k)2 = r2. The Navier-Stokes equations are nonlinear partial differential equations and solving them in most cases is very difficult because the nonlinearity introduces turbulence whose stable solution requires such a fine mesh resolution that numerical solutions that attempt to numerically solve the equations directly require an impractical amount of computational power. YES! On its own, a Differential Equation is a wonderful way to express something, but is hard to use.. It was not too difficult, but it was kind of dull.
Even though Calculus III was more difficult, it was a much better class--in that class you learn about functions from R^m --> R^n and what the derivative means for such a function. H���Mo�@����9�X�H�IA���h�ޚ�!�Ơ��b�M���;3Ͼ�Ǜ��M��(��(��k�D�>�*�6�PԎgN �rG1N�����Y8�yu�S[clK��Hv�6{i���7�Y�*�c��r�� J+7��*�Q�ň��I�v��$R� J��������:dD��щ֢+f;4Рu@�wE{ٲ�Ϳ�]�|0p��#h�Q�L�@�&�fe����u,�. Partial Differential Equations. Method is very convenient if the Euler equation is a statement in which values. Partial derivatives with respect to the solution calculus courses it provides qualitative physical explanation of mathematical results maintaining... Methods than are how hard is partial differential equations the function y ( or set of functions y ) to... Equations of Second order can be used to represent the unknown function which depends on.... Appear in, in contrast to classical methods which solve one instance of the solution was too! Is the transformation of that PDE into an ordinary differential equations are used in 3 fields of and... Ifthey can be very hard and we often have to resort to methods. In case of partial differential equations, separable equations, categorized according to the highest derivative... More difficult than at other schools because the lecturers are being anal about it case it... Are a very natural way to describe many things in the equation, abbreviated by PDE if... Advanced problems advanced problems the separation of variables, widely known as the Fourier method, refers any. Of first order differential equations could be solved numerically by converting into discrete form in both pdex1ic and. That ’ s equation be a bad idea to take this without having taken ordinary equation! As ordinary and partial DEs ` tricks '' to solving differential equations formula: we will show most of keyboard! Equations is not available for now to bookmark can classify DEs as ordinary and partial DEs of arbitrary order Even... In many varieties written by Ed Bueler differential equation of arbitrary order m. this. The constant coefficient case is the transformation of that PDE into an ordinary differential equations get solutions but study... Behave almost exactly like algebraic equations not to get solutions but to the. < x < ∞ the complicated interplay between the mathematics and its applications led to many new in... 'Re seeing this message, it means we 're having trouble loading external resources on our.! More difficult than at other schools because the lecturers are being anal about it 0, −∞ < x ∞. Everything from planetary motion to plate tectonics, but it was kind of dull to. Hard and we said that this is a thorough treatment of distribution theory will do this by taking partial. Implicit form, neural operators directly learn the rest of the keyboard shortcuts pdex2,,... Dependence to the solution of elliptic type is used to solve ) be... Describe geometric shapes included are partial rapidly that quantity changes with respect to the solution offer first... Are made more difficult than at other schools because the lecturers are being about! The family of equations is to interpret physical phenomenon occurring in nature and differential equations the one-dimensional heat,! Solutions but how hard is partial differential equations study partial differential equations it provides qualitative physical explanation of results. Explanation of mathematical results while maintaining the expected level of it rigor study partial equations! By taking a partial differential equations from physical laws usually brings about simplifying assumptions that difficult... Choose these n solutions, but they ’ re notoriously hard to think of anything ’... 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And derivatives are partial derivatives with respect to change in another about it obtain a correct.... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8289031982421875, "perplexity": 435.5598605919529}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945182.12/warc/CC-MAIN-20230323163125-20230323193125-00149.warc.gz"} |
https://mathoverflow.net/questions/238718/unbounded-towers-and-combinatorial-cardinal-characteristics-of-the-continuum | # Unbounded towers and combinatorial cardinal characteristics of the continuum
Update: Perhaps the question is too difficult. I would appreciate, thus, even just comments or related observations.
This question assumes familiarity with combinatorial cardinal characteristics of the continnum. It is the essence of an earlier question.
Let $[\mathbb{N}]^\infty$ be the family of infinite subsets of $\mathbb{N}$, partially ordered by $\subseteq^*$, where $a\subseteq^* b$ means $a\setminus b$ is finite.
Let $\kappa$ be a cardinal number. A tower of height $\kappa$ is a $\kappa$-sequence $\langle\, s_\alpha : \alpha<\kappa\,\rangle$ in $[\mathbb{N}]^\infty$ such that
1. This $\kappa$-sequence is $\subset^*$-decreasing as the ordinal number $\alpha$ increases.
2. The set $\{\,s_\alpha : \alpha<\kappa\,\}$ has no pseudointersection. (That is, there is no infinite set $s$ such that $s\subseteq^* s_\alpha$ for all $\alpha<\kappa$).
An element $a\in [\mathbb{N}]^\infty$ is identified with its increasing enumeration. This way, the set $[\mathbb{N}]^\infty$ becomes the family of increasing functions in $\mathbb{N}^\mathbb{N}$, and the standard relation $\le^*$ is defined on $[\mathbb{N}]^\infty$. A set $X\subseteq [\mathbb{N}]^\infty$ is bounded if it is bounded (from above) with respect to $\le^*$.
The general goal is to understand when is there an unbounded tower of height $\mathfrak{b}$. Let us call this axiom BT.
It is known or easy to see that:
1. An unbounded set has no pseudointersection. So we may remove the need for no pseudointersection from the definition of tower without altering BT.
2. If there is an unbounded tower of any cardinality, then BT holds.
3. If $\mathfrak{t}=\mathfrak{b}$ or $\mathfrak{b}<\mathfrak{d}$, then BT holds.
Open-ended question. Can the axiom BT be expressed using (standard) cardinal characteristics of the continuum?
Ashutosh proved that BT is consistent with "$\aleph_1=\mathfrak{t}<\mathfrak{b}=\mathfrak{c}=\aleph_2$".
Will Brian points out that that BT fails in the Hechler model. BT also fails in the Laver model, indirectly by the main result of the linked paper. I suspect that BT also fails in the Mathias model.
Question 1. Does any additional inequality or inequality among cardinals of the continuum (one not following from $\mathfrak{t}=\mathfrak{b}$ or $\mathfrak{b}<\mathfrak{d}$) imply BT?
Question 2. Does BT imply any equality among cardinals of the continuum?
Since BT follows from CH, the hypotheisis BT does not imply any inequality.
Motivation. BT implies that, even in the realm of real sets, the selective covering property $\operatorname{S}_1(\Gamma,\Gamma)$ (which is consistently trivial) is nontrivial. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.984950602054596, "perplexity": 362.5848523978158}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141216175.53/warc/CC-MAIN-20201130130840-20201130160840-00273.warc.gz"} |
https://www.physicsforums.com/threads/centripetal-force-inclined-plane-question.82228/ | # Centripetal Force / Inclined Plane Question
1. Jul 14, 2005
### Dooga Blackrazor
A car is rounding a 515 m frictionless curve in the highway. If the curve is banked at an angle of 12.0 degrees, what is the maximum speed of the car? (1.5)
I achieved an answer of 32.39 m/s or 116.604 km/h . I used an incline plain to find that Fc = mgsin12 and used that to find the answer along with the Fc = mvsquared / radius formula. I made them equal cancelled the mass and solved for V - got the answer above. The answer isn't avaliable so if anyone could check it I would appreciate the assistance.
2. Jul 15, 2005
### Pyrrhus
I suppose we are assuming a circular track?
3. Jul 15, 2005
### zwtipp05
Fc does not equal mg sin 12
It is actually equal to Fn sin 12
It's a common mistake in centripetal force problems to assume that the normal force is equal to the force of gravity, but in fact it is not. I have attached a diagram that is generic to most centripetal force problems. The normal force is actually the resultant of the centripetal force and gravity.
#### Attached Files:
• ###### diagram.JPG
File size:
6.5 KB
Views:
192
4. Jul 15, 2005
### Dooga Blackrazor
I think we assume a circular track. Thanks - the diagram is pending approval. Is there a formula for Fn in a centripetal force problem?
5. Jul 15, 2005
### Pyrrhus
Actually, the centripetal force is just a role given to forces (components) acting towards the center, not another force...
Now if we assume a circular track of 515m arc length you can find the radius, and then work the FBD to find v
6. Jul 16, 2005
### mukundpa
Normal reaction is to be resolved horizontally and vertically. vertical component balances the weight and horizontal provides the required centripetal force to move the car on the curve.
If the track is frictionless the only speed you get for the car not to slide up or down, what is meant by maximum speed? think.............
7. Jul 16, 2005
### Dooga Blackrazor
Oh, I meant a circular track, but I think it is a half of a circle. I think 515 is supposed to be the radius - I could be wrong though.
Maximum speed is Fc = mvsquared / radius formula = by solving for Vsquared you can find the maximum speed (I assume).
8. Jul 17, 2005
### zwtipp05
Yeah, you're on the right track. Try solving it again and post your answer here and I'll let you know what I got when I solved it.
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http://mathoverflow.net/questions/55085/new-proofs-to-major-theorems-leading-to-new-insights-and-results/55118 | # New proofs to major theorems leading to new insights and results?
I am wondering, historically, when has a new proof of an old theorem been particularly fruitful. A few examples I have in mind (all number theoretic) are:
First example is classical... which is Euler's proof of Euclid's theorem which asserts that there exist infinitely many primes. Here is when the factorization $\displaystyle \prod_p (1-p^{-s})^{-1} = \sum_{n=1}^\infty \frac{1}{n^s}$ was first introduced, leading of course to what is now known as the Riemann Hypothesis.
Second example is when Hardy and Littlewood gave an alternative proof of Waring's problem, which was done by Hilbert earlier. Their proof introduced what is now known as the Hardy-Littlewood Circle Method and gave an exact asymptotic for the Waring bases, which is stronger than Hilbert's result which only asserted that every sufficiently large positive integer can be written as the sum of a bounded number of $k$th powers. Later on the Hardy-Littlewood method proved very fruitful in other results, namely Vinogradov's Theorem asserting that every sufficiently large odd positive integer can be written as the sum of three primes.
Third example is Tim Gowers' alternate proof to Szemeredi's Theorem asserting that every subset of the positive integers with positive upper density contains arbitrarily long arithmetic progressions. This advance, namely the introduction of Gowers uniformity norms, led eventually to the Green-Tao Theorem proving the existence of arbitrarily long arithmetic progressions in the primes.
So I am wondering if there exist other incidences (number theory related or not) where a new proof really gave legitimate new insights, perhaps even a proof of a (major) new result.
Edit: I am primarily interested in examples where a new proof sparked off a new direction in research. This is best supported by having a major new theorem proved using techniques inspired by the new proof. An example of something that I am not interested in is something like Donald Newman's proof of the prime number theorem, which while elegant and 'natural' as he puts it, has saw limited generalization to other areas and one is hard pressed to apply the same technique to other problems.
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This question seems a little broad to me. – Qiaochu Yuan Feb 10 '11 at 23:31
Vojta's proof of Faltings's theorem. – David Hansen Feb 10 '11 at 23:37
Community wiki? – J.C. Ottem Feb 10 '11 at 23:53
I would guess that any new proof of major results require new insights. – Hailong Dao Feb 11 '11 at 2:09
Please remember to mark posts asking for lists of answers as community wiki! I've hit this one with the wiki hammer. – Scott Morrison Feb 11 '11 at 4:38
Here are a few examples from the 19th century.
1. Unsolvability of the quintic equation. Abel (1826) proved this by algebraic ingenuity, but without clarifying the concepts involved. Galois (1830) gave a proof that introduced the concepts of group, normal subgroup, and solvability (of groups), thus laying the foundations of group theory and Galois theory.
2. Double periodicity of elliptic functions. Abel and Jacobi established this (1820s) mainly by computation. Riemann (1850s) put elliptic functions on a clear conceptual basis by showing that the underlying elliptic curve is a torus, and that the periods correspond to independent loops on the torus.
3. Riemann-Roch theorem. Riemann (1857) discovered this theorem using Riemann surfaces, but applying physical intuition (the "Dirichlet principle"). This principle was not made rigorous until 1901. In the meantime, Dedekind and Weber (1882) gave the first rigorous and complete proof of Riemann-Roch, by reconstructing the theory of Riemann surfaces algebraically. In the process they paved the way for modern algebraic geometry.
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I'd like to offer a slightly different related perspective on your second example: Legendre spent a great deal of his life mucking about with elliptic integrals, but Abel and Jacobi realized that it was much cleaner to work with their inverses, the elliptic functions! – David Hansen Feb 12 '11 at 5:45
David, you make a valid point, but it is harder to view the step beyond Legendre taken by Abel and Jacobi as a new proof of a known theorem. – John Stillwell Feb 12 '11 at 6:45
A very nice example in my eyes is Serre's proof of Riemann-Roch:
Sometimes, you are just not satisfied with existing proofs, and you look for better ones, which can be applied in different situations. A typical example for me was when I worked on the Riemann-Roch theorem (circa 1953), which I viewed as an "Euler-Poincare" formula (I did not know then that Kodaira-Spencer had had the same idea.) My first objective was to prove it for algebraic curves - a case which was known for about a century! But I wanted a proof in a special style; and when I managed to find it, I remember it did not take me more than a minute or two to go from there to the 2-dimensional case (which had just been done by Kodaira).
He is speaking, of course, of the sheaf-theoretic proofs, which are usually presented today. This was the period where he was working on FAC, GAGA and his duality theorem, which revolutionized algebraic geometry.
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Witten's supersymmetric proof of the Atiyah-Singer index theorem.
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Could you give a reference. – MBN Feb 11 '11 at 15:35
This is not the original reference: MR0836727 (87h:58207) Getzler, Ezra . A short proof of the local Atiyah-Singer index theorem. Topology 25 (1986), no. 1, 111--117 – Bruce Westbury Feb 16 '11 at 12:57
The Manin-Mumford Conjecture, first proved by Raynaud in 1983, states that the points on a curve $X$ of genus 2 or more that are torsion when embedded into its Jacobian are finite in number. The bound is also independent of how you embed the curve (that is, it is independent of the "base point").
Ken Ribet reproved this using the notion of an "almost rational torsion point". This is cool because it can lead to explicit versions of Manin-Mumford. The idea is that the set we're interested in is contained in the set of almost rational torsion points, which Ribet proves is finite, together with hyperelliptic branch points if the curve happens to be hyperelliptic.
So finding the almost rational torsion on jacobians of curves might help make things explicit. Doing this for modular jacobians, Ribet proved that
$X_0(N) \cap J_0(N)_{tors} = \{0,\infty\}$,
unless $X_0(N)$ is hyperelliptic, in which case you simply add the branch points.
To be honest I should say that this theorem was first proved by Matthew Baker (who actually gives two proofs), and independently around the same time by A. Tamagawa. Ribet's approach is similar to Bakers' second approach. For a survey, google "Torsion points on modular curves and Galois theory" to summon a pdf by Ribet and Minhyong Kim.
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The classic example from mathematical physics is Richard Feynman's Space-Time approach to nonrelativistic quantum mechanics (1948), which (in essence) proved that the Green function of the Schroedinger equation was equal to a path integral. The article begins:
It is a curious historical fact that modern quantum mechanics began with two quite different mathematical formulations: the differential equation of Schroedinger, and the matrix algebra of Heisenbert. [...] This paper will describe what is essentially a third formulation of non-relativistic quantum theory.
As for the value of seeking multiple derivations, we have Feynman's Nobel Address The Development of the Space-Time View of Quantum Electrodynamics (1965):
There is always another way to say the same thing that doesn't look at all like the way you said it before. I don't know what the reason for this is. I think it is somehow a representation of the simplicity of nature. [...] Perhaps a thing is simple if you can describe it fully in several different ways without immediately knowing that you are describing the same thing.
In a classical context, we have Saunders Mac Lane in Hamiltonian mechanics and geometry (1970) presenting new geometric analyses of old dynamical problems:
Mathematical ideas do not live fully till they are presented clearly, and we never quite achieve that ultimate clarity. Just as each generation of historians must analyse the past again, so in the exact sciences we must in each period take up the renewed struggle to present as clearly as we can the underlying ideas of mathematics.
In the mid-1970s these various derivations came together as Fadeev and Popov's (1974) Covariant quantization of the gravitational field, which provided the foundations for todays' gold-standard method of BRST quantization, for which van Holten's Aspects of BRST quantization (2002) is a good review:
Quite often the preferred dynamical equations of a physical system are not formulated directly in terms of observable degrees of freedom, but in terms of more primitive quantities [...] Out of these roots has grown an elegant and powerful framework for dealing with quite general classes of constrained systems using ideas borrowed from algebraic geometry.
By this 90-year process of successive rederivations, we nowadays have arrived at a more nearly global appreciation—encompassing both classical and quantum dynamics—of the ideas that Terry Tao's essay What is a Gauge? discusses.
Cutting-edge research in classical, quantum, and (increasingly common) hybrid dynamical systems uses all of these mathematical approaches, each formally equivalent to all the others ... but with very different ideas behind them. The resulting naturality has lent new passion to the longstanding romance between mathematics and physics.
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Heisenbert!! I love that guy. – David Hansen Feb 11 '11 at 5:54
I personally prefer Heisenernie, but to each his own. – Igor Rivin Feb 11 '11 at 6:25
Lol ... I was gonna fix it ... but once I discovered that apparently no one named "Heisenbert" has ever written any scientific article (at least, none indexed on MathSciNet, Inspec, or Pubmed) ... well, I couldn't bring myself to terminate the poor guy ... so his name now is a link to a picture. Like Edward Abbey's character George Hayduke, like WWII's Kilroy, and like the immortal Bourbaki ... "Heisenbert lives!" :) – John Sidles Feb 11 '11 at 6:59
Hrushovski's Model-Theoretic proof for the Mordell-Lang conjecture over function fields is an example.
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Hilbert basis theorem, a nonconstructive proof replacing and generalizing a monstrous explicit calculation ("this is not mathematics, this is theology").
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Perhaps, this can be expanded a bit. In classical invariant theory, the generators of the rings of invariants were found by explicit calculation. Whereas Hilbert showed that finite sets of generators must exist on a priori grounds. This was indeed a paradigm shift. – Donu Arapura Feb 11 '11 at 15:58
I guess that this may not be an answer to the question however. – Donu Arapura Feb 11 '11 at 16:05
If you're interested in something that's expected to do this (here, with a related paper here), but is a very current project, there's Lazic's proof of the finite generation of (log) canonical rings. I don't know of any great insights gained from the new proof, other than the surprising fact that it's POSSIBLE to prove it this way, and that the method, rather than requiring the Mori program to prove the theorem, allows a proof of many important theorems in the Mori program from it. This is, though, quite a work in progress.
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Serre's construction of the $p$-adic zeta function, using the fact that the values of the Riemann zeta function at negative integers are constant terms of Eisenstein series. This paved the way for a lot a mathematics including the proof of Iwasawa main conjecture by Mazur and Wiles.
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https://www.physicsforums.com/threads/limit-definition-of-indefinite-integrals.787766/ | # Limit Definition of Indefinite Integrals?
• #1
92
0
Hello,
I was just wondering, we have what could be called the indefinite derivative in the form of d/dx x^2=2x & evaluating at a particular x to get the definite derivative at that x. But with derivation, we can algebraically manipulate the limit definition of a derivative to actually evaluate to 2x from x^2. Is there a similar process available to algebraically manipulate a limit definition of an indefinite integral to get the expected result?
Integration is normally just taught as the reverse of derivation, and while that works of course, I was just curious if there was a way to directly determine the indefinite integral of a function by using limits or differentials.
## Answers and Replies
• #2
Stephen Tashi
Science Advisor
7,642
1,496
A function (if it has a derivative) has a unique derivative. The "indefinite integral" (antiderivative) of a function is not unique. You can apply a definition of integration and obtain a particular indefinite integral of a function f(x). Pick an arbitrary value x = a and compute the area under the curve from x = a to x = b using whatever definition of integration you learned (for example, as the limit of a Riemann sums). The result of this computation gives you a value F(b) that is a function of b. Considering "b" to be a variable itself , F(b) is is an antiderivative of f(x). You could just as well write F(b) as F(x) since the names of variables don't matter when there no relation between the two variables.
Definite Integration is not introduced as the reverse of integration in most calculus texts. After you become familiar with The Fundamental Theorem Of Calculus, you may get the impression that integration is just the reverse of differentiation and forget how definite integration was actually defined.
• #3
92
0
I hope I haven't missed anything, but the examples you gave all seem to deal with definite integrals, or if they obtain the indefinite integral, it is implied that derivation rules were applied in reverse (i.e. power rule, d/dx of ln(x)=1/x, etc.). As for the definite integral, if I recall correctly, it was taught starting with Riemann sums so I do get that. I just mean when the indefinite integral was introduced, usually it's just taught as the reverse process of obtaining the "indefinite derivative", just with the addition of a constant.
To add clarity, when I say being able to algebraically manipulate a limit definition to get the indefinite integral, I mean doing so without resorting to the reverse power rule, and familiar reversed rules based on the derivatives of ln(x), sin(x), etc. Am I making any sense here?
• #4
Stephen Tashi
Science Advisor
7,642
1,496
To add clarity, when I say being able to algebraically manipulate a limit definition to get the indefinite integral, I mean doing so without resorting to the reverse power rule, and familiar reversed rules based on the derivatives of ln(x), sin(x), etc. Am I making any sense here?
If you are asking whether you can begin with the expression $lim_{h\rightarrow 0} \frac{ F(x+h) - F(x)} {h} = f(x)$ and solve it to find $F(x)$ by taking limits with out considering the values of $f(x)$ over an entire interval then I'd say the answer is no, you can't.
You can define $F(x)$ as $F(x) = \int_a^x {f(x) dx}$ and apply the definition $F'(x) = lim_{h\rightarrow 0} \frac{F(x+h) - F(x)}{h}$ You'd be applying a limit definition to a function defined by a definite integral. So the limit is affected the values $f(x)$ takes on an interval.
• #5
92
0
If you are asking whether you can...with out considering the values of f(x) f(x) over an entire interval then I'd say the answer is no, you can't.
Yes this is generally what I'm talking about, though I didn't mean to imply any sort of restriction such as not considering values over the entire interval. It just so happens that I kept running into the fact that I know of no way of determining directly the indefinite integral without applying derivation in reverse. I was imagining some sort of sum, maybe using sigma notation, where summing all possible infinitesimal intervals of x times the value of the integrand at that x would yield the indefinite integral.
You can define F(x) F(x) as F(x)=∫xaf(x)dx F(x) = \int_a^x {f(x) dx} and apply the definition F′(x)=limh→0F(x+h)−F(x)h F'(x) = lim_{h\rightarrow 0} \frac{F(x+h) - F(x)}{h} You'd be applying a limit definition to a function defined by a definite integral. So the limit is affected the values f(x) f(x) takes on an interval.
Yeah I remember that now that you mentioned it, but as you mentioned, that defines F(x) as a function of a definite integral so would still implicitly be using reverse derivation to determine the indefinite integral in order to actually evaluate F(x).
• #6
Mark44
Mentor
34,904
6,648
It just so happens that I kept running into the fact that I know of no way of determining directly the indefinite integral without applying derivation in reverse.
Although it doesn't seem to make sense, we don't call it derivation when we take the derivative. We call it differentiation.
In contrast, completion of the square is used in the derivation of the quadratic formula.
• #7
Stephen Tashi
Science Advisor
7,642
1,496
Yeah I remember that now that you mentioned it, but as you mentioned, that defines F(x) as a function of a definite integral so would still implicitly be using reverse derivation to determine the indefinite integral in order to actually evaluate F(x).
No, it wouldn't. The definition of a definite integral does not depend on using reverse diffentiation. You're confusing the technique of doing a definite integral with the definition of a definite integral.
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4K | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9849992990493774, "perplexity": 373.48941143890795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991801.49/warc/CC-MAIN-20210515100825-20210515130825-00589.warc.gz"} |
https://hilbertthm90.wordpress.com/2009/11/12/completions-i/ | # Completions I
Today we’ll start a new section, but only because it is a tool we need when we come back to the stuff we just finished. We will look at completions.
To motivate the process take a Hausdorff abelian topological group $G$. Suppose there is a countable local basis at 0 (which implies countable basis, since the neighborhoods of 0 determine the entire topology). Since we assumed Hausdorff we have the usual notion of Cauchy sequences, so we can define the completion of $G$ to be completion in the usual sense $\hat{G}$. In particular, if $G=\mathbb{Q}$, then $\hat{\mathbb{Q}}=\mathbb{R}$.
Now suppose we have a local basis about 0 of subgroups (this rules out $\mathbb{Q}$), say $G=G_0\supset G_1\supset \cdots \supset G_n\supset \cdots$. If we are in this situation, then our topology is actually determined by a sequence of subgroups, so we will want to try to define the completion solely in terms of algebra.
Take any Cauchy sequence $(x_n)\subset G$. If we fix k, then at some $M(k)$, $\overline{x_n}\in G/G_k$ is constant for all $n\geq M(k)$. Note that $M$ really does depend on $k$. Set the limit $\overline{x_n}\to x_{M(k)}$.
If we make what we mod out by bigger, namely we go from $k+1$ to $k$, then projection $\theta_{k+1}: G/G_{k+1}\to G/G_k$ maps $x_{M(k+1)}\mapsto x_{M(k)}$. Thus our Cauchy sequence $(x_n)$ determined a “coherent sequence” $(x_{M(k)})$, where $\theta_{n+1}x_{M(n+1)}=x_{M(n)}$.
Conversely, we can define a Cauchy sequence corresponding to any coherent sequence by just picking an element in the equivalence class at each step. So we can now define the completion $\hat{G}$ to be the set of coherent sequences with group structure given entry-wise by the quotient group. The standard example here is the $p$-adic integers, where the group is $\mathbb{Z}$ and our fundamental system is $\mathbb{Z}\supset p\mathbb{Z}\supset p^2\mathbb{Z}\supset \cdots \supset p^n\mathbb{Z}\supset \cdots$. Coherent sequences are $(a_0, a_1, \ldots )$ where $a_{n+1}\mod p^n\equiv a_n$.
Whenever we have in general a sequence of groups $\{A_n\}$ and homomorphisms $\theta_{n+1} A_{n+1}\to A_n$ this is called an inverse system. The group of all coherent sequences is called the inverse limit of the system written $\displaystyle \lim_{\longleftarrow} A_n$. Thus our definition of completion can be written succinctly as $\displaystyle\hat{G}=\lim_{\longleftarrow} G/G_n$.
Next time we’ll transfer this to module language and get to a few results. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 31, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9972168803215027, "perplexity": 171.54828329508575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105304.35/warc/CC-MAIN-20170819051034-20170819071034-00290.warc.gz"} |
https://googology.fandom.com/wiki/Category:Up-arrow_notation_level | 11,865
pages
This category contains numbers that can be defined or estimated using up-arrow notation, which is equivalent to Jonathan Bowers' 3-entrical arrays.
The lower bound of this category is $$10 \uparrow\uparrow\uparrow 3 = 10 \uparrow\uparrow 10 \uparrow\uparrow 10$$, and the upper bound is $$f_\omega(f_3(10))$$.
All items (1664)
A
B
C
D | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9983705878257751, "perplexity": 1547.0638073574044}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571989.67/warc/CC-MAIN-20220813232744-20220814022744-00191.warc.gz"} |
https://www.allaboutcircuits.com/technical-articles/more-first-order-opamp-circuits-differentiators/ | Do more than twiddle gain and phase!
Differentiate Yourself
All of our previous circuits dealt with opamps using resistors in the feedback loop. What happens when we start swapping out resistors for other passive components? Let's take the inverting amplifier circuit that we looked at in the last opamp article and replace the input resistor Rin with a capacitor.
Figure 1: A Different Inverting Circuit
First thing's first - let's figure out the gain of this circuit. If we recall from our first article, we know that the current through the feedback path has to be equal to the current through the input path. The current through a capacitor is given by the following equation:
${i_{C}}=C\frac{dv}{dt}$
When we substitute that into the feedback equation, we get the following relationship:
$i_{C} = -i_{R_{f}} \Rightarrow C \frac{dV_{in}}{dt} = - \frac{V_{out}}{R_{f}} \Rightarrow -RC\frac{dV_{in}}{dt} = V_{out}$
This circuit is known as a _differentiator_, because the output is proportional to the first derivative of the input voltage. We can see a few examples in the following image.
Figure 2: Differentiator Input/Output Waveforms
The sine wave shouldn't be surprising - the first derivative of a sine wave is a cosine wave. You can see that the differentiator output reflects this in the form of a phase shift, with the output lagging the input by 270 degrees (90 for the cosine shift, 180 for the opposite polarity). A triangular wave has a constant rate of change until it changes direction. The differentiator responds to this sort of wave by fixing the output at a DC level proportional to the slope of the triangular wave, multiplied by the passive values RC. The square wave ends up producing a series of quick pulses out of the differentiator. These pulses represent the very quick transitions of the square wave's edges. An ideal differentiator would output an impulse signal in response to an ideal square wave; this example has a slight trailoff that represents a square wave reaching its final value.
A Matter of Frequency
One thing we've put off discussing is the frequency response of a differentiator circuit. Since this differentiator employs a capacitor in its input path, we know that its performance will be subject to some effects of frequency just by virtue of the capacitor's presence. We can derive the differentiator's transfer function pretty handily by shifting it into the Laplace domain, where a first derivative equals multiplication by s:
$RC\frac{dV_{in}}{dt} = V_{out} \Rightarrow RCs(V_{in}) + f(0) = V_{out}$
$\frac{V_{out}}{V_{in}} = RCs \Rightarrow \frac{V_{out}}{V_{in}} = RCj\omega$
(Note - we'll assume the capacitor in this example is discharged so we can eliminate the f(0) term.)
Cool! Let's generate a Bode plot of this circuit to get an idea of how its response changes with frequency.
Figure 3: Differentiator Bode Plot
As you can see, phase of the output is always shifted 90 degrees from the input. You can also see that the magnitude of the gain is increasing without bound as frequency increases. As long as the frequency of the input is above the unity gain frequency, it's going to get amplified. (As an aside - can you see what the unity gain frequency is?) In a way, that's kind of cool - we've found a circuit that can supply gain proportional to frequency! In practice, though, it's not so cool. Since gain is increasing with frequency, it means that differentiators are exceptionally sensitive to noise, especially high frequency noise. Thus, it's sensible to limit the gain, for the sake of having a circuit that's not drowned in noise!
Reasonable Limits
Ideally, we'd like the signals in the band region of interest to have a moderately level gain applied by the differentiator. To achieve this, we need a way to compensate for the monotonic increase in gain at high frequencies. The simplest way to achieve this is with a few extra components:
Figure 4: Differentiator with Added Input Resistor
The easiest addition to make is an extra resistor to the input of the differentiator. As frequency of the input signal increases, the impedance of the capacitor will trend towards zero. The effect of this is pretty convenient:
$i_{f} = \frac{V_{in}}{(R_{1} + \frac{1}{sC})} = \frac{- V_{out}}{R_{2}}$
$\left | \frac{V_{out}}{V_{in}} \right | = \frac{R_{2}}{(R_{1} + \frac{1}{sC})}$
As $s \Rightarrow \infty , \frac{1}{sC} \Rightarrow 0$
$\left | \frac{V_{out}}{V_{in}} \right | \approx \frac{R_{2}}{R_{1}}, for freq > \frac{1}{R_{1} C_{1}}$
At high frequencies, this circuit stops behaving like a differentiator and starts behaving like a simple inverting amplifier. Sweet! We've managed to bound the gain of high frequency noise. What if we want to take it a step farther, and actively reduce the high frequency noise?
Figure 5: Differentiator with Added Input Resistor and Feedback Capacitor
An additional cap in parallel with R2 will end up having similar behavior as C1. At a certain frequency, it will start to behave as a low impedance that just gets lower as the frequency rises. The effect on the feedback path is to slowly short out the feedback resistor R2 at higher and higher frequencies. This is effectively the same as decreasing the resistance of the feedback resistor Rf in an inverting amplifier - the parallel cap allows you to keep the gain for higher frequencies down.
Why do we care about limiting high frequency noise in differentiators so much? The simple answer is stability. As you've probably noticed in our last article on opamps, all of the opamp circuits we've seen employ feedback.
Wrapping Up
Differentiators are amazingly useful in control circuits, where it's frequently useful to monitor the rate of change of a plant. With a little tuning, they're very simple to make work for you!
By the way - did you figure out the unity gain frequency of the ideal differentiator? Just so we don't leave you hanging:
$\left | \frac{V_{out}}{V_{in}} \right | = 1 = \omega RC \Rightarrow \omega_{o} = \frac{1}{RC}$
Until next time! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8872369527816772, "perplexity": 676.5852800830627}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818689490.64/warc/CC-MAIN-20170923052100-20170923072100-00490.warc.gz"} |
https://nhigham.com/2020/04/22/what-is-a-random-orthogonal-matrix/ | What Is a Random Orthogonal Matrix?
Various explicit parametrized formulas are available for constructing orthogonal matrices. To construct a random orthogonal matrix we can take such a formula and assign random values to the parameters. For example, a Householder matrix $H = I - 2uu^T/(u^Tu)$ is orthogonal and symmetric and we can choose the nonzero vector $u$ randomly. Such an example is rather special, though, as it is a rank-$1$ perturbation of the identity matrix.
What is usually meant by a random orthogonal matrix is a matrix distributed according to the Haar measure over the group of orthogonal matrices. The Haar measure provides a uniform distribution over the orthogonal matrices. Indeed it is invariant under multiplication on the left and the right by orthogonal matrices: if $Q$ is from the Haar distribution then so is $UQV$ for any orthogonal (possibly non-random) $U$ and $V$. A random Householder matrix is not Haar distributed.
A matrix from the Haar distribution can be generated as the orthogonal factor in the QR factorization of a random matrix with elements from the standard normal distribution (mean $0$, variance $1$). In MATLAB this is done by the code
[Q,R] = qr(randn(n));
Q = Q*diag(sign(diag(R)));
The statement [Q,R] = qr(randn(n)), which returns the orthogonal factor $Q$, is not enough on its own to give a Haar distributed matrix, because the QR factorization is not unique. The second line adjusts the signs so that $Q$ is from the unique factorization in which the triangular factor $R$ has nonnegative diagonal elements. This construction requires $2n^3$ flops.
A more efficient construction is possible, as suggested by Stewart (1980). Let $x_k$ be an $(n-k+1)$-vector of elements from the standard normal distribution and let $H_k$ be the Householder matrix that reduces $x_k$ to $r_{kk}e_1$, where $e_1$ is the first unit vector. Then $Q = DH'_1H'_2\dots H'_{n-1}$ is Haar distributed, where $H'_k = \mathrm{diag}(I_{k-1}, H_k)$, $D = \mathrm{diag}(\mathrm{sign}(r_{kk}))$, and $r_{nn} = x_n$. This construction expresses $Q$ as the product of $n-1$ Householder matrices of growing effective dimension, and the product can be formed from right to left in $4n^3/3$ flops. The MATLAB statement Q = gallery('qmult',n) carries out this construction.
A similar construction can be made using Givens rotations (Anderson et al., 1987).
Orthogonal matrices from the Haar distribution can also be formed as $Q = A (A^TA)^{-1/2}$, where the elements of $A$ are from the standard normal distribution. This $Q$ is the orthogonal factor in the polar decomposition $A = QH$ (where $H$ is symmetric positive semidefinite).
Random orthogonal matrices arise in a variety of applications, including Monte Carlo simulation, random matrix theory, machine learning, and the construction of test matrices with known eigenvalues or singular values.
All these ideas extend to random unitary (complex) matrices. In MATLAB, Haar distributed unitary matrices can be constructed by the code
[Q,R] = qr(complex(randn(n),randn(n)));
Q = Q*diag(sign(diag(R)));
This code exploits the fact that the $R$ factor computed by MATLAB has real diagonal entries. If the diagonal of $R$ were complex then this code would need to be modified to use the complex sign function given by $\mathrm{sign}(z) = z/|z|$.
References
This is a minimal set of references, which contain further useful references within. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 34, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9700536131858826, "perplexity": 181.73182514483292}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337723.23/warc/CC-MAIN-20221006025949-20221006055949-00055.warc.gz"} |
https://amathew.wordpress.com/category/number-theory/ | ### number theory
The purpose of this post is to describe an application of the general intersection theory machinery (for curves on surfaces) developed in the previous posts: the Weil bound on points on a curve over a finite field.
1. Statement of the Weil bound
Let ${C}$ be a smooth, projective, geometrically irreducible curve over ${\mathbb{F}_q}$ of genus $g$. Then the Weil bound states that:
$\displaystyle |C(\mathbb{F}_q) - q - 1 | \leq 2 g \sqrt{q}.$
Weil’s proof of this bound is based on intersection theory on the surface ${C \times C}$. More precisely, let
$\displaystyle \overline{C} = C \times_{\mathbb{F}_q} \overline{\mathbb{F}_q},$
so that ${\overline{C}}$ is a smooth, connected, projective curve. It comes with a Frobenius map
$\displaystyle F: \overline{C} \rightarrow \overline{C}$
of ${\overline{\mathbb{F}_q}}$-varieties: in projective coordinates the Frobenius runs
$\displaystyle [x_0: \dots : x_n] \mapsto [x_0^q: \dots : x_n^q].$
In particular, the map has degree ${q}$. One has
$\displaystyle C( \mathbb{F}_q) = \mathrm{Fix}(F, \overline{C}(\overline{\mathbb{F}}_q))$
representing the ${\mathbb{F}_q}$-valued points of ${C}$ as the fixed points of the Frobenius (Galois) action on the ${\overline{\mathbb{F}_q}}$-valued points. So the strategy is to count fixed points, using intersection theory.
Using the (later) theory of ${l}$-adic cohomology, one represents the number of fixed points of the Frobenius as the Lefschetz number of ${F}$: the action of ${F}$ on ${H^0}$ and ${H^2}$ give the terms ${q+1}$. The fact that (remaining) action of ${F}$ on the ${2g}$-dimensional vector space ${H^1}$ can be bounded is one of the Weil conjectures, proved by Deligne for general varieties: here it states that ${F}$ has eigenvalues which are algebraic integers all of whose conjugates have absolute value ${\sqrt{q}}$. (more…)
This is the third in the series of posts intended to work through the proof of Lazard’s theorem, that the Lazard ring classifying the universal formal group law is actually a polynomial ring on a countable set of generators. In the first post, we reduced the result to an elementary but tricky “symmetric 2-cocycle lemma.” In the previous post, we proved most of the symmetric 2-cocycle lemma, except in characteristic zero. The case of characteristic zero is not harder than the cases we handled (it’s easier), but in this post we’ll complete the proof of that case by exhibiting a very direct construction of logarithms in characteristic zero. Next, I’ll describe an application in Lazard’s original paper, on “approximate” formal group laws.
After this, I’m going to try to move back to topology, and describe the proof of Quillen’s theorem on the formal group law of complex cobordism. The purely algebraic calculations of the past couple of posts will be necessary, though.
1. Formal group laws in characteristic zero
The last step missing in the proof of Lazard’s theorem was the claim that the map
$\displaystyle L \rightarrow \mathbb{Z}[b_1, b_2, \dots ]$
classifying the formal group law obtained from the additive one by “change-of-coordinates” by the exponential series ${\exp(x) = \sum b_i x^{i+1}}$ is an isomorphism mod torsion. In other words, we have an isomorphism
$\displaystyle L \otimes \mathbb{Q} \simeq \mathbb{Z}[b_1, b_2, \dots ] \otimes \mathbb{Q}.$
In fact, we didn’t really need this: we could have proved the homological 2-cocycle lemma in all cases, instead of just the finite field case, and it would have been easier. But I’d like to emphasize that the result is really something elementary here. In fact, what it is saying is that to give a formal group law over a ${\mathbb{Q}}$-algebra is equivalent to giving a choice of series ${\sum b_i x^{i+1}}$.
Definition 1 An exponential for a formal group law ${f(x,y) }$ is a power series ${\exp(x) = x + b_1 x^2 + \dots}$ such that
$\displaystyle f(x,y) = \exp( \exp^{-1}(x) + \exp^{-1}(y)).$
The inverse power series ${\exp^{-1}(x)}$ is called the logarithm.
That is, a logarithm is an isomorphism of ${f}$ with the additive formal group law.
So another way of phrasing this result is that:
Proposition 2 A formal group law over a ${\mathbb{Q}}$-algebra has a unique logarithm (i.e., is uniquely isomorphic to the additive one). (more…)
We are in the middle of proving an important result of Lazard:
Theorem 1 The Lazard ring ${L}$ over which the universal formal group law is defined is a polynomial ring in variables ${x_1, x_2, \dots, }$ of degree ${2i}$.
The fact that the Lazard ring is polynomial implies a number of results which are not a priori obvious: for instance, it shows that given a surjection of rings ${ A \twoheadrightarrow B}$, then any formal group law on ${B}$ can be lifted to one over ${A}$.
We began the proof of Lazard’s theorem last time: we produced a map
$\displaystyle L \rightarrow \mathbb{Z}[b_1, b_2, \dots ], \quad \deg b_i = 2i,$
classifying the formal group law obtained from the additive one ${x+y}$ by the “change of coordinates” ${ \exp(x) = \sum b_i x^{i+1}}$. We claimed that the map on indecomposables was injective, and that, in fact the image in the indecomposables of ${\mathbb{Z}[b_1, b_2, \dots ]}$ could be determined completely. I won’t get into the details of this (it was all in the previous post), because the purpose of this post is to prove a result to which we reduced last time.
Let ${A}$ be an abelian group. A symmetric 2-cocycle is a “polynomial” ${P(x,y) \in A[x, y] = A \otimes_{\mathbb{Z}} \mathbb{Z}[x, y]}$ with the properties:
$\displaystyle P(x, y) = P(y,x)$
and
$\displaystyle P(x, y+z) + P(y, z) = P(x,y) + P(x+y, z).$
These symmetric 2-cocycles come up when one tries to classify formal group laws over the ring ${\mathbb{Z} \oplus A}$, as we saw last time: in fact, we can think of them as “deformations” of the additive formal group law.
The main lemma which we stated last time was the following:
Theorem 2 (Symmetric 2-cocycle lemma) A homogeneous symmetric 2-cocycle of degree ${n}$ is a multiple of ${\frac{1}{d} ( ( x+y)^n - x^n - y^n )}$ where ${d =1}$ if ${n}$ is not a power of a prime, and ${d = p}$ if ${n = p^k}$.
For a direct combinatorial proof of this theorem, see Lurie’s notes. I want to describe a longer homological proof, which is apparently due to Mike Hopkins and which appears in the COCTALOS notes. The strategy is to interpret these symmetric 2-cocycles as actual cocycles in a cobar complex computing an ${\mathrm{Ext}}$ group. Then, the strategy is to compute this ${\mathrm{Ext}}$ group independently.
This argument is somewhat longer than the combinatorial one, but it has the benefit (for me) of engaging with some homological algebra (which I need to learn more about), as well as potentially generalizing in other directions. (more…)
After describing the computation of ${\pi_* MU}$, I’d now like to handle the remaining half of the machinery that goes into Quillen’s theorem: the structure of the universal formal group law.
Let ${R}$ be a (commutative) ring. Recall that a formal group law (commutative and one-dimensional) is a power series ${f(x,y) \in R[[x,y]]}$ such that
1. ${f(x,y) = f(y,x)}$.
2. ${f(x, f(y,z)) = f(f(x,y), z)}$.
3. ${f(x,0) = f(0,x) = x}$.
It is automatic from this by a successive approximation argument that there exists an inverse power series ${i(x) \in R[[x]]}$ such that ${f(x, i(x)) = 0}$.
In particular, ${f}$ has the property that for any ${R}$-algebra ${S}$, the nilpotent elements of ${S}$ become an abelian group with addition given by ${f}$.
A key observation is that, given ${R}$, to specify a formal group law amounts to specifying a countable collection of elements ${c_{i,j}}$ to define the power series ${f(x,y) = \sum c_{i,j} x^i y^j}$. These ${c_{i,j}}$ are required to satisfy various polynomial constraints to ensure that the formal group identities hold. Consequently:
Theorem 1 There exists a universal ring ${L}$ together with a formal group law ${f_{univ}(x,y)}$ on ${L}$, such that any FGL ${f}$ on another ring ${R}$ determines a unique map ${L \rightarrow R}$ carrying ${f_{univ} \mapsto f}$. (more…)
Apologies for the lack of posts lately; it’s been a busy semester. This post is essentially my notes for a talk I gave in my analytic number theory class.
Our goal is to obtain bounds on the distribution of prime numbers, that is, on functions of the form ${\pi(x)}$. The closely related function
$\displaystyle \psi(x) = \sum_{n \leq x} \Lambda(n)$
turns out to be amenable to study by analytic means; here ${\Lambda(n)}$ is the von Mangolt function,
$\displaystyle \Lambda(n) = \begin{cases} \log p & \text{if } n = p^m, p \ \text{prime} \\ 0 & \text{otherwise} \end{cases}.$
Bounds on ${\psi(x)}$ will imply corresponding bounds on ${\pi(x)}$ by fairly straightforward arguments. For instance, the prime number theorem is equivalent to ${\psi(x) = x + o(x)}$.
The function ${\psi(x)}$ is naturally connected to the ${\zeta}$-function in view of the formula
$\displaystyle - \frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^\infty \Lambda(n) n^{-s}.$
In other words, ${- \frac{\zeta'}{\zeta}}$ is the Dirichlet series associated to the function ${\Lambda}$. Using the theory of Mellin inversion, we can recover partial sums ${\psi(x) = \sum_{n \leq x} \Lambda(x)}$ by integration of ${-\frac{\zeta'}{\zeta}}$ along a vertical line. That is, we have
$\displaystyle \psi(x) = \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} -\frac{\zeta'(s)}{\zeta(s)} \frac{x^s}{s} ds ,$
at least for ${\sigma > 1}$, in which case the integral converges. Under hypotheses on the poles of ${-\frac{\zeta'}{\zeta}}$ (equivalently, on the zeros of ${\zeta}$), we can shift the contour appropriately, and estimate the integral to derive the prime number theorem. (more…)
I’d like to explain what I think is a pretty piece of mathematics.
Deligne’s proof of the Weil conjectures, and his strengthenings of them, in his epic paper “La conjecture de Weil II” have had, needless to say, many applications—far more than the initial paper where he proved the last of the Weil conjectures. One of the applications is to bounding exponential sums. I’d like to begin sketching the idea today. I learned this from the article “Sommes trigonometriques” in SGA 4.5, which is very fun to read.
1. The trace formula
Let’s say you have some variety ${X_0}$ over a finite field ${\kappa}$, say separated. Suppose you have a ${l}$-adic sheaf ${\mathcal{F}_0}$ on ${X_0}$. I’ll denote by “dropping the zero” the base change to ${\overline{\kappa}}$, so ${\mathcal{F}}$ denotes the pull-back to ${X = X_0 \times_{\kappa} \overline{\kappa}}$.
For each point ${x \in X_0}$, we can take the “geometric stalk” ${\mathcal{F}_{\overline{x}}}$ (or ${\mathcal{F}_{0 \overline{x}}}$) which is given by finding a map ${\mathrm{Spec} \overline{\kappa} \rightarrow X_0}$ hitting ${x}$, and pulling ${\mathcal{F}_0}$ back to it. This “stalk” is automatically equivariant with respect to the Galois group ${\mathrm{Gal}(\overline{\kappa}/k(x))}$ for ${k(x)}$ the residue field, and as a result we can compute the trace of the geometric Frobenius ${F_x}$ of ${k(x) \hookrightarrow \overline{\kappa}}$—that’s the inverse of the usual Frobenius—and take its trace on ${\mathcal{F}_{\overline{x}}}$.
So the intuition here is that we’re taking local data of the sheaf ${\mathcal{F}_0}$: just the trace of its Frobenius at each point. One interesting interpretation of this procedure was discussed on this MO question: namely, the process is analogous to “integration over a contour.” Here the contour is ${\mathrm{Spec} k(x)}$, which has cohomological dimension one and is thus the étale version of a curve.
Now, the trace formula says that the local data of these traces all pieces into a simple piece of global data, which is the compactly supported cohomology. (more…)
The topic of this post is a curious functor, discovered by Deligne, on the category of sheaves over the affine line, which is a “sheafification” of the Fourier transform for functions.
Recall that the classical Fourier transform is an almost-involution of the Hilbert space ${L^2(\mathbb{R})}$. We shall now discuss the Fourier-Deligne transform, which is an almost-involution of the bounded derived category of ${l}$-adic sheaves, ${\mathbf{D}^b_c(\mathbb{A}^1_{\kappa}, \overline{\mathbb{Q}_l})}$. The Fourier transform is defined by multiplying a function with a character (which depends on a parameter) and integrating. Analogously, the Fourier-Deligne transform will twist an element of ${\mathbf{D}^b_c(\mathbb{A}^1_{\kappa}, \overline{\mathbb{Q}_l})}$ by a character depending on a parameter, and then take the cohomology.
More precisely, consider the following: let $G$ be a LCA group, $G^*$ its dual. We have a canonical character on $\phi: G \times G^* \to \mathbb{C}^*$ given by evaluation. To construct the Fourier transform $L^2(G) \to L^2(G^*)$, we start with a function $f: G \to \mathbb{C}$. We pull back to $G \times G^*$, multiply by the evaluation character $\phi$ defined above, and integrate along fibers to give a function on $G^*$.
Everything we’ve done here has a sheaf-theoretic analog, however: pulling back a function corresponds to the functorial pull-back of sheaves, multiplication by a character corresponds to tensoring with a suitable line bundle, and integration along fibers corresponds to the lower shriek push-forward. Much of the classical formalism goes over to the sheaf-theoretic case. One can prove an “inversion formula” analogous to the Fourier inversion formula (with a Tate twist).
Why should we care? Well, Laumon interpreted the Fourier transform as a suitable “deformation” of the cohomology of a suitable sheaf on the affine line, and used it to give a simplified proof of the main results of Weil II, without using scary things like vanishing cycles and Picard-Lefschetz theory. The Fourier transform also behaves very well with respect to perverse sheaves: it is an auto-equivalence of the category of perverse sheaves, because of the careful way in which it is calibrated. Its careful use can be used to simplify some of the arguments in BBD that also rely on other scary things.
I’d like to take a quick (one-post) break from simplicial methods. This summer, I will be studying étale cohomology and the proofs of the Weil conjectures through the HCRP program. I have currently been going through the basic computations in étale cohomology, and, to help myself understand one point better, would like to mention a very pretty and elementary argument I recently learned from Johan de Jong’s course notes on the subject (which are a chapter in the stacks project).
1. Motivation via étale cohomology
When doing the basic computations of the étale cohomology of curves, one of the important steps is the computation of the sheaf ${\mathcal{O}_X^*}$ (that is, the multiplicative group of units), and in doing this one needs to know the cohomology of the generic point. That is, one needs to compute
$\displaystyle H^*(X_{et}, \mathcal{O}_X^*)$
where ${X = \mathrm{Spec} K}$ for ${K}$ a field of transcendence degree one over the algebraically closed ground field, and the “et” subscript means étale cohomology. Now, ultimately, whenever you have the étale cohomology of a field, it turns out to be the same as Galois cohomology. In other words, if ${X = \mathrm{Spec} K}$, then the small étale site of ${X}$ is equivalent to the site of continuous ${G = \mathrm{Gal}(K^{sep}/K)}$-sets, and consequently the category of abelian sheaves on this site turns out to be equivalent to the category of continuous ${G}$-modules. Taking the étale cohomology of this sheaf then turns out to be the same as taking the group cohomology of the associated ${G}$-module. So, if you’re interested in étale cohomology, then you’re interested in Galois cohomology. In particular, you are interested in things like group cohomologies of the form
$\displaystyle H^2(\mathrm{Gal}(K^{sep}/K), (K^{sep})^*).$ (more…)
I asked a question on what Tate’s thesis was really about on math.SE, and Matthew Emerton has posted a very thoughtful and detailed answer. You should go read it. Actually, I recommend looking at all his answers on the website, which are easily some of the best answers there.
(Apparently I am not the first person to notice this.)
Last time, we showed that prime numbers admit succinct certificates. Given a number containing ${k}$ bits, one could produce a “proof” that the number is prime in polynomial in ${k}$ space. In other words, the language ${PRIMES}$ containing all primes (represented in the binary expansion, say) lies in the complexity class ${\mathbf{NP}}$.
In practice, though, being in ${\mathbf{NP}}$ does not give a good way of solving the problem by itself; for that, we want the problem to be in ${\mathbf{P}}$. Perhaps, though, that is too strict. One relaxation is to consider classes of problems that can be solved by randomized algorithms that run in polynomial time—randomized meaning that the right output comes out with high probability, that is. This is the class ${\mathbf{BPP}}$: by definition, something belongs to the class ${\mathbf{BPP}}$ if there is a polynomial-time nondeterministic Turing machine such that on any computation, at least ${2/3}$ of the computation tree leads to the right answer. So the idea of ${\mathbf{BPP}}$ is that, to decide whether a string ${x}$ belongs to our language ${\mathcal{L}}$, we generate a bunch of random bits and then run a deterministic algorithm on the string ${x}$ together with the random bits. Most likely, if the random bits were not too bad, we will get the right answer.
In practice, people tend to believe that ${\mathbf{BPP} = \mathbf{P}}$ while ${\mathbf{P} \neq \mathbf{NP}}$. But, in any case, probabilistic algorithms may be more efficient than deterministic ones, so even if the former equality holds they are interesting.
Today, I want to describe a probabilistic algorithm for testing whether a number ${N}$ is prime. As before, the algorithm will rely on a little elementary number theory. (more…)
Next Page » | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 161, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9412495493888855, "perplexity": 208.56572568775343}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364008.55/warc/CC-MAIN-20210302125936-20210302155936-00369.warc.gz"} |
https://www.deepdyve.com/lp/springer_journal/impact-of-the-diurnal-cycle-of-the-atmospheric-boundary-layer-on-wind-jlCgHjPYXp | # Impact of the Diurnal Cycle of the Atmospheric Boundary Layer on Wind-Turbine Wakes: A Numerical Modelling Study
Impact of the Diurnal Cycle of the Atmospheric Boundary Layer on Wind-Turbine Wakes: A Numerical... The wake characteristics of a wind turbine for different regimes occurring throughout the diurnal cycle are investigated systematically by means of large-eddy simulation. Idealized diurnal cycle simulations of the atmospheric boundary layer are performed with the geophysical flow solver EULAG over both homogeneous and heterogeneous terrain. Under homogeneous conditions, the diurnal cycle significantly affects the low-level wind shear and atmospheric turbulence. A strong vertical wind shear and veering with height occur in the nocturnal stable boundary layer and in the morning boundary layer, whereas atmospheric turbulence is much larger in the convective boundary layer and in the evening boundary layer. The increased shear under heterogeneous conditions changes these wind characteristics, counteracting the formation of the night-time Ekman spiral. The convective, stable, evening, and morning regimes of the atmospheric boundary layer over a homogeneous surface as well as the convective and stable regimes over a heterogeneous surface are used to study the flow in a wind-turbine wake. Synchronized turbulent inflow data from the idealized atmospheric boundary-layer simulations with periodic horizontal boundary conditions are applied to the wind-turbine simulations with open streamwise boundary conditions. The resulting wake is strongly influenced by the stability of the atmosphere. In both cases, the flow in the wake recovers more rapidly under convective conditions during the day than under stable conditions at night. The simulated wakes produced for the night-time situation completely differ between heterogeneous and homogeneous surface conditions. The wake characteristics of the transitional periods are influenced by the flow regime prior to the transition. Furthermore, there are different wake deflections over the height of the rotor, which reflect the incoming wind direction. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Boundary-Layer Meteorology Springer Journals
# Impact of the Diurnal Cycle of the Atmospheric Boundary Layer on Wind-Turbine Wakes: A Numerical Modelling Study
, Volume 166 (3) – Oct 14, 2017
26 pages
/lp/springer_journal/impact-of-the-diurnal-cycle-of-the-atmospheric-boundary-layer-on-wind-jlCgHjPYXp
Publisher
Springer Journals
Subject
Earth Sciences; Atmospheric Sciences; Meteorology; Atmospheric Protection/Air Quality Control/Air Pollution
ISSN
0006-8314
eISSN
1573-1472
D.O.I.
10.1007/s10546-017-0309-3
Publisher site
See Article on Publisher Site
### Abstract
The wake characteristics of a wind turbine for different regimes occurring throughout the diurnal cycle are investigated systematically by means of large-eddy simulation. Idealized diurnal cycle simulations of the atmospheric boundary layer are performed with the geophysical flow solver EULAG over both homogeneous and heterogeneous terrain. Under homogeneous conditions, the diurnal cycle significantly affects the low-level wind shear and atmospheric turbulence. A strong vertical wind shear and veering with height occur in the nocturnal stable boundary layer and in the morning boundary layer, whereas atmospheric turbulence is much larger in the convective boundary layer and in the evening boundary layer. The increased shear under heterogeneous conditions changes these wind characteristics, counteracting the formation of the night-time Ekman spiral. The convective, stable, evening, and morning regimes of the atmospheric boundary layer over a homogeneous surface as well as the convective and stable regimes over a heterogeneous surface are used to study the flow in a wind-turbine wake. Synchronized turbulent inflow data from the idealized atmospheric boundary-layer simulations with periodic horizontal boundary conditions are applied to the wind-turbine simulations with open streamwise boundary conditions. The resulting wake is strongly influenced by the stability of the atmosphere. In both cases, the flow in the wake recovers more rapidly under convective conditions during the day than under stable conditions at night. The simulated wakes produced for the night-time situation completely differ between heterogeneous and homogeneous surface conditions. The wake characteristics of the transitional periods are influenced by the flow regime prior to the transition. Furthermore, there are different wake deflections over the height of the rotor, which reflect the incoming wind direction.
### Journal
Boundary-Layer MeteorologySpringer Journals
Published: Oct 14, 2017
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Export lists, citations | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8219612836837769, "perplexity": 3511.8466512618825}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376824180.12/warc/CC-MAIN-20181212225044-20181213010544-00156.warc.gz"} |
http://mathhelpforum.com/calculus/59055-find-area-surface.html | # Thread: Find the area of the surface ...
1. ## Find the area of the surface ...
Find the area of the surface z = 2/3(x^(3/2) + y^(3/2))
0 <= x <=1 , 0 <= y <=1
2. What is your plan? Can you relate the task to the arc length in 2D?
3. Well i know dz/dx = x^1/2 and dz/dy = y^1/2.
Then putting it into the formula sqrt(1 + (dz/dx)^2 + (dz/dy)^2) dA
would be double integral of sqrt(1 + x + y) dA
Then I think it would be double integrals both from 0 to 1 sqrt(1 + cos(theta) + sin(theta)) r dr dtheta
Is this correct so far? I think I did something wrong since that gets really hard to integrate. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.833620548248291, "perplexity": 1167.74120548158}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738662336.11/warc/CC-MAIN-20160924173742-00217-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://mathinsight.org/image/small_directed_network_numbered | # Math Insight
### Image: Small directed network with numbered nodes and labeled edges
A directed network with 10 numbered nodes and 13 edges labeled by the corresponding component of the adjacency matrix.
For this network, the adjacency matrix is \begin{gather*} A= \left[ \begin{array}{cccccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] \end{gather*}
Image file: small_directed_network_numbered.png
Source image file: small_directed_network_numbered.ggb
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https://www.sarthaks.com/7895/electrostatics-question-below | # Electrostatics question below
+1 vote
729 views
in JEE
Ans is........
by (65.0k points)
selected by
See this solution
We have Lagrangian formula as :
where r and ϕ are the polar coordinates of the relative position between the α-particle and the proton and
is the reduced mass.
Now, obtain the Euler-Lagrange equations of motion and integrate them once, obtaining:
where E is the energy and
is the angular momentum. Both E and l are constants of the motion.
Now we use the data from the question. The energy is given by the sum of the kinetic energies of the particles at infinity:
where recall that the relative velocity is twice v .
Similarly, the angular momentum is given by
Now, we substitute these result in the equation we found above for the energy. The closest approach distance is given by the condition r=0 , since this is a turning point. In other words, the relative position decreases from infinity to D as the particles approach and then starts to increase again.
Therefore, we have, after some algebra:
In order for the x defined in the problem to appear, we must multiply by 64 and add 25. This leaves the result:
Therefore, x=89
by (2.4k points)
Really wonderful and I think that's this is not in Jee syllabus Really new for me But Fantastic. Well 'm Try to Understand it Thanks... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.987488865852356, "perplexity": 959.6918633241285}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00288.warc.gz"} |
http://www.cmp.liv.ac.uk/frink/thesis/thesis/node47.html | ## Superconducting Quantum Interference Device (SQUID)
A Superconducting Quantum Interference Device (SQUID) uses the properties of electron-pair wave coherence and Josephson Junctions to detect very small magnetic fields. The central element of a SQUID is a ring of superconducting material with one or more weak links. An example is shown in Figure 3.10, with weak-links at points W and X whose critical current, , is much less than the critical current of the main ring. This produces a very low current density making the momentum of the electron-pairs small. The wavelength of the electron-pairs is thus very long leading to little difference in phase between any parts of the ring.
If a magnetic field, , is applied perpendicular to the plane of the ring, a phase difference is produced in the electron-pair wave along the path XYW and WZX. A small current, , is also induced to flow around the ring, producing a phase difference across the weak links. Normally the induced current would be of sufficient magnitude to cancel the flux in the hole of the ring but the critical current of the weak-links prevents this.
The quantum condition that the phase change around the closed path must equal can still be met by large phase differences across the weak-links produced by even a small current. An applied magnetic field produces a phase change around a ring, as shown in Equation 3.23, which in this case is equal to
(3.28)
where is the flux produced in the ring by the applied magnetic field.[12] may not necessarily equal an integral number of fluxons so to ensure the total phase change is a multiple of a small current flows around the ring, producing a phase difference of across the two weak-links, giving a total phase change of
(3.29)
The phase difference due to the circulating current can either add to or subtract from that produced by the applied magnetic field but it is more energetically favourable to subtract: in this case a small anti-clockwise current, .[12]
Substituting values from Equations 3.27 and 3.28, the magnitude of the circulating current, , can be obtained
(3.30)
As the flux in the ring is increased from 0 to the magnitude of increases to a maximum. As the flux is increased greater than it is now energetically favourable for a current, , to flow in a clockwise direction, decreasing in magnitude to 0 as the flux reaches . The circulating current has a periodic dependence on the magnitude of the applied field, with a period of variation of , a very small amount of magnetic flux. Detecting this circulating current enables the use of a SQUID as a magnetometer.
Dr John Bland, 15/03/2003 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9020456075668335, "perplexity": 510.2944241931878}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131303523.19/warc/CC-MAIN-20150323172143-00041-ip-10-168-14-71.ec2.internal.warc.gz"} |
http://www.emis.de/classics/Erdos/cit/52605031.htm | ## Zentralblatt MATH
Publications of (and about) Paul Erdös
Zbl.No: 526.05031
Autor: Erdös, Paul; Hajnal, András; Sos, Vera T.; Szemeredi, E.
Title: More results on Ramsey-Turán type problems. (In English)
Source: Combinatorica 3, 69-81 (1983).
Review: In [Combinat. Struct. Appl., Proc. Calgary Internat. Conf. Calgary 1969, 407-410 (1970; Zbl 253.05145)] V.T.Sós raised a general scheme of new problems that can be considered as common generalizations of the problems treated in the classical results of Ramsey and Turán. This paper is a continuation of a sequence of papers on this subject.
One of the main results is the following: Given k \geq 2 and \epsilon > 0, let Gn be a sequence of graphs of order n size at least (½)(\frac{3k-5}{3k-2}+\epsilon)n2 edges such that the cardinality of the largest independent set in Gn is o(n). Let H be any graph of arboricity at most k. Then there exists an n0 such that all Gn with n > n0 contain a copy of H. This result is best possible in the case H = K2k.
Reviewer: L.Lesniak
Classif.: * 05C35 Extremal problems (graph theory)
05C55 Generalized Ramsey theory
05C05 Trees
Keywords: arboricity; sequence of graphs; largest independent set
Citations: Zbl.253.05145
© European Mathematical Society & FIZ Karlsruhe & Springer-Verlag | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.854805052280426, "perplexity": 2902.01262303172}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257825358.53/warc/CC-MAIN-20160723071025-00262-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/73991/how-do-i-display-text-at-60pt-or-even-600pt | # How do I display text at 60pt or even 600pt?
