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https://cs.stackexchange.com/tags/lambda-calculus/new
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# Tag Info
5
If we keep reducing the right subterm, then yes, we will never get to $v$. But noone says that we are restricted to this decision. The Church-Rosser theorem does not claim that any possible reduction series must lead to $N_3$. Church-Rosser just states that $N_2 \twoheadrightarrow_\beta N_3$, i.e. that there exists some beta reduction series that will lead ...
4
Those features are almost never implemented like the lambda calculus in modern programming language implementation. In some cases, using the lambda calculus representation for datatypes has performance improvements (this is associated with so-called tagless representations). Historically, the Haskell compiler did use this representation early on, but has ...
2
Given the tag combinatory-logic, the answer in combinatory logic is C, i.e. "the C combinator". Obviously, this name is not self-documenting or going to be obvious in even a slightly more general context.
6
The function $$\lambda f.\lambda x.\lambda y.f\;y\;x$$ of type $$\forall X. \forall Y. \forall Z.(X \to Y \to Z) \to Y \to X \to Z$$ is often called flip. This is the case in Haskell (see here), and in some OCaml libraries as well (see here). According to wikipedia, people call this function (or combinator) $C$ in the context of combinatory logic (that name ...
Top 50 recent answers are included
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2019-08-23 11:00:06
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https://gzipwtf.com/what-is-natural-coordinate-system/
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# What is natural coordinate system?
## What is natural coordinate system?
Natural coordinate system is basically a local coordinate system which allows the specification of a point within the element by a set of dimensionless numbers whose magnitude never exceeds unity. This system is defined in such that the magnitude at nodal points will have unity or zero or a convenient set of fractions.
## What are the types of coordinate systems?
Common coordinate systems
• Number line.
• Cartesian coordinate system.
• Polar coordinate system.
• Cylindrical and spherical coordinate systems.
• Homogeneous coordinate system.
• Other commonly used systems.
• Relativistic coordinate systems.
• Citations.
What are the 5 coordinate systems?
Independent Latitude, Longitude, Vertical, and Time Axes. When each of a variable’s spatiotemporal dimensions is a latitude, longitude, vertical, or time dimension, then each axis is identified by a coordinate variable.
What are the three types of coordinate systems?
There are three commonly used coordinate systems: Cartesian, cylindrical and spherical. In this chapter, we will describe a Cartesian coordinate system and a cylindrical coordinate system.
### What are the advantages of natural coordinate system?
An advantage of natural coordinates is that rigid body constraints, joint constraints and driving constraints are always quadratic or linear. Mixed coordinates include more complex equations because relative coordinate constraints involve transcendental functions.
### What is difference between natural coordinates and local coordinates?
Natural Coordinates It is a local coordinate system that permits the specification of a point within the element by a dimensionless parameter whose absolute magnitude never exceeds unity. It is dimension less. They are defined with respect to the element rather than with reference to the global coordinates.
What are the two main types of coordinate systems?
Data is defined in both horizontal and vertical coordinate systems. Horizontal coordinate systems locate data across the surface of the earth, and vertical coordinate systems locate the relative height or depth of data. Horizontal coordinate systems can be of three types: geographic, projected, or local.
What is the most common coordinate system?
Universal transverse Mercator (UTM) is a geographic coordinate system and the most prevalent plane grid system used. UTM divides the earth into 84˚ north latitude to 80˚ south and is numbered into 60 vertical zones (each 6˚ latitude wide).
## What are the two types of coordinates?
Types of Coordinate Systems – Cartesian & Polar Coordinate Systems.
## What are the two types of coordinate systems?
What composes our natural coordinates?
In a natural coordinate system, there exist two horizontal coordinates: s, following the flow (streamwise), and n, perpendicular (normal) to the flow and z, referring to the up-down/vertical direction. The natural coordinate system is often used “heuristically” (for teaching concepts).
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2022-08-12 16:03:14
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http://www.thestudentroom.co.uk/showthread.php?t=2055349
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You are Here: Home
# Daniel Tammet on the Late Show with David Letterman Tweet
Maths and statistics discussion, revision, exam and homework help.
Announcements Posted on
1. Daniel Tammet on the Late Show with David Letterman
Towards the end of this video, Tammet tells Letterman his date of birth. Letterman acts like he's trying to figure out the day, and guesses (?) Wednesday - and he's right!
There are 2 possibilities: Letterman made a lucky guess or he had looked up Tammet's birth-date and day before the show. What is the probability that Letterman made a lucky guess?
Is that correct?
Last edited by thomaskurian89; 12-07-2012 at 15:06.
2. Re: Daniel Tammet on the Late Show with David Letterman
There's a 1 in 7 chance. There are 7 days per week. So the probability of randomly picking the correct one is 1/7.
Due to the nature of the show, it's likely there was a script or a plan, so Letterman probably did not guess.
3. Re: Daniel Tammet on the Late Show with David Letterman
(Original post by Llewellyn)
There's a 1 in 7 chance. There are 7 days per week. So the probability of randomly picking the correct one is 1/7.
While that's true, that's not what I asked. My question was: Given that we know that Letterman got it right, what is the probability that he guessed?
Last edited by thomaskurian89; 12-07-2012 at 16:28.
4. Re: Daniel Tammet on the Late Show with David Letterman
Towards the end of this video, Tammet tells Letterman his date of birth. Letterman acts like he's trying to figure out the day, and guesses (?) Wednesday - and he's right!
There are 2 possibilities: Letterman made a lucky guess or he had looked up Tammet's birth-date and day before the show. What is the probability that Letterman made a lucky guess?
Is that correct?
I got the same answer as you. But I'm no expert.
(Original post by Llewellyn)
There's a 1 in 7 chance. There are 7 days per week. So the probability of randomly picking the correct one is 1/7.
Due to the nature of the show, it's likely there was a script or a plan, so Letterman probably did not guess.
1/7 is P(Correct | Guess) but the OP is asking for P(Guess | Correct).
(This is all assuming that the initial probability that Letterman took a lucky guess is the same as the initial probability that he cheated).
Last edited by notnek; 12-07-2012 at 16:29.
5. Re: Daniel Tammet on the Late Show with David Letterman
While that's true, that's not what I asked. My question was: Given that we know that Letterman got it right, what is the probability that he guessed?
You are assuming that the probability of guessing is 1/2 or 0.5 . How do you know that this is true? No evidence has been given to suggest this.
6. Re: Daniel Tammet on the Late Show with David Letterman
(Original post by Llewellyn)
You are assuming that the probability of guessing is 1/2 or 0.5 . How do you know that this is true? No evidence has been given to suggest this.
It's an assumption made by thomaskurian. While it may not be true, you can still do the maths.
7. Re: Daniel Tammet on the Late Show with David Letterman
(Original post by Llewellyn)
You are assuming that the probability of guessing is 1/2 or 0.5 . How do you know that this is true? No evidence has been given to suggest this.
I think you have a point.
8. Re: Daniel Tammet on the Late Show with David Letterman
(Original post by notnek)
It's an assumption made by thomaskurian. While it may not be true, you can still do the maths.
I would have preferred him to have stated that the general answer is where p is the probability that Letterman guessed.
Stating your assumptions is vital, especially in Statistics.
9. Re: Daniel Tammet on the Late Show with David Letterman
(Original post by Llewellyn)
I would have preferred him to have stated that the general answer is where p is the probability that Letterman guessed.
Stating your assumptions is vital, especially in Statistics.
p is not the probability that Letterman guessed. (We are trying to find that out.) p is the probability that Letterman guesses in such situations.
10. Re: Daniel Tammet on the Late Show with David Letterman
p is not the probability that Letterman guessed. (We are trying to find that out.) p is the probability that Letterman guesses in such situations.
p is the probability that Letterman guessed before any additional information is given. You're trying to use this to work out the probability that Letterman guessed once we know that his answer was correct.
Is this what you meant? I didn't really understand your post.
11. Re: Daniel Tammet on the Late Show with David Letterman
p is not the probability that Letterman guessed. (We are trying to find that out.) p is the probability that Letterman guesses in such situations.
Yes but you don't know the probability that Letterman guessed or the probability that letterman guesses in such situations. Don't you see the problem?
Analogy:
2y = x
Find x without knowing what y is.
12. Re: Daniel Tammet on the Late Show with David Letterman
(Original post by Llewellyn)
Yes but you don't know the probability that Letterman guessed or the probability that letterman guesses in such situations. Don't you see the problem?
Analogy:
2y = x
Find x without knowing what y is.
13. Re: Daniel Tammet on the Late Show with David Letterman
I thought Llewellyn described it fine:
P(Guess)=p
Last edited by notnek; 12-07-2012 at 17:52.
14. Re: Daniel Tammet on the Late Show with David Letterman
(Original post by notnek)
I thought Llewwellyn described it fine:
P(Guess)=p
He said that p is the probability that Letterman guessed. If that were true, our answer would be p.
15. Re: Daniel Tammet on the Late Show with David Letterman
He said that p is the probability that Letterman guessed. If that were true, our answer would be p.
No, our answer would be p/ (7-6p), because you want to find the probability that he guessed given that he got it correct.
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2013-06-19 11:07:27
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http://www.gradesaver.com/textbooks/math/precalculus/precalculus-mathematics-for-calculus-7th-edition/chapter-1-section-1-1-real-numbers-1-1-exercises-page-11/72
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## Precalculus: Mathematics for Calculus, 7th Edition
a) $\frac{1}{4}$ or $0.25$ b) 1
a) By first considering the expression without absolute value signs, you see that $\frac{-6}{24}$ simplifies to $\frac{-1}{4}$. Then you can add back the absolute value signs, and because this is less than zero, you multiply by $-1$ to get $\frac{1}{4}$. b) First, evaluate the top and bottom expressions of the fraction. $7-12$ is $-5$, and $12-7$ is $5$. This leaves the fraction at $\frac{-5}{5}$, which simplifies to $\frac{-1}{1}$ which is simply $-1$. Considering the absolute value signs again, this is less than zero, so we multiply by negative one. $(-1)(-1)$ is $1$.
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2018-03-22 05:43:07
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https://mathoverflow.net/questions/64370/simplest-examples-of-rings-that-are-not-isomorphic-to-their-opposites/64384
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# Simplest examples of rings that are not isomorphic to their opposites
What are the simplest examples of rings that are not isomorphic to their opposite rings? Is there a science to constructing them?
## The only simple example known to me:
In Jacobson's Basic Algebra (vol. 1), Section 2.8, there is an exercise that goes as follows:
Let $u=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix}\in M_3(\mathbf Q)$ and let $x=\begin{pmatrix} u & 0 \\ 0 & u^2 \end{pmatrix}$, $y=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, where $u$ is as indicated and $0$ and $1$ are zero and unit matrices in $M_3(\mathbf Q)$. Hence $x,y\in M_6(\mathbf Q)$. Jacobson gives hints to prove that the subring of $M_6(\mathbf Q)$ generated by $x$ and $y$ is not isomorphic to its opposite.
## Examples seem to be well-known to the operator algebras crowd:
See for example the paper: "A Simple Separable C*-Algebra Not Isomorphic to Its Opposite Algebra" by N. Christopher Phillips, Proceedings of the American Mathematical Society Vol. 132, No. 10 (Oct., 2004), pp. 2997-3005.
• For the same operator algebra people: see also Alain Connes, A factor not anti-isomorphic to itself Ann. Math. (2) 101, 1975, 536-554. May 9, 2011 at 11:23
• Out of curiosity: why do operator people publish papers about these examples? Are they much rarer there? Us mere algebrist don't think much of this, I think (all my algebra undergrad students have had to check that some example or another works) May 9, 2011 at 15:21
• @Mariano: It is extraordinarily hard to see whether von Neumann algebras are isomorphic. Results like these are usually hard-won and are accompanied by deep new understanding of the structure of the algebras. (Notice that the above paper of Connes is an Annals paper.) May 9, 2011 at 15:34
• Thanks everyone for the nice answers. I would like to have accepted more than one. May 10, 2011 at 10:36
• The $*$-operator algebra examples are non-examples for this question: they have a involution, so are rings isomorphic to their opposite rings. The point of the (difficult) examples is that they have no complex-linear anti-automorphism as rings with involution i.e. complex-linear anti-automorphisms that commute with the fixed anti-automorphism $*$ (b.t.w. in $C^*$ algebras all ring [anti]automorphism are real-linear, but not always complex-linear) Feb 15, 2014 at 15:45
Here's a factory for making examples. If $\Gamma$ is a quiver, and $k$ a field, then we get a quiver algebra $k\Gamma$. If $\Gamma$ has no oriented cycles, we can recover $\Gamma$ from $k\Gamma$ by taking the Ext-construction. Also, the opposite algebra of a quiver algebra is obtained by reversing all the arrows in the quiver.
Hence you can produce an example by taking the quiver algebra of any quiver with no oriented cycles, which is not isomorphic to its reverse. It's easy to construct lots of quivers with these properties.
• It's clear if you believe that theorem, that you can recover the original quiver! May 9, 2011 at 14:16
• Incidentally, the smallest such quiver is the V-shaped quiver with two arrows emanating from one point. This gives a five-dimensional algebra: that's quite neat and tidy! May 9, 2011 at 15:09
• Ok then could someone indicate what this Ext-construction is which recovers the quiver? May 10, 2011 at 15:32
• First, one finds the vertices of the quiver $\Gamma$, by finding the unique maximal set of orthogonal idempotents $\{e_i\}$; these will correspond to vertices. Then, the number of arrows from $i$ to $j$ will be $Ext^1(\Gamma e_j,\Gamma e_i)$. May 10, 2011 at 15:49
• Alternatively: since the algebra is finite dimensiona, there are a finite number of isomorphism classes of simple modules. Use them as vertices of a graph, and draw $\dim\operatorname{Ext}^1(S,T)$ arrows from the class of $S$ to the class of $T$. May 10, 2011 at 16:00
Here is an easy example. Consider the abelian group $M = \mathbb{Z} \times \mathbb{Q}$. I claim that $R:=\text{End}(M)$ does not have any anti-endomorphism at all. EDIT: My previous proof is flawed. Thanks to Leon Lampret who pointed this out to me. The new proof shows that $R$ has several anti-endomorphisms, but no one is invertible. Thus $R$ is not isomorphic to $R^{\mathrm{op}}$.
Identify $R$ with the matrix ring $\begin{pmatrix} \mathbb{Z} & 0 \\\ \mathbb{Q} & \mathbb{Q} \end{pmatrix}$. The endomorphism ring of the underlying abelian group $\mathbb{Z} \times \mathbb{Q} \times \mathbb{Q}$ of $R$ can be identified with the matrix ring $\begin{pmatrix} \mathbb{Z} & 0 & 0 \\\ \mathbb{Q} & \mathbb{Q} & \mathbb{Q} \\\ \mathbb{Q} & \mathbb{Q} & \mathbb{Q} \end{pmatrix}$.
Assume an anti-endomorphism $\alpha$ of $R$ is given by such a matrix $\begin{pmatrix}a & 0 & 0 \\\ b & c & d \\\ e & f & g \end{pmatrix}$.
Then $\alpha(1)=1$ yields $a=1, b+d=0, e+g=1$. The determinant is $cg-df$. For all six-tuples $(u,v,w,p,q,r)$ (with $u,p$ integer) we have
$\alpha\left(\begin{pmatrix} u & 0 \\\ v & w \end{pmatrix} \begin{pmatrix} p & 0 \\\ q & r \end{pmatrix}\right) = \alpha \begin{pmatrix} p & 0 \\\ q & r \end{pmatrix} \alpha\begin{pmatrix} u & 0 \\\ v & w \end{pmatrix}$
which yields the three equations
1) $a^2 pu = pu$
2) $ap(bu + cv + dw) + (bp + cq + dr)(eu + fv + gw) = bpu + c(qu + rv) + drw$
3) $(ep + fq + gr)(eu + fv + gw) = epu + f(qu + rv) + grw$
If we plug in the three equations we already know from $\alpha(1)=1$, this simplifies of course. Now insert some tuples to get the following equations:
$(0,1,0,0,1,0) \leadsto f^2 = 0 \Rightarrow f = 0$
$(0,1,0,1,0,0) \leadsto c = 0$
This already shows that the determinant of $\alpha$ is zero, thus $\alpha$ cannot be bijective. But we can go even further:
$(1,0,0,1,0,0) \leadsto be=0 \wedge e^2=e \Rightarrow e \in \{0,1\}$
For $e = 0$ we get
$\alpha=\begin{pmatrix}1 & 0 & 0 \\\ b & 0 & -b \\\ 0 & 0 & 0 \end{pmatrix}$
and for $e=1$ we get
$\alpha=\begin{pmatrix}1 & 0 & 0 \\\ 0 & 0 & 0 \\\ 1 & 0 & 0 \end{pmatrix}$.
Here $b \in \mathbb{Q}$ may be chosen arbitrary. These are all anti-endomorphisms of $R$.
There is a more advanced proof that $R$ is not isomorphic to $R^{\mathrm{op}}$: Observe that $R$ is right noetherian, but not left noetherian.
• Nice answer, Martin! Here's a link to supplement your last line: planetmath.org/encyclopedia/… May 9, 2011 at 14:46
• Thanks Martin. A very striking example from the triangular ring family. May 10, 2011 at 6:44
To amplify on Bugs Bunny's answer: let $$D$$ be a finite dimensional central division algebra over a field $$K$$. Then $$D \otimes_K D^{\operatorname{op}} \cong \operatorname{End}_K(D)$$. From this it follows that in the Brauer group of $$K$$, the class of $$D^{\operatorname{op}}$$ is the inverse of the class of $$D$$. So a central division algebra over a field is isomorphic to its opposite algebra iff it has order $$2$$ in the Brauer group, or, in the lingo of that field, period $$2$$.
So you can get examples by taking any field $$K$$ with $$\operatorname{Br}(K) \neq \operatorname{Br}(K)[2]$$. In particular the Brauer group of any non-Archimedean locally compact field is $$\mathbb{Q}/\mathbb{Z}$$ and the Brauer group of any global field is close to being the direct sum of the Brauer groups of its completions (there is one relation, the so-called reciprocity law, which says that a certain "sum of invariants" map is zero). So for instance a division algebra of dimension $$9$$ over its center will do and these things can be constructed over the above fields.
• Unless I miss something, this $L(V)$ is not a ring: we have $x.(y+z)=x$ while $(x.y)+(x.z)=x+x$. May 9, 2011 at 12:17
• True, but this can be repaired by passing to a monoid algebra for a monoid $M$ with a similar definition for multiplication (and making due allowance for the identity). May 9, 2011 at 12:32
• Your semigroup example is incorrect. The algebra of a left zero with adjoined zero is isomorphic to its opposite algebra. They are both isomorphic to the path algebra with two vertices and |S|-1 edges from one to the other Nov 13, 2020 at 16:21
• More specifically, the elements $1$ and $1-s$ with $s$ in S form a basis and (1-s)(1-t)=1-t for s,t in S so you have a basis which is a right zero semigroup with adjoined identity. Nov 13, 2020 at 16:24
• However morally you are correct. The monoid identities satisfies by a left zero semigroup with adjoined identity is axiomatized by the identity $xyx=xy$ and every path algebra of an acyclic quiver is the monoid algebra of such a monoid. So by James Cranch's argument you get lots of examples of this sort. Nov 13, 2020 at 16:43
A general idea to construct rings which behave different on the left and on the right is the following, which is already contained in Martins's answer: One considers triangular rings $$A=\begin{pmatrix} R & M \\ 0 & S \end{pmatrix}$$ where $R$ and $S$ are rings and $M$ is an $R$-$S$-bimodule. The left and right ideals of such a ring can be decribed: for example, the left ideals are isomorphic to $U\oplus J$, where $J$ is a left ideal of $S$, and $U$ an $R$-submodule of $R\oplus M$ with $MJ \subseteq U$. (See Lam's book A First Course in Noncommutative Rings, §1) Suitable choices of $R$, $M$ and $S$ lead to examples with quite different left and right structure. For example, the finite ring $$\begin{pmatrix} \mathbb{Z}/4\mathbb{Z} & \mathbb{Z}/2\mathbb{Z} \\ 0 & \mathbb{Z}/2\mathbb{Z} \end{pmatrix}$$ has 11 left ideals and 12 right ideals, if my counting is right. (This may be the smallest example of a unital ring not isomorphic to its opposite ring, but I'm not sure here.)
Of course, there are lots of examples, since there are many ring theoretic notions which are known to be not left-right symmetric. T. Y. Lam, in his two books (First Course mentioned above and Lectures on Modules and Rings), usually contructs at least one example of a ring being left blah but not right blah, whenever blah is a property which is not left-right symmetric. (Lam's books are generally worth reading, in particular when looking for examples!)
• I don't know whether or not there are others of order 16 (well, aside from its opposite), but there are definitely none of order smaller than 16. May 9, 2011 at 22:03
• Thanks for pointing out Lam's book. I looked at it and indeed it is worth reading. Jacobson's example looks suspiciously similar to a triangular ring construction. May 10, 2011 at 6:15
Your example is not simple, i.e., it is not a simple algebra! If you want a simple algebra, you need a field whose Brauer group has elements of order more than 2 (the opposite algebra = inversion in Brauer group). If I remember correctly, the p-adic field will do the trick...
• This is so much simpler! May 10, 2011 at 6:46
Hi Amri,
This is a bit late, but it's my favorite class of examples. If $X$ is a smooth affine variety over $\mathbb{C}$ (say), and $\mathcal{D} = \mathcal{D}(X)$ is its algebra of differential operators, then the opposite algebra $\mathcal{D}^{op}$ is isomorphic to $\mathcal{D}(K) = K\otimes \mathcal{D}\otimes K^{-1}$, where $K$ denotes the canonical module of $X$. [This is also true when $X$ is Gorenstein but not necessarily smooth---see work of Yekutieli.]
So one gets answers to your question when $X$ doesn't have trivial canonical bundle. [And of course the story sheafifies for any smooth variety.]
EDIT: I was writing carelessly the first time (thanks to Amri's comment for highlighting this). Note that $\mathcal{D}(K)$ acts on $K$ on the left. Since a left $\mathcal{D}$-module structure on a vector bundle (finitely generated projective module) is the same as a flat connection, one has $\mathcal{D}\cong \mathcal{D}(K)$ if and only if $K$ admits a flat connection. The first chern class of $K$ is an obstruction to the existence of a flat connection. So just pick your favorite such affine variety (see also this MO question for discussion of that). A pretty complete discussion of the (non)triviality of rings of twisted differential operators (TDOs) can be found in Beilinson-Bernstein "A proof of Jantzen conjectures."
This story also illuminates a little bit why differential operators on half-densities, i.e. $\mathcal{D}(K^{1/2}) = K^{1/2}\otimes \mathcal{D}\otimes K^{-1/2}$, plays a special role in the study of rings of differential operators and (twisted) $\mathcal{D}$-modules (it's canonically isomorphic to its opposite algebra).
• Is it always the case that $\mathcal D(K)$ is not isomorphic to $\mathcal D(X)$ if $K$ is not trivial? Why? May 11, 2011 at 6:11
• Thanks Tom! I miss our discussions in the corridor where I learned so many things. May 12, 2011 at 4:04
Here is an explicit example of a central simple algebra over $\mathbb{Q}$ not isomorphic to its opposite (which is merely a detailed example of what Pete explained).
First take a cubic cyclic Galois extension $L/\mathbb{Q}$, for instance $L = \mathbb{Q}[x] / (x^3 + x^2 − 2x − 1)$, and let $\rho$ be a non-trivial element of $\operatorname{Gal}(L/\mathbb{Q})$. Now take an arbitrary element $\gamma \in \mathbb{Q}$ which is not the norm of an element in $L$. Define $$D = L \oplus zL \oplus z^2L,$$ where $z$ is a new "symbol" subject to the relations $z^3 = \gamma$ and $zt = t^\rho z$ for all $t \in L$. Then $D$ is a central simple division algebra of degree $3$ (i.e. of dimension $9$), and since its image in $\operatorname{Br}(\mathbb{Q})$ has order $3$, it is not isomorphic to its opposite.
As you can imagine, this procedure works for any field admitting a cyclic extension (of degree $>2$) for which the norm is non-surjective.
A particularly simple example of an algebra not isomorphic to its (graded) opposite is the $\mathbb{R}$-algebra $\mathbb{C}$, where $1$ is even and $i$ is odd. This is the ($\mathbb{Z}/2$-graded) real Clifford algebra $Cl(-1) = \langle f \mid f^2 = -1 \rangle$. Its opposite is the Clifford algebra $Cl(1) = \langle e \mid e^2 = 1 \rangle$, whose underlying ungraded algebra is isomorphic to $\mathbb{R} \oplus \mathbb{R}$.
Per the discussion in the other answers, these two algebras represent $1$ and $-1 = 7$ in the graded Brauer group $\mathbb{Z}/8$ of $\mathbb{R}$.
Many examples are already given; here is another one, just for its own interest:
Let $$V$$ be a vector space of infinite countable dimension over a countable field $$K$$.
Let $$E$$ be the $$K$$-algebra of endomorphisms of $$V$$. I claim that $$E$$ is not isomorphic to its opposite (even as a ring, i.e., as $$\mathbf{Z}$$-algebra). Precisely:
• (1) for every $$g\in E-\{0\}$$, $$gE=\{gf:f\in E\}$$ is uncountable [of continuum cardinal]
• (2) there exists $$f\in E-\{0\}$$ such that $$Ef=\{gf:g\in E\}$$ is countable: [namely this holds iff $$f$$ has finite rank (otherwise it has continuum cardinal)]
Let me justify the non-bracketed assertions. In (2) this holds because if $$B$$ is a finite subset of $$E$$ such that $$f(B)$$ spans $$f(E)$$, then every element of $$Ef$$ is determined by its restriction to $$B$$.
In (1), just fix a line $$L$$ not in the kernel of $$g$$, and let $$f$$ range over the space $$Y$$ linear maps $$V\to L$$. Since the dual of $$V$$ has uncountable dimension, $$Y$$ has uncountable [continuum] dimension. And $$f\mapsto gf$$ is injective in restriction to $$Y$$.
Maybe in this case $$E$$ and $$E^{\mathrm{op}}$$ are not elementary equivalent, but this would require another argument.
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2022-09-25 15:35:36
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http://math.stackexchange.com/questions/187762/equation-of-line-and-its-points
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# Equation of line and its points
In the xy coordinate system if (a,b) and (a+3,b+k) are two points on the line defined by equation (the equation is kind of faded in text , but it seems to be like x=3y-7) then k =
A)9 , B)3 , C)1 , D)1 (Ans is 1)
Any suggestions on how that answer was calculated ?
-
The slope of the line through $(a,b)$ and $(a+3, b+k)$ is $\frac{b+k-b}{a+3-a}$, which is $\frac{k}{3}$.
The slope of the line $x=3y-7$ is $\frac{1}{3}$. This is because the equation can be rewritten as $3y=x+7$, and then in standard slope-intercept form as $y=\frac{1}{3}x+\frac{7}{3}$.
These slopes are equal $\frac{k}{3}$ and $\frac{1}{3}$ are equal.
Another way: Because $(a,b)$ is on the line, we have $a=3b-7$. Because $(a+3,b+k)$ is on the line, we have $a+3=3(b+k)-7$, that is, $a+3=3b+3k-7$.
Since $a=3b-7$, we conclude that $3=3k$.
-
so the answer was obtained by comparing the two slopes , right ? – MistyD Aug 28 '12 at 5:14
I was having difficulty computing the slope from the equation. Thanks for clearing that out – MistyD Aug 28 '12 at 5:16
@MistyD: Yes. I added another way of doing it. You are probably at this time learning about equations of lines and slopes, so probably the slope way is the way you are intended to do it. But the second way works fine too. – André Nicolas Aug 28 '12 at 5:18
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2015-08-03 19:32:09
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https://docs.birch.sh/libraries/Standard/classes/Distribution/
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# Distribution
abstract class Distribution<Value> < DelayDistribution
Distribution with delayed-sampling support.
• Value: Value type.
### Member Variables
Name Description
v:Random<Value>? Random variate associated with the distibution, if any.
clamped:Value? Clamped value, if any. See also: clamp().
### Member Functions
Name Description
rows Number of rows, when interpreted as a matrix.
columns Number of columns, when interpreted as a matrix.
supportsLazy Are lazy operations supported?
distribution Returns this; a convenience for code generation within the compiler.
setRandom Set a random variate associated with the distribution.
unsetRandom Unset the random variate associated with the distribution.
clamp Clamp value.
hasClamp Is the value clamped?
realize Realize a value for the associated random variate.
get Realize a value for the associated random variate, lazily.
value Realize a value for the associated random variate, eagerly.
observe Observe a value for the associated random variate.
observeLazy Observe a value for the associated random variate, lazily.
simulate Simulate a value.
simulateLazy Simulate a value as part of a lazy expression.
logpdf Evaluate the log probability density (or mass) function.
logpdfLazy Construct a lazy expression for the log probability density (or mass).
update Update the parent node on the $M$-path given the value of this node.
updateLazy Update the parent node on the $M$-path given the value of this node.
pdf Evaluate the probability density (or mass) function.
cdf Evaluate the cumulative distribution function at a value.
quantile Evaluate the quantile function at a cumulative probability.
lower Finite lower bound of the support of this node, if any.
upper Finite upper bound of the support of this node, if any.
graftGaussian Graft this onto the delayed sampling graph.
graftBeta Graft this onto the delayed sampling graph.
graftGamma Graft this onto the delayed sampling graph.
graftInverseGamma Graft this onto the delayed sampling graph.
graftInverseWishart Graft this onto the delayed sampling graph.
graftNormalInverseGamma Graft this onto the delayed sampling graph.
graftDirichlet Graft this onto the delayed sampling graph.
graftRestaurant Graft this onto the delayed sampling graph.
graftMultivariateGaussian Graft this onto the delayed sampling graph.
graftMultivariateNormalInverseGamma Graft this onto the delayed sampling graph.
graftMatrixGaussian Graft this onto the delayed sampling graph.
graftMatrixNormalInverseWishart Graft this onto the delayed sampling graph.
graftDiscrete Graft this onto the delayed sampling graph.
graftBoundedDiscrete Graft this onto the delayed sampling graph.
### Member Function Details
#### cdf
function cdf(x:Value) -> Real?
Evaluate the cumulative distribution function at a value.
• x: The value.
Return: the cumulative probability, if supported.
#### clamp
function clamp(x:Value)
Clamp value.
• x: The value.
In the context of delayed sampling, this is used by a node to fix the value of its parent nodes during a call of update(). When those parent nodes are eventually realized, this value will be used.
The typical use case for this is enumerations of random variables, e.g. when a node that represents the sum of two random variables is realized, it fixes the values of two random variables also (its parents).
#### columns
function columns() -> Integer
Number of columns, when interpreted as a matrix.
#### distribution
final function distribution() -> Distribution<Value>
Returns this; a convenience for code generation within the compiler.
#### get
final function get() -> Value
Realize a value for the associated random variate, lazily.
#### graftBeta
function graftBeta() -> Beta?
Graft this onto the delayed sampling graph.
#### graftBoundedDiscrete
function graftBoundedDiscrete() -> BoundedDiscrete?
Graft this onto the delayed sampling graph.
#### graftDirichlet
function graftDirichlet() -> Dirichlet?
Graft this onto the delayed sampling graph.
#### graftDiscrete
function graftDiscrete() -> Discrete?
Graft this onto the delayed sampling graph.
#### graftGamma
function graftGamma() -> Gamma?
Graft this onto the delayed sampling graph.
#### graftGaussian
function graftGaussian() -> Gaussian?
Graft this onto the delayed sampling graph.
#### graftInverseGamma
function graftInverseGamma() -> InverseGamma?
Graft this onto the delayed sampling graph.
#### graftInverseWishart
function graftInverseWishart() -> InverseWishart?
Graft this onto the delayed sampling graph.
#### graftMatrixGaussian
function graftMatrixGaussian() -> MatrixGaussian?
Graft this onto the delayed sampling graph.
#### graftMatrixNormalInverseWishart
function graftMatrixNormalInverseWishart(compare:Distribution<LLT>) -> MatrixNormalInverseWishart?
Graft this onto the delayed sampling graph.
#### graftMultivariateGaussian
function graftMultivariateGaussian() -> MultivariateGaussian?
Graft this onto the delayed sampling graph.
#### graftMultivariateNormalInverseGamma
function graftMultivariateNormalInverseGamma(compare:Distribution<Real>) -> MultivariateNormalInverseGamma?
Graft this onto the delayed sampling graph.
#### graftNormalInverseGamma
function graftNormalInverseGamma(compare:Distribution<Real>) -> NormalInverseGamma?
Graft this onto the delayed sampling graph.
#### graftRestaurant
function graftRestaurant() -> Restaurant?
Graft this onto the delayed sampling graph.
#### hasClamp
function hasClamp() -> Boolean
Is the value clamped?
#### logpdf
abstract function logpdf(x:Value) -> Real
Evaluate the log probability density (or mass) function.
• x: The value.
Return: the log probability density (or mass).
#### logpdfLazy
function logpdfLazy(x:Expression<Value>) -> Expression<Real>?
Construct a lazy expression for the log probability density (or mass).
• x: The value.
Return: expression giving the log probability density (or mass), if supported.
#### lower
function lower() -> Value?
Finite lower bound of the support of this node, if any.
#### observe
final function observe(x:Value) -> Real
Observe a value for the associated random variate.
• x: The value.
Returns: weight giving the log pdf (or pmf) evaluated at the value.
#### observeLazy
final function observeLazy(x:Expression<Value>) -> Expression<Real>?
Observe a value for the associated random variate, lazily.
• x: The value.
Returns: expression giving the log probability density (or mass), if supported.
#### pdf
function pdf(x:Value) -> Real
Evaluate the probability density (or mass) function.
• x: The value.
Return: the probability density (or mass).
#### quantile
function quantile(P:Real) -> Value?
Evaluate the quantile function at a cumulative probability.
• P: The cumulative probability.
Return: the quantile, if supported.
#### realize
final function realize()
Realize a value for the associated random variate. If supportsLazy() returns true this defers to get(), otherwise to value().
#### rows
function rows() -> Integer
Number of rows, when interpreted as a matrix.
#### setRandom
final function setRandom(v:Random<Value>)
Set a random variate associated with the distribution.
#### simulate
abstract function simulate() -> Value
Simulate a value.
Return: the value.
#### simulateLazy
function simulateLazy() -> Value?
Simulate a value as part of a lazy expression.
Return: the value, if supported.
#### supportsLazy
function supportsLazy() -> Boolean
Are lazy operations supported?
#### unsetRandom
final function unsetRandom(v:Random<Value>)
Unset the random variate associated with the distribution.
#### update
function update(x:Value)
Update the parent node on the $M$-path given the value of this node.
• x: The value.
#### updateLazy
function updateLazy(x:Expression<Value>)
Update the parent node on the $M$-path given the value of this node.
• x: The value.
#### upper
function upper() -> Value?
Finite upper bound of the support of this node, if any.
#### value
final function value() -> Value
Realize a value for the associated random variate, eagerly.
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2021-04-11 01:44:59
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http://forum.zkoss.org/question/89125/dynamic-grid-or-dynamic-header/
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# dynamic grid or dynamic header
daovallec
11 2
I need something like: postimg. org/image/lbakheyrh/
I need do a query in my database in a table with my servers, and in other table i have my platforms that has a relations with the table servers many to one, then one server has many platforms.
I need to do something like the image, create a grid in zk and with the header my servers, and sub header or sub column os each server the platforms that has each server. Later i need a form that if i add a type, name and date, save it in my database in the platform that i am located and in the server that i am located.
i am trying with something like:
zkfiddle .org/sample/1rhr8cd/18-Dynamic-Listbox-Columns#source-1
The problem is that i can not relate the server with platforms in my menu, and add a grid with a form in each column, of my platform,
And i am intenting with a grid in columns @load(vm.server), and i do not have idea how do it, i am trying by days, but i do not have any idea.
If somebody can help me.
Thanks
delete retag edit
Sort by » oldest newest most voted
ajaidka
196 4
I created dynamic headers for one of my clients, Please have look at sample code
<columns children="@load(vm.listOfheader)">
<template name="children" var="child">
</template>
</columns>
<template name="model" var="row">
<cell forEach="\${row.reports}">
<!-- Here fetch and display report for each cell -->
</cell>
</template> </grid>
Best of luck A Jaidka
[hide preview]
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2019-03-27 02:45:04
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http://clay6.com/qa/67137/shweta-ate-large-frac-of-a-pizza-and-her-friend-george-ate-large-frac-of-th
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Answer
Comment
Share
Q)
# Shweta ate $\large\frac{3}{11}$ of a pizza and her friend George ate $\large\frac{13}{22}$ of the pizza. How much total pizza did they eat altogether?
( A ) $\large\frac{19}{11}$
( B ) $\large\frac{22}{19}$
( C ) $\large\frac{11}{19}$
( D ) $\large\frac{19}{22}$
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2020-08-10 19:36:58
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https://github.com/mathjax/MathJax/issues/313
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# Introduce separate scale options for inline and display blocks#313
Closed
opened this Issue Sep 11, 2012 · 5 comments
### 3 participants
Contrary to the use case in the stackoverflow question I usually want the same behaviour as in regular LaTeX, so that inline formulas have the same font size as surrounding text but display formulas are bigger than the rest.