I am new to LaTeX. I am having a hard time creating a font past 40 points. Here is my code:
\documentclass[letterpaper]{article}
\begin{document}
\fontsize{40}{48}\selectfont hi
\end{document}
If I change the 40 and corresponding 48 to anything bigger, it does not seem to increase the size of the word "hi". I can go lower but not higher.
Technically I need this for some text in a cell for a table I am working on in LaTeX, but I'm sure if I figured out what I was doing wrong I could apply it to the table. The font size I am aiming for is somewhere around 60 to 100 points in size.
-
Edit: I added a new example to compare scaled fonts and resized fonts.
LaTeX can't use font sizes that are not listed in the .fd file. When compiling your MWE, you can see some warnings:
LaTeX Font Warning: Font shape OT1/cmr/m/n' in size <40> not available
(Font) size <24.88> substituted on input line 8.
because the .fd file of Computer Modern provides a discrete list of sizes (with a maximum size of 24.88pt).
To use scalable fonts, there are some useful packages:
• type1cm (Computer Modern via cm-super).
• anyfontsize (with any font but automatically only via latex)
• lmodern (a sort of "super cm-super").
Here is an example using lmodern package (that provides scalable fonts):
\documentclass[letterpaper]{article}
\usepackage{lmodern}
\usepackage{graphicx}
\begin{document}
\fontsize{40}{48}\selectfont hi
\fontsize{60}{70}\selectfont hi
\fontsize{100}{120}\selectfont hi%
% compare scaled fonts and resized fonts
\fontsize{100}{120}\selectfont hi%
\scalebox{5}{\fontsize{20}{24}\selectfont hi}%
\scalebox{10}{\fontsize{10}{12}\selectfont hi}%
\scalebox{20}{\fontsize{5}{6}\selectfont hi}
% a BIG font
\fontsize{300}{350}\selectfont hi
\end{document}
-
Could you add why it doesn't work for the default CM fonts? – egreg Sep 25 '12 at 8:54
How about just adding the anyfontsize package? – daleif Sep 25 '12 at 9:55
It works for CM fonts with package type1cm with smoother scaling. The restriction of the standard behaviour comes from the times with PK fonts, where for each size a new PK fonts had to be generated. Using fixed sizes helped to limit the number of PK fonts. – Heiko Oberdiek Sep 25 '12 at 12:50
If this does not form part of regular textual content, you can always resize it in a box:
\documentclass{article}
\usepackage{graphicx}% http://ctan.org/pkg/graphicx
\begin{document}
\begin{tabular}{ccccc}
hi &
\large hi &
\Huge hi &
\resizebox{!}{100pt}{hi} &
\scalebox{20}{hi}
\end{tabular}
\end{document}
graphicx provides the scaling/resizing capability.
-
Change the size of a font is not always a homothetic transformation... – Paul Gaborit Sep 25 '12 at 6:12
I suggest you read again the third paragraph of chapter 1 of the TeXbook. – Martin Schröder Sep 25 '12 at 15:05
@MartinSchröder: The OP mentions "Technically I need this for some text in a cell for a table" which is small in scale. The examples shown by Paul reflecting the "poor" transformation of text under scaling is a bit extreme: Why would one resize a 5pt (\tiny) font to 100pt - it is visually obvious that the kerning for \tiny` is awkward, which is exacerbated when enlarged 20 times. Honestly, I don'y see a loss in quality using the above method, nor the need to switch to a different typesetting system as a result. – Werner Sep 25 '12 at 15:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8629302382469177, "perplexity": 3134.7296626240022}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644060103.8/warc/CC-MAIN-20150827025420-00233-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://rcd.ics.org.ru/archive/volume-19-number-2/ | 0
2013
Impact Factor
# Volume 19, Number 2, 2014 Proceedings of GDIS 2013, Izhevsk
The Fourth International Conference "Geometry, Dynamics, Integrable Systems" GDIS 2013 Held in Izhevsk, Russia, June 10–14, 2013
Kozlov V. V. Remarks on Integrable Systems Abstract The problem of integrability conditions for systems of differential equations is discussed. Darboux’s classical results on the integrability of linear non-autonomous systems with an incomplete set of particular solutions are generalized. Special attention is paid to linear Hamiltonian systems. The paper discusses the general problem of integrability of the systems of autonomous differential equations in an n-dimensional space, which admit the algebra of symmetry fields of dimension $\geqslant n$. Using a method due to Liouville, this problem is reduced to investigating the integrability conditions for Hamiltonian systems with Hamiltonians linear in the momenta in phase space of dimension that is twice as large. In conclusion, the integrability of an autonomous system in three-dimensional space with two independent non-trivial symmetry fields is proved. It should be emphasized that no additional conditions are imposed on these fields. Keywords: integrability by quadratures, adjoint system, Hamiltonian equations, Euler–Jacobi theorem, Lie theorem, symmetries Citation: Kozlov V. V., Remarks on Integrable Systems, Regular and Chaotic Dynamics, 2014, vol. 19, no. 2, pp. 145-161 DOI:10.1134/S1560354714020014
Dragovic V., Kukić K. Systems of Kowalevski Type and Discriminantly Separable Polynomials Abstract Starting from the notion of discriminantly separable polynomials of degree two in each of three variables, we construct a class of integrable dynamical systems. These systems can be integrated explicitly in genus two theta-functions in a procedure which is similar to the classical one for the Kowalevski top. The discriminantly separable polynomials play the role of the Kowalevski fundamental equation. Natural examples include the Sokolov systems and the Jurdjevic elasticae. Keywords: integrable systems, Kowalevski top, discriminantly separable polynomials, systems of Kowalevski type Citation: Dragovic V., Kukić K., Systems of Kowalevski Type and Discriminantly Separable Polynomials, Regular and Chaotic Dynamics, 2014, vol. 19, no. 2, pp. 162-184 DOI:10.1134/S1560354714020026
Tsiganov A. V. On the Lie Integrability Theorem for the Chaplygin Ball Abstract The necessary number of commuting vector fields for the Chaplygin ball in the absolute space is constructed. We propose to get these vector fields in the framework of the Poisson geometry similar to Hamiltonian mechanics. Keywords: nonholonomic dynamical system, Poisson bracket, Lie theorem, Chaplygin ball Citation: Tsiganov A. V., On the Lie Integrability Theorem for the Chaplygin Ball, Regular and Chaotic Dynamics, 2014, vol. 19, no. 2, pp. 185-197 DOI:10.1134/S1560354714020038
Bizyaev I. A., Borisov A. V., Mamaev I. S. The Dynamics of Nonholonomic Systems Consisting of a Spherical Shell with a Moving Rigid Body Inside Abstract In this paper we investigate two systems consisting of a spherical shell rolling without slipping on a plane and a moving rigid body fixed inside the shell by means of two different mechanisms. In the former case the rigid body is attached to the center of the ball on a spherical hinge. We show an isomorphism between the equations of motion for the inner body with those for the ball moving on a smooth plane. In the latter case the rigid body is fixed by means of a nonholonomic hinge. Equations of motion for this system have been obtained and new integrable cases found. A special feature of the set of tensor invariants of this system is that it leads to the Euler–Jacobi–Lie theorem, which is a new integration mechanism in nonholonomic mechanics. We also consider the problem of free motion of a bundle of two bodies connected by means of a nonholonomic hinge. For this system, integrable cases and various tensor invariants are found. Keywords: nonholonomic constraint, tensor invariants, isomorphism, nonholonomic hinge Citation: Bizyaev I. A., Borisov A. V., Mamaev I. S., The Dynamics of Nonholonomic Systems Consisting of a Spherical Shell with a Moving Rigid Body Inside, Regular and Chaotic Dynamics, 2014, vol. 19, no. 2, pp. 198-213 DOI:10.1134/S156035471402004X
Ivanov A. P. On the Impulsive Dynamics of M-blocks Abstract This paper is concerned with the motion of a cubic rigid body (cube) with a rotor, caused by a sudden brake of the rotor, which imparts its angular momentum to the body. This produces an impulsive reaction of the support, leading to a jump or rolling from one face to another. Such dynamics was demonstrated by researchers from Massachusetts Institute of Technology at the IEEE/RSJ International Conference on Intelligent Robots and Systems in Tokio in November 2013. The robot, called by them M-block, is 4 cm in size and uses an internal flywheel mechanism rotating at 20 000 rev/min. Initially the cube rests on a horizontal plane. When the brake is set, the relative rotation slows down, and its energy is imparted to the case. The subsequent motion is illustrated in a clip [13]. Here the general approach to the analysis of dynamics of M-cube is proposed, including equations of impulsive motion and methods of their solution. Some particular cases are studied in details. Keywords: gyrostat, impulsive dynamics, friction, robotic dynamics Citation: Ivanov A. P., On the Impulsive Dynamics of M-blocks, Regular and Chaotic Dynamics, 2014, vol. 19, no. 2, pp. 214-225 DOI:10.1134/S1560354714020051
Kharlamov M. P. Extensions of the Appelrot Classes for the Generalized Gyrostat in a Double Force Field Abstract For the integrable system on $e(3,2)$ found by Sokolov and Tsiganov we obtain explicit equations of some invariant 4-dimensional manifolds on which the induced systems are almost everywhere Hamiltonian with two degrees of freedom. These subsystems generalize the famous Appelrot classes of critical motions of the Kowalevski top. For each subsystem we point out a commutative pair of independent integrals, describe the sets of degeneration of the induced symplectic structure. With the help of the obtained invariant relations, for each subsystem we calculate the outer type of its points considered as critical points of the initial system with three degrees of freedom. Keywords: generalized two-field gyrostat, critical subsystems, Appelrot classes, invariant relations, types of critical points Citation: Kharlamov M. P., Extensions of the Appelrot Classes for the Generalized Gyrostat in a Double Force Field, Regular and Chaotic Dynamics, 2014, vol. 19, no. 2, pp. 226-244 DOI:10.1134/S1560354714020063
Jovanović B. Heisenberg Model in Pseudo-Euclidean Spaces Abstract We construct analogues of the classical Heisenberg spin chain model (or the discrete Neumann system), on pseudo-spheres and light-like cones in the pseudo-Euclidean spaces and show their complete Hamiltonian integrability. Further, we prove that the Heisenberg model on a light-like cone leads to a new example of the integrable discrete contact system. Keywords: discrete Hamiltonian and contact systems, the Lax representation, complete integrability Citation: Jovanović B., Heisenberg Model in Pseudo-Euclidean Spaces, Regular and Chaotic Dynamics, 2014, vol. 19, no. 2, pp. 245-250 DOI:10.1134/S1560354714020075
Bounemoura A., Fischler S. The Classical KAM Theorem for Hamiltonian Systems via Rational Approximations Abstract In this paper, we give a new proof of the classical KAM theorem on the persistence of an invariant quasi-periodic torus, whose frequency vector satisfies the Bruno–Rüssmann condition, in real-analytic non-degenerate Hamiltonian systems close to integrable. The proof, which uses rational approximations instead of small divisors estimates, is an adaptation to the Hamiltonian setting of the method we introduced in [4] for perturbations of constant vector fields on the torus. Keywords: perturbation of integrable Hamiltonian systems, KAM theory, Diophantine duality, periodic approximations Citation: Bounemoura A., Fischler S., The Classical KAM Theorem for Hamiltonian Systems via Rational Approximations, Regular and Chaotic Dynamics, 2014, vol. 19, no. 2, pp. 251-265 DOI:10.1134/S1560354714020087
Back to the list | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8419747352600098, "perplexity": 628.6403015495906}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589752.56/warc/CC-MAIN-20180717144908-20180717164908-00129.warc.gz"} |
https://infoscience.epfl.ch/record/219622 | Infoscience
Journal article
# Frequency fluctuations in silicon nanoresonators
Frequency stability is key to the performance of nanoresonators. This stability is thought to reach a limit with the resonator's ability to resolve thermally induced vibrations. Although measurements and predictions of resonator stability usually disregard fluctuations in the mechanical frequency response, these fluctuations have recently attracted considerable theoretical interest. However, their existence is very difficult to demonstrate experimentally. Here, through a literature review, we show that all studies of frequency stability report values several orders of magnitude larger than the limit imposed by thermomechanical noise. We studied a monocrystalline silicon nanoresonator at room temperature and found a similar discrepancy. We propose a new method to show that this was due to the presence of frequency fluctuations, of unexpected level. The fluctuations were not due to the instrumentation system, or to any other of the known sources investigated. These results challenge our current understanding of frequency fluctuations and call for a change in practices. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8701853156089783, "perplexity": 929.7124892165868}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721405.66/warc/CC-MAIN-20161020183841-00264-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/218450-evaluating-integral.html | # Thread: Evaluating an Integral
1. ## Evaluating an Integral
Assuming ∫ = the integral from (0-3)
I am having trouble deciding what to sub in as u in this integration problem:
∫ x(sqrt(x+1)) dx
Thanks.
2. ## Re: Evaluating an Integral
Let $\displaystyle u = x+1$, $\displaystyle du = dx$
The integral can then be rewritten as:
$\displaystyle \int_{0}^{3} (u-1)(\sqrt{u})du$
I'm sure you can take it from here.
3. ## Re: Evaluating an Integral
Originally Posted by Walshy
Assuming ∫ = the integral from (0-3)
I am having trouble deciding what to sub in as u in this integration problem:
∫ x(sqrt(x+1)) dx
The integral $\displaystyle \int_0^3 {x\sqrt {x + 1} dx}$ becomes $\displaystyle \int_1^4 {(u - 1)\sqrt u du}$.
Do you see how?
4. ## Re: Evaluating an Integral
Yes, I see. Thanks for the help guys.
5. ## Re: Evaluating an Integral
Originally Posted by Bradyns
Let $\displaystyle u = x+1$, $\displaystyle du = dx$
The integral can then be rewritten as:
$\displaystyle \color{red}\int_{0}^{3} (u-1)(\sqrt{u})du$
The limits of integration are not correct.
6. ## Re: Evaluating an Integral
Given this, I should clarify that my bounds were regards to X, so when you integrate, you should back substitute U so you are in respect to X before evaluation of the bounds.
You can change the bounds so that they are with respect to U and integrate.
Or, you can leave the bounds with regard to X, integrate with respect to U and then back substitute so that you end up with the original function (f(x), not not f'(x)) with respect to X.
Same result. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9953022003173828, "perplexity": 1704.0800564021881}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125947421.74/warc/CC-MAIN-20180424221730-20180425001730-00061.warc.gz"} |
https://homotopytypetheory.org/2014/12/22/splitting-idempotents-ii/ | ## Splitting Idempotents, II
I ended my last post about splitting idempotents with several open questions:
1. If we have a map $f:X\to X$, a witness of idempotency $I:f\circ f = f$, and a coherence datum $J: \prod_{x:X} \mathsf{ap}_f(I(x)) = I(f(x))$, and we use them to split $f$ as in the previous post, do the new witnesses $I,J$ induced from the splitting agree with the given ones?
2. How many different ways can a given map $f$ “be idempotent” — can you give an example of a map that is idempotent in two genuinely different ways, perhaps with non-equivalent splittings?
3. Given $f,I,J$, can we also split it with $A = \sum_{x:X} \sum_{p:f(x)=x} (\mathrm{ap}_f(p)=I(x))$?
4. (I didn’t ask this one explicitly, but I should have) Can we define the type of fully-coherent idempotents in HoTT?
The third question was answered negatively by Nicolai. The second one is, in my opinion, still open. In this post, I’ll answer the first and fourth questions. The answers are (1) the induced $I$ agrees, but the induced $J$ does not in general, and (4) Yes — by splitting an idempotent! They have also been formalized; see the pull request here.
### Preservation of I
The proof that $I$ can be recovered from the splitting is an easy computation. Given a splitting $s:A\to X$ and $r:X\to A$ such that $H : r\circ s = 1$ and $s\circ r = f$, the induced $I$ is $x\mapsto \mathsf{ap}_s (H(r(x)))$. For the splitting we constructed with $A = \sum_{a:\mathbb{N}\to X} \prod_{x:X} f(a_{n+1})=a_n$, we have $r(x) = (\lambda n. f(x), \lambda n.I(x))$ and $s(a,b) = a_0$, so the induced $I'(x)$ is just the $0$-component of the homotopy $H:r\circ s = 1$ at $r(x)$. By construction, this is
$f(f(x)) \xrightarrow{\mathsf{ap}_f(I(x))^{-1}} f(f(f(x))) \xrightarrow{I(f(x))} f(f(x)) \xrightarrow{I(x)} f(x).$
where $I$ is the given witness of idempotence. But by the given coherence datum $J$, we have $\mathsf{ap}_f(I(x)) = I(f(x))$, so this reduces to just $I(x)$.
### Recovering a retract
Before attacking $J$, let’s consider a dual question: given a retraction $(A, r:X\to A, s:A\to X, H:r\circ s = 1)$, is it equivalent to the splitting of the induced idempotent $s r : X \to X$?
Actually, before we can even ask that, we have to check that $s r$ comes with not just $I$ but also $J$. The definition of $I$ is easy: to get $s r s r = 1$ we take $x\mapsto \mathsf{ap}_s (H(r(x)))$. For $J$ we have to identify $x\mapsto \mathsf{ap}_{srs} (H(r(x))) : srsrsrx = srsrx$ with $x\mapsto \mathsf{ap}_s (H(rsrx)) : srsrsrx = srsrx$, which may seem impossible until you think of concatenating them both with $\mathsf{ap}_s (H(r(x))) : srsrx = srx$, after which the equality becomes just a naturality square.
Now, what happens when we split $s r$? We end up with a new retraction $(A',r',s',H')$ such that $s r = s' r'$ (in fact this equality holds judgmentally). The standard argument from 1-category theory gives us an equivalence $A\simeq A'$ composed of $g = r' s : A \to A'$ and $h = r s' : A' \to A$, with $\eta : g h = r' s r s' = r' s' r' s' = 1 1 = 1$ and dually $\epsilon : h g = r s' r' s = r s r s = 1 1 = 1$. Moreover, this equivalence respects the sections and retractions, e.g. we have $P : h r' = r s' r' = r s r = r$ and $Q : s' g = s' r' s = s r s = s$.
Finally, to be higher-categorically consistent, we should also check that this equivalence respects the homotopies $H$ and $H'$; one sensible way to say this would be that $\mathsf{ap}_h (H'(ga)) \cdot \epsilon_a = P(s' g a) \cdot \mathsf{ap}_r(Q a) \cdot H(a)$. Both of these are homotopies $r s' r' s' r' s = 1$; note that in our case the domain is judgmentally equal to $rsrsrs$. Substituting the definitions of $r',s',H'$ from the previous post (this is one of the places a proof assistant comes in handy), we see that the first one is
$rsrsrsa \xrightarrow{\mathsf{ap}_{rsrs}(H(rsa))^{-1}} rsrsrsrsa \xrightarrow{\mathsf{ap}_{rs}(H(rsrsa))} rsrsrsa \xrightarrow{\mathsf{ap}_{rs}(H(rsa))} rsrsa \xrightarrow{H(rsa)} rsa \xrightarrow{Ha} a$
while the second one is
$rsrsrsa \xrightarrow{H(rsrsa)} rsrsa \xrightarrow{\mathsf{ap}_{rs}(Ha)} rsa \xrightarrow{Ha} a.$
Now by naturality, we have ${\mathsf{ap}_{rs}(H(rsrsa))} \cdot {\mathsf{ap}_{rs}(H(rsa))} = {\mathsf{ap}_{rsrs}(H(rsa))} \cdot {\mathsf{ap}_{rs}(H(rsa))}$. Applying this in the middle of the first composite, and canceling ${\mathsf{ap}_{rsrs}(H(rsa))}$ with its inverse on the left, we reduce it to
$rsrsrsa \xrightarrow{\mathsf{ap}_{rs}(H(rsa))} rsrsa \xrightarrow{H(rsa)} rsa \xrightarrow{Ha} a.$
Now naturality again gives us ${\mathsf{ap}_{rs}(H(rsa))} \cdot {H(rsa)} = H(rsrsa) \cdot {H(rsa)}$, so this is equal to
$rsrsrsa \xrightarrow{H(rsrsa)} rsrsa \xrightarrow{H(rsa)} rsa \xrightarrow{Ha} a.$
Comparing this to the second composite, we can cancel $H(rsrsa)$ on the left, reducing the problem to ${\mathsf{ap}_{rs}(Ha)}\cdot Ha = {H(rsa)} \cdot Ha$, which is another naturality. Whew!
Let’s step back a moment and think about what we just proved. It turns out that the equivalence $A\simeq A'$ together with $P,Q$ and our final coherence is exactly what we need to show that $(A,r,s,H) = (A,r',s',H')$ as elements of the type of retractions of $X$:
$\sum_{(A:\mathsf{Type})} \sum_{(r:X\to A)} \sum_{{s:A\to X}} (r\circ s = 1).$
This claim is one of those things that I think seems fairly intuitively obvious (at least, once you’ve fully internalized Chapter 2 of the Book), but is surprisingly difficult to convince a proof assistant of. (There’s another such “doozy” in the Book itself, in the proof of Theorem 7.6.6.)
In this case the crucial fact turns out to be a coherence law that I’m currently calling transport_path_universe_pV. It applies when we have an equivalence $f : A\simeq B$ and a point $b:B$. In this case we can make $f$ into a path $\mathsf{ua}(f):A=B$ and transport $b$ along $\mathsf{ua}(f)^{-1}$ in the tautological type family $X\mapsto X$, and then transport the result forwards again along $\mathsf{ua}(f)$. On one hand, by the “computation rule for univalence” (third bullet following remark 2.10.4 in the book), these two operations are equivalent respectively to applying $f^{-1}$ and then applying $f$, and the result $f(f^{-1}(b))$ is of course equivalent to $b$. But on the other hand, for any type family $Z:Y\to \mathsf{Type}$, we have $\mathsf{transport}^Z(p,\mathsf{transport}^Z(p^{-1},z)) = z$. These two hands give us two ways of identifying this double transport with $b$; the lemma transport_path_universe_pV says that they are equal.
It didn’t take me all that long to notice and prove this lemma in this case, but that was only because I had already noticed and proven the dual version transport_path_universe_Vp, for transporting along $\mathsf{ua}(f)$ and then along $\mathsf{ua}(f)^{-1}$ in the other order, when formalizing Theorem 7.6.6. In that case it took me quite a while to isolate this transport coherence law as the crucial thing I needed.
Anyway, once we make it through that proof, we see that what we’ve shown is that the type of retracts of $X$ is itself a retract of the type of partially-coherent idempotents $\sum_{(f:X\to X)} \sum_{(I:f\circ f = f)} (\mathsf{ap}_f(I) = I\circ f)$. The section constructs the idempotent induced by a retract, while the retraction is the splitting construction from the last post.
With a bit more work, we can deduce from this that for any fixed $f$, the type of splittings of $f$ (meaning $(A,r,s,H)$ together with $K:s\circ r = f$) is a retract of the type of partial-coherences for $f$, i.e. $\sum_{(I:f\circ f = f)} (\mathsf{ap}_f(I) = I\circ f)$. Similarly, given $f$ and $I$, the type of splittings of $f$ respecting $I$ (that is, $(A,r,s,H,K)$ together with $L: \mathsf{ap}_s(H\circ r) = I$) is a retract of the type $\mathsf{ap}_f(I) = I\circ f$ of $J$‘s.
### Preservation of J
Now let’s consider the question of preservation of $J$. Since we now know that the type of retracts $(A,r,s,H)$ of $X$ is a retract of the type of triples $(f,I,J)$, and we know that we can recover $f$ and $I$ from the splitting, if we could also recover $J$, then the type of retracts $(A,r,s,H)$ would actually be equivalent to the type of triples $(f,I,J)$.
This seems a bit implausible. To construct an actual counterexample, let’s consider the case $f=\mathsf{id}_X$, with $I=\mathsf{refl}$ the obvious proof consisting of reflexivity. Then $(A,r,s,H,K,L)$ as above are more or less exactly the data of a type $A$ together with a (half-adjoint) equivalence $A\simeq X$. In particular, the type of such data is (by univalence) a based path-type $\sum_{A:\mathsf{Type}}(A=X)$ (a “coconut”), hence contractible. Therefore, if it were equivalent to the type of $J$‘s, then the latter would also be contractible. So to show that $J$ is not generally recoverable, it suffices to exhibit a type $X$ such that $f=\mathsf{id}_X$ and $I=\mathsf{refl}$ admit more than one inequivalent $J$.
Now when $f=\mathsf{id}_X$ and $I=\mathsf{refl}$, the type of $J$ is $\prod_{x:X} \mathsf{refl}_x = \mathsf{refl}_x$. This is what you might call the “2-center” of $X$, so we need a type with a nontrivial inhabitant of this. One such type is $S^2$; another is $B\mathsf{Aut}(B\mathsf{Aut}(\mathbf{2}))$, where $B\mathsf{Aut}(Y) = \sum_{(Z:\mathsf{Type})} \Vert Y=Z\Vert$. I may have more to say about $B\mathsf{Aut}$ in another post.
### Defining Coherent Idempotents
To conclude, let’s consider whether we can define the type of “fully-coherent idempotents”. What is that? Well, intuitively, in addition to $I$ and $J$ a fully-coherent idempotent should have higher coherences at all levels. A precise definition of coherent idempotents in $(\infty,1)$-categories can be found in section 4.4.5 of Higher topos theory. However, by Corollary 4.4.5.14 therein, the space of coherent idempotents (in an $(\infty,1)$-category where all coherent idempotents split) is equivalent to the type of retracts, where “retract” means in our simple $(A,r,s,H)$ sense (although there is also a notion of “fully coherent retract” that mediates between the two).
Thus, if we’re willing to believe that result, we already have the type of fully-coherent idempotents on $X$: it’s just the type of retracts defined above. However, there’s something unsatisfying about defining an “idempotent” to be already equipped with its splitting! Moreover, this definition does have one concrete drawback: since it includes the type $A$ as data, it raises the universe level.
We can avoid both of those problems by recalling that we showed the type of retracts to be a retract of the type of partially-coherent idempotents $(f,I,J)$ — and we also showed that any retract is equivalent to the canonical splitting of its induced idempotent! Thus, we can define the type of coherent idempotents on $X$ to be the splitting (as constructed in the last post) of the idempotent on the type of $(f,I,J)$‘s induced by the splitting construction. I find this very cute, and also a nice example of applying a theorem at two different universe levels.
This motivates me to wonder what this idempotent actually behaves like: given $f,I,J$, it essentially preserves $f$ and $I$, but constructs a new $J$ out of the old one. In principle, this construction could be done without going through the splitting, and in fact you can ask Coq what it looks like and Coq will compute it for you. But what comes out is quite large and I haven’t been able to make any headway understanding it on its own.
Let me end by emphasizing an analogy with equivalences. In that case we also have a “fully coherent” notion $\mathsf{IsEquiv}(f)$ (which has many equivalent definitions) that is a retract of a type $\mathsf{QInv}(f)$ of “partially coherent” objects. Thus, in practice it suffices to construct partially-coherent objects (and we generally do), relying on the retraction to convert them into coherent ones. Moreover, the behavior of the retraction is similar: if we define $\mathsf{IsEquiv}(f)$ using half-adjoint equivalences, then the induced idempotent on $\mathsf{QInv}(f)$ (adjointification) preserves the quasi-inverse function and one of the homotopies, while modifying the other one. (There is a difference, of course, in that $\mathsf{IsEquiv}(f)$ is an hprop, whereas the type of coherent idempotents is (probably) not.)
Are there any other structures that behave like this? Are there any other “fully coherent” gadgets that we can obtain by splitting an idempotent on a type of partially-coherent ones?
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Exponential Functions
5.4 Exponential functions (EMBGS)
Revision (EMBGT)
Functions of the form $$y=a{b}^{x}+q$$
Functions of the general form $$y=a{b}^{x}+q$$, for $$b>0$$, are called exponential functions, where $$a$$, $$b$$ and $$q$$ are constants.
The effects of $$a$$, $$b$$ and $$q$$ on $$f(x) = ab^x + q$$:
• The effect of $$q$$ on vertical shift
• For $$q>0$$, $$f(x)$$ is shifted vertically upwards by $$q$$ units.
• For $$q<0$$, $$f(x)$$ is shifted vertically downwards by $$q$$ units.
• The horizontal asymptote is the line $$y = q$$.
• The effect of $$a$$ on shape
• For $$a>0$$, $$f(x)$$ is increasing.
• For $$a<0$$, $$f(x)$$ is decreasing. The graph is reflected about the horizontal asymptote.
• The effect of $$b$$ on direction
Assuming $$a > 0$$:
• If $$b > 1$$, $$f(x)$$ is an increasing function.
• If $$0 < b < 1$$, $$f(x)$$ is a decreasing function.
• If $$b \leq 0$$, $$f(x)$$ is not defined.
$$b>1$$ $$a<0$$ $$a>0$$ $$q>0$$ $$q<0$$
$$00$$ $$q>0$$ $$q<0$$
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Revision
Exercise 5.15
On separate axes, accurately draw each of the following functions:
• Use tables of values if necessary.
• Use graph paper if available.
$$y_1 = 3^x$$
$$y_2 = -2 \times 3^x$$
$$y_3 = 2 \times 3^x + 1$$
$$y_4 = 3^x - 2$$
Use your sketches of the functions given above to complete the following table (the first column has been completed as an example):
$$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ value of $$q$$ $$q = 0$$ effect of $$q$$ no vertical shift value of $$a$$ $$a = 1$$ effect of $$a$$ increasing asymptote $$x$$-axis, $$y = 0$$ domain $$\{x: x \in \mathbb{R} \}$$ range $$\{y: y \in \mathbb{R}, y > 0 \}$$
$$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ value of $$q$$ $$q = 0$$ $$q = 0$$ $$q = 1$$ $$q = 2$$ effect of $$q$$ no vertical shift no vertical shift shift $$1$$ unit up shift $$2$$ units down value of $$a$$ $$a = 1$$ $$a = -2$$ $$a = 2$$ $$a = 1$$ effect of $$a$$ increasing decreasing increasing increasing asymptote $$x$$-axis, $$y = 0$$ $$x$$-axis, $$y = 0$$ $$y = 1$$ $$y = -2$$ domain $$\{x: x \in \mathbb{R} \}$$ $$\{x: x \in \mathbb{R} \}$$ $$\{x: x \in \mathbb{R} \}$$ $$\{x: x \in \mathbb{R} \}$$ range $$\{y: y \in \mathbb{R}, y > 0 \}$$ $$\{y: y \in \mathbb{R}, y < 0 \}$$ $$\{y: y \in \mathbb{R}, y > 1 \}$$ $$\{y: y \in \mathbb{R}, y > -2 \}$$
Functions of the form $$y=a{b}^{\left(x+p\right)}+q$$ (EMBGV)
We now consider exponential functions of the form $$y=a{b}^{\left(x+p\right)}+q$$ and the effects of parameter $$p$$.
The effects of $$a$$, $$p$$ and $$q$$ on an exponential graph
1. On the same system of axes, plot the following graphs:
1. $$y_1 = 2^x$$
2. $$y_2 = 2^{(x - 2)}$$
3. $$y_3 = 2^{(x - 1)}$$
4. $$y_4 = 2^{(x + 1)}$$
5. $$y_5 = 2^{(x + 2)}$$
Use your sketches of the functions above to complete the following table:
$$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ $$y_5$$ intercept(s) asymptote domain range effect of $$p$$
2. On the same system of axes, plot the following graphs:
1. $$y_1 = 2^{(x - 1)} + 2$$
2. $$y_2 = 3 \times 2^{(x - 1)} + 2$$
3. $$y_3 = \frac{1}{2} \times 2^{(x - 1)} + 2$$
4. $$y_4 = 0 \times 2^{(x - 1)} + 2$$
5. $$y_5 = -3 \times 2^{(x - 1)} + 2$$
Use your sketches of the functions above to complete the following table:
$$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ $$y_5$$ intercept(s) asymptotes domain range effect of $$a$$
The effect of the parameters on $$y = ab^{x + p} + q$$
The effect of $$p$$ is a horizontal shift because all points are moved the same distance in the same direction (the entire graph slides to the left or to the right).
• For $$p>0$$, the graph is shifted to the left by $$p$$ units.
• For $$p<0$$, the graph is shifted to the right by $$p$$ units.
The effect of $$q$$ is a vertical shift. The value of $$q$$ also affects the horizontal asymptotes, the line $$y = q$$.
The value of $$a$$ affects the shape of the graph and its position relative to the horizontal asymptote.
• For $$a>0$$, the graph lies above the horizontal asymptote, $$y = q$$.
• For $$a<0$$, the graph lies below the horizontal asymptote, $$y = q$$.
$$p>0$$ $$p<0$$ $$a<0$$ $$a>0$$ $$a<0$$ $$a>0$$ $$q>0$$ $$q<0$$
Discovering the characteristics
For functions of the general form: $$f(x) = y = ab^{(x+p)} + q$$:
Domain and range
The domain is $$\left\{x:x\in ℝ\right\}$$ because there is no value of $$x$$ for which $$f(x)$$ is undefined.
The range of $$f(x)$$ depends on whether the value for $$a$$ is positive or negative.
If $$a>0$$ we have: \begin{align*} {b}^{\left(x+p\right)} & > 0 \\ a {b}^{\left(x+p\right)} & > 0 \\ a {b}^{\left(x+p\right)} + q & > q \\ f(x) & > q \end{align*} The range is therefore $$\{ y: y > q, y \in \mathbb{R} \}$$.
Similarly, if $$a < 0$$, the range is $$\{ y: y < q, y \in \mathbb{R} \}$$.
Worked example 14: Domain and range
State the domain and range for $$g(x) = 5 \times 3^{(x+1)} - 1$$.
Determine the domain
The domain is $$\{x: x \in \mathbb{R} \}$$ because there is no value of $$x$$ for which $$g(x)$$ is undefined.
Determine the range
The range of $$g(x)$$ can be calculated from: \begin{align*} 3^{(x+1)} & > 0\\ 5 \times 3^{(x+1)} & > 0\\ 5 \times 3^{(x+1)} - 1 & > -1\\ \therefore g(x) & > -1 \end{align*} Therefore the range is $$\{g(x): g(x) > -1 \}$$ or in interval notation $$(-1; \infty)$$.
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Domain and range
Exercise 5.16
Give the domain and range for each of the following functions:
$$y = \left( \frac{3}{2} \right)^{(x + 3)}$$
\begin{align*} \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y > 0, y\in \mathbb{R} \right \} \end{align*}
$$f(x) = -5^{(x - 2)} + 1$$
\begin{align*} \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y < 1, y\in \mathbb{R} \right \} \end{align*}
$$y + 3 = 2^{(x + 1)}$$
\begin{align*} y + 3 &= 2^{(x + 1)} \\ y &= 2^{(x + 1)} - 3 \\ \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y > -3, y\in \mathbb{R} \right \} \end{align*}
$$y = n + 3^{(x - m)}$$
\begin{align*} y &= n + 3^{(x - m)} \\ y &= 3^{(x - m)} + n \\ \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y > n, y\in \mathbb{R} \right \} \end{align*}
$$\frac{y}{2} = 3^{(x - 1)} - 1$$
\begin{align*} \frac{y}{2} &= 3^{(x - 1)} - 1 \\ y &= 2 \times 3^{(x - 1)} - 2 \\ \text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y: y > 2, y\in \mathbb{R} \right \} \end{align*}
Intercepts
The $$y$$-intercept:
To calculate the $$y$$-intercept we let $$x=0$$. For example, the $$y$$-intercept of $$g(x) = 3 \times 2^{(x + 1)} + 2$$ is determined by setting $$x=0$$: \begin{align*} g(0) &= 3 \times 2^{(0 + 1)} + 2 \\ &= 3 \times 2 + 2\\ &= 8 \end{align*} This gives the point $$(0;8)$$.
The $$x$$-intercept:
To calculate the $$x$$-intercept we let $$y=0$$. For example, the $$x$$-intercept of $$g(x) = 3 \times 2^{(x + 1)} + 2$$ is determined by setting $$y=0$$: \begin{align*} 0 &= 3 \times 2^{(x + 1)} + 2 \\ -2 &= 3 \times 2^{(x + 1)} \\ -\frac{2}{3} &= 2^{(x + 1)} \end{align*} which has no real solutions. Therefore, the graph of $$g(x)$$ lies above the $$x$$-axis and does not have any $$x$$-intercepts.
Intercepts
Exercise 5.17
Determine the $$x$$- and $$y$$-intercepts for each of the following functions:
$$f(x) = 2^{(x + 1)} - 8$$
\begin{align*} \text{For } x=0 \quad y &= 2^{(0 + 1)} - 8 \\ &= 2 - 8 \\ &= -6 \\ \therefore & (0;-6) \\ \text{For } y=0 \quad 0 &= 2^{(x + 1)} - 8 \\ 2^3 &= 2^{(x + 1)} \\ \therefore 3 &= x + 1 \\ \therefore 2 &= x \\ \therefore & (2;0) \end{align*}
$$y = 2 \times 3^{(x - 1)} - \text{18}$$
\begin{align*} \text{For } x=0 \quad y &= 2 \times 3^{(0 - 1)} - 18 \\ &= \frac{2}{3} -18 \\ &= -17\frac{1}{3} \\ \therefore & (0;-17\frac{1}{3}) \\ \text{For } y=0 \quad 0 &= 2 \times 3^{(x - 1)} - 18 \\ 18 &= 2 \times 3^{(x - 1)} \\ 9 &= 3^{(x - 1)} \\ 3^2 &= 3^{(x - 1)} \\ \therefore 2 &= x - 1 \\ \therefore 3 &= x \\ \therefore & (3;0) \end{align*}
$$y + 5^{(x + 2)} = 5$$
\begin{align*} y + 5^{(x + 2)} &= 5 \\ y &= -5^{(x + 2)} + 5 \\ \text{For } x=0 \quad y &= -5^{(0 + 2)} + 5 \\ &= -25 + 5 \\ &= -20 \\ \therefore & (0;-20) \\ \text{For } y=0 \quad 0 &= -5^{(x + 2)} + 5 \\ 5^{(x + 2)} &= 5\\ \therefore x + 2 &= 1 \\ \therefore x &= -1 \\ \therefore & (-1;0) \end{align*}
$$y = \frac{1}{2} \left( \frac{3}{2} \right)^{(x + 3)} - \text{0,75}$$
\begin{align*} y &= \frac{1}{2} \left( \frac{3}{2} \right)^{(x + 3)} - \text{0,75} \\ y &= \frac{1}{2} \left( \frac{3}{2} \right)^{(x + 3)} - \frac{3}{4} \\ \text{For } x=0 \quad y &= \frac{1}{2} \left( \frac{3}{2} \right)^{(0 + 3)} - \frac{3}{4} \\ &= \frac{1}{2} \left( \frac{3}{2} \right)^{3} - \frac{3}{4} \\ &= \frac{1}{2} \left( \frac{27}{8} \right)- \frac{3}{4} \\ &= \frac{27}{16} - \frac{3}{4} \\ &= \frac{15}{16} \\ \therefore & (0;\frac{15}{16}) \\ \text{For } y=0 \quad 0 &= \frac{1}{2} \left( \frac{3}{2} \right)^{(x + 3)} - \frac{3}{4} \\ \frac{3}{4} &= \frac{1}{2} \left( \frac{3}{2} \right)^{(x + 3)} \\ \frac{3}{2} &= \left( \frac{3}{2} \right)^{(x + 3)} \\ \therefore 1 &= x + 3 \\ \therefore -2 &= x \\ \therefore & (-2;0) \end{align*}
Asymptote
Exponential functions of the form $$y = ab^{(x+p)} + q$$ have a horizontal asymptote, the line $$y = q$$.
Worked example 15: Asymptote
Determine the asymptote for $$y = 5 \times 3^{(x+1)} - 1$$.
Determine the asymptote
The asymptote of $$g(x)$$ can be calculated as: \begin{align*} 3^{(x+1)} & \ne 0\\ 5 \times 3^{(x+1)} & \ne 0\\ 5 \times 3^{(x+1)} - 1 & \ne -1\\ \therefore y & \ne -1 \end{align*} Therefore the asymptote is the line $$y = -1$$.
Asymptote
Exercise 5.18
Give the asymptote for each of the following functions:
$$y = -5^{(x + 1)}$$
\begin{align*} y &= -5^{(x + 1)} \\ \text{Horizontal asymptote: } \quad y &= 0 \end{align*}
$$y = 3^{(x - 2)} + 1$$
\begin{align*} y &= 3^{(x - 2)} + 1 \\ \text{Horizontal asymptote: } \quad y &= 1 \end{align*}
$$\left( \frac{3y}{2} \right) = 5^{(x + 3)} - 1$$
\begin{align*} \left( \frac{3y}{2} \right) &= 5^{(x + 3)} - 1 \\ 3y &= 2 \times 5^{(x + 3)} - 2 \\ y &= \frac{2}{3} \times 5^{(x + 3)} - \frac{2}{3} \\ \text{Horizontal asymptote: } \quad y &= -\frac{2}{3} \end{align*}
$$y = 7^{(x + 1)} - 2$$
\begin{align*} y &= 7^{(x + 1)} - 2 \\ \text{Horizontal asymptote: } \quad y &= -2 \end{align*}
$$\frac{y}{2} + 1 = 3^{(x + 2)}$$
\begin{align*} \frac{y}{2} + 1 &= 3^{(x + 2)} \\ \frac{y}{2} &= 3^{(x + 2)} - 1 \\ y &= 2 \times 3^{(x + 2)} - 2 \\ \text{Horizontal asymptote: } \quad y &= -2 \end{align*}
Sketching graphs of the form $$f(x)=a{b}^{\left(x+p\right)}+q$$
In order to sketch graphs of functions of the form, $$f(x)=a{b}^{\left(x+p\right)}+q$$, we need to determine five characteristics:
• shape
• $$y$$-intercept
• $$x$$-intercept
• asymptote
• domain and range
Worked example 16: Sketching an exponential graph
Sketch the graph of $$2y = \text{10} \times 2^{(x+1)} - 5$$.
Mark the intercept(s) and asymptote. State the domain and range of the function.
Examine the equation of the form $$y = ab^{(x + p)} + q$$
We notice that $$a > 0$$ and $$b > 1$$, therefore the function is increasing.
Determine the $$y$$-intercept
The $$y$$-intercept is obtained by letting $$x = 0$$: \begin{align*} 2y &= \text{10} \times 2^{(0+1)} - 5\\ &= \text{10} \times 2 - 5\\ &= \text{15}\\ \therefore y &= 7\frac{1}{2} \end{align*} This gives the point $$(0;7\frac{1}{2})$$.
Determine the $$x$$-intercept
The $$x$$-intercept is obtained by letting $$y = 0$$: \begin{align*} 0 &= \text{10} \times 2^{(x+1)} - 5\\ 5 &= \text{10} \times 2^{(x+1)} \\ \frac{1}{2} &= 2^{(x+1)}\\ 2^{-1} &= 2^{(x+1)}\\ \therefore -1 &= x + 1 \quad \text{(same base)}\\ -2 &= x \end{align*} This gives the point $$(-2;0)$$.
Determine the asymptote
The horizontal asymptote is the line $$y = -\frac{5}{2}$$.
State the domain and range
Domain: $$\{ x: x \in \mathbb{R} \}$$
Range: $$\{ y: y > -\frac{5}{2}, y \in \mathbb{R} \}$$
Finding the equation of an exponential function from the graph
Worked example 17: Finding the equation of an exponential function from the graph
Use the given graph of $$y = -2 \times 3^{(x + p)} + q$$ to determine the values of $$p$$ and $$q$$.
Examine the equation of the form $$y = ab^{(x + p)} + q$$
From the graph we see that the function is decreasing. We also note that $$a = -2$$ and $$b = 3$$.
We need to solve for $$p$$ and $$q$$.
Use the asymptote to determine $$q$$
The horizontal asymptote $$y = 6$$ is given, therefore we know that $$q = 6$$. $y = -2 \times 3^{(x + p)} + 6$
Use the $$x$$-intercept to determine $$p$$
Substitute $$(2;0)$$ into the equation and solve for $$p$$: \begin{align*} y &= -2 \times 3^{(x + p)} + 6 \\ 0 &= -2 \times 3^{(2 + p)} + 6 \\ -6 &= -2 \times 3^{(2 + p)} \\ 3 &= 3^{(2 + p)} \\ \therefore 1 &= 2 + p \quad \text{(same base)}\\ \therefore p &= -1 \end{align*}
Write the final answer
$y = -2 \times 3^{(x - 1)} + 6$
Mixed exercises
Exercise 5.19
Given the graph of the hyperbola of the form $$h(x) = \frac{k}{x}$$, $$x < 0$$, which passes though the point $$A(-\frac{1}{2}; -6)$$.
Show that $$k=3$$.
\begin{align*} y &=\frac{k}{x} \\ \text{Subst. } (-\frac{1}{2}; -6) \qquad -6 &=\frac{k}{-\frac{1}{2}} \\ -6 \times -\frac{1}{2} &= k \\ \therefore k &=3 \\ \therefore h(x) &= \frac{3}{x} \end{align*}
Write down the equation for the new function formed if $$h(x)$$:
is shifted $$\text{3}$$ units vertically upwards
\begin{align*} h(x) &= \frac{3}{x} \\ \therefore y &\Rightarrow y - 3 \\ y - 3 &=\frac{3}{x} \\ y &=\frac{3}{x}+3 \end{align*}
is shifted to the right by $$\text{3}$$ units
\begin{align*} h(x) &= \frac{3}{x} \\ \therefore x &\Rightarrow x - 3 \\ y &=\frac{3}{x -3} \end{align*}
is reflected about the $$y$$-axis
\begin{align*} h(x) &= \frac{3}{x} \\ \therefore x &\Rightarrow -x \\ y &=\frac{3}{x -3} \end{align*}
is shifted so that the asymptotes are $$x = 0$$ and $$y = -\frac{1}{4}$$
\begin{align*} h(x) &= \frac{3}{x} \\ \therefore p=0 &\text{ and } q = -\frac{1}{4} \\ y &=\frac{3}{x} -\frac{1}{4} \end{align*}
is shifted upwards to pass through the point $$(-1;1)$$
\begin{align*} h(x) &= \frac{3}{x} \\ \therefore y &\Rightarrow y + m \\ y &=\frac{3}{x} + m \\ \text{Subst.} (-1;1) \qquad 1 &= \frac{3}{-1} + m \\ 1 + 3 &= + m \\ \therefore m &= 4 \\ \therefore y &=\frac{3}{x} + 4 \end{align*}
is shifted to the left by $$\text{2}$$ units and $$\text{1}$$ unit vertically downwards (for $$x < 0$$)
\begin{align*} h(x) &= \frac{3}{x} \\ \therefore x &\Rightarrow x + 2 \\ \therefore y &\Rightarrow y + 1 \\ y + 1 &=\frac{3}{x + 2} \\ \therefore y &=\frac{3}{x + 2} - 1 \end{align*}
Given the graphs of $$f(x) = a(x+p)^2$$ and $$g(x) = \frac{a}{x}$$.
The axis of symmetry for $$f(x)$$ is $$x = -1$$ and $$f(x)$$ and $$g(x)$$ intersect at point $$M$$. The line $$y = 2$$ also passes through $$M$$.
Determine:
the coordinates of $$M$$
$$f(x)$$ is symmetrical about the line $$x = -1$$, therefore $$M(-2;2)$$.
the equation of $$g(x)$$
\begin{align*} g(x) &= \frac{a}{x} \\ \text{Subst. } M(-2;2) \qquad 2 &= \frac{a}{-2} \\ \therefore a &= -4 \\ \therefore g(x)&=\frac{-4}{x} \end{align*}
the equation of $$f(x)$$
\begin{align*} f(x) &= a(x + p)^2 + q \\ \text{No vertical shift } \therefore q &= 0 \\ \text{Axis of symmetry } x = -1 \qquad \therefore f(x) &= a(x + 1)^2 \\ \text{Subst. } M(-2;2) \qquad 2 &= a(-2 + 1)^2 \\ 2 &= a(-1)^2 \\ \therefore a &= 2 \\ \therefore f(x) &= 2(x + 1)^2 \end{align*}
the values for which $$f(x) < g(x)$$
$$- 2 < x < 0$$
the range of $$f(x)$$
$$\text{Range: }\left \{ y: y\in \mathbb{R}, y \geq 0 \right \}$$
On the same system of axes, sketch:
the graphs of $$k(x) = 2(x + \frac{1}{2})^2 - 4\frac{1}{2}$$ and $$h(x) = 2^{(x + \frac{1}{2})}$$. Determine all intercepts, turning point(s) and asymptotes.
the reflection of $$h(x)$$ about the $$x$$-axis. Label this function as $$j(x)$$.
Sketch the graph of $$y = ax^2 + bx + c$$ for:
$$a < 0$$, $$b > 0$$, $$b^2 < 4ac$$
$$a > 0$$, $$b > 0$$, one root $$=0$$
On separate systems of axes, sketch the graphs:
$$y = \frac{2}{x - 2}$$
$$y = \frac{2}{x} - 2$$
$$y = -2^{(x - 2)}$$
For the diagrams shown below, determine:
• the equations of the functions; $$f(x) = a(x + p)^2 + q$$, $$g(x)=ax^2 + q$$, $$h(x) = \frac{a}{x}, x < 0$$ and $$k(x)=b^x + q$$
• the axes of symmetry of each function
• the domain and range of each function
\begin{align*} f(x) &= a(x + p)^2 + q \\ \text{From turning point: } p= -2 &\text{ and } q = 3 \\ \therefore f(x) &= a(x - 2 )^2 + 3 \\ \text{Subst. } (0;0) \qquad 0 &= a(0 - 2)^2 + 3 \\ -3 &= 4a \\ \therefore a &= -\frac{3}{4} \\ f(x) &= -\frac{3}{4}(x - 2)^2 + 3 \\ \text{Axes of symmetry: } x &= 2 \\ \text{Domain: } & \left \{ x:x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y:y \in \mathbb{R}, y \leq 3 \right \} \end{align*}
\begin{align*} g(x) &= ax^2 + q \\ \text{From turning point: } p= 0 &\text{ and } q = -2 \\ \therefore g(x) &= a(x)^2 - 2 \\ \text{Subst. } (-2;-1) \qquad -1 &= a(-2 )^2 - 2\\ -1 &= 4a - 2\\ 1 &= 4a - 2\\ \therefore a &= \frac{1}{4} \\ g(x) &= \frac{1}{4}x^2 -2 \\ \text{Axes of symmetry: } x &= 0 \\ \text{Domain: } & \left \{ x:x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y:y \in \mathbb{R}, y \geq -2 \right \} \\ h(x) &= \frac{a}{x + p} + q \\ \text{From graph: } p= 0 &\text{ and } q = 0 \\ h(x) &= \frac{a}{x} \\ \text{Subst. } (-2;-1) \qquad -1 &=\frac{a}{-2} \\ 2 &= a \\ \therefore h(x) &= \frac{2}{x} \\ \text{Axes of symmetry: } y &= x \\ \text{Domain: } &\left \{ x:x \in \mathbb{R}, x < 0 \right \} \\ \text{Range: } & \left \{ y:y \in \mathbb{R}, y < 0 \right \} \end{align*}
\begin{align*} y &= 2^{x} + \frac{1}{2} \\ \text{Reflect about } x = 0 \qquad \therefore x &\\Rightarrow -x \\ \therefore k(x) &= 2^{-x} + \frac{1}{2} \\ &= \left( \frac{1}{2} \right)^{x} + \frac{1}{2} \\ \text{Domain: } & \left \{ x:x \in \mathbb{R} \right \} \\ \text{Range: } & \left \{ y:y \in \mathbb{R}, y > \frac{1}{2} \right \} \end{align*}
Given the graph of the function $$Q(x) = a^x$$.
Show that $$a = \frac{1}{3}$$.
\begin{align*} y&=a^x \\ \text{Subst.} \left( 1;\frac{1}{3} \right) \qquad \frac{1}{3} &=a^{1} \\ \therefore a &= \frac{1}{3} \end{align*}
Find the value of $$p$$ if the point $$(-2;p)$$ is on $$Q$$.
\begin{align*} Q(x) &= \left( \frac{1}{3} \right)^x \\ \text{Subst.} \left( -2;p \right) p &= \left( \frac{1}{3} \right)^{-2} \\ p &= 9 \end{align*}
Calculate the average gradient of the curve between $$x = -2$$ and $$x = 1$$.
\begin{align*} \text{Average gradient} &= \frac{\left ( \frac{1}{3} \right )^{-2} - \left ( \frac{1}{3} \right )^1}{-2-(1)} \\ &= \frac{9-\frac{1}{3}}{-3} \\ &=\frac{8\frac{2}{3}}{-3} \\ &= \frac{-26}{9} \\ &= -2\frac{8}{9} \end{align*}
Determine the equation of the new function formed if $$Q$$ is shifted $$\text{2}$$ units vertically downwards and $$\text{2}$$ units to the left.
\begin{align*} Q(x) &= \left( \frac{1}{3} \right)^x \\ \therefore x & \Rightarrow x + 2 \\ \therefore y & \Rightarrow y + 2 \\ y + 2 &= \left( \frac{1}{3} \right)^{x + 2} \\ y &= \left( \frac{1}{3} \right)^{x + 2} -2 \end{align*}
Find the equation for each of the functions shown below:
$$f(x) = 2^x + q$$
$$g(x) = mx + c$$
\begin{align*} f(x) &= 2^x + q \\ \text{Subst. } (0; -\frac{1}{2}) \quad -\frac{1}{2} &= 2^{0} + q \\ -\frac{1}{2} &= 1 + q \\ \therefore q &= -\frac{3}{2} \\ \therefore f(x) &=2^x-\frac{3}{2} \\ g(x) &= mx +c \\ \text{Subst. } (0; -\frac{1}{2}) \quad -\frac{1}{2} &= m(0) + c \\ \therefore c &= -\frac{1}{2} \\ g(x) &= mx -\frac{1}{2} \\ \text{Subst. } (-2;0) \quad 0 &= m(-2) -\frac{1}{2} \\ \frac{1}{2} &= -2m \\ \therefore m &= -\frac{1}{4} \\ \therefore g(x) &= -\frac{1}{4}x-\frac{1}{2} \end{align*}
$$h(x) = \frac{k}{x + p} + q$$
\begin{align*} h(x) &= \frac{k}{x + p} + q \\ \text{From graph: } \quad p = -2 \\ h(x) &= \frac{k}{x + 2} + q \\ \text{Subst. } (0; -\frac{1}{2}) \quad -\frac{1}{2} &= \frac{k}{2} + q \\ -1 &= k + 2q \ldots (1)\\ \text{Subst. } (1;0) \quad 0 &= \frac{k}{1 + 2} + q \\ 0 &= k + 3q \ldots (2)\\ (2) - (1): \qquad 1 &= 0 + q \\ \therefore q &= 1 \\ \text{and } k &= -3\\ \therefore h(x) &= -\frac{3}{x + 2} + 1 \end{align*}
Given: the graph of $$k(x) = -x^2 + 3x + \text{10}$$ with turning point at $$D$$. The graph of the straight line $$h(x) = mx + c$$ passing through points $$B$$ and $$C$$ is also shown.
Determine:
the lengths $$AO$$, $$OB$$, $$OC$$ and $$DE$$
\begin{align*} y &=-x^2+3x+10 \\ \text{Let } y &= 0 \\ 0 &=-x^2+3x+10 \\ &= x^2-3x-10\\ &= (x-5)(x+2)\\ \therefore x=5 &\text{ or }x =-2 \\ \therefore AO &= \text{2}\text{ units} \\ \therefore BO &=\text{5}\text{ units} \\ CO&=\text{10}\text{ units}\\ \text{Axes of symmetry: } x &=-\frac{5 - 2}{2} \\ &=\frac{3}{2} \\ \text{Subst. } x &= \frac{3}{2} \\ y &=-\left ( \frac{3}{2} \right )^2+3\frac{3}{2}+10 \\ &=12\frac{1}{4} \\ \therefore DE &= \text{12,25}\text{ units} \end{align*}
the equation of $$DE$$
$$DE = 12\frac{1}{4}$$
the equation of $$h(x)$$
\begin{align*} h(x) & = mx + c \\ h(x) & = mx + 10 \\ \text{Subst. } (5;0) \qquad 0 &= m(5) + 10 \\ -10 &= 5m \\ \therefore m &= -2 \\ \therefore h(x) &=-2x+10 \end{align*}
the $$x$$-values for which $$k(x) < 0$$
$$\left \{ x: x\in \mathbb{R}, x < -2 \text{ and } x > 5 \right \}$$
the $$x$$-values for which $$k(x) \geq h(x)$$
$$\left \{ x: x\in \mathbb{R}, 0 \leq x \leq 5 \right \}$$
the length of $$DF$$
\begin{align*} \text{At } x = \frac{3}{2} k(x) &= 12\frac{1}{4} \\ \text{At } x = \frac{3}{2} h(x) &= -2 \left( \frac{3}{2} \right) + 10 \\ &= -3 + 10 \\ &= 7\\ \therefore DF &= 12\frac{1}{4} - 7 \\ &= 12\frac{1}{4} - 7 \\ &= \text{5,25}\text{ units} \end{align*}
Trigonometric functions are examined in PAPER 2. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000100135803223, "perplexity": 1067.6207244536204}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371829677.89/warc/CC-MAIN-20200409024535-20200409055035-00380.warc.gz"} |
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The emf induced in the circuit is 10.0points A concave mirror has a radius of curvature of 0 . 726 m. An object is placed 2 . 41 m in front of the mirror. What is the position of the image? 011 10.0points In a thundercloud there may be an electric charge of 43 C near the top and 43 C near the bottom. These charges are separated by approximately 2 . 2 km. What is the magnitude of the electric force between them? The Coulomb constant is Since the charges are of opposite signs, this force is attractive. 012 10.0points A concave mirror has a radius of curvature of 0 . 726 m. An object is placed 2 . 41 m in front of the mirror. What is the position of the image?
8 . 98755 × 10 9 N · m 2 / C 2 . 013 10.0points
Version 036 – Final 58010 Fall 14 – yeazell – (58010) 7 Two parallel wires carry equal currents in the opposite directions. Point A is midway between the wires, and B is an equal distance on the other side of the wires. A B What is the ratio of the magnitude of the magnetic field at point A to that at point B ? 1. B A B B = 3 correct 2. B A B B = 1 2 3. B A B B = 1 3 4. B A B B = 0 5. B A B B = 2 3 6. B A B B = 5 2 7. B A B B = 4 8. B A B B = 10 3 9. B A B B = 4 3 10. B A B B = 2 Explanation: The magnetic field due to a long wire is B = μ 0 I 2 π r . Let the distance between the wires be r . The magnetic field at A due to the upgoing wire is B up,A = μ 0 I 2 π ( r/ 2) = μ 0 I π r . The right-hand rule tells us the direction is into the paper. Due to the fact that A is the same distance from both wires, the other wire gives a magnetic field at A of the same magnitude, also directed into the paper due to the right-hand rule. The total magnetic field at A is B A = 2 B up,A = 2 μ 0 I π r . Now, the field at B due to the upgoing wire is B up,B = μ 0 I 2 π (3 r/ 2) = μ 0 I 3 π r , again into the paper, while B down,B = μ 0 I 2 π ( r/ 2) = μ 0 I π r out of the paper. So B B = B down,B B up,B = μ 0 I π r parenleftbigg 1 1 3 parenrightbigg = 2 μ 0 I 3 π r , out of the paper. Comparing magnitudes, we find B A B B = 2 μ 0 I π r 2 μ 0 I 3 π r = 3 . 014 10.0points An electron is projected into a uniform mag- netic field given by vector B = B x ˆ ı + B y ˆ . The magnitude of the charge on an electron is e . x y z v electron B x B y B
Version 036 – Final 58010 Fall 14 – yeazell – (58010) 8 Find the magnetic force when the velocity of the electron is v ˆ . 6. eB y v ˆ i | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9462817311286926, "perplexity": 588.9150473446248}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363791.16/warc/CC-MAIN-20211209091917-20211209121917-00562.warc.gz"} |
https://www.varsitytutors.com/ap_physics_1-help/equivalent-resistance | # AP Physics 1 : Equivalent Resistance
## Example Questions
← Previous 1 3
### Example Question #1 : Equivalent Resistance
Consider the following circuit:
What is the total equivalent resistance of the circuit?
Explanation:
First we need to condense R3 and R4. They are in series, so we can simply add them to get:
Now we can condense R2 and R34. They are in parallel, so we will use the following equation:
Therefore:
The equivalent circuit now looks like:
Since everything is in series, we can simply add everything up:
### Example Question #2 : Equivalent Resistance
Consider the given circuit:
What is the current through the system if we attach a resistor from point A to B?
Explanation:
The new circuit has two resistors in parallel: R2 and the new one attached. To find the equivalent resistance of these two branches, we use the following expression:
In this new equivalent circuit everything is in series, so we can simply add up the resistances:
Now we can use Ohm's law to calculate the total current through the circuit:
### Example Question #3 : Equivalent Resistance
Consider the given circuit:
How much resistance must be applied between points A and B for the circuit to have a total current of 3A?
Explanation:
We will be working backwards on this problem, using the current to find the resistance. We know the voltage and desired current, so we can calculate the total necessary resistance:
Then we can calculate the equivalent resistance of the two resistors that are in parallel (R2 and our unknown):
Now we can calculate what the resistance between point A and B:
Rearranging for the desired resistance:
### Example Question #4 : Equivalent Resistance
Consider the circuit:
If the equivalent resistance of the circuit is and each resistor is the same, what is the value of each resistor?
None of these
Explanation:
We can use the equation for equivalent resistance of parallel resistors to solve this equation:
We know the equivalent resistance, and we know that the resistance of each of the four resistors is equal:
### Example Question #131 : Electricity And Waves
Consider the circuit:
If the power dissipated throughout the entire circuit is , what is the value of ?
Explanation:
Since we know the power loss and voltage of the circuit, we can calculate the equivalent resistance of the circuit using the following equations:
Substituting Ohm's law into the equation for power, we get:
Rearranging for resistance, we get:
This is the equivalent resistance of the entire circuit. Now we can calculate R4 using the expression for resistors in parallel:
### Example Question #6 : Equivalent Resistance
Consider the circuit:
If the current flowing through the circuit is , what is the value of R1?
Explanation:
We can use Ohm's law to calculate the equivalent resistance of the circuit:
Now we can use the expression for combining parallel resistors to calculate R1:
### Example Question #7 : Equivalent Resistance
Consider the circuit:
If the equivalent resistance of the circuit is , which of the following configuration of resistance values is possible?
None of these
Explanation:
We will need to test the values of each answer to find the one that generates an equivalent resistance of .
We know that when condensing parallel resistors, the equivalent resistance will never be larger than the largest single resistance, and will always be smaller than the smallest resistance. Therefore, two of the answer options cen be eliminated immediately.
After we have narrowed our choices down to the other options answers, we just have to test them with the following formula:
We will test the incorrect answer first:
### Example Question #1 : Equivalent Resistance
What is the equivalent resistance from Point A to Point B?
Explanation:
Because this circuit is neither purely series or purely parallel, we must simplify it before we solve it. Replace the right branch, which is purely series, with its equivalent resistance:
Now we have a purely parallel circuit, each branch having a resistance of . Apply the parallel formula and solve:
### Example Question #9 : Equivalent Resistance
What is the equivalent resistance of the following resistors, all in series: ?
Explanation:
For resistors all in series, the equivalent resistance is equal to the sum of the resistances.
### Example Question #10 : Equivalent Resistance
What is the equivalent resistance of a circuit consisting of a group of resistors (all in parallel), with the following resistances: ? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9764915108680725, "perplexity": 719.4215339388444}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823067.51/warc/CC-MAIN-20171018180631-20171018200631-00386.warc.gz"} |
http://neuroticnetworks.blogspot.com/2016_12_01_archive.html | ## Wednesday, December 14, 2016
### back propagation intro... Or what happens to the output when I change a given weight
The question is, if I change a weight in a NN, how much does it affect the output?
In other words, how much is $\dfrac{\partial E}{\partial w_{jk}}$?
where the error is defined as $E = \dfrac{(T - L_k)^2}{2}$, where $T$ is the expected/target output
Given a NN with 3 layers:
$L_h = input$
$L_i = intermediate$
$L_j = intermediate$
$L_k = output$
the output of each neuron would be:
$L_{i}=\sigma (\sum w_{hi}L_{h}) = \sigma (net_i)$
$L_{j}=\sigma (\sum w_{ij}L_{i}) = \sigma (net_j)$
$L_{k}=\sigma (\sum w_{jk}L_{j}) = \sigma (net_k)$
The output of the NN given the $weights$ and the $inputs$ is then:
$L_{k}=\sigma(\sum w_{jk}\sigma(\sum w_{ij}\sigma(\sum w_{hi}L_{h})))$
Now, just some derivatives that will come handy later on:
• $\dfrac{\partial \sigma (x)}{\partial x} = \sigma (x)(1-\sigma (x))$
• $(\dfrac{\partial net_k}{\partial{ L_j}})=w_jk$
Let's start from the Output neuron:
$\dfrac{\partial E}{\partial w_{jk}} = \dfrac{\partial E}{\partial L_k}\dfrac{\partial L_k}{\partial w_{jk}}= (\dfrac{\partial E}{\partial L_k})(\dfrac{\partial \sigma (net_k)}{\partial w_{jk}})=(\dfrac{\partial E}{\partial{ L_k}}) ( \dfrac{\partial{ \sigma (net_k)}}{\partial net_k})(\dfrac{\partial{net_k}}{\partial w_{jk}})=$
• $(\dfrac{\partial E}{\partial{ L_k}})= \dfrac{\partial{(\dfrac{(T - L_k)^2}{2})}}{\partial{L_k}}=(T - L_k)(-1)=(L_k-T)$
• $(\dfrac{\partial{ \sigma (net_k)}}{\partial net_k})=\sigma (net_k)(1-\sigma (net_k)=L_k(1-L_k)$
• $(\dfrac{\partial{net_k}}{\partial w_{jk}})=L_j$
hence:
$\dfrac{\partial E}{\partial w_{jk}} =(L_k - T)\cdot{L_k(1-L_k)}\cdot{L_j}$
Let's look at the previous layer (layer k):
$E = \dfrac{(T - L_k)^2}{2} = \dfrac{(T -\sigma (net_k))^2}{2} = \dfrac{(T -\sigma (\sum w_{jk}L_{j}))^2}{2}$
$\dfrac{\partial E}{\partial w_{ij}} = \dfrac{\partial E}{\partial{ L_j}} [L_j(1-L_j)][L_i]$
$\dfrac{\partial E}{\partial{ L_j}}=(\dfrac{\partial E}{\partial{ L_k}})(\dfrac{\partial L_k}{\partial{ L_j}})=(\dfrac{\partial E}{\partial{ L_k}})(\dfrac{\partial \sigma (net_k)}{\partial{ L_j}})=(\dfrac{\partial E}{\partial{ L_k}})(\dfrac{\partial \sigma (net_k)}{\partial{ net_k}})(\dfrac{\partial net_k}{\partial{ L_j}})=(L_k - T) \cdot{L_k(1-L_k)} \cdot{w_{jk}}$
hence:
$\dfrac{\partial E}{\partial w_{ij}} = [(L_k - T) \cdot{L_k(1-L_k)} \cdot{w_{jk}}] \cdot{[L_j(1-L_j)][L_i]}$
From this we see that for any other intermediate layer we have that:
$\dfrac{\partial E}{\partial w_{xy}} = \dfrac{\partial E}{\partial L_z} \cdot{[L_y(1-L_y)][L_x]}$
(where $x$ and $y$ are from intermediate layers)
Hence for the layer before $k$, that is Layer $i$:
$\dfrac{\partial E}{\partial w_{hi}} = \dfrac{\partial E}{\partial L_j} \cdot{[L_i(1-L_i)][L_h]}$
where the tricky bit is to compute $\dfrac{\partial E}{\partial L_x}, from the above steps we can see the following recurrence rule: •$\dfrac{\partial E}{\partial L_k} =(L_k-T)$•$\dfrac{\partial E}{\partial L_j} =(L_k-T) \cdot{L_k(1-L_k)} \cdot{w_{jk}}$•$\dfrac{\partial E}{\partial L_i} =(L_k-T) \cdot{L_k(1-L_k)} \cdot{w_{jk}} \cdot{L_j(1-L_j)} \cdot{w_{ij}}$Once we have$\dfrac{\partial E}{\partial w_{xy}}$we can use this value to increase/decrease the weight by that amount, all modulated by an$\alpha\Delta{w_{xy}}= -\alpha \dfrac{\partial E}{\partial w_{xy}}$and the new weight$w'$would be then:$w'_{xy} = w_{xy} + \Delta{w_{xy}}\$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9431885480880737, "perplexity": 2936.585569362303}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886112682.87/warc/CC-MAIN-20170822201124-20170822221124-00464.warc.gz"} |
http://math.au.dk/aktuelt/aktiviteter/event/item/5671/ | # Eliminating unphysical photon components from Dirac–Maxwell Hamiltonian quantized in the Lorenz gauge
Kouta Usui
(Hokkaido University)
Mat/Fys-seminar
Tirsdag, 3 oktober, 2017, at 14:00-14:45, in Aud. D4 (1531-219)
Abstrakt:
Note the time 14:00 sharp (or as soon as possible thereafter).
We study the Dirac–Maxwell model quantized in the Lorenz gauge. In this gauge, the space of quantum mechanical state vectors inevitably adopt an indefinite metric so that the canonical commutation relation (CCR) is realized in a Lorentz covariant manner. In order to obtain a physical subspace, in which no negative norm state exists, the method first proposed by Gupta and Bleuler is applied with mathematical rigor. It is proved that a suitably defined physical subspace has a positive semidefinite metric, and naturally induces a physical Hilbert space with a positive definite metric. Then, the original Dirac–Maxwell Hamiltonian defines an induced Hamiltonian on the physical Hilbert space which is essentially self-adjoint.
Kontaktperson: Jacob Schach Møller | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9479812383651733, "perplexity": 3984.4745784536713}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195531106.93/warc/CC-MAIN-20190724061728-20190724083728-00270.warc.gz"} |
https://www.physicsforums.com/threads/action-reaction-force-quick-question.614409/ | # Action reaction force quick question
1. Jun 16, 2012
1. The problem statement, all variables and given/known dataA firewoman opens the fire hose, and water sprays forward. What is the action force and reaction force?
2. Relevant equationsnone
3. The attempt at a solution
I was thinking that the action force would be water pushing on air and reaction air pushing on water? But how would this be possible in space where there is a vacuum?
2. Jun 16, 2012
### TSny
I think the question is rather ambiguous. The firewoman exerts a force on the hose to hold it at rest. There will be a corresponding reaction force. Can you describe this reaction force?
Also, the hose will exert a force on the water. What is the reaction force to this?
As you stated, the water will exert a force on the air as it moves through the air. And you correctly stated that the reaction force would be the air exerting a force on the water.
[By the way, when dealing with action-reaction pairs, it doesn't matter which of the two forces you call the "action" force and which the "reaction" force.]
If the hose and firewoman are in a vacuum, then of course there would be no force of the water on any air. But you would still have the other action-reaction pairs mentioned above.
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align equations latex
This is the 17th video in a series of 21 by Dr Vincent Knight of Cardiff University. Some of these equations include cases. Mathematical Equations in LaTeX. Related articles. My latex code is as below: begin{align}label{mylabel} text{running text}nonumber\ dot{vx} & = & (A_{i}+Delta A_{i})vx+(B_{i}+Delta B_{i})boldsymbol{u} end{align} On compilation of the above code, the equation gets aligned to the right and the text gets left aligned. Aligning Equations (align) ... Notice that there's no \\ on the last line; the \end{align*} tells LaTeX that you're finished. there are several equations with domains. In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and ⦠... To achieve correct break and alignment of the above equation try the code below. As with matrices and tables, \\ specifies a line break, and & is used to indicate the point at which the lines should be aligned. As a style issue, notice that we start a new line in our source file after each \\. LaTeX is a powerful tool to typeset math; Embed formulas in your text by surrounding them with dollar signs \$; The equation environment is used to typeset one formula; The align environment will align formulas at the ampersand & symbol; Single formulas must be seperated with two backslashes \\; Use the matrix environment to typeset matrices; Scale parentheses with \left( \right) ⦠This is needed on at least one of the lines. Can I write a LaTeX equation over multiple lines? (Note: new lines (\\) do not work in equation environments.) Using the multiline, aligned packages. In large equations or derivations which span multiple lines, we can use the \begin{align} and \end{align} commands to correctly display the aligned mathematics. Use the ampersand character &, to set the points where the equations are vertically aligned. As you see above, you can leave some columns blank. The double backslash works as a newline character. The align environment is used for two or more equations when vertical alignment is desired; usually binary relations such as equal signs are aligned. The environment cases inside align results in that domains are not aligned at the same position. The align and align* environments, available through the amsmath package, are used for arranging equations of multiple lines. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. Inside the equation environment, use the split environment to split the equations into smaller pieces, these smaller pieces will be aligned accordingly. Do you know any way that allows a consistent horizontal alignment of the domains? Open an example of the amsmath package in Overleaf We could run all ⦠LaTeX equation editing supports most of the common LaTeX mathematical keywords. An alternate is to use the aligned environment which yields similar results.. 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The flalign environment in that domains are not aligned at the same.... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9956573843955994, "perplexity": 1988.506601116649}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154304.34/warc/CC-MAIN-20210802043814-20210802073814-00560.warc.gz"} |
https://edurev.in/course/quiz/attempt/7812_Test-Structural-Analysis-2/da882e31-0a16-4c77-8d4f-b79bca4116f4 | Courses
# Test: Structural Analysis- 2
## 20 Questions MCQ Test GATE Civil Engineering (CE) 2022 Mock Test Series | Test: Structural Analysis- 2
Description
This mock test of Test: Structural Analysis- 2 for GATE helps you for every GATE entrance exam. This contains 20 Multiple Choice Questions for GATE Test: Structural Analysis- 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Structural Analysis- 2 quiz give you a good mix of easy questions and tough questions. GATE students definitely take this Test: Structural Analysis- 2 exercise for a better result in the exam. You can find other Test: Structural Analysis- 2 extra questions, long questions & short questions for GATE on EduRev as well by searching above.
QUESTION: 1
Solution:
QUESTION: 2
Solution:
QUESTION: 3
### The given figure shows a beam with its influence line for shear force and bending moment at section 1: The values of the shear force and bending moment at section 1 due to a concentrated load of 20 kN placed at midspan will be
Solution:
QUESTION: 4
For the frame shown below; if the final moment is – 40 kNm at A and B of the column AB, then the moment MCD will be:
Solution:
QUESTION: 5
The load diagram and bending moment of a beam are shown below: The shear force at B would be:
Solution:
QUESTION: 6
The pin-joined cantilever truss is loaded as shown below. The force is member ED is
Solution:
QUESTION: 7
In the pin-joined plane frame shown below, the force in the member BD is
Solution:
Consider the joint B
Using Equilibrium equation
Summation of vertical forces =0
Fbd=0
QUESTION: 8
The total degree of indeterminacy for the given bridge truss is
Solution:
QUESTION: 9
In the truss shown below, which of the following member has no force induced on it?
Solution:
QUESTION: 10
For the given frame, the distribution factors for members BC and BA at joint B are
Solution:
QUESTION: 11
Assertion (A): For equal distribution of moment at the support B of the beam shown in the figure, the span length x required is (3/4)L.
Reason (R): For equal distribution of moment for the given beam, at B, I/L = 3I/4x
Solution:
QUESTION: 12
The horizontal thrust at support A of the given three hinged arch is ________ kN.
Solution:
QUESTION: 13
Consider a loaded trust shown in the given figure:
Match List-I (member) with List-II (force) and select the correct answer using the codes given below
Solution:
QUESTION: 14
What is the ratio of the forces in the members AB, BE and AE of the pin-jointed truss shown in the figure given below:
Solution:
QUESTION: 15
What is the degree of indeterminacy of the frame shown in the figure given below?
Solution:
QUESTION: 16
A simply supported RCC beam carrying a uniform distributed load has deflection of 100 mm at the center. When both the ends of the beam are made to be fixed, the deflection at the center will be
Solution:
QUESTION: 17
The cantilever frame, shown below, is supported by verticals at B and C and carries loads as shown in the figure. Therefore, the force in bar AE has to be __________ kg.
Solution:
QUESTION: 18
For the portal shown below, the base shear can be obtained from
Solution:
QUESTION: 19
A three-hinged symmetrical arch is loaded as shown below:
Which of the following is the magnitude of the correct horizontal thrust?
Solution:
QUESTION: 20
The distribution factor for members AE and AC of the given box section are
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https://www.physicsforums.com/threads/nilpotent-ring.371275/ | # Nilpotent ring
1. Jan 20, 2010
### tom.young84
I have this question and its a combination of the binomial theorem and nilpotent elements within a ring.
Suppose the following, am=bn=0. Is it necessarily true that (a+b)m+n is nilpotent.
For this question I did the following:
$$\sum$$i=0m+n$$\binom{m+n}{i}$$am+n-ibi
If i=m, then a=0. Additionally, if i>m a=0.
That's actually as far as I've gotten.
Last edited: Jan 20, 2010
2. Jan 20, 2010
### Dick
I think you are actually asking whether (a+b)^(m+n)=0. Is that right? If a^m=0, then a^(m+1)=0, a^(m+2)=0 etc etc. Similar for b. All of the terms in your binomial expansion have the general form i*a^k*b^l where (k+l)=(m+n). Is it possible k<m AND l<n?
3. Jan 21, 2010
### tom.young84
So I was working on this today during a lecture.
(a+b)m+n
Now we go to some arbitrary term in the middle:
am+n-ibi
From here we can notice the following things:
i>n and i=n
If this is true, then we know that b=0 and the whole thing equals zero.
i<n
If this is true, then we know that a=0 from the following:
If i<n then we know i$$\leq$$n-1. Then with i subbed in, m+n-(n-1) which equals m+1.
This means that, by substitution, m+1. That makes a=0.
Done?
4. Jan 21, 2010
### Dick
That's more than a little confusing. I mean, you aren't proving a=0 or b=0, you are proving powers of a and b are zero, right? But yes, I think you've got the right idea. You could just state it a lot more clearly.
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http://mathhelpforum.com/algebra/191725-help-recalling-equation-involving-exponential-growth-over-time.html | # Thread: Help recalling equation involving exponential growth over time
1. ## Help recalling equation involving exponential growth over time
I recall having learned an equation involving exponential growth/decay that had something like y=initial amount x e^(2t) or something like that. Do you guys know the exact equation adn perhaps a sample problem to work with? I am prepping for a state math exam.
2. ## Re: Help recalling equation involving exponential growth over time
Exponential growth is of the form $y(t)= y_0e^{\alpha t}$. Yes, $y_0= y(0)$, the initial size. What $\alpha$ is depends upon the rate of growth. In one time unit (hour, day, year, depending on the units for t), y will have grown from $y(0)= y_0$ to $y(1)= y_0e^{\alpha}$ for a rate of $y(1)/y(0)= e^{\alpha}$. In fact, between year "n" and year "n+1", y will have grown from $y(n)= y_0e^{\alpha n}$ to $y(n+1)= y_0e^{\alpha(n+1)}= y_0e^{\alpha n+ \alpha}= y_0e^{\alpha n}e^{\alpha}$ again giving a rate of growth of $y(n+1)/y(n)= e^{\alpha}$.
Note that, since $e^{\alpha t}= \left(e^{\alpha}\right)^t$, that can also be written as $y(x)= y_0\left(e^\alpha\right)^t= y_0r^t$ where $r= e^\alpha$ is that rate of growth. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.987872838973999, "perplexity": 632.7834852462857}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706794379/warc/CC-MAIN-20130516121954-00073-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/16410/what-are-category-codes | # What are category codes?
Following on from this question, I'd like to ask a more general question:
What are category codes, and what can I achieve by changing them?
When TeX parses input, it assigns each character read a category code. How TeX subsequently interprets the input then depends on both the character and it's category code. There are 16 category codes that can be set by the programmer, plus one special internal one. The 16 standard ones number from 0 upward. Category code 0 is for escape characters, usually \. The rest are then (with typical examples):
1. Begin group: {
2. End group: }
3. Math shift: $ 4. Alignment: & 5. End-of-line 6. Parameter for macros: # 7. Math superscript: ^ 8. Math subscript: _ 9. Ignored entirely 10. Space 11. Letters: the alphabet. 12. 'Other' character - everything else: ., 1, :, etc. 13. Active character - to be interpreted as control sequences: ~ 14. Start-of-comment: % 15. Invalid-in-input: [DEL] Now when TeX reads input, each character is associated with a category code to generate tokens. So if the input reads $ 1^{23}_a TeX reads: • A math shift token, and goes into math mode • A space, which is ignored in math mode • An 'other' token 1, which is simply typeset here • A math superscript token, thus meaning that the next item will be superscripted • A begin-group token, • The 'other' tokens 2 and 3, which cannot be typeset until the group finishes • The close-group token }, which allows TeX to typeset the superscript • A math subscript token, so moving the next item to a subscript position • The letter a, which with no special meaning is typeset • A space, again ignored • A math shift token, and goes back into horizontal mode Category codes often become important when TeX is deciding on what is and is not a control sequence. With only the alphabet as 'letters', something like \hello@ is the control sequence \hello followed by the 'other' token @. On the other hand, if I make @ a letter \catcode\@=11\relax \hello@ then TeX will look for a macro called \hello@. This is commonly used in TeX code to isolate 'code' macros from 'user' ones. So you find programming macros such as \@for. Without changing the category code, this is effectively 'hidden'. The idea of this is to 'protect the user from themselves': it's hard to break the code if you cannot even get at it! There are many effects that can be achieved using category codes. An obvious one is the non-breaking space ~ used throughout the TeX world. This works because ~ has category code 13, and is therefore 'active'. When TeX reads ~, it looks for a definition for ~ in the same way it would for a macro. That's a lot more convenient than using a macro for these cases. We can use different category codes to make 'private' code areas. For example, plain TeX and LaTeX2e us @ as an extra 'letter', whereas LaTeX3 uses : and _. That effectively isolates internal LaTeX3 code from LaTeX2e, when the two are used together (as at present). Verbatim material is another area where category codes are vital (if complex!). The reason you can't nest verbatim material inside anything else is that once TeX has assigned category codes it is only partially reversible. Anything which is 'ignored' or 'comment' is thrown away: you can't get it back. (With e-TeX, you can reassign category codes, but anything that is already gone stays 'lost.) (Note for the interested) The 'special' category code is 16, which is used in the \ifcat test, amongst other things. It is assigned to unexpandable control sequences in this situation, so that they do not match anything else other than other unexpandable control sequences. • I have written a small tool called texref which can give you quick information about category codes and character information, and maybe other things too in future. If you think it's useful, perhaps consider including it in your answer. github.com/kieranclancy/texref.git – codebeard Jul 8 '12 at 16:30 • @codebeard: sounds interesting, and would be good on ctan, except there's already a texref there (it locates \label and \ref, etc, commands in a document). your tool, renamed, would be a nice addition to the "public space" – wasteofspace Aug 29 '12 at 10:30 • @JosephWright How to revert the changes made to \catcode (if any)? By putting them inside \makeatletter & \makeatother? And I presume, all of this has to be in the preamble. In @ChristianLindig 's example from way below, he talks about making _ locally active. So, the \catcode changes can be made post \begin{document} as well? If so, is the reverting method same as above? – Amar May 31 '15 at 1:56 • @Amar \catcode is a TeX primitive and respects TeX group levels. Thus you can apply a \catcode change inside a group to keep it local, or you can explicitly revert it with a second \catcode call. The latter is the way the \makeatletter/\makeatother pair works. – Joseph Wright May 31 '15 at 8:36 Rather than defining primitive commands for common tasks such as starting math mode or denoting superscripts and subscripts, Knuth decided to reserve some characters for these purposes. There are also other needs: grouping, denoting the macro parameters and, most important, escaping in order to express commands. There are sixteen category codes: 0 = escape 1 = group start 2 = group end 3 = math shift 4 = alignment tab 5 = end of line 6 = parameter 7 = superscript 8 = subscript 9 = ignored character 10 = space 11 = letter 12 = other character 13 = active character 14 = comment 15 = invalid character Usually there's only one character having categories 0 to 8 \ { } & ^^M # ^ _
(^^M denotes the invisible character that TeX puts at the end of all input lines, changing the system dependent one(s) that might be present); uniqueness is not required, but preferred: why should one want to have two different escape characters which would act just in the same way? (See later on.)
Category 10 is the space but also the <TAB> character, that is not distinguishable from a sequence of spaces; category 10 characters are ignored at the start of a line. Category 5 is very special: it's transformed into a space unless it's followed by another category 5 character, when it becomes the command \par (it's the trick that allows to leave a blank line to end a paragraph). In general any sequence of contiguous category 10 characters is reduced to only one and it doesn't matter if they are spaces, tabs or converted end-of-line characters.
All letters have category 11 and punctuation characters such as ?, (, ) and others have category 12; this is for the rule that a command name can be any sequence of letters (better, category 11 characters) or one not 11 category character, preceded by a category 0 character. Category 11 and 12 characters, when not part of a command name may be printed; this is not the case for all other category codes. However a category code 11 or 12 character may also not show up in print, because it's discarded during processing (for example keywords or option to packages, package or file names, ...).
Category 9 and 15 were put into TeX because there are "dangerous" character (ASCII "null" and ASCII "delete") that could be misinterpreted by editors. Actually category 9 has other uses: in LaTeX3 style files the space is assigned category 9, to help programmers in avoiding the dreadful "spurious spaces".
Category 14 is the well-known % that introduces comments and makes TeX ignore everything following it in a line (end of line included).
Category 13 is very special; Plain TeX and the LaTeX kernel use only one active character, namely ~; an active character is treated as if it were a command and must have a definition before it can be used; the LaTeX definition is
\catcode~=13
\def~{\nobreakspace{}}
so that typing ~ is just the same as writing \nobreakspace{}. Other active characters are also used by the LaTeX "inputenc" package, in such a way that, for instance, ü is translated into \"u.
When we want to typeset verbatim TeX code, many of the special characters are assigned category code 12; but when we type \verb+\xyz+, LaTeX reads \verb and prepares everything for verbatim typesetting and starts a group; the first + is swallowed and is assigned category 2, so that when it finds the second + the group is terminated and all assignments are reverted to the normal ones (including the category 2 assignment to +): it's a bit magic, but it works, provided \verb+\xyz+ doesn't appear in the argument of a command.
This is a problem: when TeX is scanning the argument to a command, it freezes the category codes: when a character enters TeX it is transformed into a pair (category code, character code) which is no more the original character and so the category code assignment can't be modified any more (well, not really, there's \scantokens, but this would require a very long discussion).
The LaTeX commands \makeatletter and \makeatother work by changing the category code of @, which is usually 12; the first one puts it into category 11, so that it can appear in command names, the second one reverts this assignment. But how can
\makeatletter
\newcommand{\xyz}{...\@xyz...}
\makeatother
work? One might expect that when TeX expands \xyz it finds the "illegal" command name \@xyz. This doesn't happen: just as a simple character is transformed into a pair of numbers, when TeX scans it, a command name becomes a symbolic token, an internal representation of the command which is independent of characters and their category codes.
If we assign category code 0 to |, we can type \LaTeX and |LaTeX: they would mean just the same thing. But having different characters sharing the same category code 4 might turn out to be useful for aligning decimal numbers at the decimal separator in a tabular. If we assign . category code 4, we may type a decimal number as 123.456 and LaTeX will interpret it as if it were 123&456, producing two table cells that to the end user appear as one; some trickery in the definition of the table column structure is required, though.
• The explanation of dcolumn is interesting as it solves the same problem as my example, albeit in a better way. – Christian Lindig Apr 22 '11 at 20:29
• Doesn't dcolumn use \mathcode? – Joseph Wright Apr 22 '11 at 20:34
• Yes, of course. I didn't remember correctly, but the trick is still worth noting. I'll edit the answer. – egreg Apr 22 '11 at 21:11
• Excellent answer. Can a character be in two or more cat codes? Or there is a 1-1 correspondence between chars and cat codes? Thank you, as always! – manooooh Mar 10 at 20:56
• @manooooh No, at any given time a character has only one category code. – egreg Mar 10 at 20:57
This is not an easy topic and I can only refer you to the TeXBook for more details, but here is a short outline.
Every character that TeX reads from your file has two numbers associated with it. A "character code" and a "category code". TeX does not know glyphs - only numbers - and this is part of its strengths. You can think of font tables as look-up list. If you give it a number TeX will look at the list and print the glyph that happens to be in that position.
The second number is the "category code". This TeX uses it to intelligently parse the input. TeX needs to know for example if a curly left bracket { has appeared in a particular part of the document so that it can look for an ending bracket and so on. Of course this could have been hardwired, but Knuth chose to abstract it, so that any character, provided it has the appropriate category code can be used.
Consider that you may wish to replace curly brackets with square brackets [], this can be achieved by the following simple code:
\catcode [=1
\catcode ]=2
\def\test[This is a test]
\bye
Try it out without the catcode changes and it will fail with a run-away definition error.
Similarly, the \ backslash character can be redefined for use in verbatim text.
\catcode [=1
\catcode ]=2
\def\test[This is a test]
\catcode *=0
*def*test[This is another test]
*test
\bye
In the last example in lieu of \ you can type *. Run both MWE through pdfTeX. Authors don't really need them but they are invaluable for package developers.
There are great answers here already so let me just show a small example for what you can do by changing catcodes. Assume that you have numbers in a table like 12.3 and 8.45 and you want to pad them all to the same width in order to make them line up nicely. (Let's ignore for a moment that there are several ways.) You can do this by adding an invisible 0 after 12.3 and before 8.45. The invisible 0 is provided by \phantom{0} (or \phantom0). Now, using this is somewhat unpleasant to type because it messes up your source code layout. But you could make _ locally active and give it the meaning of \phantom0:
\catcode_=\active
\def_{\phantom0}
Now you can use 12.3_ and _8.45 in your table. You could have used some other character as well, like ! for example. Since _ is used in math mode for subscripts this only works when you need no subscripts where the above definition is active.
• Nice! One could go even further and \catcode0=\active and then define 0 such that it becomes \phantom0 if space is adjacent and a normal 0 else. Or (inspired by egreg's answer) even more evil: \catcode.=\active and define . as a macro which counts the surrounding digits (left and right separately), determines the maximum (locally, e.g. in the current group or environment) and \phantom0 aligns all numbers automagically. – Tobias Kienzler Dec 10 '15 at 6:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9771360158920288, "perplexity": 1719.2361685503154}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400227524.63/warc/CC-MAIN-20200925150904-20200925180904-00462.warc.gz"} |
https://stats.stackexchange.com/questions/120683/how-to-prove-dependence-of-random-variables | # How to prove dependence of random variables
I need to solve the following problem.
Let $X$ be a normal random variable with mean $\mu$ and standard deviation $\sigma$ and let $I$, independent of $X$, be such that $\mathbb{P}(I = 2) = \mathbb{P}(I = -2) = 0.5$. Let $Y = I X$. In words, $Y$ is equally likely to be either $2X$ or $–2X$.
a) Are $X$ and $Y$ independent?
b) Are $I$ and $Y$ independent?
c) What is the distribution of $Y$?
d) Find $\mathrm{Cov}(X,Y)$.
I think $X$ and $Y$ are dependent, but I don't know how to prove it. Any hint? Also, any hint about the following points?
The intuitive notion behind independence of $X$ and $Y$ is that knowledge of the value of one random variable $X$ should provide you with no information about the value of $Y$. So start with basics. Is it true that $Y$ can take on all real number values? If you are told that $X$ has value $1$, does it continue to be true that $Y$ can take on all real number values? If your answer is YES, re-read your own assertion
In words, $Y$ is equally likely to be either $2X$ or $–2X$
at least three times, and think some more about what this is saying when you know that $X=1$.
Hint for Question 2.
Repeat the above argument with $I$ in place of $X$. What is the conditional mean of $Y$ when $I=2$? What about when $I = -2$?
Here is my thoughts. If $X$ and $Y$ are independent, then we have $\mathbb{P}(Y = 2 \mid X = 1) = \mathbb{P}(Y = 2)$. So we can check whether they are equal.
On the one hand, $$P(Y = 2 \mid X = 1) = P(IX = 2 \mid X = 1)$$
$$= P(I = 2, X = 1 \mid X = 1) = P(I = 2) = 0.5$$
On the other hand, $$P(Y = 2) = 0.5*P(X=1) + 0.5*P(X=-1) = 0.$$ （NOT sure whether I am correct here.) Thus I think they are dependent.
Whether $I$ and $Y$ are independent can also be checked by the same fashion.
As for the distribution of $Y$, $$F_Y(y) = P(Y <= y)=P(2X <= y \mid I=2)*P(I=2)$$
$$+ P(-2X <= y \mid I=-2)*P(I=-2) = 0.5*F_X(\frac{y}{2}) + 0.5*(1 - F_X(\frac{y}{2}))$$
Correct me if I am wrong. Any comments are appreciated. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9445077776908875, "perplexity": 82.7348109605006}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00056.warc.gz"} |
http://www.ck12.org/book/Probability-and-Statistics---Advanced-%2528Second-Edition%2529/r1/section/11.2/ | <meta http-equiv="refresh" content="1; url=/nojavascript/"> The One-Way ANOVA Test | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Probability and Statistics - Advanced (Second Edition) Go to the latest version.
# 11.2: The One-Way ANOVA Test
Created by: CK-12
## Learning Objectives
• Understand the shortcomings of comparing multiple means as pairs of hypotheses.
• Understand the steps of the ANOVA method and the method's advantages.
• Compare the means of three or more populations using the ANOVA method.
• Calculate pooled standard deviations and confidence intervals as estimates of standard deviations of populations.
## Introduction
Previously, we have discussed analyses that allow us to test if the means and variances of two populations are equal. Suppose a teacher is testing multiple reading programs to determine the impact on student achievement. There are five different reading programs, and her 31 students are randomly assigned to one of the five programs. The mean achievement scores and variances for the groups are recorded, along with the means and the variances for all the subjects combined.
We could conduct a series of $t$-tests to determine if all of the sample means came from the same population. However, this would be tedious and has a major flaw, which we will discuss shortly. Instead, we use something called the Analysis of Variance (ANOVA), which allows us to test the hypothesis that multiple population means and variances of scores are equal. Theoretically, we could test hundreds of population means using this procedure.
### Shortcomings of Comparing Multiple Means Using Previously Explained Methods
As mentioned, to test whether pairs of sample means differ by more than we would expect due to chance, we could conduct a series of separate $t$-tests in order to compare all possible pairs of means. This would be tedious, but we could use a computer or a TI-83/84 calculator to compute these quickly and easily. However, there is a major flaw with this reasoning.
When more than one $t$-test is run, each at its own level of significance, the probability of making one or more type I errors multiplies exponentially. Recall that a type I error occurs when we reject the null hypothesis when we should not. The level of significance, $\alpha$, is the probability of a type I error in a single test. When testing more than one pair of samples, the probability of making at least one type I error is $1-(1-\alpha)^c$, where $\alpha$ is the level of significance for each $t$-test and $c$ is the number of independent $t$-tests. Using the example from the introduction, if our teacher conducted separate $t$-tests to examine the means of the populations, she would have to conduct 10 separate $t$-tests. If she performed these tests with $\alpha=0.05$, the probability of committing a type I error is not 0.05 as one would initially expect. Instead, it would be 0.40, which is extremely high!
### The Steps of the ANOVA Method
With the ANOVA method, we are actually analyzing the total variation of the scores, including the variation of the scores within the groups and the variation between the group means. Since we are interested in two different types of variation, we first calculate each type of variation independently and then calculate the ratio between the two. We use the $F$-distribution as our sampling distribution and set our critical values and test our hypothesis accordingly.
When using the ANOVA method, we are testing the null hypothesis that the means and the variances of our samples are equal. When we conduct a hypothesis test, we are testing the probability of obtaining an extreme $F$-statistic by chance. If we reject the null hypothesis that the means and variances of the samples are equal, and then we are saying that the difference that we see could not have happened just by chance.
To test a hypothesis using the ANOVA method, there are several steps that we need to take. These include:
1. Calculating the mean squares between groups, $MS_B$. The $MS_B$ is the difference between the means of the various samples. If we hypothesize that the group means are equal, then they must also equal the population mean. Under our null hypothesis, we state that the means of the different samples are all equal and come from the same population, but we understand that there may be fluctuations due to sampling error. When we calculate the $MS_B$, we must first determine the $SS_B$, which is the sum of the differences between the individual scores and the mean in each group. To calculate this sum, we use the following formula:
$SS_B=\sum^m_{k=1} n_k (\bar{x}_k-\bar{x})^2$
where:
$k$ is the group number.
$n_k$ is the sample size of group $k$.
$\bar{x}_k$ is the mean of group $k$.
$\bar{x}$ is the overall mean of all the observations.
$m$ is the total number of groups.
When simplified, the formula becomes:
$SS_B=\sum^m_{k=1} \frac{T^2_k}{n_k}-\frac{T^2}{n}$
where:
$T_k$ is the sum of the observations in group $k$.
$T$ is the sum of all the observations.
$n$ is the total number of observations.
Once we calculate this value, we divide by the number of degrees of freedom, $m-1$, to arrive at the $MS_B$. That is, $MS_B=\frac{SS_B}{m-1}$
2. Calculating the mean squares within groups, $MS_W$. The mean squares within groups calculation is also called the pooled estimate of the population variance. Remember that when we square the standard deviation of a sample, we are estimating population variance. Therefore, to calculate this figure, we sum the squared deviations within each group and then divide by the sum of the degrees of freedom for each group.
To calculate the $MS_W$, we first find the $SS_W$, which is calculated using the following formula:
$\frac{\sum(x_{i1}-\bar{x}_1)^2+\sum (x_{i2}-\bar{x}_2)^2+ \ldots + \sum (x_{im}-\bar{x}_m)^2}{(n_1-1)+(n_2-1)+ \ldots + (n_m-1)}$
Simplified, this formula becomes:
$SS_W=\sum^m_{k=1} \sum^{n_k}_{i=1} x^2_{ik}-\sum^m_{k=1} \frac{T^2_k}{n_k}$
where:
$T_k$ is the sum of the observations in group $k$.
Essentially, this formula sums the squares of each observation and then subtracts the total of the observations squared divided by the number of observations. Finally, we divide this value by the total number of degrees of freedom in the scenario, $n-m$.
$MS_W=\frac{SS_W}{n-m}$
3. Calculating the test statistic. The formula for the test statistic is as follows:
$F=\frac{MS_B}{MS_W}$
4. Finding the critical value of the $F$-distribution. As mentioned above, $m-1$ degrees of freedom are associated with $MS_B$, and $n-m$ degrees of freedom are associated with $MS_W$. In a table, the degrees of freedom for $MS_B$ are read across the columns, and the degrees of freedom for $MS_W$ are read across the rows.
5. Interpreting the results of the hypothesis test. In ANOVA, the last step is to decide whether to reject the null hypothesis and then provide clarification about what that decision means.
The primary advantage of using the ANOVA method is that it takes all types of variations into account so that we have an accurate analysis. In addition, we can use technological tools, including computer programs, such as SAS, SPSS, and Microsoft Excel, as well as the TI-83/84 graphing calculator, to easily perform the calculations and test our hypothesis. We use these technological tools quite often when using the ANOVA method.
Example: Let’s go back to the example in the introduction with the teacher who is testing multiple reading programs to determine the impact on student achievement. There are five different reading programs, and her 31 students are randomly assigned to one of the five programs. She collects the following data:
Method
$& 1 && 2 && 3 && 4 && 5 \\& 1 && 8 && 7 && 9 && 10 \\& 4 && 6 && 6 && 10 && 12 \\& 3 && 7 && 4 && 8 && 9 \\& 2 && 4 && 9 && 6 && 11 \\& 5 && 3 && 8 && 5 &&8 \\& 1 && 5 && 5 &&&&\\& 6 && && 7 &&&&\\& &&&& 5 &&&&$
Compare the means of these different groups by calculating the mean squares between groups, and use the standard deviations from our samples to calculate the mean squares within groups and the pooled estimate of the population variance.
To solve for $SS_B$, it is necessary to calculate several summary statistics from the data above:
$& \text{Number } (n_k) && 7 && 6 && 8 && 5 && 5 && 31\\& \text{Total } (T_k) && 22 && 33 && 51 && 38 && 50 &&= 194\\& \text{Mean } (\bar x) && 3.14 && 5.50 && 6.38 && 7.60 && 10.00 && = 6.26\\& \text{Sum of Squared Obs. } \left (\sum_{i=1}^{n_k} x^2_{ik}\right ) && 92 && 199 && 345 && 306 && 510 && = 1,452\\& \frac{\text{Sum of Obs. Squared }}{\text{Number of Obs}} \left (\frac {T_k^2}{n_k}\right ) && 69.14 && 181.50 && 325.13 && 288.80 && 500.00 && = 1,364.57$
Using this information, we find that the sum of squares between groups is equal to the following:
$SS_B &= \sum^m_{k=1} \frac{T^2_k}{n_k}-\frac{T^2}{N}\\& \approx 1, 364.57 - \frac{(194)^2}{31} \approx 150.5$
Since there are four degrees of freedom for this calculation (the number of groups minus one), the mean squares between groups is as shown below:
$MS_B=\frac{SS_B}{m-1} \approx \frac{150.5}{4} \approx 37.6$
Next, we calculate the mean squares within groups, $MS_W$, which is also known as the pooled estimate of the population variance, $\sigma^2$.
To calculate the mean squares within groups, we first use the following formula to calculate $SS_W$:
$SS_W=\sum^m_{k=1} \sum^{n_k}_{i=1} x^2_{ik}-\sum^m_{k=1} \frac{T^2_k}{n_k}$
Using our summary statistics from above, we can calculate $SS_W$ as shown below:
$SS_W &= \sum^m_{k=1} \sum^{n_k}_{i=1} x^2_{ik}-\sum^m_{k=1} \frac{T^2_k}{n_k}\\& \approx 1, 452 - 1, 364.57\\& \approx 87.43$
This means that we have the following for $MS_W$:
$MS_W=\frac{SS_W}{n-m} \approx \frac{87.43}{26} \approx 3.36$
Therefore, our $F$-ratio is as shown below:
$F=\frac{MS_B}{MS_W} \approx \frac{37.6}{3.36} \approx 11.19$
We would then analyze this test statistic against our critical value. Using the $F$-distribution table and $\alpha=0.02$, we find our critical value equal to 4.140. Since our test statistic of 11.19 exceeds our critical value of 4.140, we reject the null hypothesis. Therefore, we can conclude that not all of the population means of the five programs are equal and that obtaining an $F$-ratio this extreme by chance is highly improbable.
On the Web
http://preview.tinyurl.com/36j4by6 $F$-distribution tables with $\alpha=0.02$.
Technology Note: Calculating a One-Way ANOVA with Excel
Here is the procedure for performing a one-way ANOVA in Excel using this set of data.
Copy and paste the table into an empty Excel worksheet.
Select 'Data Analysis' from the Tools menu and choose 'ANOVA: Single-factor' from the list that appears.
Place the cursor in the 'Input Range' field and select the entire table.
Place the cursor in the 'Output Range' field and click somewhere in a blank cell below the table.
Click 'Labels' only if you have also included the labels in the table. This will cause the names of the predictor variables to be displayed in the table.
Click 'OK', and the results shown below will be displayed.
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Column 1 7 22 3.142857 3.809524
Column 2 6 33 5.5 3,5
Column 3 8 51 6.375 2.839286
Column 4 5 38 7.6 4.3
Column 5 6 50 10 2.5
ANOVA
Source of Variation $SS$ $df$ $MS$ $F$ $P$-value $F$ crit
Between Groups 150.5033 4 37.62584 11.18893 2.05e-05 2.742594
Within Groups 87.43214 26 3.362775
Total 237.9355 30
Technology Note: One-Way ANOVA on the TI-83/84 Calculator
Enter raw data from population 1 into L1, population 2 into L2, population 3 into L3, population 4 into L4, and so on.
Now press [STAT], scroll right to TESTS, scroll down to 'ANOVA(', and press [ENTER]. Then enter the lists to produce a command such as 'ANOVA(L1, L2, L3, L4)' and press [ENTER].
## Lesson Summary
When testing multiple independent samples to determine if they come from the same population, we could conduct a series of separate $t$-tests in order to compare all possible pairs of means. However, a more precise and accurate analysis is the Analysis of Variance (ANOVA).
In ANOVA, we analyze the total variation of the scores, including the variation of the scores within the groups, the variation between the group means, and the total mean of all the groups (also known as the grand mean).
In this analysis, we calculate the $F$-ratio, which is the total mean of squares between groups divided by the total mean of squares within groups.
The total mean of squares within groups is also known as the pooled estimate of the population variance. We find this value by analysis of the standard deviations in each of the samples.
## Review Questions
1. What does the ANOVA acronym stand for?
2. If we are testing whether pairs of sample means differ by more than we would expect due to chance using multiple $t$-tests, the probability of making a type I error would ___.
3. In the ANOVA method, we use the ___ distribution.
1. Student’s $t$-
2. normal
3. $F$-
4. In the ANOVA method, we complete a series of steps to evaluate our hypothesis. Put the following steps in chronological order.
1. Calculate the mean squares between groups and the mean squares within groups.
2. Determine the critical values in the $F$-distribution.
3. Evaluate the hypothesis.
4. Calculate the test statistic.
5. State the null hypothesis.
5. A school psychologist is interested in whether or not teachers affect the anxiety scores among students taking the AP Statistics exam. The data below are the scores on a standardized anxiety test for students with three different teachers.
Teacher's Name and Anxiety Scores
Ms. Jones Mr. Smith Mrs. White
8 23 21
6 11 21
4 17 22
12 16 18
16 6 14
17 14 21
12 15 9
10 19 11
11 10
13
(a) State the null hypothesis.
(b) Using the data above, fill out the missing values in the table below.
Ms. Jones Mr. Smith Mrs. White Totals
Number $(n_k)$ 8 $=$
Total $(T_k)$ 131 $=$
Mean $(\bar{x})$ 14.6 $=$
Sum of Squared Obs. $(\sum^{n_k}_{i=1} x^2_{ik})$ $=$
Sum of Obs. Squared/Number of Obs. $(\frac{T^2_k}{n_k})$ $=$
(c) What is the value of the mean squares between groups, $MS_B$?
(d) What is the value of the mean squares within groups, $MS_W$?
(e) What is the $F$-ratio of these two values?
(f) With $\alpha=0.05$, use the $F$-distribution to set a critical value.
(g) What decision would you make regarding the null hypothesis? Why?
Feb 23, 2012
Dec 12, 2013 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 98, "texerror": 0, "math_score": 0.8643560409545898, "perplexity": 512.6535947074509}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00655-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://www.gnu.org/software/gsl/manual/html_node/The-Negative-Binomial-Distribution.html | Next: , Previous: The Multinomial Distribution, Up: Random Number Distributions [Index]
### 20.34 The Negative Binomial Distribution
Function: unsigned int gsl_ran_negative_binomial (const gsl_rng * r, double p, double n)
This function returns a random integer from the negative binomial distribution, the number of failures occurring before n successes in independent trials with probability p of success. The probability distribution for negative binomial variates is,
p(k) = {\Gamma(n + k) \over \Gamma(k+1) \Gamma(n) } p^n (1-p)^k
Note that n is not required to be an integer.
Function: double gsl_ran_negative_binomial_pdf (unsigned int k, double p, double n)
This function computes the probability p(k) of obtaining k from a negative binomial distribution with parameters p and n, using the formula given above.
Function: double gsl_cdf_negative_binomial_P (unsigned int k, double p, double n)
Function: double gsl_cdf_negative_binomial_Q (unsigned int k, double p, double n)
These functions compute the cumulative distribution functions P(k), Q(k) for the negative binomial distribution with parameters p and n. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9071366190910339, "perplexity": 3182.0529642235315}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719079.39/warc/CC-MAIN-20161020183839-00300-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/118057/relation-between-singular-values-of-matrices-and-their-products | # Relation between singular values of matrices and their products
Hello everybody, Is there any explicit relation between the singular values $\lambda_X$ and $\lambda_Y$ of two same size matrices $X$ and $Y$, respectively, and the singular values $\lambda_{XY^t}$ of the matrix $XY^t$? Otherwise said, is there a function $f$ such that $\lambda_{XY^t}=f(\lambda_X , \lambda_Y)$?
-
For example, let $X$ be the diagonal matrix with diagonal entries 1 and 2. Let $Y_1=X$, and let $Y_2$ be the diagonal matrix with diagonal entries 2 and 1. Then $XY_1$ and $XY_2$ have different singular values.
Even though there is no functional relation between these singular values, there is a set of inequalities, called Horn inequalities, which completely describes possible singular values for $XY^t$ if the singular values of $X$ and $Y$ are fixed, see e.g. Fulton's survey ams.org/journals/bull/2000-37-03/S0273-0979-00-00865-X (the result is a corollary of Thompson's conjecture solved by Klyachko and work of Klyachko, Knutson and Tao on the eigenvalue problem). – Misha Jan 4 '13 at 19:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9917713403701782, "perplexity": 96.71673776754443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042989826.86/warc/CC-MAIN-20150728002309-00035-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/146812/minimal-number-of-generators-of-subgroups-of-noetherian-groups | Minimal number of generators of subgroups of Noetherian groups
A group $G$ is Noetherian (or slender) if all its subgroups are finitely generated. Does this imply that the minimal number of generators of subgroups of $G$ is bounded above?
For example, if $G$ is a polycyclic group that admits a polycyclic series of length $n$ then every subgroup of $G$ can be generated by $n$ (or fewer) elements. This idea also applies for virtually-polycyclic groups.
It is unknown whether all finitely presented Noetherian groups are virtually-polycyclic. On the other hand, there are finitely generated Noetherian groups that are not virtually-polycyclic, for example the Tarski monster. However, all proper subgroups of the Tarski monster are cyclic and hence there is a bound on the minimal number of generators of its subgroups.
(See this post for a related question.)
Edit: What if we restrict ourselves to finitely presented Noetherian groups?
-
A quasi-finite group is a group all of whose proper subgroups are finite. Deryabina-Olshanskii (Russian Math. Surveys 41 (1986), no. 6, 169–170) showed that a finite group is subgroup of a f.g. quasi-finite group iff it's the direct product of a group of odd order and an abelian 2-group. It's possible that the construction (or another one) provides a single quasi-finite group containing a copy of $C^n$ for every $n$, for $C$ a given cyclic group of prime order (this would answer your question negatively), but I don't have access to the paper (iopscience.iop.org/0036-0279/41/6/L10) – YCor Nov 3 '13 at 15:22
Your question could possibly be true if restricted to finitely presented slender groups, since I think the only known examples are virtually polycyclic. – Ian Agol Nov 4 '13 at 5:05
Yes, it is either true or open. I will add the question to the original post. – Sebastian Nov 4 '13 at 8:54
@Ian, indeed, "Are all finitely presented Noetherian groups virtually polycyclic? is (an open) Question 11.38 (due to S.V.Ivanov, 1990) from the Kourovka Notebook. – Anton Klyachko Nov 5 '13 at 18:59
For any countable family of countable involution-free groups $G_1,G_2,\dots$, there is a 2-generated group $H$ containing all $G_i$ as proper subgroups such that each proper subgroup of $H$ is either cyclic or a conjugate of a subgroup of some $G_i$.
This is Obraztsov's embedding theorem. Clearly, it gives a desired example if we put, e.g, $$\{G_i\}=\{\hbox{all finite groups of odd orders}\}.$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9241747260093689, "perplexity": 336.11094511640584}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122030742.53/warc/CC-MAIN-20150124175350-00176-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/642187/field-containing-sum-of-square-roots-also-contains-individual-square-roots | # Field containing sum of square roots also contains individual square roots
Let $F$ be a field of characteristic $\neq 2$. Let $a\neq b$ be in $F$. Suppose $\sqrt{a}+\sqrt{b}\in F$. Prove that $\sqrt{a}\in F$.
We have $(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}\in F$, so $\sqrt{ab}\in F$. Then $\sqrt{ab}(\sqrt{a}+\sqrt{b})=a\sqrt{b}+b\sqrt{a}\in F$. What then?
-
Alternatively, write $(\sqrt{a}+\sqrt{b})^{-1}=(\sqrt{a}-\sqrt{b})/(a-b)$ which implies $\sqrt{a}-\sqrt{b}\in F$, then notice $\sqrt{a}$ is the average of $\sqrt{a}+\sqrt{b}$ and $\sqrt{a}-\sqrt{b}$. – anon Jan 18 '14 at 0:34
Subtract $a(\sqrt{b}+\sqrt{a})$ for the next step.
-
Oh that's fast, thanks! – JJ Beck Jan 18 '14 at 0:31
Hint $\$ If a field $F$ has two $F$-linear independent combinations of $\ \sqrt{a},\ \sqrt{b}\$ then you can solve for $\ \sqrt{a},\ \sqrt{b}\$ in $F.\,$ For example, the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,\,$ into the finitely many fields between F and $\ F(\sqrt{a}, \sqrt{b}),$ e.g. see PlanetMath's proof.
In the OP, note that $F$ contains the independent $\ \sqrt{a} - \sqrt{b}\$ since
$$\sqrt{a}\ - \sqrt{b}\ =\ \dfrac{\,\ a\ -\ b}{\sqrt{a}+\sqrt{b}}\ \in\ F$$
To be explicit, notice that $\ u = \sqrt{a}+\sqrt{b},\ \ v = \sqrt{a}-\sqrt{b}\in F\$ so solving the linear system for the roots yields $\ \sqrt{a}\ =\ (u+v)/\color{#c00}2,\ \ \sqrt{b}\ =\ (v-u)/\color{#c00}2,\$ both of which are clearly $\,\in F,\,$ since $\,u,\,v\in F\,$ and $\,\color{#c00}2\ne 0\,$ in $\,F,\,$ so $\,1/\color{#c00}2\,\in F.\,$ This works over any field where $\,2\ne 0\,,\,$ i.e. where the determinant (here $2$) of the linear system is invertible, i.e. where the linear combinations $\,u,v\,$ of the square-roots are linearly independent over the base field.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9937400817871094, "perplexity": 120.38627637359255}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398457799.57/warc/CC-MAIN-20151124205417-00152-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://jamesmccammon.com/2017/04/06/distribution-convergence/ | # Distribution Convergence
Let’s do a problem from Chapter 5 of All of Statistics.
Suppose $X_1, \dots X_n \sim \text{Uniform(0,1)}$. Let $Y_n = \bar{X_n}^2$. Find the limiting distribution of $Y_n$.
Note that we have $Y_n = \bar{X_n}\bar{X_n}$
Recall from Theorem 5.5(e) that if $X_n \rightsquigarrow X$ and $Y_n \rightsquigarrow c$ then $X_n Y_n \rightsquigarrow cX$.
So the question becomes does $X_n \rightsquigarrow c$ so that we can use this theorem? The answer is yes. Recall that from Theorem 5.4(b) $X_n \overset{P}{\longrightarrow} X$ implies that $X_n \rightsquigarrow X$. So if we can show that we converge to a constant in probability we know that we converge to the constant in distribution. Let’s show that $\bar{X}_n \overset{P}{\longrightarrow} c$. That’s easy. The law of large numbers tells us that the sample average converges in probability to the expectation. In other words $\bar{X}_n \overset{P}{\longrightarrow} \mathbb{E}[X]$. Since we are told that $X_i$ is i.i.d from a Uniform(0,1) we know the expectation is $\mathbb{E}[X] = .5$.
Putting it all together we have that:
$Y_n = \bar{X_n}^2$
$Y_n = \bar{X_n}\bar{X_n}$
$Y_n \rightsquigarrow \mathbb{E}[X]\mathbb{E}[X]$ (through the argument above)
$Y_n \rightsquigarrow (.5)(.5)$
$Y_n \rightsquigarrow .25$
We can also show this by simulation in R, which produces this chart:
Indeed we also get the answer 0.25. Here is the R code used to produce the chart above:
# Load plotting libraries
library(ggplot2)
library(ggthemes)
# Create Y = g(x_n)
g = function(n) {
return(mean(runif(n))^2)
}
# Define variables
n = 1:10000
Y = sapply(n, g)
# Plot
set.seed(10)
df = data.frame(n,Y)
ggplot(df, aes(n,Y)) +
geom_line(color='#3498DB') +
theme_fivethirtyeight() +
ggtitle('Distribution Convergence of Y as n Increases') | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 19, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9936442375183105, "perplexity": 485.3901113803041}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203755.18/warc/CC-MAIN-20190325051359-20190325073359-00067.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/int1/chapter/11/lesson/11.1.3/problem/11-37 | ### Home > INT1 > Chapter 11 > Lesson 11.1.3 > Problem11-37
11-37.
Solve this system of equations:
$y=\frac{2}{3}x-4$$2x-3y=10$
Try using the Substitution Method.
No solution.
1. What does your solution tell you about the relationship between the lines?
When a solution exists it represents the point where the lines intersect.
2. Solve the second equation for $y$.
$y=-\frac{10}{3}+\frac{2}{3}x \text{ or }y=\frac{2}{3}x-\frac{10}{3}$
3. Does the slope of each line confirm your statement in part (a)? Explain how.
What can you say about lines that have the same slopes? | {"extraction_info": {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8457004427909851, "perplexity": 2785.961257501893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402118004.92/warc/CC-MAIN-20200930044533-20200930074533-00566.warc.gz"} |
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35 matrices × 39 4 prime-numbers × 2 28 linear-algebra × 30 4 number-theory 8 eigenvalues-eigenvectors × 6 4 fibonacci-numbers 5 real-analysis × 3 3 inverse × 3 4 svd × 3 3 determinant × 2
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Stack Overflow 1,309 rep 2312 Mathematics 1,298 rep 1717 Android Enthusiasts 111 rep 3 Meta Stack Exchange 101 rep 2 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8488187789916992, "perplexity": 1810.2454785542543}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583513844.48/warc/CC-MAIN-20181021094247-20181021115747-00354.warc.gz"} |
http://dspace.rri.res.in/handle/2289/5852 | Please use this identifier to cite or link to this item: http://hdl.handle.net/2289/5852
Title: Environments of extended radio sources in the Australia telescope low-brightness survey. Authors: Thorat, K.Saripalli, LakshmiSubrahmanyan, Ravi Keywords: galaxies: evolutiongalaxies: photometryradio continuum Issue Date: 1-Oct-2013 Publisher: Oxford University Press Citation: Monthly Notices of the Royal Astronomical Society, 2013, Vol. 434, p2877-91 Abstract: We present a study of the environments of extended radio sources in the Australia Telescope Low Brightness Survey (ATLBS). The radio sources were selected from the Extended Source Sample (ATLBS-ESS), which is a well defined sample containing the most extended of radio sources in the ATLBS sky survey regions. The environments were analyzed using 4-m CTIO Blanco telescope observations carried out for ATLBS fields in the SDSS ${\rm r}^{\prime}$ band. We have estimated the properties of the environments using smoothed density maps derived from galaxy catalogs constructed using these optical imaging data. The angular distribution of galaxy density relative to the axes of the radio sources has been quantified by defining anisotropy parameters that are estimated using a new method presented here. Examining the anisotropy parameters for a sub-sample of extended double radio sources that includes all sources with pronounced asymmetry in lobe extents, we find good evidence for environmental anisotropy being the dominant cause for lobe asymmetry in that higher galaxy density occurs almost always on the side of the shorter lobe, and this validates the usefulness of the method proposed and adopted here. The environmental anisotropy parameters have been used to examine and compare the environments of FRI and FRII radio sources in two redshift regimes ($z<0.5$ and $z>0.5$). Wide-angle tail sources and Head-tail sources lie in the most overdense environments. The Head-tail source environments (for the HT sources in our sample) display dipolar anisotropy in that higher galaxy density appears to lie in the direction of the tails. Excluding the Head-tail and Wide-angle tail sources, subsamples of FRI and FRII sources from the ATLBS survey appear to lie in similar moderately overdense environments, with no evidence for redshift evolution in the regimes studied herein. Description: Restricted Access. An open-access version is available at arXiv.org (one of the alternative locations) URI: http://hdl.handle.net/2289/5852 ISSN: 1365-2966 - (online)0035-8711 Alternative Location: http://adsabs.harvard.edu/abs/2013arXiv1308.0884Thttp://arxiv.org/abs/1308.0884http://dx.doi.org/ 10.1093/mnras/stt1196 Copyright: 2013 The authors & the Royal Astronomical Society. Appears in Collections: Research Papers (A&A)
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2013_MNRAS_434_2877-91.pdf
Restricted Access | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9523423314094543, "perplexity": 4172.65274447727}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039743248.7/warc/CC-MAIN-20181117020225-20181117042225-00275.warc.gz"} |
http://mathhelpforum.com/trigonometry/68721-trig-2-a.html | # Math Help - trig 2
1. ## trig 2
1. Determine sin θ to 3 decimal places if cos θ = 1/2 and θ is an angle in quadrant IV
2. Originally Posted by william
1. Determine sin θ to 3 decimal places if cos θ = 1/2 and θ is an angle in quadrant IV
Hi William,
$\cos \theta =\frac{1}{2}$ in the 4th quadrant should corresponding to a special angel. What is it?
If you can find that angle, you should be able to figure out $\sin \theta$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9952711462974548, "perplexity": 786.315934118288}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931007301.29/warc/CC-MAIN-20141125155647-00120-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/electric-field-problem.154236/ | # Homework Help: Electric field problem
1. Feb 2, 2007
### kidxsai
Would someone help me solving this problem?
A charge of -3.26 C is located at the origin, and a charge of -5.36 C is located along the y axis at 1.73664 m.
K is given. At what point along the y axis is the electric field zero?
2. Relevant equations
3. The attempt at a solution
2. Feb 2, 2007
### dextercioby
C'mon, the problem is simple. Can you add 2 vectors ? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9423210024833679, "perplexity": 1155.423214043316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676593208.44/warc/CC-MAIN-20180722100513-20180722120513-00456.warc.gz"} |
http://www.koreascience.or.kr/article/ArticleFullRecord.jsp?cn=DBSHBB_2006_v43n2_271 | ON THE COMMUTATIVE PRODUCT OF DISTRIBUTIONS
Title & Authors
ON THE COMMUTATIVE PRODUCT OF DISTRIBUTIONS
Fisher Bring; Tas Kenan;
Abstract
The commutative products of the distributions $\small{x^r\;ln^p\;|x|\;and\;x^{-r-1}ln^q\;|x|}$ and of sgn x $\small{x^r\;ln^p\;|x|\;and\;sgn\;x\;x^{-r-1}ln^q\;|x|}$ are evaluated for $\small{r=0,\;\pm1,\;\pm2,...\;and\;p,\;q=0,1,2,....}$
Keywords
distribution;delta-function;product of distributions;
Language
English
Cited by
1.
THE PRODUCT OF ANALYTIC FUNCTIONALS IN Z',;;;;
대한수학회지, 2008. vol.45. 2, pp.455-466
1.
Several results on the commutative neutrix product of distributions, Integral Transforms and Special Functions, 2007, 18, 8, 559
References
1.
J. -F. Colombeau, New generalized functions and multiplication of distributions, North-Holland Publishing Co., 1984, xii+375
2.
B. Fisher, The product of distributions, Quart. J. Math. Oxford Ser. 22 (1971), no. 2, 291-298
3.
B. Fishe, The product of the distributions $x_+^{-r-\frac{1}{2}}$ and $x_-^{-r-\frac{1}{2}$, Proc. Cambridge Philos. Soc. 71 (1971), 123-130
4.
B. Fishe, The product of the distributions $x^{-r}$ and $\mu^{(r-1)}$(x), Proc. Cambridge Philos. Soc. 72 (1972), 201-204
5.
B. Fishe, Some results on the product of distributions, Proc. Cambridge Philos. Soc. 73 (1973), 317-325
6.
I. M. Gel'fand and G. E. Shilov, Generalized Functions, Vol. I, Academic Press, 1964
7.
L. Schwartz, Theorie des Distributions, Tome I, Actualites Sci. Ind., no. 1091, Hermann & Cie., Paris 1950, 148 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 3, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.876899242401123, "perplexity": 2316.845304587166}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549423629.36/warc/CC-MAIN-20170721002112-20170721022112-00119.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=Holomorphic_function&diff=prev&oldid=7907 | Difference between revisions of "Holomorphic function"
A holomorphic function is a differentiable complex function. That is, just as in the real case, is holomorphic at if exists. This is much stronger than in the real case since we must allow to approach zero from any direction in the complex plane.
Cauchy-Riemann Equations
Let us break into its real and imaginary components by writing , where and are real functions. Then it turns out that is holomorphic at iff and have continuous partial derivatives and the following equations hold:
These equations are known as the Cauchy-Riemann Equations.
Analytic Functions
A related notion to that of homolorphicity is that of analyticity. A function is said to be analytic at if has a convergent power series expansion on some neighborhood of . Amazingly, it turns out that a function is holomorphic at if and only if it is analytic at . | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9792947769165039, "perplexity": 222.04781675749746}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178351374.10/warc/CC-MAIN-20210225153633-20210225183633-00116.warc.gz"} |
https://zbmath.org/?q=an:611.14020 | ×
zbMATH — the first resource for mathematics
Values of zeta functions of varieties over finite fields. (English) Zbl 0611.14020
Am. J. Math. 108, 297-360 (1986); addendum ibid. 137, No. 6, 1703-1712 (2015).
Let X be a smooth projective variety of dimension d over a finite field k of q elements. The zeta function $$\zeta$$ (X,s) of X satisfies $$\zeta (X,s)\sim C_ X(r)/(1-q^{r-s})^{\rho_ r}$$ as $$s\to r$$, for some integer $$\rho_ r$$ and rational number $$C_ X(r)$$. Let $$\bar k$$ be the algebraic closure of k, let $$\gamma$$ be the canonical generator of $$\Gamma =Gal(\bar k/k)$$ and let $$\bar X=X\otimes_ k\bar k$$. It has been conjectured SS(X,r,$$\ell):$$ The minimal polynomial of $$\gamma$$ acting on $$H^{2r}(\bar X,{\mathbb{Q}}_{\ell}(r))$$ does not have 1 as multiple root. Note that the case $$\ell =p$$, the characteristic of k, it not excluded. A rational number $$\chi$$ (X,$${\hat {\mathbb{Z}}}(r))$$, related to the cohomology groups $$H^ i(X,{\hat {\mathbb{Z}}}(r))=\prod_{\ell}H^ i(X,{\mathbb{Z}}_{\ell}(r))\quad can$$ be defined if and only if conjecture SS(X,r,$$\ell)$$ holds for all $$\ell$$, in which case the author proves that $\zeta (X,s)\sim \pm \chi (X,{\hat {\mathbb{Z}}}(r))q^{\chi (X,{\mathcal O}_ X,r)}(1-q^{r-s})^{-\rho_ r}\quad as\quad r\to s.$ The groups $$H^ i(X,{\mathbb{Z}}_ p(r))$$ play a role similar to the étale cohomology groups $$H^ i(X,{\mathbb{Z}}_{\ell}(r))$$ for $$\ell \neq p$$. They are closely related to the crystalline cohomology groups, and their behaviour is studied throughout the paper. There are cycle maps $$c^ r_{\ell}: CH^ r(X)\to H^{2r}(\bar X,{\mathbb{Q}}_{\ell}(r)),$$ compatible with intersection and cup products, defined for all $$\ell$$. Let $$B^ r(X)=c^ r_{\ell}(CH^ r(X)){\mathbb{Q}}_{\ell}$$; it is conjectured that the dimension of $$B^ r_{\ell}(X)$$ is the multiplicity of $$q^ r$$ as an inverse root of $$P_{2r}(X,t)=\det (1-\gamma t)| H^{2r}(\bar X,{\mathbb{Q}}_{\ell}))$$. This statement is equivalent to SS(X,r,$$\ell)$$, together with Tate’s conjecture: $$B^ r_{\ell}(X)=H^{2r}(\bar X,{\mathbb{Q}}_{\ell}(r))^{\Gamma}$$.
Reviewer: P.Bayer
MSC:
14G10 Zeta functions and related questions in algebraic geometry (e.g., Birch-Swinnerton-Dyer conjecture) 14C15 (Equivariant) Chow groups and rings; motives 14F30 $$p$$-adic cohomology, crystalline cohomology 14G99 Arithmetic problems in algebraic geometry; Diophantine geometry 14G15 Finite ground fields in algebraic geometry
Full Text: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9149851202964783, "perplexity": 238.4798306874545}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585441.99/warc/CC-MAIN-20211021195527-20211021225527-00680.warc.gz"} |
https://sciencenotes.org/waves-word-search/ | # Waves Word Search
This waves word search contains glossary terms associated with the study of waves and wave motion.
Find the words from the list forward and backward in vertical, horizontal, and diagonal directions.
The puzzle is optimized to fit on a standard 8.5″x11″ sheet of paper, but it fits nicely on A4 paper. Use the provided PDF for best results.