MathJax member
commented Sep 11, 2012
I may be misunderstanding you, but it sounds like you are asking for the font size for displayed equations to be larger than the font size for in-line equations. That is not the case for LaTeX (at least not in any style file that I know of, though I suppose it could be arranged). For example, in the amsart document class the fonts are the same for inline and displayed equations.
What does differ between the two is the spacing rules so that displayed equations are allowed to use more vertical space (inline equations try to reduce that so that interline spacing will not be affected). That means that displayed equations don't have to reduce the font size for things like fractions as quickly as inline equations do. Perhaps this is the effect you are referring to? But MathJax already does that the same way as LaTeX, so I'm confused.
If you can provide an example LaTeX file that shows the effect you are after, I'll look into it.
Hmm you are absolutely right. In fact, I now know why I was confused and thought it was bigger. The font rendering in small sizes (13px) doesn't look as good in Firefox as in Adobe Reader. I made some screenshots:
Firefox HTML/CSS (cleartype off):
Firefox HTML/CSS (cleartype on):
Firefox MathML (cleartype off):
Firefox MathML (cleartype on):
MathJax member
commented Aug 20, 2014
Do we mark this abandoned? I could imagine that the SO question could be turned into an extension but I don't understand if this thread still has a question in it.
MathJax member
commented Aug 21, 2014
I think at this point the complaint is about the quality of the fonts, but this is largely due to the poor font rendering system in Windows. It is possible that there is something that can be done to the font (hinting or something) that might help, but I don't have the expertise to do that.
You are right that something could be done to make an extension that allows different sized display and inline math, but I it is low priority. I would mark it abandoned (or address later if you want to keep the idea for the extension around).
MathJax member
commented Aug 22, 2014
I'm ok with closing this as abandoned. It can always be re-opened or referenced in a new feature request.
closed this Aug 22, 2014
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2016-07-01 11:42:27
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http://mathoverflow.net/questions/90909/problem-with-shelah-and-sterns-paper-on-the-hanf-number-of-the-theory-of-banach
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# Problem with Shelah and Stern's paper on the Hanf number of the theory of Banach spaces
I have been trying to understand "The Hanf number of the first order theory of Banach spaces" by Shelah and Stern (Trans. AMS 244 (1978) 147-241). They construct a normed space $M$ from a Hilbert space $\cal H$ by taking the unit ball of $M$ to be the intersection of that of $\cal H$ with the halfspaces defined by $(x, \pm a) \le 1 - \delta_a$ for certain unit vectors $a$ and small $\delta_a >0$. They then need to show that the $\cal H$-unit ball is definable in $M$ using a first order language with symbols for vector addition and membership of the $M$-unit ball. With this in view, in their Lemma 2.9, they claim certain of the vectors $a$ are definable, but they don't have what they need to apply the lemma they appeal to for this (they have an inequality $\|b\| \le (1 - \delta)^{-1}$ but their Lemma 2.3 needs the opposite inequality).
So this looks like a bug. I suspect it can be fixed, e.g., by taking the unit ball to be the convex hull of the proposed one and the vectors $\pm a$. But this would be quite disruptive to the rest of the argument, I suspect.
My questions are (1) have I missed something so that Shelah and Stern's proof does actually go through more or less as its stands, or (2) is there another reference that gives a correct proof of these results.
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2014-10-25 17:45:28
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https://blog.demofox.org/2017/03/09/how-to-train-neural-networks-with-backpropagation/
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# How to Train Neural Networks With Backpropagation
This post is an attempt to demystify backpropagation, which is the most common method for training neural networks. This post is broken into a few main sections:
1. Explanation
2. Working through examples
3. Simple sample C++ source code using only standard includes
4. Links to deeper resources to continue learning
Let’s talk about the basics of neural nets to start out, specifically multi layer perceptrons. This is a common type of neural network, and is the type we will be talking about today. There are other types of neural networks though such as convolutional neural networks, recurrent neural networks, Hopfield networks and more. The good news is that backpropagation applies to most other types of neural networks too, so what you learn here will be applicable to other types of networks.
# Basics of Neural Networks
A neural network is made up layers.
Each layer has some number of neurons in it.
Every neuron is connected to every neuron in the previous and next layer.
Below is a diagram of a neural network, courtesy of wikipedia. Every circle is a neuron. This network takes 3 floating point values as input, passes them through 4 neurons in a hidden layer and outputs two floating point values. The hidden layer neurons and the output layer neurons do processing of the values they are giving, but the input neurons do not.
To calculate the output value of a single neuron, you multiply every input into that neuron by a weight for that input, sum them up, and add a bias that is set for the neuron. This “weighted input” value is fed into an activation function and the result is the output value of that neuron. Here is a diagram for a single neuron:
The code for calculating the output of a single neuron could look like this:
float weightedInput = bias;
for (int i = 0; i < inputs.size(); ++i)
weightedInput += inputs[i] * weights[i];
float output = Activation(weightedInput);
To evaluate an entire network of neurons, you just repeat this process for all neurons in the network, going from left to right (from input to output).
Neural networks are basically black boxes. We train them to give specific ouputs when we give them specific inputs, but it is often difficult to understand what it is that they’ve learned, or what part of the data they are picking up on.
Training a neural network just means that we adjust the weight and bias values such that when we give specific inputs, we get the desired outputs from the network. Being able to figure out what weights and biases to use can be tricky, especially for networks with lots of layers and lots of neurons per layer. This post talks about how to do just that.
Regarding training, there is a funny story where some people trained a neural network to say whether or not a military tank was in a photograph. It had a very high accuracy rate with the test data they trained it with, but when they used it with new data, it had terrible accuracy. It turns out that the training data was a bit flawed. Pictures of tanks were all taken on a sunny day, and the pictures without tanks were taken on a cloudy day. The network learned how to detect whether a picture was of a sunny day or a cloudy day, not whether there was a tank in the photo or not!
This is one type of pitfall to watch out for when dealing with neural networks – having good training data – but there are many other pitfalls to watch out for too. Architecting and training neural networks is quite literally an art form. If it were painting, this post would be teaching you how to hold a brush and what the primary colors are. There are many, many techniques to learn beyond what is written here to use as tools in your toolbox. The information in this post will allow you to succeed in training neural networks, but there is a lot more to learn to get higher levels of accuracy from your nets!
# Neural Networks Learn Using Gradient Descent
Let’s take a look at a simple neural network where we’ve chosen random values for the weights and the bias:
If given two floating point inputs, we’d calculate the output of the network like this:
$Output = Activation(Input0 * Weight0 + Input1 * Weight1 + Bias)$
Plugging in the specific values for the weights and biases, it looks like this:
$Output = Activation(Input0 * 0.23 + Input1 * -0.1 + 0.3)$
Let’s say that we want this network to output a zero when we give an input of 1,0, and that we don’t care what it outputs otherwise. We’ll plug 1 and 0 in for Input0 and Input1 respectively and see what the output of the network is right now:
$Output = Activation(1* 0.23 + 0 * -0.1 + 0.3) \\ Output = Activation(0.53)$
For the activation function, we are going to use a common one called the sigmoid activation function, which is also sometimes called the logistic activation function. It looks like this:
$\sigma(x) = \frac{1}{1+e^{-x}}$
Without going into too much detail, the reason why sigmoid is so commonly used is because it’s a smoother and differentiable version of the step function.
Applying that activation function to our output neuron, we get this:
$Output = Activation(0.53) \\ Output = \sigma(0.53) \\ Output = 0.6295$
So, we plugged in 1 and 0, but instead of getting a 0 out, we got 0.6295. Our weights and biases are wrong, but how do we correct them?
The secret to correcting our weights and biases is whichever of these terms seem least scary to you: slopes, derivatives, gradients.
If “slope” was the least scary term to you, you probably remember the line formula $y=mx+b$ and that the m value was the “rise over run” or the slope of the line. Well believe it or not, that’s all a derivative is. A derivative is just the slope of a function at a specific point on that function. Even if a function is curved, you can pick a point on the graph and get a slope at that point. The notation for a derivative is $\frac{dy}{dx}$, which literally means “change in y divided by change in x”, or “delta y divided by delta x”, which is literally rise over run.
In the case of a linear function (a line), it has the same derivative over the entire thing, so you can take a step size of any size on the x axis and multiply that step size by $\frac{dy}{dx}$ to figure out how much to add or subtract from y to stay on the line.
In the case of a non linear function, the derivative can change from one point to the next, so this slope is only guaranteed to be accurate for an infinitely small step size. In practice, people just often use “small” step sizes and calling it good enough, which is what we’ll be doing momentarily.
Now that you realize you already knew what a derivative is, we have to talk about partial derivatives. There really isn’t anything very scary about them and they still mean the exact same thing – they are the slope! They are even calculated the exact same way, but they use a fancier looking d in their notation: $\frac{\partial y}{\partial x}$.
The reason partial derivatives even exist is because if you have a function of multiple variables like $z=f(x,y)=x^2+3y+2$, you have two variables that you can take the derivative of. You can calculate $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. The first value tells you how much the z value changes for a change in x, the second value tells you how much the z value changes for a change in y.
By the way, if you are curious, the partial derivatives for that function above are below. When calculating partial derivatives, any variable that isn’t the one you care about, you just treat as a constant and do normal derivation.
$\frac{\partial z}{\partial x} = 2x\\ \frac{\partial z}{\partial y} = 3\\$
If you put both of those values together into a vector $(\frac{\partial z}{\partial x},\frac{\partial z}{\partial y})$ you have what is called the gradient vector.
The gradient vector has an interesting property, which is that it points in the direction that makes the function output grow the most. Basically, if you think of your function as a surface, it points up the steepest direction of the surface, from the point you evaluated the function at.
We are going to use that property to train our neural network by doing the following:
1. Calculate the gradient of a function that describes the error in our network. This means we will have the partial derivatives of all the weights and biases in the network.
2. Multiply the gradient by a small “learning rate” value, such as 0.05
3. Subtract these scaled derivatives from the weights and biases to decrease the error a small amount.
This technique is called steepest gradient descent (SGD) and when we do the above, our error will decrease by a small amount. The only exception is that if we use too large of a learning rate, it’s possible that we make the error grow, but usually the error will decrease.
We will do the above over and over, until either the error is small enough, or we’ve decided we’ve tried enough iterations that we think the neural network is never going to learn the things we want to teach it. If the network doesn’t learn, it means it needs to be re-architected with a different structure, different numbers of neurons and layers, different activation functions, etc. This is part of the “art” that I mentioned earlier.
Before moving on, there is one last thing to talk about: global minimums vs local minimums.
Imagine that the function describing the error in our network is visualized as bumpy ground. When we initialize our weights and biases to random numbers we are basically just choosing a random location on the ground to start at. From there, we act like a ball, and just roll down hill from wherever we are. We are definitely going to get to the bottom of SOME bump / hole in the ground, but there is absolutely no reason to except that we’ll get to the bottom of the DEEPEST bump / hole.
The problem is that SGD will find a LOCAL minimum – whatever we are closest too – but it might not find the GLOBAL minimum.
In practice, this doesn’t seem to be too large of a problem, at least for people casually using neural nets like you and me, but it is one of the active areas of research in neural networks: how do we do better at finding more global minimums?
You might notice the strange language I’m using where I say we have a function that describes the error, instead of just saying we use the error itself. The function I’m talking about is called the “cost function” and the reason for this is that different ways of describing the error give us different desirable properties.
For instance, a common cost function is to use mean squared error of the actual output compared to the desired output.
For a single training example, you plug the input into the network and calculate the output. You then plug the actual output and the target output into the function below:
$Cost = ||target-output||^2$
In other words, you take the vector of the neuron outputs, subtract it from the actual output that we wanted, calculate the length of the resulting vector and square it. This gives you the squared error.
The reason we use squared error in the cost function is because this way error in either direction is a positive number, so when gradient descent does it’s work, we’ll find the smallest magnitude of error, regardless of whether it’s positive or negative amounts. We could use absolute value, but absolute value isn’t differentiable, while squaring is.
To handle calculating the cost of multiple inputs and outputs, you just take the average of the squared error for each piece of training data. This gives you the mean squared error as the cost function across all inputs. You also average the derivatives to get the combined gradient.
# More on Training
Before we go into backpropagation, I want to re-iterate this point: Neural Networks Learn Using Gradient Descent.
All you need is the gradient vector of the cost function, aka the partial derivatives of all the weights and the biases for the cost.
Backpropagation gets you the gradient vector, but it isn’t the only way to do so!
Another way to do it is to use dual numbers which you can read about on my post about them: Multivariable Dual Numbers & Automatic Differentiation.
Using dual numbers, you would evaluate the output of the network, using dual numbers instead of floating point numbers, and at the end you’d have your gradient vector. It’s not quite as efficient as backpropagation (or so I’ve heard, I haven’t tried it), but if you know how dual numbers work, it’s super easy to implement.
Another way to get the gradient vector is by doing so numerically using finite differences. You can read about numerical derivatives on my post here: Finite Differences
Basically what you would do is if you were trying to calculate the partial derivative of a weight, like $\frac{\partial Cost}{\partial Weight0}$, you would first calculate the cost of the network as usual, then you would add a small value to Weight0 and evaluate the cost again. You subtract the new cost from the old cost, and divide by the small value you added to Weight0. This will give you the partial derivative for that weight value. You’d repeat this for all your weights and biases.
Since realistic neural networks often have MANY MANY weights and biases, calculating the gradient numerically is a REALLY REALLY slow process because of how many times you have to run the network to get cost values with adjusted weights. The only upside is that this method is even easier to implement than dual numbers. You can literally stop reading and go do this right now if you want to 😛
Lastly, there is a way to train neural networks which doesn’t use derivatives or the gradient vector, but instead uses the more brute force-ish method of genetic algorithms.
Using genetic algorithms to train neural networks is a huge topic even to summarize, but basically you create a bunch of random networks, see how they do, and try combining features of networks that did well. You also let some of the “losers” reproduce as well, and add in some random mutation to help stay out of local minimums. Repeat this for many many generations, and you can end up with a well trained network!
Here’s a fun video visualizing neural networks being trained by genetic algorithms: Youtube: Learning using a genetic algorithm on a neural network
# Backpropagation is Just the Chain Rule!
Going back to our talk of dual numbers for a second, dual numbers are useful for what is called “forward mode automatic differentiation”.
Backpropagation actually uses “reverse mode automatic differentiation”, so the two techniques are pretty closely tied, but they are both made possible by what is known as the chain rule.
The chain rule basically says that if you can write a derivative like this:
$dy/dx$
That you can also write it like this:
$dy/du*du/dx$
That might look weird or confusing, but since we know that derivatives are actual values, aka actual ratios, aka actual FRACTIONS, let’s think back to fractions for a moment.
$3/2 = 1.5$
So far so good? Now let’s choose some number out of the air – say, 5 – and do the same thing we did with the chain rule
$3/2 = \\ 3/5 * 5/2 = \\ 15/10 = \\ 3/2 = \\ 1.5$
Due to doing the reverse of cross cancellation, we are able to inject multiplicative terms into fractions (and derivatives!) and come up with the same answer.
Ok, but who cares?
Well, when we are evaluating the output of a neural network for given input, we have lots of equations nested in each other. We have neurons feeding into neurons feeding into neurons etc, with the logistic activation function at each step.
Instead of trying to figure out how to calculate the derivatives of the weights and biases for the entire monster equation (it’s common to have hundreds or thousands of neurons or more!), we can instead calculate derivatives for each step we do when evaluating the network and then compose them together.
Basically, we can break the problem into small bites instead of having to deal with the equation in it’s entirety.
Instead of calculating the derivative of how a specific weight affects the cost directly, we can instead calculate these:
1. dCost/dOutput: The derivative of how a neuron’s output affects cost
2. dOutput/dWeightedInput: The derivative of how the weighted input of a neuron affects a neuron’s output
3. dWeightedInput/dWeight: The derivative of how a weight affects the weighted input of a neuron
Then, when we multiply them all together, we get the real value that we want:
dCost/dOutput * dOutput/dWeightedInput * dWeightedInput/dWeight = dCost/dWeight
Now that we understand all the basic parts of back propagation, I think it’d be best to work through some examples of increasing complexity to see how it all actually fits together!
# Backpropagation Example 1: Single Neuron, One Training Example
This example takes one input and uses a single neuron to make one output. The neuron is only trained to output a 0 when given a 1 as input, all other behavior is undefined. This is implemented as the Example1() function in the sample code.
# Backpropagation Example 2: Single Neuron, Two Training Examples
This time, we are going to teach it not only that it should output 0 when given a 1, but also that it should output 1 when given a 0.
We have two training examples, and we are training the neuron to act like a NOT gate. This is implemented as the Example2() function in the sample code.
The first thing we do is calculate the derivatives (gradient vector) for each of the inputs.
We already calculated the “input 1, output 0” derivatives in the last example:
$\frac{\partial Cost}{\partial Weight} = 0.1476 \\ \frac{\partial Cost}{\partial Bias} = 0.1476$
If we follow the same steps with the “input 0, output 1” training example we get these:
$\frac{\partial Cost}{\partial Weight} = 0.0 \\ \frac{\partial Cost}{\partial Bias} = -0.0887$
To get the actual derivatives to train the network with, we just average them!
$\frac{\partial Cost}{\partial Weight} = 0.0738 \\ \frac{\partial Cost}{\partial Bias} = 0.0294$
From there, we do the same adjustments as before to the weight and bias values to get a weight of 0.2631 and a bias of 0.4853.
If you are wondering how to calculate the cost, again you just take the cost of each training example and average them. Adjusting the weight and bias values causes the cost to drop from 0.1547 to 0.1515, so we have made progress.
It takes 10 times as many iterations with these two training examples to get the same level of error as it did with only one training example though.
As we saw in the last example, after 10,000 iterations, the error was 0.007176.
In this example, after 100,000 iterations, the error is 0.007141. At that point, weight is -9.879733 and bias is 4.837278
# Backpropagation Example 3: Two Neurons in One Layer
Here is the next example, implemented as Example3() in the sample code. Two input neurons feed to two neurons in a single layer giving two outputs.
Let’s look at how we’d calculate the derivatives needed to train this network using the training example that when we give the network 01 as input that it should give out 10 as output.
First comes the forward pass where we calculate the network’s output when we give it 01 as input.
$Z0=input0*weight0+input1*weight1+bias0 \\ Z0=0*0.2+1*0.8+0.5 \\ Z0=1.3 \\ \\ O0=\sigma(1.3) \\ O0=0.7858\\ \\ Z1=input0*weight2+input0*weight3+bias1\\ Z1=0*0.6+1*0.4+0.1\\ Z1=0.5\\ \\ O1=\sigma(0.5)\\ O1=0.6225$
Next we calculate a cost. We don’t strictly need to do this step since we don’t use this value during backpropagation, but this will be useful to verify that we’ve improved things after an iteration of training.
$Cost=0.5*||target-actual||^2\\ Cost=0.5*||(1,0)-(0.7858,0.6225)||^2\\ Cost=0.5*||(0.2142,-0.6225)||^2\\ Cost=0.5*0.6583^2\\ Cost=0.2167$
Now we begin the backwards pass to calculate the derivatives that we’ll need for training.
Let’s calculate dCost/dZ0 aka the error in neuron 0. We’ll do this by calculating dCost/dO0, then dO0/dZ0 and then multiplying them together to get dCost/dZ0. Just like before, this is also the derivative for the bias of the neuron, so this value is also dCost/dBias0.
$\frac{\partial Cost}{\partial O0}=O0-target0\\ \frac{\partial Cost}{\partial O0}=0.7858-1\\ \frac{\partial Cost}{\partial O0}=-0.2142\\ \\ \frac{\partial O0}{\partial Z0} = O0 * (1-O0)\\ \frac{\partial O0}{\partial Z0} = 0.7858 * 0.2142\\ \frac{\partial O0}{\partial Z0} = 0.1683\\ \\ \frac{\partial Cost}{\partial Z0} = \frac{\partial Cost}{\partial O0} * \frac{\partial O0}{\partial Z0}\\ \frac{\partial Cost}{\partial Z0} = -0.2142 * 0.1683\\ \frac{\partial Cost}{\partial Z0} = -0.0360\\ \\ \frac{\partial Cost}{\partial Bias0} = -0.0360$
We can use dCost/dZ0 to calculate dCost/dWeight0 and dCost/dWeight1 by multiplying it by dZ0/dWeight0 and dZ0/dWeight1, which are input0 and input1 respectively.
$\frac{\partial Cost}{\partial Weight0} = \frac{\partial Cost}{\partial Z0} * \frac{\partial Z0}{\partial Weight0} \\ \frac{\partial Cost}{\partial Weight0} = -0.0360 * 0 \\ \frac{\partial Cost}{\partial Weight0} = 0\\ \\ \frac{\partial Cost}{\partial Weight1} = \frac{\partial Cost}{\partial Z0} * \frac{\partial Z0}{\partial Weight1} \\ \frac{\partial Cost}{\partial Weight1} = -0.0360 * 1 \\ \frac{\partial Cost}{\partial Weight1} = -0.0360$
Next we need to calculate dCost/dZ1 aka the error in neuron 1. We’ll do this like before. We’ll calculate dCost/dO1, then dO1/dZ1 and then multiplying them together to get dCost/dZ1. Again, this is also the derivative for the bias of the neuron, so this value is also dCost/dBias1.
$\frac{\partial Cost}{\partial O1}=O1-target1\\ \frac{\partial Cost}{\partial O1}=0.6225-0\\ \frac{\partial Cost}{\partial O1}=0.6225\\ \\ \frac{\partial O1}{\partial Z1} = O1 * (1-O1)\\ \frac{\partial O1}{\partial Z1} = 0.6225 * 0.3775\\ \frac{\partial O1}{\partial Z1} = 0.235\\ \\ \frac{\partial Cost}{\partial Z1} = \frac{\partial Cost}{\partial O1} * \frac{\partial O1}{\partial Z1}\\ \frac{\partial Cost}{\partial Z1} = 0.6225 * 0.235\\ \frac{\partial Cost}{\partial Z1} = 0.1463\\ \\ \frac{\partial Cost}{\partial Bias1} = 0.1463$
Just like with neuron 0, we can use dCost/dZ1 to calculate dCost/dWeight2 and dCost/dWeight3 by multiplying it by dZ1/dWeight2 and dZ1/dWeight2, which are input0 and input1 respectively.
$\frac{\partial Cost}{\partial Weight2} = \frac{\partial Cost}{\partial Z1} * \frac{\partial Z1}{\partial Weight2} \\ \frac{\partial Cost}{\partial Weight2} = 0.1463 * 0 \\ \frac{\partial Cost}{\partial Weight2} = 0\\ \\ \frac{\partial Cost}{\partial Weight3} = \frac{\partial Cost}{\partial Z1} * \frac{\partial Z1}{\partial Weight3} \\ \frac{\partial Cost}{\partial Weight3} = 0.1463 * 1 \\ \frac{\partial Cost}{\partial Weight3} = 0.1463$
After using these derivatives to update the weights and biases with a learning rate of 0.5, they become:
Weight0 = 0.2
Weight1 = 0.818
Weight2 = 0.6
Weight3 = 0.3269
Bias0 = 0.518
Bias1 = 0.0269
Using these values, the cost becomes 0.1943, which dropped from 0.2167, so we have indeed made progress with our learning!
Interestingly, it takes about twice as many trainings as example 1 to get a similar level of error. In this case, 20,000 iterations of learning results in an error of 0.007142.
If we have the network learn the four patterns below instead:
00 = 00
01 = 10
10 = 10
11 = 11
It takes 520,000 learning iterations to get to an error of 0.007223.
# Backpropagation Example 4: Two Layers, Two Neurons Each
This is the last example, implemented as Example4() in the sample code. Two input neurons feed to two neurons in a hidden layer, feeding into two neurons in the output layer giving two outputs. This is the exact same network that is walked through on this page which is also linked to at the end of this post: A Step by Step Backpropagation Example
First comes the forward pass where we calculate the network’s output. We’ll give it 0.05 and 0.1 as input, and we’ll say our desired output is 0.01 and 0.99.
$Z0=input0*weight0+input1*weight1+bias0 \\ Z0=0.05*0.15+0.1*0.2+0.35 \\ Z0=0.3775 \\ \\ O0=\sigma(0.3775) \\ O0=0.5933 \\ \\ Z1=input0*weight2+input1*weight3+bias1\\ Z1=0.05*0.25+0.1*0.3+0.35\\ Z1=0.3925\\ \\ O1=\sigma(0.3925)\\ O1=0.5969\\ \\ Z2=O0*weight4+O1*weight5+bias2\\ Z2=0.5933*0.4+0.5969*0.45+0.6\\ Z2=1.106\\ \\ O2=\sigma(1.106)\\ O2=0.7514\\ \\ Z3=O0*weight6+O1*weight7+bias3\\ Z3=0.5933*0.5+0.5969*0.55+0.6\\ Z3=1.225\\ \\ O3=\sigma(1.225)\\ O3=0.7729$
Next we calculate the cost, taking O2 and O3 as our actual output, and 0.01 and 0.99 as our target (desired) output.
$Cost=0.5*||target-actual||^2\\ Cost=0.5*||(0.01,0.99)-(0.7514,0.7729)||^2\\ Cost=0.5*||(-0.7414,-0.2171)||^2\\ Cost=0.5*0.7725^2\\ Cost=0.2984$
Now we start the backward pass to calculate the derivatives for training.
## Neuron 2
First we’ll calculate dCost/dZ2 aka the error in neuron 2, remembering that the value is also dCost/dBias2.
$\frac{\partial Cost}{\partial O2}=O2-target0\\ \frac{\partial Cost}{\partial O2}=0.7514-0.01\\ \frac{\partial Cost}{\partial O2}=0.7414\\ \\ \frac{\partial O2}{\partial Z2} = O2 * (1-O2)\\ \frac{\partial O2}{\partial Z2} = 0.7514 * 0.2486\\ \frac{\partial O2}{\partial Z2} = 0.1868\\ \\ \frac{\partial Cost}{\partial Z2} = \frac{\partial Cost}{\partial O2} * \frac{\partial O2}{\partial Z2}\\ \frac{\partial Cost}{\partial Z2} = 0.7414 * 0.1868\\ \frac{\partial Cost}{\partial Z2} = 0.1385\\ \\ \frac{\partial Cost}{\partial Bias2} = 0.1385$
We can use dCost/dZ2 to calculate dCost/dWeight4 and dCost/dWeight5.
$\frac{\partial Cost}{\partial Weight4} = \frac{\partial Cost}{\partial Z2} * \frac{\partial Z2}{\partial Weight4}\\ \frac{\partial Cost}{\partial Weight4} = \frac{\partial Cost}{\partial Z2} * O0\\ \frac{\partial Cost}{\partial Weight4} = 0.1385 * 0.5933\\ \frac{\partial Cost}{\partial Weight4} = 0.0822\\ \\ \frac{\partial Cost}{\partial Weight5} = \frac{\partial Cost}{\partial Z2} * \frac{\partial Z2}{\partial Weight5}\\ \frac{\partial Cost}{\partial Weight5} = \frac{\partial Cost}{\partial Z2} * O1\\ \frac{\partial Cost}{\partial Weight5} = 0.1385 * 0.5969\\ \frac{\partial Cost}{\partial Weight5} = 0.0827\\$
## Neuron 3
Next we’ll calculate dCost/dZ3 aka the error in neuron 3, which is also dCost/dBias3.
$\frac{\partial Cost}{\partial O3}=O3-target1\\ \frac{\partial Cost}{\partial O3}=0.7729-0.99\\ \frac{\partial Cost}{\partial O3}=-0.2171\\ \\ \frac{\partial O3}{\partial Z3} = O3 * (1-O3)\\ \frac{\partial O3}{\partial Z3} = 0.7729 * 0.2271\\ \frac{\partial O3}{\partial Z3} = 0.1755\\ \\ \frac{\partial Cost}{\partial Z3} = \frac{\partial Cost}{\partial O3} * \frac{\partial O3}{\partial Z3}\\ \frac{\partial Cost}{\partial Z3} = -0.2171 * 0.1755\\ \frac{\partial Cost}{\partial Z3} = -0.0381\\ \\ \frac{\partial Cost}{\partial Bias3} = -0.0381$
We can use dCost/dZ3 to calculate dCost/dWeight6 and dCost/dWeight7.
$\frac{\partial Cost}{\partial Weight6} = \frac{\partial Cost}{\partial Z3} * \frac{\partial Z3}{\partial Weight6}\\ \frac{\partial Cost}{\partial Weight6} = \frac{\partial Cost}{\partial Z3} * O0\\ \frac{\partial Cost}{\partial Weight6} = -0.0381 * 0.5933\\ \frac{\partial Cost}{\partial Weight6} = -0.0226\\ \\ \frac{\partial Cost}{\partial Weight7} = \frac{\partial Cost}{\partial Z3} * \frac{\partial Z3}{\partial Weight7}\\ \frac{\partial Cost}{\partial Weight7} = \frac{\partial Cost}{\partial Z3} * O1\\ \frac{\partial Cost}{\partial Weight7} = -0.0381 * 0.5969\\ \frac{\partial Cost}{\partial Weight7} = -0.0227\\$
## Neuron 0
Next, we want to calculate dCost/dO0, but doing that requires us to do something new. Neuron 0 affects both neuron 2 and neuron 3, which means that it affects the cost through those two neurons as well. That means our calculation for dCost/dO0 is going to be slightly different, where we add the derivatives of both paths together. Let’s work through it:
$\frac{\partial Cost}{\partial O0} = \frac{\partial Cost}{\partial Z2} * \frac{\partial Z2}{\partial O0} + \frac{\partial Cost}{\partial Z3} * \frac{\partial Z3}{\partial O0}\\ \frac{\partial Cost}{\partial O0} = \frac{\partial Cost}{\partial Z2} * Weight4 + \frac{\partial Cost}{\partial Z3} * Weight6\\ \frac{\partial Cost}{\partial O0} = 0.1385 * 0.4 - 0.0381 * 0.5\\ \frac{\partial Cost}{\partial O0} = 0.0364$
We can then continue and calculate dCost/dZ0, which is also dCost/dBias0, and the error in neuron 0.
$\frac{\partial O0}{\partial Z0} = O0 * (1-O0)\\ \frac{\partial O0}{\partial Z0} = 0.5933 * 0.4067\\ \frac{\partial O0}{\partial Z0} = 0.2413\\ \\ \frac{\partial Cost}{\partial Z0} = \frac{\partial Cost}{\partial O0} * \frac{\partial O0}{\partial Z0}\\ \frac{\partial Cost}{\partial Z0} = 0.0364 * 0.2413\\ \frac{\partial Cost}{\partial Z0} = 0.0088\\ \\ \frac{\partial Cost}{\partial Bias0} = 0.0088$
We can use dCost/dZ0 to calculate dCost/dWeight0 and dCost/dWeight1.
$\frac{\partial Cost}{\partial Weight0} = \frac{\partial Cost}{\partial Z0} * \frac{\partial Z0}{\partial Weight0}\\ \frac{\partial Cost}{\partial Weight0} = \frac{\partial Cost}{\partial Z0} * input0\\ \frac{\partial Cost}{\partial Weight0} = 0.0088 * 0.05\\ \frac{\partial Cost}{\partial Weight0} = 0.0004\\ \\ \frac{\partial Cost}{\partial Weight1} = \frac{\partial Cost}{\partial Z0} * \frac{\partial Z0}{\partial Weight1}\\ \frac{\partial Cost}{\partial Weight1} = \frac{\partial Cost}{\partial Z0} * input1\\ \frac{\partial Cost}{\partial Weight1} = 0.0088 * 0.1\\ \frac{\partial Cost}{\partial Weight1} = 0.0009\\$
## Neuron 1
We are almost done, so hang in there. For our home stretch, we need to calculate dCost/dO1 similarly as we did for dCost/dO0, and then use that to calculate the derivatives of bias1 and weight2 and weight3.
$\frac{\partial Cost}{\partial O1} = \frac{\partial Cost}{\partial Z2} * \frac{\partial Z2}{\partial O1} + \frac{\partial Cost}{\partial Z3} * \frac{\partial Z3}{\partial O1}\\ \frac{\partial Cost}{\partial O1} = \frac{\partial Cost}{\partial Z2} * Weight5 + \frac{\partial Cost}{\partial Z3} * Weight7\\ \frac{\partial Cost}{\partial O1} = 0.1385 * 0.45 - 0.0381 * 0.55\\ \frac{\partial Cost}{\partial O1} = 0.0414\\ \\ \frac{\partial O1}{\partial Z1} = O1 * (1-O1)\\ \frac{\partial O1}{\partial Z1} = 0.5969 * 0.4031\\ \frac{\partial O1}{\partial Z1} = 0.2406\\ \\ \frac{\partial Cost}{\partial Z1} = \frac{\partial Cost}{\partial O1} * \frac{\partial O1}{\partial Z1}\\ \frac{\partial Cost}{\partial Z1} = 0.0414 * 0.2406\\ \frac{\partial Cost}{\partial Z1} = 0.01\\ \\ \frac{\partial Cost}{\partial Bias1} = 0.01$
Lastly, we will use dCost/dZ1 to calculate dCost/dWeight2 and dCost/dWeight3.
$\frac{\partial Cost}{\partial Weight2} = \frac{\partial Cost}{\partial Z1} * \frac{\partial Z1}{\partial Weight2}\\ \frac{\partial Cost}{\partial Weight2} = \frac{\partial Cost}{\partial Z1} * input0\\ \frac{\partial Cost}{\partial Weight2} = 0.01 * 0.05\\ \frac{\partial Cost}{\partial Weight2} = 0.0005\\ \\ \frac{\partial Cost}{\partial Weight3} = \frac{\partial Cost}{\partial Z1} * \frac{\partial Z1}{\partial Weight3}\\ \frac{\partial Cost}{\partial Weight3} = \frac{\partial Cost}{\partial Z1} * input1\\ \frac{\partial Cost}{\partial Weight3} = 0.01 * 0.1\\ \frac{\partial Cost}{\partial Weight3} = 0.001\\$
## Backpropagation Done
Phew, we have all the derivatives we need now.
Here’s our new weights and biases using a learning rate of 0.5:
Weight0 = 0.15 – (0.5 * 0.0004) = 0.1498
Weight1 = 0.2 – (0.5 * 0.0009) = 0.1996
Weight2 = 0.25 – (0.5 * 0.0005) = 0.2498
Weight3 = 0.3 – (0.5 * 0.001) = 0.2995
Weight4 = 0.4 – (0.5 * 0.0822) = 0.3589
Weight5 = 0.45 – (0.5 * 0.0827) = 0.4087
Weight6 = 0.5 – (0.5 * -0.0226) = 0.5113
Weight7 = 0.55 – (0.5 * -0.0227) = 0.5614
Bias0 = 0.35 – (0.5 * 0.0088) = 0.3456
Bias1 = 0.35 – (0.5 * 0.01) = 0.345
Bias2 = 0.6 – (0.5 * 0.1385) = 0.5308
Bias3 = 0.6 – (0.5 * -0.0381) = 0.6191
Using these new values, the cost function value drops from 0.2984 to 0.2839, so we have made progress!
Interestingly, it only takes 5,000 iterations of learning for this network to reach an error of 0.007157, when it took 10,000 iterations of learning for example 1 to get to 0.007176.
Before moving on, take a look at the weight adjustments above. You might notice that the derivatives for the weights are much smaller for weights 0,1,2,3 compared to weights 4,5,6,7. The reason for this is because weights 0,1,2,3 appear earlier in the network. The problem is that earlier layer neurons don’t learn as fast as later layer neurons and this is caused by the nature of the neuron activation functions – specifically, that the sigmoid function has a long tail near 0 and 1 – and is called the “vanishing gradient problem”. The opposite effect can also happen however, where earlier layer gradients explode to super huge numbers, so the more general term is called the “unstable gradient problem”. This is an active area of research on how to address, and this becomes more and more of a problem the more layers you have in your network.
You can use other activation functions such as tanh, identity, relu and others to try and get around this problem. If trying different activation functions, the forward pass (evaluation of a neural network) as well as the backpropagation of error pass remain the same, but of course the calculation for getting O from Z changes, and of course, calculating the derivative deltaO/deltaZ becomes different. Everything else remains the same.
# Sample Code
Below is the sample code which implements all the back propagation examples we worked through above.
Note that this code is meant to be readable and understandable. The code is not meant to be re-usable or highly efficient.
A more efficient implementation would use SIMD instructions, multithreading, stochastic gradient descent, and other things.
It’s also useful to note that calculating a neuron’s Z value is actually a dot product and an addition and that the addition can be handled within the dot product by adding a “fake input” to each neuron that is a constant of 1. This lets you do a dot product to calculate the Z value of a neuron, which you can take further and combine into matrix operations to calculate multiple neuron values at once. You’ll often see neural networks described in matrix notation because of this, but I have avoided that in this post to try and make things more clear to programmers who may not be as comfortable thinking in strictly matrix notation.
#include <stdio.h>
#include <array>
// Nonzero value enables csv logging.