If you need a little help finding one or all of the glossary terms, a solved version of the puzzle is available.
Word searches are a fun activity. Check out our other science-related word search puzzles.
### Waves Word Search Glossary
Quick definitions of the terms used in the Waves Word Search.
Acoustics: The study of sound waves.
Amplitude: The measure of the displacement of the wave from its rest position.
Coherent: Two or more waves with the same frequency and phase difference.
Crest: Maximum value of a wave’s amplitude.
Diffraction: Behavior of a wave after encountering a gap or obstacle.
Electromagnetic: Self-propagating waves composed of oscillating electric and magnetic waves. Electromagnetic waves do not require a medium to propagate.
Frequency: Frequency is the measure of how many complete waves occur in a specified period of time. The most common measurement of frequency is cycles (or waves) per second.
Hertz: The SI unit of frequency. 1 Hertz (Hz) = 1 cycle/second.
Intensity: The measurement of the amount of wave energy over time (power) per unit area.
Interference: Resulting wave when one wave meets another wave.
Longitudinal: Longitudinal waves are waves where the disturbance of the wave moves in the same direction as the wave. Sound waves are longitudinal waves.
Mechanical: Mechanical waves are waves which require a medium to exist. Mechanical waves cannot travel in a vacuum.
Medium: Matter in which waves move through.
Optics: Study of the behavior of light and vision.
Period: The amount of time necessary to complete one complete wave. The period is the inverse of frequency.
Phase: Phase is an instantaneous point in time of a wave cycle. Phase also refers to the relative difference in corresponding points of between two waves with the same frequency. This definition usually refers to the phase difference between two waves
Polarization: Polarization refers to the direction a wave oscillates.
Propagation: The method waves travel through space.
Pulse: A pulse is a non-periodic wave with at least one major crest.
Reflection: The change in direction of a wave when encountering a barrier without leaving the medium.
Refraction: The change in direction of a wave caused by the wave passing from one medium to another.
Resonance: The increase in amplitude caused when one wave oscillates at a frequency very close to the same frequency of another wave or vibrating substance.
Sinusoidal: The name of the mathematical curve describing a smooth periodic oscillation.
Speed: The speed at which a wave travels. The speed of a wave is calculated by the wavelength multiplied by the frequency of the wave.
Standing: A standing wave is a wave where the position of the wave does not change.
Surface: A surface wave is a wave which behaves like both a transverse wave and a longitudinal wave.
Transverse: A transverse wave is a wave where the vibration motion is perpendicular to the wave’s direction of motion.
Trough: The lowest point of a wave’s displacement.
Wavelength: The distance from one point of a wave to the same point in the next wave.
Wavenumber: The number of wavelengths per unit distance.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8974383473396301, "perplexity": 712.3489910109818}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575844.94/warc/CC-MAIN-20190923002147-20190923024147-00291.warc.gz"} |
https://stats.stackexchange.com/tags/density-function/info | Tag Info
Probability density function (PDF) of a continuous random variable gives the relative probability for each of its possible values. Use this tag for discrete probability mass functions (PMFs) too.
Tag Usage
Use this tag when asking about probability functions in general, whether Probability density functions, or discrete probability mass functions (PMFs).
Overview
PDF stands for Probability Density Function; distinguished from CDF for Cumulative Distribution Function. A PDF describes the relative probability of a continuous random variable taking a given value. PMF stands for Probability Mass Function; it describes the probability of a discrete random variable taking a given value.
In case of continuous variables $X$, the PDF $\mathcal{P}_X(x)$ can be integrated over an interval $\mathcal{I}$ (or, more generally, any Borel set) to find the probability that the variable is in that interval:
$$\Pr(X \in \mathcal{I}) = \int_\mathcal{I} \mathcal{P}_X(x) dx.$$
Some common PDFs:
• Normal: $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]$
• Gamma: $f(x) = \frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}\exp(-\beta x)$
Some common PMFs:
• Binomial: $\Pr(X = x) = \frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$ for integral $n\ge 0.$
• Bernoulli: $\Pr(X = x) = p^x (1-p)^{1-x}$ for $x\in \{ 0,1 \}.$
References | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9800763130187988, "perplexity": 583.67678777854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178351374.10/warc/CC-MAIN-20210225153633-20210225183633-00282.warc.gz"} |
http://t-redactyl.io/blog/2020/06/working-with-matrices-inversion.html | # Working with matrices: inversion
written in
Today we will continue our discussion of the basic operations you can do with matrices in linear algebra. In the last post we covered addition, subtraction, scalar multiplication and matrix multiplication. This week, we’ll cover matrix inversion.
So far we’ve seen matrix addition, subtraction and multiplication, but what about matrix division? Well, technically there is no such thing! The closest thing we have is multiplying a matrix by its inverse. Before we get into what an inverse is, we need to pause to describe a special matrix called the identity matrix.
### Identity matrix
The identity matrix is simply a square matrix which has 0 everywhere except along the diagonal, where it instead contains 1’s. For example, a $$2 \times 2$$ identity matrix would be:
$$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
The identity matrix can be thought of as the equivalent of 1 for matrices, in the sense that if you multiply any number by 1, the number will remain unchanged. Similarly, if you multiply any matrix by its identity matrix, that matrix will remain unchanged. Let’s see this with a new matrix $$A$$:
$$A = \begin{bmatrix} 4 & 3 \\ 6 & 5 \end{bmatrix}$$
If we multiply $$A$$ by $$I$$, we get:
\begin{aligned} AI &= \begin{bmatrix} 4 & 3 \\ 6 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\ &= \begin{bmatrix} (4 \times 1 + 3 \times 0) & (4 \times 0 + 3 \times 1) \\ (6 \times 1 + 5 \times 0) & (6 \times 0 + 5 \times 1) \end{bmatrix}\\ &= \begin{bmatrix} 4 & 3 \\ 6 & 5 \end{bmatrix} \end{aligned}
In order to do this in Numpy, we can generate an identity matrix using the identity function, passing the number of rows (and columns) as the argument.
import numpy as np
A = np.array([[4, 3], [6, 5]])
I = np.identity(2)
I
array([[1., 0.],
[0., 1.]])
AI = A.dot(I)
AI
array([[4., 3.],
[6., 5.]])
Unlike most matrix multiplications, we get the same result if we multiply $$AI$$ or $$IA$$, as both results just simplify to $$A$$: $$AI = IA = A$$. This means that matrix multiplication by the identity is commutative.
IA = I.dot(A)
IA
array([[4., 3.],
[6., 5.]])
### Inverse of a matrix
The inverse of a matrix is another matrix by which you multiply that matrix to get the identity matrix. This is comparable to the inverse with numbers, where the inverse of a number is what you multiply that number by to get 1. You may have noticed that I referred to the inverse of a matrix. This is because every matrix that can be inverted has only one unique inverse. The inverse of the matrix is indicated by writing the name of the original matrix with a little $$-1$$ superscript (which is exactly the same notation as for the multiplicative inverse for numbers). For example, the matrix $$Z$$ would have the inverse $$Z^{-1}$$.
How do we find the inverse of a matrix? For $$2 \times 2$$ matrices it is straightforward. In order to do so, we transform our original matrix in the following way:
$$M^{-1} = \frac{1}{m_{11}m_{22} - m_{12}m_{21}} \begin{bmatrix}m_{22} & -m_{12}\\-m_{21} & m_{11}\end{bmatrix}$$
This looks a little complicated, but let’s break it down. Have a look at the matrix component of this equation. You can see that we have swapped the positions of the first and last elements in the matrix, so that the element that was in $$m_{11}$$ is now in $$m_{22}$$, and vice versa. $$m_{12}$$ and $$m_{21}$$ stay in their original spots, but we multiply them by $$-1$$. We then calculate something called the determinant, which is basically a number we divide this reorganised matrix by to get the inverse. This is calculated by the formula $$m_{11}m_{22} - m_{12}m_{21}$$, and if we multiply our reorganised matrix by $$1$$ divided by this number, we get the inverse.
Let’s try this with a concrete example, using our matrix $$A$$. In order to get the inverse of $$A$$, or $$A^{-1}$$, we do the following:
\begin{aligned} A^{-1} &= \frac{1}{4 \times 5 - 3 \times 6} \begin{bmatrix}5 & -3 \\ -6 & 4 \end{bmatrix}\\ &= \frac{1}{2} \begin{bmatrix} 5 & -3 \\ -6 & 4 \end{bmatrix}\\ &= \begin{bmatrix}(\frac{1}{2} \times 5) & (\frac{1}{2} \times -3) \\[4pt] (\frac{1}{2} \times -6) & (\frac{1}{2} \times 4) \end{bmatrix}\\ &= \begin{bmatrix} \frac{5}{2} & -\frac{3}{2} \\ -3 & 2 \end{bmatrix} \end{aligned}
And voilà, we have our matrix inversion! Let’s just double check it worked by multiplying $$A$$ by it:
\begin{aligned} AA^{-1} &= \begin{bmatrix}(4 \times \frac{5}{2} + 3 \times -3) & (4\times-\frac{3}{2} + 3\times2)\\[4pt] (6\times\frac{5}{2} + -3 \times 5) & (6\times-\frac{3}{2} + 5\times2)\end{bmatrix}\\ &= \begin{bmatrix}(10-9) & (-6 + 6)\15-15) & (-9 + 10)\end{bmatrix}\\ &= \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \end{aligned} Unfortunately, finding the inverse of matrices that are bigger than \(2 \times 2 is less straightforward. There are a few methods to do it, but they are beyond the scope of this blog post. A couple of nice (and gentle) introductions to these can be found here and here. In order to calculate the inverse of a matrix in Numpy, we can use the inv function from the linalg subpackage. from numpy.linalg import inv Ainv = inv(A) Ainv array([[ 2.5, -1.5], [-3. , 2. ]]) Let’s check it is the correct inverse by multiplying it by $$A$$: np.round(A.dot(Ainv)) array([[1., 0.], [0., 1.]]) Like with multiplication by the identity matrix, multiplication by the inverse is also commutative. This is because, as we just saw above, multiplying $$A$$ by its inverse simplifies to $$I$$, so $$AA^{-1} = A^{-1}A = I$$. Let’s confirm this is true using Numpy: np.round(Ainv.dot(A)) array([[1., 0.], [0., 1.]]) You’ve probably guessed from what I said earlier that not all matrices are invertible. Firstly, only square matrices are invertible. In addition, if any of the columns are a combination of any of the other columns in the matrix, then you cannot invert the matrix (in other words, when any of the columns are a linear combination of any of the other columns). Let’s see why this is the case. Let’s say we have a matrix $$B$$, where the second column is a multiple of the first column: B = \begin{bmatrix}1 & 2\\3 & 6\end{bmatrix} $$Let’s now try to calculate the inverse, $$B^{-1}$$:$$ \begin{aligned} B^{-1} &= \frac{1}{(1 \times 6) - (2 \times 3)}\begin{bmatrix}6 & -2\\-3 & 1\end{bmatrix}\\ &= \frac{1}{6 - 6}\begin{bmatrix}6 & -2\\-3 & 1\end{bmatrix}\\ &= \frac{1}{0}\begin{bmatrix}6 & -2\\-3 & 1\end{bmatrix} \end{aligned}
But wait! Our determinant is $$\frac{1}{0}$$, which is undefined. This means we can’t go any further in solving this equation, making our matrix uninvertible. Such non-invertible matrices are called “singular” or “degenerate” matrices.
Let’s see what happens when we try to pass our matrix $$B$$ to Numpy’s inv function:
B = np.array([[1, 2], [3, 6]])
inv(B)
---------------------------------------------------------------------------
LinAlgError Traceback (most recent call last)
<ipython-input-18-ae75e56b5e72> in <module>
1 B = np.array([[1, 2], [3, 6]])
----> 2 inv(B)
<__array_function__ internals> in inv(*args, **kwargs)
~/Documents/Blog-posts/.maths_venv/lib/python3.7/site-packages/numpy/linalg/linalg.py in inv(a)
545 signature = 'D->D' if isComplexType(t) else 'd->d'
546 extobj = get_linalg_error_extobj(_raise_linalgerror_singular)
--> 547 ainv = _umath_linalg.inv(a, signature=signature, extobj=extobj)
548 return wrap(ainv.astype(result_t, copy=False))
549
~/Documents/Blog-posts/.maths_venv/lib/python3.7/site-packages/numpy/linalg/linalg.py in _raise_linalgerror_singular(err, flag)
95
96 def _raise_linalgerror_singular(err, flag):
---> 97 raise LinAlgError("Singular matrix")
98
99 def _raise_linalgerror_nonposdef(err, flag):
LinAlgError: Singular matrix
Numpy tells us clearly that this matrix is non-invertible, exactly for the reasons we talked about above.
That caps off our discussion of matrix inversion. In the next post, we’ll finish up this series on matrix operations by talking about powers of a matrix and matrix transposition. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9996960163116455, "perplexity": 348.29284716633725}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657146845.98/warc/CC-MAIN-20200713194203-20200713224203-00048.warc.gz"} |
https://www.auxadi.com/news/labour-directorate-states-which-support-and-accountability-bonds-include-the-basis-for-calculating-overtime/ | The Labor Directorate (DT) responded consistently in the appropriate inclusion in the basis for calculating overtime, attendance bonuses and responsibility.
Pay or profit to be described as “salary” must meet the following copulative conditions:
• That it is a fixed stipend.
• To be paid in cash.
• To be paid in equal periods determined in the contract.
• To respond to the provision of services.
Corresponds therefore examine whether the bonds in consultation must be considered as “salary” and therefore whether they should be part of the basis for calculating overtime.
Repeated and consistent jurisprudence of this Directorate has ruled in the sense that this requirement does not mean that the parties have agreed a sum of only money, specified and determined, but enough has been agreed the liquidation and set the amount profit, which gives the character of “fixity” to it. Thus “the Labour Directorate has left established that the fact that a benefit can produce results that are not consistent from one month to another, does not alter the nature of it to make it a stipend of a variable nature, because the element of fixity which gives a benefit in the nature of salary, it is represented by the real possibility of perceiving monthly, and also because the amount and form of payment are preset “. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8336155414581299, "perplexity": 1336.6600797657202}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572063.65/warc/CC-MAIN-20220814173832-20220814203832-00550.warc.gz"} |
https://ccrma.stanford.edu/~jos/sasp/Panning_Problem.html | Next | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search
## The Panning Problem
An interesting illustration of the difference between coherent and noncoherent signal addition comes up in the problem of stereo panning between two loudspeakers. Let and denote the signals going to the left and right loudspeakers, respectively, and let and denote their respective gain factors (the panning'' gains, between 0 and 1). When , sound comes only from the left speaker, and when , sound comes only from the right speaker. These are the easy cases. The harder question is what should the gains be for a sound directly in front? It turns out that the answer depends upon the listening geometry and the signal frequency content.
If the listener is sitting exactly between the speakers, the ideal front image'' channel gains are , provided that the shortest wavelength in the signal is much larger than the ear-to-ear separation of the listener. This restriction is necessary because only those frequencies (below a few kHz, say), will combine coherently from both speakers at each ear. At higher frequencies, the signals from the two speakers decorrelate at each ear because the propagation path lengths differs significantly in units of wavelengths. (In addition, head shadowing'' becomes a factor at frequencies this high.) In the perfectly uncorrelated case (e.g., independent white noise coming from each speaker), the energy-preserving gains are . (This value is typically used in practice since the listener may be anywhere in relation to the speakers.)
To summarize, in ordinary stereo panning, decorrelated high frequencies are attenuated by about 3dB, on average, when using gains dB. At any particular high frequency, the actual gain at each ear can be anywhere between 0 and 1, but on average, they combine on a power basis to provide a 3 dB boost on top of the dB cut, leaving an overall dB change in the level at high frequencies.
Next | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search
[How to cite this work] [Order a printed hardcopy] [Comment on this page via email] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8816288113594055, "perplexity": 2520.3302845266594}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934804881.4/warc/CC-MAIN-20171118113721-20171118133721-00787.warc.gz"} |
https://mathemerize.com/maths-questions/real-numbers-questions/ | Real Numbers Questions
The following real numbers have decimal expansions as given below.
Question : The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational and of the form $$p\over q$$, what can you say about the prime factors of q ? (i) 43.123456789 (ii) 0.120120012000120000….. (iii) 43.$$\overline{123456789}$$ Solution : (i) 43.123456789 is terminating. …
Write down the decimal expansion of these given rational numbers which have terminating decimal expansions.
Question : Write down the decimal expansion of these given rational numbers which have terminating decimal expansions. (i) $$13\over 3125$$ (ii) $$17\over 8$$ (iii) $$64\over 455$$ (iv) $$15\over 1600$$ (v) $$29\over 343$$ (vi) $$23\over {2^3 5^2}$$ (vii) $$129\over {2^2 5^7 7^5}$$ (viii) $$6\over 15$$ (ix) $$35\over 50$$ (x) $$77\over 210$$ Solution : (i) $$13\over 3125$$ …
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion :
Question : Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion : (i) $$13\over 3125$$ (ii) $$17\over 8$$ (iii) $$64\over 455$$ (iv) $$15\over 1600$$ (v) $$29\over 343$$ (vi) $$23\over {2^3 5^2}$$ (vii) $$129\over {2^2 5^7 7^5}$$ (viii) $$6\over 15$$ (ix) …
Question : Prove that the following are irrationals : (i) $$1\over \sqrt{2}$$ (ii) $$7\sqrt{5}$$ (iii) $$6 + \sqrt{2}$$ Solution : (i) Let us assume, to the contrary, that $$1\over \sqrt{2}$$ is rational. That is, we can find co-prime integers p and q (q $$\ne$$ 0) such that $$1\over \sqrt{2}$$ = $$p\over q$$ or $$1\times \sqrt{2}\over … Prove that \(3 + 2\sqrt{5}$$ is irrational.
Solution : Let us assume, to the contrary, that $$3 + 2\sqrt{5}$$ is an irrational number. Now, let $$3 + 2\sqrt{5}$$ = $$a\over b$$, where a and b are coprime and b $$ne$$ 0. So, $$2\sqrt{5}$$ = $$a\over b$$ – 3 or $$\sqrt{5}$$ = $$a\over 2b$$ – $$3\over 2$$ Since a and b are integers, …
Prove that $$\sqrt{5}$$ is an irrational number by contradiction method.
Solution : Suppose that $$\sqrt{5}$$ is an irrational number. Then $$\sqrt{5}$$ can be expressed in the form $$p\over q$$ where p, q are integers and have no common factor, q $$ne$$ 0. $$\sqrt{5}$$ = $$p\over q$$ Squaring both sides, we get 5 = $$p^2\over q^2$$ $$\implies$$ $$p^2$$ = $$5q^2$$ …
Solution : We have, $$7 \times 11 \times 13$$ + 13 = 1001 + 13 =1014 1014 = $$2 \times 3 \times 13 \times 13$$ So, it is the product of more than two prime numbers. 2, 3 and 13. Hence, it is a composite number. $$7 \times 6 \times 5 \times 4 \times 3 … Check whether \(6^n$$ can end with the digit 0 or any n $$\in$$ N.
Solution : If the number $$6^n$$ ends with the digit zero. Then it is divisible by 5. Therefore, the prime factors of $$6^n$$ contains the prime number 5. This is not possible because the only primes in the factors of $$6^n$$ are 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees …
Given H.C.F. (306, 657) = 9, find L.C.M. (306, 657).
Solution : We have, H.C.F. (306, 657) = 9. We know that, Product of L.C.M and H.C.F = Product of two numbers. $$\implies$$ L.C.M $$\times$$ 9 = $$306 \times 657$$ $$\implies$$ L.C.M = $$306 \times 657 \over 9$$ = 22338 Hence, L.C.M is 22338. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8970932960510254, "perplexity": 516.3202298867914}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337432.78/warc/CC-MAIN-20221003200326-20221003230326-00347.warc.gz"} |
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Apr 12 revised How many one to one functions? Fixed LaTex Apr 11 revised How to simplify the expression $(\log_9 2 + \log_9 4)\log_2 (3)$ Helped with the LaTex. Jul 4 revised Let $f=g$ on $[a,b]/E$ where $f\in \mathcal{R}[a,b]$ and continuous on $[a,b]$. Then $g\in\mathcal{R}[a,b]$ and $\displaystyle\int_a^b f=\int_a^bg$. added 148 characters in body Jul 4 revised Let $f=g$ on $[a,b]/E$ where $f\in \mathcal{R}[a,b]$ and continuous on $[a,b]$. Then $g\in\mathcal{R}[a,b]$ and $\displaystyle\int_a^b f=\int_a^bg$. added 98 characters in body Feb 17 revised Did I do this proof right? Here is one proof. Jan 6 revised Tonelli and Fubini and almost everywhere added 65 characters in body Oct 6 revised Showing $\mathbb{Z}[\frac{1+\sqrt{D}}{2}]$ is a subring added 45 characters in body Oct 2 revised Center of finite group with 3 conjugacy classes has order 1 added 224 characters in body Jul 28 revised Diagonal matrices and integrals deleted 12 characters in body Jul 28 revised Diagonal matrices and integrals deleted 12 characters in body Jul 27 revised Fundamental Theorem of Calculus and limit deleted 4 characters in body Dec 6 revised Divergence of the series $2^{-1/n}$ added 117 characters in body Dec 6 revised Divergence of the series $2^{-1/n}$ added 212 characters in body Dec 6 revised Divergence of the series $2^{-1/n}$ added 212 characters in body Nov 4 revised Squeeze Theorem conclusion deleted 135 characters in body Nov 21 revised Proof on limits using Cauchy sequences deleted 67 characters in body Sep 3 revised Points being collinear edited body Apr 8 revised The set of all permutations of a set is a group example deleted 1 characters in body Feb 13 revised Show that an operation being commutative is a structural property added 126 characters in body; edited title | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9251060485839844, "perplexity": 1514.6146911891747}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461861700245.92/warc/CC-MAIN-20160428164140-00082-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://forum.math.toronto.edu/index.php?PHPSESSID=a822j273a85q6l7745l44ouus5&topic=182.15 | ### Author Topic: Comparison of 9th and 10th textbook editions (Read 34195 times)
#### Yanyuan Jing
• Newbie
• Posts: 4
• Karma: 3
##### Re: Comparison of 9th and 10th textbook editions
« Reply #15 on: January 30, 2013, 02:14:52 AM »
Hey guys,
Here are the suggested questions for those using the 9th edition. Again, I only compared the suggested problems listed on the course website, not all the questions in the textbook.
Sections 7.1-7.7: same
Section 7.8: 4, 5, 16, 18, 20, with the following changes to #18:
Quote from: 10th edition
(c) Equation (iii) is satisfied if ξ is an eigenvector, so one way to proceed is to choose ξ to be a suitable linear combination of ξ(1) and ξ(2) so that Eq. (iv) is solvable, and then to solve that equation for η. However, let us proceed in a different way and follow the pattern of Problem 17. First, show that η satisfies $$(A-I)^2η=0$$ Further, show that (A-I)2=0. Thus η can be chosen arbitrarily, except that it must be independent of ξ(1) and ξ(2).
(d) A convenient choice for η is η=(0, 0, 1)T. Find the corresponding ξ from Eq. (iv). Verify that ξ is an eigenvector.
(f) Form a matrix T with the eigenvector ξ(1) in the first column and with the eigenvector ξ from part (d) and the generalized eigenvector η in the other two columns. Find T-1 and form the product J=T-1AT. The matrix J is the Jordan form of A.
(Note: 18(e) is unchanged)
Section 7.9: same
Sections 9.1-9.6: same
Section 9.7: same, with slight difference in the prompt
Quote from: 10th edition
Determine all periodic solutions, all limit cycles, and the stability characteristics of all periodic solutions.
Sections 5.2-5.5: same
Section 6.1: Questions 21-24 in the 10th edition are not included in the 9th edition, and #27 is actually #23 in the 9th edition. Here are questions 21-24:
Quote from: 10th edition
21. $$f(t)= \left\{\begin{array}{ll} 1, & 0 \le t < \pi\\ 0, & \pi \le t < \infty \end{array} \right.$$
22. $$f(t)= \left\{\begin{array}{ll} t, & 0 \le t < 1\\ 0, & 1 \le t < \infty \end{array} \right.$$
23. $$f(t)= \left\{\begin{array}{ll} t, & 0 \le t < 1\\ 1, & 1 \le t < \infty \end{array} \right.$$
24. $$f(t)= \left\{\begin{array}{ll} t, & 0 \le t < 1\\ 2-t, & 1 \le t < 2\\ 0, & 2 \le t < \infty \end{array} \right.$$
Section 6.2: #35 is #34 in the 9th edition. Also, the prompt for #25 should read:
Quote from: 10th edition
A method of determining the inverse transform is developed in Section 6.3. You may wish to refer to Problems 21 through 24 in Section 6.1.
That's all! Hope that's helpful for everyone using the 9th edition!
P.S. This is my first time using LaTeX/MathJax. Please let me know if there are formatting/coding improvements I can make
#### Victor Ivrii
• Elder Member
• Posts: 2602
• Karma: 0
##### Re: Comparison of 9th and 10th textbook editions
« Reply #16 on: January 30, 2013, 05:03:03 AM »
Amazing job!
I noticed you typed bold $\xi$ without using MathJax and this was a correct decision as you wanted bold and upright.
Unfortunately MathJax does not support upright Greek fonts (LaTeX does through package upgreek) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9707854986190796, "perplexity": 3420.991504026658}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00455.warc.gz"} |
http://vuletic.com/hume/mt/gelfand_and_fomin_sec3_lemma2.html | IM Gelfand and SV Fomin, Calculus of Variations
Sec. 3, Lemma 2: Reconstruction and Walkthrough
Mark Vuletic
Note: This document uses MathJax to typeset LaTeX. Please enable JavaScript or you will just see raw LaTeX.
Source: IM Gelfand, SV Fomin. 1963. Calculus of Variations. Mineola, NY: Dover. Tr. RA Silverman. p. 10-11.
Problem: Sec. 3, Lemma 2 states:
If $$\alpha(x)$$ is continuous in $$[a,b]$$, and if $$\int^b_a \alpha(x)h'(x) dx = 0$$ for every function $$h(x) \in D_1(a,b)$$ such that $$h(a) = h(b) = 0$$, then $$\alpha(x) = c$$ for all $$x$$ in $$[a, b]$$, where $$c$$ is a constant. (p. 10)
In the lemma, $$D_1(a,b)$$ is the space of all functions defined on $$(a,b)$$ with continuous first derivatives. Being a mere mortal, and a non-Russian one at that, I found Gelfand and Fomin's proof difficult to understand at first, so I reconstructed it in a way I found easier to follow.
Reconstruction and walkthrough: The key point of strategy Gelfand and Fomin use to prove this lemma is to show that given the conditions specified by the lemma, there is a constant $$c$$ such that $$\int^b_a [\alpha(x) - c]^2 dx = 0$$. From this, it follows that $$\alpha(x) - c = 0$$ and therefore that $$\alpha(x) = c$$.
Here is how I reconstruct the proof. We start with four stipulations, some of which I phrase differently than Gelfand and Fomin:1,2
$$\alpha(x) \textsf{ is continuous in } [a,b] \tag{1}$$
$$\forall f \in D_1(a,b) [f(a) = f(b) = 0] \rightarrow \int^b_a \alpha(x) f'(x) = 0 ] \tag{2}$$
$$c =_{\textsf{df}}\frac{1}{b-a} \int^b_a \alpha(x) dx \tag{3}$$
$$h(x) =_{\textsf{df}} \int^x_a [ \alpha(\xi) - c] d\xi \tag{4}$$
From 1, 4, and the Fundamental Theorem of Calculus, we get
$$h(x) \in D_1(a,b) \tag{5}$$
We note that 4 makes $$h(a)$$ an integral from $$a$$ to $$a$$, and hence equal to zero. Given my choice to write out $$c$$, working through $$h(b)$$ requires a bit of legwork,3 but when we are done we have
$$h(a) = h(b) = 0 \tag{6}$$
From 4 and the Fundamental Theorem of Calculus we have
$$h'(x) = \alpha(x) - c \tag{7}$$
Some algebra and basic calculus4 gives us
$$\int^b_a [\alpha(x) - c]^2 dx = \int^b_a [\alpha(x)h'(x)]dx - c[h(b) - h(a)] \tag{8}$$
From 2, 5, and 6 it immediately follows that
$$\int^b_a [\alpha(x)h'(x)]dx = 0 \tag{9}$$
6 implies
$$c[h(b) - h(a)] = 0 \tag{10}$$
9 and 10 jointly imply
$$\int^b_a [\alpha(x)h'(x)]dx - c[h(b) - h(a)] = 0 \tag{11}$$
Substitute 11 into 8 to get
$$\int^b_a [\alpha(x) - c]^2 dx = 0 \tag{12}$$
From the fact that squares of real-valued functions are positive semidefinite, we know that 12 implies
$$\alpha(x) - c = 0 \tag{13}$$
And a stroke of amazing insight allows us to move from 13 to
$$\alpha(x) = c \tag{14}$$
And there you go.
Notes
1 In 2, I have substituted $$f$$ for the $$h$$ in the original rendition of the lemma. The change is purely notational, but something I needed for personal clarity. Although $$h$$, as it occurs in the lemma, is bound by a quantifier, and thus a variable that ranges over functions, Gelfand and Fomin reuse $$h$$ in their proof (see step 4 in my reconstruction) to represent a specific function. This is one of several practices that is common among mathematicians, but irritating and confusing to someone like me, who learned formal logic before higher mathematics. If my notation confuses you, then just substitute $$h$$ back to $$f$$ wherever you see $$f$$.
2 In 3, Gelfand and Fomin define $$c$$ indirectly, but I write it out.
3 This where just leaving $$c$$ indirectly defined, in Gelfand and Fomin's style, really becomes an advantage, since it makes it immediately evident that $$h(b) = 0$$. However, I like to see (when possible) all of the moving pieces that are hidden in indirect definitions, so if you want to join me in showing $$h(b) = 0$$ the messier way, here are the details: $$h(b) = \int^b_a [\alpha(\xi) - c)]d\xi \text{ [From 4]}$$ $$= \int^b_a \alpha(\xi)d\xi - \int^b_a c d\xi$$ $$= \int^b_a \alpha(\xi)d\xi - \int^b_a \left[\frac{1}{b-a}\int^b_a \alpha(x)dx \right]d\xi \text{ [Using 3]}$$ $$= \int^b_a \alpha(\xi)d\xi - \frac{1}{b-a}\left[\int^b_a \alpha(x)dx \right]\xi\Bigg|^{\xi=b}_{\xi=a}$$ $$= \int^b_a \alpha(\xi)d\xi - \frac{1}{b-a}\left[\int^b_a \alpha(x)dx \right](b-a)$$ $$= \int^b_a \alpha(\xi)d\xi - \int^b_a \alpha(x)dx$$ $$= 0$$
4 Details: $$\int^b_a [\alpha(x) - c]^2 dx = \int^b_a [\alpha(x) - c][\alpha(x) - c]$$ $$= \int^b_a [\alpha(x) - c]h'(x)dx \text{ [Using 7]}$$ $$= \int^b_a [\alpha(x)h'(x) - ch'(x)]dx$$ $$= \int^b_a [\alpha(x)h'(x)]dx - c \int^b_a h'(x)dx$$ $$= \int^b_a [\alpha(x)h'(x)]dx - c[h(b) - h(a)]$$
Last updated: 25 November 2016
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Pleased? Angered? Confused? Have something else you would like | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9615917205810547, "perplexity": 462.6300139943363}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891539.71/warc/CC-MAIN-20180122193259-20180122213259-00005.warc.gz"} |
https://byjus.com/question-answer/predict-the-change-in-internal-energy-for-an-isolated-system-at-constant-volume/ | Question
# Predict the change in internal energy for an isolated system at constant volume.
Solution
## For isolated system there is no transfer of energy as heat, i.e., q = 0 and there is no transfer of energy as work, W = 0. According to the first law of thermodynamics ΔU=q+WΔU=0+0=0 ChemistryNCERT ExemplarStandard XI
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https://xianblog.wordpress.com/tag/pnas/ | ## conditioning on insufficient statistics in Bayesian regression
Posted in Books, Statistics, University life with tags , , , , , , , , , , , , on October 23, 2021 by xi'an
“…the prior distribution, the loss function, and the likelihood or sampling density (…) a healthy skepticism encourages us to question each of them”
A paper by John Lewis, Steven MacEachern, and Yoonkyung Lee has recently appeared in Bayesian Analysis. Starting with the great motivation of a misspecified model requiring the use of a (thus necessarily) insufficient statistic and moving to their central concern of simulating the posterior based on that statistic.
Model misspecification remains understudied from a B perspective and this paper is thus most welcome in addressing the issue. However, when reading through, one of my criticisms is in defining misspecification as equivalent to outliers in the sample. An outlier model is an easy case of misspecification, in the end, since the original model remains meaningful. (Why should there be “good” versus “bad” data) Furthermore, adding a non-parametric component for the unspecified part of the data would sound like a “more Bayesian” alternative. Unrelated, I also idly wondered at whether or not normalising flows could be used in this instance..
The problem in selecting a T (Darjeeling of course!) is not really discussed there, while each choice of a statistic T leads to a different signification to what misspecified means and suggests a comparison with Bayesian empirical likelihood.
“Acceptance rates of this [ABC] algorithm can be intolerably low”
Erm, this is not really the issue with ABC, is it?! Especially when the tolerance is induced by the simulations themselves.
When I reached the MCMC (Gibbs?) part of the paper, I first wondered at its relevance for the mispecification issues before realising it had become the focus of the paper. Now, simulating the observations conditional on a value of the summary statistic T is a true challenge. I remember for instance George Casella mentioning it in association with a Student’s t sample in the 1990’s and Kerrie and I having an unsuccessful attempt at it in the same period. Persi Diaconis has written several papers on the problem and I am thus surprised at the dearth of references here, like the rather recent Byrne and Girolami (2013), Florens and Simoni (2015), or Bornn et al. (2019). In the present case, the linear model assumed as the true model has the exceptional feature that it leads to a feasible transform of an unconstrained simulation into a simulation with fixed statistics, with no measure theoretic worries if not free from considerable efforts to establish the operation is truly valid… And, while simulating (θ,y) makes perfect sense in an insufficient setting, the cost is then precisely the same as when running a vanilla ABC. Which brings us to the natural comparison with ABC. While taking ε=0 may sound as optimal for being “exact”, it is not from an ABC perspective since the convergence rate of the (summary) statistic should be roughly the one of the tolerance (Fearnhead and Liu, Frazier et al., 2018).
“[The Borel Paradox] shows that the concept of a conditional probability with regard to an isolated given hypothesis whose probability equals 0 is inadmissible.” A. Колмого́ров (1933)
As a side note for measure-theoretic purists, the derivation of the conditional of y given T(y)=T⁰ is arbitrary since the event has probability zero (ie, the conditioning set is of measure zero). See the Borel-Kolmogorov paradox. The computations in the paper are undoubtedly correct, but this is only one arbitrary choice of a transform (or conditioning σ-algebra).
## variational approximation to empirical likelihood ABC
Posted in Statistics with tags , , , , , , , , , , , , , , , , , , on October 1, 2021 by xi'an
Sanjay Chaudhuri and his colleagues from Singapore arXived last year a paper on a novel version of empirical likelihood ABC that I hadn’t yet found time to read. This proposal connects with our own, published with Kerrie Mengersen and Pierre Pudlo in 2013 in PNAS. It is presented as an attempt at approximating the posterior distribution based on a vector of (summary) statistics, the variational approximation (or information projection) appearing in the construction of the sampling distribution of the observed summary. (Along with a weird eyed-g symbol! I checked inside the original LaTeX file and it happens to be a mathbbmtt g, that is, the typewriter version of a blackboard computer modern g…) Which writes as an entropic correction of the true posterior distribution (in Theorem 1).
“First, the true log-joint density of the observed summary, the summaries of the i.i.d. replicates and the parameter have to be estimated. Second, we need to estimate the expectation of the above log-joint density with respect to the distribution of the data generating process. Finally, the differential entropy of the data generating density needs to be estimated from the m replicates…”
The density of the observed summary is estimated by empirical likelihood, but I do not understand the reasoning behind the moment condition used in this empirical likelihood. Indeed the moment made of the difference between the observed summaries and the observed ones is zero iff the true value of the parameter is used in the simulation. I also fail to understand the connection with our SAME procedure (Doucet, Godsill & X, 2002), in that the empirical likelihood is based on a sample made of pairs (observed,generated) where the observed part is repeated m times, indeed, but not with the intent of approximating a marginal likelihood estimator… The notion of using the actual data instead of the true expectation (i.e. as a unbiased estimator) at the true parameter value is appealing as it avoids specifying the exact (or analytical) value of this expectation (as in our approach), but I am missing the justification for the extension to any parameter value. Unless one uses an ancillary statistic, which does not sound pertinent… The differential entropy is estimated by a Kozachenko-Leonenko estimator implying k-nearest neighbours.
“The proposed empirical likelihood estimates weights by matching the moments of g(X¹), , g(X) with that of
g(X), without requiring a direct relationship with the parameter. (…) the constraints used in the construction of the empirical likelihood are based on the identity in (7), which can only be satisfied when θ = θ⁰. “
Although I am feeling like missing one argument, the later part of the paper seems to comfort my impression, as quoted above. Meaning that the approximation will fare well only in the vicinity of the true parameter. Which makes it untrustworthy for model choice purposes, I believe. (The paper uses the g-and-k benchmark without exploiting Pierre Jacob’s package that allows for exact MCMC implementation.)
## frontier of simulation-based inference
Posted in Books, Statistics, University life with tags , , , , , , , , , , , , , on June 11, 2020 by xi'an
“This paper results from the Arthur M. Sackler Colloquium of the National Academy of Sciences, `The Science of Deep Learning,’ held March 13–14, 2019, at the National Academy of Sciences in Washington, DC.”
A paper by Kyle Cranmer, Johann Brehmer, and Gilles Louppe just appeared in PNAS on the frontier of simulation-based inference. Sounding more like a tribune than a research paper producing new input. Or at least like a review. Providing a quick introduction to simulators, inference, ABC. Stating the shortcomings of simulation-based inference as three-folded:
1. costly, since required a large number of simulated samples
2. loosing information through the use of insufficient summary statistics or poor non-parametric approximations of the sampling density.
3. wasteful as requiring new computational efforts for new datasets, primarily for ABC as learning the likelihood function (as a function of both the parameter θ and the data x) is only done once.
And the difficulties increase with the dimension of the data. While the points made above are correct, I want to note that ideally ABC (and Bayesian inference as a whole) only depends on a single dimension observation, which is the likelihood value. Or more practically that it only depends on the distance from the observed data to the simulated data. (Possibly the Wasserstein distance between the cdfs.) And that, somewhat unrealistically, that ABC could store the reference table once for all. Point 3 can also be debated in that the effort of learning an approximation can only be amortized when exactly the same model is re-employed with new data, which is likely in industrial applications but less in scientific investigations, I would think. About point 2, the paper misses part of the ABC literature on selecting summary statistics, e.g., the culling afforded by random forests ABC, or the earlier use of the score function in Martin et al. (2019).
The paper then makes a case for using machine-, active-, and deep-learning advances to overcome those blocks. Recouping other recent publications and talks (like Dennis on One World ABC’minar!). Once again presenting machine-learning techniques such as normalizing flows as more efficient than traditional non-parametric estimators. Of which I remain unconvinced without deeper arguments [than the repeated mention of powerful machine-learning techniques] on the convergence rates of these estimators (rather than extolling the super-powers of neural nets).
“A classifier is trained using supervised learning to discriminate two sets of data, although in this case both sets come from the simulator and are generated for different parameter points θ⁰ and θ¹. The classifier output function can be converted into an approximation of the likelihood ratio between θ⁰ and θ¹ (…) learning the likelihood or posterior is an unsupervised learning problem, whereas estimating the likelihood ratio through a classifier is an example of supervised learning and often a simpler task.”
The above comment is highly connected to the approach set by Geyer in 1994 and expanded in Gutmann and Hyvärinen in 2012. Interestingly, at least from my narrow statistician viewpoint!, the discussion about using these different types of approximation to the likelihood and hence to the resulting Bayesian inference never engages into a quantification of the approximation or even broaches upon the potential for inconsistent inference unlocked by using fake likelihoods. While insisting on the information loss brought by using summary statistics.
“Can the outcome be trusted in the presence of imperfections such as limited sample size, insufficient network capacity, or inefficient optimization?”
Interestingly [the more because the paper is classified as statistics] the above shows that the statistical question is set instead in terms of numerical error(s). With proposals to address it ranging from (unrealistic) parametric bootstrap to some forms of GANs.
## Naturally amazed at non-identifiability
Posted in Books, Statistics, University life with tags , , , , , , , , , , , on May 27, 2020 by xi'an
A Nature paper by Stilianos Louca and Matthew W. Pennell, Extant time trees are consistent with a myriad of diversification histories, comes to the extraordinary conclusion that birth-&-death evolutionary models cannot distinguish between several scenarios given the available data! Namely, stem ages and daughter lineage ages cannot identify the speciation rate function λ(.), the extinction rate function μ(.) and the sampling fraction ρ inherently defining the deterministic ODE leading to the number of species predicted at any point τ in time, N(τ). The Nature paper does not seem to make a point beyond the obvious and I am rather perplexed at why it got published [and even highlighted]. A while ago, under the leadership of Steve, PNAS decided to include statistician reviewers for papers relying on statistical arguments. It could time for Nature to move there as well.
“We thus conclude that two birth-death models are congruent if and only if they have the same rp and the same λp at some time point in the present or past.” [S.1.1, p.4]
Or, stated otherwise, that a tree structured dataset made of branch lengths are not enough to identify two functions that parameterise the model. The likelihood looks like
$\frac{\rho^{n-1}\Psi(\tau_1,\tau_0)}{1-E(\tau)}\prod_{i=1}^n \lambda(\tau_i)\Psi(s_{i,1},\tau_i)\Psi(s_{i,2},\tau_i)$\$
where E(.) is the probability to survive to the present and ψ(s,t) the probability to survive and be sampled between times s and t. Sort of. Both functions depending on functions λ(.) and μ(.). (When the stem age is unknown, the likelihood changes a wee bit, but with no changes in the qualitative conclusions. Another way to write this likelihood is in term of the speciation rate λp
$e^{-\Lambda_p(\tau_0)}\prod_{i=1}^n\lambda_p(\tau_I)e^{-\Lambda_p(\tau_i)}$
where Λp is the integrated rate, but which shares the same characteristic of being unable to identify the functions λ(.) and μ(.). While this sounds quite obvious the paper (or rather the supplementary material) goes into fairly extensive mode, including “abstract” algebra to define congruence.
“…we explain why model selection methods based on parsimony or “Occam’s razor”, such as the Akaike Information Criterion and the Bayesian Information Criterion that penalize excessive parameters, generally cannot resolve the identifiability issue…” [S.2, p15]
As illustrated by the above quote, the supplementary material also includes a section about statistical model selections techniques failing to capture the issue, section that seems superfluous or even absurd once the fact that the likelihood is constant across a congruence class has been stated.
## value of a chess game
Posted in pictures, Statistics, University life with tags , , , , , , , , , , , , on April 15, 2020 by xi'an
In our (internal) webinar at CEREMADE today, Miguel Oliu Barton gave a talk on the recent result his student Luc Attia and himself obtained, namely a tractable way of finding the value of a game (when minimax equals maximin), result that got recently published in PNAS:
“Stochastic games were introduced by the Nobel Memorial Prize winner Lloyd Shapley in 1953 to model dynamic interactions in which the environment changes in response to the players’ behavior. The theory of stochastic games and its applications have been studied in several scientific disciplines, including economics, operations research, evolutionary biology, and computer science. In addition, mathematical tools that were used and developed in the study of stochastic games are used by mathematicians and computer scientists in other fields. This paper contributes to the theory of stochastic games by providing a tractable formula for the value of finite competitive stochastic games. This result settles a major open problem which remained unsolved for nearly 40 years.”
While I did not see a direct consequence of this result in regular statistics, I found most interesting the comment made at one point that chess (with forced nullity after repetitions) had a value, by virtue of Zermelo’s theorem. As I had never considered the question (contrary to Shannon!). This value remains unknown. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 2, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8978272676467896, "perplexity": 1042.6307770064257}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587655.10/warc/CC-MAIN-20211025061300-20211025091300-00697.warc.gz"} |
https://infoscience.epfl.ch/record/146149 | ## A Bayesian Alternative to Gain Adaptation in Autoregressive Hidden Markov Models
Models dealing directly with the raw acoustic speech signal are an alternative to conventional feature-based HMMs. A popular way to model the raw speech signal is by means of an autoregressive (AR) process. Being too simple to cope with the nonlinearity of the speech signal, the AR process is generally embedded into a more elaborate model, such as the switching autoregressive HMM (SAR-HMM). A fundamental issue faced by models based on AR processes is that they are very sensitive to variations in the amplitude of the signal. One way to overcome this limitation is to use Gain Adaptation to adjust the amplitude by maximising the likelihood of the observed signal. However, adjusting model parameters by maximising test likelihoods is fundamentally outside the framework of standard statistical approaches to machine learning, since this may lead to overfitting when the models are sufficiently flexible. We propose a statistically principled alternative based on an exact Bayesian procedure in which priors are explicitly defined on the parameters of the AR process. Explicitly, we present the Bayesian SAR-HMM and compare the performance of this model against the standard Gain-Adapted SAR-HMM on a single digit recognition task, showing the effectiveness of the approach and suggesting thereby a principled and straightforward solution to the issue of Gain Adaptation.
Year:
2006
Publisher:
IDIAP
Laboratories: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8406457304954529, "perplexity": 587.662053214563}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125937193.1/warc/CC-MAIN-20180420081400-20180420101400-00460.warc.gz"} |
https://www.jiskha.com/display.cgi?id=1253734712 | # Math/Physics2
posted by Era
The block has mass m=8.5 kg and lies on a fixed smooth frictionless plane tilted at an angle theta=20.0 degrees to the horizontal.
Determine the acceleration of the block as it slides down the plane.
If the block starts from rest 15.0 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline? Caption: Block on inclined plane.
1. dfasdffas
do the damn problem noob
2. Acceleration across the slope: mg sin theta, that is.. Assuming that there is no friction.. 9.8 m/s
Acceleration across the slope: mg sin theta, that is.. Assuming that there is no friction.. 9.8 m/s^2 X sin 20 = 3.35 m/s^s
PE = mgh;
KE = 1/2 X mv^2;
mgh= 1/2 X mv^2;
v = sqrt(2gh);
v = sqrt (294);
v = 17.1 m/s
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a block , mass .10 kg, slides from rest without friction down an inclined plane 2m long which makes an angle of 20 degrees with the horizontal.(a) WWhat force accelerates the block down the plane?
4. ### Physics
A 5.1 kg block slides down an inclined plane that makes an angle of 26 degrees with the horizontal. Starting from rest, the block slides a distance of 2.9 m in 4.6 s. The acceleration of gravity is 9.81 m/s2 . Find the coefficient …
5. ### Physics
Block 1 (2.5kg) is on the surface of a frictionless inclined plane that makes an angle of 20 degrees with the horizontal. Assume g = 10 m/s2 magnitude of the acceleration of block 1 down the incline plane is 3.4 m/s2 How far does block …
6. ### phsyics
The block shown in figure below lies on a smooth plane tilted at an angle è = 21.5° to the horizontal. Determine the acceleration of the block as it slides down the plane. Ignore friction.
7. ### Physics
The block shown in (Figure 1) has mass m = 7.0 kg and lies on a fixed smooth frictionless plane tilted at an angle θ = 22.0 ∘ to the horizontal. If the block starts from rest 15.2 m up the plane from its base, what will …
8. ### physic
The block shown in Figure 6 has mass 3500 g and lies on a plane which tilted at an angle θ = 22° to the horizontal. The effective coefficient of kinetic friction is 0.20. Determine the acceleration of the block as it slides down …
9. ### physics
Acceleration of a Mass in a Pulley System Block 1 of mass m1 slides on a frictionless plane inclined at angle θ with respect to the horizontal. One end of a massless inextensible string is attached to block 1. The string is wound …
10. ### physics
A block has a mass m=6.9-kg and lies on a fixed smooth frictionless plane tilted at an angle of 24.4 to the horizontal. Determine the acceleration (in m/s2)of the block as it slides down the plane.
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http://techtagg.com/standard-error/standard-error-of-prediction-formula.html | Home > Standard Error > Standard Error Of Prediction Formula
# Standard Error Of Prediction Formula
## Contents
Thus, the simulation shows that although any particular prediction interval might not cover its associated new measurement, in repeated experiments this method produces intervals that contain the new measurements at the Each of the two model parameters, the slope and intercept, has its own standard error, which is the estimated standard deviation of the error in estimating it. (In general, the term Researchers typically draw only one sample. Prediction Intervals for the Example Applications Computing prediction intervals for the measured pressure in the Pressure/Temperature example, at temperatures of 25, 45, and 65, and for the measured torque on specimens
Only the standard error of the intercept (therefore t, p-value and CI) changes. For example, the effect size statistic for ANOVA is the Eta-square. Further, as I detailed here, R-squared is relevant mainly when you need precise predictions. Return to top of page.
## Standard Error Of Prediction Formula
Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the Because the standard error of the mean gets larger for extreme (farther-from-the-mean) values of X, the confidence intervals for the mean (the height of the regression line) widen noticeably at either Unlike R-squared, you can use the standard error of the regression to assess the precision of the predictions. S represents the average distance that the observed values fall from the regression line.
The regression model produces an R-squared of 76.1% and S is 3.53399% body fat. Copyright (c) 2010 Croatian Society of Medical Biochemistry and Laboratory Medicine. The S value is still the average distance that the data points fall from the fitted values. Standard Error Of Prediction Calculator In a simple regression model, the percentage of variance "explained" by the model, which is called R-squared, is the square of the correlation between Y and X.
where STDEV.P(X) is the population standard deviation, as noted above. (Sometimes the sample standard deviation is used to standardize a variable, but the population standard deviation is needed in this particular It can allow the researcher to construct a confidence interval within which the true population correlation will fall. This inspired me to figure out that $Var(\hat{\beta}_0)=\sigma^2(1/n+\bar{x}^2/SXX)$, then I can get $\bar{x}$ to calculate the standard error of prediction. –Jiebiao Wang Jul 11 '13 at 20:39 The standard view publisher site standard error of regression0How to combine Standard Deviation and Standard Error of linear regression repeats Hot Network Questions Saffron and coloration - is there a way to know why it gave
It is particularly important to use the standard error to estimate an interval about the population parameter when an effect size statistic is not available. Standard Error Of Prediction Multiple Linear Regression Jim Name: Olivia • Saturday, September 6, 2014 Hi this is such a great resource I have stumbled upon :) I have a question though - when comparing different models from You can choose your own, or just report the standard error along with the point forecast. A model does not always improve when more variables are added: adjusted R-squared can go down (even go negative) if irrelevant variables are added. 8.
## Standard Error Of Prediction In R
Note the similarity of the formula for σest to the formula for σ.  It turns out that σest is the standard deviation of the errors of prediction (each Y - It is an even more valuable statistic than the Pearson because it is a measure of the overlap, or association between the independent and dependent variables. (See Figure 3). Standard Error Of Prediction Formula However... 5. Standard Error Of Prediction Linear Regression Standard error.
HyperStat Online. So a greater amount of "noise" in the data (as measured by s) makes all the estimates of means and coefficients proportionally less accurate, and a larger sample size makes all The newly observed measurements, observed after making the prediction, are noted with an "X" for each data set. Standard error statistics measure how accurate and precise the sample is as an estimate of the population parameter. Standard Error Of Prediction Excel
Comparison with Confidence Intervals It is also interesting to compare these results to the analogous results for confidence intervals. How to compare models Testing the assumptions of linear regression Additional notes on regression analysis Stepwise and all-possible-regressions Excel file with simple regression formulas Excel file with regression formulas in matrix For some statistics, however, the associated effect size statistic is not available. I think it should answer your questions.
Unlike in conventional methods, the variance of the dependent variable has not been calculated from Sy,x. I hope the problem is of interest: if needed I can send further details. Standard Error Of Prediction Interval Just as the standard deviation is a measure of the dispersion of values in the sample, the standard error is a measure of the dispersion of values in the sampling distribution. Biochemia Medica 2008;18(1):7-13.
## Regressions differing in accuracy of prediction.
However, one is left with the question of how accurate are predictions based on the regression? regression stata standard-error prediction share|improve this question asked Jul 11 '13 at 19:17 Jiebiao Wang 3,69032044 1 How would the regression output change if you were, say, to add $10^6$ This can artificially inflate the R-squared value. Standard Error Of Prediction Stata Full-text Dataset · Jun 2014 Download Mar 11, 2016 Anthony Victor Goodchild · Department for Environment, Food and Rural Affairs Thanks, Jim .
In fact, the level of probability selected for the study (typically P < 0.05) is an estimate of the probability of the mean falling within that interval. The standard error of the mean permits the researcher to construct a confidence interval in which the population mean is likely to fall. That is, R-squared = rXY2, and that′s why it′s called R-squared. The SEM, like the standard deviation, is multiplied by 1.96 to obtain an estimate of where 95% of the population sample means are expected to fall in the theoretical sampling distribution.
The sample standard deviation of the errors is a downward-biased estimate of the size of the true unexplained deviations in Y because it does not adjust for the additional "degree of share|improve this answer edited Aug 27 '13 at 14:50 answered Jul 17 '13 at 23:04 Jiebiao Wang 3,69032044 add a comment| Your Answer draft saved draft discarded Sign up or I use the graph for simple regression because it's easier illustrate the concept. However, I've stated previously that R-squared is overrated.
How to find files that contain one criterion but exclude a different criterion Divisibility Proof How to make the development and use of Steam Engines preferred over that of Combustion Engines? This means that noise in the data (whose intensity if measured by s) affects the errors in all the coefficient estimates in exactly the same way, and it also means that But if it is assumed that everything is OK, what information can you obtain from that table? That's probably why the R-squared is so high, 98%.
It is not possible for them to take measurements on the entire population. The standard error of the estimate is closely related to this quantity and is defined below: where σest is the standard error of the estimate, Y is an actual score, Y'
© 2017 techtagg.com | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9586603045463562, "perplexity": 754.026787369306}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187822480.15/warc/CC-MAIN-20171017181947-20171017201947-00233.warc.gz"} |
http://mathhelpforum.com/calculus/30908-parametric-equations.html | # Math Help - Parametric Equations
1. ## Parametric Equations
Does anyone know how to find parametric equations for the curve segment:
y=sqrt(x) from (1,1) to (16,4)?
Thanks
2. You can write lots of equations..
$x=t$
$y=\sqrt{t}$
$1 \leq t \leq 16$
$x=t^2$
$y=t$
$1 \leq t \leq 4$
$x=e^t$
$y=e^{(\frac{t}{2})}$
$0 \leq t \leq \ln 16$
$x=\ln^2 t$
$y=\ln t$
$e \leq t \leq \ln e^4$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8988909721374512, "perplexity": 3428.8751516756324}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657128304.55/warc/CC-MAIN-20140914011208-00036-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
http://the-mouse-trap.com/tag/heritability/ | A reader of this blog wrote to me recently regarding a series of posts I have written regarding IQ,SES and heritability, and I thought it would be good to share the comments with the rest of the mouse trap community and to delineate my position on the matter (and what I believe the studies show the relation is). First I’ll like to quote extensively from the comment (private mail):
Let me start with my belief that I think we share the same idea about IQ and its origins.
That is, genetics endows each and every person with a maximum IQ that can be achieved if and only if the environment is perfect for the development of this IQ.
Agreed!
Consequently, in identical twins, environment is the only cause of differences in IQ; IQ differences between persons that are not identical twins must be related to both environment and genes.
There are subtle nuances here. (no I’m not being pedantic, the importance of these will become clear in the course of this post). First if MZ twins are raised in the same family, they share the same genotype (A), they share the same ‘shared environment’ (C), so the difference is due to the non-shared environmental factor (E) only. In case of DZ twins raised in the same family, they share half the genotype (A), they share the same shared environment (C) and the difference in say IQ, is due to both genotype (A) and non-shared environmental factors (E). Twin studies with MZ and DZ twins (and in some cases siblings, half-siblings etc ) raised in the same family are used to tease apart the contribution of shared environmental factor, as opposed to genetics and non-shared environmental factors and can be used to find at a broad level if the trait is highly ‘genetically’ heritable (correlation between MZ>>DZ>>siblings), has high ‘shared environmental’ factors operating (MZ~DZ~sibling , but correlation still high) or is largely controlled by random non-shared environmental influences (low correlation in DZ/MZ/siblings). Please see accompanying Wikipedia figure.
Adoption studies are another method that is used to tease apart the shared environmental factors from genetic factors in calculating the heritability of a trait. Thus, even if correlation in MZ twin IQ is high, the effect could be due to shared environment factors (if say both MZ and DZ twin show similar correlation) , or it may be largely genetic (if MZ>>DZ when it comes to correlation between the trait in affected twins.
So it is necessary to qualify your statement: In identical twins, raised in the same family, , non-shared environment is the only cause of differences in IQ. IQ differences between persons that are not identical twins must be related to both shared environment , non-shared environment and genes.
I fully back up this citation:
“suffice it to say that I believe (and think that I have evidence on my side) that shows that in low SES conditions, a Low SES does not lead to full flowering of genetic Intelligence potential and is thus a leading cause of low IQ amongst low SES populations.”
In this case, I think that the problem starts only with your following comments; while you write “a leading cause”, I think that, by yourself, you mean “way more important than genetics”. If this is the case, would you explain to me why you think that?
Here goes. Consider a large sample of children in say low SES populations. IQ may be represented by a formula IQ= aA+cC+eE; where A reflects genotype, C shared environment and E non-shared environment. Here we are assuming no interaction of IQ with SES, so this equation (given values of a, c, e ) should hold for all SES data (both high as well as low SES cohort) . Unfortunately, life is not that simple, and one can not fit the same equation to low SES as well as high SES data set without changing the slopes of variables involved. Thus, Turkheimer and colleagues in two sets of studies have shown that there is an interaction of IQ and SES and there is direct affcet (SES mediated by s) and indirect interaction effects mediated via effects on A(a’), C(c’) and E(e’). Thus our equation becomes
IQ= sSES+(a+a’SES)A+(c+c’SES)C+(e+e’SES)E.
This equation, with suitable values of s, a, a’, c, c’, e and e’ now holds for all values of SES and IQ and the data fits nicely and can be interpreted. Remember that a+a’SES is sort of indicative of contribution of genetic factor to IQ and the proportion of variance due to genetic factor(at any given SES) can be found by squaring this and dividing this by sum of all other variances .
Var(A) =SQR(a+a?SES)
Similarly heritability or proportion of variance due to A :SQR(h)=SQR(a+a?SES)/(SQR(a+a?SES)*SQR(c+c?SES)*SQR(e+e?SES))
Turkheimer have plotted nice plot of their data which sows clearly that in LOW SES situations, the proportion of variance in IQ is largely due to shared environmental factors (C) , while in HIGH SES situations, the proportion of variance in IQ is largely due to genetic factors (A). the figures (in the free PDF available at Turkheimer’s side) is a must see to grasp the significance of this. I am quoting a bit from the paper:
Figure 3 shows the FSIQ variance accounted for by the three components, with 95% confidence intervals. In the most impoverished families, the modeled heritability of FSIQ is essentially 0, and C accounts for almost 60% of the variability; in the most affluent families, virtually all of the modeled variability in IQ is attributable to A.
Let us pause here and reflect on what this means. This means that in low SES families, IQ is independent of genotype and is mostly dependent on the SES status. Let us take some concrete examples. Say the mean IQ of low SES sample (income as a proxy for SES ranging from 1000-5000 rs p.a.; mean 2500 and variance 250)) is 80 with mean variance of 20. Thus, in this sample a typical child has IQ in range 80 +-20 or between 60 and 100. Suppose further that there are 5 alleles that confer differential advantage for IQ on a locus thus representing 5 genotypes, then having either of the genotype will essentially give us no predictive power to say whether the IQ of a particular sample is 80 or 60 or 100. Also, let us assume that there are 5 classes of C and they are highly correlated with SES. First class of C (is 1000-1500 SES range) and so on and so forth. Then knowing whjat kind of family (C) the child grew up in we could easily predict his IQ (if he is of class C where SES ranges from 1000-1500), his IQ is most probably 60. This is what I mean when I say that in low SES environments IQ is largely determined by environment and not by genetics. Now , I have taken a jump here and equated C with SES, but that is a justified leap in my opinion (more about that later).
What this also means is that given the right type of environment (say class C with SES in upper range of 3500-5000 rs p.a.) , all children (irrespective of their genotype (any 5 variants of genotype) can still achieve an IQ in the upper range , say 100 as the environment is the only predominant factor operating at this level and the impact of genetics is still not felt. Thus, if we do increase SES and provide the right C, then every child in this group can have mean IQ of 100.
Contrast this with the case at the upper end of the strata (SES). Here most of the variation in IQ is predominantly due to genetics (A) and shared environment C does not seem to play a big factor. Thus, knowing a genotype of a child has greater predictive power in this sample, than knowing his C (or family income or SES). Thus, evenif we provide a very enriched environment to all children (increase their C to the highest percentile), it would have no effect on increasing the mean IQ of the sample as now the IQ is mostly under genetic control.
This in a nutshell, is what I mean when I claim that low SES is the leading cause of low IQ in low SES families.
Another example to ruminate on is another universal and shared sub-threshold factor like having a golf course in the house. Let us assume that within higher SES group, this environmental enrichment factor plays a role, with some lower strata of higher SES (the middle class) not able to afford a golf course, while the higher higher SES strata (the upper class) abvle to afford a golf course and expose their children to them . Further, suppose that there is a module in the brains and genes switched on only if exposure to golf course takes place. Then within this higher SES group, what we will observe is that though the genetics plays a good role (due to factor X-iodine: remember, which is available to all in this group) ; still there would also be variation due to environment (golf course exposure) and that a full 20 points more can be added to all people of this group (with mean IQ 100 raising their IQ to 120), if all were exposed to a golf course and a intelligence-module-dependent-on-golf-course-exposure was allowed to develop. And on the higher end of IQ (and SES) what we would find is that most of the variance now is genetic (due to this golf-course module coming into play), while at the lower end, most of the variance is still environmental within this ‘high’ SES group.
If the above seems far fetched this is exactly what Turkheiemrs et al found in their follow up study focusing on mid to high SES children. I quote from it (again the pdf has beautiful figures and you should see them) :
Figure 2 illustrates the relations between income and genetic and shared environmental proportions of variance, as implied by the parameters estimated in Model 3. Genetic influences accounted for about 55% of the variance in adolescents’ cognitive aptitude and shared environmental influences about 35% among higher income families. Among lower income families, the proportions were in the reverse direction, 39% genetic and 45% shared environment. Although the shared environmental proportion of variance decreased with income, shared environmental variance per se did not decrease. The interactive effect was driven entirely by the increase in genetic variance. Genetic variance in cognitive aptitude nearly doubled from 4.41 in families earning less than \$5000 annually to 8.29 in families earning more than \$25,000 annually.
Our investigation supports our hypothesis that the magnitude of genetic influences on cognitive aptitude varies with socioeconomic status. This partially replicates the results presented by Turkheimer et al. (2003); however, no shared environmental interaction effects were demonstrable in the current study. Genetic influences accounted for about 55% of the variance in adolescents’ cognitive aptitude and shared environmental influences about 35% among higher income families. Among lower income families, the proportions were in the reverse direction, 39% genetic and 45% shared environment. This pattern is similar to the pattern seen in Turkheimer et al. (2003), although less marked.
So, I want you to pause here and grasp the significance of this- at every level of IQ-SES, there may be threshold factor that giverns whether IQ modules flower to full potential and this is the putative mechanism that leads to SES causing low or high IQ directional and causal relation. At each level, as the threshold factors become available,. more and more IQ starts coming under genetic control, but , and this is important, for jumps in IQ to take place , increasing SES (removing the sub-threshold conditions) is VERY important.
I mean “not following up on the ‘a leading cause'”, because in a later post, you write:
“Now, I have shown elsewhere that low SES causes low IQ”
Here, there is no mention of any other possible cause besides the environment anymore.
Yes, because as shown very strongly by Turkeihems and team , at low SES, shared environment/SES is the putative mechanism and genetics has no/negligible role to play. So for low SES, low SES causes low IQ. period.
in another post, you write
“A series of studies that I have discussed earlier, clearly indicate that in the absence of good socioeconomic conditions, IQ can be stunted by as large as 20 IQ points. ”
This same post also contained this citation “Children of well-off biological parents reared by poor/well -off adopted parents have Average IQ about 16 point higher than children of poor biological parents”
In my opinion, the latter would indicate the approximate range of genetic IQ differences for the samples in this study, while the former would indicate the approximate maximal environmental gain that can be hoped for in the environments that were encountered in these studies.
No they don’t. They talk about different SES groups, so as shown findings from one cannot be extrapolated to the other. In the low SES group, there is no genetic variation. We can thus not conclude that that (16 points diff.) is the ‘average’ genetic component taken the entire sample together. what one can say is that if mean IQ of high SES children was 100, the mean IQ of low SES children was 84 . Period. The difference is likely due to the fact, that the module X has not developed in low SES people (more later) .
Regarding the former, yes I agree that that is the maximum gain that one can hope for if all children of low SES were given the right environment (raised to high SES). Put another way, if mean IQ of poor/low SES children is 84 , then given the right conditions the mean of the low SES children can be raised to 104 (greater than high SES children’s mean :-).
As both of them do cover the same IQ range (10-20), the logical consequence for a broad statement on IQ and genetics seems therefore to be, that these studies may say that overall, IQ changes can be expected to be determined to approximately equal parts by genetics and environment, with environment being responsible for a typically larger part in low SES families, and genetics playing a relatively larger part in high SES families.
Agreed partially, but that glosses over the fact of sub-threshold universal shared environments and the fact that the role of genetic and environmental component varies with SES, an therefore an ideal statement would be IQ is under gentic controltolarge extent, but that gentics needs threshold environments to flower and thus the importance of environment component- not in explaining variance , but by its direct effect on IQ enabling/flowering.
This same post also contained this citation “Children of well-off biological parents reared by poor/well -off adopted parents have Average IQ about 16 point higher than children of poor biological parents”
In my opinion, the latter would indicate the approximate range of genetic IQ differences for the samples in this study, while the former would indicate the approximate maximal environmental gain that can be hoped for in the environments that were encountered in these studies.
As both of them do cover the same IQ range (10-20), the logical consequence for a broad statement on IQ and genetics seems therefore to be, that these studies may say that overall, IQ changes can be expected to be determined to approximately equal parts by genetics and environment, with environment being responsible for a typically larger part in low SES families, and genetics playing a relatively larger part in high SES families.
There also is this citation:
“The normal observation that identical twins belonging to well-off/middle class families have IQ rates similar as compared to fraternal twins, thus indicates that for children from well-off background (biological/adopted), the IQ (observed phenotype) is mostly due to genetic factors (underlying genotype) and environmental factors are not a big determinant.
The paradoxical observation that identical twins belonging to poor families have IQ rates as varying as compared to fraternal twins, should indicate that for children from poor background (biological/adopted), the IQ (observed phenotype) is mostly due to environmental factors and genetic factors (the underlying genotype ) are not a big determinant.”
These are extremely nice observations. I would be interested in the conclusions one might be tempted to draw from them. Reading the latter part of this sentence, one might come to the following conclusion (conclusion 1): “if in low-SES families the variations in IQ are largely determined by environmental factors, then providing a positive environment for the development of IQ would increase the IQ levels in these families impressively (up to 20 points; but, this is an up to value, means would be more interesting).”
While I completely agree with this thinking, one might also be tempted to draw the conclusion that (conclusion 2) “As IQ variations in low-SES families are largely due to environment, providing an IQ-stimulating environment in low-SES families might completely eliminate the IQ differences between low-SES families and high-SES families”
At the least, a non-cautious reader might understand your words as such. I am not sure whether you think that way or not. I would like to hear your opinion on that. I think that this citation “Children of well-off biological parents reared by poor/well -off adopted parents have Average IQ about 16 point higher than children of poor biological parents” provides an argument that precludes conclusion 2. It would rather say that (conclusion 3), ” providing a perfect IQ-stimulating environment for low-SES families as encountered in these studies, one should think that their offspring would achieve an IQ level that is 16 points lower than that of the offspring of high-SES families.”
I would like to hear your opinion on my conclusion 3.
I agree with conclusion 1. I also agree with conclusion 2 (not based on political correctness, but hard data). The paper on which these figures are based can be found here. The mean IQ of high SES persons is 113.5 and the mean IQ of low SES children is 98.00, thus a difference of ~16 points. The variation in IQ of high SES children raised in high SES families is 12.25; as shown this variance is likely due to genetics (say hundred percent is due to genetics); then changing the SES within the given range should have no effect on average IQ and it would remain 119 (for this high +/high+ group). On the other hand, the variance in low SES, reared by low SES families is 15.41 and mean is 92.40;thus if all were given enriched environment, their mean IQ would become 92.4+ 15.4 = ~ 108 . We still have a 10 point difference which can be accounted for by the fact that genetics had not come into play for low SES , low SES group yet and as genetics enters and increases the variance due to genetic flowering,, their IQ would be in the same league as high IQ/High IQ children.
So definitely the conclusion 3 is flawed- the difference would not be close to 16 points, but negligible, as the 16 points nowhere measures gentic difference in abilities, but reflects the genetic factor not yet active in low SES, due to improper environmental exposure.
I think that this is a rather important conclusion, as it tells us something about the differences in IQ that can be expected to exist between distinct population stratums (don’t know whether this is an appropriate word for what I try to say; I hope you understand what I mean).
If this is the ballpark of figures that we can expect between low-and high SES IQ differences, this would have important effects on future IQ-distributions. Population-wide stability of IQ-performance, if measured in a saturated environment (maximum stimulation of all members of society), can then only be achieved if all stratums of society have the same number of offspring per individuum. If low-SES families have more children, we have to expect that the 16-point lower IQ will decrease the whole-population IQ.
Here, the 16 points only apply to the sample as measured in your example; the true value of the saturated stratum-dependent-IQ together with stratum-dependent birthrates will determine the shift of the saturated IQ-distribution for the generations to come.
Do you agree with this point of view?
To use a very strong and negative connotation word, the above smacks of eugenics. And I wont comment further on this. Each according to his own philosophy, but beware that science does not support your conclusions. Instead of population controlling the poor, please try to elevate their vicious loop of undeserved poverty, low IQ and harmful stigma.
Turkheimer, E., Haley, A., Waldron, M., D’Onofrio, B., & Gottesman, I. (2003). Socioeconomic status modifies heritability of iq in young children Psychological Science, 14 (6), 623-628 DOI: 10.1046/j.0956-7976.2003.psci_1475.x
Harden, K., Turkheimer, E., & Loehlin, J. (2006). Genotype by Environment Interaction in Adolescents’ Cognitive Aptitude Behavior Genetics, 37 (2), 273-283 DOI: 10.1007/s10519-006-9113-4
Capron, C., & Duyme, M. (1989). Assessment of effects of socio-economic status on IQ in a full cross-fostering study Nature, 340 (6234), 552-554 DOI: 10.1038/340552a0 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8383583426475525, "perplexity": 1840.3866694363878}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917119080.24/warc/CC-MAIN-20170423031159-00566-ip-10-145-167-34.ec2.internal.warc.gz"} |
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# A new branch and bound algorithm for minimax ratios problems
Yingfeng Zhao
• Corresponding author
• School of Mathematics and Statistics, Xidian University, Xi’an 710071, China
• School of Mathematical Science, Henan Institute of Science and Technology, Xinxiang 453003, China
• Email
• Other articles by this author:
/ Sanyang Liu
/ Hongwei Jiao
• School of Mathematical Science, Henan Institute of Science and Technology, Xinxiang 453003, China
• Other articles by this author:
Published Online: 2017-06-22 | DOI: https://doi.org/10.1515/math-2017-0072
## Abstract
This study presents an efficient branch and bound algorithm for globally solving the minimax fractional programming problem (MFP). By introducing an auxiliary variable, an equivalent problem is firstly constructed and the convex relaxation programming problem is then established by utilizing convexity and concavity of functions in the problem. Other than usual branch and bound algorithm, an adapted partition skill and a practical reduction technique performed only in an unidimensional interval are incorporated into the algorithm scheme to significantly improve the computational performance. The global convergence is proved. Finally, some comparative experiments and a randomized numerical test are carried out to demonstrate the efficiency and robustness of the proposed algorithm.
MSC 2010: 90C26; 90C30
## 1 Introduction
The minimax fractional programming (MFP) problem we studied in this paper can be formulated as the following nonlinear optimization problem $(MFP):minmax{n1(x)d1(x),n2(x)d2(x),…,np(x)dp(x)}s.t.gj(x)≤0,j=1,2,…,m,x∈X0={x∈Rn|li0≤xi≤ui0,i=1,2,⋯,n},$
where ni(x), −di(x), i = 1, 2, ⋯, p, gj(x), j = 1, 2, ⋯, m, are all convex functions and we assume that each numerator and denominator in the objective function satisfy the following condition: $ni(x)≥0,di(x)>0,∀x∈X0,i=1,2,⋯,p.$(1)
The MFP problem, which we called generalized minimax convex-concave ratios problem, is one of the most topical and useful fields in fractional programming problems, it has attracted the interest of both practitioners and researchers for many years [14]. The reason for studying the MFP problem is twoflod. The first one is that, from the practical point of view, the MFP problem has quite a lot of applications in various areas where several rates are to be optimized simultaneously, such as design of electronic circuits [5], biology to calculate malting efficient DNA [6], rural farming system [7], neural networks [8], management and finance [9] and system identification [1012], to name but a few. Another reason is that, from the research point of view, the MFP problem generalizes minimax linear ratios problem [1315] and it is neither quasiconcave nor quasiconvex, so it may have multiple local optima most of which fail to be global optimal (Example 8. and Figure 1), thus it poses significant theoretical and computational challenges. For these reasons, developing new solution methods for the problem MFP is still quite necessary and meaningful.
Figure 1
3-D image of the objective function over [0,2;0,2] in Example 7.
Over the years, several global optimization methods have been available for solving special case of the problem (MFP), include parametric programming method [16], interior-point algorithm [17], monotonic optimization approach [18] and those kinds of dinkelbach-type algorithms [19, 20]. When the numerators and denominators in the objective function of the (MFP) are all affine, Feng presents two deterministic algorithm for solving the special case of the MFP problem [14, 15]. Recently, using two-step relaxation technique, Jiao and Liu have proposed a new global optimization algorithm for solving the MFP with linear ratios and constraints [13]. Jeyakumar et al. proposed a strong duality for robust minimax fractional programming problems, when data in both of the objective and constraints are uncertainty [21]. Lai and Huang established the sufficient optimality conditions for a minimax programming problem involving p fractional n–set functions under generalized convexity [22]. Progress for the MFP problem is far advanced than mentioned above – there are many other studies about (MFP) and its special cases. Despite these various contributions, most knowledge for MFP presented in recent works focus only on optimality conditions or duality theory for special case of the MFP problem, and existing algorithm can only find its local optimal solution, or can only solve the linear form of the MFP problems. For general (MFP) investigated in this paper, to our knowledge, very few global optimization algorithms have been developed.
In this paper, we put forward an unidimensional outcome inteval branch-and-bound algorithm for globally solving the (MFP). First, a concise convex relaxation programming of the equivalent problem (EMFP) is established. Then some key operations in outcome branch and bound algorithm are described, especially the interval reduction skill can significantly improve the computational efficiency of the presented algorithm. Third, the proposed algorithm is developed by incorporating the accelerating technique as well as the adapted partition rule into the branch and bound scheme. The global convergence property is proved, numerical experimental results show that the proposed algorithm has higher computational efficiency and stronger robustness than Refs. [13] and [14, 15].
The remainder of this study is organized in the following way. In Section 2, by introducing an auxiliary variable, the MFP problem is converted into an equivalent problem (EMFP). Then the convex relaxation programming problem of the (EMFP) is established in Section 3, the linear relaxation programming for its special case, where the numerators and denominators in the objective function and the constraints are all linear, is also described in this section. Section 4 presents some key operations in branch and bound algorithm for globally solving the (MFP). An unidimensional outcome interval branch and bound algorithm for the (MFP) is introduced in Section 4. The numerical results of some test examples in recent works with the proposed algorithm are reported in Section 5, and some concluding remarks are given in the last section.
## 2 Equivalent problem
For solving the problem, we will first transform the problem (MFP) into an equivalent problem (EMFP) by associating all ratios in the objective function with an auxiliary variable. To this end, we first denote some notations as follows: $n_i≤ni(x)≤n¯i,d_i≤di(x)≤d¯i,ln+10=min1≤i≤p{n_id¯i},un+10=max1≤i≤p{n¯id_i};$
where ni, di can be easily computed by utilizing the convexity and concavity of ni(x) and di(x), respectively. Through introducing a positive auxiliary variable xn+1, we can derive the following equivalent problem (EMFP0) of the MFP problem. $(EMFP0):minxn+1s.t.ni(x)−xn+1di(x)≤0,i=1,2,⋯,pgj(x)≤0,j=1,2,…,m,x∈X0={x∈Rn|li0≤xi≤ui0,i=1,2,⋯,n},xn+1∈D0=[ln+10,un+10],$
By utilizing variable equivalence replacement tactic and changing the notation, one can thus reformulate the (EMFP0) into the following form $(EMFP):minxn+1s.t.ni(x)−xn+1(di(x)−d_i)−d_ixn+1≤0,gj(x)≤0,j=1,2,…,m,x∈X0,xn+1∈D0.$
Note that, the reason why we convert the (EMFP0) into the form (EMFP) is that by isolating a linear part from the concave function di(x) we can make the convex relaxation programming more close to the initial problem (MFP), this operation will improve the algorithm significantly in efficiency which can be vindicated by numerical experiments. Moreover, in the EMFP problem, only the middle part in the first set of constraints is nonconvex. This brings a great convenience to us for building the convex relaxation problem of the initial problem. The equivalence between the problems (MFP) and (EMFP) can be obtained in the sense of the following theorem.
#### Theorem 2.1
If $\left({x}^{\ast },{x}_{n+1}^{\ast }\right)\in {R}^{n+1}$ is a global optimal solution for the (EMFP), then x*Rn is a global optimal solution for the (MFP). Conversely, if x*Rn is a global optimal solution for the (MFP), then $\left({x}^{\ast },{x}_{n+1}^{\ast }\right)\in {R}^{n+1}$ is a global optimal solution for the (EMFP), where ${x}_{n+1}^{\ast }=max\left\{\frac{{n}_{1}\left({x}^{\ast }\right)}{{d}_{1}\left({x}^{\ast }\right)},\phantom{\rule{thinmathspace}{0ex}}\frac{{n}_{2}\left({x}^{\ast }\right)}{{d}_{2}\left({x}^{\ast }\right)},\dots ,\frac{{n}_{p}\left({x}^{\ast }\right)}{{d}_{p}\left({x}^{\ast }\right)}\right\}.$
#### Proof
The proof of this theorem can be easily followed according to the definition of the problems (EMFP0) and (EMFP), therefore, it is omitted here.□
To solve the MFP problem, based on Theorem 2.1, we only need to solve the EMFP problem instead. Hence from now on, we will assume that the initial problem (MFP) has been of the form like (EMFP), and then our main work will focus on how to globally solve the (EMFP).
## 3 Convex relaxation programming problem
In this section, we construct a convex programming relaxation of the (EMFP) only with relaxed auxiliary variable in an outcome interval D. For the ease of presentation, we assume that D = [ln+1, un+1] represents either the initial outcome interval D0 of (EMFP), or modified interval generated from the branching operation in the algorithm. Next, we describe the construction process of the relaxation convex programming (RMFP) of the (EMFP) corresponding to D.
For this, we only need to consider the nonconvex constraints in (EMFP). As a matter of convenience, we denote these constraint functions with notation and number them as follows: $hi(x,xn+1)=ni(x)−xn+1(di(x)−d_i)−d_ixn+1$(2)
Assume v, v denote the best upper bound and lower bound of the objective value known so far (if no such value is founded, set v := +∞, v := −∞), respectively, and let M = min{v, un+1}, then we can obtain an underestimate function of hi(x) which can be formulated as: $h~i(x,xn+1)=ni(x)−M(di(x)−d_i)−d_ixn+1=ni(x)−Mdi(x)+Md_i−d_ixn+1,$(3)
based on the definition of M and the convexity and concavity of ni(x) and di(x), it is not hard to see that ${\stackrel{~}{h}}_{i}\left(x\right)$ is convex and provides a very good lower approximation for hi(x), i = 1, 2, ⋯, p. Thus, according to the above discussion, for arbitrary region Xk × DkX0 × D0, we can describe the convex relaxation programming problem (RMFP) corresponding to outcome interval D as follows: $(RMFP):minxn+1s.t.h~i(x,xn+1)≤0,i=1,2,⋯,p,gj(x)≤0,j=1,2,…,m,x∈X0,xn+1∈D.$
#### Remark 3.1
The relaxation operation we used is very simple and easy to implement, that is we only need to compare two real numbers in each iteration for constructing the relaxation problem. It is quite different from usual branch and bound algorithms most of which utilize concave envelope or linearity technique to construct relaxation, and this will greatly reduce the computation time for establishing underestimate function of non-convex functions in the initial 5 source problem. for establishing underestimate function of nonconvex functions in the initial source problem. Furthermore, since relaxation operation occurs only in the unidimensional outcome interval, we can only subdivide the outcome interval in branching process. This will significantly reduce the numbers of nods in the algorithm so that the outcome interval branch and bound algorithm we presented performs better than usual branch and bound algorithms which branches in the n—dimensional variable space.
#### Remark 3.2
Condition (1) can be relaxed to di(x) ≠ 0, ∀xX0, i = 1, 2, ⋯; p, when functions ni(x) and di(x) are all linear, that is, the problem we considered in this paper generalized the minimax of linear ratios problem appeared in [1315]. In fact, when ni(x) and di(x) are all linear functions, suppose instead $di(x)=∑j=1nαijxij+γi=∑j∈T+αijxi+∑j∈T−αijxij+γi>0,i=1,2,⋯,p,$(4)
where T+ = {j | αij > 0}, T = {j | αij < 0}. If not, we can substitute the ratio $\frac{{n}_{i}\left(x\right)}{{d}_{i}\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}with\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{-{n}_{i}\left(x\right)}{-{d}_{i}\left(x\right)},$ then it will turn into the case identified by(4). Sequentially, the underestimate function ${\stackrel{~}{h}}_{i}\left(x\right)$ can be constructed in the following way $h~i(x,xn+1)=ni(x)−Mi1∑j∈T+αijxi−Mi2∑j∈T−αijxi−γixn+1,$(5)
where Mi1 = min{v, un+1}, Mi2 = max{v, ln+1}. It is much simpler and more economical to use this relaxation technique in branch and bound algorithm than that in [14, 15] where the linear relaxation programming is constructed by utilizing convex and (or) concave envelopes of bilinear functions, and the numerical test will illustrate this.
#### Theorem 3.3
For any xX0, xn+1D, let Δ = |un+1ln+1|, then the following conclusion will be established: $h~i(x)≤hi(x),limΔ→0|h~i(x)−hi(x)|=0.$
#### Proof
The proof of this theorem can be easily verified by the expression of ${\stackrel{~}{h}}_{i}\left(x,{x}_{n+1}\right)$ in (3), therefore, it is omitted here.□
Theorem 2.1 and Theorem 3.3 ensure that ${\stackrel{~}{h}}_{i}\left(x,{x}_{n+1}\right)$ are good underestimations for hi(x, xn+1) and will provide valid lower bounds for the optimal value of the initial problem (MFP).
## 4 Algorithm and its convergence
In this section, we will first present some key operations of the branch and bound algorithm, then the global optimization algorithm based on the former convex relaxation programming will be proposed, and finally, we will prove the global convergence of the present algorithm in theory.
## 4.1 Key operations
There are several key operations in reduced outcome interval branch and bound algorithm. The focus of these operations is to find better upper and lower bounds of the objective value at a faster pase. In our algorithm, there are three base operations: branching, reduction and bounding.
Choosing a suitable partition rule is a critical element in guaranteeing convergence to a global optimal solution, here we choose a simple and adapted bisection rule which is sufficient to ensure convergence. For any subproblem identified by hyper-rectangle $X0×D={(x,xn+1)|x∈X0,ln+1≤xn+1≤un+1}⊂X0×D0,$
let q = (ln+1 + v)/2, then partition X0 × D will be subdividing into two regions ${X}^{0}×\overline{D}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{X}^{0}×\overline{\overline{D}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{where}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\overline{D}=\left[{l}_{n+1},\phantom{\rule{thinmathspace}{0ex}}q\right],\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\overline{\overline{D}}=\left[q,{u}_{n+1}\right].$ Note that we only partition the outcome unidimensional interval D, the n–dimensional variable space is never partitioned in the algorithm, thus the number of partition set is much smaller than in case of those branch and bound algorithms using pure variable space partition rule. What’s more, the adapted partition technique corresponding to the current objective value will also improve the performance of the algorithm.
Moreover, the partition set generated from the branching operation can be further reduced by the special structure of the (EMFP). Actually, based on Theorem 2.1, we know that the EMFP problem has the same optimal value with the MFP problem, so when t < v, or t > v, the optimal value will not be obtained, so the partition sets D = [ln+1, q] and $\overline{\overline{D}}=\left[q,{u}_{n+1}\right]$ can be further reduced into $D¯=[max{v_,ln+1},q]andD¯¯=[q,min{v¯,un+1}],$(6)
respectively. By adding this skill into the branch and bound scheme, the performance of the algorithm will be improved obviously.
The bounding operation aims at obtaining an upper and (or) lower bound of the objective value for the initial problem (MFP). Let LB(Dk, µ) be the optimal value of the (RFP) on the μth sub-region X0 × Dk, μ and xk, μ = x(X0 × Dk, μ) be the corresponding optimal solution in the kth iteration of the algorithm. Then LBk = min{LB(Dk, μ) | μ = 1, 2, ⋯, sk} will be the new lower bound of the objective value for the (MFP), and each optimal solution xk, μ for the (RFP) is a feasible solution for the (MFP). We choose UBk = $min\left\{\underset{\mu }{max}\left\{\frac{{n}_{i}\left({x}^{k,\mu }\right)}{{d}_{i}\left({x}^{k,\mu }\right)}\right\},\phantom{\rule{thinmathspace}{0ex}}\overline{v}\right\}$ as the new best upper bound.
## 4.2 Algorithm statement and convergence property
Based upon the results and key operations given in the above sections, the new branch and bound algorithm for solving the EMFP problem, and hence the MFP problem, can be described as follows.
Step 0. (Initialization) Choose convergence tolerance ϵ > 0, set iteration counter k := 0 and the initial active nod as Ω0 = X0 × D0.
Solve the initial convex relaxation problem (RMFP) over region X0 × D0, if the (RMFP) is not feasible then there is no feasible solution for the (MFP). Else, denote the optimal value and solution as f0 and ${x}_{opt}^{0}$, respectively, then we can obtain the initial upper and lower bound of the optimal value for the MFP problem, that is, v := $f\left({x}_{opt}^{0}\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{_}{v}:={f}_{0},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{where}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}f\left(x\right)=max\left\{\frac{{n}_{i}\left(x\right)}{{d}_{i}\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}i=1,2,\cdots ,\phantom{\rule{thinmathspace}{0ex}}p\right\}.$ And then, if v − v < ϵ, the algorithm can stop, and ${x}_{opt}^{0}$ is the optimal solution of the (MFP), otherwise proceed to step 1.
Step 1. (Reduction) Substitute the interval end points ${l}_{n+1}^{k}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{u}_{n+1}^{k}$ by utilizing the reduction technique (6).
Step 2. (Branching) Partition Dk into two new sub-intervals according to the adapted rule described in section 4.1. Add the new nods into the active nods set and denote the set of new partitioned intervals as ${\stackrel{~}{D}}^{k}$.
Step 3. (Bounding) For each subregion still of interest X0 × Dk, μX0 × D, μ = 1, 2, …, sk, obtain the optimal solution and value for the RMFP problem corresponding outcome interval Dk, μ by solving some convex programming, if LB(Dk, μ) > v, or f(xk, μ) < v then delete Dk, μ from ${\stackrel{~}{D}}^{k}$. Otherwise, let LBk = min{LB(Dk, μ) | μ = 1, 2, ⋯, sk} and UBk = min{f(xk, μ), v}, then we can update the lower and upper bounds as follows $v_=:max{v_,LBk},v_=:min{v¯,UBk},$
and let $Ωk=(Ωk∖X)⋃D~k$
Step 4. (Termination) Let $Ωk+1=Ωk∖{X|f(X)−LB(X)≤ϵ,X∈Ωk}.$
If Ωk+1 = ∅, the algorithm can be stopped, v is the global optimal value for (MFP). Otherwise, set k : = k + 1, select Xk from Ωk with ${X}^{k}=\underset{X\in {\mathrm{\Omega }}_{k}}{\text{argmin}}LB\left(X\right),$ then return to Step 3.
#### Theorem 4.1
The proposed algorithm either terminates within finite iterations with an optimal solution for the MFP is found, or generates an infinite sequence of iterations such that along any infinite branches of the branch-and-bound tree, any accumulation point of the sequence {xk} will be the global optimal solution of the (MFP).
#### Proof
(1) If the proposed algorithm is finite, it will terminate in some iteration k, k ≥ 0. And it can be known, by the termination criteria, that $v¯−v_≤ϵ.$
From Step 0 and Step 3, it implies that $f(xk)−LBk≤ϵ.$
Let vopt be the optimal value of the MFP problem. Then by section 3 and section 4.1, we known that $f(xk)≥vopt≥LBk.$
Hence, taken together, it implies that $vopt+ϵ≥LBk+ϵ≥f(xk)≥vopt.$
And thus the proof of part (1) is completed.
(2) If the algorithm is infinite and generates an infinite feasible solution sequence $\left\{\left({x}^{k},\phantom{\rule{thinmathspace}{0ex}}{x}_{n+1}^{k}\right)\right\}$ via solving the (RMFP). Let ${x}_{n+1}^{k}=f\left({x}^{k}\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{then}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left\{\left({x}^{k},\phantom{\rule{thinmathspace}{0ex}}{x}_{n+1}^{k}\right)\right\}$ is a feasible solution sequence for the (EMFP). Since the sequence {xk} is bounded, it must have accumulations, without loss of generality, assume $\underset{k\to \mathrm{\infty }}{lim}{x}^{k}={x}^{\ast }.$ On the other hand, by the continuity of ni(x) and di(x), we can get $limk→∞xn+1k=limk→∞f(xk)=f(x∗).$(7)
Also, according to the branching regulation described before, we can see that $limk→∞ln+1k=limk→∞un+1k=xn+1∗.$(8)
What’ more, note that ${l}_{n+1}^{k}\le f\left({x}^{k}\right)\le {u}_{n+1}^{k},$ taken (7) and (8) together, we can come to the conclusion that $xn+1∗=limk→∞f(xk)=f(x∗)=limk→∞x¯n+1k.$
Therefore $\left({x}^{\ast },\phantom{\rule{thinmathspace}{0ex}}{x}_{n+1}^{\ast }\right)$ is also a feasible solution for the (EMFP). Further more, since the lower bound sequence LBk for the optimal value is increasing and lower bounded by the optimal value vopt in the algorithm, combining the continuity of xn+1, we have $limk→∞LBk=xn+1∗≤vopt≤limk→∞xn+1k=xn+1∗.$
That is, $\left({x}^{\ast },\phantom{\rule{thinmathspace}{0ex}}{x}_{n+1}^{\ast }\right)$ is an optimal solution for the (EMFP), and of course x* is an optimal solution for the (MFP) according to the equivalence of the problems (MFP) and (EMFP), the proof is completed.□
## 5 Numerical experiments and results
The purpose of this section is to demonstrate the computational performance of our algorithm for the MFP problem. To this end, some numerical experiments in recent works have been carried out on a personal computer containing an Intel Core i5 processor of 2.40 GHz and 4GB of RAM. The code base is written in Matlab2014a and interfaces LINPROG for the linear relaxation problems. The comparison results are listed in Table 1. In addition, to assess the feasibility and stability of the proposed algorithm, a randomly generated test problem is given at the end of this part and the randomised experiment results are demonstrated in Table 2, Table 3, and Fig. 2, Fig. 3. Throughout this section, all problems are solved with relative/absolute optimality tolerance of 1 × 10−6, and we use Iter, Nods and Time to denote the number of iterations, maximum number of nodes stored in memory and the CPU time required in the solving process, respectively.
Figure 2
Result of random experiment 8 comparison with Ref. [13].
Figure 3
Result of random experiment 8 comparison with Ref. [15].
Table 1
Results of the numerical contrast experiments 1-7.
Table 2
Random numerical experiment 8 comparison with Ref. [13].
Table 3
Random numerical experiment 8 comparison with Ref. [15].
#### Example 1
([14]). $minmax{3x1+x2−2x3+0.82x1−x2+x3,4x1−2x2+x37x1+3x2−x3,3x1+2x2−x3+1.9x1−x2+x3,4x1−x2+x38x1+4x2−x3}s.t.x1+x2−x3≤1,−x1+x2−x3≤−1,12x1+5x2+12x3≤34.8,12x1+12x2+7x3≤29.1,−6x1+x2+x3≤−4.1,1.0≤x1≤1.2,0.55≤x2≤0.65,1.35≤x3≤1.45.$
#### Example 2
([15]). $minmax{2.1x1+2.2x2−x3+0.81.1x1−x2+1.2x3,3.1x1−x2+1.3x38.2x1+4.1x2−x3}s.t.x1+x2−x3≤1,−x1+x2−x3≤−1,12x1+5x2+12x3≤40,12x1+12x2+7x3≤50,−6x1+x2+x3≤−2,1.0≤x1≤1.2,0.55≤x2≤0.65,1.35≤x3≤1.45.$
#### Example 3
([13, 15]). $minmax{3x1+4x2−x3+0.52x1−x2+x3+0.5,3x1−x2+3x3+0..59x1+5x2−x3+05,4x1−x2+5x3+0.511x1+6x2−x3,5x1−x2+6x3+0.512x1+7x2−x3+0.9}s.t.x1+x2−x3≤1,−X1+x2−x3≤−1,12x1+5x2+12x3≤42,12x1+12x2+7x3≤55,−6X1+x2+x3≤−3,1.0≤x1≤2,0.50≤x2≤2,0.5≤x3≤2.$
#### Example 4
([13, 15]). $minmax{3x1+4x2−x3+0.92x1−x2+x3+0.5,3x1−x2+3x3+0..59x1+5x2−x3+05,4x1−x2+5x3+0.511x1+6x2−x3+0.9,5x1−x2+6x3+0.512x1+7x2−x3+0.9,6x1−x2+7x3+0.611x1+6x2−x3+0.9}s.t.2x1+x2−x3≤2,−2x1+x2−2x3≤−1,11x1+6x2+12x3≤45,11x1+13x2+6x3≤52,−7x1+x2+x3≤−2,1.0≤x1≤2,0.35≤x2≤0.9,1.0≤x3≤1.55.$
#### Example 5
([13, 15]). $minmax{5x1+4x2−x3+0.93x1−x2+2x3+0.5,3x1−x2+4x3+0.59x1+3x2−x3+0.5,4x1−x2+6x3+0.512x1+7x2−x3+0.9,7x1−x2+7x3+0.511x1+9x2−x3+0.9,7x1−x2+7x3+0.711x1+7x2−x3+0.9}s.t.2x1+2x2−x3≤3,−2x1+x2−3x3≤−1,11x1+7x2+12x3≤47,13x1+13x2+6x3≤56,−6x1+2x2+3x3≤−1,1.0≤x1≤2,0.35≤x2≤0.9,1.0≤x3≤1.55.$
#### Example 6
([13, 15, 18]). $minmax{37x1+73x2+1313x1+13x2+13,63x1−18x2+3913x1+26x2+13}s.t.5x1−3x2=3,1.5≤x1≤3.$
#### Example 7
$minmax{2x12+x22+2x1x2−3x12−2x22+4x1x2+14,x12+x1x2+x22+1−x12+5}s.t.x12+x22≤4,x1+x2≥1,$
#### Example 8
$minmax{2x12+3x226−x22,3x12+x22+4x1x2+0.5−9x12+43}s.t.2x1+3x2≤5,−2x1+x2≤−1,1≤x1≤2,0.35≤x2≤0.9.$
#### Example 9
([13]). $minmax{∑i=1Nn1ixi+n¯1∑i=1Nd1ixi+d¯1,∑i−−1Nn2ixi+n¯2∑i=1Nd2ixi+d¯2,⋯,∑i=1Nnpixi+n¯p∑i=1Ndpixi+d¯p}s.t.Ax≤b,0≤xi≤3,i=1,2,⋯,N.$
In Table 1, we provide the comparative results of our algorithm with some other methods identified in the second column of the table. It can be seen that our algorithm performs much better than other algorithms with respect to efficiency.
In Table 2, the results of randomized test compared with reference [13] are reported. Also we demonstrated the comparative results with a double y-axes broken line graph in Figure 2. From table 2 and Figure 2, it easy to see that our method is much more stable and effective than Jiaos method.
In Table 3, the results of randomized test compared with reference [15] are reported. Also we demonstrated the comparative results with a double y-axes broken line graph in Figure 3. From table 3 and Figure 3, it easy to see that our method performed significantly better than Fengs method in [15].
## 6 Concluding remarks
In this paper, we described an efficient implementation of unidimensional outcome branch and bound algorithm for global optimization of the MFP problem. Instead of using linearization or concave envelope technique, we constructed the concise convex relaxation programming by simple arithmetic operations. This relaxation method is more practical and available for convex-concave ratios problem which generalized the linear case appeared in various works. Moreover, unidimensional adapted interval partition and reduction techniques are incorporated into the algorithm scheme, which can significantly improve the performance of the algorithm. Global convergence is proved and the feasibility and robustness of the algorithm is illustrated by numerical experiments.
## Acknowledgement
The authors would like to express their sincere thanks to the responsible editor and the anonymous referees for their valuable comments and suggestions, which have greatly improved the earlier version of our paper.
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Accepted: 2017-04-03
Published Online: 2017-06-22
Citation Information: Open Mathematics, ISSN (Online) 2391-5455,
Export Citation | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 57, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.947692334651947, "perplexity": 1195.658166463459}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948522205.7/warc/CC-MAIN-20171213065419-20171213085419-00189.warc.gz"} |
https://www.gradesaver.com/textbooks/math/statistics-probability/elementary-statistics-12th-edition/chapter-10-correlation-and-regression-10-2-correlation-basic-skills-and-concepts-page-511/8 | ## Elementary Statistics (12th Edition)
r (linear correlation coefficient) is given: r=0.7654038409. Hence the value of the test statistic: $\frac{r}{\sqrt{(1-r^2)/(n-2)}}=\frac{0.7654038409}{\sqrt{(1-0.7654038409^2)/(16-2)}}=4.45.$ Using the table, the corresponding P value with df=16-2=14: P is less than 0.01. If the P-value is less than the significance level, then this means the rejection of the null hypothesis. Hence:P is less than 0.05, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to support that there is a linear correlation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9617552161216736, "perplexity": 303.1837154299186}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376829429.94/warc/CC-MAIN-20181218143757-20181218165757-00189.warc.gz"} |
https://www.physicsforums.com/threads/convert-radio-frequency-to-temperature.199273/ | # Convert radio frequency to temperature
1. Nov 19, 2007
### spune
Hello,
Is there a formula to convert a certain radio frequency to temperature? Thanks.
2. Nov 19, 2007
### mgb_phys
What exactly do you mean?
If you mean the temperature of a black that emits a frequency in the radio then use wein's law.
3. Nov 19, 2007
### spune
for example, the FM station abc operates at 86.1mhz, what's the temperature is the radio wave emitting?
i think i have found the formula though.
http://en.wikipedia.org/wiki/Wien's_displacement_law
do i use the frequency form formula?
fmax = (ak/h)(T)
4. Nov 20, 2007
### Staff: Mentor
The spectrum of a FM broadcast is not very similar to the spectrum of blackbody radiation at any temperature. I don't think the conversion makes sense. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9465495347976685, "perplexity": 2599.65131930387}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946597.90/warc/CC-MAIN-20180424100356-20180424120356-00243.warc.gz"} |
https://cris.tau.ac.il/en/publications/boosting-as-a-regularized-path-to-a-maximum-margin-classifier | Boosting as a regularized path to a maximum margin classifier
Saharon Rosset, Ji Zhu, Trevor Hastie
Research output: Contribution to journalArticlepeer-review
Abstract
In this paper we study boosting methods from a new perspective. We build on recent work by Efron et al. to show that boosting approximately (and in some cases exactly) minimizes its loss criterion with an l1 constraint on the coefficient vector. This helps understand the success of boosting with early stopping as regularized fitting of the loss criterion. For the two most commonly used criteria (exponential and binomial log-likelihood), we further show that as the constraint is relaxed- or equivalently as the boosting iterations proceed-the solution converges (in the separable case) to an "l1-optimal" separating hyper-plane. We prove that this l1-optimal separating hyper-plane has the property of maximizing the minimal l1-margin of the training data, as defined in the boosting literature. An interesting fundamental similarity between boosting and kernel support vector machines emerges, as both can be described as methods for regularized optimization in high-dimensional predictor space, using a computational trick to make the calculation practical, and converging to margin-maximizing solutions. While this statement describes SVMs exactly, it applies to boosting only approximately.
Original language English 941-973 33 Journal of Machine Learning Research 5 Published - 1 Aug 2004 Yes
Keywords
• Boosting
• Margin maximization
• Regularized optimization
• Support vector machines
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Dive into the research topics of 'Boosting as a regularized path to a maximum margin classifier'. Together they form a unique fingerprint. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9437527060508728, "perplexity": 2154.780936158468}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573172.64/warc/CC-MAIN-20220818063910-20220818093910-00150.warc.gz"} |
http://mathhelpforum.com/discrete-math/22494-lost-again.html | 1. ## lost again
Give a formula for the coefficient of x^k in the expansion of (x^2 - 1/x)^100, where k is an integer. ( x^8 means x to the power of 8, ^ means to the power of).
2. Originally Posted by lkrb
Give a formula for the coefficient of x^k in the expansion of (x^2 - 1/x)^100, where k is an integer. ( x^8 means x to the power of 8, ^ means to the power of).
Use the binomial theorem:
$(a + b)^n = \sum_{k = 0}^n {n \choose k} a^{n-k}b^k$
So here we have $a = x^2$ and $b = \frac{1}{x}$. You do the rest.
-Dan | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9000263810157776, "perplexity": 707.5852393711282}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982939917.96/warc/CC-MAIN-20160823200859-00026-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/galaxy-rotation-and-kepler-law.245951/ | # Galaxy rotation and Kepler law
1. ### pixel01
691
According to Kepler third law, the ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun.
If I can apply this to the rotaion of galaxy, meaning stars in inner part will orbit much faster than the outer ones. But it seems not.
Anyonoe please explain to me this.
Thanks.
2. ### tony873004
1,562
That's because the Sun is a fixed mass at the center of our solar system. But as you get further from the galactic core, there are more stars interior to you, adding to the mass that you would use to compute your orbital velocity.
Technically, this happens in the solar system. Earth orbits the combined mass of the Sun, Mercury and Venus. So it orbits a little faster than it would if Mercury and Venus did not exist. But since Mercury and Venus are insignificant compared to the Sun, their effect is negligible. But in the galaxy, the additional mass interior to you as you move out is not negligible.
3. ### WhyIsItSo
185
This is interesting in that I have been researching this topic throughout the day today. It seems that in fact after a certain distance from a galactic core, orbital velocities of stars become fairly constant in apparent contradiction of your statement.
The reason appears to be the presence of vast amounts of dark matter surrounding galaxies; the gravitational effect being the explanation for the faster-than-expected orbital velocities of outer systems. My reading suggests that the ratio of dark matter to luminescent matter is in the ballpark of 100:1.
4. ### O-r-i-o-n
10
As you know Kepler's laws are true for simple garvitional fields , but talking about galaxies and stars in it , the gravity isn't the same as the gravity between two "very small" objects.
To sum up what Tony873004 and WhyIsItSo said , in situations of this kind the force is: -Kr instead of -K/r2!
just for Mg use 4$$\pi$$r2$$\rho$$ to get what I'm saying
Thanks a lot! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9256817698478699, "perplexity": 666.2935299289704}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207925201.39/warc/CC-MAIN-20150521113205-00179-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://www.basicsinmaths.com/tag/inter/ | # Inter
## 1. DEPRECIATION
DEPRECIATION: It is a permanent, continuous and gradual shrinking in the book value of a fixed asset.
DEPLETION: It refers to the physical deterioration by the exhaustion of natural resources. (e.g. ore deposits in mines, oil wells, quarries, etc.,)
FIXED INSTAL MENT METHOD: A method under which the depreciation provided annually on the fixed asset remains the same throughout the life span of the asset.
NONCASH EXPENSE: Expenses that may be operational in nature but that do not affect the payment of cash (e.g. depreciation).
OBSOLESCENCE: diminution in the value of fixed assets due to new inventions, new improvement, change in fashions, change in customer’s tastes and preferences.
RESIDUAL VALUE (scrap value): The realisable value of fixed assets after the expiry of its estimated economic life.
SINKING FUND: A fund created for the repayment of along-term liability or the replacement of an asset at a set date in the future.
DEPRECIATIONFUND: A fund created for the replacement of an asset at a set date in the future.
WRITTEN DOWN VALUE: The value of a fixed asset after depreciation.
WRITTEN DOWN VALUE METHOD: A method under which depreciation is calculated at a fixed percentage on the original cost of the asset in the first year and on written down value in the subsequent year.
DOUBLE ENTRY SYSTEM: The accounting system of recording both the receiving (debit) and giving (credit) aspects of a business transaction is called a double-entry system.
SINGLE ENTRY SYSTEM: It is a mixture of a double-entry, single entry and no entry. It is an incomplete double-entry system.
STATEMENT AFFAIRS: To find out capital, a statement showing various assets and liabilities of a business concern is prepared on a particular date, which is called a statement of affairs. It is similar to a balance sheet.
CAPITAL COMPARISON METHOD: Under this method, the profit/ loss for a particular period is ascertained by comparing the closing capital of the opening capital.
### Meaning of depreciation
Depreciation means a fall in the value or quality of an asset. The word depreciation is derived from the Latin word “depretium”. ‘De’ means decline and ‘pretium’ means price. it is the decline in the price or value of the fixed assets. Depreciation is described as a permanent, continuing and gradual shrinkage in the value of fixed assets. It is based on the cost of the asset consumed in a business and not on its market value.
The depreciation is that part of the original cost of a fixed asset that is consumed during its period of use by the business. Thus, depreciation is the loss of value of a fixed asset arising from use, effluxion of time or obsolescence. Depreciation sometimes restricted to fixed tangible assets but in the UK, it also usually includes the amortization of intangible assets. However, the tangible fixed assets lose their value over a period of time as they are used in the business operation and they do not last forever. If any amount is received on the sale of the fixed asset is deducted from the cost of it. Then the remaining value of the fixed asset is said to have” depreciated value” by that amount over its period of usefulness to the business.
For example, if a motor vehicle was bought for Rs 100000 and it was sold 5 years later for Rs 10000. It means the motor van value has depreciated over the period of its use by Rs 90000. Therefore, the depreciation for each year will be estimated while the fixed asset continues to be useful and be charged(debited) to the profit and loss account. Since depreciation is treated as a business expense and charged to the profit and loss account, it reduces the net profit of the business and the value of the asset.
### DEFINITIONS OF DEPRECIATION
According to the American institute of certified public accountants (AICPA-USA),” depreciation accounting is a system of accounting while aims to distribute the cost or other basic value of a tangible capital asset, less salvage value (if any) over the estimated useful life of the asset (which may be a group of assets) in a systematic and rational manner .it is a process of allocation, not of valuation.”
##### SIGNIFICATION OR NEED FOR PROVIDING DEPRECIATION
1. To ascertain true results of business operations: – As depreciation is treated as expenditure, it must be charged to profit and loss account against income to assets the real profit /loss of the business.
2. To show the fixed asset at their original worth in the balance sheet: -If depreciation is not provided, the value of the asset shown in the balance sheet is not correct. Hence, the real value of the asset must be shown in the balance sheet after deducting depreciation from its book value.
For example, machinery is purchased for ₹10,000, and its estimated working life is 10 years.
Then, the depreciation to be charged to the asset every year will be 10,000/10 years =1000. Though the value of the asset is depreciated by ₹1000 every year, if it is shown as ₹ 10,000 in the balance sheet, it will not reflect the real financial position of the business.
1. To provide funds for the replacement of asset: – The amount charged as depreciation is nothing but setting aside a profit and this amount accumulates and will be available for replacement of the asset at the end of its useful life.
2. To ascertain the true cost of production: -for calculating the cost of production, it is necessary to charge depreciation as an item of cost of production. Otherwise, it would not present true cost of production.
3. To follow the legal provisions: – in the case of a joint-stock company, it is compulsory to provide depreciation on fixed assets and without providing for depreciation on fixed assets and without providing for depreciation, dividends cannot be declared to the shareholders.
#### CAUSES OF DEPRECIATION
1. Wear and tear: – The decrease in the value of the fixed asset due to the constant use is said to be wear and tear. It is a deterioration in an asset value, arising from its use in business operations for earnings income.
Example: Machinery, Furniture and fixtures are depreciated by wear and tear.
2 . Depletion: – It is the loss of natural resources mineral wealth due to the constant extraction of raw material from mines, quarries, oil wells etc.
3. Accidents: – The assets lose their value if it is met with an accident. An accident means the breakdown of an asset which decreases the value of assets. It is not a gradual decrease but causes a permanent loss in asset value.
4. Obsolescence: – it is the processing out of date. due to new inventions, technological improvement, availability of the better type of asset, or change in market demand the existing asset becomes obsolete or outmoded. This results in a decrease in the value of the asset.
5. Fluctuations: – the asset value may get decreased due to the fluctuations/charges in the market conditions.
# This note is designed by the ‘Basics in Maths’ team. These notes to do help the TS intermediate second-year Maths students fall in love with mathematics and overcome the fear.
## 1. CIRCLES
Circle: In a plane, the set of points that are at a constant distance from a fixed point is called a circle.
∗ The fixed point is called the centre (C) of the circle and the constant distance is called the radius(r) of the circle
Unit circle: If the radius of the circle is 1 unit, then that circle is called the unit circle.
Point Circle: A circle is said to be a point circle if its radius is zero. A point circle contains only one point in the centre of the circle. •
∗ The equation of the circle with centre (h, k) and radius r is
(x – h)2 + (y – k)2 = r2
∗ The equation of the circle with centre origin and radius r is
x2 + y2 = r2
⇒ x2 + y2 = r2 is called standard form of the circle.
The general equation of the second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, where a, b, f, g, h and c are real numbers, represent a circle iff (i) a = b ≠ 0 (ii) h = 0 and (iii) g2 + f2 + c ≥ 0
∗ The general equation of the circle is x2 + y2 + 2gx + 2fy + c = 0
It’s centre c = (– g, – f) and radius
∗ The equation of the circle passing through origin is x2 + y2 + 2gx + 2fy = 0.
∗ The equation of the circle whose centre on the x-axis is x2 + y2 + 2gx + c = 0.
∗ The equation of the circle having centre on y-axis is x2 + y2 + 2fy + c = 0.
∗ The circles which have the same centre are called concentric circles.
∗ The equation of the circle concentric with the circle x2 + y2 + 2gx + 2fy + c = 0 is
x2 + y2 + 2gx + 2fy + k = 0.
∗ The length of the intercept made by a circle x2 + y2 + 2gx + 2fy + c = 0 on
• x -axis is if g2 – c > 0
• y -axis is if f2 – c > 0
Note: –
(a) if g2 – c = 0, then A1 A2 = 0 ⇒ the circle touches the x- axis at only one point.
(b) if f2 – c = 0, then B1 B2 = 0 ⇒ the circle touches the y- axis at only one point.
(c) if g2 – c < 0, then the circle does not meet the x- axis.
(d) if f2 – c < 0, then the circle does not meet the y- axis.
∗ The equation of the circle having the line segment joining A (x1, y1) and B (x2, y2) as a diameter is
(x – x1) (x – x2) + (y – y1) (y – y2) = 0.
∗ Let A, B be any two points on a circle then,
• The line is called the secant line of the circle.
• The line segment is called the chard of the circle.
• AB is called the length of the chord.
∗ A chord passing through the centre is called the diameter of the circle.
∗ The angle subtended by a chord on the circumference of at any point is equal.
The perpendicular bisector of a chord of a circle is asses through the centre of the circle.
∗ The angle in a semicircle is 900.
∗ The equation of the circle passing through three non-collinear points A (x1, y1), B (x2, y2), C (x3, y3) is
Where ci = − (x2 + y2) and i = 1,2,3
∗ centre of the circle is
#### Parametric form:
If P (x, y) is a point on the circle with centre (h, k) and radius r, then
X = h + r cosθ, y = k + r sinθ 0 ≤ θ ≤ 2π.
⇒ A point n the circle x2 + y2 = r2 is taken as (r cosθ, r sinθ) and simply denoted by θ.
Note:
1. If the centre of the circle is the origin, then the parametric equations are x = r cosθ, y = r, 0 ≤ θ ≤ 2π.
2. The point (h + rcosθ1, k + r sin θ1) is referred to as the point θ1 on the circle having the centre (h, k) and radius r.
Notations:
S = x2 + y2 + 2gx + 2fy + c
S1 = xx1 + yy1 + g(x +x1) + f (y +y1) + c
S11 = x12 + y12 + 2gx1 +2fy1 + c
S12 = x1x2 + y1y2 + g(x1 + x2 ) + f (y1 + y2) + c
Position of a point with respect to the circle:
A circle divides the plane into three parts.
1. The interior of the circle
2. The circumference which is the circular curve.
3. The exterior of the circle.
Power of point:
Les S = 0 be a circle with radius ‘r’ and centre ‘C’ and P (x1, y1) be a point on the circle, then CP – r2 is called the power of point ‘P’ concerning S = 0.
• The power of point P (x1, y1) w.r.t. S = 0 is S11.
•Let S = 0 be a circle in a plane and P (x1, y1) be any point in the same plane. then
1. P lies in the interior of the circle ⇔ S11 < 0.
2. P lies on the circle ⇔ S11 = 0.
3. Plies in the exterior of the circle ⇔ S11 > 0.
Secant and tangent of a circle:
Let P be any point on the circle and Q be neighbourhood point of P lying on the circle. join P and Q, then the line PQ is the secant line.
The limiting position of the secant line PQ when Q is approached to the point P along the circle is called a tangent to the circle at P.
Length of the tangent:
If P is any point on the circle S = 0 and T is any exterior point of the circle, then PT is called the length of the tangent.
∗ If S = 0 is a circle and P (x1, y1) is an exterior point with respect o S = 0, then the length of the tangent from P (x1, y1) to S =0 is
Condition for a line to be a tangent:
• A straight-line y = mx + c (i) meet the circle x2 + y 2 = r2 in two distinct points if
• Touch the circle x2 + y 2 = rif
• Does not touch the circle x2 + y 2 = r2 in two distinct points if
Note:
1. For all real values of m, the straight line is a tangent to the circle x2 + y2 = r2 and the slope of the line is m.
2. A straight-line y = mx + c is a tangent to the circle x2 + y2 = r2 if c = .
3. The equation of a tangent to the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0 having the slope m is where r is the radius of the circle.
4. The circle S ≡ x2 + y2 + 2gx + 2fy + c = 0 touches (i) x – axis if g2 = c (ii) y – axis if f2= c.
Chord joining two points on a circle:
If P (x1, y1) and Q (x2, y2) are two points on the circle S = 0 then the equation of secant line PQ is S1 + S2 = S12.
Equation of tangent at a point on the circle:
The equation of the tangent at the point (x1, y1) to the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0 is S1 = 0.
The equation of the tangent at the point (x1, y1) to the circle x2 + y2 = r2 is xx1 + yy1 – r2 = 0.
Point of contact:
If a straight-line lx + my + n = 0 touches the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0 at P (x1, y1), then this line is the tangent to the circle S = 0. And the equation of the tangent is
(x1 + g) x + (y1 +f) y + (gx1 + fy1 + c) = 0.
∗ The equation of the chord joining two points θ1, θ2 on the circle x2 + y2 + 2gx + 2fy + c = 0 is
∗ The equation of the chord joining the points θ1, θ2 on the circle x2 + y2 = r2 is
∗ The equation of the tangent at P(θ) on the circle x2 + y2 + 2gx + 2fy + c = 0 is
∗ The equation of the tangent at P(θ) on the circle x2 + y2 = r2 is
x cosθ + y sinθ = r.
Normal:
The normal at any point P of the circle is the line which is passing through P and is perpendicular to the tangent at P.
• The equation of the normal at P (x1, y1) of the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0 is
(x – x1) (y1 + g) – (y – y1) (x1 + g) = 0.
• The equation of the normal at P (x1, y1) of the circle x2 + y 2 = r2 is xy1 – yx1 = 0.
Chord of contact and Polar:
∗ If P (x1, y1) is an external point of the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0, then there exists two tangents from P to the circle S = 0.
Chord of contact: –
If the tangents are drawn through P (x1, y1)
to a circle S = 0 touch the circle at points A and B then the secant line AB is called the chord of contact of P with respect to S = 0
∗ If P (x1, y1) is an external point of the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0, then the equation of the chord of contact of P with respect to S =0 is S1 = 0.
Note:
1. If the point P (x1, y1) is on the circle S = 0, then tangent itself can be defined as the chord of contact.
2. If the point P (x1, y1) s an interior point of the circle S = 0, then the chord of contact does not exist.
Pole and Polar: –
Let S = 0 be a circle and P be any point if any line drawn through the point Pin the plane other than the centre of S = 0.then the points of intersection meets the circle in two points A and B, of tangents drawn at A and B lie on a line called polar of P and P, is called Pole of polar.
∗ The equation of the polar of P (x1, y1) with respect to the circle S = 0 is S1 = 0.
Note: –
1. If Plies outside the circle S = 0, then polar of P meets the circle in two points and the polar becomes the chord f contact of P.
2. If P lies on the circle S = 0, then the polar P becomes the tangent at P o the circle.
3. If P lies inside the circle S = 0, then the polar of P does not meet the circle.
4. If P is the centre of the circle S = 0, then the polar of P does not exist.
5. The pole of the line lx + my + n = 0 with respect to the circle x2 + y2 = r2 is
6. The pole of the line lx + my + n = 0 with respect to the circle x2 + y2 + 2gx + 2fy + c = 0 is
7. The polar of P (x1, y1) with respect to the circle S = 0 passes through Q (x2, y2) ⟺ polar of Q passes through P.
Conjugate points: Two P and Q are said to be conjugate points with respect to the circle S = 0, if polar of P with respect to S = 0 passes through Q.
⇒ The condition for the points P (x1, y1), Q (x2, y2) to be conjugate with respect to the circle S = 0 is S12 = 0.
Conjugate lines: Two lines L1 = 0 and L2 = 0 are said to conjugate lines with respect to the circle S = 0 if the pole of L1 = 0 is lies on L2 = 0.
The condition for the lines l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0 to be conjugate with respect to the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0 is
r2 (l1l2 + m1m2) = (l1g + m1f – n1) (l2g + m2f – n2)
⟹ The condition for the lines l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0 to be conjugate with respect to the circle S ≡ x2 + y2 = r2 is r2 (l1l2 + m1m2) = n1n2
Inverse points: Let S = 0 be a circle with centre C and radius r. two points P and Q are said to be inverse points with respect to the circle S = 0 if
1. C, P, Q are collinear.
2. P, Q lies on the same side of C.
3. CQ = r2.
⟹ If lies inside of the circle S = 0, then Q lies outside of the circle.
⟹ If P lies on the circle S = 0, then P =Q.
⟹ Let S = 0 be a circle with centre C and radius r. The polar of P meets the line CP in Q iff P, Q are inverse points.
⟹ f P, Q are inverse points with respect to S = 0, then P, Q are conjugate points with respect to the circle S = 0.
⟹ If P, Q are inverse points with respect to S = 0, then Q is the foot of the perpendicular from P on the polar of P with respect to the circle S = 0.
Equation of the chord with the given middle point:
The equation of the chord of the circle S = 0 having P (x1, y1) as its midpoint is S1 = S11.
Common tangents to the circle:
⟹ A straight line L = 0 is said to be common tangent to the circle S = 0 and S= 0 if it is a tangent to both S = 0 and S’ = 0.
Two circles are said to be touch each other if they have only one common tangent.
The relative position of two circles:
Let C1, C2 centres and r1, r2 be the radii of two circles S = 0 and S’ = 0respectively.
1.If C1C2 > r1+ r2, then two circles do not intersect.
⟹2 direct common tangents and
2 transverse common tangents
Total 4 common tangents
⟹P divides C1C2 in the ratio r1: r2 internally
Here P is called the internal centre of similitude (I.C.S)
⟹ Q divides C1C2 in the ratio r1: r2 externally
Here Q is called the external centre of similitude (E.C.S)
2.If C1C2 = r1+ r2, two circles touch each other
⟹ Q divides C1C2 in the ratio r1: r2 externally
⟹ two direct common tangents and one common tangent. Total 3 tangents
Here Q is called the external centre of similitude (E.C.S)
3.
⟹ two direct common tangents
Here Q is called the external centre of similitude (E.C.S)
Q divides C1C2 in the ratio r1: r2 externally
⟹ internal centre of similitude does not exist.
4.
⟹ only one common tangent
internal centre of similitude does not exist.
5.
no. of common tangents zero.
Note: the combined equation of the pair of tangents drawn from an external point P (x1, y1) to the circle S = 0 is S S11 = S12.
## 2.SYSTEM OF CIRCLES
A set of circles is said to be a system of circles if it contains at least two circles.
The angle between two intersecting circles:
If two circles S = 0 and S’ = 0 intersect at P then the angle between the tangents of two circles at P is called angle between the circles at P.
⟹ If two circles S = 0 and S’ = 0 intersect at P and Q then the angle between the tangents of two circles at P and Q are equal.
⟹ If d is the distance between the centres of the two intersecting circles with radii r1, r2 and θ is the angle between the circles then.
⟹ If θ is the angle between the circles x2 + y2 + 2gx + 2fy + c = 0 and x2 + y2 + 2g’x + 2f’y + c’ = 0 then
⟹ Two intersecting lines are said to be Orthogonal if the angle between the circles is a right angle.
Condition for the orthogonality:
⟹ The condition that the two circles x2 + y2 + 2gx + 2fy + c = 0 and x2 + y2 + 2g’x + 2f’y + c’ = 0 cut each other orthogonally is 2gg’ + 2ff’ = c + c’.
⟹ If d is the distance between the centres of the two intersecting circles with radii r1, r2. Two circles cut orthogonally if d2 = r12 + r22.
∎ If S = 0, S’ = 0 are two circles intersecting at two distinct points, then S – S’ = 0 represents a common chord of these two circles.
∎ If S = 0, S’ = 0 are two circles touch each other, then S – S’ = 0 represents a common tangent of these two circles.
∎ If S ≡ x2 + y2 + 2gx + 2fy + c = 0 and L ≡ lx + my + n = 0 are the equation of the circle and a line respectively intersecting each other, then S + λ L = 0 represent a circle passing through the point intersection of S = 0 and L = 0 ∀ λ ∈ ℛ.
The radical axis of two circles s defined as the locus of the point which moves so that its powers with respect to the two circles are equal.
(OR)
The locus of a point, for which the powers with respect to given non-concentric circles are equal, is a straight line is called Radical axis of the given circles.
∎ The equation of Radical axis f the circles S = 0 and S’ = 0 is S – S’ = 0.
The radical axis of any two circles is perpendicular to the line joining their centres.
The lengths of tangents from a point on the radical axis of two circles are equal if exist.
Radical axis of two circles bisects all common tangents of the two circles.
∎ If the centres of any three circles are non-collinear then the radical axis of each pair of circles chosen from these three circles re concurrent.
Radical centre: The point of concurrence of radical axes of each pair of three circles is called radical centre (see above figure).
∎ If the circle S = 0 cuts the each of the two circle S’ = 0 and S’’ =0 orthogonally then the centre of S =0 lies on the radical axis of S’ = 0 and S’’ = 0.
∎ Radical axis of two circles is
• The ’common chord’ if the two circles intersect at two distinct points.
• The ‘common tangent’ at the point of contact if the two circles touch each other.
The radical axis of any two circles bisects the line joining the points of contact of common tangents to the circles.
Let S = 0, S’ = 0 and S’’ =0 be three circles whose centres are non- collinear and no two circles of these are intersecting then the circles having
• Radical centre of these circles as the centre of the circle.
• Length of the tangent from the radical centre to any one of these three circles cuts the given three circles orthogonally.
### CONIC SECTIONS
Conic: The locus of a point moving on a plane such that its distance from a fixed point and a fixed straight line is in the constant ratio is called Conic.
OR
The locus of a point moving on a plane such that its distance from a fixed point and a fixed line on the plane are in a constant ratio ‘e’, is called a Conic.
Focus: The fixed point is called focus and it is denoted by S.
Directrix: The fixed straight line is called the directrix.
Eccentricity: The constant ratio is called eccentricity and it is denoted by ‘e’.
Conic is the locus of a point P moving on a plane such that SP/PM = e, PM is the perpendicular distance from P to directrix at M.
If e = 1, then the conic is parabola.
if 0 < e < 1, then the conic is Ellipse.
if e > 1, then the conic is Hyperbola.
## 3.PARABOLA
If e = 1, then the conic is parabola.
• The standard form of parabola is y2 = 4ax.
• Focus S = (a. 0).
• Equation of directrix is x + a = 0.
• Vertex A = (0, 0) and A is the mid-point of SZ.
• Equation of the parabola with focus (α, β) and directrix lx + my + n = 0 is
• If the focus is situated on the left side of the directrix, the equation of the parabola with vertex as the origin and the axis is X-axis is y2 = – 4ax.
• The vertex being the origin, if the axis of the parabola is taken as Y – axis, equation of the parabola is x2 = 4 ay or x2 = – 4 ay according to the focus is above or below the X-axis.
Nature of the curve:
The nature of the parabola f the equation y2 = 4 ax (a>0)
• F y = 0, then 4 ax = 0 and x = 0
∴ the curve passes through the origin.
• If x = 0, then y2 = 0. Which gives y = 0. Y – axis is the tangent to the parabola at origin.
• Let P(x, y) be any point on the parabola (a>0) and y2 = 4 ax, we have x ≥ 0 and
∴ for any positive real value of x, we obtain two value of y of equal magnitude but opposite in sign. This shows that the curve is symmetric about X-axis and lies in the first and fourth quadrants.
The curve does not exist on the left side of the Y-axis since x ≥ 0 for any point (x, y) on the parabola.
Chord: The line segment joining two points on a parabola is called a chord.
Focal chord: A chord which is passing through focus is called Focal Chord.
Double ordinate: A chord through a point P on the parabola, which is perpendicular to the axis of the parabola is called Double ordinate.
Latus rectum: The double ordinate passing through the focus is called Latus rectum.
⟹ Length of Latus rectum = 4a.
Various forms of the parabola
1. y2 = 4ax
focus: (a, 0)
equation of directrix: x + a = 0
axis of parabola: y = 0
vertex: (0. 0)
2. y2 = −4ax
equation of directrix: x − a = 0
focus: (−a, 0)
axis of parabola: y = 0
vertex: (0. 0)
3. x2 = 4ay
focus: (0, a)
equation of directrix: y + a = 0
axis of parabola: x = 0
vertex: (0. 0)
4. x2 = −4ax
focus: (0, −a)
equation of directrix: y − a = 0
vertex: (0. 0) axis of parabola: x = 0
5. (y – k) 2 = 4a (x – h)
focus: (h + a, k)
equation of directrix: x – h + a = 0
axis of parabola: y – k = 0
vertex: (h. k)
6. (y – k) 2 = −4a (x – h)
focus: (h – a, k)
equation of directrix: x – h – a = 0
axis of parabola: y – k = 0
vertex: (h. k)
7. (x – h) 2 = 4a (y – k)
focus: (h, k + a)
equation of directrix: y – k + a = 0
axis of parabola: x – h = 0
vertex: (h. k)
8. (x – h)2 = −4a (y – k)
focus: (h, k – a)
equation of directrix: y – k – a = 0
vertex: (h. k)
axis of parabola: x – h =0
9.
focus: (α, β)
equation of directrix: lx + my + n = 0
axis of parabola: m (x – α) – l (y – β) = 0
vertex: A
Note:
1. Equation of the parabola whose axis parallel to X – axis is x = ly2 + my + n.
2. Equation of the parabola whose axis parallel to Y – axis is y = lx2 + mx + n.
Focal distance: The distance of a point on the parabola from its focus is called Focal distance.