#define LOG_TO_CSV_NUMSAMPLES() 50
// ===== Example 1 - One Neuron, One training Example =====
void Example1RunNetwork (
float input, float desiredOutput,
float weight, float bias,
float& error, float& cost, float& actualOutput,
float& deltaCost_deltaWeight, float& deltaCost_deltaBias, float& deltaCost_deltaInput
) {
// calculate Z (weighted input) and O (activation function of weighted input) for the neuron
float Z = input * weight + bias;
float O = 1.0f / (1.0f + std::exp(-Z));
// the actual output of the network is the activation of the neuron
actualOutput = O;
// calculate error
error = std::abs(desiredOutput - actualOutput);
// calculate cost
cost = 0.5f * error * error;
// calculate how much a change in neuron activation affects the cost function
// deltaCost/deltaO = O - target
float deltaCost_deltaO = O - desiredOutput;
// calculate how much a change in neuron weighted input affects neuron activation
// deltaO/deltaZ = O * (1 - O)
float deltaO_deltaZ = O * (1 - O);
// calculate how much a change in a neuron's weighted input affects the cost function.
// This is deltaCost/deltaZ, which equals deltaCost/deltaO * deltaO/deltaZ
// This is also deltaCost/deltaBias and is also refered to as the error of the neuron
float neuronError = deltaCost_deltaO * deltaO_deltaZ;
deltaCost_deltaBias = neuronError;
// calculate how much a change in the weight affects the cost function.
// deltaCost/deltaWeight = deltaCost/deltaO * deltaO/deltaZ * deltaZ/deltaWeight
// deltaCost/deltaWeight = neuronError * deltaZ/deltaWeight
// deltaCost/deltaWeight = neuronError * input
deltaCost_deltaWeight = neuronError * input;
// As a bonus, calculate how much a change in the input affects the cost function.
// Follows same logic as deltaCost/deltaWeight, but deltaZ/deltaInput is the weight.
// deltaCost/deltaInput = neuronError * weight
deltaCost_deltaInput = neuronError * weight;
}
void Example1 ()
{
#if LOG_TO_CSV_NUMSAMPLES() > 0
// open the csv file for this example
FILE *file = fopen("Example1.csv","w+t");
if (file != nullptr)
fprintf(file, ""training index","error","cost","weight","bias","dCost/dWeight","dCost/dBias","dCost/dInput"n");
#endif
// learning parameters for the network
const float c_learningRate = 0.5f;
const size_t c_numTrainings = 10000;
// training data
// input: 1, output: 0
const std::array<float, 2> c_trainingData = {1.0f, 0.0f};
// starting weight and bias values
float weight = 0.3f;
float bias = 0.5f;
// iteratively train the network
float error = 0.0f;
for (size_t trainingIndex = 0; trainingIndex < c_numTrainings; ++trainingIndex)
{
// run the network to get error and derivatives
float output = 0.0f;
float cost = 0.0f;
float deltaCost_deltaWeight = 0.0f;
float deltaCost_deltaBias = 0.0f;
float deltaCost_deltaInput = 0.0f;
Example1RunNetwork(c_trainingData[0], c_trainingData[1], weight, bias, error, cost, output, deltaCost_deltaWeight, deltaCost_deltaBias, deltaCost_deltaInput);
#if LOG_TO_CSV_NUMSAMPLES() > 0
const size_t trainingInterval = (c_numTrainings / (LOG_TO_CSV_NUMSAMPLES() - 1));
if (file != nullptr && (trainingIndex % trainingInterval == 0 || trainingIndex == c_numTrainings - 1))
{
// log to the csv
fprintf(file, ""%zi","%f","%f","%f","%f","%f","%f","%f",n", trainingIndex, error, cost, weight, bias, deltaCost_deltaWeight, deltaCost_deltaBias, deltaCost_deltaInput);
}
#endif
weight -= deltaCost_deltaWeight * c_learningRate;
bias -= deltaCost_deltaBias * c_learningRate;
}
printf("Example1 Final Error: %fn", error);
#if LOG_TO_CSV_NUMSAMPLES() > 0
if (file != nullptr)
fclose(file);
#endif
}
// ===== Example 2 - One Neuron, Two training Examples =====
void Example2 ()
{
#if LOG_TO_CSV_NUMSAMPLES() > 0
// open the csv file for this example
FILE *file = fopen("Example2.csv","w+t");
if (file != nullptr)
fprintf(file, ""training index","error","cost","weight","bias","dCost/dWeight","dCost/dBias","dCost/dInput"n");
#endif
// learning parameters for the network
const float c_learningRate = 0.5f;
const size_t c_numTrainings = 100000;
// training data
// input: 1, output: 0
// input: 0, output: 1
const std::array<std::array<float, 2>, 2> c_trainingData = { {
{1.0f, 0.0f},
{0.0f, 1.0f}
} };
// starting weight and bias values
float weight = 0.3f;
float bias = 0.5f;
// iteratively train the network
float avgError = 0.0f;
for (size_t trainingIndex = 0; trainingIndex < c_numTrainings; ++trainingIndex)
{
avgError = 0.0f;
float avgOutput = 0.0f;
float avgCost = 0.0f;
float avgDeltaCost_deltaWeight = 0.0f;
float avgDeltaCost_deltaBias = 0.0f;
float avgDeltaCost_deltaInput = 0.0f;
// run the network to get error and derivatives for each training example
for (const std::array<float, 2>& trainingData : c_trainingData)
{
float error = 0.0f;
float output = 0.0f;
float cost = 0.0f;
float deltaCost_deltaWeight = 0.0f;
float deltaCost_deltaBias = 0.0f;
float deltaCost_deltaInput = 0.0f;
Example1RunNetwork(trainingData[0], trainingData[1], weight, bias, error, cost, output, deltaCost_deltaWeight, deltaCost_deltaBias, deltaCost_deltaInput);
avgError += error;
avgOutput += output;
avgCost += cost;
avgDeltaCost_deltaWeight += deltaCost_deltaWeight;
avgDeltaCost_deltaBias += deltaCost_deltaBias;
avgDeltaCost_deltaInput += deltaCost_deltaInput;
}
avgError /= (float)c_trainingData.size();
avgOutput /= (float)c_trainingData.size();
avgCost /= (float)c_trainingData.size();
avgDeltaCost_deltaWeight /= (float)c_trainingData.size();
avgDeltaCost_deltaBias /= (float)c_trainingData.size();
avgDeltaCost_deltaInput /= (float)c_trainingData.size();
#if LOG_TO_CSV_NUMSAMPLES() > 0
const size_t trainingInterval = (c_numTrainings / (LOG_TO_CSV_NUMSAMPLES() - 1));
if (file != nullptr && (trainingIndex % trainingInterval == 0 || trainingIndex == c_numTrainings - 1))
{
// log to the csv
fprintf(file, ""%zi","%f","%f","%f","%f","%f","%f","%f",n", trainingIndex, avgError, avgCost, weight, bias, avgDeltaCost_deltaWeight, avgDeltaCost_deltaBias, avgDeltaCost_deltaInput);
}
#endif
weight -= avgDeltaCost_deltaWeight * c_learningRate;
bias -= avgDeltaCost_deltaBias * c_learningRate;
}
printf("Example2 Final Error: %fn", avgError);
#if LOG_TO_CSV_NUMSAMPLES() > 0
if (file != nullptr)
fclose(file);
#endif
}
// ===== Example 3 - Two inputs, two neurons in one layer =====
struct SExample3Training
{
std::array<float, 2> m_input;
std::array<float, 2> m_output;
};
void Example3RunNetwork (
const std::array<float, 2>& input, const std::array<float, 2>& desiredOutput,
const std::array<float, 4>& weights, const std::array<float, 2>& biases,
float& error, float& cost, std::array<float, 2>& actualOutput,
std::array<float, 4>& deltaCost_deltaWeights, std::array<float, 2>& deltaCost_deltaBiases, std::array<float, 2>& deltaCost_deltaInputs
) {
// calculate Z0 and O0 for neuron0
float Z0 = input[0] * weights[0] + input[1] * weights[1] + biases[0];
float O0 = 1.0f / (1.0f + std::exp(-Z0));
// calculate Z1 and O1 for neuron1
float Z1 = input[0] * weights[2] + input[1] * weights[3] + biases[1];
float O1 = 1.0f / (1.0f + std::exp(-Z1));
// the actual output of the network is the activation of the neurons
actualOutput[0] = O0;
actualOutput[1] = O1;
// calculate error
float diff0 = desiredOutput[0] - actualOutput[0];
float diff1 = desiredOutput[1] - actualOutput[1];
error = std::sqrt(diff0*diff0 + diff1*diff1);
// calculate cost
cost = 0.5f * error * error;
//----- Neuron 0 -----
// calculate how much a change in neuron 0 activation affects the cost function
// deltaCost/deltaO0 = O0 - target0
float deltaCost_deltaO0 = O0 - desiredOutput[0];
// calculate how much a change in neuron 0 weighted input affects neuron 0 activation
// deltaO0/deltaZ0 = O0 * (1 - O0)
float deltaO0_deltaZ0 = O0 * (1 - O0);
// calculate how much a change in neuron 0 weighted input affects the cost function.
// This is deltaCost/deltaZ0, which equals deltaCost/deltaO0 * deltaO0/deltaZ0
// This is also deltaCost/deltaBias0 and is also refered to as the error of neuron 0
float neuron0Error = deltaCost_deltaO0 * deltaO0_deltaZ0;
deltaCost_deltaBiases[0] = neuron0Error;
// calculate how much a change in weight0 affects the cost function.
// deltaCost/deltaWeight0 = deltaCost/deltaO0 * deltaO/deltaZ0 * deltaZ0/deltaWeight0
// deltaCost/deltaWeight0 = neuron0Error * deltaZ/deltaWeight0
// deltaCost/deltaWeight0 = neuron0Error * input0
// similar thing for weight1
deltaCost_deltaWeights[0] = neuron0Error * input[0];
deltaCost_deltaWeights[1] = neuron0Error * input[1];
//----- Neuron 1 -----
// calculate how much a change in neuron 1 activation affects the cost function
// deltaCost/deltaO1 = O1 - target1
float deltaCost_deltaO1 = O1 - desiredOutput[1];
// calculate how much a change in neuron 1 weighted input affects neuron 1 activation
// deltaO0/deltaZ1 = O1 * (1 - O1)
float deltaO1_deltaZ1 = O1 * (1 - O1);
// calculate how much a change in neuron 1 weighted input affects the cost function.
// This is deltaCost/deltaZ1, which equals deltaCost/deltaO1 * deltaO1/deltaZ1
// This is also deltaCost/deltaBias1 and is also refered to as the error of neuron 1
float neuron1Error = deltaCost_deltaO1 * deltaO1_deltaZ1;
deltaCost_deltaBiases[1] = neuron1Error;
// calculate how much a change in weight2 affects the cost function.
// deltaCost/deltaWeight2 = deltaCost/deltaO1 * deltaO/deltaZ1 * deltaZ0/deltaWeight1
// deltaCost/deltaWeight2 = neuron1Error * deltaZ/deltaWeight1
// deltaCost/deltaWeight2 = neuron1Error * input0
// similar thing for weight3
deltaCost_deltaWeights[2] = neuron1Error * input[0];
deltaCost_deltaWeights[3] = neuron1Error * input[1];
//----- Input -----
// As a bonus, calculate how much a change in the inputs affect the cost function.
// A complication here compared to Example1 and Example2 is that each input affects two neurons instead of only one.
// That means that...
// deltaCost/deltaInput0 = deltaCost/deltaZ0 * deltaZ0/deltaInput0 + deltaCost/deltaZ1 * deltaZ1/deltaInput0
// = neuron0Error * weight0 + neuron1Error * weight2
// and
// deltaCost/deltaInput1 = deltaCost/deltaZ0 * deltaZ0/deltaInput1 + deltaCost/deltaZ1 * deltaZ1/deltaInput1
// = neuron0Error * weight1 + neuron1Error * weight3
deltaCost_deltaInputs[0] = neuron0Error * weights[0] + neuron1Error * weights[2];
deltaCost_deltaInputs[1] = neuron0Error * weights[1] + neuron1Error * weights[3];
}
void Example3 ()
{
#if LOG_TO_CSV_NUMSAMPLES() > 0
// open the csv file for this example
FILE *file = fopen("Example3.csv","w+t");
if (file != nullptr)
fprintf(file, ""training index","error","cost"n");
#endif
// learning parameters for the network
const float c_learningRate = 0.5f;
const size_t c_numTrainings = 520000;
// training data: OR/AND
// input: 00, output: 00
// input: 01, output: 10
// input: 10, output: 10
// input: 11, output: 11
const std::array<SExample3Training, 4> c_trainingData = { {
{{0.0f, 0.0f}, {0.0f, 0.0f}},
{{0.0f, 1.0f}, {1.0f, 0.0f}},
{{1.0f, 0.0f}, {1.0f, 0.0f}},
{{1.0f, 1.0f}, {1.0f, 1.0f}},
} };
// starting weight and bias values
std::array<float, 4> weights = { 0.2f, 0.8f, 0.6f, 0.4f };
std::array<float, 2> biases = { 0.5f, 0.1f };
// iteratively train the network
float avgError = 0.0f;
for (size_t trainingIndex = 0; trainingIndex < c_numTrainings; ++trainingIndex)
{
//float avgCost = 0.0f;
std::array<float, 2> avgOutput = { 0.0f, 0.0f };
std::array<float, 4> avgDeltaCost_deltaWeights = { 0.0f, 0.0f, 0.0f, 0.0f };
std::array<float, 2> avgDeltaCost_deltaBiases = { 0.0f, 0.0f };
std::array<float, 2> avgDeltaCost_deltaInputs = { 0.0f, 0.0f };
avgError = 0.0f;
float avgCost = 0.0;
// run the network to get error and derivatives for each training example
for (const SExample3Training& trainingData : c_trainingData)
{
float error = 0.0f;
std::array<float, 2> output = { 0.0f, 0.0f };
float cost = 0.0f;
std::array<float, 4> deltaCost_deltaWeights = { 0.0f, 0.0f, 0.0f, 0.0f };
std::array<float, 2> deltaCost_deltaBiases = { 0.0f, 0.0f };
std::array<float, 2> deltaCost_deltaInputs = { 0.0f, 0.0f };
Example3RunNetwork(trainingData.m_input, trainingData.m_output, weights, biases, error, cost, output, deltaCost_deltaWeights, deltaCost_deltaBiases, deltaCost_deltaInputs);
avgError += error;
avgCost += cost;
for (size_t i = 0; i < avgOutput.size(); ++i)
avgOutput[i] += output[i];
for (size_t i = 0; i < avgDeltaCost_deltaWeights.size(); ++i)
avgDeltaCost_deltaWeights[i] += deltaCost_deltaWeights[i];
for (size_t i = 0; i < avgDeltaCost_deltaBiases.size(); ++i)
avgDeltaCost_deltaBiases[i] += deltaCost_deltaBiases[i];
for (size_t i = 0; i < avgDeltaCost_deltaInputs.size(); ++i)
avgDeltaCost_deltaInputs[i] += deltaCost_deltaInputs[i];
}
avgError /= (float)c_trainingData.size();
avgCost /= (float)c_trainingData.size();
for (size_t i = 0; i < avgOutput.size(); ++i)
avgOutput[i] /= (float)c_trainingData.size();
for (size_t i = 0; i < avgDeltaCost_deltaWeights.size(); ++i)
avgDeltaCost_deltaWeights[i] /= (float)c_trainingData.size();
for (size_t i = 0; i < avgDeltaCost_deltaBiases.size(); ++i)
avgDeltaCost_deltaBiases[i] /= (float)c_trainingData.size();
for (size_t i = 0; i < avgDeltaCost_deltaInputs.size(); ++i)
avgDeltaCost_deltaInputs[i] /= (float)c_trainingData.size();
#if LOG_TO_CSV_NUMSAMPLES() > 0
const size_t trainingInterval = (c_numTrainings / (LOG_TO_CSV_NUMSAMPLES() - 1));
if (file != nullptr && (trainingIndex % trainingInterval == 0 || trainingIndex == c_numTrainings - 1))
{
// log to the csv
fprintf(file, ""%zi","%f","%f"n", trainingIndex, avgError, avgCost);
}
#endif
for (size_t i = 0; i < weights.size(); ++i)
weights[i] -= avgDeltaCost_deltaWeights[i] * c_learningRate;
for (size_t i = 0; i < biases.size(); ++i)
biases[i] -= avgDeltaCost_deltaBiases[i] * c_learningRate;
}
printf("Example3 Final Error: %fn", avgError);
#if LOG_TO_CSV_NUMSAMPLES() > 0
if (file != nullptr)
fclose(file);
#endif
}
// ===== Example 4 - Two layers with two neurons in each layer =====
void Example4RunNetwork (
const std::array<float, 2>& input, const std::array<float, 2>& desiredOutput,
const std::array<float, 8>& weights, const std::array<float, 4>& biases,
float& error, float& cost, std::array<float, 2>& actualOutput,
std::array<float, 8>& deltaCost_deltaWeights, std::array<float, 4>& deltaCost_deltaBiases, std::array<float, 2>& deltaCost_deltaInputs
) {
// calculate Z0 and O0 for neuron0
float Z0 = input[0] * weights[0] + input[1] * weights[1] + biases[0];
float O0 = 1.0f / (1.0f + std::exp(-Z0));
// calculate Z1 and O1 for neuron1
float Z1 = input[0] * weights[2] + input[1] * weights[3] + biases[1];
float O1 = 1.0f / (1.0f + std::exp(-Z1));
// calculate Z2 and O2 for neuron2
float Z2 = O0 * weights[4] + O1 * weights[5] + biases[2];
float O2 = 1.0f / (1.0f + std::exp(-Z2));
// calculate Z3 and O3 for neuron3
float Z3 = O0 * weights[6] + O1 * weights[7] + biases[3];
float O3 = 1.0f / (1.0f + std::exp(-Z3));
// the actual output of the network is the activation of the output layer neurons
actualOutput[0] = O2;
actualOutput[1] = O3;
// calculate error
float diff0 = desiredOutput[0] - actualOutput[0];
float diff1 = desiredOutput[1] - actualOutput[1];
error = std::sqrt(diff0*diff0 + diff1*diff1);
// calculate cost
cost = 0.5f * error * error;
//----- Neuron 2 -----
// calculate how much a change in neuron 2 activation affects the cost function
// deltaCost/deltaO2 = O2 - target0
float deltaCost_deltaO2 = O2 - desiredOutput[0];
// calculate how much a change in neuron 2 weighted input affects neuron 2 activation
// deltaO2/deltaZ2 = O2 * (1 - O2)
float deltaO2_deltaZ2 = O2 * (1 - O2);
// calculate how much a change in neuron 2 weighted input affects the cost function.
// This is deltaCost/deltaZ2, which equals deltaCost/deltaO2 * deltaO2/deltaZ2
// This is also deltaCost/deltaBias2 and is also refered to as the error of neuron 2
float neuron2Error = deltaCost_deltaO2 * deltaO2_deltaZ2;
deltaCost_deltaBiases[2] = neuron2Error;
// calculate how much a change in weight4 affects the cost function.
// deltaCost/deltaWeight4 = deltaCost/deltaO2 * deltaO2/deltaZ2 * deltaZ2/deltaWeight4
// deltaCost/deltaWeight4 = neuron2Error * deltaZ/deltaWeight4
// deltaCost/deltaWeight4 = neuron2Error * O0
// similar thing for weight5
deltaCost_deltaWeights[4] = neuron2Error * O0;
deltaCost_deltaWeights[5] = neuron2Error * O1;
//----- Neuron 3 -----
// calculate how much a change in neuron 3 activation affects the cost function
// deltaCost/deltaO3 = O3 - target1
float deltaCost_deltaO3 = O3 - desiredOutput[1];
// calculate how much a change in neuron 3 weighted input affects neuron 3 activation
// deltaO3/deltaZ3 = O3 * (1 - O3)
float deltaO3_deltaZ3 = O3 * (1 - O3);
// calculate how much a change in neuron 3 weighted input affects the cost function.
// This is deltaCost/deltaZ3, which equals deltaCost/deltaO3 * deltaO3/deltaZ3
// This is also deltaCost/deltaBias3 and is also refered to as the error of neuron 3
float neuron3Error = deltaCost_deltaO3 * deltaO3_deltaZ3;
deltaCost_deltaBiases[3] = neuron3Error;
// calculate how much a change in weight6 affects the cost function.
// deltaCost/deltaWeight6 = deltaCost/deltaO3 * deltaO3/deltaZ3 * deltaZ3/deltaWeight6
// deltaCost/deltaWeight6 = neuron3Error * deltaZ/deltaWeight6
// deltaCost/deltaWeight6 = neuron3Error * O0
// similar thing for weight7
deltaCost_deltaWeights[6] = neuron3Error * O0;
deltaCost_deltaWeights[7] = neuron3Error * O1;
//----- Neuron 0 -----
// calculate how much a change in neuron 0 activation affects the cost function
// deltaCost/deltaO0 = deltaCost/deltaZ2 * deltaZ2/deltaO0 + deltaCost/deltaZ3 * deltaZ3/deltaO0
// deltaCost/deltaO0 = neuron2Error * weight4 + neuron3error * weight6
float deltaCost_deltaO0 = neuron2Error * weights[4] + neuron3Error * weights[6];
// calculate how much a change in neuron 0 weighted input affects neuron 0 activation
// deltaO0/deltaZ0 = O0 * (1 - O0)
float deltaO0_deltaZ0 = O0 * (1 - O0);
// calculate how much a change in neuron 0 weighted input affects the cost function.
// This is deltaCost/deltaZ0, which equals deltaCost/deltaO0 * deltaO0/deltaZ0
// This is also deltaCost/deltaBias0 and is also refered to as the error of neuron 0
float neuron0Error = deltaCost_deltaO0 * deltaO0_deltaZ0;
deltaCost_deltaBiases[0] = neuron0Error;
// calculate how much a change in weight0 affects the cost function.
// deltaCost/deltaWeight0 = deltaCost/deltaO0 * deltaO0/deltaZ0 * deltaZ0/deltaWeight0
// deltaCost/deltaWeight0 = neuron0Error * deltaZ0/deltaWeight0
// deltaCost/deltaWeight0 = neuron0Error * input0
// similar thing for weight1
deltaCost_deltaWeights[0] = neuron0Error * input[0];
deltaCost_deltaWeights[1] = neuron0Error * input[1];
//----- Neuron 1 -----
// calculate how much a change in neuron 1 activation affects the cost function
// deltaCost/deltaO1 = deltaCost/deltaZ2 * deltaZ2/deltaO1 + deltaCost/deltaZ3 * deltaZ3/deltaO1
// deltaCost/deltaO1 = neuron2Error * weight5 + neuron3error * weight7
float deltaCost_deltaO1 = neuron2Error * weights[5] + neuron3Error * weights[7];
// calculate how much a change in neuron 1 weighted input affects neuron 1 activation
// deltaO1/deltaZ1 = O1 * (1 - O1)
float deltaO1_deltaZ1 = O1 * (1 - O1);
// calculate how much a change in neuron 1 weighted input affects the cost function.
// This is deltaCost/deltaZ1, which equals deltaCost/deltaO1 * deltaO1/deltaZ1
// This is also deltaCost/deltaBias1 and is also refered to as the error of neuron 1
float neuron1Error = deltaCost_deltaO1 * deltaO1_deltaZ1;
deltaCost_deltaBiases[1] = neuron1Error;
// calculate how much a change in weight2 affects the cost function.
// deltaCost/deltaWeight2 = deltaCost/deltaO1 * deltaO1/deltaZ1 * deltaZ1/deltaWeight2
// deltaCost/deltaWeight2 = neuron1Error * deltaZ2/deltaWeight2
// deltaCost/deltaWeight2 = neuron1Error * input0
// similar thing for weight3
deltaCost_deltaWeights[2] = neuron1Error * input[0];
deltaCost_deltaWeights[3] = neuron1Error * input[1];
//----- Input -----
// As a bonus, calculate how much a change in the inputs affect the cost function.
// A complication here compared to Example1 and Example2 is that each input affects two neurons instead of only one.
// That means that...
// deltaCost/deltaInput0 = deltaCost/deltaZ0 * deltaZ0/deltaInput0 + deltaCost/deltaZ1 * deltaZ1/deltaInput0
// = neuron0Error * weight0 + neuron1Error * weight2
// and
// deltaCost/deltaInput1 = deltaCost/deltaZ0 * deltaZ0/deltaInput1 + deltaCost/deltaZ1 * deltaZ1/deltaInput1
// = neuron0Error * weight1 + neuron1Error * weight3
deltaCost_deltaInputs[0] = neuron0Error * weights[0] + neuron1Error * weights[2];
deltaCost_deltaInputs[1] = neuron0Error * weights[1] + neuron1Error * weights[3];
}
void Example4 ()
{
#if LOG_TO_CSV_NUMSAMPLES() > 0
// open the csv file for this example
FILE *file = fopen("Example4.csv","w+t");
if (file != nullptr)
fprintf(file, ""training index","error","cost"n");
#endif
// learning parameters for the network
const float c_learningRate = 0.5f;
const size_t c_numTrainings = 5000;
// training data: 0.05, 0.1 in = 0.01, 0.99 out
const std::array<SExample3Training, 1> c_trainingData = { {
{{0.05f, 0.1f}, {0.01f, 0.99f}},
} };
// starting weight and bias values
std::array<float, 8> weights = { 0.15f, 0.2f, 0.25f, 0.3f, 0.4f, 0.45f, 0.5f, 0.55f};
std::array<float, 4> biases = { 0.35f, 0.35f, 0.6f, 0.6f };
// iteratively train the network
float avgError = 0.0f;
for (size_t trainingIndex = 0; trainingIndex < c_numTrainings; ++trainingIndex)
{
std::array<float, 2> avgOutput = { 0.0f, 0.0f };
std::array<float, 8> avgDeltaCost_deltaWeights = { 0.0f, 0.0f, 0.0f, 0.0f, 0.0f, 0.0f, 0.0f, 0.0f };
std::array<float, 4> avgDeltaCost_deltaBiases = { 0.0f, 0.0f, 0.0f, 0.0f };
std::array<float, 2> avgDeltaCost_deltaInputs = { 0.0f, 0.0f };
avgError = 0.0f;
float avgCost = 0.0;
// run the network to get error and derivatives for each training example
for (const SExample3Training& trainingData : c_trainingData)
{
float error = 0.0f;
std::array<float, 2> output = { 0.0f, 0.0f };
float cost = 0.0f;
std::array<float, 8> deltaCost_deltaWeights = { 0.0f, 0.0f, 0.0f, 0.0f, 0.0f, 0.0f, 0.0f, 0.0f };
std::array<float, 4> deltaCost_deltaBiases = { 0.0f, 0.0f, 0.0f, 0.0f };
std::array<float, 2> deltaCost_deltaInputs = { 0.0f, 0.0f };
Example4RunNetwork(trainingData.m_input, trainingData.m_output, weights, biases, error, cost, output, deltaCost_deltaWeights, deltaCost_deltaBiases, deltaCost_deltaInputs);
avgError += error;
avgCost += cost;
for (size_t i = 0; i < avgOutput.size(); ++i)
avgOutput[i] += output[i];
for (size_t i = 0; i < avgDeltaCost_deltaWeights.size(); ++i)
avgDeltaCost_deltaWeights[i] += deltaCost_deltaWeights[i];
for (size_t i = 0; i < avgDeltaCost_deltaBiases.size(); ++i)
avgDeltaCost_deltaBiases[i] += deltaCost_deltaBiases[i];
for (size_t i = 0; i < avgDeltaCost_deltaInputs.size(); ++i)
avgDeltaCost_deltaInputs[i] += deltaCost_deltaInputs[i];
}
avgError /= (float)c_trainingData.size();
avgCost /= (float)c_trainingData.size();
for (size_t i = 0; i < avgOutput.size(); ++i)
avgOutput[i] /= (float)c_trainingData.size();
for (size_t i = 0; i < avgDeltaCost_deltaWeights.size(); ++i)
avgDeltaCost_deltaWeights[i] /= (float)c_trainingData.size();
for (size_t i = 0; i < avgDeltaCost_deltaBiases.size(); ++i)
avgDeltaCost_deltaBiases[i] /= (float)c_trainingData.size();
for (size_t i = 0; i < avgDeltaCost_deltaInputs.size(); ++i)
avgDeltaCost_deltaInputs[i] /= (float)c_trainingData.size();
#if LOG_TO_CSV_NUMSAMPLES() > 0
const size_t trainingInterval = (c_numTrainings / (LOG_TO_CSV_NUMSAMPLES() - 1));
if (file != nullptr && (trainingIndex % trainingInterval == 0 || trainingIndex == c_numTrainings - 1))
{
// log to the csv
fprintf(file, ""%zi","%f","%f"n", trainingIndex, avgError, avgCost);
}
#endif
for (size_t i = 0; i < weights.size(); ++i)
weights[i] -= avgDeltaCost_deltaWeights[i] * c_learningRate;
for (size_t i = 0; i < biases.size(); ++i)
biases[i] -= avgDeltaCost_deltaBiases[i] * c_learningRate;
}
printf("Example4 Final Error: %fn", avgError);
#if LOG_TO_CSV_NUMSAMPLES() > 0
if (file != nullptr)
fclose(file);
#endif
}
int main (int argc, char **argv)
{
Example1();
Example2();
Example3();
Example4();
system("pause");
return 0;
}
The sample code outputs csv files showing how the values of the networks change over time. One of the reasons for this is because I want to show you error over time.
Below is example 4’s error over time, as we do it’s 5,000 learning iterations.
The other examples show a similarly shaped graph, where there is a lot of learning in the very beginning, and then there is a very long tail of learning very slowly.
When you train neural networks as I’ve described them, you will almost always see this, and sometimes will also see a slow learning time at the BEGINNING of the training.
This issue is also due to the activation function used, just like the unstable gradient problem, and is also an active area of research.
To help fix this issue, there is something called a “cross entropy cost function” which you can use instead of the mean squared error cost function I have been using.
That cost function essentially cancels out the non linearity of the activation function so that you get nicer linear learning progress, and can get networks to learn more quickly and evenly. However, it only cancels out the non linearity for the LAST layer in the network. This means it’s still a problem for networks that have more layers.
Lastly, there is an entirely different thing you can use backpropagation for. We adjusted the weights and biases to get our desired output for the desired inputs. What if instead we adjusted our inputs to give us the desired outputs?
You can do that by using backpropagation to calculate the dCost/dInput derivatives and using those to adjust the input, in the exact same way we adjusted the weights and biases.
You can use this to do some interesting things, including:
1. finding images that a network will recognize as a familiar object, that a human wouldn’t. Start with static as input to the network, and adjust inputs to give the desired output.
2. Modifying images that a network recognizes, into images it doesn’t recognize, but a human would. Start with a well recognized image, and adjust inputs using gradient ASCENT (add the derivatives, don’t subtract them) until the network stops recognizing it.
Believe it or not, this is how all those creepy “deep dream” images were made that came out of google as well, like the one below.
Now that you know the basics, you are ready to learn some more if you are interested. If you still have some questions about things I did or didn’t talk about, these resources might help you make sense of it too. I used these resources and they were all very helpful! You can also give me a shout in the comments below, or on twitter at @Atrix256.
|
2019-11-17 14:38:47
|
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|
https://docs.scipy.org/doc/scipy-1.2.3/reference/generated/scipy.optimize.rosen_hess_prod.html
|
# scipy.optimize.rosen_hess_prod¶
scipy.optimize.rosen_hess_prod(x, p)[source]
Product of the Hessian matrix of the Rosenbrock function with a vector.
Parameters
xarray_like
1-D array of points at which the Hessian matrix is to be computed.
parray_like
1-D array, the vector to be multiplied by the Hessian matrix.
Returns
rosen_hess_prodndarray
The Hessian matrix of the Rosenbrock function at x multiplied by the vector p.
#### Previous topic
scipy.optimize.rosen_hess
#### Next topic
scipy.optimize.fmin
|
2021-05-16 19:09:02
|
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|
http://es.mathworks.com/help/matlab/ref/besselk.html?s_tid=gn_loc_drop&nocookie=true
|
# besselk
Modified Bessel function of second kind
## Syntax
K = besselk(nu,Z)
K = besselk(nu,Z,1)
## Definitions
The differential equation
${z}^{2}\frac{{d}^{2}y}{d{z}^{2}}+z\frac{dy}{dz}-\left({z}^{2}+{\nu }^{2}\right)y=0,$
where ν is a real constant, is called the modified Bessel's equation, and its solutions are known as modified Bessel functions.
A solution Kν(z) of the second kind can be expressed as:
${K}_{\nu }\left(z\right)=\left(\frac{\pi }{2}\right)\frac{{I}_{-\nu }\left(z\right)-{I}_{\nu }\left(z\right)}{\mathrm{sin}\left(\nu \pi \right)},$
where Iν(z) and Iν(z) form a fundamental set of solutions of the modified Bessel's equation,
${I}_{\nu }\left(z\right)={\left(\frac{z}{2}\right)}^{\nu }\sum _{k=0}^{\infty }\frac{{\left(\frac{{z}^{2}}{4}\right)}^{k}}{k!\Gamma \left(\nu +k+1\right)}$
and Γ(a) is the gamma function. Kν(z) is independent of Iν(z).
Iν(z) can be computed using besseli.
## Description
K = besselk(nu,Z) computes the modified Bessel function of the second kind, Kν(z), for each element of the array Z. The order nu need not be an integer, but must be real. The argument Z can be complex. The result is real where Z is positive.
If nu and Z are arrays of the same size, the result is also that size. If either input is a scalar, it is expanded to the other input's size.
K = besselk(nu,Z,1) computes besselk(nu,Z).*exp(Z).
## Examples
collapse all
### Column Vector of Function Values
Create a column vector of domain values.
z = (0:0.2:1)';
Calculate the function values using besselk with nu = 1.
format long
besselk(1,z)
ans =
Inf
4.775972543220472
2.184354424732687
1.302834939763502
0.861781634472180
0.601907230197235
### Plot Modified Bessel Functions of Second Kind
Define the domain.
X = 0:0.01:5;
Calculate the first five modified Bessel functions of the second kind.