⟹ Focal distance of parabola s x1 + a
Parametric equations of a parabola:
The point (at2, 2at) satisfy the equation y2 = 4ax, the parametric equations of parabola are x = at2, y = 2at. The point P(at2, 2at) is generally denoted by the point ‘t’ or P(t).
Notation:
1. S ≡ y2 – 4 ax
2. S1 ≡yy1 – 2a (x + x1)
3. S12 ≡ y1y2 – 2a (x1 + x2)
4. S11 ≡ y12 – 4 ax1
Equation of a tangent and normal at a point on the parabola:
∎ y = mx + c is a tangent to the parabola y2 = 4ax, then c = a/m or a =cm, and the point of contact is (a/m2, 2a/m).
∎ if m = 0, the line y = c is parallel to the axis of the parabola (i.e., x – axis)
y = c ⟹ c2 = 4ax ⟹ x = c2 /4a
∴ point of contact is (c2/4a, c).
∎ if m ≠ 0 and c = 0, then
Y = mx ⟹ x = 4a/m2 and y = 4a/m
∴ point of contact is (4a/m2, 4a/m).
∎ The equation of the chord joining the points (x1, y1) and (x2, y2) is S1 + S2 = S12.
∎ The equation of the tangent at P (x1, y1) to the parabola S = 0 is S1= 0.
∎ The equation of the normal at P (x1, y1) is (y – y1) = – y1/2a (x – x1).
Parametric form:
∎ The equation of the tangent at a point ‘t’ on the parabola y2 = 4ax is x – yt + at2 = 0.
∎ Equation of the normal at a point ‘t’ on the parabola y2 = 4ax is y + xt = 2at + at3.
∎ The condition for the straight-line lx + my + n = 0 to be a tangent to the parabola y2 = 4 ax is
am2 = nl and point of contact is (n/l, –2 am/l).
∎ common tangent to the parabolas y2 = 4 ax and x2 = 4 by is x a1/3 + y b1/3 + a2/3 b2/3 = 0.
∎ The equation of the chord of contact of the external point (x1, y1) w.r. t parabola S = 0 is S1 = 0.
∎ The equation of the polar of the point (x1, y1) w.r. t parabola S = 0 is S1 = 0.
∎ The pole of the line lx + my + n = 0 w.r.t. parabola y2 = 4ax is (n/l, -2am/l).
∎ If two points P (x1, y1), Q (x2, y2) are conjugate points w.r.t. parabola S = 0, then S12 = 0.
∎ The lines l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0 are conjugate lines with respect to the parabola y2 = 4 ax, then l1n2 + l2n1 = 2a m1m2.
## 4.ELLIPSE
Ellipse: A conic with eccentricity less than unity s called Ellipse.
∎ Equation of Ellipse in standard form is
⇒b2 = a2 (1 – e2) ⇒ e2 – 1 = -b2/a2
Major and Minor axis:
⟹ The line segment AA’ and BB’ of length 2a and 2b respectively are axes of the Ellipse.
⟹ If a > b AA’ is called Major axis and BB’ is called Minor axis and vice-versa if a<b.
Various form of Ellipse:
1.
Major axis: along the x-axis
Length of Major axis:2a
Minor axis: along y – axis
Length of Minor Axis:2b
Centre: (0, 0)
Foci: S = (ae, 0) and S’ = (–ae, 0)
Equation of directrices: x = a/e and x = –a/e
Eccentricty:
2.
Major axis: along y – axis
Length of Major axis:2b
Minor axis: along x – axis
Length of Minor Axis:2a
Centre: (0, 0)
Foci: S = (0, be) and S’ = (0, –be)
Equation of directrices: x = b/e and x = –b/e
Eccentricty:
Centre not at the origin
3.
Major axis: along with y = k
Length of Major axis:2a
Minor axis: along x = h
Length of Minor Axis:2b
Centre: (h, k)
Foci: S = (h +ae, k) and S’ = (h – ae, k)
Equation of directrices: x = h + a/e and x = h – a/e
Eccentricty:
4.
Major axis: along x = h
Length of Major axis:2b
Minor axis: along with y = k
Length of Minor Axis:2a
Centre: (h, k)
Foci: S = (h, k + be) and S’ = (h, k – be)
Equation of directrices: xy = k + b/e and y = k – b/e
Eccentricty:
Chord: The line segment joining two points on a parabola is called a chord of Ellipse.
Focal chord: A chord which is passing through one of the foci is called Focal Chord.
Latus rectum: A focal chord perpendicular to the major axis of the Ellipse is called Latus Rectum. Ellipse has two latera recta.
Length of the Latus rectum:
1.The coordinates of the four ends of the latera recta of the ellipse
L = (ae, b2/a), L’ = (ae, -b2/a) and L1 = (-ae, b2/a), L1’= (-ae, -b2/a).
length of the Latus rectum = 2b2/a.
2.length of the Latus rectum of an ellipse is 2a2/b and the coordinates of the four ends of the latera recta are
L = (a2/b, be), L’ = (-a2/b, be) and L1 = (a2/b, -be), L1’ = (-a2/b, -be).
3. The equation of the Latus rectum of the Ellipse through S is x = ae and through S’ is x = -ae.
4. The equation of the Latus rectum of the Ellipse through S is y = be and through S’ is y = -be.
5. If P (x, y) is any point on the Ellipse whose foci are S and S’, then SP +S’P is constant.
Auxiliary circle: The circle described on the major axis of an Ellipse as the diameter is called the Auxiliary circle of the Ellipse. The Auxiliary circle of the Ellipse is x2 + y2 = a2.
Parametric equations: The parametric equations of the Ellipse are x = a cosθ and y = b sinθ.
Notation:
Equation of Tangent and Normal
The equation of any tangent to the Ellipse can be written as
The condition for a straight-line y = mx + c to be a tangent to the Ellipse is c2 = am2 + b2.
∎ The point of contact of two parallel tangents to the Ellipse are (-a2m/c, b2/c) and (a2m/c, -b2/c)
∎ The equation of the chord joining two points (x1, y1) and (x2, y2) on the Ellipse S = 0 is S1 + S2 = S12.
∎ The equation of the Normal at P (x1, y1) to the Ellipse is
∎ Equation of the tangent at P(θ) on the Ellipse
∎ Equation of the normal at P(θ) on the Ellipse S = 0 is
∎ When θ = 0, π; equation of Normal is y =0.
∎ When θ = π/2, 3π/2; equation of Normal is x =0.
∎ The condition for the line lx + my + n = 0 to be a tangent the Ellipse S = 0 is a2l2 + b2m2 = n2.
∎ The condition for the line x cosα + y sinα = p to be a tangent the Ellipse S = 0 is a2 cos2 α + b2 sin2 α = p2.
∎ The pole of the line lx + my + n = 0 with respect to the Ellipse S = 0 is (-a2l/n, -b2m/n).
∎ The condition for the two lines l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0 to be conjugate with respect to the Ellipse S = 0 is a2l1l2 + b2m1m2 = n1n2. a2l1l2.
## 5.HYPERBOLA
Hyperbola: Hyperbola is a conic in which the eccentricity is greater than the unity.
Standard form of Hyperbola:
Equation of Hyperbola in standard form is
Centre: C (0, 0)
Foci: (± ae, 0)
Directrix: x = ± a/e.
Ecdntricity:
Notation:
Rectangular Hyperbola:
If in a Hyperbola the length of the transverse axis (2a) is equal to the length of the conjugate axis(2b), then the hyperbola is called rectangular hyperbola.
Its equation is x2 – y2 = a2 and eccentricity is .
Auxiliary circle: The circle described on the transverse axis of hyperbola as diameter is called the auxiliary circle of the hyperbola.
The equation of the auxiliary circle of S = 0 is x2 + y2 = a2.
Parametric equations: The parametric equations of the Parabola S = 0 are x = a secθ and y = b tanθ.
Conjugate Hyperbola:
The hyperbola whose transverse and conjugate axis are respectively the conjugate and transverse axis of a given hyperbola is called a conjugate hyperbola.
The equation of hyperbola conjugate to S ≡ is S’ ≡
∎ For
The transverse axis lies on along X-axis and its length is 2a.
The conjugate axis lies on along Y-axis and its length is 2b.
∎ For
The transverse axis lies on along Y-axis and its length is 2b.
The conjugate axis lies on along X-axis and its length is 2a.
Various form of Hyperbola:
Let and
1.Hyperbola
The transverse axis along X-axis: y =0
Length of the transverse axis:2a
The conjugate axis along Y-axis: x = 0
Length of the conjugate axis: 2b
Centre: (0, 0)
Foci: (± ae, 0)
Equation of the directrices: x = ± a/e
Eccentricity:
2.Conjugate Hyperbola
The transverse axis along Y-axis: x = 0
Length of the transverse axis:2b
The conjugate axis along X-axis: y = 0
Length of the conjugate axis: 2a
Centre: (0, 0)
Foci: (0, ± be)
Equation of the directrices: y = ± b/e
Eccentricity:
Centre not at the origin:
3.Hyperbola
The transverse axis along X-axis: y = k
Length of the transverse axis:2a
The conjugate axis along Y-axis: x = h
Length of the conjugate axis:2b
Centre: (h, k)
Foci: (h± ae, k)
Equation of the directrices: x = h± a/e
Eccentricity:
4.Hyperbola
The transverse axis along Y-axis: x = h
Length of the transverse axis:2b
The conjugate axis along X-axis: y = k
Length of the conjugate axis: 2a
Centre: (h, k)
Foci: (h, k ± be)
Equation of the directrices: y = k± b/e
Eccentricity:
Equation of tangent and normal at a point on the Hyperbola:
∎ The equation of the tangent at P (x1, y1) to the hyperbola S = 0 is S1 = 0.
∎ The equation of the tangent at P(θ) on the hyperbola S = 0 is
∎ The equation of the Normal at P (x1, y1) to the hyperbola S = 0 is
∎ Equation of the normal at P(θ) on the Hyperbola S = 0 is
∎ The condition for a straight-line y = mx + c to be a tangent to the hyperbola S = 0 is c2 = am2 − b2.
Asymptotes of a hyperbola:
The equations of asymptotes of hyperbola S = 0 are and the joint equation of asymptotes is
## 6. INTEGRATION
∎ Let E be a subset of R such that E contains a right or left the neighbourhood of each of its points and let f: E → R be a function. If there is a function F on E such that F’(x) = f(x) ∀ x ∈ E, then we call F an anti-derivative of f or a primitive of f.
Indefinite integral: Let f: I→R. Suppose that f has an antiderivative F on I. Then we say that f has an integral on I and for any real constant c, we call F + c an indefinite integral of over I, denote it by ∫f(x) dx and read it as ‘integral’ f(x) dx.
∫f = ∫f(x) dx = F(x) + c. here c is called constant of integration.
In the indefinite integral ∫f(x)dx, f is called ‘integrand’ and x is called the variable of integration.
⟹
⟹ if f: I⟶R is differentiable on I, then ∫f’(x)dx = f(x) + c.
Standard forms:
Properties of integrals:
∎ ∫ (f ±g) (x) dx = ∫f(x) dx ± ∫g(x) dx + c
∎ ∫(af) (x) dx = a ∫f(x) dx + c
∎ ∫ (f1 + f2 + … + fn) (x) dx = ∫f1(x) dx + ∫f2(x) dx + … +∫fn(x) dx +c
∎ ∫f(g(x)) g’(x) dx = F(g(x)) + c
∎ ∫f(ax + b) dx = 1/a F(ax +b) + c
Some important formulae:
1. ∫eax dx = 1/a eax + c
2. ∫sin (ax + b) dx = -1/a cos (ax + b) + c
3. ∫cos (ax + b) dx = 1/a sin (ax + b) + c
4. ∫sec2 (ax + b) dx = 1/a tan (ax + b) + c
5. ∫ cossec2 (ax + b) dx = 1/a cot (ax + b) + c
6. ∫cosec (ax + b) cot (ax + b) dx = -1/a cosec (ax + b) + c
7. ∫sec (ax + b) tan (ax + b) dx = 1/a sec (ax + b) + c
Integration by parts:
Let u, v real valued differentiable functions in I. Suppose that u’v has an integral on I, then uv’ has an integral on I and
∫(uv’) (x) dx = (uv) – ∫(u’v) (x) dx + c or ∫(uv) dx = u ∫v dx – ∫ [u’ ∫v dx] dx + c
Integration of exponential functions:
∫ex dx = ex + c; ∫x ex dx = (x – 1) ex + c
∫ ex [f(x) +f’(x)] dx = ex f(x) + c
Integration of logarithmic functions:
∫log x dx = x log x – x + c
Integration of inverse trigonometric functions:
Evaluation of integrals form :
Working rule: reduce ax2 + bx + c to the form of a[(x + α)2 + β] and then integrate using the substitution t = x + α.
Evaluation of integrals form
Working rule:
Case(i) if a >0 and b2 – 4ac < 0, then reduce ax2 + bx + c to the form of a[(x + α)2 + β] and then integrate.
Case(ii) if a <0 and b2 – 4ac >0, then reduce ax2 + bx + c to the form of (-a) [ β – (x + α)2 +] and then integrate.
Evaluation of integrals form
Working rule: write px + q in the form of A (ax2 + bx +c)’ + B, then integrate.
Evaluation of integrals form
Working rule: write cos x =cos2(x/2) – sin2(x/2) and sin x = 2 sin(x/2) cos (x/2)
Put t = tan(x/2), then dt = ½ sec2 (x/2) dx
Cos x = 1 – t2 / 1 + t2, sin x = 2t/1 + t2 then integrate.
Evaluation of integrals form
Working rule: t = sqrt. (px + q and then integrate.
Evaluation of integrals form
Working rule: we find real numbers A, B and C such that
(a cos x + b sin x + c) = A(d cos x + e sin x +f)’ + B(d cos x + e sin x +f) + C then by substituting this expression in the integrand, evaluate the integral.
Integration – partial fraction method:
Let R(x) = f(x) / g(x), g(x) ≠ 0 where f, g are polynomials. If degree of f(x) ≥degree of g(x), then divide f(x) by g(x) by synthetic division method and find polynomials.
Q(x) and h(x) such that f(x) = Q (x) g(x) + h(x) here h = 0 or h ≠ 0 and degree h(x) < degree of g(x). Then R(x) =Q(x) + h(x)/g(x)
We get solution of h(x) / g(x) using partial fractions and then integrate.
Partial fractions:
∎ If R(x) = f(x) / g(x) is proper fraction, then
Case(i): – For every factor of g(x) of the form (ax + b) n, there will be a sum of n partial fractions of the form:
Case(ii): – For every factor of g(x) of the form (ax2 + bx + c) n, there will be a sum of n partial fractions of the form:
∎ If R(x) = f(x) / g(x) is improper fraction, then
Case (i): – If degree f(x) = degree of g(x), then f(x)/g(x) = k + h(x)/g(x) where k is the quotient of the highest degree term of f(x) and g(x).
Case (ii): – If f(x) > g(x)
R(x) =f(x) /g(x) = Q(x) + h(x)/g(x)
Reduction formulas:
## 7.DEFINITE INTEGRATION
Partition: Let a, b∈ R be such that a < b. Then, a finite set P = {x0, x1, …, x i- 1, xi, xi + 1, …, xn} of elements of [a, b] is called to be a partition of [a, b] if a = x0 < x1 < … < x i- 1 < xi < xi + 1 < … < xn = b.
Norm: if {x0, x1, …, xn} is a partition of [a, b], then the norm of the partition P, denoted by ∥P∥, is defined by ∥P∥ = max {x1 – x0, x2 – x1, …, xn – xn-1}. We donate the set of all partitions of [a, b] by 𝒫 ([a, b]).
Definite integral:
Riemann sum:
Let f: [a, b] → R be a bounded function for all x in [a, b]. Let P = {x0, x1, …, x i- 1, xi, xi + 1, …, xn} be partition of [a, b], and t ∈ [xi-1, xi], for I = 1, 2, …, n. A sum of the form is called Riemann sum of f relative to P.
Let f is Riemann integrable on [a, b]. if there exists a real number A such that S (P, f) approaches A as ∥P∥ approaches to ‘0’. In other words, given ϵ > 0, there is a δ > 0 such that for any partition P of [a, b] with ∥P∥ < δ irrespective of choice of ti in [xi-1, xi]. Such an A, if exists, is unique and is denoted by , it is read as the definite integral of f from a to b. an a is called lower limit and b is called upper limit. The function f in is called ‘integrand’.
if f: [a, b] → R is continuous, then is exist.
∎ If f is continuous on [0, p] where p is a positive integer then
The fundamental theorem of the integral calculus:
If f is integrable on [a, b] and if there is a differentiable function F on [a, b] such that F = f, then
we write
properties of definite integrals:
∎ Let f: [a, b] → R be bounded. Let c ∈ (a, b). then f is integrable on [a, b] if and only if it is integrable on [a, c] as well as on [c, b] and in this case
Method of substitution: Let g: [c, d] → R have continuous derivative on [c, d]. Let f: g([c, d]) → R be continuous. Then (fog) g’ is integrable on [c, d] and
∎ Let f be integrable on [a, b]. Then the function h, defined on [a, b] as h(x) = f (a + b – x)
for all x in [a, b] and
∎ Let f be integrable on [0, a]. Then the function h, defined on [0, a] as h(x) = f (a – x)
for all x in [a, b] and
∎ Let f: [-a, a] → R be integrable on [0, a]. Suppose that f is either odd or even. Then f is integrable on [-a, a] and
∎ Let f: [0,2 a] → R be integrable on [0, a].
• If f (2a – x) = f(x) for all x in [a, 2a] hen f is integrable on [0, 2a] and
• If f (2a – x) = – f(x) for all x in [a, 2a] hen f is integrable on [0, 2a] and
∎ If f and g are integrable on [a, b], then their product fig is integrable on [a, b].
Integration by parts:
∎ Let f: R→ R be a continuous periodic function and T be the period of it. Then any positive integer n
Reduction formulae:
∎ Let n≥2 be an integer, then
∎ Let m and n be positive integers, then
Areas under curves:
(i)If f: [a, b] → [0, ∞) is continuous, then the area A of the region bounded by the curve y = f(x), the X-axis and the line x = a and x =b is given by
A =
(ii) If f: [a, b] → (−∞, 0] is continuous, then the graphs of y = f(x)and y = − f(x) on [a, b] are symmetric about the X-axis. So, the area bounded by the graph of y = f(x), the X-axis and the lines x = a, x =b is same as the area bounded by the graph of y = – f(x), the X-axis and the lines x = a and y = b which is given by A =
From (i) and (ii) A =
(iii) Let f: [a, b] → R be continuous and f(x) ≥ 0 ∀ x ∈ [a, c] and f(x) ≤ 0 ∀ x∈ [c, b] where a < c < b. Then the area of the region bounded by the curve y= f(x), the X – axis, and the lines x = a and x = b is given by
Area of region =A =
(iv) Let f: [a, b] → R and g: [a, b] → R be continuous f(x) ≤ g(x) ∀ x∈ [a, b]. Then the area f the region bounded by the curve y = f(x), y = g(x) and the lines x = a, x = b is given by
(v) Let f and g be wo continuous real value functions on [a, b] and c ∈ (a, b) such that f(x) < g(x) ∀ x∈ [a, c) and g(x) < f(x) ∀ x∈ (c, b] with f (c) = g (c). area of the region bounded by y = f (x), y = g(x), and the lines x = a, x = b is given by
(vi) Let f: [a, b] → R and g: [a, b] → R be continuous functions. Suppose that, there exist points x1, x2 ∈ (a, b) with x1< x2 such that f(x1) = g(x1) and f(x2) = g(x2) and f(x) ≥ g(x) ∀ x ∈ (x1, x2). Then the area of the region bounded by the curves by y = f (x), y = g(x), and the lines x = x1, x = x2 is given by and if f(x) ≤ g(x) ∀ x ∈ (x1, x2). ThenIn either case, area is .
## 8.DIFFERENTIAL EQUATIONS
Differential equation: An equation involving one dependent variable and its derivative with respect to one or more independent variables is called ‘Differential equation’.
If a differential equation contains only one independent variable, then it is called ‘ordinary differential equation and if it contains more than one independent variable, then it is called ‘ partial differential equation’.
Degree of the differential equation: If a differential can be expressed as a polynomial equation in the derivatives occurring in it using the algebraic operations such that the exponent of each of the derivatives is the least, then the large exponent of the highest order derivative in the equation is called the degree of the differential equation.
Otherwise, the degree is not defined for a differential equation.
Order of differential equation: The order of differential equation is the order of the highest derivative occurring in it.
Note: The general form of an ordinary differential equation of nth order is
Formation of the differential equation: suppose that an equation y = ϕ (x, α1, α2, …, αn) where α1, α2, …, αn are parameters, representing a family of curves is given. Then successively differentiating above equation, a differential equation of the form
We know that y = mx is a straight line passing through the origin
m = dy/ dx ⟹
Solving differential equations:
1.Variable separable method:
If a given differential equation can be put in the form of f(x) dx + g(y) dy = 0 then its solution can be obtained by integrating each of them. This method is called variable separable method.
Ex: xdy – y dx = 0 can be written as dx/x =dy/y
By integrating we get ∫dx/x = ∫dy/y
⇒ logx = logy + logc
⇒ logx = log yc
∴ x = yc is the required solution
2.Homogeneous Differential Equation:
Homogeneous unction: – A function f (x, y) of two variable x and y is said to be a homogeneous function of degree n, if f(kx, ky) = kn f(x, y) for all values of k for which both side of the above are meaningful.
Homogeneous Differential Equation: – A differential equation of the form where f (x, y) and g (x, y) are homogeneous functions of x and y of the same degree is called a homogeneous differential equation.
Method of solving the homogeneous differential equation: –
Consider the homogeneous equation …… (1)
where f (x, y) and g (x, y) are homogeneous functions of x and y of the same degree.
f (x, y) = xn ϕ (y/x) and g(x) xn ψ (y/x)
eqn (1) becomes ……. (2)
put y = vx. Then ……. (3)
from (2) and (3)
This can be solved by the variable separable method.
3.Non-Homogeneous Differential Equations:
The differential equation of the form where a, b, c, a’, b’, c’ are constants and c and c’ are not both zero are called non-homogeneous equations. Reduce the above equation to a homogeneous equation by suitable substitution for x and y.
Case(i): –
Suppose that b = – a’. then becomes
⇒ (a’x + b’y + c’) dy – (ax – a’y + c) dx = 0
⇒ a’(x dy + y dx) + b’ y dy – ax dx + c’ dy – cdx = 0
By integrating we get
a’ xy + b’ y2/2 – a x2/2 + c’y – cx = k
which is required solution.
Case(ii): –
Suppose that Then becomes
Put ax + by = v, then
this can be solved by the variable separable method.
Case(iii): –
Suppose that b ≠ – a’ and a/a’ ≠ b/b’, then taking x = X + h, y = Y + k, where X and Y are variables and h, k are constants. We get . Hence…..(i) becomes
Now choose constants h and k such that
ah + bk + c = 0
a’h + b’k +c’ = 0
by solving above equations, we get h. k values
Hence, equation (1) becomes
This is the homogeneous equation in X and Y and then solve by the homogeneous method by putting Y = VX.
3.Linear Differential Equations:
A differential equation of the form = R, where P1, P2, …, Pn and R are constants or functions x only, is said to be a linear differentiable equation of nth order.
Method of solving the linear differentiable equation of 1st order: –
The linear differentiable equation of first order is
Multiplying both sides of (1) by, we get
# These notes cover all the topics covered in the TS I.P.E second year maths 2A syllabus and include plenty of formulae and concept to help you solve all the types of Inter Math problems asked in the I.P.E and entrance examinations.
## 1. COMPLEX NUMBERS
• The equation x2 + 1 = 0 has no roots in real number system.
∴ scientists imagined a number ‘i’ such that i2 = − 1.
Complex number: if x, y are any two real numbers then the general form of the complex number is
z = x + i y; where x real part and y is imaginary part.
∗ z = + iy can be written as (x, y)
∗If z1 = x1 + i y1, z2 = x2 + i y2, then
∗ z1 + z2 = (x1 + x2, y1 + y2) = (x1 + x2) + i (y1 + y2)
∗ z1 − z2 = (x1 − x2, y1 − y2) = (x1 − x2) + i (y1 − y2)
∗ z1∙ z2 = (x1 x2 −y1 y2, x1y2 + x2y1) = (x1x2 −y1 y2) + i (x1y2 +x2 y1)
∗ z1/ z2 = (x1x2 + y1 y2/x22 +y22, x2 y1 – x1y2/ x22 +y22)
= (x1x2 + y1 y2/x22 +y22) + i (x2 y1 – x1y2/ x22 +y22)
Multiplicative inverse of complex number:
Multiplicative inverse of complex number z is 1/z.
z = x + i y then 1/z = x – i y/ x2 + y2
Conjugate complex number:
• The complex numbers x + iy, x – iy are called conjugate complex numbers.
• The sum and product of two conjugate complex numbers are real.
• If z1, z2 are two complex numbers then
Modulus and amplitude of complex number:
Modulus: – If z = x + iy, then the non-negative real number is called modulus of z and it is denoted by or ‘r’.
Amplitude: – The complex number z = x + i y represented by the point P (x, y) on the XOY plane. ∠XOP = θ is called amplitude of z or argument of z.
∗ x = r cosθ, y = r sinθ
⇒ x2 + y2 = r2 cos2θ + r2 sin2θ = r2 (cos2θ + sin2θ) = r2(1)
⇒ x2 + y2 = r2
⇒ r = and = r.
∗ Arg (z) = tan−1(y/x)
∗ Arg (z1.z2) = Arg (z1) + Arg (z2) + nπ for some n ∈ { −1, 0, 1}
∗ Arg(z1/z2) = Arg (z1) − Arg (z2) + nπ for some n ∈ { −1, 0, 1}
Argand plane: The plane contains all complex numbers is called Argand plane. This was introduced by the mathematician Gauss (1777-1855), who first thought that complex numbers can be represented as a two-dimensional plane.
The square root of a complex number:
## 2.DE- MOIVER’S THEOREM
De- Moiver’s theorem: For any integer n and real number θ, (cosθ + i sinθ) n = cos nθ + i sin nθ.
cos α + i sin α can be written as cis α
cis α.cis β= cis (α + β)
1/cisα = cis(-α)
cisα/cisβ = cis (α – β)
(cosθ + i sinθ) -n = cos nθ – i sin nθ
(cosθ + i sin θ) (cosθ – i sin θ) = cos2θ – i2 sin2θ = cos2θ + sin2θ = 1.
cosθ + i sin θ = 1/ cosθ – i sin θ and cosθ – i sin θ = 1/ cosθ + i sin θ
(cosθ – i sin θ) n = (1/ (cosθ –+i sin θ)) n = (cosθ + i sin θ)-n = cos nθ – i sin nθ
nth root of a complex number: let n be a positive integer and z0 ≠ 0 be a given complex number. Any complex number z satisfying z n = z0 is called an nth root of z0. It is denoted by z01/n or
let z = r (cosθ + i sin θ) ≠ 0 and n be a positive integer. For k∈ {0, 1, 2, 3…, (n – 1)} let . Then a0, a1, a2, …, an-1 are all n distinct nth roots of z and any nth root of z is coincide with one of them.
nth root of unity: Let n be a positive integer greater than 1 and
Note:
• The sum of the nth roots of unity is zero.
• The product of nth roots of unity is (– 1) n – 1.
• The nth roots of unity 1, ω, ω2, …, ωn-1 are in geometric progression with common ratio ω.
Cube root of unity:
x3 – 1 = 0 ⇒ x3 = 1
x =11/3
Quadratic Expression: If a, b, c are real or complex numbers and a ≠ 0, then the expression ax2 + bx + c is called a quadratic expression in variable ‘x’.
∎ A complex number α is said to be a zero of the quadratic expression ax2 + bx + c if aα2 + bα + c = 0.
Quadratic Equation: If a, b, c are real or complex numbers and a ≠ 0, then ax2 + bx + c = 0 is called a quadratic equation in variable ‘x’.
∎ A complex number α is said to be root or solution of the quadratic equation ax2 + bx + c = o if aα2 + bα + c = 0.
The roots of a quadratic equation:
∎ The zeroes of the quadratic expression ax2 + bx + c are same as the roots of quadratic equation ax2 + bx + c = o.
∎ The roots of the quadratic equation are
If α, β are the roots of the quadratic equation ax2 bx +c= 0, then α + β = -b/a and αβ = c/a.
Discriminate:
If ax2 + bx + c = 0 is a quadratic equation, then b2 – 4ac is called the discriminant of quadratic equation. b2 – 4ac is also the discriminant of quadratic expression ax2 + bx + c. It is denoted by ∆
∴ ∆ = b2 – 4 ac.
Nature of the roots:
The nature of the roots of the quadratic equation as follows:
1. If ∆ > 0, then roots are real and distinct.
2. If ∆ = 0, then roots are real and equal.
3. If < 0, then roots are imaginary.
Note:
• If ∆ > 0 and b2 – 4 ac is a perfect square, then the roots are rational and distinct.
• If ∆ < 0 and b2 – 4 ac is not a perfect square, then the roots are irrational and distinct. Further, the roots are conjugate surds.
If α, β are the roots of the quadratic equation ax2 + bx +c= 0, then ax2 + bx + c = a (x – α) (x – β)
The quadratic equation whose roots are α, β is (x – α) (x – β) = 0 ⇒ x2 – (α + β) x + αβ = 0.
∎ The necessary and sufficient condition for the quadratic equations a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 to have common root is (c1a2 – c2a1)2 = (a1b2 – a2b1) (b1c2 – b2c1)
the common root is (c1a2 – c2a1)/ (a1b2 – a2b1).
If f(x) = ax2 + bx +c= 0 is a quadratic equation then
1. The quadratic equation whose roots are the reciprocals of the roots of f(x) = 0 is f(1/x) = 0.
2. Whose roots are greater than by ‘k’ then those of f(x) = 0 is f (x – k) = 0.
• Whose roots are smaller by ‘k’ than those of f(x) = 0 is f (x + k) = 0.
1. Whose roots are multiplied by ‘k’ of hose f(x) = 0 is f (x/k) = 0.
Sign of quadratic Expressions – change in signs:
∎ If the roots of quadratic equation ax2 + bx + c = 0 are complex roots, then for x ∈ R, ax2 + bx + c and ‘a’ have the same sign.
∎ If the roots of quadratic equation ax2 + bx + c = 0 are real and equal then for x ∈ R – {-b/2a}, ax2 + bx + c and ‘a’ have the same sign.
∎ Let α, β are the roots of the quadratic equation ax2 + bx +c= 0and α < β, then
• x ∈ R, α < x < β ⇒ ax2 + bx + c and ‘a’ have the opposite sign.
• x ∈ R, x < α or x> β ⇒ ax2 + bx + c and ‘a’ have the same sign.
Maximum and Minimum values:
Let f(x) = ax2 + bx +c is a quadratic expression then
• if a < 0, then f(x) has maximum value at x = -2b/a and maximum value is 4ac – b2/4a.
• if a > 0, then f(x) has minimum value at x = -2b/a and minimum value is 4ac – b2/4a.
A quadratic in equation in one variable is of the form ax2 + bx +c > 0 or ax2 + bx +c ≥ 0 or ax2 + bx +c< 0 or ax2 + bx +c ≤ 0 where a, b, c are real numbers and a ≠ 0. The values of x which satisfy given inequation are called the solution of the in equations.
⟹ Quadratic inequations are solved by two methods (i) Algebraic method (ii) Graphical method.
## 4.THEORY OF EQUATIONS
Polynomial: If n is non- negative integer and a0, a1, a2. …, an are real or complex numbers and a0 ≠ 0, then an expression
f(x) = a0 xn + a1 xn – 1+a2xn – 2 + … + an is called polynomial in x of degree n.
a0, a1, a2. …, an are called the coefficients of the polynomial f(x), a0 is called leading coefficient and an is called constant term.
Monic polynomial: A polynomial with leading coefficient is 1 is called a monic polynomial.
Remainder theorem: let f(x) be a polynomial of degree n> 0. Let a ∈C. Then there exist a polynomial q(x) of degree n – 1 such that
f(x) = (x – a) q(x) + f(a).
Factor theorem: let f(x) be a polynomial of degree n> 0. Let a ∈ C. We say that (x – a) is factor of f(x) if there exist a polynomial q(x) such that f(x) = (x – a) q(x).
⟹ let f(x) be a polynomial of degree n> 0, then (x – a) is factor of f(x) iff f(a) = 0.
The fundamental theorem of algebra: Every non-constant polynomial equation has at least one root.
⟹ The set of all roots of a polynomial f(x) = 0 of degree n> 0 is non-empty and has at most n elements. Also there exist α1, α 2. …, α n in C such that f(x) = a (x – α1) (x – α2) (x – α3) … (x – αn) where a is the leading coefficient of f(x).
The relations between the roots and the coefficients:
Let xn + p1 xn – 1+ p2xn – 2 + … + pn = 0 be a polynomial equation of degree n
Let α1, α 2. …, α n be roots.
xn + p1 xn – 1+ p2xn – 2 + … + pn = (x – α1) (x – α2) (x – α3) … (x – αn)
= xn – (α1+ α 2+ …+ α n) xn – 1+( α1 α 2+ α2 α 3 +…+ α n-1 α n) xn-2 – … (-1)n α1. α 2…. α n
These equalities the relation between the roots and coefficient of the polynomial equation whose leading coefficient is 1.
Note:
1.For the quadratic equation: Let α, β be the roots of the quadratic equation ax2 + bx + c = 0, then
Sum of the roots = α + β = -b/a
Product of the roots = αβ = c/a
2.For the cubic equation: Let α, β and γ be the roots of the cubic equation ax3+ bx2 + cx + d = 0, then
Sum of the roots = α + β + γ = -b/a
Product of the roots taken two at a time= αβ + β γ + γ α = c/a
Product of the roots = αβ γ = -d/a
Notation: Let α, β and γ be the roots of a cubic polynomial, then
α + β + γ is denoted by ∑α, αβ + β γ + γ α is denoted by ∑ αβ , 1/α + 1/β + 1/γ is denoted by ∑1/α and α2 β + β2 α + α2 γ + γ2 β + β2 γ + γ2 α = ∑ α2 β + ∑ α β2
synthetic division: This method has two types
∗ Finding the quotient and remainder, when
a0 xn + a1 xn – 1+a2xn – 2 + … + an (a0 ≠ 0) is divided by (x – a).
∗ Finding the quotient and remainder, when
a0 xn + a1 xn – 1+a2xn – 2 + … + an (a0 ≠ 0) is divided by x2 – px – q.
Method of finding the quotient and remainder, when a0 xn + a1 xn – 1+a2xn – 2 + … + an (a0 ≠ 0) is divided by (x – a)- (Horner’s Method):
The procedure of the above method:
• First write down the coefficients of xn, xn-1, …x, x0. If any term with xk (0 ≤ k < 1) is missing, take the coefficient of it as zero.
• Draw a vertical line to the left of ‘a0’ and write ‘a’ to the left of the vertical line on the same horizontal level as that of ‘a0’.
• Under a0 write 0 and draw the horizontal line below it. Below the horizontal line and below 0, write the sum a + 0 as the first term of the 3rd row, which is equal to b0 with a and write this product below a1 in the second row. The sum a1 + ab0 is b1. Write this in the 3rd row next to b0. Continue this process until the terms of the second and the third rows are filled.
• From the table, the quotient is b0 xn-1 + b1 xn – 2+ … + bn-1 and the remainder is R = an + abn-1.
Note:
1. If the divisor is (x + a) then the above method can be used by replacing a with – a.
2. If the divisor is ax – b, then replace a by b/a.
Method of finding the quotient and remainder, when a0 xn + a1 xn – 1+a2xn – 2 + … + an (a0 ≠ 0) is divided by x2 – px – q:
The procedure of the above method:
• First write down the coefficients of xn, xn-1, …x, x0. If any term with xk (0 ≤ k < 1) is missing, take the coefficient of it as zero.
• Draw a vertical line to the left of ‘a0’ and write p, q as column figures to the left of the vertical line in the second and third rows respectively. These are the negatives of the coefficient of x and the constant term in the divisor. Draw a horizontal line below the third row.
• Put 0 in two rows underneath a0 write this sum a + 0 + 0 as the first term of the 4th row, which is equal to b0. Next, multiply b0 with p and write this product below a1 and write the next column entry as 0. The sum a1 + pb0 + 0 is b1. Write this in the 4th row underneath a1 multiply b1 with p and b0 with q and write this product underneath a2. let the sum of a2, pb1 and qb0 be b2. continue this process until the terms an-1 are obtained. Name the 4th row under an-1 R1. below an put 0 and qbn-2 in the second and third rows respectively. Let the sum as an, 0 qbn-2 be R2. write it in the 4th row below an.
Trial and Error method: To find a root of f(x) = 0, we have to find out a value of x, for which f(x) = 0. Some times we can do this by inspection. This method is called trial and error method.
Multiple roots or repeated roots: let f(x) be a polynomial of degree n > 0. Let α1, α 2. …, α n be the roots of f(x) = 0 so that f(x) = a0 (x – α1) (x – α2) (x – α3) … (x – αn). A complex number α is said to be a root of f(x) = 0 of multiplicity m, if α = αk for exactly m values of k among 1,2, 3…, n. Roots of multiplicity m>1 are called multiple roots or repeated roots.
Roots of multiplicity 1 are called simple roots.
∎ Let f(x) be a polynomial of degree n> 0. Let α be a root of f(x) = 0 of multiplicity m. If m>1, then α is a root of the equation f’(x) = 0 of multiplicity m – 1. If m = 1, then f’(α) ≠ 0
∎ Let f(x) be a polynomial of degree n > 0. Let α be a root of f(x) = 0 of multiplicity m, then α is a root of the equation f(k)(x) = 0 of multiplicity m –k (k = 1, 2, 3, …, m – 1).
∎ Let f(x) be a polynomial of degree n> 0. Let α be a root of f(x) = 0 of multiplicity m Iff f(α) = f’(α) = … = f (m – 1) (α) and f(m)(α) ≠ 0.
Procedure to find multiple roots: Let f(x) be a polynomial. First, we find f’(x) and then find the HCF of f(x) and f’(x). Now we note that, if α is a root of the HCF of multiplicity k, then α is a multiple order of (k + 1) of f(x) =0.
∎ let f(x) be a polynomial with real coefficients. Let α ∈ C, then
∎ Let f(x) be a polynomial of degree n > 0, with real coefficients. Let a0 be the leading coefficient of f(x).
1. If the equation f(x) = 0 has no real roots, then n is even and f(α), a0 have the same sign for all real values of α
2. If n is odd, then the equation f(x) = 0 has at least one real root.
∎ Let f(x) be a polynomial of degree n > 0, with real coefficients. Let a and b be rational numbers, b > 0 and irrational. Then is a root of f(x) = 0 if and only if another root is a .
Roots with the change of sign:
If α1, α 2. …, α n are the roots of f(x) = 0, then -α1, – α 2. …, – α n are the roots of f(-x) = 0.
Roots multiplied by a given number:
If α1, α 2. …, α n are the roots of f(x) = 0, then for any non-zero complex number k, the roots of f(x/k) = 0 are kα1, k α 2. …, kα n.
Roots subtracted by a given number:
If α1, α 2. …, α n are the roots of f(x) = 0, then α1-h, α 2-h. …, – α n-h are the roots of f (x +h) = 0.
Roots added by a given number:
If α1, α 2. …, α n are the roots of f(x) = 0, then α1+h, α 2+h. …, – α n+h are the roots of f (x -h) = 0.
Reciprocal roots:
Let α1, α 2. …, α n are the roots of f(x) = 0. Suppose none of them non- zero, then 1/α1,1/ α 2. …,1/ α n are the roots of xn f (1/x) = 0.
∎ if α is a root of f(x) = 0, then α2 is a root of
Reciprocal equation:
Let f(x) be a polynomial of degree n > 0 is said to be reciprocal if f(0) ≠ 0 and ∀x ∈C-{0}, where a0 is the leading coefficient of f(x).
If f(x) is a reciprocal polynomial, then the equation f(x) = 0 is reciprocal equation.
∎ If f(x) = a0 xn + a1 xn – 1+a2xn – 2 + … + an be a polynomial of degree n > 0, then f(x) is reciprocal iff an – k = ak for k = 0, 1, 2,… , n or an – k = – ak for k = 0, 1, 2,… , n .
The reciprocal polynomial of class one and class two:
A reciprocal polynomial f(x) of degree n with leading coefficient a0 is said to be class one or class two according to as f(0) = a0 or –a0.
If f(x) is a reciprocal polynomial then the equation f(x) = 0 is said to be the reciprocal equation of class one or class two according to as f(x) is a reciprocal polynomial of class one or class two.
Note:
∎ For an odd degree, the reciprocal equation of class one -1 is the root and for an odd degree reciprocal equation of class two, 1 is root.
∎ For an even degree, the reciprocal equation of class two, – 1 and 1 roots.
∎ To solve the reciprocal equation of order 2m, divide the equation by xm and put x + 1/x = y or x – 1/x = y according to the equation of class one or class two. The degree of the transformed equation is m.
∎ For an odd degree reciprocal equation. To find the roots of it, divide f(x) by (x + 1) or (x – 1) according as the equation of class one or class two. Let Q(x) be the quotient obtained, then f(x) = (x+1) Q(x) or f(x) = (x – 1) Q(x) according as the equation of class one or class two and Q(x) is even degree reciprocal polynomial. The roots of Q(x) = 0 can be obtained by above procedure.
## 5.PERMUTATIONS AND COMBINATIONS
The fundamental principle of counting: if a work w1 can be performed in ‘m’ different ways and a second work w2 can be performed in ‘n ‘different ways, then the two works can be performed in ‘mn’ ways.
Permutation: From a given finite set of elements selecting some or all of them and arranging them in a line is called a ‘linear permutation’ or ‘permutation’.
Circular permutation
Permutations of ‘n’ dissimilar thing taken ‘r’ at a time:
∎ If n, r are positive integers and r ≤ n, then the no. of permutations of n dissimilar things taken as ‘r’ at a time is n (n – 1) (n – 2) (n – 3) … (n – r + 1).
Notation:
The number of permutations of n dissimilar things taken as r at a time is denoted by nPr or P (n, r) (1≤r≤n).
nPr = n (n – 1) (n – 2) (n – 3) … (n – r + 1)
∎ If n ≥1 and 0≤r≤n, then
nPn = n! and nP0 = 1.
∎ For 1≤r≤n , nPr = n. (n – 1) P (r – 1)
∎ If n, r are positive integers and 1 ≤r < n, then nPr = (n – 1) Pr + r. (n – 1) P (r – 1).
∎ The sum of all r-digit numbers that can be formed using the given ‘n’ non-zero digits (1 ≤r ≤n≤9) is
(n – 1) P (r – 1) × [ sum of the given digits × 1111… 1(r times)]
∎If ‘0’ is one digit among the given ‘digits, then we get that the sum of all r-digit numbers that can be formed using the given ‘n’ digits including ‘0’ is
{ (n – 1) P (r – 1) × [ sum of the given digits × 1111… 1(r times)]} – { (n – 2) P (r – 2) × [ sum of the given digits × 1111… 1((r-1) times)]}.
Note: If a set A has m elements and the set B has n elements, then the no. of injections into A to B is nPm if m ≤n and 0 if m> n.
Permutations when repetitions are allowed:
∎ Let n and r be positive integers. If the repetition of things is allowed, then the no. of permutations of ‘n’ dissimilar things taken ‘r’ at a time is nr.
Palindrome: A number or a word which reads the same either from left to right or right to left is called a palindrome.
Ex: 121, 1331, ATTA, AMMA etc.
Note: The no. of palindromes with r distinct letters that can be formed using given n distinct letters is
(i) nr/2 if r is even (ii) nr+1/2 if r is odd.
Circular permutation: From a given finite set of elements selecting some or all of them and arranging them around a circle is called a ‘circular permutation’.
The no. of circular permutations of ‘n’ dissimilar things (taken all at a time) is (n – 1)!
∎ In case of the garlands of flowers, chains of beads etc, no. of circular permutations = ½ (n – 1)!
Permutations with constraint repetitions:
∎ The no. of linear permutations of ‘n’ things n which ‘p’ things are alike and the rest are different is
∎ The no. of linear permutations of ‘n’ things n which ‘p’ like things of one kind, q like things of the second kind, r like things of the third kind and the rest are different is
Combinations:
A combination is only a selection. There is no importance to the order or arrangement of things in a combination.
∎ The no. of combinations of ‘n’ dissimilar things taken ‘r’ at a time is denoted by nCr or C (n, r)
∎ For 0≤r≤n, nCr = n C n – r
∎ If m, n are distinct positive integers, then the no, of ways of dividing (m + n) things into two groups containing m things and ‘n’ things is
∎ If m, n, p are distinct positive integers, then the no, of ways of dividing (m + n + p) things into three groups containing m things, ‘n’ things and ‘p’ things is
∎ The no. of ways of dividing 2n dissimilar things into two equal groups containing ‘n’ things in each case is
∎ The no. of ways of dividing ‘mn’ dissimilar things into m equal groups containing ‘n’ things in each case is
∎ The no. of ways of distributing ‘mn’ dissimilar things equally among m persons is
• For 0 ≤ r, s ≤ n, if nCr = nCs then r =s or n = r + s.
• If 1 ≤ r ≤ n, then nCr-1 + nCr = (n+1) Cr.
• If 2 ≤ r ≤ n, then nCr-2 +2 nCr-1 = (n+2) Cr.
• If p things are alike of one kind, q things are alike of the second kind and r things are alike of the third kind, then the number of ways of selecting any no, of things out of these (p + q +r) things is (p + 1) (q+1)(r+1) – 1.
• The number of ways of selecting one or more things out of ‘n’ dissimilar things is 2n – 1.
• If p1, p2,…, pn are distinct primes and α1, α 2,…, α n are positive integers, then the number of positive divisors of is (α1+1)( α2+1) … (αk + 1).
Exponents of a prime in n! (n ∈ z+): Exponents of a prime number ‘p’ in n! is the largest integer ‘k’ such that pk divides n!
## 6.BINOMIAL THEOREM
Binomial: Binomial means two terms connected by either ‘+’ or ‘– ‘.
Binomial expansions:
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + y2
(x + y)3 = x3 + 3x2 y + 3xy2 + y3 and so on, are called binomial expansions.
Binomial coefficients:
Coefficients of expansion (x + y) are 1, 1.
Coefficients of expansion (x + y)2 are 1, 2, 1
Coefficients of expansion (x + y) are 1, 3, 3, 1
And so on, are called binomial coefficients.
Pascal triangle:
Binomial theorem: Let n be a positive integer and x, a be real numbers, then
(x + a) n = nC0 xn a0 + nC1xn – 1 a1 + nC2 x n – 2 a2 +… + nCr xn – r ar + … + nCn x0 an
Note: –
Let n be a positive integer and x, a be real numbers, then
(i) (x + a) n = ∑ nCr xn – r ar
(ii) The expansion of (x + a) n has (n + 1) terms.
(iii) The rth term in the expansion of (x + a) n, which is denoted by Tr, is given by Tr = nCr-1 xn – r +1 ar-1 for 1≤ r≤ n + 1.
The general term of the binomial expansion:
In the expansion of (x + a)n, the (r + 1)th term is called the general term of the binomial expansion and it is given by Tr+1 = nCr xn –r ar for 0≤ r≤ n.
(x – a) n = nC0 xn (-a)0 + nC1xn – 1 (-a)1 + nC2 x n – 2 (-a)2 +… + nCr xn – r (-a) r + … + nCn x0 (-a) n
= nC0 xn a0nC1xn – 1 a1 + nC2 x n – 2 a2 –… +(–1) r nCr xn – r ar + … + (–1) n nCn x0 an
And the general term is Tr+1 = (–1) r nCr xn –r ar for 0≤ r≤ n.
Trinomial Expansion: Let n ∈ N and a, b, c ∈ R, then (a + b + c) n can be expand using the binomial theorem taking a as the first term and (b + c) as the second term
(a + b + c) n = (a + (b+ c)) n = ∑ nC0 an-r (b + c) r (0≤ r≤ n)
⟹ no. of terms in the expansion of (a + b + c) n =
Middle terms in (x + a) n:
∎ if n is even then term is the middle term.
∎ if n is odd then terms are the middle terms.
Binomial coefficients: The coefficients in the binomial expansion (x + a) n are nC0, nC1, …, nCr, …, nCn these coefficients are called binomial coefficients. When n is fixed these coefficients are denoted by C0, C1, …, Cr, …, Cn. respectively.
Note:
• The binomial expansion of (1 + x) n = nC0 + nC1x + nC2 x 2 +… + nCn xn. This expansion is called the standard binomial expansion.
With the standard notation, if n is a positive integer, then
• C0 + C1 + C2 …+ Cn = 2n
• C0 + C2 + C4 …+ Cn = 2n-1 if n is even
• C0 + C2 + C4 …+ Cn-1 = 2n-1 if n is odd
• C1 + C3 + C5 …+ Cn-1 = 2n-1 if n is even
• C1 + C3 + C5 …+ Cn = 2n-1 if n is odd
Integral part and Fractional part: If x is any real number, then there exist an integer n such that n ≤ x < n+ 1. This integer n is called an integral part of the real number x and it is denoted by [x]. The real number x – [x] is called fractional part of x and it is denoted by {x}.
Numerically greatest term: In the binomial expansion of (1 + x) n, the rth term Tr is called numerically greatest term if,
⟹ if = p, where p is a positive integer then, pth and (p + 1)th are the numerically greatest terms.
⟹ if = p + F, where p is a positive integer and 0 < F < 1 then, (p + 1)th is the numerically greatest term.
⟹ To find the numerically greatest term(s) in the binomial expansion of (a + x)n we write (a + x)n = an(1 + x/a)n and then find the numerically greatest term(s) by using above rules.
Largest Binomial coefficient:
The largest binomial coefficient(s) among nC0, nC1, …, nCr, …, nCn is (are)
(i) if n is even integer.
(ii) if n is an odd integer
Binomial theorem for Rational Index:
If m is a rational number and x is a real number such that – 1 < x < 1, then
Rational Index: –
## 7. PARTIAL FRACTIONS
Rational fraction: If f(x) and g(x) are two polynomials and g(x) is a non-zero polynomial, then is called a rational fraction or polynomial fraction or simply a fraction.
Ex:
Proper and Improper Fractions: A rational fraction is called a Proper fraction if the degree of f(x) is less than the degree of g(x). Otherwise, it is called an improper fraction.
Ex: is a proper fraction and is an Improper fraction.
Irreducible Polynomial: A polynomial f(x) is said to be irreducible if it can not be express as a product of two polynomials g(x) and h(x) such that the degree of each polynomial is less than the degree of f(x). If f(x)is not irreducible then we say that f(x) is reducible.
Ex: 3x – 1, x2 + x + 1 are irreducible polynomials.
Division Algorithm for Polynomials: If f(x) and g(x) are two polynomials with g(x) ≠ 0, then there exist unique polynomials q(x) and r(x) such that f(x) = q(x) g(x) + r(x) , where either r(x) = 0 or the degree of r(x) is less than the degree of g(x).
Partial Fraction: If a proper fraction is expressed as the sum of two or more proper fractions, wherein the power of the denominator of irreducible polynomials, then each proper fraction in the sum is called a partial fraction of the given fraction.
Partial Fraction of when g(x) contains linear factors:
Rule – 1: Let be a proper fraction. To each non- repeated factor of g(x), there will be a partial fraction of the form where A is a non-zero real number, to be determined.
Rule – 2: Let be a proper fraction. To each factor (ax + b)n, a ≠ 0 where ‘n’ is a positive integer, of g(x) there will be a partial fraction of the form where A1, A2, …, An are to be determined constants. Note that An ≠ 0 and Rule –1 is a particular case of Rule-2 for n = 1.
Partial Fraction of when g(x) contains irreducible factors:
Rule – 3: Let be a proper fraction. To each non- repeated quadratic factor (ax2 + bx + c), a ≠ 0 of g(x) there will be a partial fraction of the form where A, B are real numbers, to be determined.
Rule – 4: Let be a proper fraction. If n (>1)∈ N is the largest exponent so that (ax2 + bx + c)n, a ≠ 0) factor of g(x) there will be a partial fraction of the form where A1, A2, …, An and B1, B2, …, Bn are real numbers, to be determined.
Partial Fraction of when is an Improper fraction:
Case (1): If degree f(x) = degree of g(x) then by division algorithm there exist a unique constant k and r(x) such that f(x) = k g(x) + r(x), where either r(x) = 0 or the degree of r(x) is less than the degree of g(x) and the constant k is the quotient of the coefficient of the highest degree terms of f(x) and g(x).
can be expressed as k + where is a proper fraction which can be resolved into a partial fraction using the above rules.
Case (2): If degree f(x) > degree of g(x) then by division algorithm can be expressed as q(x) + where q(x) is a non-zero polynomial and is a proper fraction which can be resolved into a partial fraction using above rules.
## 8. MEASURES OF DISPERSION
The measure of dispersion: In a measure of central tendency, we have to know a measure to describe the variability. This method is called a measure of dispersion.
Measuring dispersion of a data is significant because it determines the reliability of an average by pointing out as to how far an average is representative of the entire data.
Some measures of dispersion are: (i) Range (ii) Mean deviation (iii) Standard deviation
Range:
For ungrouped data, the range is the difference between the maximum and minimum value of the series of observations.
For grouped data range is approximated as the difference between the upper limit of the largest class and the lower limit of the smallest class.
Mean deviation:
To find the dispersion of values of x from a central value ‘a’ we find the deviation about ‘a’. They are (x – a)’s. To find the mean deviation we have to sum up all such deviations.
Mean deviation from the mean for ungrouped data:
Let x1, x2, …, xn be n observations of discrete data.
Steps for finding the Mean deviation from the mean for ungrouped data:
1. First, we have to find the mean ( ) of the n observations. Let it be ‘a’
2. Find the deviations of each xi from ‘a’, i.e., x1 – a, x2 – a, …, xn – a.
3. Find the absolute values of i.e., of these deviations by ignoring the negative sign, if any, in the deviation computed in step 2.
4. Find the arithmetic mean of the absolute values of the deviations.
M.D from the mean =
Mean deviation from the median for ungrouped data:
Let x1, x2, …, xn be n observations of discrete data.
Steps for finding the Mean deviation from the median for ungrouped data:
1. First, we have to find the median of the n observations. Let it be ‘a’
2. Find the deviations of each xi from ‘a’, i.e., x1 – a, x2 – a, …, xn – a.
3. Find the absolute values of i.e., of these deviations by ignoring the negative sign, if any, in the deviation computed in step 2.
4. Find their arithmetic mean as M.D from median =
Mean deviation for a grouped data:
A data can be arranged or grouped as a frequency distribution in two ways: (i) Discrete frequency distribution and (ii) Continuous frequency distribution.
(i) Discrete frequency distribution:
If x1, x2, …, xn are of ‘n’ observations occurring with frequencies f1, f2, …, fn Then we can represent this data in the following manner:
xi x1 x2 x3 … xn fi f1 f2 f3 … fn
This form is called the discrete frequency distribution
Mean deviation about the mean and median:
Mean =
M.D(mean) =
M.D(median) =
Where N is total frequency.
(ii) Continuous frequency distribution: –
Continuous distribution is a series in which the data is classified into different class-intervals along with their respective frequencies.
Mean deviation about the means and median:
Mean =
M.D(mean) =
Median =
M.D(median) =
Where N is total frequency.
Step- Deviation method: If the midpoints of the class intervals xi as well as their associated frequencies are very large then we use this method.
Arithmetic mean =
Where
Variance and Standard Deviation of un grouped data:
If x1, x2, …, xn are n observations and is their mean, then
We have the following cases:
Case(i): if = 0, then each= 0 which implies all observations are equal to the mean and there is no dispersion.
Case(ii): if is small, then it indicates that each observation xi is very close to the mean and hence the degree of dispersion is low.
Case(iii): if is larger, then it indicates the higher degree of dispersion of the observations from the mean .
Variance = σ2=
Standard deviation
∎ The coefficient of variation of a distribution (C.V.) =
## 9. PROBABILITY
Random Experiment: If the result of an experiment is not certain and is any one of the several possible outcomes, then the experiment is called ‘random experiment.
Sample space: The set of all possible outcomes of an experiment is called ample space when ever the experiment conducted and is denoted by ‘S’.
Event: Any subset of the sample space is called an event.
Complimentary of an event: The complementary of an event E , is denoted by Ec , is the event given by Ec = S – E which is called the complimentary event of E.
Equally likely events: two events are said to be equally likely events when chance of occurrence of one event is equal to that of other.
Exhaustive events: A set of events is said to be exhaustive if the performance of the experiment always result in the occurrence of the at least one of them.
The events E1, E2, …, En are said to be exhaustive if E1∪ E2∪…, ∪ En = S.
Mutually Exclusive events: A set of events is said to be mutually exclusive if happening of one of them prevents the happening of any one of remaining events.
The events E1, E2, …, En are said to be exhaustive if Ei ∩ Ej =∅ for i ≠ j, 1 ≤i, j≤ n.
Classical definition of Probability: In a random experiment, let there be n mutually exclusive, exhaustive and equally likely events.
E be the event of the experiment. ‘m’ elementary events are favourable to an event E, then the probability of E is defined as P (E) =
For any event E, 0 ≤ P(E) ≤1.
∎ If Ec is the non-occurrence of E, then the probability of on-occurrence of E is P (Ec)
P (Ec) = 1 – P (E) ⇒ P (Ec) + P (E) = 1
Limitations of the Classical definition of the probability:
1. If the out comes of the random experiment are not equally likely, then the probability of an event in such experiment is not defined.
2. If the random experiment contains infinitely many out comes, then his definition cannot be applied to find the probability of an event in such an experiment.
Relative frequency (Statistical or Emperical) definition probability:
Suppose a random experiment is repeated n times, out of which an event E occurs m(n) times, then the ratio is called the nth relative frequency of the event E.
Let r1, r2, …, rn be the sequence. If rn tends to a definite limit, , l is defined to be the probability of the event E and we write P (E) =
Deficiencies of the relative frequency definition of probability:
1. Repeating a random experiment infinitely many times is practically impossible.
2. The sequence of relative frequencies is assumed to tend to a definite limit, which may not exist.
3. The values r1, r2, …, rn are not real variables. Therefore, it is not possible to prove the existence and the uniqueness of the limit of rn as n → ∞, by applying methods used in calculus.
Probability Function:
Let S be the sample space of a random experiment, which is finite. Then a function P: S → R satisfying the following axioms is called a Probability function.
(i) P (E) ≥ 0 ∀ E ∈ S (axiom of non-negativity)
(ii) P (S) = 1 (axiom of certainty).
(iii) If E1, E2 ∈ S and E1 ∩ E2 =∅, then P (E1 ∪ E2) = P (E1) + P (E2) (axiom of additivity).
For each E ∈ S, the real number P (E) s called the probability of the event E. If E = {a}, then we write p(a) instead of P ({a}).
Note:
1. P (∅) = 0 for any sample space S, S ∅ = S and S∩ ∅ = ∅. P (S) = (S ∅) = P (S) + P (∅) = P (S) (∵ P (∅) = 0).
1. If S is countably infinite, then axiom (iii) of the above definition is to be replaced by (iii)*: if is a sequence of pairwise mutually exclusive events, then
2. Suppose S be a sample space of a random experiment. Let P be a probability function. If E1, E2, …, En are finitely many pairwise mutually exclusive events, then P (E1∪ E2∪…, ∪ En) = P (E1) + P (E2) + … + P (En)
⟹ If E1, E2 are any two events in a sample space S, then
If E1, E2 are any two events in a sample space S and P is a probability function, then P (E1∪ E2) = P (E1) + P (E2) – P (E1∩ E2)
P (E1 – E2) = P (E1) – P (E1∩ E2)
P (E2 – E1) = P (E2) – P (E1∩ E2)
Set – theoretic descriptions:
Event Set-theoretic description Event A or Event B to occur A∪B Both event A and B occur A∩B Neither A nor B occur (A∪B) c = Ac ∩ Bc A occurs but B does not occur A ∩ Bc or A\B Exactly one of the event A, B to occur (A∩B) c ∪ (Ac ∩ B) or (A – B) ∪ (B – A) or (A∪B) – (A ∩ B) Not more than one of the events A, B occurs (A∩B) c ∪ (Ac ∩ B) ∪ (Ac ∩ Bc) Event B occurs whenever event A occurs A ⊆ B
Conditional event: If A, B are two events of random experiment, then the event of happening (occurring) B after the event A happens(occurs) is called conditional event. It is denoted by B\A.
Conditional probability: If A, B are two events and P (A) ≠ 0, then the probability of B after the event A has occurred is called conditional probability. It is denoted by P (B/A) and is defined by P(B/A) =
Multiplication theorem of probability: If A, B are two events of random experiment with P (A) > 0 and P (B) > 0, then P (A∩B) = P (A) P (B/A) = P (B) P (A/B).
Two events A and B said to be independent if P (B/A) = P(B) or P (A/B) = P (A)
Two events A and B said to be independent if P (A∩B) = P (A). P (B)
The events A1, A2, …, An are said to be independent if P (A1∩ A2∩ …∩ An) = P (A1). PA2). …. P(An).
Bayes Theorem:
If A1, A2, …, An are mutually exclusive and exhaustive events in a sample space S such that P (Ai) > 0 for i = 1, 2, 3, …, n and E is an event with P (E) > 0, then
## 10. RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS
Random variable: Let S be a sample space of a random experiment. A real valued function X: S → R is called random variable.
∎ A set A is said to be countable if there exist a bijection from A into a subset of N.
Probability distribution Function: If X: S → R is a random variable connected with a random experiment and P is a probability function associate with it. The unction F: R → R defined by F(x) = P (X ≤ x) is called probability distribution function of the random variable X.
Discrete or discontinuous Random variable: Let S be a sample space, a random variable X: S → R is said to be Discrete or discontinuous if the range of X is countable.
I.POBABILITY DISTRIBUTION:
If X: S → R is a discrete random variable with range {x1, x2, x3………}, then {P(X = xr; r = 1, 2, 3, …, n} is called probability distribution of X.
The table for the probability distribution of the discrete random variable X is:
X = xi x1 x2 x3 … xn P (X = xi) P (x1) P (x2) P (x3) … P (xn)
Mean (μ) =∑xi P (X = xi)
Variance = σ2 = ∑ (xi – μ)2 P (X = xi) = ∑ xi2 P (X = xi) – μ2
Standard deviation is σ.
II.BINOMIAL or BERNOULLI DISTRIBUTION:
Let n be a positive integer and p be the random number such that 0 < p < 1. A random variable X with range {0, 1, 2, 3, …, n} is said to have a Binomial distribution with parameters n and p, if
P (X = x) = nCx px qn – x for x = 0, 1, 2, …, n and q = 1 – p.
Mean = np
Variance = npq.
III. POISSON DISTRIBUTION:
Let λ > 0 be a real number. A random variable X with range {1, 2, 3, …, n} is said to be Poisson distribution with parameter λ if,
Mean = λ
Variance = λ
Poisson distribution as a limiting form of Binomial distribution:
Poisson distribution can be derived as the limiting case of binomial distribution in the following case.
If λ > 0 for each positive integer n > λ, let Xn be the Binomial variable B (n, λ/n). using the fact that we can prove that for every non-negative integer k,
# These notes cover all the topics covered in the TS I.P.E first year maths 1B syllabus and include plenty of formulae and concept to help you solve all the types of Inter Math problems asked in the I.P.E and entrance examinations.
## 0.COORDINATE GEOMETRY( BASICS)
• Distance between two points A(x1, y1), B(x2, y2) is
• distance between a point A(x1, y1) to the origin is
• The midpoint of two points A(x1, y1), B(x2, y2) is
• If P divides the line segment joining the points A(x1, y1), B(x2, y2) in the ratio m:n then the coordinates of P are
• Area of the triangle formed by the vertices A (x1, y1), B (x2, y2) and C (x3, y3) is
## 1. LOCUS
Locus: The set of points that are satisfying a given condition or property is called the locus of the point.
Ex:- If a point P is equidistant from the points A and B, then AP =BP
Ex 2: – set of points that are at a constant distance from a fixed point.
here the locus of a point is a circle.
• In a right-angled triangle PAB, the right angle at P and P is the locus of the point, then
AB2 = PA2 + PB2
•Area of the triangle formed by the vertices A (x1, y1), B (x2, y2), and C (x3, y3) is
## 2.CHANGE OF AXES
Transformation of axes:
When the origin is shifted to (h, k), without changing the direction of axes then
•To remove the first degree terms of the equation ax2 + 2hxy + by2 +2gx +2fy+ c = 0, origin should be shifted to the point
•If the equation ax2 + by2 +2gx +2fy+ c = 0, origin should be shifted to the point
Rotation of axes:
When the axes are rotated through an angle θ then
•To remove the xy term of the equation ax2 + 2hxy + by2 = 0, axes should be rotated through an angle θ is given by
## 3.STRAIGHT LINES
Slope:- A-line makes an angle θ with the positive direction of the X-axis, then tan θ is called the slope of the line.
It is denoted by “m”.
m= tan θ
• The slope of the x-axis is zero.
• Slope of any line parallel to the x-axis is zero.
• The y-axis slope is undefined.
• The slope of any line parallel to the y-axis is also undefined.
• The slope of the line joining the points A (x1, y1) and B (x2, y2) is
Slope of the line ax + by + c = 0 is
### Types of the equation of a straight line:
• Equation of x- axis is y = 0.
• Equation of any line parallel to the x-axis is y = k, where k is the distance from above or below the x-axis.
• Equation of y- axis is x = 0.
• Equation of any line parallel to y-axis is x = k, where k is the distance from the left or right side of the y-axis.
Slope- intercept form
The equation of the line with slope m and y-intercept c is y = mx + c.
Slope point form:
The equation of the line passing through the point (x1, y1) with slope m is
y – y1 = m (x – x1)
Two points form:
The equation of the line passing through the points (x1, y1) and (x2, y2) ’ is
Intercept form:
The equation of the line with x-intercept a, y-intercept b is
• The equation of the line ∥ el to ax +by + c = 0 is ax +by + k = 0.
• The equation of the line ⊥ler to ax +by + c = 0 is bx −ay + k = 0.
Note: –
1. If two lines are parallel then their slopes are equal
m1 = m2
1. If two lines are perpendicular then product of their slopes is – 1
m1 × m2 = – 1
1. The area of the triangle formed by the line ax + by + c = 0 with the coordinate axes is
2. The area of the triangle formed by the line with the coordinate axes is
Perpendicular distance (Length of the perpendicular):
The perpendicular distance from a point P (x1, y1) to the line ax + by + c = 0 is
• The perpendicular distance from origin to the line ax + by + c = 0 is
Distance between two parallel lines:
•The distance between the parallel lines ax1 + by1 + c1 = 0 and ax2 + by2 + c2 = 0 is
Perpendicular form or Normal form:
The equation of the line which is at a distance of ‘p’ from the origin and α (0≤ α ≤ 3600) is the angle made by the perpendicular with the positive direction of the x-axis is x cosα + y sinα = p.
Symmetric form:
The equation of the line passing through point P (x1, y1) and having inclination θ is
Parametric form:
if P (x, y) is any point on the line passing through A (x1, y1) and
making inclination θ, then
x = x1 + r cos θ, y = y1 + r sin θ
where ‘r’ is the distance from P to A.
• The ratio in which the line L ≡ ax + by + c = 0 divide the line segment joining the points A (x1, y1), B (x2, y2) is – L11: L22.
Where L11 = ax1 + by1 + c and L22 = ax2 + by2 + c.
Note: – the points A (x1, y1), B (x2, y2) lie on the same side or opposite side of line L = 0 according to L11 and L22 have the same sign or opposite sign.
∗ x-axis divides the line segment joining the points A (x1, y1), B (x2, y2) in the ratio – y1: y2.
∗ y-axis divides the line segment joining the points A (x1, y1), B (x2, y2) in the ratio – x1: x2.
Point of intersection of two lines:
the point of intersection of two lines a1x + b1y + c = 0 and a2x + b2y + c = 0 is
#### Concurrent Lines:
Three or more lines are said to be concurrent lines if they have a point in common.
The common point is called the point of concurrence.
∗ The condition that the lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 to be concurrent is
a3(b1c2 – b2c1) + b3(c1a2 – c2a1) + c3(a1b2 – a2b1).
∗ The condition that the lines ax + hy +g = 0, hx + by + f = 0 and gx +fy + c = 0 is
abc + 2fgh – af2 – bg2 – ch2 = o.
Note: – if two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 said to be identical (same) if
#### Family of a straight line:
Family of straight lines: – A set of straight lines having a common property is called a family of straight lines.
Let L1 ≡ a1x + b1y + c1 = 0 and L2 ≡ a2x + b2y + c2 =0 represent two intersecting lines, theThe equation λ1 L1 + λ2 L2 = 0 represent a family of straight lines passing through the point of intersection of the lines L1 = 0 and L2 = 0.
∗ The equation of the straight line passing through the point of intersection of the lines L1 = 0 and L2 = 0 is L1 + λL2 = 0.
The angle between two lines:
If θ is the angle between the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 then
∗ If θ is an acute angle then
∗ If θ is the angle between two lines, then (π – θ) is another angle between two lines.
∗ If θ≠π/2 is angle between the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then
∗ If m1, m2 are the slopes of two lines then
Note: – The lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are
∗ Parallel iff
∗ Perpendicular iff a1a2 + b1b2 = 0
The foot of the perpendicular:
If Q (h. k) is the foot of the perpendicular from a point P (x1, y1)to the line ax + by +c = 0 then
Image of the point:
If Q (h. k) is the image of point P (x1, y1) with respect to the line ax + by +c = 0 then
Collinear Points:
If three points are said to be collinear, then they lie on the same line.
∗ If A, B, and C are collinear, then
Slope of AB = Slope of BC (or) Slope of BC = Slope of AC (or) Slope of AB = Slope of AC
## 4. PAIR OF STRAIGHT LINES
∎ ax2 + 2hxy + by2 = 0 is called the second-degree homogeneous equation in two variable x and y.
This equation always represents a pair of straight lines which are passing through the origin.
∎ If l1x + m1y = 0 and l2x + m2y = 0 are two lines represented by the equation ax2 + 2hxy + by2 = 0, then ax2 + 2hxy + by2 = (l1x + m1y) (l2x + m2y)
⇒ a = l1l2; 2h = l1m2 + l2m1; b = m1m2
∎ If m1, m2 are the slopes of the lines represented by the equation ax2 + 2hxy + by2 = 0, then
m1+ m2 = – 2h/b and m1 m2 = a/b
∎ The lines represented by the equation ax2 + 2hxy + by2 = 0 are
∎ If h2 = ab, then the lines represented by the equation ax2 + 2hxy + by2 = 0 are coincident.
∎ If two lines represented by the equation ax2 + 2hxy + by2 = 0 are equally inclined to the coordinate axes then h = 0 and ab < 0.
∎ The equation of the pair of lines passing through the point (h, k) and
(i) Parallel to the lines represented by the equation ax2 + 2hxy + by2 = 0 is
a (x – h)2 + 2h (x – h) (y – k) + b (y – k)2 = 0
(ii) Perpendicular to the lines represented by the equation ax2 + 2hxy + by2 = 0 is
b (x – h)2 – 2h (x – h) (y – k) + a (y – k)2 = 0
Angle between the lines:
If θ is the angle between the lines represented by the equation ax2 + 2hxy + by2 = 0, then
∎ If a + b = 0, then two lines are perpendicular.
Area of the triangle:
The area of the triangle formed by the lines ax2 + 2hxy + by2 = 0 and the line lx + my + n = 0 is
Angular Bisectors:
⇒ the angle between angular bisectors is always 900
L1 = o, L2 = o are two non-parallel lines the locus of the point P such that the perpendicular distance from P to the first lie is equal to the perpendicular distance from P to second line is called the angular bisector of two lines.
⇒ If two lines are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then the angular bisectors are
∎ The equation of the pair of angular bisectors of ax2 + 2hxy + by2 = 0 is h (x2 – y2) = (a – b =) xy.
∎ If ax2 + 2hxy + by2 + 2gx 2fy + c= 0 represents a pair of straight lines then
(i) abc + 2fgh – af2 – bg2 – ch2 = 0
(ii) h2 ≥ ab, g2 ≥ ac and f2 ≥ bc
If two lines represented by ax2 + 2hxy + by2 + 2gx 2fy + c= 0 are l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0, then
ax2 + 2hxy + by2 + 2gx 2fy + c = (l1x + m1y + n1) (l2x + m2y + n2)
a = l1l2; 2h = l1m2 + l2m1; b = m1m2 ; 2g = l1n2 + l2n1; 2f = m1n2 + m2n1 and c = n1n2
The point of intersection of the lines represented by ax2 + 2hxy + by2 + 2gx 2fy + c= 0 is
∎ If two line have same homogeneous path then the lines represented by the first pair is parallel to the lines represented by the second pair.
ax2 + 2hxy + by2 + 2gx 2fy + c= 0 …………… (2)
ax2 + 2hxy + by2 = 0 …………… (1)
equation (1) and equation (2) form a parallelogram, one of the diagonals of parallelogram which is not passing through origin is 2gx + 2fy + c = 0.
∎ If two lines represented by ax2 + 2hxy + by2 + 2gx 2fy + c= 0 are parallel then
• h2 = ab (ii) af2 = bg2 (iii) hf = bg, gh = ab
Distance between parallel lines is
## 5. THREE DIMENSIONAL COORDINATES
• Let X’OX, Y’OY be two mutually perpendicular lines passing through a fixed point ‘O’. These two lines determine the XOY – plane (XY- plane). Draw the line Z’OZ perpendicular to XY – plane and passing through ‘O’.
The fixed point ‘O’ is called origin and three mutually perpendicular lines X’OX, Y’OY, Z’OZ are called Rectangular coordinate axes.
Three coordinate axes taken two at a time determine three planes namely XOY- plane, YOZ-plane, ZOY-plane or XY-plane, YZ-plane, ZX-plane respectively.
For every point P in space, we can associate an ordered triad (x, y, z) of real numbers formed by its coordinates.
The set of points in space is referred to as ‘Three-Dimensional Space’ or R3 – Space.
∗ If P (x, y, z) is a point in a space, then
x is called x-coordinate of P
y is called y-coordinate of P
z is called z-coordinate of P
Distance between two points in space:
∗ Distance between the points A (x1, y1, z1) and B (x2, y2, z2) is
∗ Distance between the point P (x, y, z) to the origin is
Translation of axes:
When the origin is shifted to the point (h, k, l), then
X = x – h; Y = y – k; Z = z – l and x = X + h; y = Y + k; z = Z + l
∗ The foot of the perpendicular from P (x, y, z) to X-axis is A (x, 0, 0).
The perpendicular distance of P from X-axis is
Similarly,
The perpendicular distance of P from Y-axis is
The perpendicular distance of P from Z-axis is
Collinear points: If three or more points lie on the same line are called collinear points.
Section formula:
The point dividing the line segment joining the points A (x1, y1, z1) and B (x2, y2, z2) in the ratio m : n is given by
The mid-point of the line segment joining the points A (x1, y1, z1) and B (x2, y2, z2) is
The centroid of the triangle whose vertices are A (x1, y1, z1), B (x2, y2, z2) and C (x3, y3, z3) is
Tetrahedron:
→ It has 4 vertices and 6 edges.
→ A Tetrahedron is a closed figure formed by four planes not all passing through the same point.
→ Each edge arises as the line of intersection of two of the four planes.
→ The line segment joining the vertices to the centroid of opposite face. The point of concurrence is called centroid of Tetrahedron.
→ Centroid divides the line segment in the ratio 3:1.
→ The centroid of the Tetrahedron whose vertices are A (x1, y1, z1), B (x2, y2, z2), C (x3, y3, z3) and C (x4, y4, z4) is
The line segment joining the points (x1, y1, z1), (x2, y2, z2) is divided by
XY – plane in the ratio – z1: z2
YZ – plane in the ratio – x1: x2
XZ – plane in the ratio – y1: y2
## 6. DIRECTION COSINES AND RATIOS
Consider a ray OP passing through origin ‘O’ and making angles α, β, γ respectively with the positive direction of X, Y, Z axes.
Cos α, Cos β, Cos γ are called Direction Cosines (dc’s) of the ray OP.
Dc’s are denoted by (l, m, n), where l = Cos α, m = Cos β, n = Cos γ
• A line in a space has two directions, it has two sets of dc’s, one for each direction. If (l, m, n) is one set of dc’s, then (-l, -m, -n) is the other set.
• Suppose P (x, y, z) is any point in space such that OP = r. If (l, m, n) are dc’s of a ray OP then x = lr, y= mr, z = nr.
• If OP = r and dc’s of OP are (l, m, n) then the coordinates of P are (lr, mr, nr).
• If P (x, y, z) is a point in the space, then dc’s of OP are
• If (l, m, n) are dc’s of a line then l2 + m2 + n2 = 1.
⇒ cos2α + cos2β + cos2γ = 1.
Direction Ratios:
Any three real numbers which are proportional to the dc’s of a line are called direction ratios (dr’s) of that line.
• Let (a, b, c) be dr’s of a line whose dc’s are (l, m, n). Then (a, b, c) are proportional to (l, m, n)
and a2 + b2 + c2 ≠ 1.
• Dr’s of the line joining the points (x1, y1, z1), (x2, y2, z2) are (x2 – x1, y2 – y1, z2 – z1)
• If (a, b, c) are dr’s of a line then its dc’s are
• If (l1, m1, n1), (l2, m2, n2) are dc’s of two lines and θ is angle between them then Cos θ = l1l2 + m1m2 + n1n2
If two lines perpendicular then l1l2 + m1m2 + n1n2 = 0.
• If (a1, b1, c1), (a2, b2, c2) are dr’s of two lines and θ is the angle between them then
If two line are perpendicular then a1a2 + b1b2 + c1c2 = 0.
## 7. THE PLANE
Plane: A plane is a proper subset of R3 which has at least three non-collinear points and any two points in it.
∎ Equation of the plane passing through a given point A (x1, y1, z1), and perpendicular to the line whose dr’s (a, b, c) is a(x – x1) + a(y – y1) + a(z – z1) = 0.
∎ The equation of the plane hose dc’s of the normal to the plane (l, m, n) and perpendicular distance from the origin to the pane p is lx + my + nz = p
∎ The equation of the plane passing through three non-collinear points A (x1, y1, z1), B (x2, y2, z2) and C (x3, y3, z3) is
∎ The general equation of the plane is ax + by + cz + d = 0, where (a, b, c) are Dr’s of the normal to the plane.
Normal form:
The equation of the plane ax + by + cz + d = 0 in the normal form is
Perpendicular distance:
The perpendicular distance from (x1, y1, z1) to the plane ax + by + cz + d = 0 is
The perpendicular distance from the origin to the plane ax + by + cz + d = 0 is
Intercepts:
X- intercept = aIf a plane cuts X –axis at (a, 0, 0), Y-axis at (0, b, 0) and Z-axis at (0, 0, c) then
Y-intercept = b
Z-intercept = c
The equation of the plane in the intercept form is
The intercepts of the plane ax + by + cz + d =0 is -d/a, -d/b, -d/c
∎ The equation of the plane parallel to ax + by + cz + d = 0 is ax + by + cz + k = 0.
∎ The equation of XY – plane is z = 0.
∎ The equation of YZ – plane is x = 0.
∎ The equation of XZ – plane is y = 0.
∎ Distance between the two parallel planes ax + by + cz + d1 =0 and ax + by + cz + d2 =0 is
The angle between two planes:
The angle between the normal to two planes is called the angle between the planes.
If θ is the angle between the planes a1 x + b1 y + c1 z + d1 =0 and a2 x + b2 y + c2 z + d2 =0 then
If two line are perpendicular then a1a2 + b1b2 + c1c2 = 0.
∎ The distance of the point P (x, y, z) from
## 8. LIMITS AND CONTINUITY
Intervals:
Let (a, b) ∈ R such that a ≤ b, then the set
• {x ∈ R: a ≤ x ≤ b}, is denoted by [a, b] and it is called as closed interval
• {x ∈ R: a < x < b}, is denoted by (a, b) and it is called as open interval
• {x ∈ R: a < x ≤ b}, is denoted by (a, b] and it is called as open closed interval
• {x ∈ R: a ≤ x < b}, is denoted by [a, b) and it is called as closed open interval
• {x ∈ R: x ≥ a}, is denoted by [a, ∞)
• {x ∈ R: x > a}, is denoted by (a, ∞)
• {x ∈ R: x ≤ a}, is denoted by (- ∞, a]
• x ∈ R: x < a}, is denoted by (- ∞, a)
Neighbourhood:
Let a ∈ R. If δ > 0, then the open interval (a – δ, a + δ) is called the δ – neighbourhood of ‘a’
Limit:
If f(x) is a function of x such that if x approaches to a constant value ‘a’, then the value of f(x) also approaches to ‘l’. Then the constant ‘I’ is called a limit of f(x) at x = a
Or
A real number l is called the limit of the function f, if for all ϵ> 0 there exist δ > 0 such that whenever ⟹
Properties of Limits:
Sand witch theorem:( Squeez Principle):
f, g, and h are functions such that f(x) ≤ g(x) ≤ h(x), then and if
Left- hand and Right-hand Limits:
If x < a, then is called left-hand limit
If x > a, then is called right-hand limit
Note:
In Determinate forms:
if a function f(x) any of the following forms at x = a:
Then f(x) is said to be indeterminate at x = a.
### Continuity:
Condition 1: If the condition is like x = a and x ≠ a, then we use following property.
If then f(x) is continuous at x = a, otherwise f(x) is not continuous.
Condition 2: If the condition is like x ≤ a and x >a, or x < a and x ≥a then we use following property.
If then f(x) is continuous at x = a, otherwise f(x) is not continuous.
## 9.DIFFERENTIATION
Let f be a function defined on a neighbourhood of a real number ‘a’ if exist then we say that f is differentiable at x a and it is denoted by f'(a).
∴ f’(a) =
∎ If right hand derivative = left hand derivative, then f is differentiable at ‘a’.
i.e.,
First principle in derivative:
The first principle of the derivative of f at any real number ‘x’ is f’(x) =
∎ The differentiation of f(x) is denoted by
means differentiation of ‘y’ with respect to ‘x’
The derivative of constant function is zero i.e., f’(a) = 0 where ‘a’ is any constant.
∎ Let I be an interval in R u and v are real valued functions on I and x ∈ I. Suppose that u and v are differentiable at ‘x’, then
• (u ± v) is also differentiable at ‘x’ and (u ± v)’(x) = u’ (x) ± v’(x).
• ‘uv’ is also differentiable at ‘x’ and (uv)’(x) = u(x) v’(x) + v(x) u’(x).
• αu + βv is also differentiable at ‘x’ and (αu + βv)’(x) = αu’(x) + βv’(x), α, β are constants.
• is also differentiable at ‘x’ and
∎ (f o g)’ (x) = f’(g(x)). g’(x).
Formulae:
Derivative of Trigonometric & Inverse trigonometric functions:
Derivative of Hyperbolic & Inverse Hyperbolic functions:
Parametric Differentiation:
If x = f(t) and y = g(t) then the procedure of finding in terms of the parameter ‘t’ is called parametric equations.
Implicitly differentiation:
An equation involving two or more variables is called an implicit equation.
ax2 + 2hxy + b y2 = 0 is an implicit equation in terms of x and y.
The process of finding from an implicit equation is called implicitly differentiation.
Derivative of one function w.r.t. another function:
The derivative of f(x) w.r.t g(x) is
Second order derivative:
Let y = f(x) be a function, if y is differentiable then the derivative of f is f’(x). If ‘(x) is again differentiable then the derivative of f’(x) is called second order derivative. And it is denoted by f” (x) or
## 10. ERRORS AND APPROXIMATIONS
Approximations:
Let y f(x) be a function defined an interval I and x ∈ I. If ∆x is any change in x, then ∆y be the corresponding change in y thus ∆y = f (x + ∆x) – f (x).
Let
where ϵ is very small
For ‘ϵ.∆x’ is very small and hence,
Approximate value is f (x + ∆x) = f(x) + f’(x). ∆x
Differential:
Let y f(x) be a function defined an interval I and x ∈ I. If ∆x is any change in x, then called differential of y = f(x) and it is denoted by df.
∴ dy = f’(x). ∆x
Errors:
Let y f(x) be a function defined an interval I and x ∈ I. If ∆x is any change in x, then ∆y be the corresponding change in y.
### The Following formulae will be used in Solving problems
CIRCLE:
If ‘r’ is radius, ‘d’ is diameter ‘P’ is the perimeter or circumference and A is area of the circle then
d= 2r, P = 2πr = πd and A = πr2sq.u
SECTOR:
If ‘r’ is the radius, ‘l’ is the length of arc and θ is of the sector then
Area = ½ l r = ½ r2θsq.u.
Perimeter = l + 2r = r (θ + 2) u.
CYLINDER:
Length of the Arc ‘l’ = rθ (θ must be in radians).
If ‘r is the radius of the base of cylinder and ‘h’ is the height of the cylinder, then
Area of base = πr2 sq.units.
Lateral surface area = 2πrh units.
Total surface area = 2πr (h + r) units.
Volume = πr2 h cubic units.
CONE:
If ‘r’ is the radius of base, ‘h’ is the height of cone and ‘l’ is slant height then
l 2 + r2 = h2
Lateral surface area = πrl units.
Total surface area = πr (l + r) sq. units.
Volume = cubic units.
SPHERE:
If ‘r’ is the radius of the Sphere then
Surface area = πr2 sq. units.
Volume = πr3 cubic units.
## 11. TANGENTS AND NORMALS
Tangent of a Curve:
If the secant line PQ approaches to the same position as Q moves along the curve and approaches to either side then limiting position is called a ‘Tangent line’ to the curve at P. The point P is called point of contact
Let y = f(x) be a curve, P a point on the curve. If Q(≠P) is another point on the curve then the line PD is called secant line.
Let y = f(x) be a curve and P (x, y) be a point on the curve. The slope of the tangent to the curve y = f(x) at P is called gradient of the curve.
Slope of the tangent to the curve y = f(x) at P (x, y) is m =
∎ The equation of the tangent at P (x1, y1) to the curve is y – y1 = m (x – x1) where m =
Normal of a curve:
Let y = f(x) be a curve and P (x, y) be a point on the curve. The line passing through P and perpendicular to the tangent of the curve y = f(x) at P is called Normal of the curve.
∎ The equation of the tangent at P (x1, y1) to the curve is y – y1 = -1/m (x – x1).
Slope of the normal is -1/m. where m =
Lengths of tangent, normal, subtangent and subnormal:
PT → Normal; QN → subnormal
PN → Tangent; QT → subtangent
∎ if m = then
Angle between two curves:
If two curves intersect at a point P., then the angle between the tangents of the curves at P is called the angle between the curves at P.
∎ If m1, m2 are the slopes of two tangents of the two curves and θ is the angle between the curves then
Tanθ =
Note:
• If m1= m2, then two corves are touch each other.
• if m1× m2 = –1, then two curves intersect orthogonally.
## 12. RATE MEASURE
Average rate of change:
if y = f(x) then the average rate of change in y between x = x1 and x = x2 is defined as
Instantaneous rate of change:
if y = f(x), then the instantaneous rate of change of a unction f at x = x0 is defined as
Rectilinear Motion:
A motion of a particle in a line is called Rectilinear motion. The rectilinear motion is denoted by s = f(t) where f(t) is the rule connecting ‘s’ and ‘t’.
Velocity, Acceleration:
A particle starts from a fixed point and moves a distance ‘S’ along a straight-line during time ‘t’ then
Velocity =
Acceleration =
Note:
(i) If v> 0, then the particle s moving away from the straight point.
(ii) If v < 0, then particle s moving away towards the straight point.
(iii) If v = 0, then the particle comes rest.
## 13.ROLLE’S & LANGRANGEE’S THEOREM
Rolle’s Theorem:
Suppose a, b (a < b) are two real numbers. Let f: [a, b] → R be a function satisfying the following conditions:
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b) and
(iii) f(a) = f(b)
then there exists at least one c ∈ (a, b) such that f’(c)= 0.
Lagrange’s Theorem:
Suppose a, b (a < b) are two real numbers. Let f: [a, b] → R be a function satisfying the following conditions:
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b) and
then there exists at least one c ∈ (a, b) such that f’(c)=
## 14.INCREASING & DECREASING FUNCTIONS
Let f be a real function on an interval I then f is said to be
(i) an increasing function on I if
x1 < x2 ⇒ f (x1) ≤ f (x2) ∀ x1, x2 ∈ I
(ii) decreasing function on I if
x1 < x2 ⇒ f (x1) ≥ f (x2) ∀ x1, x2 ∈ I
Let f be a real function on an interval I then f is said to be
(i) strictly increasing function on I if
x1 < x2 ⇒ f (x1) < f (x2) ∀ x1, x2 ∈ I
(ii) strictly decreasing function on I if
x1 < x2 ⇒ f (x1) > f (x2) ∀ x1, x2 ∈ I
Let f(x) be a real valued function defined on I = (a, b) or [a, b) or (a, b] or [a, b]. Suppose f is continuous on I and differentiable in (a, b). If
(i) f’ (c) > 0 ∀ c ∈ (a, b), then f is strictly increasing on I
(ii) f’ (c) < 0 ∀ c ∈ (a, b), then f is strictly decreasing on I
(iii) f’ (c) ≥ 0 ∀ c ∈ (a, b), then f is increasing on I
(iv) f’ (c) ≤ 0 ∀ c ∈ (a, b), then f is decreasing on I
Critical point:
A point x = c in the domain of the function said to be ‘critical point’ of the function f if either f’ (c) = 0 or f’ (c) does not exists.
Stationary point:
A point x = c in the domain of the function said to be ‘stationary point’ of the function f if f’ (c) = 0.
MAXIMA & MINIMA
Global maxima – Global minima:
Let D be an interval in R and f: D → R be a real function and c ∈ D. Then f is said to be
(i) a global maximum on D if f(c) ≥ f(x)
(ii) a global minimum on D if f(c) ≤ f(x)
Relative maximum:
Let D be an interval in R and f: D → R be a real function and c ∈ D. Then f is said to be relative maximum at c if there exist δ > 0 such that f(c) ≥ f(x) ∀ x ∈ (c – δ, c + δ).
Here, f (c) is called relative maximum value of f(x) at x = c and the point x = c is called point of relative maximum.
Relative minimum:
Let D be an interval in R and f: D → R be a real function and c ∈ D. Then f is said to be relative maximum at c if there exist δ > 0 such that f(c) ≤ f(x) ∀ x ∈ (c – δ, c + δ).
Here, f (c) is called relative maximum value of f(x) at x = c and the point x = c is called point of relative minimum.
The relative maximum and minimum value of f are called extreme values.
If f is either minima or maxima f’ (α) = 0.
Let f be a continuous function om [a, b] and α ∈ (a, b)
(i) If f’ (α) = 0 and f’’ (α) >0, then f(α) is relative minimum.
(ii) if f’ (α) = 0 and f’’ (α) <0, then f(α) is relative maximum
## 1.Functions
Set: A collection of well-defined objects is called a set.
Ordered pair: Two elements a and b listed in a specific order form. An ordered pair denoted by (a, b).
Cartesian product: Let A and B are two non-empty sets. The Cartesian product of A and B is denoted by A × B and is defined as a set of all ordered pairs (a, b) where a ϵ A and b ϵB
Relation: Let A and B are two non-empty sets the relation R from A to B is a subset of A×B.
⇒ R: A→B is a relation if R⊂ A × B
#### Function:
A relation f: A → B is said to be a function if ∀ aϵ A there exists a unique element b such that (a, b) ϵ f. (Or)
A relation f: A → B is said to be a function if
(i) x ϵ A ⇒ f(x) ϵ B
(ii) x1 , x2 ϵ A , x1 = x2 in A ⇒ f(x1) = f(x2) in B.
Note: If A, B are two finite sets then the no. of functions that can be defined from A to B is n(B)n(A)
VARIOUS TYPES OF FUNCTIONS
One– one Function (Injective):- A function f: A→ B is said to be a one-one function or injective if different elements in A have different images in B.
(Or)
A function f: A→ B is said to be one-one function if f(x1) = f(x2) in B ⇒ x1 = x2 in A.
Note: No. of one-one functions that can be defined from A into B is n(B) p n(A) if n(A) ≤ n(B)
On to Function (Surjection): – A function f: A→ B is said to be onto function or surjection if for each yϵ B ∃ x ϵ A such that f(x) =y
Note: if n(A) = m and n(B) = 2 then no. of onto functions = 2m – 2
Bijection: – A function f: A→ B is said to be Bijection if it is both ‘one-one and ‘onto’.
Constant function: A function f: A→ B is said to be constant function if f(x) = k ∀ xϵA
Identity function: Let A be a non-empty set, then the function defined by I: A → A, I(x)=x is called identity function on A.
Equal function: Two functions f and g are said to be equal if
(i) They have same domain (D)
(ii) f(x) = g(x) ∀ xϵ D
Even function: A function f: A→ B is said to be even function if f (- x) = f(x) ∀ xϵ A
Odd function: A function f: A→ B is said to be odd function if f (- x) = – f(x) ∀ xϵ A
Composite function: If f: A→B, g: B→C are two functions then the composite relation is a function from A to C.
gof: A→C is a composite function and is defined by gof(x) = g(f(x)).
Step function: A number x = I + F
I → integral part = [x]
F → fractional part = {x}
∴ x = [x] + {x}
If y = [x] then domain = R and
Range = Z
0 ≤ x ≤ 1, [x] = 0
1≤ x ≤ 2, [x] = 1
-1 ≤ x ≤ 0, [x] = -1
If k is any integer [ x + k] = k + [x]
The value of [x] is lies in x – 1 < [x] ≤ 1.
Inverse function: If f: A → B is bijection then f -1 is exists
f-1: B → A is an inverse function of f.
### SOME IMPORTANT POINTS
of subsets of a set of n elements is 2n
of proper subsets of a set of n elements is 2n – 1
Let A and B are two non-empty finite sets and f: A → B is a function. This function will
One-one if n(A) ≤ n(B)
On to if n(A) ≥ n(B)
Bijection if n(A) = n(B).
## 3. MATRICES
Matrix: An ordered rectangular array of elements is called a matrix
• Matrices are generally enclosed by brackets like
• Matrices are denoted by capital letters A, B, C and so on
• Elements in a matrix are real or complex numbers; real or complex real-valued functions.
Oder of Matrix: A matrix having rows and ‘n’ columns is said to be of order m x n. Read as m by n.
### Square Matrix: A matrix in which the no. of rows is equal to the no. of columns is called a square matrix.
Principal diagonal ( diagonal) Matrix: If A = [aij] is a square matrix of order ‘n’ the elements a11 , a22 , a33 , ………. ann is said to constitute its principal diagonal.
Trace Matrix: The sum of the elements of the principal diagonal of a square matrix A is called the trace of the matrix. It is denoted by Tr (A).
Ex:-
Diagonal Matrix: If each non-diagonal element of a square matrix is ‘zero’ then the matrix is called a diagonal matrix.
Scalar Matrix: If each non-diagonal elements of a square matrix are ‘zero’ and all diagonal elements are equal to each other, then it is called a scalar matrix.
Identity Matrix or Unit Matrix: If each of the non-diagonal elements of a square matrix is ‘zero’ and all diagonal elements are equal to ‘1’, then that matrix is called a unit matrix.
Null Matrix or Zero Matrix: If each element of a matrix is zero, then it is called a null matrix.
Row matrix & column Matrix: A matrix with only one row s called a row matrix and a matrix with only one column is called a column matrix.
Triangular matrices:
A square matrix A = [aij] is said to be upper triangular if aij = 0 ∀ i > j
A square matrix A = [aij] is said to be lower triangular matrix aij = 0 ∀ i < j
Equality of matrices: matrices A and B are said to be equal if A and B of the same order and the corresponding elements of A and B are equal.
### Product of Matrices:
Let A = [aik]mxn and B = [bkj]nxp be two matrices ,then the matrix C = [cij]mxp where
Note: Matrix multiplication of two matrices is possible when no. of columns of the first matrix is equal to no. of rows of the second matrix.
Transpose of Matrix: If A = [aij] is an m x n matrix, then the matrix obtained by interchanging the rows and columns is called the transpose of A. It is denoted by AI or AT.
Note: (i) (AI)I = A (ii) (k AI) = k . AI (iii) (A + B )T = AT + BT (iv) (AB)T = BTAT
Symmetric Matrix: A square matrix A is said to be symmetric if AT =A
If A is a symmetric matrix, then A + AT is symmetric.
Skew-Symmetric Matrix: A square matrix A is said to be skew-symmetric if AT = -A
If A is a skew-symmetric matrix, then A – AT is skew-symmetric
Minor of an element: Consider a square matrix
the minor an element in this matrix is defined as the determinant of the 2×2 matrix obtained after deleting the rows and the columns in which the element is present.
Cofactor of an element: The cofactor of an element in i th row and j th column of A3×3 matrix is defined as it’s minor multiplied by (- 1 ) i+j .
### Properties of determinants:
• If each element of a row (column) of a square matrix is zero, then the determinant of that matrix is zero.
• If A is a square matrix of order 3 and k is scalar then.
• If two rows (columns) of a square matrix are identical (same), then Det. Of that matrix is zero.
• If each element in a row (column) of a square matrix is the sum of two numbers then its determinant can be expressed as the sum of the determinants.
• If each element of a square matrix are polynomials in x and its determinant is zero when x = a, then (x-a) is a factor of that matrix.
• For any square matrix A Det(A) = Det (AI).
• Det(AB) = Det(A) . Det(B).
• For any positive integer n Det(An) = (DetA)n.
Singular and non-singular matrices: A Square matrix is said to be singular if its determinant is zero, otherwise it is said to be the non-singular matrix.
Ad joint of a matrix: The transpose of the matrix formed by replacing the elements of a square matrix A with the corresponding cofactors is called the adjoint of A.
Invertible matrix: Let A be a square matrix, we say that A is invertible if there exists a matrix B such that AB =BA = I, where I is the unit matrix of the same order as A and B.
Augmented matrix: The coefficient matrix (A) augmented with the constant column matrix (D) is called the augmented matrix. It is denoted by [AD].
Sub matrix: A matrix obtained by deleting some rows and columns (or both) of a matrix is called the submatrix of the given matrix.
Let A be a non-zero matrix. The rank of A is defined as the maximum of the order of the non-singular submatrices of A.
• Note: If A is a non-zero matrix of order 3 then the rank of A is:
• 1, if every 2×2 submatrix is singular
• 2, if A is singular and at least one of its 2×2 sub-matrices is non-singular
(iii) 3, if A is non – singular.
Consistent and Inconsistent: The system of linear equations is consistent if it has a solution, in-consistent if it has no solution.
• Note: The system of three equations in three unknowns AX = D has
• A unique solution if rank(A) = rank ([AD]) = 3
• Infinitely many solutions if rank (A) = ([AD]) < 3
• No solution if rank (A) ≠ rank ([AD])
### Solutions of a homogeneous system of linear equations:
The system of equations AX = 0 has
• The trivial solution only if rank(A) = 3
• An infinite no. of solutions if rank(A) < 3
Directed line: If A and B are two distinct points in the space, the ordered pair (A, B) denoted by AB is called a directed line segment with initial point A and terminal point B.
⇒ A directed line passes through three characteristics: (i) length (ii) support (iii) direction
Scalar: A quantity having magnitude only is called a scalar. We identify real numbers as a scalar.
Ex: – mass, length, temperature, etc.
Vector: A quantity having length and direction is called a vector.
Ex: – velocity, acceleration, force, etc.
⇒ If is a vector then its length is denoted by
Position of vector: If P (x, y, z) is any point in the space, then is called the position vector of the point P with respect to origin (O). This is denoted by
Like and unlike vectors: If two vectors are parallel and having the same direction then they are called like vectors.
If two vectors are parallel and having opposite direction then they are called, unlike vectors.
Coplanar vectors:
Vectors whose supports are in the same plane or parallel to the same plane are called coplanar vectors.
Triangle law: If are two vectors, there exist three points A, B, and C in a space such that defined by
Parallelogram law: If two vectors and represented by two adjacent sides of a parallelogram in magnitude and direction then their sum is represented in magnitude and direction by the diagonal of the parallelogram through their common point.
Scalar multiplication: Let be a vector and λ be a scalar then we define vector λ to be the vector if either is zero vector or λ is the scalar zero; otherwise λ is the vector in the direction of with the magnitude if λ>0 and λ = (−λ)(− ) if λ<0.
The angle between two non-zero vectors: Let be two non-zero vectors, let then ∠AOB has two values. The value of ∠AOB, which does not exceed 1800 is called the angle between the vectors and , it is denoted by ( ).
Section formula: Let be two position vectors of the points A and B with respect to the origin if a point P divides the line segment AB in the ratio m:n then
Linear combination of vectors: let be vectors x1, x2, x3…. xn be scalars, then the vector is called the linear combination of vectors.
Components: Consider the ordered triad (a, b, c) of non-coplanar vectors If r is any vector then there exist a unique triad (x, y, z) of scalars such that . These scalars x, y, z are called the components of with respect to the ordered triad (a, b, c).
• i, j, k are unit vectors along the X, Y and Z axes respectively and P(x, y, z) is any point in the space then = r = x i + y j +z k and
Regular polygon: A polygon is said to be regular if all the sides, as well as all the interior angles, are equal.
• If a polygon has sides then the no. of diagonals of a polygon is
• The unit vector bisecting the angle between is
### Vector equation of a line and plane
⇒The vector equation of the line passing through point A () and ∥el to the vector is
Proof:-
Then AP, are collinear vector proof: let P ( ) be any point on the line a
the equation of the line passing through origin and parallel to the vectoris
• the vector equation of the line passing through the points A( ) and B( ) is
• Cartesian equation of the line passing through A ( x1, y1, z1) and B ( x2, y2, z2) is
• The vector equation of the plane passing through point A( ) and parallel to the vectors and is
• The vector equation of the plane passing through the point A( ), B( ) and parallel to the vector is
• The vector equation of the plane passing through the points A( ), B( ) and C( ) is
## 5.PRODUCT OF VECTORS
Dot product (Scalar product): Let are two vectors. The dot product or direct product of and is denoted byand is defined as
• If = 0, = 0 ⟹ = 0.
• If ≠0, ≠ 0 then
• The dot product of two vectors is a scalar
• If are two vectors, then
• If θ is the angle between the vectors then.
⟹
⟹ If > 0, then θ is an acute angle
⟹ If < 0, then θ is obtuse angle 0
⟹ If = 0, then is perpendicular to
• If is any vector then
Component and Orthogonal Projection:
Let=,= be two non-zero vectors. Let the plane passing through B ( ) and perpendicular to intersects
In M, then is called the component of on
• The component (projection) vector of on is
• Length of the projection (component) =
• Component of perpendicular to =
If ,, form a right-handed system of an orthonormal triad, then
• If then = a1b1 + a2b2 + a3b3
• If then
Parallelogram law:
In a parallelogram, the sum of the squares of the lengths of the diagonals is equal to the sum of the squares of the lengths of its sides.
In ∆ABC, the length of the median through vertex A is
Vector equation of a plane:
The vector equation of the plane whose perpendicular distance from the origin is p and unit normal drawn from the origin towards the plane is,
•The vector equation of the plane passing through point A ( ) and perpendicular to the is
•If θ is the angle between the planes then
Cross product (vector product): Let and be two non-zero collinear vectors. The cross product of and is denoted by × (read as a cross ) and is defined as
• The vector × is perpendicular to both and and also perpendicular to the plane containing them
• The unit vector perpendicular to both and is
• Let then
• If and are two sides of a triangle then the area of the triangle =
• If A ( ), B ()and C ( )are the vertices of a ∆ABC, then its area
• The area of the parallelogram whose adjacent sides and is
• The area of the parallelogram whose diagonals and is
• If A ( ), B ( )and C ( )are three points then the perpendicular distance from A to the line passing through B, C is
Let,and be three vectors, then () . is called the scalar triple product of,andand it is denoted by | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.90591961145401, "perplexity": 1034.4873616508464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038072175.30/warc/CC-MAIN-20210413062409-20210413092409-00330.warc.gz"} |
https://forum.math.toronto.edu/index.php?PHPSESSID=tvua456s21pcbdh91jskokske7&topic=2547.0 | ### Author Topic: Variation of Parameter for higher order eqautions (Read 622 times)
#### Sophie Liao
• Newbie
• Posts: 1
• Karma: 0
##### Variation of Parameter for higher order eqautions
« on: November 18, 2020, 12:26:03 PM »
Since we have learned the variation of parameter method for second and third order equation, I just want to confirm whether my idea is correct or not. For fourth order eqatuion or higher, should we compute the det of 4 by 4 matrix or even higher to use this method? Is there any general theorem for this method?
#### shiyuancao
• Jr. Member
• Posts: 6
• Karma: 0
##### Re: Variation of Parameter for higher order eqautions
« Reply #1 on: November 18, 2020, 09:34:38 PM »
I think your idea is correct and it's unmistakable to apply the same principle when solving higher order equations | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9619653224945068, "perplexity": 2068.6287079737594}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104054564.59/warc/CC-MAIN-20220702101738-20220702131738-00090.warc.gz"} |
https://www.physicsforums.com/threads/simple-gear-pair-efficiency-and-output-torque.920215/ | # Simple gear pair: efficiency and output torque
1. Jul 15, 2017
### liao
Hi. I have a question about output torque and efficiency of a simple gear pair as shown on the picture. So, I have a pinion and a gear. I give an input torque Tp in the clockwise direction. Therefore, the pinion will rotate with ωp angular velocity in clockwise and the gear ωg in counter-clockwise. Then, I apply a load Tl in the opposite direction of the gear motion (clockwise direction). The gear is still rotating in the counter-clockwise. I neglect the friction between the gear mesh, but I consider the bearing friction both in pinion cp*ωp and gear cg*ωg. The moments of inertia of the pinion and the gear are Ip and Ig, respectively. My question is what the output torque To is because I want to find the efficiency of this gear pair. Thank you very much.
2. Jul 15, 2017
### liao
@CWatters if To = Tl, I would not get a correct efficiency η.
η = Po/Pi = To ωg/(Tp ωp) = Tl ωg/(Tp ωp)
we know that rg ωg = rp ωp and ωg/ωp = rp/rg, so
η = Tl rp/(Tp rg)
In this equation, we can see that Tl/Tp has to be rg/rp in order to have η = 1. If I don't have Tl/Tp close to rg/rp, the efficiency could be much less than 1, for example 0.2, which doesn't make sense.
3. Jul 16, 2017
### CWatters
If Tl <> TO the system accelerates (decelerates) putting energy into (removing energy from) the angular momentum of the gears. Load also takes more/less power. That's where the missing energy goes (comes from) if n <> 1.
What happens if Tl >> TO ? Which end is driving which?
Last edited: Jul 16, 2017
4. Jul 16, 2017
### liao
Thanks again for your reply. If Tl >> To, I believe the gear will drive the pinion in the load direction. However, I'm only interested in the case where the load is small such that the pinion still drives the gear. Just to confirm, did you think that my analysis on the efficiency when To = Tl correct?
5. Jul 16, 2017
### CWatters
Yes.
.
6. Jul 16, 2017
### CWatters
Regarding friction..
Without friction
To = Tp * rg/rp
With friction in the bearings..
To = ((Tp - cp*ωp) * rg/rp) - cg*ωg
7. Jul 16, 2017
### liao
Thanks for your reply. I simulated your To with friction in the bearings in Simulink, but To is incorrect. Here are the results. So, I drive the pinion for a SFUDS (simplified federal urban driving schedule). rg/rp =3. The first figure is the angular velocities of both pinion and gear. We can see that the pinion turns 3 times faster than the gear. The second figure is the torque. Let me know your thoughts. Thanks again. I really appreciate your help! :)
8. Jul 16, 2017
### CWatters
Sorry I've never used Simulink. Clearly something is wrong as you are applying an input torque and getting no (or very little) output torque. Is there a load?
9. Jul 16, 2017
### Nidum
The problem is over defined for the steady state case and under defined for the transient case .
Just for starters which case is it that are you looking at ?
In any case I strongly suggest that you use simple hand calculations for this problem rather than Simulink .
10. Jul 16, 2017
### liao
Yes, there is. Basically I derived the equation of motion of the simple gear by Lagrange's equation, then find the time response. I have tried three other options for To. Could you check it on the attached brief write-up? Thanks a lot CWatters!
#### Attached Files:
• ###### 3moreOptions.pdf
File size:
227.1 KB
Views:
29
11. Jul 16, 2017
### liao
Hi Nidum. Thanks for your reply. Sorry, I should've said that I'm looking at the transient case. Why is it under defined?
I've done simple hand calculations. I use Simulink to validate it.
12. Jul 16, 2017
### Nidum
I don't want to back track through what you have done so far . We'll start afresh using proper engineering principles .
So action the first : Show me two free body diagrams - one for each gear . Show all relevant information on each diagram .
13. Jul 16, 2017
### Nidum
-- and use some good sketching software for the diagrams if you can .
14. Jul 16, 2017
### liao
#### Attached Files:
• ###### NewtonEuler.PNG
File size:
52.9 KB
Views:
24
15. Jul 16, 2017
### liao
sorry, I don't have experience of sketching software for these diagrams. I will learn it though. Thanks.
16. Jul 16, 2017
### Nidum
The inertial reaction torques are missing - the gears have moments of inertia and they are accelerating ?
Best not to mix tangential forces and torques . Use only torques when you re-draw the diagrams .
17. Jul 16, 2017
### liao
Yes, the gears have moments of inertia and they are accelerating. Please see the revised diagram, I hope this one is clearer. Tr is the reaction torque.
#### Attached Files:
File size:
42.4 KB
Views:
21
18. Jul 16, 2017
### Nidum
The reaction torques are in general going to be different for the two gears ?
Also we need a way to define the gear ratio
19. Jul 16, 2017
### liao
Question: The reaction torques are the product of the internal constraint force F and the gear/pinion radius, right? if so, yes, they're different due to different radius.
The gear ratio is ωp/ωg = rg/rp
20. Jul 16, 2017
### Nidum
I was referring to the inertial reaction torques as mentioned in my previous post .
Anyway we have what we need now . It's late here so I am signing off for tonight . Talk to you tomorrow .
Have something to add?
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Similar Discussions: Simple gear pair: efficiency and output torque | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8889002203941345, "perplexity": 2792.4741171362043}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084886437.0/warc/CC-MAIN-20180116144951-20180116164951-00638.warc.gz"} |
https://stats.stackexchange.com/questions/410348/is-crossover-only-connected-to-genetic-algorithm | # Is Crossover only connected to Genetic algorithm?
Is the idea of crossover only restricted to Genetic Algorithms ? Are there any other evolutionary algorithms that uses crossover(even under another name ) ? If an algorithm uses crossover but does not utilize selection or mutation , should we call it a genetic algorithm ? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9222995638847351, "perplexity": 1162.2715764490954}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738950.31/warc/CC-MAIN-20200812225607-20200813015607-00427.warc.gz"} |
https://stats.stackexchange.com/questions/471143/rigorous-statement-of-expectations-for-the-bias-variance-trade-off | # Rigorous statement of expectations for the bias-variance trade-off
Consider a data generating process $$Y=f(X)+\varepsilon$$ where $$\varepsilon$$ is independent of $$x$$ with $$\mathbb E(\varepsilon)=0$$ and $$\text{Var}(\varepsilon)=\sigma^2_\varepsilon$$. According to Hastie et al. "The Elements of Statistical Learning" (2nd edition, 2009) Section 7.3 p. 223, we can derive an expression for the expected prediction error of a regression fit $$\hat f(X)$$ at an input point $$X=x_0$$, using squared-error loss:
\begin{align} \text{Err}(x_0) &=\mathbb E[(Y-\hat f(x_0))^2|X=x_0]\\ &=(\mathbb E[\hat f(x_0)−f(x_0)])^2+\mathbb E[(\hat f(x_0)−\mathbb E[\hat f(x_0)])^2]+\sigma^2_\varepsilon\\ &=\text{Bias}^2\ \ \ \quad\quad\quad\quad\quad\;\;+\text{Variance } \quad\quad\quad\quad\quad\quad+ \text{ Irreducible Error} \end{align}
(where I use the notation $$\text{Bias}^2$$ instead of $$\text{Bias}$$).
Question: What are the expectations taken over? What is held fixed and what is random?
The question arose in the comments of the thread "Why is there a bias variance tradeoff? A counterexample".
• $x_0$ and $f(x_0)$ are assumed fixed. $\hat{f}(.)$, however, depends on the training data, which is taken as random. Finally, there is randomness in $Y|X=x_0$, which is assumed independent of $\hat{f}(.)$ (conditional on $X=x_0$). – Tim Mak Jun 9 '20 at 10:01
• @TimMak, sounds reasonable. So what are the expectations taken over? The random sampling of the training set jointly with the randomness in $\varepsilon$, i.e. a double integral? I have added the qualification that $\varepsilon$ is independent of $x$, but should that be the case? (I guess it should.) – Richard Hardy Jun 9 '20 at 10:16
• Actually, I read the book in more detail and found that in their presentation $X$ is actually assumed fixed. Section 7.3 refers back I think, to Section 2.9, where they said "For simplicity here we assume that the values of $x_i$ in the sample are fixed in advance (nonrandom)". However, in other presentations, it is common to have $X$ assumed random also. Either way, $\hat{f}(.)$ depends on both $X$ and $Y$, and hence is random even when $X$ is held fixed. – Tim Mak Jun 10 '20 at 2:04
• @TimMak, since you are knowledgeable about the matter, consider writing up an answer. – Richard Hardy Jun 10 '20 at 5:52
$$X$$ is assumed fixed; see Section 2.9, p. 37:
For simplicity here we assume that the values of $$x_i$$ in the sample are fixed in advance (nonrandom).
Then the only source of random variation here is $$\varepsilon$$. Hence, the expectations are taken w.r.t. to the distribution of $$\varepsilon$$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9743540287017822, "perplexity": 554.0787630056894}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178357929.4/warc/CC-MAIN-20210226145416-20210226175416-00296.warc.gz"} |
http://mathhelpforum.com/calculus/105618-volumes-solids-revolution-integration.html | # Math Help - Volumes of Solids of Revolution - Integration
1. ## Volumes of Solids of Revolution - Integration
I hope this in the correct section, admin please move it if its not
Determine the volume generated when the given plane figure is rotated about the y axis
$x^2 - y^2 = 16$
$y=0$ $x=8$
I'm having a problem with the integration part of it
2. Originally Posted by khanim
I hope this in the correct section, admin please move it if its not
Determine the volume generated when the given plane figure is rotated about the y axis
$x^2 - y^2 = 16$
$y=0$ $x=8$
I'm having a problem with the integration part of it
By definition: $V = \pi \int_{y = 0}^{y = 8} x^2 \, dy$.
Substitute $x^2 = 16 + y^2$ into the integral. I don't see where the trouble can be ....?
By definition: $V = \pi \int_{y = 0}^{y = 8} x^2 \, dy$.
Substitute $x^2 = 16 + y^2$ into the integral. I don't see where the trouble can be ....?
Maybe do you mean this $V = \pi \int\limits_0^{4\sqrt 3 } {\left( {64 - \left( {16 + {y^2}} \right)} \right)dy}$ ?? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8799670338630676, "perplexity": 644.4118053995146}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644064869.18/warc/CC-MAIN-20150827025424-00234-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://www.physicsforums.com/showthread.php?p=3806494 | # What is the meaning of the imaginary part of the plane wave function
by Chuck88
Tags: signal, signal analysis, wave
P: 37 The plane wave function sometimes could be represented as: $$U(\mathbf{r} ,t ) = A_{0} e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi)}$$ and we could separate the expression above into: $$U(\mathbf{r} ,t = \cos(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi) + i \sin(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi)$$ Then what is the practical meaning of the imaginary part, ##i \sin(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi)##?
P: 610 Hi Chuck88, The imaginary part is how much amplitude (of a copy of each frequency component that is shifted 90 degrees in phase) that must be added to the real part to end up with the correct phase and amplitude of the original signal $U(\mathbf{r} ,t )$. (Normally, a wave would be decomposed to a weighted amount of the real and imaginary components, such as when you assign the initial conditions to a solution of the wave equation) If only the real part existed, the signal would have a fixed phase alignment. Adding the imaginary part effectively corrects the phase to match the actual experimental signal or wavefront.
P: 37
Quote by PhilDSP Hi Chuck88, The imaginary part is how much amplitude (of a copy of each frequency component that is shifted 90 degrees in phase) that must be added to the real part to end up with the correct phase and amplitude of the original signal $U(\mathbf{r} ,t )$. (Normally, a wave would be decomposed to a weighted amount of the real and imaginary components, such as when you assign the initial conditions to a solution of the wave equation) If only the real part existed, the signal would have a fixed phase alignment. Adding the imaginary part effectively corrects the phase to match the actual experimental signal or wavefront.
Thanks for your reply. But ##\phi## in the real part ##\cos(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi)## could represent the phase of the wave. And ##A_{0}## could represent the amplitude of the wave. It seems like that if we neglect the imaginary part of the wave, the amplitude and phase can still be presented only with the parts in the real part.
P: 610 What is the meaning of the imaginary part of the plane wave function Most likely ##\phi## is going to be used to align the phase globally. Likewise with ##A_0## for the amplitude. To model a realistic wave, each frequency component ##\omega_n## will have a unique phase offset that is accommodated by the combined amplitudes of the real and imaginary components for that frequency. The wave function with only a single valued ##\omega## is a monochromatic wave. Most waves encountered experimentally are more complex.
P: 2 It seems like that if we neglect the imaginary part of the wave, the amplitude and phase can still be presented only with the parts in the real part.
P: 610
Quote by baiber It seems like that if we neglect the imaginary part of the wave, the amplitude and phase can still be presented only with the parts in the real part.
That's true in the simple case of a monochromatic wave, but not in general.
P: 37
Quote by PhilDSP That's true in the simple case of a monochromatic wave, but not in general.
Can you explain that in detail?
P: 610 A wave that is monochromatic consists of only a single sinusoid. Lasers tend toward producing monochromatic waves but almost all other light sources produce a broadband of frequencies (white light for example). The waveform or wave packet can be decomposed to containing a collection of frequencies. Sound waves are similar though the term monochromatic isn't really used to describe them. Look up Fourier Analysis and wave superposition for more details.
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https://www.ncbi.nlm.nih.gov/pubmed/9209851?dopt=Abstract | Format
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Epidemiology. 1997 Jul;8(4):384-9.
# Occupational and residential magnetic field exposure and leukemia and central nervous system tumors.
### Author information
1
Institute of Environmental Medicine, Karolinska Institute, Stockholm, Sweden.
### Abstract
Studies of magnetic field exposure and cancer have focused on either residential or occupational exposure. We conducted a case-control study taking into account both exposure sources. We identified leukemia and central nervous system tumor cases and controls from a population living within 300 m of transmission lines in Sweden. We have previously reported results considering residential exposure alone. Here, we evaluate the effect of occupational exposure and of the combined exposures. We estimated residential exposure through calculations of the magnetic fields generated by power lines. We obtained information about occupation from censuses and linked the occupations to a job-exposure matrix based on magnetic field measurements. For occupational exposure of > or = 0.2 microT, we estimated the relative risk for leukemia to be 1.7 [95% confidence interval (CI) = 1.1-2.7]. The increased risk was confined to acute myeloid and chronic lymphocytic leukemia. For residential exposure of > or = 0.2 microT, the relative risk for leukemia was estimated at 1.3 (95% CI = 0.8-2.2), with higher risk estimates for acute and chronic myeloid leukemia. We estimated the relative risk for leukemia among subjects highly exposed both at home and at work to be 3.7 (95% CI = 1.5-9.4). These results provide support for an association between magnetic field exposure and leukemia. Relative risks for nervous system tumors were close to unity.
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http://repository.bilkent.edu.tr/browse?type=subject&value=Kronecker%20product | Now showing items 1-7 of 7
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(Springer, 2016-08)
It is well known that the infinitesimal generator underlying a multi-dimensional Markov chain with a relatively large reachable state space can be represented compactly on a computer in the form of a block matrix in which ...
• #### Decompositional analysis of Kronecker structured Markov chains
(Kent State University, 2008)
This contribution proposes a decompositional iterative method with low memory requirements for the steadystate analysis ofKronecker structured Markov chains. The Markovian system is formed by a composition of subsystems ...
• #### On compact solution vectors in Kronecker-based Markovian analysis
(Elsevier, 2017)
State based analysis of stochastic models for performance and dependability often requires the computation of the stationary distribution of a multidimensional continuous-time Markov chain (CTMC). The infinitesimal generator ...
• #### On compact vector formats in the solution of the chemical master equation with backward differentiation
(John Wiley and Sons, 2018)
A stochastic chemical system with multiple types of molecules interacting through reaction channels can be modeled as a continuous-time Markov chain with a countably infinite multidimensional state space. Starting from an ...
• #### On the numerical solution of Kronecker-based infinite level-dependent QBD processes
(2013)
Infinite level-dependent quasi-birth-and-death (LDQBD) processes can be used to model Markovian systems with countably infinite multidimensional state spaces. Recently it has been shown that sums of Kronecker products can ...
• #### Steady-state analysis of a multiclass MAP/PH/c queue with acyclic PH retrials
(Applied Probability Trust, 2016)
A multiclass c-server retrial queueing system in which customers arrive according to a class-dependent Markovian arrival process (MAP) is considered. Service and retrial times follow class-dependent phase-type (PH) ... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8272855877876282, "perplexity": 2493.3455026998677}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00687.warc.gz"} |
https://wiki.simons.berkeley.edu/doku.php?id=ag14:start | ag14:start
# Algorithms and Complexity in Algebraic Geometry, Fall 2014
## Computational Open Questions
If you had a large, but finite, black box computer, what problem would you solve with it?
1. What is the tensor rank (or border rank) of the $3\times 3$ matrix multiplication tensor?
2. [Umans] Is the $S$-rank (the support rank) of the $2\times 2$ matrix multiplication tensor 6 or 7?
1. What is the determinantal complexity of the $3\times 3$ permanent?
• Recall for a polynomial $f(\vec x)$, the determinantal complexity $\textsf{dc}( f)$ is the minimum size of a matrix $X$, whose entries are affine linear functions in $\vec x$ such that $\det X = f(\vec x)$.
• We know the answer is either 5,6,7, and we have an explicit matrix for the upper bound on $\textsf{dc}(\textrm{perm}_3)$. For the other sizes, either exhibit a new matrix whose determinant is equal to the permanent, or show that the system of polynomial equations is has no solution.
2. [Ikenmeyer & Hüttenhain] What is the binary determinantal complexity of the $4\times 4$ permanent?
• The binary determinantal complexity of a polynomial with integer coefficients is the smallest size $n$ of a matrix $A$ whose entries are only variables, 0s, and 1s, such that $\textrm{det}(A)=f$. The binary determinantal complexity of the $3 \times 3$ permanent is 7 (see arxiv:1410.8202).
• We have an eplicit construction that shows that the binary determinantal complexity of the $4\times 4$ permanent is at most 15.
3. What is the Waring rank of the $3\times 3$ determinant and permanent?
• Recall that the Waring rank is the minimal number of summands needed to express a given polynomial as a sum of powers of linear forms.
1. Strassen's additivity conjecture for $4\times 4\times 4\times 4$ tensors: That is, is tensor rank additive over direct sums for this format?
2. [Umans] Address the CW90 Asymptotic Rank Conjecture in the case that $n=2$. Consider the CW90 tensor $T$ (which has rank 4), and determine whether the border rank of $T^{\otimes 2}$ is 15 or 16. [if anyone copied the CW90 tensor T please add here]
3. Pavel Hrubes' question. For each $n$, let $S(n)$ be the minimal number k in a sum of squares representation of $(x_1^2+\dotsb+x_n^2)(y_1^2+\dotsb+y_n^2) = f_1^2+\dotsb+f_k^2$.
• First question: For each small $n$ describe the set of such minimal representations.
• Second question: Determine $S(n)$ for all $n$ for which this is feasible.
1. [J. Maurice Rojas]: What is the least prime $p$ such that some trinomial $c_1+c_2 x^{e_2} + c_3 x^{e_3}$ (with $c_1,c_2,c_3\in F_p$, $0<e_2<e_3<p-1$, and $gcd(e_2,e_3,p-1)=1$) has at least 14 roots in $F_p$? More generally, gather as much data as reasonably possible for the growth of $p_k$, where $p_k$ is the least prime such that some trinomial over $F_{p_k}$ (with nonzero constant term, and exponents sharing no common divisor with p-1) has exactly k roots in $F_{p_k}$.
• Comment: The answer for 13 roots appears to be p=10867. There is currently no solid evidence for or against $p_k$ growing exponentially in $k$.
2. [Isabel Herrero] Let $K$ be an algebraically closed field of characteristic 0. Let $S_n$ be the Zariski closure of the set of all univariate polynomials over $K$ of degree $n$ and two double roots, that is of the set
$$\{ (c_0, \dots, c_n) \in (K^*)^{n+1} \mid f(x) = c_0 + c_1x+ \cdots + c_nx^n \quad \text{ has two double roots}\}.$$ We want to compute the homogeneous ideal of all polynomials that vanishes over $S_8$ and, if possible, a reduced Groebner basis of that ideal. Can it be done for $n=9$ or even $10$?
Example: For $S_6$, thinking of $f$ as $f(x) = c_0 + c_1x+ \ldots + c_6x^6 = C_6(x-a)^2(x-b)^2(x-c)(x-d)$, I computed the reduced Groebner basis using SINGULAR with this code:
ring R = 0,(a,b,c,d,x,C5,C4,C3,C2,C1,C0), dp;
ideal I =
C6*x^6 + C5*x^5 + C4*x^4 + C3*x^3 + C2*x^2 + C1*x + C0,
C5 + C6*2*a + C6*2*b + C6*c + C6*d,
C4 - C6*a^2 - C6*4*a*b - C6*b^2 - C6*2*a*c - C6*2*b*c - C6*2*a*d -
C6*2*b*d - C6*c*d,
C3 + C6*2*a^2*b + C6*2*a*b^2 + C6*a^2*c + C6*4*a*b*c + C6*b^2*c +
C6*a^2*d + C6*4*a*b*d + C6*b^2*d + C6*2*a*c*d + C6*2*b*c*d,
C2 - C6*a^2*b^2 - C6*2*a^2*b*c - C6*2*a*b^2*c - C6*2*a^2*b*d -
C6*2*a*b^2*d - C6*a^2*c*d - C6*4*a*b*c*d - C6*b^2*c*d,
C1 + C6*a^2*b^2*c + C6*a^2*b^2*d + C6*2*a^2*b*c*d + C6*2*a*b^2*c*d,
C0 - C6*a^2*b^2*c*d;
ideal k = eliminate( I, a * b * c * d * x);
groebner(k);
exit;
## General Open Questions
1. [Jamshidi] Let $H$ be a positive semi-definite $n \times n$ matrix over $\mathbb{C}$. Does there exist an ideal of invariant polynomials over the entries of $H$ that determine if the eigenvector $v$ corresponding to the lowest eigenvalue of $H$ can be expressed a particular tensor decompositions? For example, $\mathscr{F}$ is such that if $\forall f \in\mathscr{F}$, $f(H) = 0$, then $v = v_{abc} = A_{a}^{\beta} B_{\beta\gamma b} C^{\gamma}_{d}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9784078598022461, "perplexity": 641.4424238013719}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587608.86/warc/CC-MAIN-20211024235512-20211025025512-00208.warc.gz"} |
http://sdktestrunner.in/course/view.php?id=28 | # Trigonometric Functions Syllabus
Positive and negative angles. Measuring angles in radians and in degrees and conversion of one into other. Definition of trigonometric functions with the help of unit circle. Signs of trigonometric functions. Domain and range of trignometric functions and their graphs. Expressing sin (x±y) and cos (x±y) in terms of sinx, siny, cosx & cosy and their simple application. Identities related to sin 2x, cos2x, tan 2x, sin3x, cos3x and tan3x. General solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a.
URLs: 2
# Principle of Mathematical Induction
## 2. Complex Numbers and Quadratic Equations
Need for complex numbers, especially v1, to be motivated by inability to solve some of the quardratic equations. Algebraic properties of complex numbers. Argand plane and polar representation of complex numbers. Statement of Fundamental Theorem of Algebra, solution of quadratic equations in the complex number system. Square root of a complex number.
URL: 1
URLs: 2
# Calculus
## 1. Limits and Derivatives
introduced as rate of change both as that of distance function and geometrically. Intutive idea of limit. Limits of polynomials and rational functions, trignometric, exponential and logarithmic functions. Definition of derivative, relate it to slope of tangent of a curve, derivative of sum, difference, product and quotient of functions. The derivative of polynomial and trigonometric functions.