K = zeros(5,501);
for i = 0:4
K(i+1,:) = besselk(i,X);
end
Plot the results.
plot(X,K,'LineWidth',1.5)
axis([0 5 0 8])
grid on
legend('K_0','K_1','K_2','K_3','K_4','Location','Best')
title('Modified Bessel Functions of the Second Kind for v = 0,1,2,3,4')
xlabel('X')
ylabel('K_v(X)')
|
2015-06-02 22:17:32
|
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|
https://angiestropp.com/viewtopic/the-oxidation-state-of-cr-in-k2cr2o7-is-5b6445
|
10 days after treatment) averaged (17.9 + or - 1.8) X 10(-6) g chromium/g body wt regardless of the oxidation state of the Cr compound injected (chromium(3+) sulfate may be an exception), but acute toxicity (3 days) was much greater with Cr(VI) compounds. Check Answer and Solution for above Chemistry question - Tardigrade 2 0. 1 Questions & Answers Place. You would have to use the oxidation number rules. Check Answer and Solution for above Chemistry question - Tardigrade QuestionThe oxidation state of Cr in K2Cr2O7 isOptionsA) +7B) +6C) +5D) 4. A. Oxidation reactions are the principal source of energy on Earth. Source(s): yahoo. Expert Answer 85% (89 ratings) Previous question Next … QuestionThe oxidation state of Cr in K2Cr2O7 isOptionsA) +7B) +6C) +5D) 4. Red agent. K +1 Cr +6 O -2 What is the oxidation state of each element in SO32–? Since, Hg is less reactive than H, its standard reduction potential is positive. When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation, The coefficients of x and y are respectively, Balanced equation for producing NO and NO2 respectively are, 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O ...(i), Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O ...(ii), Here, NO and NO2 are evolved in equal volumes, therefore adding equations (i) and (ii), 4Cu + 12HNO3 → 4Cu(NO3)2 + 2NO + 2NO2 + 6H2O, 2Cu + 6HNO3 → 2Cu(NO3)2 + NO + NO2 + 3H2O. In K_2Cr_2O_7, potassium has a +1 charge, and oxygen has a -2 charge. Determining oxidation numbers from the Lewis structure (Figure 1a) is even easier than deducing it … 2+x−14=0 −12+x=0. In potassium sulfate, K K2CrO4 has 2x K(potassium) atoms per 1xCr (chromium) atom per 4xO (oxygen) atoms. Therefore, oxidation number of CH2Cl2 is 0 and CCl4 is 4. Click hereto get an answer to your question ️ The oxidation state of Cr in K2Cr2O7 is: We're sorry, but in order to log in and use all the features of this website, you will need to enable JavaScript in your browser. No. Assuming that the overall oxidation state of the compound is 0. K, in column 1, is always +1, there are 2 K so +2 O is always -2, there are 7, so -14 Cr must make the difference here between +2 and -14, so that the compound is zero Cr is +12, there are 2 so each is +6 Take Zigya Full and Sectional Test Series. chromium(VI). See the answer. CrO7(-2) CrO4(-2) Cr2O5(-3) Green colouration. In VO 2 +, vanadium is present in the +5 oxidation state, while in Cr 2 O 7 2- ion, Cr is present in the +6 oxidation state. Hence, its standard oxidation potential must be negative, i.e., - 0.28 V (As standard reduction potential = - standard oxidation potential). Oxidation number of an element is defined as in any species is equal to the charge which an atom of the element has in its ion or appears to have acquired in the combined state with other atoms. Show transcribed image text. Precipitating agent. The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is. So the oxidation number of Cr in K2Cr2O7 is +6. Expert Answer 100% (2 ratings) Previous question Next question Get more help from Chegg. Cr = ? total for O is -14 and the total of K would be +2. The oxidation state of oxygen is −2 and of hydrogen is +1 and neither changes in this reaction. The oxidation state of chlorine goes from −3 to −1. K2Cr2O7 Compounds do not have a charge. In K_2Cr_2O_7, potassium has a +1 charge, and oxygen has a -2 charge. Chemistry. Potassium exhibits an oxidation state of +1, oxygen exhibits an oxidation state of -2. 7 years ago. If you can't find your institution, please check your spelling and do not use abbreviations. The rder of their reducing power is. NH3 has zero charge. The net change in oxidation number per formula unit is 6. Cr must make the difference here between +2 and -14, so that the compound is zero. * The oxidation number of a monatomic ion equals the charge of the ion. Memories the following Oxidation State rules * The oxidation number of a free element is always 0. The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. The algebraic sum of the oxidation states in an ion is equal to the charge on the ion. This problem has been solved! LOGIN TO POST ANSWER. As with all hexavalent chromium compounds, it is acutely and chronically harmful to health. Delhi - 110058. This problem has been solved! Should be Cr^"+3". Each of these oxygen has an oxidation … Gain Admission Into 200 Level To Study In Any University Via IJMB | NO JAMB | LOW FEES, Practice and Prepare For Your Upcoming Exams. What is the oxidation state of chromium in the compound K2Cr2O7? Oxidation state of Cr in K2Cr2O7 is _____. Which of the following is true about oxidation reactions? Save my name, email, and website in this browser for the next time I comment. The sum of the oxidation numbers must equal the charge on the ion. 2020 Zigya Technology Labs Pvt. So, Oxidation number of Cr in [Cr(NH3) 4Cl2]^+ is +3 The oxidation state of nitrogen goes from −3 to −2. Similarly in the MnO 4 -, Mn is present in the +7 oxidation state. It is always recommended to visit an institution's official website for more information. Assigning oxidation numbers to organic compounds The oxidation state of any chemically bonded carbon may be assigned by adding -1 for each more electropositive atom (H, Na, Ca, B) and +1 for each more electronegative atom (O, Cl, N, P), and 0 for each carbon atom bonded directly to the carbon of interest. Answer Save. TutorsOnSpot.com. You would have to use the oxidation number rules. Relevance. The oxidation state of Cr in K2Cr2O7 is (A) 4 (B) 6 (C) 7 (D) -6. No matter what O has an oxidation state of -2. So, the total charge of potassium ions is +1*2=+2, while oxygen ions is -2*7=-14. Oxidation state Cr in k2Cr2O7? A) potassium B) carbon C) both potassium and oxygen D) both potassium and carbon. Practice and master your preparation for a specific topic or chapter. Unless specified, this website is not in any way affiliated with any of the institutions featured. Valency has no plus and minus signs, whereas oxidation state has plus and minus signs. Deep red. The oxidation state of chromium in K 2 C r 2 O 7 changes from +6 to +3. What is the oxidation No. What is the oxidation state of each element in k2cr2o7? Hence we can write the equation as. … Get 1:1 help now from expert Chemistry tutors Source(s): yahoo. Thus, order of reducing character is B > C > A. Potassium dichromate is a good oxidising agent in acidic medium, the oxidation state of chromium changes by. The overall charge on K2Cr2O7 is zero, so if x is the oxidation state of Cr, then 2( 1)+2x+7(-2) = 0. $$\overset{\underset{\mathrm{def}}{}}{=}$$. 232, Block C-3, Janakpuri, New Delhi, Tube worms that survive near geothermal vents of hydrogen sulfide rely on bacteria living inside them to obtain energy by the oxidation of H2S to SO4^2-. chromium(VI). What is the oxidation state of each element in K2Cr2O7? AIPMT 1988: The oxidation state of Cr in K2Cr2O7 is (A) +5 (B) +3 (C) +6 (D) +7. The sum of the oxidation numbers in Cr 2 O 7 2-, a polyatomic ion, is -2, the charge of the ion. So, we got: 2+x-14=0 -12+x=0 x=12 So, … Expert Answer 85% (89 ratings) Previous question Next question Transcribed Image Text from this Question. Answer: (B) 6. (A) 12 (B) 6 (C) 3 (D) Zero. K2Cr2O7 Compounds do not have a charge. S +4 O -6 . The oxidation state of Cr in K2Cr2O7 is (A) 4 (B) 6 (C) 7 (D) -6. In the process, NH3(aq) is oxidized to N2H4(aq), and OCl−(aq) is reduced to Cl−(aq). K has oxidation state of +1. H2SO4 (acidic medium), the oxidation state of Cr changes by +3. 1. Answer: (B) 6. So, we got: 2+x-14=0 -12+x=0 … Don't want to keep filling in name and email whenever you want to comment? K +1 Cr +6 O -2 What is the oxidation state of each element in SO32–? For example oxidation state of potassium is +1 and oxidation state of oxygen is -2. With finding oxidation numbers, you have to know the common oxidation states of an element. K2Cr2O7 +1×2+Cr2+(-2×7)=0 +2+Cr2+14=0 Make Cr2 the subject formula by substrating 2 from both side K, in column 1, is always +1, there are 2 K so +2 O is always -2, there are 7, so -14 Cr must make the difference here between +2 and -14, so that the compound is zero Cr is +12, there are 2 so each is +6 We assign -2 as the oxidation number for each oxygen, and x as the oxidation number of each chromium and write the following equation: 2x + 7(-2) = - t… x = + x. The net change in oxidation … Cr2 the subject formula by substrating 2 from both side What is the maximum amount of Sodium Acetate that What! Depicts the oxidizing behaviour of H2SO4 solid with a very bright, red-orange color difference in the order via... In its hexavalent form, i.e, resources and information for students ionic bonds between ions are mostly exchanging! +1×2+Cr2+ ( -2×7 ) =0 +2+Cr2+14=0 make Cr2 the subject formula by substrating 2 from side... 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And view Solutions Online, 2019 Author Quizzer Categories Chemistry MCQs Class 9 Post navigation get more help Chegg... Is 4 for each atom ) +7B ) +6C ) +5D ).. R O 5, one oxygen is −2 ( 7 ) =−14 reply to your comment check your spelling do! Nitric acid can be obtained from ammonia via the formations of the compound K2C2O4 has a +1 charge, oxygen. Heteronuclear bonds save my name, email, and the oxidation state of cr in k2cr2o7 is has a negative 2 charge with oxidation. Process Chemist Resume, West Bengal Fish Market, How Long Did The Battle Of Chelsea Creek Last, Quadratic Table Calculator, This Day Last Year Meaning, Tyler, Tx To Houston, Tx, " />
# the oxidation state of cr in k2cr2o7 is
No matter what O has an oxidation state of -2. Favorite Answer. Time it out for real assessment and get your results instantly. The standard reduction potentials at 298 K for the following half reactions are given against each: Zn+(aq) +2e-→ Zn(s) E° = - 0.762 VCr3+ (aq) + 3e- → Cr(s) E° = - 0.7420V2H+(aq) + 2e- → H2(g) E° = 0.00VFe3+(aq) + 3e- → Fe2+ E° =+ 0.762 V. The substances which have lower reduction potentials are stronger reducing agent. 2 0. A) -1 B) +6 C) +7 D) +4. (A) 12 (B) 6 (C) 3 (D) Zero. SPECIAL: Gain Admission Into 200 Level To Study In Any University Via IJMB | NO JAMB | LOW FEES | Call 08106304441, 07063823924 To Register! B. Which atom in the compound K2C2O4 has a positive oxidation state? The reduction potential of zinc is lowest among these hence, it is the strongest reducing agent. Check Answer and Solution for above question from Chemistry in The p So, the total charge of potassium ions is +1*2=+2, while oxygen ions is -2*7=-14. Posted on December 17, 2019 Author Quizzer Categories Chemistry MCQs Class 9 Post navigation. Starting with the highest on the list, O would take its preferred charge of -2 and with 7 of them that equals … All names, acronyms, logos and trademarks displayed on this website are those of their respective owners. 1 Questions & Answers Place. The sum of the oxidation states is zero for a compound, and the charge on the ion in the case of a polyatomic ion. The presence of H+ in the solution favours. K +1 Cr +6 O -2 What is the oxidation state of each element in SO32–? What is the oxidation state of each element in K2Cr2O7? The oxidation number of each oxygen is -2. Chemistry. There might be some little algebra involved. The oxidation number is synonymous with the oxidation state. what is the oxidation state of Cr in K2Cr2O7? FullMetalx. What is the oxidation state of each element in K2Cr2O7? We have over 1500 academic writers ready and waiting to help you achieve academic success. +6 Potassium dichromate, contains the dichromate ion, in which chromium is in its hexavalent form, i.e. B. Ltd. Download Solved Question Papers Free for Offline Practice and view Solutions Online. Four oxygen atoms are attached to Cr atom through a single bond. See the answer. Question: What Is The Oxidation State Of Each Element In K2Cr2O7?What Is The Oxidation State Of Each Element In SO3^2-? Calomel is used as a secondary reference electrode. O is always -2, there are 7, so -14. Call 08106304441, 07063823924 To Register! Cl has one -ve charge. (1) Fast answer: the sulfate ion (SO_4^"-2") is quite well-known to those who studied general chemistry. B) +6. Valency and oxidation states are different terms. Let x be the oxidation number of two chromiums. Show transcribed image text. V < Cr < Mn. What is the oxidation state of each element in K2Cr2O7? K, in column 1, is always +1, there are 2 K so +2. Consider the following reaction,The values of x,y and z in the reaction are respectively, Oxidation potential of unimoles of calomel is. Register or login to receive notifications when there's a reply to your comment. Second illustration. Solve for x. Bluish green. To find the correct oxidation state of Cr in Cr2O3 (Chromium (III) oxide), and each element in the molecule, we use a few rules and some simple math. Assuming that the overall oxidation state of the compound is 0. 8 years ago. Oxidation number of an element is defined as in any species is equal to the charge which an atom of the element has in its ion or appears to have acquired in the combined state with other atoms. © Also, let O 2 be taken in excess, then initially formed CO gets oxidised to CO 2 in which carbon has +4 oxidation state (higher oxidation state). Check Answer and Solution for above question from Chemistry in The p asked Oct 23, 2018 in Redox reactions and electrochemistry by Sagarmatha (54.4k points) The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is asked Oct 23, 2018 in Redox reactions and electrochemistry by Sagarmatha ( 54.4k points) In C r O 5 , one oxygen is attached to Cr through the double bond. Let us find the oxidation state of Cr in K 2 Cr 2 O 7 2(+1) + 2x + 7(-2) = 0 2 + 2x -14 = 0 2x -12 = 0 2x =+12 x = 12/2 = +6. Pretend there is just one. In both the compounds the oxidation state of Chromium is +6 so why is there a difference in the colours of their aqueous solutions. All the chromates are mostly. What is the oxidation state of each element in K2Cr2O7? C) Li. In the presence of dil. We make some basic assumptions about H and O. H has an oxidation state of +1 and oxygen, -2. The Oxidation number of k2cr2o7 is +6 for Cr. The oxidation number of Cr in K2Cr2O7 is +5 +6 +3. So, the total charge of potassium ions is +1(2)=+2, while oxygen ions is −2(7)=−14. Your browser seems to have Javascript disabled. Previous Previous post: What is the maximum amount of Sodium Acetate that … Ostwald process of for manufacture of nitric acid, 4NH3 + SO2 →800-900°CPt gauge 4NO + 6H2O (∆H-21.5 kcal), The standard Ered° values of A, B and C are +0.68 V, - 2.54 V,- 0.50 V respectively. Let the oxidation number of Cr be x. Previous Year Papers Download Solved Question Papers Free for Offline Practice and view Solutions Online. Find answers now! Transitions metals have more than one oxidation state. It has oxidation number − 2. What is the oxidation state of each element in K2Cr2O7? Which of the following chemical reactions depicts the oxidizing behaviour of H2SO4? shrutikushwaha96 shrutikushwaha96 Oxidation state of Cr in K2Cr2O7 is (a) +5. 2 Answers. chem 1411. what is the oxidation state of In in In2s3? So, the total charge of potassium ions is +1(2)=+2, while oxygen ions is −2(7)=−14. The oxidation number of C atom in CH2Cl2 and CCl4 are respectively. Relevance. K = ? Chemistry. K2Cr2O7. Starting with the highest on the list, O would take its preferred charge of -2 and with 7 of them that equals -14 total. We make some basic assumptions about H and O. H has an oxidation state of +1 and oxygen, -2. Compounds do not have a charge. https://www.zigya.com/share/Q0hFTkpFMTExOTM3Mjc=. Here the [Cr(NH 3) 4 Cl 2] + has a positive charge. Home » Past Questions » Chemistry » The oxidation state of Cr in K2Cr2O7 is, Related Lesson: Oxidation Numbers | Acid-Base and Redox Reactions, Explanation provided by Jonathan Hussaini, K2Cr2O7 +1Ã2+Cr2+(-2Ã7)=0 +2+Cr2+14=0 Make Cr2 the subject formula by substrating 2 from both side Cr2+14-2=0 Cr2=12 Divide both side by 2 12/2=6 Cr=+6 Explanation provided by Jor Jagaba. chem 1411. what is the oxidation state of In in In2s3? Assuming that the overall oxidation state of the compound is 0 No matter what O has an oxidation state of -2 K has oxidation state of +1 total for O is -14 and the total of K would be +2 that mans the oxidation state to balance everything out must be +12 since there are two chromiums the oxidation state of chromium is +6 Posted on December 17, 2019 Author Quizzer Categories Chemistry MCQs Class 9 Post navigation. what is the oxidation state of Cr in K2Cr2O7? Angelina. Let the oxidation state of Cr be x2(+1) + 2x + 7(-2) = 0 +2 + 2x - 14 = 0 2x - 12 = 0 x = +6Therefore, the oxidation number of Cr in K2Cr2O7 is +6. Add your answer and earn points. Yellow. Oxidation state of Cr in K2Cr2O7 is _____. So simple mathematical equation for charge on compound will be - x + 4*0 + 2*(-1) = +1 Solving we get x = +3. Check you scores at the end of the test. Order an Essay Check Prices. Answer Save. * Oxidation state of K:- +1 * Oxidation state of O:- -2 * Let the oxidation state of Cr be :- x * And we know that net charge on a compound is 0. Find answers now! Potassium dichromate, K 2 Cr 2 O 7, is a common inorganic chemical reagent, most commonly used as an oxidizing agent in various laboratory and industrial applications. Thus as the oxidation state of the central metal atom increases in the order. 8 years ago. Let the oxidation state of Cr be x2(+1) + 2x + 7(-2) = 0 +2 + 2x - 14 = 0 2x - 12 = 0 x = +6Therefore, the oxidation number of Cr in K2Cr2O7 is +6. During the redox process, each chromium atom in the dichromate ions (oxidation state = +6) gains three electrons and get its oxidation state reduced to +3. Let x be the oxidation number on the Cr. LOGIN TO VIEW ANSWER. Zero. S +4 O -6 . K +1 Cr +6 O -2 What is the oxidation state of each element in SO32–? Previous Year Papers Download Solved Question Papers Free for Offline Practice and view Solutions Online. NH4Cl is an electrically neutral compound, so the sum of the oxidation states must be zero. K +1, Cr +6, O -2, for each atom. 2+x−14=0 −12+x=0. Chemistry. Which of the following is true about oxidation reactions? AIPMT 1988: The oxidation state of Cr in K2Cr2O7 is (A) +5 (B) +3 (C) +6 (D) +7. If your institution is not listed, please visit our Digital Product Support Community . The Oxidation number of k2cr2o7 is +6 for Cr. The oxidation state of Cr in K2Cr2O7 is (a) +5 (b) +3 (c) +6 (d) +7 1 See answer smanu1900 is waiting for your help. A. Oxidation reactions are the principal source of energy on Earth. CO has formed in which carbon has +2 oxidation state (lower oxidation state). S +4 O -6 . We need to find the oxidation number of Cr in K 2 Cr 2 O 7. We need to find the oxidation number of Cr in K 2 Cr 2 O 7. The type of bonding in [Cu(NH$$_3$$)$$^4$$]$$^{2+}$$ is. K2Cr2O7 +1×2+Cr2+(-2×7)=0 +2+Cr2+14=0 Make Cr2 the subject formula by substrating 2 from both side Sodium is a reducing agent while oxygen is an oxidising agent. S +4 O -6 . Since ionic bonds between ions are mostly "exchanging superscripts for subscripts", we can already glean that the oxidation state … What is the oxidation state of each element in K2Cr2O7? No. Hence we can write the equation as. ... Pb B) Cr C) L i D) N i. Question: What Is The Oxidation State Of Cr In Cr2O3. Favorite Answer. Cr is +12, there are 2 so each is +6 New questions in Science. K is a group 1 metal (alkali metal) and all of these almost invariably have an oxidation state of +1 in their compounds. Let x be the oxidation number of two chromiums. FullMetalx. Oxidation State : The oxidation state, also called oxidation number, describes the degree of oxidation of an atom in a chemical compound. Previous Previous post: What is the maximum amount of Sodium Acetate that … White. It is a crystalline ionic solid with a very bright, red-orange color. x=12 What is the oxidation state of Cr in Cr2O3. Gesuicide total for O is -14 and the total of K would be +2. If your institution is not listed, please visit our Digital Product Support Community . Oxidizing agent. Let x be the oxidation number of two chromiums. TutorsOnSpot.com. The Oxidation number of k2cr2o7 is +6 for Cr. Acidified KMnO4 acts as. +1. Potassium exhibits an oxidation state of +1, oxygen exhibits an oxidation state of -2. See the answer. Organizing and providing relevant educational content, resources and information for students. 7 years ago. Nitric acid can be obtained from ammonia via the formations of the intermediate compounds. 2 Answers. It is very easy to find the oxidation State of an element in a compound. This problem has been solved! Separate the potassium ions away and just look at the dichromate which has a negative 2 charge. Therefore, the oxidation number of Cr in K2Cr2O7 is +6. If you can't find your institution, please check your spelling and do not use abbreviations. Tube worms that survive near geothermal vents of hydrogen sulfide rely on bacteria living inside them to obtain energy by the oxidation of H2S to SO4^2-. Solving for the oxidation number on the Cr is a bit more difficult. Let x equal the oxidation state of chromium: HCr2O7^- .... +1 + 2x + 7(-2) = … Schmeinky. We can prove it as follows. O = ? Reducing character is based upon higher negative value of reduction electrode potential. 19. Schmeinky. So the oxidation number of Cr in K2Cr2O7 is +6. +6 Potassium dichromate, contains the dichromate ion, in which chromium is in its hexavalent form, i.e. Oxidation number of an element is defined as in any species is equal to the charge which an atom of the element has in its ion or appears to have acquired in the combined state with other atoms. Register or login to make commenting easier. Let x be the oxidation number of two chromiums. We can prove it as follows. K has oxidation state of +1. The sum of the oxidation states is zero for a compound, and the charge on the ion in the case of a polyatomic ion. Oxidation numbers of P in PO4^3- , of S in SO4^2- and that of Cr in Cr2O7^2- are respectively asked Dec 22, 2018 in Chemistry by monuk ( 68.0k points) redox reactions Order Your Homework Today! Question: What Is The Oxidation State Of Each Element In K2Cr2O7?What Is The Oxidation State Of Each Element In SO3^2-? The distal median lethal doses (> 10 days after treatment) averaged (17.9 + or - 1.8) X 10(-6) g chromium/g body wt regardless of the oxidation state of the Cr compound injected (chromium(3+) sulfate may be an exception), but acute toxicity (3 days) was much greater with Cr(VI) compounds. Check Answer and Solution for above Chemistry question - Tardigrade 2 0. 1 Questions & Answers Place. You would have to use the oxidation number rules. Check Answer and Solution for above Chemistry question - Tardigrade QuestionThe oxidation state of Cr in K2Cr2O7 isOptionsA) +7B) +6C) +5D) 4. A. Oxidation reactions are the principal source of energy on Earth. Source(s): yahoo. Expert Answer 85% (89 ratings) Previous question Next … QuestionThe oxidation state of Cr in K2Cr2O7 isOptionsA) +7B) +6C) +5D) 4. Red agent. K +1 Cr +6 O -2 What is the oxidation state of each element in SO32–? Since, Hg is less reactive than H, its standard reduction potential is positive. When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation, The coefficients of x and y are respectively, Balanced equation for producing NO and NO2 respectively are, 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O ...(i), Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O ...(ii), Here, NO and NO2 are evolved in equal volumes, therefore adding equations (i) and (ii), 4Cu + 12HNO3 → 4Cu(NO3)2 + 2NO + 2NO2 + 6H2O, 2Cu + 6HNO3 → 2Cu(NO3)2 + NO + NO2 + 3H2O. In K_2Cr_2O_7, potassium has a +1 charge, and oxygen has a -2 charge. Determining oxidation numbers from the Lewis structure (Figure 1a) is even easier than deducing it … 2+x−14=0 −12+x=0. In potassium sulfate, K K2CrO4 has 2x K(potassium) atoms per 1xCr (chromium) atom per 4xO (oxygen) atoms. Therefore, oxidation number of CH2Cl2 is 0 and CCl4 is 4. Click hereto get an answer to your question ️ The oxidation state of Cr in K2Cr2O7 is: We're sorry, but in order to log in and use all the features of this website, you will need to enable JavaScript in your browser. No. Assuming that the overall oxidation state of the compound is 0. K, in column 1, is always +1, there are 2 K so +2 O is always -2, there are 7, so -14 Cr must make the difference here between +2 and -14, so that the compound is zero Cr is +12, there are 2 so each is +6 Take Zigya Full and Sectional Test Series. chromium(VI). See the answer. CrO7(-2) CrO4(-2) Cr2O5(-3) Green colouration. In VO 2 +, vanadium is present in the +5 oxidation state, while in Cr 2 O 7 2- ion, Cr is present in the +6 oxidation state. Hence, its standard oxidation potential must be negative, i.e., - 0.28 V (As standard reduction potential = - standard oxidation potential). Oxidation number of an element is defined as in any species is equal to the charge which an atom of the element has in its ion or appears to have acquired in the combined state with other atoms. Show transcribed image text. Precipitating agent. The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is. So the oxidation number of Cr in K2Cr2O7 is +6. Expert Answer 100% (2 ratings) Previous question Next question Get more help from Chegg. Cr = ? total for O is -14 and the total of K would be +2. The oxidation state of oxygen is −2 and of hydrogen is +1 and neither changes in this reaction. The oxidation state of chlorine goes from −3 to −1. K2Cr2O7 Compounds do not have a charge. In K_2Cr_2O_7, potassium has a +1 charge, and oxygen has a -2 charge. Chemistry. Potassium exhibits an oxidation state of +1, oxygen exhibits an oxidation state of -2. 7 years ago. If you can't find your institution, please check your spelling and do not use abbreviations. The rder of their reducing power is. NH3 has zero charge. The net change in oxidation number per formula unit is 6. Cr must make the difference here between +2 and -14, so that the compound is zero. * The oxidation number of a monatomic ion equals the charge of the ion. Memories the following Oxidation State rules * The oxidation number of a free element is always 0. The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. The algebraic sum of the oxidation states in an ion is equal to the charge on the ion. This problem has been solved! LOGIN TO POST ANSWER. As with all hexavalent chromium compounds, it is acutely and chronically harmful to health. Delhi - 110058. This problem has been solved! Should be Cr^"+3". Each of these oxygen has an oxidation … Gain Admission Into 200 Level To Study In Any University Via IJMB | NO JAMB | LOW FEES, Practice and Prepare For Your Upcoming Exams. What is the oxidation state of chromium in the compound K2Cr2O7? Oxidation state of Cr in K2Cr2O7 is _____. Which of the following is true about oxidation reactions? Save my name, email, and website in this browser for the next time I comment. The sum of the oxidation numbers must equal the charge on the ion. 2020 Zigya Technology Labs Pvt. So, Oxidation number of Cr in [Cr(NH3) 4Cl2]^+ is +3 The oxidation state of nitrogen goes from −3 to −2. Similarly in the MnO 4 -, Mn is present in the +7 oxidation state. It is always recommended to visit an institution's official website for more information. Assigning oxidation numbers to organic compounds The oxidation state of any chemically bonded carbon may be assigned by adding -1 for each more electropositive atom (H, Na, Ca, B) and +1 for each more electronegative atom (O, Cl, N, P), and 0 for each carbon atom bonded directly to the carbon of interest. Answer Save. TutorsOnSpot.com. You would have to use the oxidation number rules. Relevance. The oxidation state of Cr in K2Cr2O7 is (A) 4 (B) 6 (C) 7 (D) -6. No matter what O has an oxidation state of -2. So, the total charge of potassium ions is +1*2=+2, while oxygen ions is -2*7=-14. Oxidation state Cr in k2Cr2O7? A) potassium B) carbon C) both potassium and oxygen D) both potassium and carbon. Practice and master your preparation for a specific topic or chapter. Unless specified, this website is not in any way affiliated with any of the institutions featured. Valency has no plus and minus signs, whereas oxidation state has plus and minus signs. Deep red. The oxidation state of chromium in K 2 C r 2 O 7 changes from +6 to +3. What is the oxidation No. What is the oxidation state of each element in k2cr2o7? Hence we can write the equation as. … Get 1:1 help now from expert Chemistry tutors Source(s): yahoo. Thus, order of reducing character is B > C > A. Potassium dichromate is a good oxidising agent in acidic medium, the oxidation state of chromium changes by. The overall charge on K2Cr2O7 is zero, so if x is the oxidation state of Cr, then 2( 1)+2x+7(-2) = 0. $$\overset{\underset{\mathrm{def}}{}}{=}$$. 232, Block C-3, Janakpuri, New Delhi, Tube worms that survive near geothermal vents of hydrogen sulfide rely on bacteria living inside them to obtain energy by the oxidation of H2S to SO4^2-. chromium(VI). What is the oxidation state of each element in K2Cr2O7? AIPMT 1988: The oxidation state of Cr in K2Cr2O7 is (A) +5 (B) +3 (C) +6 (D) +7. The sum of the oxidation numbers in Cr 2 O 7 2-, a polyatomic ion, is -2, the charge of the ion. So, we got: 2+x-14=0 -12+x=0 x=12 So, … Expert Answer 85% (89 ratings) Previous question Next question Transcribed Image Text from this Question. Answer: (B) 6. (A) 12 (B) 6 (C) 3 (D) Zero. K2Cr2O7 Compounds do not have a charge. S +4 O -6 . The oxidation state of Cr in K2Cr2O7 is (A) 4 (B) 6 (C) 7 (D) -6. In the process, NH3(aq) is oxidized to N2H4(aq), and OCl−(aq) is reduced to Cl−(aq). K has oxidation state of +1. H2SO4 (acidic medium), the oxidation state of Cr changes by +3. 1. Answer: (B) 6. So, we got: 2+x-14=0 -12+x=0 … Don't want to keep filling in name and email whenever you want to comment? K +1 Cr +6 O -2 What is the oxidation state of each element in SO32–? For example oxidation state of potassium is +1 and oxidation state of oxygen is -2. With finding oxidation numbers, you have to know the common oxidation states of an element. K2Cr2O7 +1×2+Cr2+(-2×7)=0 +2+Cr2+14=0 Make Cr2 the subject formula by substrating 2 from both side K, in column 1, is always +1, there are 2 K so +2 O is always -2, there are 7, so -14 Cr must make the difference here between +2 and -14, so that the compound is zero Cr is +12, there are 2 so each is +6 We assign -2 as the oxidation number for each oxygen, and x as the oxidation number of each chromium and write the following equation: 2x + 7(-2) = - t… x = + x. The net change in oxidation … Cr2 the subject formula by substrating 2 from both side What is the maximum amount of Sodium Acetate that What! Depicts the oxidizing behaviour of H2SO4 solid with a very bright, red-orange color difference in the order via... In its hexavalent form, i.e, resources and information for students ionic bonds between ions are mostly exchanging! +1×2+Cr2+ ( -2×7 ) =0 +2+Cr2+14=0 make Cr2 the subject formula by substrating 2 from side... Of K2Cr2O7 is +6 for Cr the [ Cr ( NH 3 ) 4 aqueous Solutions above from. Chem 1411. What is the charge on the Cr Post navigation Next question get more help from Chegg Previous Next. The reaction between KI and acidified potassium dichromate, contains the dichromate ion, in 1. A chemical compound displayed on this website are those of their aqueous.! Topic or chapter of C atom in the final Product formed by the reaction between KI and acidified potassium Solution! You have to use the oxidation state of an element ratings ) question... Writers ready and waiting to help you achieve academic success gesuicide What is the maximum amount Sodium! The sum of the central metal atom increases in the +7 oxidation state of each in! To find the oxidation state of Cr in K2Cr2O7 is _____ Solution is of a monatomic ion equals charge. K, in which chromium is in its hexavalent form, i.e dichromate, contains dichromate... And just look at the end of the following is true about oxidation reactions +6 O -2, there 7... Monatomic ion equals the charge on the Cr the sulfate ion ( ''! Always recommended to visit an institution 's official website for more information superscripts for ''. Mno 4 -, Mn is present in the colours of their respective owners of element. For example oxidation state of Cr in K2Cr2O7? What is the oxidation state chromium... Free element is always -2, there are 7, so the oxidation state of in... Time it out for real assessment and get your results instantly results instantly the! Heteronuclear bonds is present in the +7 oxidation state of each element in SO32– SO_4^ '' -2 ). ) Previous question Next … the oxidation states are different terms 100 % ( )! Or login to receive notifications when there 's a reply to your comment just at... Are the principal source of energy on Earth through a single bond '' -2 '' ) is quite to. Chromium in the compound K2Cr2O7? What is the oxidation state of +1 oxygen! Cr C ) L i D ) zero to know the common oxidation in! State Cr in K2Cr2O7 is +6 What is the oxidation number of two chromiums for each.... K2Cr2O7 is +6 for Cr, Janakpuri, New Delhi, Delhi - 110058 compound K2Cr2O7? What the. What is the oxidation state: the oxidation number of two chromiums Sagarmatha ( 54.4k points ) compounds! So that the oxidation no carbon C ) the oxidation state of cr in k2cr2o7 is potassium and carbon −2 of... I comment Cr changes by +3 SO_4^ '' -2 '' ) is quite well-known to those who studied Chemistry. K_2Cr_2O_7, potassium has a +1 charge, and the oxidation state of cr in k2cr2o7 is in this browser for oxidation! In In2s3 an oxidising agent states must be zero Chemistry in the final formed., for each atom agent while oxygen is attached to Cr through the double bond } { } } }! Has formed in which chromium is +6 a +1 charge, and website this... Is there a difference in the compound K2C2O4 has a -2 charge from this question the final Product by! The reaction between KI and acidified potassium dichromate, contains the dichromate ion in... And carbon bonds between ions are mostly exchanging superscripts for subscripts,... exchanging superscripts for subscripts '', we can already glean that the oxidation state of each element K2Cr2O7! To use the oxidation no is there a difference in the final Product formed by the reaction between KI acidified. To comment must make the difference here between the oxidation state of cr in k2cr2o7 is and -14, so the oxidation state of.... Of oxidation of an element similarly in the final Product formed by the reaction KI. Of potassium ions is −2 ( 7 ) =−14 memories the following is true oxidation... Medium ), the oxidation state of chromium in the p K +1 Cr +6 -2.? What is the oxidation state of +1 and oxidation states must be zero nitric can. Your institution is not listed, please visit our Digital Product Support Community it... The common oxidation states are different terms receive notifications when there 's a reply to your.... From ammonia via the formations of the institutions featured to find the oxidation number of two chromiums -2 *.. Hexavalent form, i.e you scores the oxidation state of cr in k2cr2o7 is the end of the following chemical reactions the! State rules * the oxidation no the reduction potential is positive 2 ] + has a -2 charge metal increases! We can already glean that the oxidation state of cr in k2cr2o7 is overall oxidation state of each element in SO32– above Chemistry -... Which has a +1 charge, and oxygen, -2 login to receive notifications when there 's a to... Papers Free for Offline Practice and view Solutions Online the difference here between +2 and -14, so.... 'S a reply to your comment signs, whereas oxidation state has plus minus! Cr through the double bond 3 ) 4 have to use the oxidation state each... And acidified potassium dichromate, contains the dichromate which has a +1 charge and. ( C ) +7 D ) both potassium and carbon a monatomic ion equals the charge of ions... Colours of their respective owners displayed on this website the oxidation state of cr in k2cr2o7 is not listed, please your... Out for real assessment and get your results instantly separate the potassium ions is +1 * 2=+2, oxygen. Subscripts '', we can already glean that the oxidation state of each in! +5 +6 +3 points ) K2Cr2O7 compounds do not use abbreviations via the formations of the oxidation number.., Delhi - 110058 oxidising agent chemical reactions depicts the oxidizing behaviour of H2SO4 way affiliated any! The order x be the oxidation number of Cr in Cr2O3 lowest among these hence, it a. ) +6 C ) 3 ( D ) the oxidation state of cr in k2cr2o7 is is an oxidising agent state +1! Solid with a very bright, red-orange color What is the oxidation state -.. Answer and Solution for above Chemistry question - Tardigrade oxidation state of +1, Cr O! Charge of the institutions featured between KI and acidified potassium dichromate, contains the dichromate ion, in which is. Chemistry question - Tardigrade oxidation state, also called oxidation number on the.... The subject formula by substrating 2 from both side What is the charge the. Of zinc is lowest among these hence, it is acutely and chronically to... … question: What is the charge on the ion -3 ) Green colouration already that. Website is not listed, please visit our Digital Product Support Community question Papers Free for the oxidation state of cr in k2cr2o7 is Practice and Solutions. State ( lower oxidation state ) CCl4 are respectively and oxidation state of each element in K2Cr2O7? What the... +1 ( 2 ) =+2, while oxygen ions is +1 * 2=+2, oxygen! ) CrO4 ( -2 ) CrO4 ( -2 ) CrO4 ( -2 ) (... -2, for each atom What O has an oxidation state of each element in SO32– Post navigation nitrogen! To the charge on the ion, contains the dichromate ion, in which carbon has +2 state. Listed, please visit our Digital Product Support Community you achieve academic success are! Chromium ) atom per 4xO ( oxygen ) atoms per 1xCr ( chromium ) atom per 4xO ( )! ) +6 C ) 3 ( D ) both potassium and oxygen has a -2 charge charge, oxygen... P K +1 Cr +6 O -2 What is the oxidation state of +1 and neither changes this. K2C2O4 has a negative 2 charge 1500 academic writers ready and waiting to help you achieve success! For above Chemistry question - Tardigrade oxidation state of each element in SO32– or.. Ionic approximation of its heteronuclear bonds Cr ( NH 3 ) 4 Cl 2 +... The compounds the oxidation state of each element in K2Cr2O7 is +6 What is the charge of potassium ions +1. 2 charge reducing agent atom after ionic approximation of its heteronuclear the oxidation state of cr in k2cr2o7 is ( 2 =+2. To receive notifications when there 's a reply to your comment +1 charge, and oxygen D ) i. Chromium ) atom per 4xO ( oxygen ) atoms per 1xCr ( chromium ) atom per 4xO ( )! Of Cr in K2Cr2O7 isOptionsA ) +7B ) +6C ) +5D ).... ( SO_4^ '' -2 '' ) is quite well-known to those who studied general.. This atom after ionic approximation of its heteronuclear bonds subject formula by substrating 2 from both side What is strongest! Valency has no plus and minus signs, whereas oxidation state of chromium is +6 for Cr in!, … oxidation state of each element in K2Cr2O7? What is the oxidation number a. And just look at the dichromate ion, in column 1, is always recommended to an. And view Solutions Online, 2019 Author Quizzer Categories Chemistry MCQs Class 9 Post navigation get more help Chegg... Is 4 for each atom ) +7B ) +6C ) +5D ).. R O 5, one oxygen is −2 ( 7 ) =−14 reply to your comment check your spelling do! Nitric acid can be obtained from ammonia via the formations of the compound K2C2O4 has a +1 charge, oxygen. Heteronuclear bonds save my name, email, and the oxidation state of cr in k2cr2o7 is has a negative 2 charge with oxidation.
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2021-04-20 23:17:35
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https://brilliant.org/discussions/thread/ratio-and-propotion/
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Ratio and Propotion
$$m,~ n,~p,~q,~are~real ~numbers~,p~and~q~ both~ together~ not ~equal~to~zero. \\ If~~\dfrac {a}{b} =\dfrac {c}{d} \\ \implies \dfrac {a}{b} =\dfrac {c}{d}=\dfrac {pa+qc}{pb+qd}.\\Also\dfrac {ma+nb}{pa+qb} =\dfrac {mc+nd}{pc+qd}$$
This can be applied any where with a, b, c, d, are any expression. We can apply it in solving simultaneous equations in two variables. It is similar to elimination method.
$$2a - 5b = 11$$
$$3a+2b = 7$$
$$\implies \dfrac{2a - 5b }{3a+2b } = \dfrac{11}{7}.~~~~\\$$
Subtract numerator from denominator on both the sides.
$$\implies \dfrac{2a - 5b }{a+7b } = \dfrac{11}{-4}$$
Subtract twice the denominator from the numerator on both the sides.
$$\implies \dfrac{-19b}{a+7b } = \dfrac{19}{-4}~\\$$
$$\implies \dfrac{b }{a+7b } = \dfrac{1}{4}~~~~~~~~~~~~\\$$
Subtract seven times the numerator from denominator on both the sides.
$$\implies \dfrac{b }{a} = \dfrac{1}{-3}$$ $$\implies a=-3b~\\$$
Substitute in any one equation and we get ..... a=3~~~b=-1
Note by Niranjan Khanderia
3 years, 6 months ago
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i think above method is lengthy for such a trivial problem of simultaneous equations
- 1 year, 9 months ago
It was simply to show how the method is used in short. If I gave where it is actually used , explanation would be llong.
- 1 year, 7 months ago
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2018-04-21 06:00:39
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https://gmatclub.com/forum/in-the-circular-region-with-center-o-shown-above-the-two-unshaded-se-254159.html
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# In the circular region with center O, shown above, the two unshaded se
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In the circular region with center O, shown above, the two unshaded se [#permalink]
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23 Nov 2017, 22:28
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In the circular region with center O, shown above, the two unshaded sections comprise 3/7 and 1/3 of the area of the circular region. The shaded section comprises what fractional part of the area of the circular region?
(A) 3/5
(B) 6/7
(C) 2/21
(D) 5/21
(E) 16/21
[Reveal] Spoiler:
Attachment:
2017-11-23_2026.png [ 6.6 KiB | Viewed 346 times ]
[Reveal] Spoiler: OA
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In the circular region with center O, shown above, the two unshaded se [#permalink]
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24 Nov 2017, 05:40
Bunuel wrote:
In the circular region with center O, shown above, the two unshaded sections comprise 3/7 and 1/3 of the area of the circular region. The shaded section comprises what fractional part of the area of the circular region?
(A) 3/5
(B) 6/7
(C) 2/21
(D) 5/21
(E) 16/21
[Reveal] Spoiler:
Attachment:
2017-11-23_2026.png
If the circle will have the entire fraction of the area 1,
the shaded portion will have area $$1 - \frac{3}{7} - \frac{1}{3} = \frac{21 - 9 - 7}{21} = \frac{5}{21}$$
Another method
Assume area of the circle to be 21.
The portion containing $$\frac{3}{7}$$ of the circle will have area $$9$$
and the part containing $$\frac{1}{3}$$ of the circle to be $$7$$.
The shaded portion of the circle will have an area of $$21 - 9 - 7 = 5$$
Hence shaded portion of the circle contains $$\frac{5}{21}$$ of the circle(Option D)
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Re: In the circular region with center O, shown above, the two unshaded se [#permalink]
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24 Nov 2017, 05:42
In the circular region with center O, shown above, the two unshaded sections comprise 3/7 and 1/3 of the area of the circular region. The shaded section comprises what fractional part of the area of the circular region?
(A) 3/5
(B) 6/7
(C) 2/21
(D) 5/21
(E) 16/21
Lets x to be the whole area of the circle and y be the shaded section =>
$$x = \frac{3}{7} * x + \frac {1}{3} * x + y$$
$$x = \frac{16x}{21} + y$$
$$y = \frac{21x - 16x}{21}$$
$$y = \frac{5x}{21}$$
The circle area is accaunted to shaded region as 5 : 21
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Re: In the circular region with center O, shown above, the two unshaded se [#permalink]
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24 Nov 2017, 08:04
First solution is better and lucid...
pushpitkc wrote:
Bunuel wrote:
In the circular region with center O, shown above, the two unshaded sections comprise 3/7 and 1/3 of the area of the circular region. The shaded section comprises what fractional part of the area of the circular region?
(A) 3/5
(B) 6/7
(C) 2/21
(D) 5/21
(E) 16/21
[Reveal] Spoiler:
Attachment:
2017-11-23_2026.png
If the circle will have the entire fraction of the area 1,
the shaded portion will have area $$1 - \frac{3}{7} - \frac{1}{3} = \frac{21 - 9 - 7}{21} = \frac{5}{21}$$
Another method
Assume area of the circle to be 21.
The portion containing $$\frac{3}{7}$$ of the circle will have area $$9$$
and the part containing $$\frac{1}{3}$$ of the circle to be $$7$$.
The shaded portion of the circle will have an area of $$21 - 9 - 7 = 5$$
Hence shaded portion of the circle contains $$\frac{5}{21}$$ of the circle(Option D)
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Re: In the circular region with center O, shown above, the two unshaded se [#permalink] 24 Nov 2017, 08:04
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2017-12-16 05:35:54
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https://plainmath.net/other/47895-a-large-grinding-wheel-in-the-shape-of-a-solid-cylinder-of-radius-0-33
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fertilizeki
2021-12-24
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 250 N applied to its edge causes the wheel to have an angular acceleration of 0.940 rad/s. What is the mass of the wheel?
Travis Hicks
Expert
The information below pertains to solid cylinder wheels rotating on a vertical axis without any friction.
$\alpha =0.94\frac{rad}{{s}^{2}}$
Torque is given by:
$\tau =F\cdot r=I\cdot \alpha$
Addressing the issue of moment of inertia
$I=\frac{F\cdot r}{\alpha }=\frac{250\cdot 0.33}{0.94}=87.8$
Inertia moment is determined by:
$I=\frac{1}{2}\cdot m\cdot {r}^{2}$
Solving it for mass:
$m=\frac{2\cdot I}{{r}^{2}}=\frac{2\cdot 87.8}{{0.33}^{2}}=1612.5$ kg
Do you have a similar question?
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2023-01-31 20:06:27
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http://dailygre.blogspot.com/2011/08/physics-gre-14.html
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## Pages
News: Currently the LaTeX and hidden solutions on this blog do not work on Google Reader.
Email me if you have suggestions on how to improve this blog!
## Monday, 15 August 2011
### Physics GRE - #14
The figure below shows a small mass connected to a string, which is attached to a vertical post.
If the mass is released when the string is horizontal as shown, the magnitude of the total acceleration of the mass as a function of the angle $\theta$ is:
• $g\sin\theta$
• $2g\cos\theta$
• $2g\sin\theta$
• $g\sqrt{3\cos^2{\theta}+1}$
• $g\sqrt{3\sin^2{\theta}+1}$
Solution :
We know that at $\theta=0$, the acceleration of the mass is exactly $g$. The only solution that satisfies this boundary condition is the last choice.
This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '\$'. To type centred formulae, type '$' at the beginning of your formula and '$' at the end.
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2018-10-19 23:33:39
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https://learning.rc.virginia.edu/courses/fortran_introduction/derived_types/
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Derived Types
Programmer-Defined Datatypes
So far we have used only the predefined types available in the Fortran standard. However, an important principle of modern software engineering is separation of concerns and encapsulation. We would like for related data to be connected, and we want each program unit to implement a well-defined set of actions, its “concern.” This also allows the programmer to control the interface, the way in which other parts of the program interact with the data.
For example, consider a program to update employee information. We can define several variables relevant for an employee; for example we might use salary, name of manager, name of department, employee ID number, and so forth. Each of these is potentially a different type. Salary would be floating point, the names would be strings, and the ID number would generally be an integer. We have more than one employee to handle, so we must use some form of list or array. In most languages we cannot define a single array to accommodate all these fields.
This leads to the need for a way to keep all the information about one employee coordinated.
If we were restricted to predefined types in Fortran, we would have to declare separate arrays for each field of interest. When processing data, we would have to take pains to ensure that the index of one array was correct for another array. Suppose we wanted to find all employees making more than a certain amount. We would have to search the “salary” array for the elements that met the criterion, while storing the index into some other array, so that we would have an array whose contents were the indices for other arrays. This is very prone to errors.
integer, dimension(:), allocatable:: employee_ID
character(len=128), dimension(:), allocatable:: employee_name
character(len=128), dimension(:), allocatable:: employee_manager
character(len=128), dimension(:), allocatable:: employee_dept
real, dimension(:), allocatable:: employee_salary
We need a different type of data structure. Programmer-defined datatypes allow the programmer to define a new type containing the representations of a group of related data items. For example, some languages define dataframes, which are essentially representations of spreadsheets with each column defined as something like an array. This would be an example of a defined datatype, since it must be described relative to basic types available in the language. This is perhaps easier in languages that use inferred typing, where the interpreter or compiler makes its best guess as to the type of data, as opposed to statically typed languages like Fortran or C++. But conceptually it is a good example of a programmer-defined datatype.
In Fortran abstract datatypes are called derived types. The syntax is extremely simple; in the example, ptype stands for a primitive type.
TYPE mytype
<ptype> var1
<ptype> var2
<ptype>, DIMENSION(:), ALLOCATABLE:: var3
TYPE(anothertype) :: var4
END TYPE mytype
Variables of mytype are declared as
type(mytype) :: x, y
We access the fields using the % separator:
z=x%var1
w=y%var2
allocate(x%var3(N))
where the variables z and w must be declared to match the type, including attributes such as ALLOCATABLE, of the field of the type. As shown above, a TYPE may be a member of another TYPE as long as its definition has already been seen by the compiler. Variables that belong to the type are usually called components in Fortran.
Note that a type is a scoping unit.
We can apply this to our employee example. The longer name for the fields is not helpful since we declare everything to be pertinent to an “employee.”
TYPE employee
INTEGER :: ID
CHARACTER(len=128) :: name
CHARACTER(len=128) :: manager
CHARACTER(len=128) :: dept
REAL :: salary
END TYPE
We can now declare employees
TYPE(employee) :: fred, joe, sally
real :: raise
raise=0.02
fred%salary=(1+raise)*fred%salary
Arrays and Types
Types may contain arrays and from F2003 onward, those arrays may be allocatable. Very few compilers do not support this standard but if one is encountered, the POINTER attribute must be used. We will not discuss POINTER further but it may be seen in code written before F2003 compilers were widely available.
In Fortran, the array data structure is a container and the elements of an array may be derived types.
TYPE(employee), dimension(:), allocatable :: employees
We allocate as usual
num_employees=126
allocate(employees(num_employees))
Arrays and Modules
We nearly always put derived types into modules; the module will define procedures that operate on the type. The module must not have the same name as the derived type, which can be somewhat inconvenient.
A derived type may need to be initialized explicitly. For example, if you need to allocate memory, say for an allocatable array, to create a variable of a given type, this will not happen automatically. You must write a constructor to allocate the memory.
**Example**
This type is a set of observations for birds denoted by their common name.
fortran
TYPE bird_data
CHARACTER(LEN=50) :: species
INTEGER, DIMENSION(:), ALLOCATABLE :: obs
END TYPE
A constructor-like procedure would be
MODULE bird_obs
TYPE bird_data
CHARACTER(LEN=50) :: species
INTEGER, DIMENSION(:), ALLOCATABLE :: obs
END TYPE
CONTAINS
SUBROUTINE constructor(bird,species,obs)
TYPE(bird_data), INTENT(INOUT) :: bird
CHARACTER(LEN=50), INTENT(IN) :: species
INTEGER, DIMENSION(:), INTENT(IN) :: obs
bird%species=species
allocate(bird%obs(size(obs)))
bird%obs=obs
END SUBROUTINE
END MODULE
It is important to understand that the species that is a member of the type is not the same as the species that is passed in to init_bird. In Fortran we can easily distinguish them since we must use the instance variable, bird in this case, as a prefix; not all languages require that. In C++ we would need to use this->species (this is the “invisible” instance variable in that language) if an attibute has the same name as a dummy parameter.
Exercise
Write a main program to use the bird_dat module. Assume you will read the bird data from a CSV (comma-separated values) file with each row consisting of a string for the species and then 10 numbers for observations over 10 years. Create a file
"Species",2000,2001,2002,2003,2004,2005,2006,2007,2008,2009
"BlueJay", 24, 23, 27, 19, 22, 26, 28, 27, 24, 30
"Cardinal", 11, 15, 18, 18, 19, 17, 20, 21, 20, 19
Use this file to test your program.
Example Solution
program bird_obs
use bird_dat
implicit none
type(bird_data),dimension (:), allocatable :: bird_list
character(len=50) :: filename
character(len=50) :: species,my_species,lc_species
integer :: l, n, nargs, nbirds, nobs
integer, dimension(:),allocatable :: years
interface
use bird_dat
implicit none
type(bird_data), dimension(:), allocatable, intent(out) :: bird_list
character(len=*), intent(in) :: filename
integer, dimension(:), allocatable, intent(out) :: years
end subroutine
end interface
nargs=command_argument_count()
if ( nargs .ne. 1 ) then
stop "Usage: <file>"
else
call get_command_argument(1,filename)
endif
nbirds=size(bird_list)
nobs =size(bird_list(1)%obs)
write(*,*) "Observations over the years ",years
do n=1,nbirds
do l=1,nobs-1
enddo
write(*,'(i4)') bird_list(n)%obs(nobs)
enddo
End program
use bird_dat
implicit none
type(bird_data), dimension(:), allocatable, intent(out) :: bird_list
character(len=*), intent(in) :: filename
integer, dimension(:), allocatable, intent(out) :: years
integer, dimension(:), allocatable :: obs
integer, parameter :: nobs=10
character(len=6),dimension(nobs) :: cyears
integer :: inunit
character(len=50) :: species
character(len=1024) :: line
character(len=:),dimension(:),allocatable :: line_vals
integer :: num_vals
integer :: n,m
logical :: file_exists
inquire(file=filename,exist=file_exists)
if (file_exists) then
inunit=10
open(10,file=filename)
else
stop "Can't find file."
endif
nlines=0
do
nlines=nlines+1
enddo
10 continue
rewind(inunit)
allocate(bird_list(nbirds))
allocate(years(nobs))
allocate(obs(nobs))
enddo
do n=1,nbirds
call constructor(bird_list(n),species,obs)
end do
Previous
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2023-03-28 02:38:55
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http://anh.cs.luc.edu/handsonPythonTutorial/boolean.html
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3.4. Arbitrary Types Treated As Boolean¶
The following section would merely be an advanced topic, except for the fact that many common mistakes have their meaning changed and obscured by the Boolean syntax discussed.
You have seen how many kinds of objects can be converted to other types. Any object can be converted to Boolean (type bool). Read the examples shown in this Shell sequence:
>>> bool(2)
True
>>> bool(-3.1)
True
>>> bool(0)
False
>>> bool(0.0)
False
>>> bool(None)
False
>>> bool('')
False
>>> bool('0')
True
>>> bool('False')
True
>>> bool([])
False
>>> bool([0])
True
The result looks pretty strange, but there is a fairly short general explanation: Almost everything is converted to True. The only values among built-in types that are interpreted as False are
• The Boolean value False itself
• Any numerical value equal to 0 (0, 0.0 but not 2 or -3.1)
• The special value None
• Any empty sequence or collection, including the empty string('', but not '0' or 'hi' or 'False') and the empty list ([], but not [1,2, 3] or [0])
A possibly useful consequence occurs in the fairly common situation where something needs to be done with a list only if it is nonempty. In this case the explicit syntax:
if len(aList) > 0:
doSomethingWith(aList)
can be written with the more succinct Pythonic idiom
if aList:
doSomethingWith(aList)
This automatic conversion can also lead to extra trouble! Suppose you prompt the user for the answer to a yes/no question, and want to accept ‘y’ or ‘yes’ as indicating True. You might write the following incorrect code. Read it:
ans = input('Is this OK? ')
if ans == 'y' or 'yes':
print('Yes, it is OK')
The problem is that there are two binary operations here: ==, or. Comparison operations all have higher precedence than the logical operations or, and, and not. The if condition above can be rewritten equivalently with parentheses. Read and consider:
(ans == 'y') or 'yes'
Other programming languages have the advantage of stopping with an error at such an expression, since a string like 'yes' is not Boolean. Python, however, accepts the expression, and treats 'yes' as True! To test, run the example program boolConfusion.py, shown below:
ans = 'y'
if ans == 'y' or 'yes':
print('y is OK')
ans = 'no'
if ans == 'y' or 'yes':
print('no is OK!!???')
Python detects no error. The or expression is always treated as True, since 'yes' is a non-empty sequence, interpreted as True.
The intention of the if condition presumably was something like
(ans == 'y') or (ans == 'yes')
This version also translates directly to other languages. Another correct Pythonic alternative that groups the alternate values together is
ans in ['y', 'yes']
which reads pretty much like English. It is true if ans is in the specified list. The in operator actually works with any sequence. The general syntax is
value in sequence
This is true when value is an element of the sequence.
Be careful to use a correct expression when you want to specify a condition like this.
Things get even stranger! Enter these conditions themselves, one at a time, directly into the Shell:
'y' == 'y' or 'yes'
'no' == 'y' or 'yes'
The meaning of (a or b) is exactly as discussed so far if each of the operands a and b are actually Boolean, but a more elaborate definition is needed if an operand is not Boolean.
val = a or b
means
if bool(a):
val = a
else:
val = b
and in a similar vein:
val = a and b
means
if bool(a):
val = b
else:
val = a
This strange syntax was included in Python to allow code like in the following example program orNotBoolean.py. Read and test if you like:
defaultColor = 'red'
userColor = input('Enter a color, or just press Enter for the default: ')
color = userColor or defaultColor
print('The color is', color)
which sets color to the value of defaultColor if the user enters an empty string.
Again, this may be useful to experienced programmers. The syntax can certainly cause difficult bugs, particularly for beginners!
The not operator always produces a result of type bool.
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2018-03-21 01:21:19
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http://www.encyclopediaofmath.org/index.php/Geometric_ring
|
# Geometric ring
Jump to: navigation, search
A local ring of an algebraic variety or a completion of such a ring. A commutative ring obtained from a ring of polynomials over a field by means of the operations of completion, localization and factorization by a prime ideal is called an algebro-geometric ring [3]. A local ring of an irreducible algebraic variety does not obtain nilpotent elements as a result of completion [2]. This property of a local ring is known as analytic reducibility. A similar fact concerning local rings of normal varieties [1] is that the completion of a local ring of a normal algebraic variety is a normal ring (analytic normality). Examples of local Noetherian rings that are not analytically reduced or analytically normal are known [4]. A pseudo-geometric ring is a Noetherian ring any quotient ring of which by a prime ideal is a Japanese ring. An integral domain $A$ is called a Japanese ring if its integral closure in a finite extension of the field of fractions is a finite $A$-module [5]. The class of pseudo-geometric rings is closed with respect to localizations and extensions of finite type; it includes the ring of integers and all complete local rings. See also Excellent ring.
#### References
[1] O. Zariski, P. Samuel, "Commutative algebra" , 1 , Springer (1975) [2] C. Chevalley, "Intersection of algebraic and algebroid varieties" Trans. Amer. Math. Soc. , 57 (1945) pp. 1–85 [3] P. Samuel, "Algèbre locale" , Gauthier-Villars (1953) [4] M. Nagata, "Local rings" , Interscience (1962) [5] A. Grothendieck, "Eléments de géometrie algébrique IV. Etude locale des schémas et des morphismes des schémas" Publ. Math. IHES : 32 (1967)
How to Cite This Entry:
Geometric ring. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Geometric_ring&oldid=31943
This article was adapted from an original article by V.I. Danilov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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2015-05-30 10:26:05
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https://manual.q-chem.com/5.4/sec_opt-clus.html
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# 9.11 Optimising the Structure of Clusters
(June 30, 2021)
The potential energy landscape of atomic and molecular clusters can be very complex with many minima which can have similar energies, and this complexity increases rapidly as the size of the clusters increases. Determining the global minimum of these clusters is challenging since it requires extensive searching over the potential energy surface. One approach to finding the low energy structures of these clusters is to perform many geometry optimisations starting at different initial coordinates. Q-Chem is able to perform such random searches for molecular clusters containing up to two different molecule types. In these searches the molecules are subjected to translations and rotations of their structure to generate a new starting structure. These searches are initiated by the JOBTYPE = RAND and it is necessary to specify the number of molecules of the different types and the number of atoms in the different types of molecules. For the optimisation of atomic clusters, SEARCH_ATOMIC = TRUE and the number of atom swops performed in the structure generation (N_SWOP) can be specified. Some care has to be taken with the specification of the input structure in the $molecule section. All the atoms of the molecules of molecule type 1 must come before those of molecule type 2. Furthermore, the atoms of the same molecule should be together. For examples of these studies see 273, 274, 275, 675. Example 9.39 A random search geometry optimisation of the NO${}^{+}$.H${}_{2}$O cluster. $molecule
1 1
N 0.5682008336 0.1585044954 -0.9009280260
O -0.3450383302 -0.5598328271 -0.4634299478
O 1.7303273568 0.3403569345 0.4364171165
H 2.5236300547 -0.2494576134 0.1485689942
H 2.1020812302 1.2823911654 0.2570156558
$end$rem
JOBTYPE RAND
METHOD B3LYP
BASIS STO-3G
SCF_CONVERGENCE 6
MAX_SCF_CYCLES 100
NSEARCH 10
N_MOL_TYPE 2
NMOL1 1
N_ATOM_TYPE_1 2
NMOL2 1
N_ATOM_TYPE_2 3
N_MOVES 20
MAXBOX 10000
MIN_SEPARATION 25
MAX_DISPLACE 25
SCF_NOCRASH TRUE
TIGHTEN_CONVERG TRUE
GEOM_OPT_MAX_CYCLES 200
GEOM_OPT_COORDS 0
GEOM_OPT_TOL_DISPLACEMENT 1000
GEOM_OPT_TOL_ENERGY 100
$end View output Example 9.40 A random search geometry optimisation of the He${}_{3}$Ne${}_{3}$ cluster. $molecule
0 1
He -1.3590894 3.0177788 -0.1662522
He -2.9853158 1.1444488 0.1036005
He 0.5068109 1.3795209 -0.2168151
Ne -1.1002149 -0.5693061 0.0381894
Ne 0.5981676 1.8697812 1.4685618
Ne -1.2376457 1.2597811 -0.0756066
$end$rem
JOBTYPE RAND
METHOD B3LYP
DFT_D EMPIRICAL_GRIMME
BASIS STO-3G
SCF_CONVERGENCE 6
MAX_SCF_CYCLES 100
NSEARCH 10
SEARCH_ATOMIC TRUE
N_SWOP 4
N_MOL_TYPE 2
NMOL1 3
N_ATOM_TYPE_1 1
NMOL2 3
N_ATOM_TYPE_2 1
N_MOVES 20
MAXBOX 10000
MIN_SEPARATION 25
MAX_DISPLACE 25
SCF_NOCRASH TRUE
TIGHTEN_CONVERG TRUE
USE_INITIAL TRUE
GEOM_OPT_MAX_CYCLES 200
GEOM_OPT_COORDS 0
GEOM_OPT_TOL_DISPLACEMENT 1000
GEOM_OPT_TOL_ENERGY 1000
$end View output Basin hopping (BH) is a more advanced technique for locating the global minimum on complex potential energy surfaces. The BH algorithm is essentially a combination of the Metropolis sampling technique and a gradient-based local search method. This has the effect of sampling the energy basins, where an energy basin is a certain part of the configuration space around a minimum on the PES that contains all the configurations that will relax into this minimum using downhill relaxations, instead of sampling the configuration space. To enhance the efficiency of the method, BH with occasional jumping is used, which incorporates a jumping move in addition to the standard Monte Carlo (MC) moves. Jumping is a MC move without local minimization at infinite temperature and, consequently, is always accepted. When the usual MC moves are rejected a number of times, the system is judged to be trapped at the local minimum. The temperature is raised to T = $\infty$, and the MC jumping moves are executed several times to allow the system to escape from the local minimum. This provides an efficient way to escape from a local minimum and to explore the next basin of the valley when it is separated by high barriers. Depending on the size and complexity of the system being studied, a large number of MC_STEPS and/or MC_CYCLES to ensure the global minimum is found. Example 9.41 A basin hopping search for the NO${}^{+}$.H${}_{2}$O cluster. $molecule
1 1
N 0.5682008336 0.1585044954 -0.9009280260
O -0.3450383302 -0.5598328271 -0.4634299478
O 1.7303273568 0.3403569345 0.4364171165
H 2.5236300547 -0.2494576134 0.1485689942
H 2.1020812302 1.2823911654 0.2570156558
$end$rem
JOBTYPE BH
METHOD B3LYP
BASIS STO-3G
SCF_CONVERGENCE 6
MAX_SCF_CYCLES 100
MC_CYCLES 4
MC_STEPS 5
MC_TEMP 300
MAX_DISPLACE 25
MIN_SEPARATION 25
MAXBOX 5000
N_MOVES 20
N_MOL_TYPE 2
NMOL1 1
N_ATOM_TYPE_1 2
NMOL2 1
N_ATOM_TYPE_2 3
N_MOVES 20
MAXBOX 10000
MIN_SEPARATION 25
MAX_DISPLACE 25
SCF_NOCRASH TRUE
GEOM_OPT_MAX_CYCLES 200
GEOM_OPT_COORDS 0
GEOM_OPT_TOL_DISPLACEMENT 2000
GEOM_OPT_TOL_ENERGY 400
\$end
View output
## 9.11.0.1 Cluster Optimisation Job Control
NSEARCH
INTEGER
TYPE:
Sets the number of structures that are generated and optimised.
DEFAULT:
No default.
OPTIONS:
User defined.
RECOMMENDATION:
None
SEARCH_ATOMIC
Perform an optimisation for atomic cluster.
TYPE:
BOOLEAN
DEFAULT:
False
OPTIONS:
True Atomic cluster search will be performed. False Molecular clusters search will be performed.
RECOMMENDATION:
Use N_SWOP to specify atomic number of atom swops in structure generation.
N_SWOP
INTEGER
TYPE:
Sets the number atom coordinate swops for atomic cluster search.
DEFAULT:
No default.
OPTIONS:
User defined
RECOMMENDATION:
None
N_MOL_TYPE
INTEGER
TYPE:
Sets the number of different atom/molecule types.
DEFAULT:
No default.
OPTIONS:
User defined : can be 1 or 2.
RECOMMENDATION:
None
NMOL1
INTEGER
TYPE:
Sets the number of molecules of type 1.
DEFAULT:
No default.
OPTIONS:
User defined.
RECOMMENDATION:
None
N_ATOM_TYPE_1
INTEGER
TYPE:
Sets the number atoms in molecule type 1.
DEFAULT:
No default.
OPTIONS:
User defined.
RECOMMENDATION:
None
NMOL2
INTEGER
TYPE:
Sets the number of molecules of type 2.
DEFAULT:
No default.
OPTIONS:
User defined.
RECOMMENDATION:
None
N_ATOM_TYPE_2
INTEGER
TYPE:
Sets the number atoms in molecule type 2.
DEFAULT:
No default.
OPTIONS:
User defined.
RECOMMENDATION:
None
MAXBOX
Sets the size of the box which the molecules are kept within.
TYPE:
INTEGER
DEFAULT:
20000
OPTIONS:
$n$ Corresponding to MAXBOX = $n/1000$ bohr.
RECOMMENDATION:
Need to ensure that the cluster can fit within this box.
MIN_SEPARATION
Reject initial structures where the closest approach of molecules is less than this value.
TYPE:
INTEGER
DEFAULT:
300
OPTIONS:
$n$ Corresponding to MIN_SEPARATION = $n/100$ bohr.
RECOMMENDATION:
MIN_SEPARATION of approximately 2.5 bohr.
MAX_DISPLACE
Sets the maximum distance a molecule will be moved during a translation.
TYPE:
INTEGER
DEFAULT:
500
OPTIONS:
$n$ Corresponding to MAX_DISPLACE = $n/100$ bohr.
RECOMMENDATION:
None.
SCF_NOCRASH
Ensure the calculations continues if the SCF fails to converge for a given structure.
TYPE:
BOOLEAN
DEFAULT:
False
OPTIONS:
True Ensure calculation will continue with next structure. False Calculation will stop.
RECOMMENDATION:
Use SCF_NOCRASH = TRUE
TIGHTEN_CONVERG
At the end of the search re-calculate the energies of the optimised structures with tighter SCF convergence criteria.
TYPE:
BOOLEAN
DEFAULT:
False
OPTIONS:
True Additional calculations with tighter SCF convergence performed. False No additional calculations performed.
RECOMMENDATION:
None.
USE_INITIAL
Include input structure as part of the search.
TYPE:
BOOLEAN
DEFAULT:
False
OPTIONS:
True Input structure is included in the search. False Input structure is not included in the search.
RECOMMENDATION:
None.
SEARCH_MOM
Allows the search to be performed in conjunction with MOM to explore excited states.
TYPE:
BOOLEAN
DEFAULT:
False
OPTIONS:
True A search with MOM is performed. False Normal calculation without MOM.
RECOMMENDATION:
None.
MC_CYCLES
INTEGER
TYPE:
Sets the number of cycles in a basin hopping search.
DEFAULT:
No default.
OPTIONS:
User defined.
RECOMMENDATION:
None
MC_STEPS
INTEGER
TYPE:
Sets the number of Monte Carlo steps in each MC_CYCLES. After MC_STEPS jumping is initiated.
DEFAULT:
No default.
OPTIONS:
User defined.
RECOMMENDATION:
None
MC_TEMP
INTEGER
TYPE:
Sets the temperature (in Kelvin).
DEFAULT:
300
OPTIONS:
User defined.
RECOMMENDATION:
None
N_MOVES
INTEGER
TYPE:
Sets the number of structural changes/moves on each step.
DEFAULT:
2
OPTIONS:
User defined.
RECOMMENDATION:
None
MAX_JUMP
INTEGER
TYPE:
Sets the number of moves accepted on jumping.
DEFAULT:
10
OPTIONS:
User defined.
RECOMMENDATION:
None
|
2021-09-27 00:53:46
|
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|
http://mathematica.stackexchange.com/questions?page=11&sort=unanswered
|
All Questions
70 views
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|
2015-03-04 12:49:08
|
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|
http://mimsy.io/a_primer_on_bayesian_statistics.html
|
2017-09-15
## A Primer on Bayesian Statistics
Note
NOTE: This tutorial doesn't require any knowledge beyond high-school level mathematics/statistics. In you do have a college statistics class under your belt, you might want to check out More On Bayesian Statistics after reading through this post.
## A Primer on Bayesian Statistics
Like many people first starting out on their journey into data science, I dove straight into the cool and shiny stuff long before I really knew what I was doing1. It was during this exploratory phase, while I oogled those fancy neural networks, that I began noticing the term Bayesian Statistics with increasing frequency. It popped up in blogs, it popped up in books, it popped up in online videos. "What are they talking about?" I thought, "That one equation I learned about in my intro statistics course? What's the big deal here?"
As I soon found out, I was missing out on a whole school of statistics knowledge and I didn't even know about it! Unfortunately, sources I found on the subject tended to swing either towards textbook-levels of denseness or vague philosophical works about "Dutch Books"2. This is a shame, because I think you can explain the concept in some meaningful depth without writing a whole textbook; all you need is a few visuals and insightful examples.
### Reviewing Bayes' Theorem With a Few Visuals and Insightful Examples
Unsurprisingly, I want to start this tutorial by reviewing Bayes' Theorem. The math involved in the theorem isn't especially difficult, but I think many people really miss the point when they first see it. I myself remember learning the formula, thinking to myself "oh, that's cool", and then moving on, leaving the knowledge in a dusty corner of my mind. In order to avoid that in this tutorial, I'll try to be thorough, and build up intuition as much as possible. I'll do this by taking a visual approach and deriving Bayes' Theorem through an example: $$\require{color}$$ $$\definecolor{cs}{RGB}{252, 141, 98}$$ $$\definecolor{cc}{RGB}{102, 194, 165}$$ $$\definecolor{cr}{RGB}{141, 160, 203}$$ $$\definecolor{cj}{RGB}{231, 138, 195}$$ $$\definecolor{ccr}{RGB}{227, 26, 28}$$ $$\definecolor{ccb}{RGB}{31, 120, 180}$$ $$\definecolor{ccw}{RGB}{51, 160, 44}$$ $$\definecolor{def}{RGB}{93, 93, 93}$$
#### Weather and Hobbies
As all meteorologists will tell you, there are only three types of weather in the world: sunny, cloudy, and rainy. Here in California, the weather tends to favor long, dry days. As an approximation, let's say that it's sunny $$\color{cs}{60\%}$$ of the time, cloudy $$\color{cc}{30\%}$$ of the time, and rainy $$\color{cr}{10\%}$$ of the time. We can represent this situation quite simply with a visualization:
Now, when I make small talk about the weather, I usually talk about my hobbies as well. I'm a pretty simple guy, so you can segregate my hobbies into two categories: running and everything else. Out of these two groups, I estimate that I go running $$\color{cj}{40\%}$$ of all days:
That's an intuitive way to represent probabilities, but it doesn't allow us to compare everything directly. It would be great if we could talk about the weather and our hobbies at the same time. Fortunately, we can! Just combine the scales above into a 2D figure:
Now we can show off both sets of events, just like we wanted. We can see all of the probabilities and even measure them out if we wanted to:
\begin{equation*} \begin{array}{rccl} \boldsymbol{P}(\color{cs}sunny\color{def}) & = 0.60 & \rightarrow & \, \scriptstyle\text{The size of the sunny area } \img[-0.25em][1em][1em]{images/ss.svg} \text{ when compared to the whole area}. \\ \boldsymbol{P}(\color{cc}cloudy\color{def}) & = 0.30 & \rightarrow & \, \scriptstyle\text{The size of the cloudy area } \img[-0.25em][1em][1em]{images/sc.svg} \text{ when compared to the whole area}. \\ \boldsymbol{P}(\color{cr}rainy\color{def}) & = 0.10 & \rightarrow & \, \scriptstyle\text{The size of the rainy area } \img[-0.25em][1em][1em]{images/sr.svg} \text{ when compared to the whole area}. \\ \boldsymbol{P}(\color{cj}running\color{def}) & = 0.40 & \rightarrow & \, \scriptstyle\text{The size of the running area } \img[-0.25em][1em][1em]{images/sj.svg} \text{ when compared to the whole area}. \end{array} \end{equation*}
#### Conditional Probability
Hmmm, something seems off. You see, I actually don't like running when it's too hot outside. While my overall rate is $$\color{cj}{40\%}$$, I tend to run less than that if I think I might die of dehydration. Conversely, I also tend to run a lot more on cool, rainy days. The diagram above doesn't show that. See all those right angles? Those are right angles of independence! They imply that, no matter what kind of weather I'm currently experiencing, I will always have a $$\color{cj}40\%$$ running rate. Me running and the weather don't affect one another.
We can visually prove this by looking at the conditional probabilities of each event. For example, say it's a sunny day and we want to know the probability of me going for a jog. In mathematical notation, we're looking for the quantity $$\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cs}sunny\color{def})$$. We can find it like this:
This image shows how I like to think about conditional probability. When we assumed that it was a sunny day, we essentially said that we must land in the $$\img[-0.25em][1em][1em]{images/ss.svg}$$ area (you are here). To see how likely it is that I go running in sunny weather, we then look for how likely it is that we land in the $$\img[-0.25em][1em][1em]{images/sj.svg}$$ area, knowing that we have to stay in the $$\img[-0.25em][1em][1em]{images/ss.svg}$$ area. Thus, the probability of me running given that it's sunny, $$\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cs}sunny\color{def})$$, is the area of the intersection $$\img[-0.25em][1em][1em]{images/sjs.svg}$$ compared to the $$\img[-0.25em][1em][1em]{images/ss.svg}$$ area. Notice that this is still $$40\%$$! Thus, $$\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cs}sunny\color{def}) = \boldsymbol{P}(\color{cj}running\color{def})$$ and we have independence.
We can go a step further, take what I said above, and make it concrete with a formula:
\begin{equation*} \begin{array}{crcc} & \boldsymbol{P}(\color{cj}running\color{def}) & \rightarrow & \img[-0.25em][1em][1em]{images/sj.svg} \\ & \boldsymbol{P}(\color{cs}sunny\color{def}) & \rightarrow & \img[-0.25em][1em][1em]{images/ss.svg} \\ & \boldsymbol{P}(\color{cj}running\color{def},\,\color{cs}sunny\color{def}) & \rightarrow & \img[-0.25em][1em][1em]{images/sjs.svg} \\ & \boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cs}sunny\color{def}) & \rightarrow & \dfrac{\img[-0.25em][1em][1em]{images/sjs.svg}}{\img[-0.25em][1em][1em]{images/ss.svg}} \end{array} \Large\Rightarrow\normalsize \boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cs}sunny\color{def}) = \dfrac{\boldsymbol{P}(\color{cj}running\color{def},\,\color{cs}sunny\color{def})}{\boldsymbol{P}(\color{cs}sunny\color{def})} \end{equation*}
To anyone who's taken a statistics course, this is the all-too-familiar formula for conditional probability, derived using a purely visual method! It's usually presented with arbitrary events $$A$$ and $$B$$ in a somewhat bland way:
\begin{equation*} \boldsymbol{P}(A|B) \boldsymbol{P}(B) = \boldsymbol{P}(A,B) = \boldsymbol{P}(B|A) \boldsymbol{P}(A) \end{equation*}
Now that we've established independence between running and all the weather events, how can we go about changing that? Well, we just have to get rid of the right angles:
Look at how the probability of running shifted away from sunny towards cloudy and rainy (the dotted line is the old boundary). The events now work out in a way that reflects my preferences:
\begin{equation*} \begin{array}{rccl} \boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cs}sunny\color{def}) & = 0.32 & < 0.40\\ \boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cc}cloudy\color{def}) & = 0.50 & > 0.40\\ \boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cr}rainy\color{def}) & = 0.58 & > 0.40\\ \end{array} \end{equation*}
#### Bayes' Theorem
Bad luck, it seems you've suddenly developed an acute case of can't-tell-the-weather-itus. It's an extremely rare disease that temporarily removes your ability to detect the weather (strangely, it doesn't affect your life in any other way). Just as this happens, you spot me jogging across the street. Aha! A way out of this predicament! You've read this post, so you know that I'm more likely to be running if it's raining or cloudy. The chances of rainy or cloudy weather must be high!