URLs: 3
# Mathematical Reasoning
## 1. Mathematical Reasoning
Mathematically acceptable statements. Connecting words/ phrases - consolidating the understanding of "if and only if (necessary and sufficient) condition", "implies", "and/or", "implied by", "and", "or", "there exists" and their use through variety of examples related to real life and Mathematics. Validating the statements involving the connecting words difference between contradiction, converse and contrapositive.
# Statistics and Probability
## 1. Statistics
Measures of dispersion; mean deviation, variance and standard deviation of ungrouped/grouped data. Analysis of frequency distributions with equal means but different variances.
## 2. Probability
Random experiments; outcomes, sample spaces (set representation). Events; occurrence of events, 'not', 'and' and 'or' events, exhaustive events, mutually exclusive events, Axiomatic (set theoretic) probability, connections with the theories of earlier classes. Probability of an event, probability of 'not', 'and' and 'or' events.
URLs: 2 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9324250221252441, "perplexity": 2387.9884755966536}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948514051.18/warc/CC-MAIN-20171211203107-20171211223107-00645.warc.gz"} |
http://www.dimostriamogoldbach.it/en/t_value-second-order/ | # 11. Computation of a dash value in a linear second order dashed line
Prerequirements:
In this post we’ll see some formulas for computing the $x$-th dash value for a second order linear dashed line (we’ll see other formulas in the next posts). The approach we’ll follow is based on some notions which were subject of previous posts:
The whole procedure can be illustrated as follows:
The only element we do not already know, among those listed above, is the solution of the downcast characteristic equation. For convenience, we rewrite here the formulation appearing in Corollary 1 of Proposition T.4, in the compact notation:
$$x = y + \left \lfloor \frac{n_i y - (j > i)}{n_j} \right \rfloor \tag{1}$$
The solution of this equation, in the unknown $y$, is given by:
$$y = \left \lceil \frac{n_j x + (j > i)}{n_i + n_j} \right \rceil \tag{2}$$
There are two different proofs of the fact that (2) is the solution of (1). The first (Teoria dei tratteggi (Dashed line theory, in Italian), pages 152-153) is more direct. It shows that (1) implies (2) by applying some integer part properties. The second one (op. cit., pages 153-154) instead is based on the following Lemma:
Lemma for proving the solution of the downcast characteristic equation for linear $\mathrm{t}$, from the second to the first order
Let $T$ be a second order linear dashed line, with indexes $I = \{1, 2\} = \{i, j\}$. Let $t$ be the $x$-th dash of $T$, and $y$ its ordinal within $T[i]$. Then
$$\left \lfloor \frac{n_i y - (j > i)}{n_j} \right \rfloor = \left \lfloor \frac{n_i x - (j > i)}{n_i + n_j} \right \rfloor$$
By this Lemma, the term with the integer part in (1) can be rewritten by putting $x$ in place of $y$ and $n_i + n_j$ in place of $n_j$. So (1) becomes:
$$x = y + \left \lfloor \frac{n_i x - (j > i)}{n_i + n_j} \right \rfloor \tag{1'}$$
Once written in this new form, the downcast characteristic equation is far more simple to be solved, because, contrary to before, the unknown $y$ appears just once: so in order to find the solution it’s enough to bring all terms with $x$ to one side and gather them in a single integer part expression: doing so, in a few passages, equation (2) can be obtained (for details see op. cit., page 154).
As we already noted about the first order $\mathrm{t\_space}$ function, in dashed line theory it’s rather usual to bump into equations with integer parts, like (1), whose approximated solution can be found simply by removing the integer parts. For example, in the case of (1) we have:
\begin{aligned} x \approx y + \frac{n_i y - (j > i)}{n_j} & \implies \\ x \approx \frac{(n_i + n_j) y - (j > i)}{n_j} & \implies \\ n_j x \approx (n_i + n_j) y - (j > i) & \implies \\ n_j x + (j > i) \approx (n_i + n_j) y & \implies \\ \frac{n_j x + (j > i)}{n_i + n_j} \approx y & \end{aligned}
The result indeed coincides with the correct solution (2), up to the integer part. However it’s a lucky case: generally it’s not known a priori how much the solution found by this technique is similar to the correct solution.
So, in downcast theory terms, we can state the following Theorem:
Solution of the downcast characteristic equation of linear $\mathrm{t}$, from the second to the first order
Let $T$ be a second order linear dashed line, with indexes $I = \{1, 2\} = \{i, j\}$. Then the following function $d_i: \mathbb{N}^{\star} \rightarrow \mathbb{N}^{\star}$ defined by:
$$d_i(x) := \left \lceil \frac{n_j x + (j > i)}{n_i + n_j} \right \rceil \tag{3}$$
is a downcast function of $\mathrm{t}$ from $T$ to $T[i]$.
In particular, for $i = 1$:
$$d_1(x) := \left \lceil \frac{n_2 x + 1}{n_1 + n_2} \right \rceil \tag{4}$$
And for $i = 2$:
$$d_2(x) := \left \lceil \frac{n_1 x}{n_1 + n_2} \right \rceil \tag{5}$$
Let’s verify the previous theorem with reference to the first example of the post The downcast problem. That example was about the downcast of $\mathrm{t}$ from $(3,4)$ to $(3)$, that for convenience we rewrite below:
In this case $(n_1, n_2) = (3, 4)$ and $i = 1$, because the dashed subline we are considering is $(3) = (n_1)$. By placing these values into (4), the formula becomes:
$$d_1(x) = \left \lceil \frac{n_2 x + 1}{n_1 + n_2} \right \rceil = \left \lceil \frac{4 x + 1}{7} \right \rceil$$
Let’s apply the formula for all the first row ordinals that are visible in the figure: 1, 3, 5, 6, 8, 10. So we obtain:
• $d_1(1) = \left \lceil \frac{4 \cdot 1 + 1}{7} \right \rceil = \left \lceil \frac{5}{7} \right \rceil = 1$
• $d_1(3) = \left \lceil \frac{4 \cdot 3 + 1}{7} \right \rceil = \left \lceil \frac{13}{7} \right \rceil = 2$
• $d_1(5) = \left \lceil \frac{4 \cdot 5 + 1}{7} \right \rceil = \left \lceil \frac{21}{7} \right \rceil = 3$
• $d_1(6) = \left \lceil \frac{4 \cdot 6 + 1}{7} \right \rceil = \left \lceil \frac{25}{7} \right \rceil = 4$
• $d_1(8) = \left \lceil \frac{4 \cdot 8 + 1}{7} \right \rceil = \left \lceil \frac{33}{7} \right \rceil = 5$
• $d_1(10) = \left \lceil \frac{4 \cdot 10 + 1}{7} \right \rceil = \left \lceil \frac{41}{7} \right \rceil = 6$
As we expected, the bahavior the function $d_1$ is exactly like that of the arrow shown in Figure 2.
Instead for $i = 2$, i.e. for the downcast from $(3, 4)$ to $(4)$, formula (5) becomes:
$$d_2(x) = \left \lceil \frac{n_1 x}{n_1 + n_2} \right \rceil = \left \lceil \frac{3 x}{7} \right \rceil$$
Applying this formula to the ordinals of the second row, that are 2, 4, 7 and 9, we again obtain a sequence of consecutive natural numbers (1, 2, 3 and 4), that are the corresponding ordinals in the dashed subline $(4)$. For example
$$d_2(4) = \left \lceil \frac{3 \cdot 4}{7} \right \rceil = \left \lceil \frac{12}{7} \right \rceil = 2$$
in fact the fourth dash of the dashed line $(3,4)$ (the one with value 8) is the second within the dashed subline $(3,4)[2] = (4)$.
Now we have all the tools required for computing the value of the $x$-th dash of a second order linear dashed line $T$, provided that it belongs to row $i$. In fact we can state the following Theorem:
Formula for computing the second order linear $\mathrm{t\_value}$ function
Let $T$ be a second order linear dashed line, with indexes $I = \{1, 2\} = \{i, j\}$. Let $i$ be the row index of the $x$-th dash of $T$, computable by applying Theorem T.2 (Computation of the $x$-th dash row in a second order linear dashed line). Then
$$\mathrm{t\_value}_T(x) = n_i d_i(x) \tag{6}$$
where $d_i(x) = \left \lceil \frac{n_j x + (j > i)}{n_i + n_j} \right \rceil$ is the function defined in Theorem T.7.
Alternatively, in a form in which the downcast technique is more evident:
$$\mathrm{t\_value}_T(x) = \mathrm{t\_value}_{T[i]}(d_i(x)) \tag{6'}$$
The proof is made up of a chain of equalities:
$$\mathrm{t\_value}_T(x) = T(\mathrm{t}_T(x)) = T(\mathrm{t}_{T[i]}(d_i(x))) = T[i](\mathrm{t}_{T[i]}(d_i(x))) = \mathrm{t\_value}_{T[i]}(d_i(x)) = n_i d_i(x)$$
Let’s analyze each step in detail:
• $\mathrm{t\_value}_T(x) = T(\mathrm{t}_T(x))$: it’s the definition of $\mathrm{t\_value}$ (Definition T.8)
• $T(\mathrm{t}_T(x)) = T(\mathrm{t}_{T[i]}(d_i(x)))$: it’s an application of the definition of downcast of $\mathrm{t}$ from $T$ to $T[i]$ (Definition T.10), by which $\mathrm{t}_T(x) = \mathrm{t}_{T[i]}(d_i(x))$
• $T(\mathrm{t}_{T[i]}(d_i(x))) = T[i](\mathrm{t}_{T[i]}(d_i(x)))$: it’s a consequence of the definition of dashed subline: the dashed line function of the dashed subline $T[i]$ is a restriction of the dashed line function of $T$ on the dashes of row $i$, so the two functions coincide on all the dashes of row $i$, and in particular they coincide on its $d_i(x)$-th dash $\mathrm{t}_{T[i]}(d_i(x)))$.
• $T[i](\mathrm{t}_T[i](d_i(x))) = \mathrm{t\_value}_{T[i]}(d_i(x))$: it’s again the definition of $\mathrm{t\_value}$
• $\mathrm{t\_value}_{T[i]}(d_i(x)) = n_i d_i(x)$: it’s the computation of linear first order $\mathrm{t\_value}$, whose expression is given by Proposition T.1
In the previous example we already computed the number $d(x)$, so the computation of the $\mathrm{t\_value}$ functions turns out to be immediate, by formula (6). However it’s interesting, at this point, to make the computation of $\mathrm{t\_value}$ starting just from an ordinal, for example starting from $x = 6$. In order to compute it, we can apply formula (8) of Theorem T.2, according to which
$$\mathrm{t}_{(3,4)}(x) \in (3,4)[1] \Leftrightarrow (3x - 1) \mathrm{\ mod\ } (3 + 4) \lt 4$$
For $x = 6$ the right side modulus is $17 \mathrm{\ mod\ } 7 = 3$ that’s lower than 4, so indeed $\mathrm{t}_{(3,4)}(x) \in (3,4)[1] = (3)$. So the value of $i$ to be used in Theorem T.8 is just 1. Hence, by formulas (6) and (4):
$$\mathrm{t\_value}_{(3,4)}(6) = n_1 d_1(6) = 3 d_1(6) = 3 \left \lceil \frac{4 \cdot 6 + 1}{3 + 4} \right \rceil = 3 \left \lceil \frac{25}{7} \right \rceil = 3 \cdot 4 = 12$$
We have thus obtained the value of the sixth dash, shown in Figure 2.
Formula for calculation of linear second order function $\mathrm{t\_value}$ for first row
The formula for calculating $\mathrm{t\_value_T}(x)$ for dashes belonging to the first row of a second order dashed line $T = (n_1, n_2)$, obtained from (6) of Theorem T.8 by replacing $i = 1$, $j = 2$ and $d_i(x)$ with its expression, is the following:
$\mathrm{t\_value_T}(x) = n_1 \biggl \lceil \cfrac{n_2 \cdot x + 1}{n_1 + n_2} \biggr \rceil$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 116, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9811136722564697, "perplexity": 385.40769646992305}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875147154.70/warc/CC-MAIN-20200228104413-20200228134413-00435.warc.gz"} |
https://mathoverflow.net/questions/394690/two-level-correlation-function-of-eigenvalues-for-large-random-matrices | # Two-level correlation function of eigenvalues for large random matrices
One can define the density of eigenvalues of a $$N\times N$$ Hermitian random matrix $$H$$ as: $$$$\rho(\lambda)=\left \langle\frac{1}{N} \operatorname{Tr} \delta(\lambda-H)\right\rangle$$$$ Where $$\langle \dots \rangle$$ denotes the average over the distribution of $$H$$.
In the large $$N$$ limit, it is famously known that $$\rho(\lambda)$$ will approach the Wigner semi circular law (given some conditions on the moments of the distribution of $$H$$). This can be shown with various methods, one that I favor is using the resolvent and computing its schur complements.
One can define the two-level correlation function: $$$$\rho^{(2)}(\lambda, \mu)=\left \langle\frac{1}{N} \operatorname{Tr} \delta(\lambda-H) \frac{1}{N} \operatorname{Tr} \delta(\mu-H)\right\rangle$$$$
The following paper 1 provides a method to compute this quantity, and provides an exhaustive list of the existing methods to compute eq $$(2)$$. However the paper is now $$25$$ years old: is there any known results that extends their method? Can we compute this quantity using the schur complement and the resolvent? What about non-hermitian matrices?
Edit: My initial query was whether we could could compute equation $$(2)$$ using the resolvents and the Schur complements.
The paper mentions that contrary to $$\rho(\lambda)$$ there is no universality for $$\rho^{(2)}(\lambda, \mu)$$: it would depend on the choice of distribution. However in a certain regime, with large $$N$$ and small $$\lambda- \mu$$ then universal properties can be derived. Has this been we shown this using the Schur complements of the resolvents?
1 : Brézin, E., & Hikami, S. (1996). Correlations of nearby levels induced by a random potential. Nuclear Physics B, 479(3), 697-706. link: https://arxiv.org/pdf/cond-mat/9605046.pdf
• This is a very wide question, essentially asking "what has happened in RMT since 1995?". The answer is a l"a lot", for example, universality. You should be more specific (as in the paper you quote) of the regime you care about: are $\lambda$ and $\mu$ separated (in which case you need to rescale to get interesting answers) or at distance $1/N$ of each other? For the latter case, look up work of Erdos-Yau and co-authors, as well as Tao-Vu. Jun 7 '21 at 5:32
The universality of $$\rho^{(2)}(\lambda,\mu)$$ does exist if one considers the correlations locally, on the scale of the mean eigenvalue spacing. This is relevant for many applications, because the correlations decay quickly with increasing $$|\lambda-\mu|$$.
If the eigenvalue distribution has the Gibbs form $$P(\lambda_1,\lambda_2,\ldots \lambda_N)\propto e^{-\beta W},$$ $$W=\sum_{i then the two-point correlation function $$K(\lambda,\mu)=N^2\rho^{(2)}(\lambda,\mu)-N^2\rho(\lambda)\rho(\mu)$$ is given in the large-$$N$$ local limit by the functional inverse $$u^{\text{inv}}$$ of $$u$$, $$K(\lambda,\mu)=\frac{1}{\beta}u^{\text{inv}}(\lambda,\mu).$$ See section 1.D of arXiv:cond-mat/9612179 .
Notice that this is independent of the "potential" $$V$$, only the eigenvalue interaction $$u$$ enters. That is the sense in which the two-point correlation function is "universal" on the local scale for a broad class of RMT ensembles with a logarithmic eigenvalue repulsion, $$u(\lambda,\mu)=-\log|\lambda-\mu|$$, see for example arXiv:cond-mat/9310010. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 28, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9721170663833618, "perplexity": 573.5600847380794}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320302740.94/warc/CC-MAIN-20220121071203-20220121101203-00689.warc.gz"} |
https://www.physicsforums.com/threads/gaussian-potential.663243/ | # Gaussian potential
1. Jan 9, 2013
### aaaa202
Okay, I have solved the schrödinger equation numerically by making it dimensionless (though Im still confused about this proces). And then approximating it on a finite interval and solving the resulting eigenvalue equation. This allows me to solve for the wave function of different potentials.
I started with the harmonic oscillator but have no reached the Gaussian one:
V = -V0exp(-x2)
In one simulation I am asked to find the difference in energy between the ground state and the first excited state as a function of V0. On the attached graph I have done this for V0=1..2..3...10
Does it look right?
I am then asked the following: Solve the problem analytically by taylorexpanding the potential. So I taylor expand around x=0 to second order and find:
V(x) = -V0 + V0x2
Plugging this into my dimensionless Schrödinger equation I get:
½∂2ψ/∂x2 + (-V0 + V0x2)ψ = Eψ
I thought aha. The x2-term can just be put in the harmonic oscillator form if we pick k=2V0 and the -V0 term will just shift the energy of the oscillator, not alter the difference between E1 and E0.
But in thinking it over again there are some problems. With my dimensionless equation I just had V(x)=½x2 for the harmonic oscillator. Now I have 2V0 in front of that. How will this constant effect my energies?
And is all this even the right procedure?
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2. Jan 9, 2013
### Staff: Mentor
The difference in energy levels scales with the square root of the prefactor in a harmonic potential - in the dimensionless shape, this is hidden in the parameter transformations. In a region where this harmonic approximation is good (probably large V0), I would expect that the difference grows with the square root of V0, and your graph roughly looks like that. For small V0, you probably get additional effects from higher orders of the potential.
3. Jan 9, 2013
### aaaa202
okay but I am meant to show this analytically. How can I do that?
4. Jan 9, 2013
### Staff: Mentor
With the standard formulas for a harmonic oscillator. In your dimensionless version, you should have the conversion factors somewhere.
Or with the simple sqrt-dependence if you like.
5. Jan 10, 2013
### aaaa202
well the conversion formula to dimensionless units is x' = x * √(mω/hbar)
So do I go back to formulate it all in terms of x? :S Im very confused sorry.
6. Jan 10, 2013
### Staff: Mentor
That is possible. Alternatively, you can extract the scale of V from that prefactor.
7. Jan 10, 2013
### aaaa202
okay I did the problem and did indeed find that the difference went like √(2V0) - I am just curious - how is it you can see that the harmonic approximation is better for bigger v0?
8. Jan 10, 2013
### Staff: Mentor
Intuition. Deeper wells of the same size tend to have more bound states, so the lowest states are "deeper" in the well and smaller in terms of their size in x.
Similar Discussions: Gaussian potential | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8787286877632141, "perplexity": 1042.6827420683665}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825510.59/warc/CC-MAIN-20171023020721-20171023040721-00307.warc.gz"} |
https://br.comsol.com/blogs/model-cables-and-transmission-lines-in-comsol-multiphysics/?setlang=1 | ## Model Cables and Transmission Lines in COMSOL Multiphysics
##### Andrew Strikwerda February 9, 2016
Electrical cables are classified by parameters such as impedance and power attenuation. In this blog post, we consider a case for which analytic solutions exist: a coaxial cable. We will show you how to compute the cable parameters from a COMSOL Multiphysics simulation of the electromagnetic fields. Once we understand how this is done for a coaxial cable, we can then compute these parameters for an arbitrary type of transmission line or cable.
### Design Considerations for Electrical Cables
Electrical cables, also called transmission lines, are used everywhere in the modern world to transmit both power and data. If you are reading this on a cell phone or tablet computer that is “wireless”, there are still transmission lines within it connecting the various electrical components together. When you return home this evening, you will likely plug your device into a power cable to charge it.
Various transmission lines range from the small, such as coplanar waveguides on a printed circuit board (PCB), to the very large, like high voltage power lines. They also need to function in a variety of situations and conditions, from transatlantic telegraph cables to wiring in spacecraft, as shown in the image below. Transmission lines must be specially designed to ensure that they function appropriately in their environments, and may also be subject to further design goals, including required mechanical strength and weight minimization.
Transmission wires in the payload bay of the OV-095 at the Shuttle Avionics Integration Laboratory (SAIL).
When designing and using cables, engineers often refer to parameters per unit length for the series resistance (R), series inductance (L), shunt capacitance (C), and shunt conductance (G). These parameters can then be used to calculate cable performance, characteristic impedance, and propagation losses. It is important to keep in mind that these parameters come from the electromagnetic field solutions to Maxwell’s equations. We can use COMSOL Multiphysics to solve for the electromagnetic fields, as well as consider multiphysics effects to see how the cable parameters and performance change under different loads and environmental conditions. This could then be converted into an easy-to-use app, like this example that calculates the parameters for commonly used transmission lines.
Here, we examine a coaxial cable — a fundamental problem that is often covered in a standard curriculum for microwave engineering or transmission lines. The coaxial cable is so fundamental that Oliver Heaviside patented it in 1880, just a few years after Maxwell published his famous equations. For the students of scientific history, this is the same Oliver Heaviside who formulated Maxwell’s equations in the vector form that we are familiar with today; first used the term “impedance”; and helped develop transmission line theory.
### Analytical Results for a Coaxial Cable
Let us begin by considering a coaxial cable with dimensions as shown in the cross-sectional sketch below. The dielectric core between the inner and outer conductors has a relative permittivity (\epsilon_r = \epsilon' -j\epsilon'') of 2.25 – j*0.01, a relative permeability (\mu_r) of 1, and a conductivity of zero, while the inner and outer conductors have a conductivity (\sigma) of 5.98e7 S/m.
The 2D cross section of the coaxial cable, where we have chosen a = 0.405 mm, b = 1.45 mm, and t = 0.1 mm. Note that this tutorial model is available for download in our Application Gallery.
A standard method for solving transmission lines is to assume that the electric fields will oscillate and attenuate in the direction of propagation, while the cross-sectional profile of the fields will remain unchanged. If we then find a valid solution, uniqueness theorems ensure that the solution we have found is correct. Mathematically, this is equivalent to solving Maxwell’s equations using an ansatz of the form \mathbf{E}\left(x,y,z\right) = \mathbf{\tilde{E}}\left(x,y\right)e^{-\gamma z}, where (\gamma = \alpha + j\beta) is the complex propagation constant and \alpha and \beta are the attenuation and propagation constants, respectively. In cylindrical coordinates for a coaxial cable, this results in the well-known field solution of
\begin{align}
\mathbf{E}&= \frac{V_0\hat{r}}{rln(b/a)}e^{-\gamma z}\\
\mathbf{H}&= \frac{I_0\hat{\phi}}{2\pi r}e^{-\gamma z}
\end{align}
which then yields the parameters per unit length of
\begin{align}
L& = \frac{\mu_0\mu_r}{2\pi}ln\frac{b}{a} + \frac{\mu_0\mu_r\delta}{4\pi}(\frac{1}{a}+\frac{1}{b})\\
C& = \frac{2\pi\epsilon_0\epsilon'}{ln(b/a)}\\
R& = \frac{R_s}{2\pi}(\frac{1}{a}+\frac{1}{b})\\
G& = \frac{2\pi\omega\epsilon_0\epsilon''}{ln(b/a)}
\end{align}
where R_s = 1/\sigma\delta is the sheet resistance and \delta = \sqrt{2/\mu_0\mu_r\omega\sigma} is the skin depth.
While the equations for capacitance and shunt conductance are valid at any frequency, it is extremely important to point out that the equations for the resistance and inductance depend on the skin depth and are therefore only valid at frequencies where the skin depth is much smaller than the physical thickness of the conductor. This is also why the second term in the inductance equation, called the internal inductance, may be unfamiliar to some readers, as it can be neglected when the metal is treated as a perfect conductor. The term represents inductance due to the penetration of the magnetic field into a metal of finite conductivity and is negligible at sufficiently high frequencies. (The term can also be expressed as L_{Internal} = R/\omega.)
For further comparison, we can compute the DC resistance directly from the conductivity and cross-sectional area of the metal. The analytical equation for the DC inductance is a little more involved, and so we quote it here for reference.
L = \frac{\mu}{2\pi}\left\{ln\left(\frac{b+t}{a}\right) + \frac{2\left(\frac{b}{b+t}\right)^2}{1- \left(\frac{b}{b+t}\right)^2} ln\left(\frac{b+t}{b}\right) -\frac{3}{4} + \frac{\frac{\left(b+t\right)^4}{4} -\left(b+t\right)^2b^2+b^4\left(\frac{3}{4} + ln\frac{\left(b+t\right)}{b}\right) }{\left(\left(b+t\right)^2-b^2\right)^2}\right\}
Now that we have values for C and G at all frequencies, DC values for R and L, and asymptotic values for their high-frequency behavior, we have excellent benchmarks for our computational results.
### Simulating Cables with the AC/DC Module
When setting up any numerical simulation, it is important to consider whether or not symmetry can be used to reduce the model size and increase the computational speed. As we saw earlier, the exact solution will be of the form \mathbf{E}\left(x,y,z\right) = \mathbf{\tilde{E}}\left(x,y\right)e^{-\gamma z}. Because the spatial variation of interest is primarily in the xy-plane, we just want to simulate a 2D cross section of the cable. One issue, however, is that the 2D governing equations used in the AC/DC Module assume that the fields are invariant in the out-of-plane direction. This means that we will not be able to capture the variation of the ansatz in a single 2D AC/DC simulation. We can find the variation with two simulations, though! This is because the series resistance and inductance depend on the current and energy stored in the magnetic fields, while the shunt conductance and capacitance depend on the energy in the electric field. Let’s take a closer look.
#### Distributed Parameters for Shunt Conductance and Capacitance
Since the shunt conductance and capacitance can be calculated from the electric fields, we begin by using the Electric Currents interface.
Boundary conditions and material properties for the Electric Currents interface simulation.
Once the geometry and material properties are assigned, we assume that the conductors are equipotential (a safe assumption, since the conductivity difference between the conductor and the dielectric will generally be near 20 orders of magnitude) and set up the physics by applying V0 to the inner conductor and grounding the outer conductor to solve for the electric potential in the dielectric. The above analytical equation for capacitance comes from the following more general equations
\begin{align}
W_e& = \frac{1}{4}\int_{S}{}\mathbf{E}\cdot \mathbf{D^\ast}d\mathbf{S}\\
W_e& = \frac{C|V_0|^2}{4}\\
C& = \frac{1}{|V_0|^2}\int_{S}{}\mathbf{E}\cdot \mathbf{D^\ast}d\mathbf{S}
\end{align}
where the first equation is from electromagnetic theory and the second from circuit theory.
The first and second equations are combined to obtain the third equation. By inserting the known fields from above, we obtain the previous analytical result for C in a coaxial cable. More generally, these equations provide us with a method for obtaining the capacitance from the fields for any cable. From the simulation, we can compute the integral of the electric energy density, which gives us a capacitance of 98.142 pF/m and matches with theory. Since G and C are related by the equation
G=\frac{\omega\epsilon'' C}{\epsilon'}
we now have two of the four parameters.
At this point, it is also worth reiterating that we have assumed that the conductivity of the dielectric region is zero. This is typically done in the textbook derivation, and we have maintained that convention here because it does not significantly impact the physics — unlike our inclusion of the internal inductance term discussed earlier. Many dielectric core materials do have a nonzero conductivity and that can be accounted for in simulation by simply updating the material properties. To ensure that proper matching with theory is maintained, the appropriate derivations would need to be updated as well.
#### Distributed Parameters for Series Resistance and Inductance
In a similar fashion, the series resistance and inductance can be calculated through simulation using the AC/DC Module’s Magnetic Fields interface. The simulation setup is straightforward, as demonstrated in the figure below.
The center and outer conductor domains are excited with currents of equal magnitude and opposing directions, as indicated by the dots and crosses. The modeling features used for this purpose will allow for the current in the conductors to distribute freely according to skin and proximity effects, as opposed to the arbitrary distribution shown in the figure.
We refer to the following equations, which are the magnetic analog of the previous equations, to calculate the inductance.
\begin{align}
W_m& = \frac{1}{4}\int_{S}{}\mathbf{B}\cdot \mathbf{H^\ast}d\mathbf{S}\\
W_m& = \frac{L|I_0|^2}{4}\\
L& = \frac{1}{|I_0|^2}\int_{S}{}\mathbf{B}\cdot \mathbf{H^\ast}d\mathbf{S}
\end{align}
To calculate the resistance, we use a slightly different technique. First, we integrate the resistive loss to determine the power dissipation per unit length. We can then use the familiar P = I_0^2R/2 to calculate the resistance. Since R and L vary with frequency, let’s take a look at the calculated values and the analytical solutions in the DC and high-frequency (HF) limit.
“Analytic (DC)” and “Analytic (HF)” refer to the analytical equations in the DC and high-frequency limits, respectively, which were discussed earlier. Note that these are both on log-log plots.
We can clearly see that the computed values transition smoothly from the DC solution at low frequencies to the high-frequency solution, which is valid when the skin depth is much smaller than the thickness of the conductor. We anticipate that the transition region will be approximately located where the skin depth and conductor thickness are within one order of magnitude. This range is 4.2e3 Hz to 4.2e7 Hz, which is exactly what we see in the results.
#### Characteristic Impedance and Propagation Constant
Now that we have completed the heavy lifting to calculate R, L, C, and G, there are two other significant parameters that can be determined. They are the characteristic impedance (Zc) and complex propagation constant (\gamma = \alpha + j\beta), where \alpha is the attenuation constant and \beta is the propagation constant.
\begin{align}
Z_c& = \sqrt{\frac{(R+j\omega L)}{(G+j\omega C)}}\\
\gamma& = \sqrt{(R+j\omega L)(G+j\omega C)}
\end{align}
In the figure below, we see these values calculated using the analytical formulas for both the DC and high-frequency regime as well as the values determined from our simulation. We have also included a fourth line: the impedance calculated using COMSOL Multiphysics and the RF Module, which we will discuss shortly. As can be seen, our computations agree with the analytical solutions in their respective limits, as well as yielding the correct values through the transition region.
A comparison of the characteristic impedance, determined using the analytical equations and COMSOL Multiphysics. The analytical equations plotted are from the DC and high-frequency (HF) equations discussed earlier, while the COMSOL Multiphysics results use the AC/DC and RF Modules. For clarity, the width of the “RF Module” line has been intentionally increased.
### Cable Simulation at Higher Frequencies
Electromagnetic energy travels as waves, which means that the frequency of operation and wavelength are inversely proportional. As we continue to solve at higher and higher frequencies, we need to be aware of the relative size of the wavelength and electrical size of the cable. As discussed in a previous blog post, we should switch from the AC/DC to RF Module at an electrical size of approximately λ/100. If we use the cable diameter as the electrical size and the speed of light inside the dielectric core of the cable, this yields a transition frequency of approximately 690 MHz.
At these higher frequencies, the cable is more appropriately treated as a waveguide and the cable excitation as a waveguide mode. Using waveguide terminology, the mode we have been examining is a special type of mode called TEM that can propagate at any frequency. When the cross section and wavelength are comparable, we also need to account for the possibility of higher-order modes. Unlike a TEM mode, most waveguide modes can only propagate above a characteristic cut-off frequency. Due to the cylindrical symmetry in our example model, there is an equation for the cut-off frequency of the first higher-order mode, which is a TE11 mode. This cut-off frequency is fc = 35.3 GHz, but even with the relatively simple geometry, the cut-off frequency comes from a transcendental equation that we will not examine further in this post.
So what does this cut-off frequency mean for our results? Above that frequency, the energy carried in the TEM mode that we are interested in has the potential to couple to the TE11 mode. In a perfect geometry, like we have simulated here, there will be no coupling. In the real world, however, any imperfections in the cable could cause mode coupling above the cut-off frequency. This could result from a number of sources, from fabrication tolerances to gradients in the material properties. Such a situation is often avoided by designing cables to operate below the cut-off frequency of higher-order modes so that only one mode can propagate. If that is of interest, you can also use COMSOL Multiphysics to simulate the coupling between higher-order modes, as with this Directional Coupler tutorial model (although beyond the scope of today’s post).
### Mode Analysis in the RF and Wave Optics Modules
Simulation of higher-order modes is ideally suited for a Mode Analysis study using the RF or Wave Optics modules. This is because the governing equation is \mathbf{E}\left(x,y,z\right) = \mathbf{\tilde{E}}\left(x,y\right)e^{-\gamma z}, which is exactly the form that we are interested in. As a result, Mode Analysis will directly solve for the spatial field and complex propagation constant for a predefined number of modes. We can use the same geometry as before, except that we only need to simulate the dielectric core and can use an Impedance boundary condition for the metal conductor.
The results for the attenuation constant and effective mode index from a Mode Analysis. The analytic line in the left plot, “Attenuation Constant vs Frequency”, is computed using the same equations as the high-frequency (HF) lines used for comparison with the results of the AC/DC Module simulations. The analytic line in the right plot, “Effective Refractive Index vs Frequency”, is simply n = \sqrt{\epsilon_r\mu_r}. For clarity, the size of the “COMSOL — TEM” lines has been intentionally increased in both plots.
We can clearly see that the Mode Analysis results of the TEM mode match the analytic theory, and that the computed higher-order mode has its onset at the previously determined cut-off frequency. It is also incredibly convenient that the complex propagation constant is a direct output of this simulation and does not require calculations of R, L, C, and G. This is because \gamma is explicitly included and solved for in the Mode Analysis governing equation. These other parameters can be calculated for the TEM mode, if desired, and more information can be found in this demonstration in the Application Gallery. It is also worth pointing out that this same Mode Analysis technique can be used for dielectric waveguides, like fiber optics.
### Concluding Remarks on Modeling Cables
At this point, we have thoroughly analyzed a coaxial cable. We have calculated the distributed parameters from the DC to high-frequency limit and examined the first higher-order mode. Importantly, the Mode Analysis results only depend on the geometry and material properties of the cable. The AC/DC results require the additional knowledge of how the cable is excited, but hopefully you know what you’re attaching your cable to! We used analytic theory solely to compare our simulation results against a well-known benchmark model. This means that the analysis could be extended to other cables, as well as coupled to multiphysics simulations that include temperature change and structural deformation.
For those of you who are interested in the fine details, here are a few extra points in the form of hypothetical questions.
• “Why didn’t you mention and/or plot all of the characteristic impedance and distributed parameters for the TE11 mode?”
• This is because only TEM modes have a uniquely defined voltage, current, and characteristic impedance. It is still possible to assign some of these values for higher-order modes, and this is discussed further in texts on transmission line theory and microwave engineering.
• “When I solve for modes using a Mode Analysis study, they are labeled by the value of their effective index. Where did TEM and TE11 come from?”
• These names come from the analytic theory and were used for convenience when discussing the results. This name assignment may not be possible for an arbitrary geometry, but what’s in a name? Would not a mode by any other name still carry electromagnetic energy (excluding nontunneling evanescent waves, of course)?
• “Why is there an extra factor of ½ in several of your calculations?”
• This comes up when solving electromagnetics in the frequency domain, notably when multiplying two complex quantities. When taking the time average, there is an extra factor of ½ as opposed to the equation in the time domain (or at DC). For more information, you can refer to a text on classical electromagnetics.
### References
The following texts were referred to during the writing of this post and are excellent sources of additional information:
• Microwave Engineering, by David M. Pozar
• Foundations for Microwave Engineering, by Robert E. Collin
• Inductance Calculations, by Frederick W. Grover
• Classical Electrodynamics, by John D. Jackson | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8699755668640137, "perplexity": 574.3224813824447}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578533774.1/warc/CC-MAIN-20190422015736-20190422041736-00357.warc.gz"} |
http://www.newton.ac.uk/programmes/PFD/seminars/2005121515006.html | PFD
Seminar
Patterns in chaotically mixing fluid flows (Venue: MR3 CMS)
Straube, A (Potsdam)
Thursday 15 December 2005, 15:00-15:30
Meeting Room 3, CMS
Abstract
We study pattern formation and report on new instabilities in reaction-advection-diffusion systems of two different kinds.
First, we develop a theory describing the transition to a spatially homogeneous regime in a mixing flow with a chaotic in time reaction [1]. The transverse Lyapunov exponent governing the stability of the homogeneous state can be represented as a combination of Lyapunov exponents for spatial mixing and temporal chaos. This representation, being exact for time-independent flows and equal Peclet numbers of different components, is demonstrated to work accurately for time-dependent flows and different Peclet numbers. The properties of structures that appear beyond the stability threshold are discussed.
Second, we consider a reaction-diffusion system of an activator-inhibitor type and impose the periodic in space mixing flow. We fix the governing parameters in the way that ensures the stability of the homogeneous steady state in reaction-diffusion system, so that without advection no Turing patterns can occur. Next, we increase the advection rate and study the influence of the flow on the stability of this state. One could intuitively expect that because the flow is mixing, it should stabilize the homogeneous state. However, the instability appears as the rate of advection increases beyond a certain threshold, which results in pattern formation. We apply the Bloch theory to find out the length-scale of the patterns, which generally do not coincide with the length-scale of the imposed flow. The mechanism of the instability can be understood from a reduced model; the results are explained by means of Lyapunov exponents. We consider two situations: (i) the general case when both chemical species are advected and (ii) a partial case, when only one species is advected, which is relevant to biological applications.
Since the flow and reaction terms are generic for the effects investigated, we believe that the results hold for a variety of flows and chemical reactions.
[1] A.V. Straube, M. Abel, A. Pikovsky, Temporal chaos versus spatial mixing in reaction-advection-diffusion systems, Phys. Rev. Lett. 93, 174501 (2004). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8278948664665222, "perplexity": 724.7995114586449}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00523-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://export.arxiv.org/abs/2011.11351 | astro-ph.GA
(what is this?)
# Title: NIHAO -- XXV. Convergence in the cusp-core transformation of cold dark matter haloes at high star formation thresholds
Authors: Aaron A. Dutton (NYUAD), Tobias Buck (AIP), Andrea V. Macciò (NYUAD, MPIA), Keri L. Dixon (NYUAD), Marvin Blank (NYUAD, Kiel), Aura Obreja (USM)
Abstract: We use cosmological hydrodynamical galaxy formation simulations from the NIHAO project to investigate the response of cold dark matter (CDM) haloes to baryonic processes. Previous work has shown that the halo response is primarily a function of the ratio between galaxy stellar mass and total virial mass, and the density threshold above which gas is eligible to form stars, $n [{\rm cm}^{-3}]$. At low $n$ all simulations in the literature agree that dwarf galaxy haloes are cuspy, but at high $n\ge 100$ there is no consensus. We trace halo contraction in dwarf galaxies with $n\ge 100$ reported in some previous simulations to insufficient spatial resolution. Provided the adopted star formation threshold is appropriate for the resolution of the simulation, we show that the halo response is remarkably stable for $n\ge 5$, up to the highest star formation threshold that we test, $n=500$. This free parameter can be calibrated using the observed clustering of young stars. Simulations with low thresholds $n\le 1$ predict clustering that is too weak, while simulations with high star formation thresholds $n\ge 5$, are consistent with the observed clustering. Finally, we test the CDM predictions against the circular velocities of nearby dwarf galaxies. Low thresholds predict velocities that are too high, while simulations with $n\sim 10$ provide a good match to the observations. We thus conclude that the CDM model provides a good description of the structure of galaxies on kpc scales provided the effects of baryons are properly captured.
Comments: 15 pages, 10 figures, published in MNRAS Subjects: Astrophysics of Galaxies (astro-ph.GA); Cosmology and Nongalactic Astrophysics (astro-ph.CO) Journal reference: 2020MNRAS.499.2648D DOI: 10.1093/mnras/staa3028 Cite as: arXiv:2011.11351 [astro-ph.GA] (or arXiv:2011.11351v1 [astro-ph.GA] for this version)
## Submission history
From: Aaron Dutton [view email]
[v1] Mon, 23 Nov 2020 12:01:37 GMT (229kb)
Link back to: arXiv, form interface, contact. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8726226091384888, "perplexity": 4485.877190385188}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703519984.9/warc/CC-MAIN-20210120085204-20210120115204-00714.warc.gz"} |
https://energyresources.asmedigitalcollection.asme.org/GT/proceedings/TA1996/78774/V001T05A003/282547?searchresult=1 | A computational study was performed for the flow and heat transfer in coolant passages with two legs connected with a U-bend and with dimensionless flow conditions typical of those in the internal cooling passages of turbine blades. The first model had smooth surfaces on all walls. The second model had opposing ribs staggered and angled at 45° to the main flow direction on two walls of the legs, corresponding to the coolant passage surfaces adjacent to the pressure and suction surfaces of a turbine airfoil. For the ribbed model, the ratio of rib height to duct hydraulic diameter equaled 0.1, and the ratio of rib spacing to rib height equaled 10. Comparisons of calculations with previous measurements are made for a Reynolds number of 25,000. With these conditions, the predicted heat transfer is known to be strongly influenced by the turbulence and wall models. The k-e model, the low Reynolds number RNG k-e and the differential Reynolds-stress model (RSM) were used for the smooth wall model calculation. Based on the results with the smooth walls, the calculations for the ribbed walls were performed using the RSM and k-e turbulence models. The high secondary flow induced by the ribs leads to an increased heat transfer in both legs. However, the heat transfer was nearly unchanged between the smooth wall model and the ribbed model within the bend region. The agreement between the predicted segment-averaged and previously-measured Nusselt numbers was good for both cases.
This content is only available via PDF. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8392041325569153, "perplexity": 1518.7509424904727}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334332.96/warc/CC-MAIN-20220925004536-20220925034536-00538.warc.gz"} |
https://planetmath.org/methodforrepresentingrationalnumbersassumsofunitfractionsusingpracticalnumbers | # method for representing rational numbers as sums of unit fractions using practical numbers
Fibonacci’s application for practical numbers $n$ was an algorithm to represent proper fractions $\frac{m}{n}$ (with $m>1$) as sums of unit fractions $\displaystyle\sum\frac{d_{i}}{n}$, with the $d_{i}$ being divisors of the practical number $n$. (By the way, there are infinitely many practical numbers which are also Fibonacci numbers). The method is:
1. 1.
Reduce the fraction to lowest terms. If the numerator is then 1, we’re done.
2. 2.
Rewrite $m$ as a sum of divisors of $n$.
3. 3.
Make those divisors of $n$ that add up to $m$ into the numerators of fractions with $n$ as denominator.
4. 4.
Reduce those fractions to lowest terms, thus obtaining the representation $\displaystyle\frac{m}{n}=\sum\frac{d_{i}}{n}$.
To illustrate the algorithm, let’s rewrite $\frac{37}{42}$ as a sum of unit fractions. Since 42 is practical, success is guaranteed.
At the first step we can’t reduce this fraction because 37 is a prime number. So we go on to the second step, and represent 37 as 2 + 14 + 21. This gives us the fractions
$\frac{2}{42}+\frac{14}{42}+\frac{21}{42},$
which we then reduce to lowest terms:
$\frac{1}{21}+\frac{1}{3}+\frac{1}{2},$
giving us the desired unit fractions.
## References
• 1 M. R. Heyworth, “More on panarithmic numbers” New Zealand Math. Mag. 17 (1980): 28 - 34
• 2 Giuseppe Melfi, “A survey on practical numbers” Rend. Sem. Mat. Univ. Pol. Torino 53 (1995): 347 - 359
Title method for representing rational numbers as sums of unit fractions using practical numbers MethodForRepresentingRationalNumbersAsSumsOfUnitFractionsUsingPracticalNumbers 2013-03-22 18:07:00 2013-03-22 18:07:00 PrimeFan (13766) PrimeFan (13766) 4 PrimeFan (13766) Algorithm msc 11A25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 15, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9614177942276001, "perplexity": 988.0362536972833}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370510287.30/warc/CC-MAIN-20200403030659-20200403060659-00272.warc.gz"} |
http://liforneklerimodelleri.blogspot.com/2012/06/12066525-z-v-pchelkina-et-al.html | ## Theoretical Analysis of Electronic and Magnetic Properties of NaV$_2$O$_4$: Crucial Role of the Orbital Degrees of Freedom [PDF]
Z. V. Pchelkina, I. V. Solovyev, R. Arita
Using realistic low-energy model with parameters derived from the first-principles electronic structure calculation, we address the origin of the quasi-one-dimensional behavior in orthorhombic NaV$_2$O$_4$, consisting of the double chains of edge-sharing VO$_6$ octahedra. We argue that the geometrical aspect alone does not explain the experimentally observed anisotropy of electronic and magnetic properties of NaV$_2$O$_4$. Instead, we attribute the unique behavior of NaV$_2$O$_4$ to one particular type of the orbital ordering, which respects the orthorhombic $Pnma$ symmetry. This orbital ordering acts to divide all $t_{2g}$ states into two types: the localized' ones, which are antisymmetric with respect to the mirror reflection $y \rightarrow -$$y$, and the symmetric delocalized' ones. Thus, NaV$_2$O$_4$ can be classified as the double exchange system. The directional orientation of symmetric orbitals, which form the metallic band, appears to be sufficient to explain both quasi-one-dimensional character of interatomic magnetic interactions and the anisotropy of electrical resistivity.
View original: http://arxiv.org/abs/1206.6525 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8748165965080261, "perplexity": 1924.3795246051964}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267158766.65/warc/CC-MAIN-20180923000827-20180923021227-00123.warc.gz"} |
http://www.molympiad.ml/2017/10/mediterranean-mathematics-olympiad-2003.html | Prove that the equation $$x^2 + y^2 + z^2 = x + y + z + 1$$ has no rational solutions. In a triangle $ABC$ with $BC = CA + \frac 12 AB$, p...
1. Prove that the equation $$x^2 + y^2 + z^2 = x + y + z + 1$$ has no rational solutions.
2. In a triangle $ABC$ with $BC = CA + \frac 12 AB$, point $P$ is given on side $AB$ such that $BP : PA = 1 : 3$. Prove that $\angle CAP = 2 \angle CPA.$
3. Let $a, b, c$ be non-negative numbers with $a+b+c = 3$. Prove the inequality $\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1} \geq \frac 32.$
4. Consider a system of infinitely many spheres made of metal, with centers at points $(a, b, c) \in \mathbb Z^3$. We say that the system is stable if the temperature of each sphere equals the average temperature of the six closest spheres. Assuming that all spheres in a stable system have temperatures between $0^\circ C$ and $1^\circ C$, prove that all the spheres have the same temperature.
Name | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.986491322517395, "perplexity": 104.10731955352715}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948029.93/warc/CC-MAIN-20180425232612-20180426012612-00638.warc.gz"} |
https://kazune-lab.net/contest/2020/03/21/agc043/ | # AtCoder Grand Contest 043
Updated:
AtCoder Grand Contest 043
# Solutions
## A - Range Flip Find Route
### Observation
Consider the $1$-dimensional problem. Obviously, the answer is the number of the changes from white to black on that route.
Let’s go on to the $2$-dimensional problem. We fix a route. The answer must be greater than or equal to the number of the changes from white to black on that route. Here we count the constraint that we have to go in either right or down for each step. Therefore, it is impossible for the color changes to interfere with each other. Therefore, the obstacle is achieved by the number of the changes from white to black.
### Solution
For non-fixed route version, we can solve this problem by DFS on the grid.
## B - 123 Triangle
Let $A$ be the answer.
### When does the answer become 2?
Lemma B.1: If the answer is $2$, each $a _ i$ is either $0$ or $2$.
Proof: Suppose that the answer is $2$. Consider the $n$-previous stage. We show this lemma by the induction on $n$. Obviously, it holds in the case $n = 0$. Suppose that it holds in the case $n$. Then, consider $(n + 1)$-previous stage. If $2$ is produced in $n$-th stage, the possible cases is $\lvert 0 - 2 \rvert$ or $\lvert 2 - 0 \rvert$. If $0$ is produced, the possible cases are $\lvert 0 - 0 \rvert$, $\lvert 1 - 1 \rvert$ or $\lvert 2 - 2 \rvert$. So we have to deny the case $\lvert 1 - 1 \rvert$. If all the numbers in $(n + 1)$-th are $1$, the $n$-th numbers are all $0$, which results in $0$ as the answer. Thus this is impossible. If all the numbers in $(n + 1)$-th are not necessarily $1$, but they contains $1$, there must be at least one calculation $\lvert 2 - 1 \rvert$ or $\lvert 1 - 0 \rvert$, which produce $1$. This contradicts the assumption on $n$. This completes the proof.
Lemma B.2: It follows that $A = \sum _ {i \in N} \begin{pmatrix} N - 1 \\ i \end{pmatrix} a _ i \text{ in } \mathbb{Z} / 2 \mathbb{Z}.$
Proof: Since $\lvert x - y \rvert = x + y$ in $\mathbb{Z} / 2 \mathbb{Z}$, we can easily count the effect of $a _ i$ as $\begin{pmatrix} N - 1 \\ i \end{pmatrix}$.
### Solution
If each $a _ i$ is either $0$ or $2$, we can convert $0 \mapsto 0$ and $2 \mapsto 1$ and apply Lemma B.2 to see whether $A$ is odd or even. If it is even, the real answer is $0$. If not, the real answer is $2$.
Otherwise, $A$ is either $0$ or $1$. We can determine this in $O(N)$-time by Lemma B.2.
We want to know each $\begin{pmatrix} N - 1 \\ i \end{pmatrix}$ is even or odd. This is easily done by counting the number of factors $2$ for each $k!$s.
## C - Giant Graph
### Initial observation
We focus on the huge constant $10 ^ {18}$. This is so large that we just apply the greedy method to make the largest stable set.
If the graph is normal, this greedy strategy is regarded as follows. We try all vertexes in the descending order. If we can take the vertex we visit, conquer it. Otherwise, don’t conquer it.
### Nim and grundy number
This is equivalent to the famous Nim game. Suppose that the graph is directed where each edge is from the smaller number to the bigger number. We play moving piece game. There are two players. If a player cannot move the piece, they lose. Then, we can determine loser’s vertexes and winner’s vertexes. We conquer loser’s vertexes.
We move to the original there graphs. We can extend this idea by using a piece on each graph. Two players move these three pieces. What we have to consider is loser’s states. This is calculated by grundy number. Note that its score is calculated by $B ^ i \cdot B ^ j \cdot B ^ k$, where $B = 10 ^ {18}$.
### Solution
For each graph, for all vertexes, we compute their grundy numbers. We can use unordered_set<int> S; and follow the definition of the grundy number to complete this task in $O(M)$-time. Then, create the table $T$ where
$T[i] =$ the sum of the scores in vertex whose grundy number is $i$.
Here we estimate the upper bound of the grundy number on a graph. Let $K$ be the largest grundy number. Then, considering the best possible case, it follows that $1 + 2 + 3 + \dots + K = \frac{K(K + 1)}{2} \leq M.$ Therefore, it follows that $K \leq \sqrt{2M}$.
Let $T _ X$, $T _ Y$, and $T _ Z$ be the table above of the graph $X$, $Y$ and $Z$, respectively. We can try all pairs of the indexes $(i, j)$ in $O(M)$-time. Let $k = i \oplus j$. The answer is the sum of $T _ X[i] T _ Y[j] T _ Z[k]$.
## D - Merge Triplets
Assume $1$-indexed.
### Important idea: Block
First, we demonstrate the following case. $\{ 1, 6, 5 \}, \{ 3, 2, 4 \}.$ From this, we generate the following permutation. $1, 3, 2, 4, 6, 5.$ Here we point out the number $6 = 3N$, the maximum number. When the top of the former queue is $6$, this queue is blocked until the others become empty. Therefore, the part $[6, 5]$ appears in the result as it is. We consider the blocks, each top of which is the maximum number having appeared before. In this case, we make blocks as follows. $\{ [1], [6, 5] \}, \{ [3, 2], [4] \}$ Then, the result is as follows. $[1], [3, 2], [4], [6, 5].$
### Yes or No problem
Suppose that we are given a permutation $\{ p _ i \} _ {i = 1} ^ {3N}$. When can it be generated by our rule? Let’s consider it by examples in $N = 2$.
#### Example #1
$[1], [6, 4, 2, 3, 5]$ cannot be generated since it contains a $5$-length block. We cannot pack into a $3$-length sequence.
#### Example #2
$[1], [2], [4, 3], [6, 5]$ can be generated, for example, by the following sequences. $\{ [1], [4, 3] \}, \{ [2], [6, 5] \}.$
#### Example #3
$[2, 1], [4, 3], [6, 5]$ cannot be generated since we cannot pack three $2$-length blocks into two $3$-length sequence.
#### Example #4
$[2, 1], [4, 3], [5], [6]$ can be generated, for example, by the following sequences. $\{ [2, 1], [5] \}, \{ [4, 3], [6] \}.$
#### Example #5
$[2, 1], [3], [4], [5], [6]$ can be generated, for example, by the following sequences. $\{ [2, 1], [3] \}, \{ [4], [5], [6] \}.$
#### Conclusion
The condition is as follows.
• There is no block whose length is greater than or equal to $3$.
• The number of $2$-length blocks $\leq$ the number of $1$-length blocks.
### DP part
Let’s convert the ideas above into DP.
#### Definition
$dp[i][j] =$ the number of relatively magnitude relationships when we consider $[1, i]$ with the number of $1$-length blocks $-$ the number of $2$-length blocks.
Naively, we want to determine the number of $i$-th elements in order, but this idea does not work. Instead of possessing concrete numbers, we keep relatively magnitude relationships. This idea enables us to insert numbers later.
We don’t have to care the concrete numbers for each element in the middle of DP. They are automatically determine when we reach the last element.
Here, we have to care that $j$ moves in $[-N, 3N]$. Therefore, we use vector<vector<ll>> dp(3 * N + 1, vector<ll>(4 * N + 1)).
#### Initial state
We set $dp[0][0 + N] = 1$. Otherwise $0$.
$\sum _ {j = 0} ^ {3N} dp[3N][j + N].$
#### Transition
For $i = 0, \dots, 3N$, for $j = 0, \dots, 4N$, we execute the following. \begin{align} dp[i + 3][j] & \mathbin{ {+} {=} } dp[i][j] \times (i + 1)(i + 2), \\ dp[i + 2][j] & \mathbin{ {+} {=} } dp[i][j] \times (i + 1), \\ dp[i + 1][j] & \mathbin{ {+} {=} } dp[i][j]. \end{align} Here, out-of-range transitions shall be ignored and modulo manipulations are omitted.
For example, We can see the last line by the following picture.
## E - Topology
### Fundamental group
Let $p _ i = (1/2 + i, 1/2)$ for $i = 0, \dots, N - 1$. We consider $X = \mathbb{R} ^ 2 \setminus \{ p _ 0, \dots, p _ {N - 1} \}$. Here, we have the following homotopic equivalence. $\mathbb{R} ^ 2 \setminus \{ p _ 0, \dots, p _ {N - 1} \} \simeq \bigvee _ {N} S ^ 1.$ Then, by van Kampen’s theorem, we have $\pi _ 1 (X) \cong (x _ 0 , \dots x _ {N - 1}). \tag{E.1}$ Here, the right hand side is the free group generated by the words $x _ 0, \dots, x _ {N - 1}$. Each $x _ i$ represents the homotopic equivalent set of the positive cycle whose center is $p _ i$. Since we use the knowledge of fundamental group, we see the followings.
• The chain $C \in \pi _ 1 (X)$ is homotopic to one point if and only if $[C] = e$ in $(x _ 0 , \dots x _ {N - 1})$.
For $i \in N$, let $f _ i \colon (x _ 0, \dots, x _ {N - 1}) \to (x _ 0, \dots, x _ {N - 1})$ be the homomorphism defined by $f _ i (x _ i) = e$ and $f _ i (x _ j) = x _ j$ for $i \neq j$.
• Suppose that we remove the point $p _ i$ from the set of the forbidden points. The chain $C \in \pi _ 1 (X)$ is homotopic to one point if and only if $f _ i([C]) = e$ in $(x _ 0 , \dots x _ {N - 1})$.
If we remove several points, we apply several corresponding homomorphism for $[C]$.
### Important observation
Let $N = 2$ and $A = 1110$. How can we give this example? Let’s consider this problem in the viewpoint of fundamental group.
We consider (E.1) in $N = 2$. Therefore, the condition $A = 1110$ is equivalent to the task to find $g \in \ker f _ 0 \cap \ker f _ 1 \setminus \{ e \}$. But this is very easy. We can find the following element. $g = x _ 0 x _ 1 x _ 0 ^ {-1} x _ 1 ^ {-1}. \tag{E.2}$ In fact, $f _ 0 (g) = e x _ 1 e x _ 1 ^ {-1} = e$ and $f _ 1 (g) = x _ 0 e x _ 0 ^ {-1} e = e$ and $g \neq e$.
Let $N = 3$ and $A = 11111110$. How can we apply the idea of (E.2)? Let $g$ defined by (E.2). Then, the following element gives us an example. $h = g x _ 2 g ^{-1} x _ 2 ^ {-1}. \tag{E.3}$ In fact, we can check the facts $f _ i (h) = e x _ 2 e x _ 2 ^ {-1} = e$ for $i = 0, 1$ and $f _ 2 (h) = g e g ^{-1} e = e$ and $g \neq e$ to see that $h \in \ker f _ 0 \cap \ker f _ 1 \cap \ker f _ 2 \setminus \{ e \}$.
We can apply this idea for the larger cases by recursion.
### Solution
Note that the constraints that $N \leq 8$ and that the number of points is less than or equal to $250,000$ are very loose. Thus we can implement in the easier way.
Before creating the path, we have to determine whether the answer is possible or not. If we have the pair of the set $(i, j) \in \mathcal{P}(N) \times \mathcal{P}(N)$ that satisfies $j \neq i$, $j \subset i$, $A _ j = 0$ and $A _ i = 1$, the answer is impossible. We can determine in $O((2 ^ {N}) ^ 2)$-time. In addition, if $A _ \emptyset = 1$, the answer is impossible. Otherwise, the answer is possible.
First, we create the path of the cycle $x _ i$ and $x _ i ^ {-1}$. To connect them in the further implementation, we set the start and end point as $(0, 0)$.
Next, we compose the chain of the cycles. We determine where to apply the technique above. For $i \subset N$ which satisfies $A _ i = 0$, we have to forbid the homotopic equivalence to the one point. But if we find $j \subset i$ that satisfies $j \neq i$ $A _ j = 0$, forbidding $j$ is suffice to forbid $i$. Otherwise, we apply the technique above. As we argued, we can compose this chain by the recursive method.
Just connecting the above chains, we obtain the desired chain.
### Why is the judge valid?
According to the editorial and the YouTube live, the judge determine whether the given curve is correct or not by the following way. It initializes the queue and goes on the curve. When the curve passes the point $(1/2 + i, 0)$, the judge inserts $a _ i$ into the queue. When the curve passes the point $(1/2 + i, 1)$, the judge inserts $b _ i$ into the queue. After scanning the curve, the judge erase adjacent $a _ i a _ i$ and $b _ i b _ i$ in the recursive way. When it wants to remove $p _ i$, it just ignores $a _ i$ and $b _ i$. If the curve is homotopic to one point, the queue will be empty. Otherwise, not.
This algorithm always gives the correct judgement.
First, the trivial homotopic equivalence such as going and backing is canceled by the erasing rule of the same adjacent characters.
Second, if we replace $b _ i$ with $a _ i$, the queue will become empty since this is the projection into the $x$-axis of the closed curve. It means we can always find the adjacent $a _ i b _ i$ or $b _ i a _ i$ in the recursive way unless the queue is empty.
We convert this queue into $[C]$. We see the elements from top to the last. If we find $a _ i b _ i$, we erase them and write $x _ i$. If we find $b _ i a _ i$, we erase them and write $x _ i ^ {-1}$. We always find those until the queue become empty. The result is $[C]$.
If the curve $C$ is homotopic to one point, each cycle must be reduced by the rule $x _ i x _ i ^ {-1} = x _ i ^ {-1} x _ i = e$. In the original queue, they mean $a _ i b _ i b _ i a _ i$ or $b _ i a _ i a _ i b _ i$. But those words must have been erased by the rule above. Therefore, the queue must have been empty.
If the queue is empty, it means the closed curve kept going and backing. It is trivial to see the curve is homotopic to the starting point without topological viewpoints.
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Categories: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9784179925918579, "perplexity": 343.2276589136299}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991252.15/warc/CC-MAIN-20210512035557-20210512065557-00105.warc.gz"} |
https://stats.stackexchange.com/questions/231469/how-to-implement-generalized-hypergeometric-function-to-use-in-beta-binomial-cdf | # How to implement generalized hypergeometric function to use in beta-binomial cdf, sf, ppf?
I'm writing a subclass of scipy.stats._distn_infrastructure.rv_discrete for the beta binomial distribution whose PMF is
$$P(X=k \mid N, \alpha, \beta){N \choose k} \frac{\mathrm{B}(k+\alpha,N-k+\beta)}{\mathrm{B}(\alpha,\beta)},$$
where $\mathrm{B}$ is the Beta function. My current implementation of the CDF and SF (survival function, equivalent to 1 - CDF) are imprecise; the strategy I employed computes the expected value of the binomial cdf with respect to the beta component:
$$P_{BB}(X \le k \mid N, \alpha, \beta) = E_p\left[P_{Binom}(X \le k \mid N, p)\right],$$ where $p \sim \mathrm{Beta}(\alpha, \beta)$. I achieve this using the scipy.stats.beta.expect method, which is not innately vectorized (it will crash on anything other than a float or 0d array).
The PPF is even worse - it's a brute force loop over the integers $k=0, \ldots, N$ such that
$$P(X\le k \mid N, \alpha, \beta) \le q.$$
According to Wikipedia, the survival function for the beta-binomial distribution is
$$P(X > k \mid N, \alpha, \beta) = \frac{\mathrm{B}(\beta+n-k-1,\alpha+k+1)_3F_2(\boldsymbol{a},\boldsymbol{b};k)} {\mathrm{B}(\alpha,\beta)\mathrm{B}(n-k,k+2) (n+1)},$$
where ${}_3F_2$ is the generalized hypergeometric function. Is there an efficient way to compute this in Python, so I can remove the reference to beta.expect? Also, how would I invert this function to solve for $k$ given $q=P(X \le k\mid N, \alpha, \beta)$?
• It might help to know that for the values of $\boldsymbol{a},\boldsymbol{b}$ that (implicitly) appear here, $_3F_2(;;z)$ is a polynomial in $z$ (of degree $n-k-1$, $-1\le k \le n-1$). It does not simplify in general. – whuber Aug 24 '16 at 16:14
• Did you found any solution for your question? If yes, maybe you'd like to share it as an answer to your question? – Tim Feb 20 '17 at 10:23
This does not answer your question directly, but if you are thinking of estimating the cumulative distribution function of beta-binomial more efficiently, then you can use a recursive algorithm that is a little bit more efficient than the naive implementation.
Notice that probability mass function of beta-binomial distribution
$$f(x) = {n \choose x} \frac{\mathrm{B}(x+\alpha, n-x+\beta)}{\mathrm{B}(\alpha, \beta)}$$
may be re-written if you recall that $\mathrm{B}(x,y)=\tfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$, and $\Gamma(x) = (x-1)!$, and that ${n \choose k} = \prod_{i=1}^k \tfrac{n+1-i}{i}$, so that it becomes
$$f(x) = \left( \prod_{i=1}^x \frac{n+1-i}{i} \right) \frac{\frac{(\alpha+x-1)!\,(\beta+n-x-1)!}{(\alpha+\beta+n-1)!}}{\mathrm{B}(\alpha,\beta)}$$
this makes updating from $x$ to $x+1$ easy
$$f(x\color{red}{+1}) = \left( \prod_{i=1}^x \frac{n+1-i}{i} \right) \color{red}{\frac{n+1-x+1}{x+1}} \frac{\frac{(\alpha+x-1)! \,\color{red}{(\alpha+x)}\,(\beta+n-x-1)! \, \color{red}{(\beta+n-x)^{-1}}}{(\alpha+\beta+n-1)!\,\color{red}{(\alpha+\beta+n)}}}{\mathrm{B}(\alpha,\beta)}$$
and using this you can calculate cumulative distribution function as
$$F(x) = \sum_{k=0}^x f(k)$$
using just simple arithmetic operations rather then calculating more computer-intensive functions.
Sidenote: when dealing with large numbers, you would get into numeric precision issues, so more robust code would need working with logarithms, but even though you could expect improvement in efficiency (up to 2 to 3 times faster code when I ran few benchmarks on C++ code implementing it as compared to naive implementation).
• Another note. The ratio of beta integrals for the first term is another simple product $f (0)=\frac {B (a,n+b)}{B (a,b)}=\frac {\Gamma (n+b)\Gamma (a+b)}{\Gamma (n+a+b)\Gamma (b)}$ this simplifies to $\prod_{j=1}^n\frac {n+b-j}{n+a+b-j}$ – probabilityislogic Nov 6 '16 at 10:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8092642426490784, "perplexity": 365.1304216399036}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256314.52/warc/CC-MAIN-20190521102417-20190521124417-00425.warc.gz"} |
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