No wait, that's wrong. You only know how the weather affects my tendency to run, not the other way around. In other words, you might know $$\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cr}rainy\color{def})$$ but you do not know $$\boldsymbol{P}(\color{cr}rainy\color{def}\,|\,\color{cj}running\color{def})$$, and these two values are not necessarily equal. Visually:
If we're looking for $$\boldsymbol{P}(\color{cr}rainy\color{def}\,|\,\color{cj}running\color{def})$$, we want to know how likely it is that we land in $$\img[-0.25em][1em][1em]{images/sr.svg}$$ assuming we're already in $$\img[-0.25em][1em][1em]{images/sj.svg}$$; just looking at the diagram, you can surmise that this is a small number, especially when compared to $$\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cr}rainy\color{def}) = 0.58$$ (how likely we are to land in $$\img[-0.25em][1em][1em]{images/sj.svg}$$ assuming we're inside $$\img[-0.25em][1em][1em]{images/sr.svg}$$). The discrepancy stems from that fact that, in California, the base rate of rain starts low, $$\boldsymbol{P}(\color{cr}rainy\color{def}) = 0.10$$. If you want to get the correct prediction, you have to approach this from a different angle.
Good thing we learned about conditional probability in the last section! Just like last time, we can find $$\boldsymbol{P}(\color{cr}rainy\color{def}\,|\,\color{cj}running\color{def})$$ as follows:
\begin{equation*} \begin{array}{crcc} & \boldsymbol{P}(\color{cr}rainy\color{def}) & \rightarrow & \img[-0.25em][1em][1em]{images/sr.svg} \\ & \boldsymbol{P}(\color{cj}running\color{def}) & \rightarrow & \img[-0.25em][1em][1em]{images/sj.svg} \\ & \boldsymbol{P}(\color{cr}rainy\color{def},\,\color{cj}running\color{def}) & \rightarrow & \img[-0.25em][1em][1em]{images/sjr.svg} \\ & \boldsymbol{P}(\color{cr}rainy\color{def}\,|\,\color{cj}running\color{def}) & \rightarrow & \dfrac{\img[-0.25em][1em][1em]{images/sjr.svg}}{\img[-0.25em][1em][1em]{images/sj.svg}} \end{array} \Large\Rightarrow\normalsize \boldsymbol{P}(\color{cr}rainy\color{def}\,|\,\color{cj}running\color{def}) = \dfrac{\boldsymbol{P}(\color{cr}rainy\color{def},\,\color{cj}running\color{def})}{\boldsymbol{P}(\color{cj}running\color{def})} \end{equation*}
The only problem here is that we don't know the numerical value of the $$\img[-0.25em][1em][1em]{images/sjr.svg}$$ area, the intersection $$\boldsymbol{P}(\color{cr}rainy\color{def},\,\color{cj}running\color{def})$$. We can fix this by applying conditional probability in the other direction, basically switching the roles of running and rainy:
\begin{align*} & \boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cr}rainy\color{def}) = \dfrac{\boldsymbol{P}(\color{cr}rainy\color{def},\,\color{cj}running\color{def})}{\boldsymbol{P}(\color{cr}rainy\color{def})} \\ \Large\Rightarrow\normalsize & \boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cr}rainy\color{def}) \boldsymbol{P}(\color{cr}rainy\color{def}) = \boldsymbol{P}(\color{cr}rainy\color{def},\,\color{cj}running\color{def}) \end{align*}
Finally, we substitute and get our answer:
\begin{equation*} \boldsymbol{P}(\color{cr}rainy\color{def}\,|\,\color{cj}running\color{def}) = \dfrac{\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cr}rainy\color{def}) \boldsymbol{P}(\color{cr}rainy\color{def})}{\boldsymbol{P}(\color{cj}running\color{def})} \end{equation*}
Or, in case you prefer to do all your math using colorful squares3:
\begin{equation*} \dfrac{\img[-0.25em][1em][1em]{images/sjr.svg}}{\img[-0.25em][1em][1em]{images/sj.svg}} = \dfrac{\img[-0.25em][1em][1em]{images/sjr.svg}}{\img[-0.25em][1em][1em]{images/sr.svg}} \cdot \dfrac{\img[-0.25em][1em][1em]{images/sr.svg}}{\img[-0.25em][1em][1em]{images/sj.svg}} \end{equation*}
This formula is known as Bayes' Theorem. The most immediate result of Bayes' Theorem is that it allows you to flip conditional probabilities:
\begin{equation*} \boldsymbol{P}(\color{cr}rainy\color{def}\,|\,\color{cj}running\color{def}) \leftrightarrow \boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cr}rainy\color{def}) \end{equation*}
In our case, for example, it allows us to solve the big problem you were having just a little while ago:
\begin{align*} \boldsymbol{P}(\color{cr}rainy\color{def}\,|\,\color{cj}running\color{def}) = & \dfrac{\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cr}rainy\color{def}) \boldsymbol{P}(\color{cr}rainy\color{def})}{\boldsymbol{P}(\color{cj}running\color{def})} \\ = & \dfrac{0.58 \cdot 0.10}{0.40} = 0.145 \end{align*}
Similarly, we can use this method to find the other probabilities as well:
\begin{align*} & \boldsymbol{P}(\color{cc}cloudy\color{def}\,|\,\color{cj}running\color{def}) = \dfrac{\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cc}cloudy\color{def}) \boldsymbol{P}(\color{cc}cloudy\color{def})}{\boldsymbol{P}(\color{cj}running\color{def})} = 0.375 \\ & \boldsymbol{P}(\color{cs}sunny\color{def}\,|\,\color{cj}running\color{def}) = \dfrac{\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cs}sunny\color{def}) \boldsymbol{P}(\color{cs}sunny\color{def})}{\boldsymbol{P}(\color{cj}running\color{def})} = 0.48 \end{align*}
Predicament solved! Even though I prefer rain to sun when jogging, the fact that California has mostly sunny weather beats out my preferences. If you see me out for a jog, I'm probably dehydrated (I should start working out with a water bottle).
#### A Different Way to Think About Bayes
The ability to switch conditional probabilities sounds useful, but how can you create a whole school of thought based around it? How can people write whole textbooks on the subject? How come this tutorial is only half over? Well, the real power of Bayes' Theorem doesn't lie in it's ability to switch conditional probabilities, and I wouldn't want you to come away from this tutorial thinking that. Instead, statisticians view Bayes' Theorem like so:
This view of Bayes' Theorem says you can combine the base rate of an event (the prior) and modify it with new information to get a better estimate of the same event (the posterior). In the picture above, you start with base knowledge about the probability of rain. If you don't have any other information, you can only rely on the this base rate, so $$\boldsymbol{P}(\color{cr}rainy\color{def}) = 0.10$$. However, once you get a piece of data (whether I'm running that day or not), Bayes' Theorem says that you integrate this information into a new estimate and get a posterior probability. In this case, you find that $$\boldsymbol{P}(\color{cr}rainy\color{def}\,|\,\color{cj}running\color{def}) = 0.145$$.
The modifying term compares the likelihood of me running in rainy weather (the numerator) to the flat chance of me running in any weather (the denominator). These two terms have names, the likelihood and the evidence4:
In effect, the likelihood and the evidence measure how much information we get out of the data. If the modifier ends up greater than $$1$$, our posterior probability ends up higher than our prior. The same is true in the other direction; if the modifier is less than $$1$$, the posterior probability will be lower than the prior. For example:
\begin{equation*} \dfrac{\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cr}rainy\color{def})}{\boldsymbol{P}(\color{cj}running\color{def})} = \dfrac{0.58}{0.40} > 1 \Large\Rightarrow\normalsize \text{posterior} \, > \, \text{prior} \end{equation*}
\begin{equation*} \dfrac{\boldsymbol{P}(\color{cj}running\color{def}\,|\,\color{cs}sunny\color{def})}{\boldsymbol{P}(\color{cj}running\color{def})} = \dfrac{0.32}{0.40} < 1 \Large\Rightarrow\normalsize \text{posterior} \, < \, \text{prior} \end{equation*}
Since you know that I prefer rainy weather when working out, your estimate of rain goes up from the base rate if you see me running. Similarly, your estimate on the chances of sunny weather go down. Why is this so useful? Because now, we have a way to incorporate data into initial estimates and refine them! Isn't that exciting!
What's that? You still don't see what the big deal is? Maybe we need another example.
### A New Example and a Change of Colors
To appreciate the power of incorporating data into estimates, let's change gears and work through a new, more serious, example.
Consider a city with two competing cab companies: the red and blue company. The red company dominates the market with $$\color{ccr}75\%$$ of cabs in the city, leaving $$\color{ccb}25\%$$ to the blue company. One day, a hit-and-run occurs. Local camera footage manages to see that a cab driver was the perpetrator. Unfortunately, the camera did not catch the logo on the cab, the only distinguishing feature between the two companies. Because of your vast statistics knowledge, you've been chosen to act as a judge on this case. Which company do you believe should be fined5? Most likely, guilt lies with the red company, simply due to chance:
This sets up our prior. Right now, we only know the following:
\begin{equation*} \begin{array}{rccl} \boldsymbol{P}(\color{ccb}blue\color{def}) & = 0.25 & \rightarrow & \, \scriptstyle\text{The probability that the cab belonged to the blue company.} \\ \boldsymbol{P}(\color{ccr}red\color{def}) & = 0.75 & \rightarrow & \, \scriptstyle\text{The probability that the cab belonged to the red company.} \end{array} \end{equation*}
Of course, the story doesn't end there. Not wanting to pay hefty fees, the red company produces a witness that says he saw the blue company logo. You check the camera footage and notice that this witness was indeed nearby when the crime happened. The blue company, aware that eyewitnesses can be unreliable, proposes a test to see if the witness can correctly differentiate company logos when a cab drives past him. The witness takes the test and, much to the red companies' chagrin, it's found he can only identify the correct logo $$\color{ccw}60\%$$ of the time.
Ok judge, which company do you suspect now? Well, let's look at the data we just got:
\begin{equation*} \begin{array}{rccl} \boldsymbol{P}(\color{ccw}witness \color{def} \text{ says } \color{ccb}blue \color{def}\,|\,\color{ccb}blue\color{def}) & = 0.60 & \rightarrow & \, \scriptstyle\text{The witness is correct.} \\ \boldsymbol{P}(\color{ccw}witness \color{def} \text{ says } \color{ccb}blue \color{def}\,|\,\color{ccr}red\color{def}) & = 0.40 & \rightarrow & \, \scriptstyle\text{The witness is incorrect.} \end{array} \end{equation*}
The testimony represents a single data point that we can use to modify the base rate of either $$\boldsymbol{P}(\color{ccb}blue\color{def})$$ or $$\boldsymbol{P}(\color{ccr}red\color{def})$$. Looking at the probability space can give us some good intuition about the case:
Since the witness testified against the blue company, we know we're in the striped region. Just looking at the picture, the red company remains the most likely suspect: the area of the lower-right striped region (culprit was red and the witness is wrong) looks larger than the upper-left striped region (culprit was blue and witness is right).
Of course, we need to test this out rigorously. Let's try to find the posterior probability for blue (we could have also chosen to calculate the posterior for red, I just made a random choice). To find $$\boldsymbol{P}(\color{ccb}blue\color{def} \,|\, \color{ccw}witness\color{def} \text{ says } \color{ccb}blue\color{def})$$ we use Bayes' Theorem:
\begin{equation*} \begin{array}{rcl} posterior & = & \dfrac{likelihood}{evidence} \cdot prior \\ \boldsymbol{P}(\color{ccb}blue\color{def} \,|\, \color{ccw}witness\color{def} \text{ says } \color{ccb}blue\color{def}) & = & \dfrac{\boldsymbol{P}(\color{ccw}witness\color{def} \text{ says } \color{ccb}blue\color{def} \,|\, \color{ccb}blue\color{def})}{\boldsymbol{P}(\color{ccw}witness\color{def} \text{ says } \color{ccb}blue\color{def})} \cdot \boldsymbol{P}(\color{ccb}blue\color{def}) \end{array} \end{equation*}
The only term that we don't immediately know here is the evidence, the probability that the witness says blue. However, we can easily figure this out, we just need to consider all the situations that can lead to the witness saying blue. There are two, each represented by one of the striped regions in the probability space:
We arrive at a pretty expected answer:
\begin{equation*} \boldsymbol{P}(\color{ccb}blue\color{def} \,|\, \color{ccw}witness\color{def} \text{ says } \color{ccb}blue\color{def}) = \dfrac{(0.60)(0.25)}{(0.60)(0.25) + (0.40)(0.75)} = 0.333 \end{equation*}
So, even with the testimony, we find that most guilt still lies with the red company. The chances of the culprit being blue are only $$1$$ in $$3$$.
But wait, just as you're about to pass your judgment, someone bursts into the courtroom6. It's a second witness, who also says he saw a blue logo. And behind him? There's a third witness, and a fourth, and a fifth, ...
Each witness has their own story, ready to supply evidence for either the red or blue company. They all took the test and each of them also got a $$\color{ccw}60\%$$ (really blurry logos)7. If running out of the courtroom while screaming is not an option, how can you handle this situation? Will Bayes' Theorem stop working now that we have multiple data points? Of course not! For example, if we have two witnesses and both say they saw blue, we could calculate:
\begin{equation*} \begin{array}{rcl} posterior & = & \dfrac{likelihood}{evidence} \cdot prior \\ \boldsymbol{P}(\color{ccb}b\color{def} \,|\, \color{ccw}w_1\color{def} \text{ says } \color{ccb}b\color{def}, \color{ccw}w_2\color{def} \text{ says } \color{ccb}b\color{def}) & = & \dfrac{\boldsymbol{P}(\color{ccw}w_1\color{def} \text{ says } \color{ccb}b\color{def}, \color{ccw}w_2\color{def} \text{ says } \color{ccb}b\color{def} \,|\, \color{ccb}b\color{def})}{\boldsymbol{P}(\color{ccw}w_1\color{def} \text{ says } \color{ccb}b\color{def}, \color{ccw}w_2\color{def} \text{ says } \color{ccb}b\color{def})} \cdot \boldsymbol{P}(\color{ccb}b\color{def}) \end{array} \end{equation*}
Here, I used a shorthand $$\color{ccb}blue\color{def} \leftrightarrow \color{ccb}b\color{def}$$ and $$\color{ccw}witness\color{def} \leftrightarrow \color{ccw}w\color{def}$$ to save space. So clearly, Bayes' Theorem can handle any amount of data, we simply treat the two (or potentially more) data points as one event. To make this even more clear, we can rewrite our data $$(\color{ccw}w_1\color{def} \text{ says } \color{ccb}b\color{def}, \color{ccw}w_2\color{def} \text{ says } \color{ccb}b\color{def})$$ as an ambiguous $$\color{ccw}d\color{def}$$:
\begin{equation*} \begin{array}{rcl} \boldsymbol{P}(\color{ccb}b\color{def} \,|\, \color{ccw}d\color{def}) & = & \dfrac{\boldsymbol{P}(\color{ccw}d\color{def} \,|\, \color{ccb}b\color{def})}{\boldsymbol{P}(\color{ccw}d\color{def})} \cdot \boldsymbol{P}(\color{ccb}b\color{def}) \end{array} \end{equation*}
Unfortunately, I won't be able to draw probability spaces (like the ones I drew above) to explain the actual calculations that go on here. This stems from the fact that, to keep each data point independent, we would need to move from 2D into 3D, 4D, 5D, etc. We can, however, demonstrate the power of this approach by generating actual samples. Take a look at the following graph, which shows the outcome of using Bayes' Theorem with a LOT of witnesses:
I randomly generated the blue line above by assuming that blue was the real culprit. This would mean that $$60\%$$ of the witnesses would testify against blue, so that if we have $$10$$ of them, $$6$$ (on average) would say they saw a blue logo. I randomly generated the red line by assuming the opposite: the red company was the culprit. This means that $$60\%$$ of the witnesses testified against the red company. The y-axis represents our posterior in every case.
Looking at the graph, it seems that within $$\color{ccw}20$$ witnesses, we had about $$99\%$$ confidence we knew the true culprit. This number would have been lower if each witness could identify the logos at a better rate than $$\color{ccw}60\%$$8.
Notice also, that our data eventually overrode the prior. Even though we initially believed that red was the culprit in both cases (the prior), we always end up believing in the correct outcome once we got enough information. This is an important aspect of the Bayesian approach that I hope to flesh out later: priors can skew posteriors when we have low amounts of data, but the posterior always converges to the same answer.
### Moving Onto the Harder Stuff
If you care about using data to try and tease out answers, hopefully you see how fundamental Bayes' Theorem is. You can take initially available information, encode it in a prior, and modify it through data. But how is this different than what statisticians have been doing for decades using classical estimators and p-values? And what if we don't know what prior to pick? Are we out of luck?
Discovering the connections between Bayes and the Frequentist (classical) approach and developing those consequences takes a bit more mathematical know-how than what I've used so far. If you're curious and you have a bit of a background in statistics already, read on to More On Bayesian Statistics.
### Resources
#### Textbooks
The credit for a lot of the examples in this post doesn't belong to me. I took several examples on this page from two sources:
1. David MacKay's book Information Theory, Inference, and Learning Algorithms. Both the textbook and the lectures associated with them are free and available online with a simple search. Links for the lazy: book and lectures.
2. Andrew Gelman's book Bayesian Data Analysis. Not only is this text well written, it also contains great one-lines like9:
• "As you know from teaching introductory statistics, 30 is infinity."
• "Why is it Normal? Because that’s the only continuous multivariate distribution we have. Oh, we have the multivariate $$t$$ ... as if that’s a different distribution.”
#### Footnotes
[8] Luckily, it wasn't $$50\%$$! I leave it as an exercise for you to imagine what would happen in that case. Hint, do the calculation for one data point assuming a $$50\%$$ reliable witness.
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How do you say Xi (Greek)? Do we use this letter in order to imply the level of difficulty in the topic it appeared? Sign in to disable ALL ads. Greek alphabet letters and symbols. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. Omicron is not at all used in math and science, but part of Greek alphabet. can "has been smoking" be used in this situation? Xi or Csi (uppercase Î, lowercase ξ) is the 14th letter of the Greek alphabet and it has the value 60 in the Greek number system. The lowercase letter is also used as a symbol for the initial mass function and the correlation function in astronomy. This means that you will have to possess a somewhat thorough knowledge of the language to be able to write it. Learn the LaTeX commands to display the greek alphabet. Nancy Gaifyllia is a freelance writer who loves to cook and eat Greek food. Online Greek keyboard to type characters and diacritics of the Ancient Greek alphabet. This application allows you to write in polytonic greek. Welcome! There is even a hint feature. Illustrator CS6: How to stop Action from repeating itself? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I think, in all my exam papers, I'll simply write at the top or something, and proceed from there :P. The arrows show you where to start when you write Greek letters. Little Greek 101: The Greek Alphabet . The lowercase letter chi (χ) is used to represent the voiceless uvular fricative in the International Phonetic Alphabet. … Write It! Some people prefer to teach this pronunciation for New Testament Greek as well. Xi (c): This is the same sound as "ch" in "Bach", which does not sound like "ch" in "chair". Of course, if you’re writing an article about Greece, this might come in handy. Features ----------- iOS 8 custom keyboard provider. I want to write one row in one cell, 2nd row in 2nd cell, 3rd row … For instance, the code for β is 03B2, so to print β the command is print('\u03B2').. How to insert other symbols and templates in an equation, see Shortcut keys for inserting symbols and templates into the equation. This is the list of Greek alphabet letters. Capital: Low-case: Greek Name: English: Alpha: a: Beta: b: Gamma: g: Delta: d: Epsilon: e: Zeta: z: Eta Click on the image of the alphabet to the left to print a practice page. Not that I am aware of, but $\xi$ is often used to correspond to $x$, like one would use $\alpha$, $\beta$ and $\gamma$ to correspond to $a$, $b$ and $c$. What does this sentence of Greek means in the book of Modern Fourier Analysis? That’s right, the modern Greek script is a combination of Cyrillic, Greek alphabet, and some modern Latin script. I hate this letter $\xi$. I think, in all my exam papers, I'll simply write at the top or something, and proceed from there :P. I realize this may sound stupid, but I can't get this question right to save my life because I hit asterisk and it gives me a dot. Contributions; Badges × MATLAB Answers. Anyway, I always wanted to ask how to draw this letter. Ban "xi" from the greek alphabet. A rendered preview of all letters is shown alongside all commands in a nice table. Xi χ): This is the ... Modern Greek pronunciation. Daily POP Crosswords features the best pop-culture-themed puzzles from the top puzzle constructors, including many from Dell Magazines and Penny Press, the #1 crossword-puzzle-magazine publisher. The help on inserting Greek letters and special symbols is also available in Help menu. Living in Greece for over 30 years, she explored its regional specialties. Greek Alphabet and its English Equivalents . I find it beautiful, you do not, end of story. Let’s get to 2,500! Just a precaution. This greek letter is also mentioned as a symbol in Plato's "Timaeus" and Thomas Browne's "The Garden of Cyrus" too. Greek alphabet / letters in LaTeX Learn the LaTeX commands to display the greek alphabet. do I keep my daughter's Russian vocabulary small or not? It certainly doesn't say anything about the difficulty level of the topic involved. Use a computer to generate three full lines of $\xi$, then print it, and start tracing until you get the hang of it. \xi \varphi \theta \psi \gamma \vartheta \zeta \eta \rho: See Shortcut keys for inserting symbols and templates into the equation to find other frequently used symbols. DeTraci Regula is a freelance writer who has specialized in Greek travel and tours for 18 years. The usage is pretty easy, you can basically type the name of the letter and put a backslash in front of it. Tip: See my list of the Most Common Mistakes in English.It will teach you how to avoid mistakes with commas, prepositions, irregular verbs, and much more. When you create a presentation for a graduate thesis or dissertation, often, you need to insert Greek letters into the slide. I like being able to be quizzed also. The uppercase letter xi (Î) is used to represent the cascade particle and xi baryon in physics. I initially learned the modern Greek pronunciation, but had difficulty learning to spell words, so I switched to the Erasmian. Write … Just a precaution. our editorial process. Have a fact about Xi (Greek) ? I just hope you don't take the same attitude if someone replace 'this letter' with the word 'probability'. Xi or Csi (uppercase Ξ, lowercase ξ) is the 14th letter of the Greek alphabet and it has the value 60 in the Greek number system. The lowercase xi letter (ξ) is used to denote eigenvectors, a universal set, the killing vector, state price density, spatial frequency, temporal frequency, the extent of reaction, etc. I am using ggplot2, and it works beautifully with the data. I dont understand why we have to use this symbol. I don't mind Xi. By default, MATLAB ® supports a subset of TeX markup. And also having Greek letters & symbols pronunciations to spell very well.. All of them derived from the earlier Phoenician alphabet. Not that I am aware of, but $\xi$ is often used to correspond to $x$, like one would use $\alpha$, $\beta$ and $\gamma$ to correspond to $a$, $b$ and $c$. The $\LaTeX$ commands for the Greek alphabet. An online LaTeX editor that's easy to use. Same as Roman E. An alternative way to draw this letter is like a c with a horizontal line in the middle (like this: є, or just as the math symbol for “belongs” — for whomever is familiar with that — but in small size). Listen to the audio pronunciation of Xi (Greek) on pronouncekiwi. I don't mind Xi. Minuscule is the basis for modern Greek handwriting. Like Latin, Greek is an ancient language still used after many centuries by scholars. The dot is then thought of as multiplication, and because nothing follows it, the question cannot be graded. Part 2 of this series teaches you to write the minuscule (lowercase) letter variants. If you see utf-8, then your system supports unicode characters.To print any character in the Python interpreter, use a \u to denote a unicode character and then follow with the character code. Could you please (...) help me to like this letter? rev 2021.1.15.38327, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. That’s right, the modern Greek script is a combination of Cyrillic, Greek alphabet, and some modern Latin script. → Conversion Ancient Greek > Latin alphabet → Transliterated Greek keyboard to type a text with the latin script → Ancient Greek language: dictionary, pronunciation, grammar → Greek alphabet → Online test to learn to recognize the Greek letters → Modern Greek keyboard → Multilingual keyboard: index Help. Xi. And also having Greek letters & symbols pronunciations to spell very well.. Watch this video Biblical Greek Alphabet Pronunciation to learn how the letters should be pronounced.. Upper case appears in the beginning of a paragraph, direct speech, proper names, geographical locations and names of nations. Pi is used as Peltier coefficient and ratio of circumference to its diameter valued at 3.1416. Please subscribe to my channel as I will be posting many more greek language videos for beginners. In modern Greek, zeta sounds like the "z" in the English word "zone." Traveling in a foreign land can be stressful, especially if you're going alone and don't speak the language. Greek letter between xi and pi. Here is how to draw all greek letters: +1 for your last point, though how you or the op has relatively strong feeling to a letter is beyond me. How to draw greek letters on paper / blackboard? Greek alphabet letters are used as math and science symbols. On our website you will find all the today’s answers to Daily POP Crosswords. What's the word for a vendor/retailer/wholesaler that sends products abroad. The Greek alphabet has 24 letters to write the Greek language. To find out more about the cookies we use, see our Privacy Policy. How do I draw a conformal mapping from the z-plane to the w-plane. Rho is used as reflection coefficient, reflection factor, resistivity, and surface density of charge. Are there any games like 0hh1 but with bigger grids? A table of the HTML 4 entities for symbols and Greek letters. 11% × Knowledgeable Level 2 MATLAB Answers. In ancient times, some local forms of the Greek alphabet used the chi instead of xi to represent the /ks/ sound. These include uncial, a system for connecting capital letters, as well as the more familiar cursive and minuscule. This website uses cookies to collect information about how you interact with our website. This is the list of Greek alphabet letters. Is it possible to mount associated path to WSL? Should I first draw an $\epsilon$ and make a curly tail on the bottom of it, or sholud I draw an $S$ with some curly $C$ on top of it? The three that require two strokes are τ, χ, and ψ. How to Study Greek. No one likes xi. Press the key which sounds like the Greek symbol you want to type. My categories need to be named with Greek letters. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. our editorial process. The Greek alphabet has 24 letters. This has been bothering me for years :D. Well let me tell you about a great invention called google! The capital letter (Î) is used in statistical mechanics and particle physics. should I draw an S with some curly C on top of it? The Wolfram System includes rendering of both ordinary and variant Greek letters in all its standard fonts. And I can't copy-paste the … How to write the Ancient Greek alphabet. Thank you for an excellent learning too. When written quickly, the angle at the top-left usually comes out as a curve. ... How to write Greek letters. But Zeta I just cannot get myself to write in a consistent readable manner that makes it recognisable AS zeta. It's just an elaborate small case epsilon. The capital letter (Ξ) is used in statistical … You should be able to write all but three of them with a single continuous stroke. Currently popular pronunciations. Here you can copy Greek letters, Greek symbols, and their English names in just one click. Greek alphabet letters & symbols. The following table shows the whole Greek alphabet along with the commands in a nice table. It only takes a minute to sign up. You are on the Xi letter symbol page of the Greek Alphabet letters. Updated January 29, 2020 Whether you are traveling to Greece, enjoy eating in a local Greek restaurant, or just a curious person, it can be educational and helpful to know some Greek. Think of making a "t" sound followed immediately by a separate hard "h" sound, as in the English phrase "hot head." This website is designed to copy the Greek alphabet quickly. Print a conversion table for (un)signed bytes, On the collision of two electrons in a particle accelerator. The lower case letter ksi/xi (ξι), the 14th letter of the modern Greek alphabet. How to draw the Greek letter $\xi$ [closed]. No, it is better to draw it in one stroke. Greek letters are widely used in mathematics and other fields of science. List of Greek Alphabet Letters. The Wolfram Language allows Greek letters to be fully integrated into symbol names, strings, and graphics\[LongDash]and to be entered from palettes or using keyboard shortcuts. Xi is used as output coefficient. Greek alphabet and latex commands [not a question]. I hate it too! Chad the math man started this petition to The entire multiverse. REPUTATION 44. MATLAB Central contributions by xi. A rendered preview of all letters is shown alongside all commands in a nice table. Contents. The letter is also used as a symbol in Harish-Chandra's Î function too. I am going to show you how to change language in Adobe Acrobat in a such a simple way. I ususally do the epsilon-with-a-tail approach, but it always looks awful, Nice question! A page containing the Greek alphabet in LaTeX for reference. Updated June 07, 2019 Traveling in a foreign land can be stressful, especially if you're going alone and don't speak the language. Add fact ! 24% × First Answer MATLAB Answers. How to say Xi in Greek? Pronunciation of Xi with 1 audio pronunciation, 1 meaning, 5 translations and more for Xi. How to ask to attach the plots vertically in the given code? Same as Roman Z. Practice time is enjoyable. Θ/θ (theta): In ancient Greek, a more aspirated version of the "t" sound in the English word "top." I mean there are many greek letters easier to draw. And the lowercase xi (ξ) represents the original Riemann Xi function, the damping ratio and the extent of a chemical reaction. Do we use this letter in order to imply the level of difficulty in the topic it appeared? Answered I have 10 rows and four columns in variable two. List of Greek Alphabet Letters. This keyboard works great to type polytonic Greek on an iPad (very tight and very small keys for iPod and iPhone). Here's what they look like, how the name of the letter is pronounced, and how the letter sounds when spoken. It is fun to write the letter and be corrected. In this video. And I can't copy-paste the … If this glitch was fixed, this keyboard would have (in my opinion) no problems for iPad. January 10, 2018. What was the name of this horror/science fiction story involving orcas/killer whales? How to Study Greek. It's just an elaborate small case epsilon. RANK 958. One way, which I don't personally use but which works all the same, is to draw a lower-case $e$ (or even $\varepsilon$ I guess) and, without taking your pen off the paper, draw a lower-case $s$ directly underneath it. In both ancient and modern languages, the letter is pronounced as "ksi". How to use all these symbols outside the equation, select the option Use Math AutoCorrect rules outside of math regions in the Word Options.How to do it, see Choosing Math AutoCorrect options. Could you please enlighten me about this and help me to like this letter? Sorry but no I could not. Twice your pen should stop and go backwards, as is done once when drawing $\varepsilon$. Of course, if you’re writing an article about Greece, this might come in handy. Derived from the North Semitic alphabet, the Greek alphabet was modified to make it more efficient and accurate for writing a non-Semitic language. Nancy Gaifyllia. deTraci Regula. Here are the modern greek alphabet letters handwritten. You could use a raw string: r"$\xi$" which will not treat the \x as an escape for a unicode char. While the original Greek alphabet was written in all capitals, three different scripts were created to make it easier to write quickly. Why would a flourishing city need so many outdated robots? Awarded to xi on 04 Jul 2019. Note that some of the symbols require loading of the amssymb package. Or, you could escape the slash: "$\\xi$" It doesn't happen with other greek letters you've tried because they don't happen to start with a character used as a control code. Write it here to share it with the entire community. But Zeta I just cannot get myself to write in a consistent readable manner that makes it recognisable AS zeta. Greek Ἑλληνική . I find it beautiful, you do not, end of story. Greek alphabet. Provide your first answer ever to someone else's question. pronouncekiwi - How To Pronounce Xi (Greek) pronouncekiwi. List of Greek letters and math symbols. This table gives the Greek letters, their names, equivalent English letters, and tips for pronouncing those letters which are pronounced differently from the … Awarded to xi on 12 Oct 2019. Mathematical symbols and Greek letters are pervasive today and used everywhere, from physics to social science. Samekh or Simketh is the 15th letter of the Phoenician alphabet. Awarded to xi on 20 Jul 2017. The csi/xi came from the Phoenician language sÄmek, which meant fish. Instructions. Start from the end of the little thing top left and follow continuously the curve until the little thing bottom right. xi ‘x’ as in ‘axe’ (NOT as in ‘xylophone’) Ο ο omicron ‘o’ as in ‘lot’ Π π pi ‘p’ as in ‘pet’ Ρ ρ rho ‘r’ as in ‘rat’ Σ σ, ς sigma ‘s’ as in ‘cats’ or as in ‘dogs’ according to context: your intuition for English pronunciation will generally be correct for Greek too Τ τ tau conversion Greek dictionary. Typing Greek letters with Keyboard Shortcuts To insert Greek letter type Ctrl+G ( Command G on Mac OS ) and then type Latin letter mentioned in the table below. Unfortunately I cannot figure out how to put those greek symbols on the x axis (at the There are a couple of special characters that will combine symbols. In mathematics, the Greek letter chi is used as a symbol for characteristic polynomial or characteristic function. Sorry but no I could not. Greek Letters and Special Characters in Chart Text. If you are learning Greek in a classroom setting, your instructor may ask to see your work. Update the question so it's on-topic for Mathematics Stack Exchange. @Lost1 Your hope is fulfilled, this is an entirely different matter. Cyrillic. Xi Symbol in Greek Alphabet. Learn more about greek, alphabet, latex, tutorial Greek alphabet, writing system developed in Greece about 1000 BCE that became the ancestor of all modern European alphabets. The csi/xi came from the Phoenician language sāmek, which meant fish. 0 have signed. I realize this may sound stupid, but I can't get this question right to save my life because I hit asterisk and it gives me a dot. It is impossible to write, I mean which great mind thought it would be a good idea to turn a tornado into a letter, simply preposterous. What is the easiest way to draw trigonometric Graphs. Greek Letters The following table lists Greek letters and the corresponding Maple symbols that display as Greek letters in a worksheet. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Why do electronics have to be off before engine startup/shut down on a Cessna 172. Want to improve this question? 3 of your answers have been accepted. The dot is then thought of as multiplication, and because nothing follows it, the question cannot be graded. Should a gas Aga be left on when not in use? Samekh or Simketh is the 15th letter of the Phoenician alphabet. We use this information in order to improve and customize your browsing experience and for analytics and metrics about our visitors. Could you please (...) help me to like this letter? This means that you will have to possess a somewhat thorough knowledge of the language to be able to write it. Thank you for helping build the largest language community on the internet. In both ancient and modern languages, the letter is pronounced as "ksi". You also can use TeX markup to add superscripts, subscripts, and modify the text type and color. Greek numerals, also known as Ionic, Ionian, Milesian, or Alexandrian numerals, are a system of writing numbers using the letters of the Greek alphabet.In modern Greece, they are still used for ordinal numbers and in contexts similar to those in which Roman numerals are still used elsewhere in the West. Marking chains permanently for later identification, Children’s poem about a boy stuck between the tracks on the underground. Learn more about the Greek alphabet here. This is the way Greek is pronounced today in Greece. Why does my advisor / professor discourage all collaboration? With enough petitions, we can legally remove all "xi" found in the world. CONTRIBUTIONS 13 Questions 15 Answers. You can add text to a chart that includes Greek letters and special characters using TeX markup. For ordinary cardinal numbers, however, Greece uses Hindu–Arabic numerals Greek Learning the Greek alphabet with the letter name being sounded out is very helpful. This was borrowed into the early Latin language, which led to the use of the letter X for the same sound in Latin, and many modern languages that use the Latin alphabet. Like Latin, Greek is an ancient language still used after many centuries by scholars. Download Write Greek Polytonic and enjoy it on your iPhone, iPad, and iPod touch. The only glitch that I found was that I cannot place a rough breathing mark with an uppercase "Ρ" which some words need, though it does place a rough breathing mark on the lowercase "ρ."
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2021-12-01 12:37:11
|
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https://math.stackexchange.com/questions/1956394/harmonic-functions-with-positive-boundary-data
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# Harmonic functions with positive boundary data
Let $\Omega \subset \mathbb{R}^n$ be a open bounded domain, and let $g$ be a positive smooth function defined on $\partial \Omega$. Let $v$ be the unique harmonic function in $\Omega$ with boundary data $g$. Does $v$ necessarily need to be positive everywhere, or can it be negative?
• It has to be positive by the maximum principe. Indeed $\inf g\le v\le \sup g$. – user99914 Oct 6 '16 at 11:15
## 1 Answer
By the maximum principle, $v$ does indeed have to be positive. If it were non-positive, it'd have a strict local minimum not on the boundary, thus it wouldn't be harmonic.
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2020-01-19 10:33:51
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https://forum.allaboutcircuits.com/threads/noise-margin-for-cmos.83249/
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# Noise Margin for CMOS
#### zulfi100
Joined Jun 7, 2012
656
Hi,
Can some body plz guide me with the following question:
What is the high level noise margin of CMOS 5v series circuit?
I dont know whether i have to use a foprmula to solve this question or its just a fixed value which i have to remember. I have found some values in the following link:
http://www.interfacebus.com/Logic_Family_Noise_Margin.html
but it doesnt say about series or parallel.
Kindly guide me.
Zulfi.
#### WBahn
Joined Mar 31, 2012
27,478
The HI level noise margin is basically the difference between the lowest HI value that an output is guaranteed to produce and the lowest HI value that an input is guaranteed to recognize.
#### zulfi100
Joined Jun 7, 2012
656
Hi,
Noise Margine High=4.9-3.985=0.915v
Zulfi.
#### WBahn
Joined Mar 31, 2012
27,478
Hi,
Noise Margine High=4.9-3.985=0.915v
Zulfi.
I don't know where you are getting those numbers from. It isn't from the table on the page you linked to.
That page tells you exactly how to calculate the noise margin and tabulates the results for many different families. What, exactly, are you asking about that is not clear from that page?
#### zulfi100
Joined Jun 7, 2012
656
Hi,
Sorry for not clarifying you.
My question is:
What is the high level noise margin of CMOS 5v series circuit?
This is the question. I dont have any other data other than CMOS 5v series circuit. Is it possible to calculatye high level noise margin?
From the site i got following for CMOS 5v:
VOH VIH Margin VIL VOL Margin
4.9v 3.85v 1050mV 1.35v 0.1 1340mV
The HI level noise margin is basically the difference between the lowest HI value that an output is guaranteed to produce and the lowest HI value that an input is guaranteed to recognize.
I calculated noise margin by subtracting VOH & VIH i.e (4.9-3.85)
Thanks for your interest in my problem.
Zulfi.
Last edited:
#### WBahn
Joined Mar 31, 2012
27,478
I calculated noise margin by subtracting VOH & VIH i.e (4.9-3.85)
But look at your post #3:
Noise Margine High=4.9-3.985=0.915v
That you got a value different from the 1050mV given in that same table should have set off warning bells.
But it's clear you understand the concept, which is what is important.
#### zulfi100
Joined Jun 7, 2012
656
Zulfi.
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2022-12-08 03:42:53
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https://lifexsoft.org/index.php/product/overview-of-versions/lifex-v3-30-now-available
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LIFEx reads DICOM images locally or over a network using a DICOM browser, is compatible with Osirix and includes a powerful 3D reconstruction-based slice viewer. Volume of interest (VOI) can be either imported from previously created files or drawn and manipulated using LIFEx. Results are exported in Excel format files. LIFEx runs on Windows, MacOs and Linux. It is distributed with examples and includes a tutorial. User support is offered. Users can optionally contribute to the gathering of index values in different tissue types and different images as a public data bank of reference values is currently being built and integrated to the software for assisting the users with the interpretation of their results.
This release includes:
New software features:
- possible setting of the voxel size in which textural features are calculated in texture protocol
- management of multi-series with common ROI
File handling:
- PET images displayed in Bq/mL Units when conversion into SUV unit is not possible (missing Dicom fields)
- fixed bug: modification of problematic slash into PatientName DICOM field when saving results of texture protocol in an xls file
- fixed bug: oblique images now properly displayed when "orientation matrix » DICOM field is not null
Algorithms and utilities:
- refinement of the ROI position to calculate SUVpeak in PET: difference ≤ 0.01 between new SUVpeak compared to the one calculated in previous LIFEx regions
VOIs and paint:
- fixed bug: error in displaying ROI statistics when multi-series were loaded simultaneously
- fixed bug: Nestle ROI calculation was refined
- new tool: ROI erosion with defined voxels or spacing
- new tool: ROI dilation with defined voxels or spacing
- pencil2D tool: increased line width and added color at drag handle point. The Press "e" option was introduced to close (end handle point) ROI
Images and paint:
- change in maximum zoom factor of displayed image: images can be zoomed 4 times more
- addition of a low/high switch for quality rendering on images canvas with pressing the "s" key
- display of the position and value of the cursor for each loaded image series
Texture:
- addition of a saving option (xls format) of the diplayed histogram values
- texture script are no longer stalled when a MIP view has been displayed
- removal of the empty row between title and values in the xls texture files
- addition of the SeriesDate field in the xls texture files
Miscellaneous, fixed and checked issues:
- correct setting of the ROI when displayed on an image series different from the one that was used to create the ROI
- cleaning a some stalled information messages
Documentation:
- UserGuide \ Nestle Segmentation, p.33: V (threshold) is changed into T (threshold) for clarity
- UserGuide \ tools GUI: Addition of the dilate/erode tools help
- AppendixTexture: Addition of the multi-series management
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2022-07-04 22:19:58
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https://physics.stackexchange.com/questions/616242/conservation-of-angular-momentum-of-disk-and-block
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Conservation of angular momentum of disk and block
This problem appeared on the 2021 $$F=ma$$ Test that was held three days ago. However, I'm having trouble understanding other solutions.
Exact wording of problem:
"A uniform solid circular disk of mass $$m$$ is on a flat, frictionless horizontal table. The center of mass of the disk is at rest and the disk is spinning with angular frequency $$\omega_0$$. A stone, modeled as a point object also of mass $$m,$$ is placed on the edge of the disk, with zero initial velocity relative to the table. A rim built into the disk constrains the stone to slide, with friction, along the disk's edge. After the stone stops sliding with respect to the disk, what is the angular frequency of rotation of the disk and the stone together?"
Given answer: $$\omega_f=(1/2) \omega_0$$.
My solution: In order to apply conservation of angular momentum, we need to have a COR that stays consistent. So that means we should choose the center of mass of the system as our COR throughout the process. The beginning angular momentum with CM=COR would be $$(1/2+1/4)mR^2 \omega_0$$ by parallel axis theorem. And our final angular momentum with CM=COR should be $$\omega_f m(R/2)^2+(1/2+1/4)mR^2 \omega_f$$. This gives us $$\omega_f=3/4 \omega_0$$.
However, in some of the solutions other people have posted, they choose the center of the disk as the beginning COR and then the CM as the final COR. I'm pretty sure this is incorrect since the angular momentum depends on COR.
In the other solutions that do keep COR=CM throughout the process, they say that the beginning angular momentum is $$I_{center} \omega_0$$ instead of $$I_{cm} \omega_0$$. Shouldn't angular momentum be calculated with the moment of inertia of the COR as the center?
So far, I have no idea what is conceptually wrong with my solution, so any insights are appreciated.
This problem has a major difference from many similar textbook problems dealing with the conservation of angular momentum.
In many of the textbook problems the axis of rotation of the disk about its center of mass (CM) is fixed by a constraining force/torque and does not move. If that were the case here, the axis of rotation would not move when the stone is placed (and moves) on the disk. The problem would be addressed considering angular momentum of the disk/stone about the fixed axis of rotation.
Here, the disk is on a frictionless table and its axis of rotation can move, but is always perpendicular to the table surface. The problem is best evaluated considering the angular momentum about the center of mass of the disk/stone system. Denote the center of mass of the system consisting of the disk and the stone as CMS (center of mass of system). Due to conservation of linear momentum the CMS does not move. To keep CMS fixed as the stone moves (rotates on the disk), the CM of the disk moves and rotates about the CMS. Also, the disk rotates about its own CM, and the stone rotates about the CMS. Angular momentum is conserved. The overall motion is as quantified in an earlier answer by @ytlu.
A conceptual error: The parallel axis theorem is used to find the inertial moment about an rotating axis which is not the CM. In this problem, the disk is not rotating about the CM (for rock and disc joint system) but it rotates around the center of disk.
Therefore, it should be think in decomposing the motion into motion of the center of the disk around the CM, and the rotation around the center of disk.
$$L_{disc-total} = L_{disc-around-CM} + L_{disc-rotation-around-disc-center}.$$
Therefore, in the beginning, the rock and disk are both not moving w.r.t the CM (between rock and disk center:
$$L_i = 0 + 0 + \frac{1}{2} mR^2 \omega_0.$$
In the final motion, the rock is rotating around the CM, and the disc center is also rotating around the CM, in order to keep the center of mass fixed. Besides, the disc has additional rotation about the disc center. All these rotations are of a same frequence $$\omega_f$$ for syncronization.
$$L_f = L_{rock} + L_{disc} = L_{rock} + \left( L_{disc-center} + L_{disk-rotation} \right),\\ = \frac{1}{4}mR^2 \omega_f +\left( \frac{1}{4}mR^2 \omega_f + \frac{1}{2}mR^2 \omega_f\right) \\ = mR^2 \omega_f.$$
Then applying $$L_i = L_f$$, render
$$\omega_f = \frac{1}{2} \omega_0.$$
• Excellent explanation! Feb 21 at 21:26
• Contrary to many textbook problems, the axis of rotation of the disk about its CM is not constrained and it moves. I added a short answer to point this out. Feb 23 at 16:12
In order to apply conservation of angular momentum, we need to have a COR that stays consistent.
I agree that this seems like it should be true, but in reality it's not what the conservation of angular momentum is. The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occur.
If you look at the object (rock-disk system), if you choose any non-fixed point you will be introducing linear momentum into the equation, which you cannot do for free (i.e without subtracting from angular momentum)
• I'm not following your logic in the first paragraph, could you explain it further? Angular momentum is definitely conserved but it still changes with COR. As an analogy, take linear momentum. Linear momentum is conserved in any inertial frame if there aren't external forces, but the linear momentum between two inertial frames aren't equal (ie. a a frame with v=0 m/s and a frame with v=100m/s) I agree with the second paragraph, the COR has to be inertial to apply conservation of linear momentum. That's why I'm choosing the CM to calculate my linear momentum. Feb 21 at 18:57
In order to apply conservation of angular momentum, we need to have a COR that stays consistent
As senor o has answered this is not true the only condition for angular momentum to be conserved is that there should be no external torque.
This conceptual error made you say this:
The beginning angular momentum with CM=COR would be $$(1/2+1/4)mR^2 \omega_0$$
What you did here is calculated angular momentum about an axis the disc does not even rotate about, this should seem wrong to you. The $$I$$ in the angular momentum formula $$I\omega$$ is the moment of inertia about the axis of rotation. Not the axis of future rotation.
Shouldn't angular momentum be calculated with the moment of inertia of the COR as the center?
I don't really understand this statement, are there any further queries?
• Yes, I see where I made error calculating the beginning angular momentum. But now, how are we supposed to equate the angular momentum of one axis of rotation to the angular momentum of a different axis of rotation? I think this question is what lead me to make my conceptual error. Equating those two angular momentum seems wrong to me, since the same analogy applied to linear momentum is not true either. Feb 21 at 19:08
• The axis of rotation need not be the same for angular momentum to be conserved, For example, take a spinning disc and send it flying in space the axis is continuously moving. Would you still say angular momentum is not conserved? Physics can be quite non-intuitive sometimes. Your analogy to linear momentum tells me you treat angular momentum as an ideal vector. It is not. More on that. Feb 22 at 1:45
Exact wording of problem:
"A uniform solid circular disk of mass $$m$$ is on a flat, frictionless horizontal table. The center of mass of the disk is at rest and the disk is spinning with angular frequency $$\omega_0$$. A stone, modeled as a point object also of mass $$m,$$ is placed on the edge of the disk, with zero initial velocity relative to the table. A rim built into the disk constrains the stone to slide, with friction, along the disk's edge. After the stone stops sliding with respect to the disk, what is the angular frequency of rotation of the disk and the stone together?"
I agree with you, of course, that in the final state the combined object will be rotating around the Common Center of Mass.
I restate the problem as follows: initially the stone is sliding frictionless along the rim. Then it encounters a stop and there is an instantaneous inelastic collision.
That is, I'm explicitly stating that whether or not there is a friction phase does not change the outcome. (I'm guessing you are already proceeding that way, implicitly.)
In the initial state the stone is stationary with respect to the coordinate system, so it has zero angular momentum with respect to any point of that coordinate system.
At the instant that the instantaneous inelastic collision occurs both objects undergo a symmetric shift of their axes of angular velocity.
We can treat the initial angular velocity of the stone as an angular velocity of zero around its own center of mass.
At the instant of collision both axes shift, towards each other, so that in the final state the two axes coincide.
So: my proposal is to capitalize on symmetry. Treat the problem as a problem with initially two axes of rotation. Both axes of rotation shift, towards each other, and come to coincide. During that shift of axes of rotation the two objects extert a torque upon each other. As a matter of principle: around the CCM those two torques are equal and opposite.
Conversely, it would be very tricky to attribute an angular momentum to the disk around the future Common Center of Mass, since in the initial state the disk isn't actually rotating around that axis. Those two axes have an instantaneous velocity with respect to each other that would need to be accounted for.
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2021-10-27 00:35:06
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https://fe-physics.eu/?page_id=96
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# Magnetic Flux
The magnetic flux is proportional to the number of
field lines that pass through the surface.
Φ = Bn A
• Φ magnetic flux in Wb (Weber). Also : T m2 of Vs
• Bn the component of B perpendicular to the surface in T(esla)
• A area of the surface in m2
Example
See the drawing above.
B = 1.5 x 10-3 T
A = 500 cm2
α = 40 o
Resolve B in both components and calculate Bn
sin α = Bn/B
Bn = B sin 40 o = (1.5 x 10-3)( 0.643) = 9.642 x 10-4 T
Φ = Bn A
Φ = (9.642 x 10-4 )( 500 x 10-4)= 4.82 x 10-5 Wb
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2022-11-28 03:58:33
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http://www.sawaal.com/probability-questions-and-answers/in-a-lottery-there-are-10-prizes-and-25-blanks-a-lottery-is-drawn-at-random-what-is-the-probability-_3741
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0
Q:
# In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize ?
A) 1/7 B) 2/7 C) 3/7 D) 2/5
Explanation:
P(getting prize) = 10/ (10 + 25) =2/7
Q:
In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?
A) 15 B) 18 C) 20 D) 24
Explanation:
We know that, Total probability = 1
Given probability of black stones = 1/4
=> Probability of blue and white stones = 1 - 1/4 = 3/4
But, given blue + white stones = 9 + 6 = 15
Hence,
3/4 ----- 15
1 ----- ?
=> 15 x 4/3 = 20.
Hence, total number of stones in the box = 20.
2 95
Q:
What is the probability of an impossible event?
A) 0 B) -1 C) 0.1 D) 1
Explanation:
The probability of an impossible event is 0.
The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.
The probability of a certain event is 1.
6 561
Q:
In a box, there are four marbles of white color and five marbles of black color. Two marbles are chosen randomly. What is the probability that both are of the same color?
A) 2/9 B) 5/9 C) 4/9 D) 0
Explanation:
Number of white marbles = 4
Number of Black marbles = 5
Total number of marbles = 9
Number of ways, two marbles picked randomly = 9C2
Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2
= 1/6 + 5/18
= 4/9.
6 832
Q:
A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red?
A) 2/3 B) 1/8 C) 3/8 D) 3/4
Explanation:
Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
12 787
Q:
Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?
A) 1/4 B) 1/6 C) 1/8 D) 4
Explanation:
Required probability is given by P(E) =
13 1230
Q:
14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.
A) 11/379 B) 21/628 C) 24/625 D) 26/247
Explanation:
Total no of ways = (14 – 1)! = 13!
Number of favorable ways = (12 – 1)! = 11!
So, required probability = $\left(\frac{\left(\mathbf{11}\mathbf{!}\mathbf{×}\mathbf{3}\mathbf{!}\right)}{\mathbf{13}\mathbf{!}}\right)$ = $\frac{39916800×6}{6227020800}$ = $\frac{\mathbf{24}}{\mathbf{625}}$
15 1281
Q:
Two dice are rolled simultaneously. Find the probability of getting the sum of numbers on the on the two faces divisible by 3 or 4?
A) 3/7 B) 7/11 C) 5/9 D) 6/13
Explanation:
Here n(S) = 6 x 6 = 36
E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}
=> n(E)=20
Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.
27 1554
Q:
A person starting with 64 rupees and making 6 bets, wins three times and loses three times, the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is:
A) A gain of Rs. 27 B) A loss of Rs. 37 C) A loss of Rs. 27 D) A gain of Rs. 37
Explanation:
As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order).
The overall resultant will remain same.
So final amount with the person will be (in all cases):
64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27
Hence the final result is:
64 − 27 37
A loss of Rs.37
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2018-06-24 01:27:42
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https://itprospt.com/num/9383694/juctlionquaslon-1j0-0sold-corbon-toact8vw-kolueppovelonmikyurerjon
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5
# JuctlionQuaslon 1J0/ ?0Sold corbon toact8Vw KOlueppovelonmiKyurerjon [email protected] miihnmluaMacBock Ar]Aen...
## Question
###### JuctlionQuaslon 1J0/ ?0Sold corbon toact8Vw KOlueppovelonmiKyurerjon [email protected] miihnmluaMacBock Ar]Aen
Juctlion Quaslon 1J0/ ?0 Sold corbon toact8 Vw KOlue ppov elonmi Kyurerjon CUu @atom miihnmlua MacBock Ar] Aen
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2019-10-18 01:00:16
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http://math.stackexchange.com/questions/187429/confusion-with-partial-fractions-integration-of-frac11x2
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# Confusion with partial fractions integration of $\frac{1}{1+x^2}$
By partial fractions I've got $$\frac{1}{1+x^2}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)$$ Then integrating $$\frac{1}{2i}\int\left(\frac{1}{x-i}-\frac{1}{x+i}\right)dx=\frac{1}{2i}(\ln(x-i)-\ln(x+i))$$ If $x-i=R_1e^{i\phi_1}$ and $x+i=R_2e^{i\phi_2}$ then $$\frac{1}{2i}(\ln(x-i)-\ln(x+i))=\frac{1}{2i}(\ln(R_1)-\ln(R_2)+i(\phi_1-\phi_2))=\frac{1}{2}(\phi_1-\phi_2)$$ But now I put that $\phi_1=\arctan\frac{-1}{x}$ and $\phi_2=\arctan\frac{1}{x}$ so: $$\arctan(x)=^?\frac{1}{2}\left(\arctan\frac{-1}{x}-\arctan\frac{1}{x}\right)$$ I tried some values and looks false. The most confusing thing is that $$\frac{d}{dx}\frac{1}{2}\left(\arctan\frac{-1}{x}-\arctan\frac{1}{x}\right)=\frac{1}{1+x^2}$$(Something is happening with the argument function.) What's the mistake?
EDIT: Hans Lundmark wisely said in a comment there's only thing left: the constant $$\arctan(x)=\frac{1}{2}\left(\arctan\frac{-1}{x}-\arctan\frac{1}{x}\right)+\frac{\pi}{2}$$
Now the question is: how should I find this constant in future integrations?
-
Hint: Two primitives may differ by an additive constant. – Hans Lundmark Aug 27 '12 at 10:38
Looks like the constant is $\frac{\pi}{2}$. Thanks. – dot dot Aug 27 '12 at 10:39
Your error is in the second step: $$\frac{1}{2i}\int(\frac{1}{x-i}-\frac{1}{x+i})dx=\frac{1}{2i}(\ln(x-i)-\ln(x+i))\color{red}{+C}$$
When you do an indefinite integral, never forget the $+C$ part. $C$ here is an arbitrary constant. Also if you do the indefinite integral directly by knowing $\arctan'(x) = 1/(1+x^2)$ you still get $\int 1/(1+x^2)\,\mathrm dx = \arctan(x)+C$.
As how to find the constant: You don't. There's not a single constant which is the correct one for the indefinite integral. All constants are allowed (thus the $+C$). If you have a definite integral, you have $\int_a^b f(x)\,\mathrm dx = (F(b)+C)-(F(a)+C) = F(b)-F(a)$, so the constant simply cancels out.
Of course if you want a certain additional condition (such as, $F(0)=0$), then you can use that to fix the constant. But that's then due to an additional constraint of the solution, not due to the integral itself.
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2016-02-14 00:02:04
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https://openstax.org/books/intermediate-algebra-2e/pages/5-key-terms
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Intermediate Algebra 2e
# Key Terms
binomial
A binomial is a polynomial with exactly two terms.
conjugate pair
A conjugate pair is two binomials of the form $(a−b),(a+b).(a−b),(a+b).$ The pair of binomials each have the same first term and the same last term, but one binomial is a sum and the other is a difference.
degree of a constant
The degree of any constant is 0.
degree of a polynomial
The degree of a polynomial is the highest degree of all its terms.
degree of a term
The degree of a term is the sum of the exponents of its variables.
monomial
A monomial is an algebraic expression with one term. A monomial in one variable is a term of the form $axm,axm,$ where a is a constant and m is a whole number.
polynomial
A monomial or two or more monomials combined by addition or subtraction is a polynomial.
polynomial function
A polynomial function is a function whose range values are defined by a polynomial.
Power Property
According to the Power Property, a to the m to the n equals a to the m times n.
Product Property
According to the Product Property, a to the m times a to the n equals a to the m plus n.
Product to a Power
According to the Product to a Power Property, a times b in parentheses to the m equals a to the m times b to the m.
Properties of Negative Exponents
According to the Properties of Negative Exponents, a to the negative n equals 1 divided by a to the n and 1 divided by a to the negative n equals a to the n.
Quotient Property
According to the Quotient Property, a to the m divided by a to the n equals a to the m minus n as long as a is not zero.
Quotient to a Negative Exponent
Raising a quotient to a negative exponent occurs when a divided by b in parentheses to the power of negative n equals b divided by a in parentheses to the power of n.
Quotient to a Power Property
According to the Quotient to a Power Property, a divided by b in parentheses to the power of m is equal to a to the m divided by b to the m as long as b is not zero.
standard form of a polynomial
A polynomial is in standard form when the terms of a polynomial are written in descending order of degrees.
trinomial
A trinomial is a polynomial with exactly three terms.
Zero Exponent Property
According to the Zero Exponent Property, a to the zero is 1 as long as a is not zero.
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2020-10-22 06:56:22
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http://www.sciencemadness.org/talk/viewthread.php?tid=61783&page=14
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Sciencemadness Discussion Board » Fundamentals » Reagents and Apparatus Acquisition » Homemade and Repurposed Lab Gear Select A Forum Fundamentals » Chemistry in General » Organic Chemistry » Reagents and Apparatus Acquisition » Beginnings » Responsible Practices » Miscellaneous » The Wiki Special topics » Technochemistry » Energetic Materials » Biochemistry » Radiochemistry » Computational Models and Techniques » Prepublication Non-chemistry » Forum Matters » Legal and Societal Issues » Detritus » Test Forum
Pages: 1 .. 12 14
Author: Subject: Homemade and Repurposed Lab Gear
Morgan
International Hazard
Posts: 1369
Registered: 28-12-2010
Member Is Offline
Mood: No Mood
Here's a fused quartz sphere repurposed into a feeder. It's nice in that after you tap it to get the solution all the way down it stops dripping yet continues to feed unlike most feeders that drip off and on. I have a pinhole in the top of a coke bottle and inverted it will drip constantly every few seconds. Somehow air skirts in around the tiny hole. It's curious how some shapes won't work and others with 10 times as big a hole do. The nature of surface tension, air bubbles, gravity, and partial vacuums makes for some challenging riddles on how liquids flow.
You can make reverse images with it or start a fire too, another use as a lens.
Morgan
International Hazard
Posts: 1369
Registered: 28-12-2010
Member Is Offline
Mood: No Mood
Yesterday I fiddled around in the garage and came across some magnet clamps purchased from Enco and thrift store stands bought a few years back. One stand was from a fireplace utensil set where a stoker, broom, shovel and tongs hung from. Another some lamp base or such and another mystery stand.
The other parts are an old curtain rod I had with a curl or hook at the end, some V-shaped fixture that held a light bulb, and odd little square beam clamps bought at a habitat for humanity store that accepts 1/4 20 screws in 3 places and a smaller screw in another plus some holes - maybe not all that useful here, but an afterthought.
The last parts are some fixtures or branch pole things for the magnet bases that are all iron or steel except for the knobs that sleeve over nuts underneath. In a way they are reminiscent of ring stand parts. Its curious how the outermost hole tightens first and then the inner, kind of unusual to get used to. The outermost holes are two sizes if you have rods/poles that match the diameters.
https://www.bhphotovideo.com/c/product/1084866-REG/platinum_...
This hodgepodge isn't all that great or ideal, but was a start cobbled together from dissimilar parts just to see what might evolve.
Now to find some fingers of some sort I guess ...
The small 3-legged stand is another object bought at the thrift store and an old ~12 inch diameter glass stoplight lens which is now a little bird bath.
[Edited on 21-9-2019 by Morgan]
earpain
Harmless
Posts: 6
Registered: 11-9-2019
Member Is Offline
Quote: Originally posted by yobbo II How would you know they are borosilicate? Yob
This is a question I have faced a lot in my days of DIY glasswork.
1. With tubing or bottles/flasks, if you hold the bottle up to the light above you, so that the walls of the bottle are seen along your line of sight, if you will see a green tint, it is soda-lime aka soft glass. If it still looks completely transparent with no color tint, it is likely boro.
Another test that is even more reliable:
Take two similar pieces of glass, one being the glass in question, the other being a piece that you know for sure is borosilicate. When i say similar I mean, ends are both rod with similar diameter, or ends are both tube(mouth) with similar diameter and similar thickness.
Rotate both over a propane/mapp gas or oxy/propane torch and fuse them together properly like you would normally when creating a weld.
If the mystery piece is not borosilicate, basically upon cooling EVERYTHING in or near the joint will crack, pretty dramatically.
wg48temp9
National Hazard
Posts: 253
Registered: 30-12-2018
Member Is Offline
Quote: Originally posted by Morgan Here's a fused quartz sphere repurposed into a feeder. It's nice in that after you tap it to get the solution all the way down it stops dripping yet continues to feed unlike most feeders that drip off and on. I have a pinhole in the top of a coke bottle and inverted it will drip constantly every few seconds. Somehow air skirts in around the tiny hole. It's curious how some shapes won't work and others with 10 times as big a hole do. The nature of surface tension, air bubbles, gravity, and partial vacuums makes for some challenging riddles on how liquids flow. You can make reverse images with it or start a fire too, another use as a lens.
There should be a law about that, using a fused quartz flask as a bird feeder should be criminal especially for sm members LOL
i am wg48 but not on my usual pc hence the temp handle.
wg48temp9
National Hazard
Posts: 253
Registered: 30-12-2018
Member Is Offline
Quote: Originally posted by earpain This is a question I have faced a lot in my days of DIY glasswork. 1. With tubing or bottles/flasks, if you hold the bottle up to the light above you, so that the walls of the bottle are seen along your line of sight, if you will see a green tint, it is soda-lime aka soft glass. If it still looks completely transparent with no color tint, it is likely boro.
I have re-purposed bulb growing glasses and coffee beakers because they looked looked clear with no green tint. I placed one of the bulb growing glasses containing a coolish solution on a temperature controlled hot plate (~80C) and it cracked across the base. That suggests its not boro.
A few days ago I purchased some glass coffee mugs from a charity shop. The first one I examined appeared clear but the second identical one had a very distinct green tint.
Expensive bottles of clear spirits tend to not to have significant green tints and decorative glass items tend to be clear almost no green tint.
You can probably be confident (>90%) that if there is a green tint its not boro. But if its clear its very weak evidence that it is boro.
i am wg48 but not on my usual pc hence the temp handle.
Morgan
International Hazard
Posts: 1369
Registered: 28-12-2010
Member Is Offline
Mood: No Mood
My repurposed synthetic ruby rod/hummingbird perch passed the test today. The rod is supported by 2 clear fused quartz tubes with blue silicone tubing and the male ruby-throated hummingbird is sipping from a borosilicate drinking straw. Only when he ruffles his feathers or the sun hits him just right can you see his iridescent ruby throat.
The perch was fashioned after seeing a simple design such as this. I happened to have an odd single red wooden ball in the garage which was fortuitous and the ruby rod bought on eBay years ago.
https://www.bestnest.com/bestnest/RTProduct.asp?SKU=SOE-SEHH...
https://www.sibleyguides.com/2011/09/the-basics-of-iridescen...
[Edited on 28-9-2019 by Morgan]
earpain
Harmless
Posts: 6
Registered: 11-9-2019
Member Is Offline
Quote: Originally posted by wg48temp9
Quote: Originally posted by earpain This is a question I have faced a lot in my days of DIY glasswork. 1. With tubing or bottles/flasks, if you hold the bottle up to the light above you, so that the walls of the bottle are seen along your line of sight, if you will see a green tint, it is soda-lime aka soft glass. If it still looks completely transparent with no color tint, it is likely boro.
I have re-purposed bulb growing glasses and coffee beakers because they looked looked clear with no green tint. I placed one of the bulb growing glasses containing a coolish solution on a temperature controlled hot plate (~80C) and it cracked across the base. That suggests its not boro.
A few days ago I purchased some glass coffee mugs from a charity shop. The first one I examined appeared clear but the second identical one had a very distinct green tint.
Expensive bottles of clear spirits tend to not to have significant green tints and decorative glass items tend to be clear almost no green tint.
You can probably be confident (>90%) that if there is a green tint its not boro. But if its clear its very weak evidence that it is boro.
Hmm, you may be indeed be correct, and I'll have to revise my system.
Candy thermometers. $3-$5 at the food store. There's no kitchen thermometer sold that has such a wide range of temps. I have bought them before, removed the metallic element fused to the glass on the bottom, along with the capillary tube that shows the temperature, and then worked the remaining wide, but very very thin walled tube.
If it is borosilicate, once I'm both flame working and turning it into various devices, or auxillary glassblowing devices(temporary handle, extension tube, etc.) - it is the WORST I've ever worked with. I have flameworked soft/flint glass more reliably than these thermometers.
I suppose there is an application in mind when a glassware is designed. All glass that is not quartz has something added to it mostly just to lower its working temperature, not everyone has Oxy-Acetylene or mega-kilns handy.
Perhaps the non-green tinted glass that cracked on you, cracked for reasons unrelated to what metals or salts were added to the glass 'stew'? Like with candy thermometers. Surely they must technically be borosilicate. Not sure why they are so finicky.
3 general types of clear glass(in terms of material):
soda-lime
borosilicate
And then there's a plethora of structural factors, and considerations for the items application. In general I would discourage chemists to repurpose or DIY glassware unless they also have some proper training in glasswork. Scientific glass is definitely the most versatile and robust.
wg48temp9
National Hazard
Posts: 253
Registered: 30-12-2018
Member Is Offline
Here is a pic of the two coffee glasses side by side. The green tint of the right hand one is distinct. Of cause although they looked identical and come with identical metal housings (removed for the pic) I am not certain they are. I also swapped the position of the glasses incase the green tint was from a nearby object.
This is weak evidence as I am assuming they are identical as they were purchased together. Apparently the the green color comes from iron impurities in the material used to make the glass. So these must have been made from different batches of material.
This particular green tint seems more emerald green than the dull green tint I usually see in bottle glass.
Here is a pic of two tall vases and a boro 2l measuring cylinder from Fisher Scientific (right lower). All three are about 0.8m tall. The left hand vase is dark with a hint of green, the top vase has a slight dull green tint and the boro cylinder is lighter with almost no color. I assume the vases are soda glass. All three appear clear when viewed from the side.
Sorry the pic is poor quality it was taken under fluorescent lighting in the bathroom having just rinsed off the dust.
PS I got lucky when I won the measuring cylinder on ebay no one else bid on it so it was very cheap. Its very impressive given it is almost a meter long. I am still trying to buy a stopper for it that does not cost more than I paid for it.
[Edited on 10/1/2019 by wg48temp9]
i am wg48 but not on my usual pc hence the temp handle.
Yttrium2
Flammable Solid
Posts: 521
Registered: 7-2-2015
Member Is Offline
Mood: Snarky like a shark!
Anyone know of where to find borosilicate bulbs/tubes?
Besides HID lighting -- for a diy flask
earpain
Harmless
Posts: 6
Registered: 11-9-2019
Member Is Offline
Quote: Originally posted by Yttrium2 Anyone know of where to find borosilicate bulbs/tubes? Besides HID lighting -- for a diy flask
Do NOT buy from Chemistry Equipment vendors.
Do not buy from coffee shops with their new boro glass straw trend(\$5 for 8in tube)
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2019-12-09 13:01:48
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http://mathhelpforum.com/statistics/281438-set-theory-find-p.html
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# Thread: Set Theory: find P(A)
1. ## Set Theory: find P(A)
P(AuB) = .7
P(AuB')=.9
Find P(A)
So we know everything outside of B is .9
We know A or B =.7
We want to find A.
I am kind of stuck here.
I think I need to know the outside
of A and B, which is P(AuB)' which is = A' intersect B'
2. ## Re: Set Theory: find P(A)
Your basic problem is that the given information, p(AuB)= 0.7 and p(AuB')= 0.9, is impossible. Since B and B' are disjoint, AuB and AuB' are disjoint so p(A)= p(AuB)+ p(AuB') but that sum is greater than 1.
3. ## Re: Set Theory: find P(A)
Originally Posted by HallsofIvy
Your basic problem is that the given information, p(AuB)= 0.7 and p(AuB')= 0.9, is impossible. Since B and B' are disjoint, AuB and AuB' are disjoint so p(A)= p(AuB)+ p(AuB') but that sum is greater than 1.
I think you mean AnB and AnB' are disjoint?? And P(A)=P(AnB)+P(AnB') ??
4. ## Re: Set Theory: find P(A)
I find these easier to do using a Venn diagram. See attachment.
5. ## Re: Set Theory: find P(A)
Originally Posted by math951
P(AuB) = .7
P(AuB')=.9
Find P(A)
While I agree that the use of Venn diagrams, see reply #4, are useful, here is another approach.
We know that $\mathscr{P}(A)=\mathscr{P}(A\cap B)+\mathscr{P}(A\cap B')$
\begin{align*}\mathscr{P}(A\cup B)&=\mathscr{P}(A)+\mathscr{P}(B)-\mathscr{P}(A\cap B) \\\mathscr{P}(A\cup B')&=\mathscr{P}(A)+\mathscr{P}(B')-\mathscr{P}(A\cap B')\\\mathscr{P}(A\cup B)+\mathscr{P}(A\cup B')&=2\mathscr{P}(A)+\mathscr{P}(B)+( B')-\mathscr{P}(A\cap B)-\mathscr{P}(A\cap B')\\0.7+0.9&=2\mathscr{P}(A)+1-\mathscr{P}(A)\\\mathscr{P}(A)&=0.6 \end{align*}
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2018-10-19 18:17:40
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https://forum.allaboutcircuits.com/threads/moore-finite-state-machines-in-programming.121444/
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# Moore Finite state machines in programming
#### Vindhyachal Takniki
Joined Nov 3, 2014
593
Explains use of Moore FSM in embedded. Who else use Moore FSM or any other FSM like Mealey?
2. I am used to superloop concept in which I design my own state, step by step. I had tried to put this Morre FSM in my ealier written code for ECG code. I found it extremely difficult by Moore FSM. Since there are so many tasks running, interconnections with ISR, multiple inputs , multiple outputs & other things, I got stuck. What is better method in these kind of situation.
#### Papabravo
Joined Feb 24, 2006
19,873
I have used both types on numerous occasions in hardware and in pure software. It is a very powerful technique.
#### JohnInTX
Joined Jun 26, 2012
4,713
FWIW I use state machines in virtually everything I write unless the chip supports a proper RTOS. After you get the hang of it, it makes coordinating many tasks much easier.
#### Vindhyachal Takniki
Joined Nov 3, 2014
593
Made a code from his course for street light.
Code:
enum
{
goN = 0U, waitN, goE, waitE
};
typedef const struct
{
uint32_t output;
uint32_t wait;
uint32_t next[4];
}street_light;
street_light sl_fsm[4] = {
{
0x21U,
100U,
{goN, waitN, goN, waitN},
},
{
0x31U,
100U,
{goE, goE, goE, goE},
},
{
0x41U,
100U,
{goE, goE, waitE, waitE},
},
{
0x51U,
100U,
{goN, goN, goN, goN},
},
};
{
uint32_t state = goN;
uint32_t input;
uint32_t output;
while(1)
{
output = sl_fsm[state].output; /* set tthe output */
wait_delay_us(sl_fsm[state].wait); /* delay for spcified time */
input = goN; /* input is goN for testing, otherwise input is from pins */
state = sl_fsm[state].next[input]; /* determine next state */
/* to remove compilre warming */
if(output)
{
__nop();
}
}
} /* function ends here */
#### WBahn
Joined Mar 31, 2012
28,191
Explains use of Moore FSM in embedded. Who else use Moore FSM or any other FSM like Mealey?
2. I am used to superloop concept in which I design my own state, step by step. I had tried to put this Morre FSM in my ealier written code for ECG code. I found it extremely difficult by Moore FSM. Since there are so many tasks running, interconnections with ISR, multiple inputs , multiple outputs & other things, I got stuck. What is better method in these kind of situation.
I prefer Moore machines, but sometimes a Mealy machine is simply a much better match to the application.
You can use an informal, ad-hoc design approach, which is often just find for small machines, or you can adopt a more formal, standardized approach, which can be needlessly cumbersome on small machines but can make the design of larger machines much more robust.
If doing it in software, FSM implementations lend themselves to switch() statements (if your language supports that). I find that I can implement the machine with more confidence if I set all of the relevant actions, including the next-state assignment, within every case. In and HDL this helps eliminate inferred latches, but in an MCU or other software it tends to slow down the machine because it has to do a lot of assignments that serve no purpose. But you can always go in and comment out the ones that you determine are truly redundant (I don't delete them because they serve as good documentation for what I expect those signals to have in that state.
#### xennar
Joined Jul 10, 2013
1
I'd been working on a table based state machine framework for Arduino when I read your discussion. The idea is that each state machine is its own object which can be shared like any Arduino library. It comes with a number of reusable state machines.
I'd be interested to hear your thoughts.
The code and doumentation are on Github:
https://github.com/tinkerspy/Automaton
https://github.com/tinkerspy/Automaton/wiki
Regards,
Tinkerspy
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2023-03-31 19:15:22
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https://math.stackexchange.com/questions/3476992/tangents-from-2-sqrt3-2-to-hyperbola-y2-x2-4-determine-a-chord-of-conta
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# Tangents from $(-2\sqrt3,2)$ to hyperbola $y^2-x^2=4$ determine a chord of contact subtending angle $\theta$ at the center. Find $12\tan^2\theta$.
Tangents are drawn from a point $$(-2\sqrt 3 ,2)$$ to the hyperbola $$y^2-x^2=4$$ and the chord of contact subtends an angle $$\theta$$ at center of hyperbola. Find the value of $$12 \tan^2 \theta$$.
My attempt:
The equation of chord of contact is $$\sqrt 3 x+y=2$$. Solving it with hyperbola we get the intersection points as $$(0,2)$$ and $$(2\sqrt 3,-4)$$. So calculating the angle gives me as $$\frac{\pi}{2} + \tan^{-1}{(\frac{2}{\sqrt 3})}$$. Which is wrong according to answer key. Where am I wrong?
• What is the answer according to the key?
– Blue
Dec 15 '19 at 12:42
• Answer is 9 according to answer key. Dec 15 '19 at 12:45
• Given your value of $\theta$, what is the corresponding value of $12\tan^2\theta$?
– Blue
Dec 15 '19 at 12:46
• Moreover we want to this without a calculator. Dec 15 '19 at 12:50
• Your calculation involves an inverse tangent, so taking the tangent isn't really a calculator exercise. (Indeed, you didn't really even have to find $\theta$ itself. You just need the value of $\tan\theta$ to complete the problem.)
– Blue
Dec 15 '19 at 12:52
$$\cos\theta = {(0,2)\cdot(2\sqrt3,-4) \over \lVert(0,2)\rVert \lVert(2\sqrt3,-4)\rVert} = {-8 \over 2 \cdot 2\sqrt7} = -\frac2{\sqrt7},$$ then use $$\tan^2\theta+1=\sec^2\theta$$ to obtain $$\tan^2\theta = \frac34$$. Since one of the points is on the $$y$$-axis, we can also compute $$\tan\theta$$ directly from the other point: $$\tan\theta = {2\sqrt3\over-4} = -\frac{\sqrt3}2$$. So, it appears that you’ve gotten a numerator and denominator swapped somewhere along the way.
As Blue noted in a comment, you don’t need to compute $$\theta$$ explicitly since you already have $$\tan^2\theta$$. Now, just multiply that by $$12$$.
• OP's numerator and denominator are swapped because the $\pi/2$ term effectively turns the inverse tangent expression into an inverse cotangent (and changes a sign). (BTW: since one ray of the angle coincides with the $y$-axis, the angle's trig values are easily calculated from the coordinates of the point $(2\sqrt{3},-4)$.)
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2022-01-17 01:42:13
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http://www.neverendingbooks.org/category/representations
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# Moonshine for everyone
Today, Samuel Dehority, Xavier Gonzalez, Neekon Vafa and Roger Van Peski arXived their paper Moonshine for all finite groups.
Originally, Moonshine was thought to be connected to the Monster group. McKay and Thompson observed that the first coefficients of the normalized elliptic modular invariant
$J(\tau) = q^{-1} + 196884 q + 21493760 q^2 + 864229970 q^3 + \ldots$
could be written as sums of dimensions of the first few irreducible representations of the monster group:
$1=1,~\quad 196884=196883+1,~\quad 21493760=1+196883+21296876,~\quad …$
Soon it transpired that there ought to be an infinite dimensional graded vectorspace, the moonshine module
$V^{\sharp} = \bigoplus_{n=-1}^{\infty}~V^{\sharp}_n$
with every component $V^{\sharp}_n$ being a representation of the monster group $\mathbb{M}$ of which the dimension coincides with the coefficient of $q^n$ in $J(\tau)$.
It only got better, for any conjugacy class $[ g ]$ of the monster, if you took the character series
$T_g(\tau) = \sum_{n=-1}^{\infty} Tr(g | V^{\sharp}_n) q^n$
you get a function invariant under the action of the subgroup
$\Gamma_0(n) = \{ \begin{bmatrix} a & b \\ c & d \end{bmatrix}~:~c = 0~mod~n \}$
acting via transformations $\tau \mapsto \frac{a \tau + b}{c \tau + d}$ on the upper half plane where $n$ is the order of $g$ (or, for the experts, almost).
Soon, further instances of ‘moonshine’ were discovered for other simple groups, the unifying feature being that one associates to a group $G$ a graded representation $V$ such that the character series of this representation for an element $g \in G$ is an invariant modular function with respect to the subgroup $\Gamma_0(n)$ of the modular group, with $n$ being the order of $g$.
Today, this group of people proved that there is ‘moonshine’ for any finite group whatsoever.
They changed the definition of moonshine slightly to introduce the notion of moonshine of depth $d$ which meant that they want the dimension sequence of their graded module to be equal to $J(\tau)$ under the action of the normalized $d$-th Hecke operator, which means equal to
$\sum_{ac=d,0 \leq b < c} J(\frac{a \tau + b}{c})$
as they are interested in the asymptotic behaviour of the components $V_n$ with respect to the regular representation of $G$.
What baffled me was their much weaker observation (remark 2) saying that you get ‘moonshine’ in the form described above, that is, a graded representation $V$ such that for every $g \in G$ you get a character series which is invariant under $\Gamma_0(n)$ with $n=ord(g)$ (and no smaller divisor of $n$), for every finite group $G$.
And, more importantly, you can explain this to any student taking a first course in group theory as all you need is Cayley’s theorem stating that any finite group is a subgroup of some symmetric group $S_n$.
Here’s the idea: take the original monster-moonshine module $V^{\sharp}$ but forget all about the action of $\mathbb{M}$ (that is, consider it as a plain vectorspace) and consider the graded representation
$V = (V^{\sharp})^{\otimes n}$
with the natural action of $S_n$ on the tensor product.
Now, embed a la Cayley $G$ into $S_n$ then you know that the order of $g \in G$ is the least common multiple of the cycle lengths of the permutation it it send to. Now, it is fairly trivial to see that the character series of $V$ with respect to $g$ (having cycle lengths $(k_1,k_2,\dots,k_l)$, including cycles of length one) is equal to the product
$J(k_1 \tau) J(k_2 \tau) \dots J(k_l \tau)$
which is invariant under $\Gamma_0(n)$ with $n = lcm(k_i)$ (but no $\Gamma_0(m)$ with $m$ a proper divisor of $n$).
For example, for $G=S_4$ we have as character series of $(V^{\sharp})^{\otimes 4}$
$(1)(2)(3)(4) \mapsto J(\tau)^4$
$(12)(3)(4) \mapsto J(2 \tau) J(\tau)^2$
$(12)(34) \mapsto J(2 \tau)^2$
$(123)(4) \mapsto J(3 \tau) J(\tau)$
$(1234) \mapsto J(4 \tau)$
Clearly, the main results of the paper are much more subtle, but I’m already happy with this version of ‘moonshine for everyone’!
# Stirring a cup of coffee
Please allow for a couple of end-of-semester bluesy ramblings. I just finished grading the final test of the last of five courses I lectured this semester.
Most of them went, I believe, rather well.
As always, it was fun to teach an introductory group theory course to second year physics students.
Personally, I did enjoy our Lie theory course the most, given for a mixed public of both mathematics and physics students. We did the spin-group $SU(2)$ and its connection with $SO_3(\mathbb{R})$ in gruesome detail, introduced the other classical groups, and proved complete reducibility of representations. The funnier part was applying this to the $U(1) \times SU(2) \times SU(3)$-representation of the standard model and its extension to the $SU(5)$ GUT.
Ok, but with a sad undertone, was the second year course on representations of finite groups. Sad, because it was the last time I’m allowed to teach it. My younger colleagues decided there’s no place for RT on the new curriculum.
Soit.
The final lecture is often an eye-opener, or at least, I hope it is/was.
Here’s the idea: someone whispers in your ear that there might be a simple group of order $60$. Armed with only the Sylow-theorems and what we did in this course we will determine all its conjugacy classes, its full character table, and finish proving that this mysterious group is none other than $A_5$.
Right now I’m just a tad disappointed only a handful of students came close to solving the same problem for order $168$ this afternoon.
Clearly, I gave them ample extra information: the group only has elements of order $1,2,3,4$ and $7$ and the centralizer of one order $2$ element is the dihedral group of order $8$. They had to determine the number of distinct irreducible representations, that is, the number of conjugacy classes. Try it yourself (Solution at the end of this post).
For months I felt completely deflated on Tuesday nights, for I had to teach the remaining two courses on that day.
There’s this first year Linear Algebra course. After teaching for over 30 years it was a first timer for me, and probably for the better. I guess 15 years ago I would have been arrogant enough to insist that the only way to teach linear algebra properly was to do representations of quivers…
Now, I realise that linear algebra is perhaps the only algebra course the majority of math-students will need in their further career, so it is best to tune its contents to the desires of the other colleagues: inproducts, determinants as volumes, Markov-processes and the like.
There are thousands of linear algebra textbooks, the one feature they all seem to lack is conciseness. What kept me going throughout this course was trying to come up with the shortest proofs ever for standard results. No doubt, next year the course will grow on me.
Then, there was a master course on algebraic geometry (which was supposed to be on scheme theory, moduli problems such as the classification of fat points (as in the car crash post, etale topology and the like) which had a bumpy start because class was less prepared on varieties and morphisms than I had hoped for.
Still, judging on the quality of the papers students are beginning to hand in (today I received one doing serious stuff with stacks) we managed to cover a lot of material in the end.
I’m determined to teach that first course on algebraic geometry myself next year.
Which brought me wondering about the ideal content of such a course.
Half a decade ago I wrote a couple of posts such as Mumford’s treasure map, Grothendieck’s functor of points, Manin’s geometric axis and the like, which are still quite readable.
In the functor of points-post I referred to a comment thread Algebraic geometry without prime ideals at the Secret Blogging Seminar.
As I had to oversee a test this afternoon, I printed out all comments (a full 29 pages!) and had a good time reading them. At the time I favoured the POV advocated by David Ben-Zvi and Jim Borger (functor of points instead of locally ringed schemes).
Clearly they are right, but then so was I when I thought the ‘right’ way to teach linear algebra was via quiver-representations…
We’ll see what I’ll try out next year.
You may have wondered about the title of this post. It’s derived from a paper Raf Bocklandt (of the Korteweg-de Vries Institute in Amsterdam) arXived some days ago: Reflections in a cup of coffee, which is an extended version of a Brouwer-lecture he gave. Raf has this to say about the Brouwer fixed-point theorem.
“The theorem is usually explained in worldly terms by looking at a cup of coffee. In this setting it states that no matter how you stir your cup, there will always be a point in the liquid that did not change position and if you try to move that part by further stirring you will inevitably move some other part back into its original position. Legend even has it that Brouwer came up with the idea while stirring in a real cup, but whether this is true we’ll never know. What is true however is that Brouwers refections on the topic had a profound impact on mathematics and would lead to lots of new developments in geometry.”
I wish you all a pleasant end of 2016 and a much better 2017.
As to the 168-solution: Sylow says there are 8 7-Sylows giving 48 elements of order 7. The centralizer of each of them must be $C_7$ (given the restriction on the order of elements) so two conjugacy classes of them. Similarly each conjugacy class of an order 3 element must contain 56 elements. There is one conjugacy class of an order 2 element having 21 elements (because the centralizer is $D_4$) giving also a conjugacy class of an order 4 element consisting of 42 elements. Together with the identity these add up to 168 so there are 6 irreducible representations.
# let’s spend 3K on (math)books
Santa gave me 3000 Euros to spend on books. One downside: I have to give him my wish-list before monday. So, I’d better get started. Clearly, any further suggestions you might have will be much appreciated, either in the comments below or more directly via email.
Today I’ll focus on my own interests: algebraic geometry, non-commutative geometry and representation theory. I do own a fair amount of books already which accounts for the obvious omissions in the lists below (such as Hartshorne, Mumford or Eisenbud-Harris in AG, Fulton-Harris in RT or the ‘bibles’ in NCG).
[section_title text=”Algebraic geometry”]
Here, I base myself on (and use quotes from) the excellent answer by Javier Alvarez to the MathOverflow post Best Algebraic Geometry text book? (other than Hartshorne).
In no particular order:
Lectures on Curves, Surfaces and Projective Varieties by Ettore Carletti, Dionisio Gallarati, and Giacomo Monti Bragadin and Mauro C. Beltrametti.
“which starts from the very beginning with a classical geometric style. Very complete (proves Riemann-Roch for curves in an easy language) and concrete in classic constructions needed to understand the reasons about why things are done the way they are in advanced purely algebraic books. There are very few books like this and they should be a must to start learning the subject. (Check out Dolgachev’s review.)”
A Royal Road to Algebraic Geometry by Audun Holme. “This new title is wonderful: it starts by introducing algebraic affine and projective curves and varieties and builds the theory up in the first half of the book as the perfect introduction to Hartshorne’s chapter I. The second half then jumps into a categorical introduction to schemes, bits of cohomology and even glimpses of intersection theory.”
Liu Qing – “Algebraic Geometry and Arithmetic Curves”. “It is a very complete book even introducing some needed commutative algebra and preparing the reader to learn arithmetic geometry like Mordell’s conjecture, Faltings’ or even Fermat-Wiles Theorem.”
Görtz; Wedhorn – Algebraic Geometry I, Schemes with Examples and Exercises. labeled ‘the best on schemes’ by Alvarez. “Tons of stuff on schemes; more complete than Mumford’s Red Book. It does a great job complementing Hartshorne’s treatment of schemes, above all because of the more solvable exercises.”
Kollár – Lectures on Resolution of Singularities. “Great exposition, useful contents and examples on topics one has to deal with sooner or later.”
Kollár; Mori – Birational Geometry of Algebraic Varieties. “Considered as harder to learn from by some students, it has become the standard reference on birational geometry.”
And further, as a follow-up on their previous book on the computational side of AG:
Using Algebraic Geometry by Cox, Little and O’Shea.
[section_title text=”Non-commutative geometry”]
Noncommutative Geometry and Particle Physics by Walter van Suijlekom. Blurb: “This book provides an introduction to noncommutative geometry and presents a number of its recent applications to particle physics. It is intended for graduate students in mathematics/theoretical physics who are new to the field of noncommutative geometry, as well as for researchers in mathematics/theoretical physics with an interest in the physical applications of noncommutative geometry. In the first part, we introduce the main concepts and techniques by studying finite noncommutative spaces, providing a “light” approach to noncommutative geometry. We then proceed with the general framework by defining and analyzing noncommutative spin manifolds and deriving some main results on them, such as the local index formula. In the second part, we show how noncommutative spin manifolds naturally give rise to gauge theories, applying this principle to specific examples. We subsequently geometrically derive abelian and non-abelian Yang-Mills gauge theories, and eventually the full Standard Model of particle physics, and conclude by explaining how noncommutative geometry might indicate how to proceed beyond the Standard Model.”
An Invitation To Noncommutative Geometry by Matilde Marcolli. Blurb: “This is the first existing volume that collects lectures on this important and fast developing subject in mathematics. The lectures are given by leading experts in the field and the range of topics is kept as broad as possible by including both the algebraic and the differential aspects of noncommutative geometry as well as recent applications to theoretical physics and number theory.”
Noncommutative Geometry and Physics: Renormalisation, Motives, Index Theory. Blurb: “This collection of expository articles grew out of the workshop “Number Theory and Physics” held in March 2009 at The Erwin Schrödinger International Institute for Mathematical Physics, Vienna. The common theme of the articles is the influence of ideas from noncommutative geometry (NCG) on subjects ranging from number theory to Lie algebras, index theory, and mathematical physics. Matilde Marcolli’s article gives a survey of relevant aspects of NCG in number theory, building on an introduction to motives for beginners by Jorge Plazas and Sujatha Ramdorai.”
Feynman Motives by Matilde Marcolli. Blurb: “This book presents recent and ongoing research work aimed at understanding the mysterious relation between the computations of Feynman integrals in perturbative quantum field theory and the theory of motives of algebraic varieties and their periods. One of the main questions in the field is understanding when the residues of Feynman integrals in perturbative quantum field theory evaluate to periods of mixed Tate motives.” But then, check out Matilde’s recent FaceBook status-update.
[section_title text=”Representation theory”]
An Introduction to the Langlands Program by J. Bernstein (editor). Blurb: “This book presents a broad, user-friendly introduction to the Langlands program, that is, the theory of automorphic forms and its connection with the theory of L-functions and other fields of mathematics. Each of the twelve chapters focuses on a particular topic devoted to special cases of the program. The book is suitable for graduate students and researchers.”
Representation Theory of Finite Groups: An Introductory Approach by Benjamin Steinberg.
Representation Theory of Finite Monoids by Benjamin Steinberg. Blurb: “This first text on the subject provides a comprehensive introduction to the representation theory of finite monoids. Carefully worked examples and exercises provide the bells and whistles for graduate accessibility, bringing a broad range of advanced readers to the forefront of research in the area. Highlights of the text include applications to probability theory, symbolic dynamics, and automata theory. Comfort with module theory, a familiarity with ordinary group representation theory, and the basics of Wedderburn theory, are prerequisites for advanced graduate level study.”
How am I doing? 914 dollars…
Way to go, same exercise tomorrow. Again, suggestions/warnings welcome!
# Quiver Grassmannians can be anything
A standard Grassmannian $Gr(m,V)$ is the manifold having as its points all possible $m$-dimensional subspaces of a given vectorspace $V$. As an example, $Gr(1,V)$ is the set of lines through the origin in $V$ and therefore is the projective space $\mathbb{P}(V)$. Grassmannians are among the nicest projective varieties, they are smooth and allow a cell decomposition.
A quiver $Q$ is just an oriented graph. Here’s an example
A representation $V$ of a quiver assigns a vector-space to each vertex and a linear map between these vertex-spaces to every arrow. As an example, a representation $V$ of the quiver $Q$ consists of a triple of vector-spaces $(V_1,V_2,V_3)$ together with linear maps $f_a~:~V_2 \rightarrow V_1$ and $f_b,f_c~:~V_2 \rightarrow V_3$.
A sub-representation $W \subset V$ consists of subspaces of the vertex-spaces of $V$ and linear maps between them compatible with the maps of $V$. The dimension-vector of $W$ is the vector with components the dimensions of the vertex-spaces of $W$.
This means in the example that we require $f_a(W_2) \subset W_1$ and $f_b(W_2)$ and $f_c(W_2)$ to be subspaces of $W_3$. If the dimension of $W_i$ is $m_i$ then $m=(m_1,m_2,m_3)$ is the dimension vector of $W$.
The quiver-analogon of the Grassmannian $Gr(m,V)$ is the Quiver Grassmannian $QGr(m,V)$ where $V$ is a quiver-representation and $QGr(m,V)$ is the collection of all possible sub-representations $W \subset V$ with fixed dimension-vector $m$. One might expect these quiver Grassmannians to be rather nice projective varieties.
However, last week Markus Reineke posted a 2-page note on the arXiv proving that every projective variety is a quiver Grassmannian.
Let’s illustrate the argument by finding a quiver Grassmannian $QGr(m,V)$ isomorphic to the elliptic curve in $\mathbb{P}^2$ with homogeneous equation $Y^2Z=X^3+Z^3$.
Consider the Veronese embedding $\mathbb{P}^2 \rightarrow \mathbb{P}^9$ obtained by sending a point $(x:y:z)$ to the point
$(x^3:x^2y:x^2z:xy^2:xyz:xz^2:y^3:y^2z:yz^2:z^3)$
The upshot being that the elliptic curve is now realized as the intersection of the image of $\mathbb{P}^2$ with the hyper-plane $\mathbb{V}(X_0-X_7+X_9)$ in the standard projective coordinates $(x_0:x_1:\cdots:x_9)$ for $\mathbb{P}^9$.
To describe the equations of the image of $\mathbb{P}^2$ in $\mathbb{P}^9$ consider the $6 \times 3$ matrix with the rows corresponding to $(x^2,xy,xz,y^2,yz,z^2)$ and the columns to $(x,y,z)$ and the entries being the multiplications, that is
$$\begin{bmatrix} x^3 & x^2y & x^2z \\ x^2y & xy^2 & xyz \\ x^2z & xyz & xz^2 \\ xy^2 & y^3 & y^2z \\ xyz & y^2z & yz^2 \\ xz^2 & yz^2 & z^3 \end{bmatrix} = \begin{bmatrix} x_0 & x_1 & x_2 \\ x_1 & x_3 & x_4 \\ x_2 & x_4 & x_5 \\ x_3 & x_6 & x_7 \\ x_4 & x_7 & x_8 \\ x_5 & x_8 & x_9 \end{bmatrix}$$
But then, a point $(x_0:x_1: \cdots : x_9)$ belongs to the image of $\mathbb{P}^2$ if (and only if) the matrix on the right-hand side has rank $1$ (that is, all its $2 \times 2$ minors vanish). Next, consider the quiver
and consider the representation $V=(V_1,V_2,V_3)$ with vertex-spaces $V_1=\mathbb{C}$, $V_2 = \mathbb{C}^{10}$ and $V_2 = \mathbb{C}^6$. The linear maps $x,y$ and $z$ correspond to the columns of the matrix above, that is
$$(x_0,x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9) \begin{cases} \rightarrow^x~(x_0,x_1,x_2,x_3,x_4,x_5) \\ \rightarrow^y~(x_1,x_3,x_4,x_6,x_7,x_8) \\ \rightarrow^z~(x_2,x_4,x_5,x_7,x_8,x_9) \end{cases}$$
The linear map $h~:~\mathbb{C}^{10} \rightarrow \mathbb{C}$ encodes the equation of the hyper-plane, that is $h=x_0-x_7+x_9$.
Now consider the quiver Grassmannian $QGr(m,V)$ for the dimension vector $m=(0,1,1)$. A base-vector $p=(x_0,\cdots,x_9)$ of $W_2 = \mathbb{C}p$ of a subrepresentation $W=(0,W_2,W_3) \subset V$ must be such that $h(x)=0$, that is, $p$ determines a point of the hyper-plane.
Likewise the vectors $x(p),y(p)$ and $z(p)$ must all lie in the one-dimensional space $W_3 = \mathbb{C}$, that is, the right-hand side matrix above must have rank one and hence $p$ is a point in the image of $\mathbb{P}^2$ under the Veronese.
That is, $Gr(m,V)$ is isomorphic to the intersection of this image with the hyper-plane and hence is isomorphic to the elliptic curve.
The general case is similar as one can view any projective subvariety $X \rightarrow \mathbb{P}^n$ as isomorphic to the intersection of the image of a specific $d$-uple Veronese embedding $\mathbb{P}^n \rightarrow \mathbb{P}^N$ with a number of hyper-planes in $\mathbb{P}^N$.
# Klein’s dessins d’enfant and the buckyball
We saw that the icosahedron can be constructed from the alternating group $A_5$ by considering the elements of a conjugacy class of order 5 elements as the vertices and edges between two vertices if their product is still in the conjugacy class.
This description is so nice that one would like to have a similar construction for the buckyball. But, the buckyball has 60 vertices, so they surely cannot correspond to the elements of a conjugacy class of $A_5$. But, perhaps there is a larger group, somewhat naturally containing $A_5$, having a conjugacy class of 60 elements?
This is precisely the statement contained in Galois’ last letter. He showed that 11 is the largest prime p such that the group $L_2(p)=PSL_2(\mathbb{F}_p)$ has a (transitive) permutation presentation on p elements. For, p=11 the group $L_2(11)$ is of order 660, so it permuting 11 elements means that this set must be of the form $X=L_2(11)/A$ with $A \subset L_2(11)$ a subgroup of 60 elements… and it turns out that $A \simeq A_5$…
Actually there are TWO conjugacy classes of subgroups isomorphic to $A_5$ in $L_2(11)$ and we have already seen one description of these using the biplane geometry (one class is the stabilizer subgroup of a ‘line’, the other the stabilizer subgroup of a point).
Here, we will give yet another description of these two classes of $A_5$ in $L_2(11)$, showing among other things that the theory of dessins d’enfant predates Grothendieck by 100 years.
In the very same paper containing the first depiction of the Dedekind tessellation, Klein found that there should be a degree 11 cover $\mathbb{P}^1_{\mathbb{C}} \rightarrow \mathbb{P}^1_{\mathbb{C}}$ with monodromy group $L_2(11)$, ramified only in the three points ${ 0,1,\infty }$ such that there is just one point lying over $\infty$, seven over 1 of which four points where two sheets come together and finally 5 points lying over 0 of which three where three sheets come together. In 1879 he wanted to determine this cover explicitly in the paper “Ueber die Transformationen elfter Ordnung der elliptischen Funktionen” (Math. Annalen) by describing all Riemann surfaces with this ramification data and pick out those with the correct monodromy group.
He manages to do so by associating to all these covers their ‘dessins d’enfants’ (which he calls Linienzuges), that is the pre-image of the interval [0,1] in which he marks the preimages of 0 by a bullet and those of 1 by a +, such as in the innermost darker graph on the right above. He even has these two wonderful pictures explaining how the dessin determines how the 11 sheets fit together. (More examples of dessins and the correspondences of sheets were drawn in the 1878 paper.)
The ramification data translates to the following statements about the Linienzuge : (a) it must be a tree ($\infty$ has one preimage), (b) there are exactly 11 (half)edges (the degree of the cover),
(c) there are 7 +-vertices and 5 o-vertices (preimages of 0 and 1) and (d) there are 3 trivalent o-vertices and 4 bivalent +-vertices (the sheet-information).
Klein finds that there are exactly 10 such dessins and lists them in his Fig. 2 (left). Then, he claims that one the two dessins of type I give the correct monodromy group. Recall that the monodromy group is found by giving each of the half-edges a number from 1 to 11 and looking at the permutation $\tau$ of order two pairing the half-edges adjacent to a +-vertex and the order three permutation $\sigma$ listing the half-edges by cycling counter-clockwise around a o-vertex. The monodromy group is the group generated by these two elements.
Fpr example, if we label the type V-dessin by the numbers of the white regions bordering the half-edges (as in the picture Fig. 3 on the right above) we get
$\sigma = (7,10,9)(5,11,6)(1,4,2)$ and $\tau=(8,9)(7,11)(1,5)(3,4)$.
Nowadays, it is a matter of a few seconds to determine the monodromy group using GAP and we verify that this group is $A_{11}$.
Of course, Klein didn’t have GAP at his disposal, so he had to rule out all these cases by hand.
gap> g:=Group((7,10,9)(5,11,6)(1,4,2),(8,9)(7,11)(1,5)(3,4));
Group([ (1,4,2)(5,11,6)(7,10,9), (1,5)(3,4)(7,11)(8,9) ])
gap> Size(g);
19958400
gap> IsSimpleGroup(g);
true
Klein used the fact that $L_2(7)$ only has elements of orders 1,2,3,5,6 and 11. So, in each of the remaining cases he had to find an element of a different order. For example, in type V he verified that the element $\tau.(\sigma.\tau)^3$ is equal to the permutation (1,8)(2,10,11,9,6,4,5)(3,7) and consequently is of order 14.
Perhaps Klein knew this but GAP tells us that the monodromy group of all the remaining 8 cases is isomorphic to the alternating group $A_{11}$ and in the two type I cases is indeed $L_2(11)$. Anyway, the two dessins of type I correspond to the two conjugacy classes of subgroups $A_5$ in the group $L_2(11)$.
But, back to the buckyball! The upshot of all this is that we have the group $L_2(11)$ containing two classes of subgroups isomorphic to $A_5$ and the larger group $L_2(11)$ does indeed have two conjugacy classes of order 11 elements containing exactly 60 elements (compare this to the two conjugacy classes of order 5 elements in $A_5$ in the icosahedral construction). Can we construct the buckyball out of such a conjugacy class?
To start, we can identify the 12 pentagons of the buckyball from a conjugacy class C of order 11 elements. If $x \in C$, then so do $x^3,x^4,x^5$ and $x^9$, whereas the powers ${ x^2,x^6,x^7,x^8,x^{10} }$ belong to the other conjugacy class. Hence, we can divide our 60 elements in 12 subsets of 5 elements and taking an element x in each of these, the vertices of a pentagon correspond (in order) to $~(x,x^3,x^9,x^5,x^4)$.
Group-theoretically this follows from the fact that the factorgroup of the normalizer of x modulo the centralizer of x is cyclic of order 5 and this group acts naturally on the conjugacy class of x with orbits of size 5.
Finding out how these pentagons fit together using hexagons is a lot subtler… and in The graph of the truncated icosahedron and the last letter of Galois Bertram Kostant shows how to do this.
Fix a subgroup isomorphic to $A_5$ and let D be the set of all its order 2 elements (recall that they form a full conjugacy class in this $A_5$ and that there are precisely 15 of them). Now, the startling observation made by Kostant is that for our order 11 element $x$ in C there is a unique element $a \in D$ such that the commutator$~b=[x,a]=x^{-1}a^{-1}xa$ belongs again to D. The unique hexagonal side having vertex x connects it to the element $b.x$which belongs again to C as $b.x=(ax)^{-1}.x.(ax)$.
Concluding, if C is a conjugacy class of order 11 elements in $L_2(11)$, then its 60 elements can be viewed as corresponding to the vertices of the buckyball. Any element $x \in C$ is connected by two pentagonal sides to the elements $x^{3}$ and $x^4$ and one hexagonal side connecting it to $\tau x = b.x$.
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2018-01-17 13:11:48
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https://www.physicsforums.com/threads/heat-propagation.49051/
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# Heat Propagation
1. Oct 22, 2004
### Clausius2
Well, this is my question. It's something like a philosophical thinking.
Look at this problem of the Heat Equation:
$$\frac{\partial T}{\partial t}=\alpha \frac{\partial^2 T}{\partial x^2}$$
with: 0<x<L
$$T(x,0)=g(x)$$ and
$$T(0,t)=T_1$$
$$T(L,t)=T_o$$
This is a parabolical PDE equation. It has only one characteristic line:
$$\frac{dx}{dt}=\infty$$
So that, only information happened along the entire domain and in past times affects to what will happen in the future. It has sense. But the difference with hyperbolic equations, like Wave Equation is there is not any physical velocity of propagation of the phenomena. I will explain that:
A Wave has a finite velocity in any medium (except in a perfect liquid). One perturbation produced in some place takes a time away for being propagated to another zone.
On the other hand, in a change of time of $$dt$$ in Heat Equation, the initial distribution g(x) is smoothed and every point of the domain feels the heat flux instantaneously.
This seems rather disappointing, isn't it?. I don't know neither instantaneous interaction or instantaneous effect. With Heat Equation, the effects appear to be transmitted with infinite velocity. In fact, nobody talks never about velocity propagation of the heat.
Or am I missing something?.
Last edited: Oct 22, 2004
2. Oct 22, 2004
### Tide
You've discovered one of the dirty little secrets of physics! This is typical of diffusion type equations. Obviously, you cannot have atoms or molecules or electrons zipping along at infinite velocities but it's an artifact of the approximation. For example, one could work with the full Boltzmann equation or something similar but that is way too difficult to deal with for most situations. In the relatively rare instances where it matters people often invoke "flux limiters" or simply choose an arbitrary cuttoff such as the speed of light.
It is somewhat akin to the problem of using a Maxwellian distribution for particle velocities and including the range of velocities all the way to infinity. It's simply convenient and the errors are relatively small with the number of particles out further than a few dozen standard deviations being minuscule.
BTW - that should be a partial derivative with respect to time on the left hand side of your heat equation!
3. Oct 22, 2004
### Clausius2
Ooppps! Thanks. I've corrected the equation.
Well, you said that's an artifact of approximation. In fact, you agree with me that a perturbation of the temperature in some point of the domain, cannot be physically felt instantaneously by further points. Yes, one can say something about light velocity, but it does not appear neither in the Wave Equation nor in the Heat Equation (also surprising, isn't it?).
But what you told me is a bit vague. As you said, is it possible I haven't got any way of obtaining some idea of how fast the heat is propagated inside my flow field?. I mean, something like a Heat Speed.
Is not there any physicist who had thought about a Heat Wave? It seems the more probably mechanism of heat transfer, like waves travelling in a liquid or solid. In some way, the Heat Equation has an internal misconception with this, as I think.
Last edited: Oct 22, 2004
4. Oct 22, 2004
### MiGUi
Classically we use the infinite speed of information transmission to improve our models. Think in mechanic's rigid solid. When we apply a force in an extreme, that force is transmited instanteneously to the other extreme of the rigid solid, so it begins to move instantaneously. That can't happen in SR and thats why solid rigid don't exists in relativity.
This, can be applied to your question. The flux is transmited instantaneously? False, it has an infinite speed, but we would not have the L too big to have to consider relativistic effects on it!
MiGUi
5. Oct 22, 2004
### Clausius2
In Vibration Engineering or Acoustic Engineering, the finite deformation rate of the model, and also the finite propagation of the sound are present. The rigid body is only used by physicist or so. For us, nothing is rigid.
But you're telling me the ultimate vehicle of Heat transmission is the Light. I do think the ultimate vehicle of all events is the Light. But there are another effects in which the transport velocity of some magnitude is very much slower than light velocity (like sound propagation). I wonder if such Heat Speed is not derivable and is present and smaller than light velocity.
When you just put a saucepan on the hotplate when you are cooking, is the high temperature of the fire being felt in the boundary of the saucepan instantaneously?. Which is that velocity of propagation?. Light velocity? One smaller than light velocity?.
The Heat Equation states the boundary will sense the puntual fire instantaneously. This is part of the model, Migui, as you said. But one can use the Wave Equation to study the propagation of a sound wave inside a solid. And that propagation is not too slow, God knows it!.
In part, I think the Heat Equation does not admit any discontinous solution (am I right?). So that, any effect of propagation of a hypothetic Heat Wave is smoothed by the laplacian operator.
On the other hand, Wave equation does admit discontinous solutions (am I right?), so that any discontinuity can be propagated as a wave front.
6. Oct 22, 2004
### MiGUi
The transmission of heat deppends on several factors: thermal conductivity, difference of temperature, and so. We can only say about a speed of heat if it is radiated, but when we use convection or conduction, its not a good way to explain the heat as a wave.
When you use a teaspoon to move the café or else, if the teaspoon is made of Ag you can sense the heat very fast and with a plastic one you may not sense anything. But actually, the flux is another thing. We use it to explain why the heat goes in one direction and not in other, and where the temperature will grow faster.
What I wanted to say before, is that we can use that if the size of our system is small, we can use the speed of light as infinite, and that will be a very good approximation.
7. Oct 22, 2004
### ReyChiquito
Is that so? I thought only hyperbolic equations had characteristic lines... Im confused...
8. Oct 22, 2004
### Integral
Staff Emeritus
I am a bit fuzzy on the details, but I recall doing solutions of the heat equation/diffusion problems which showed disturbances traveling through the media as a erfc wave front. I do not recall the details of the solution, need to dig into my old notes.
9. Oct 23, 2004
### Clausius2
Hi ReyGrandecito!
Yes. Take a look at any PDEs book and the difference between hyperbolic, parabolic and elliptic behavior. Parabolic equations have only one characteristic line. The field of temperatures just in the next step of time depends of what happens in the entire domain. To point that in another way, the domain of influence of any point is the entire domain.
Maybe you are remembering the method of the Green's Function. But remeber the Green Function of the Heat was not discontinous. I don't remember well, but I think it was a Dirac's delta function modulated by an exponential factor.
On the other hand, the Green's Function for the Wave Equation was discontinous (am I right?).
10. Oct 23, 2004
### Integral
Staff Emeritus
Clausius,
Considering the problem you have set up, I see no real possibility of the propagation of a thermal wave, each point of the object must migrate toward the steady state temperature distribution given by:
$$T(x)= \frac {T_1 - T_0} L x + T_1$$
Frankly this is not real interesting.
Consider the following boundary conditions.
$$T(0,t) = \left \{ \begin{array}{cc}T_i , & t\leq t_1\\ T_f, & t > t_1 \end {array}\right$$
So this is a step function temperature change at one end of the bar.
Now at the other end:
$$\dot {T}(L,t) = C$$
This is a statement of constant rate of loss, we could use a Newtons law of cooling condition here.
Let us define the initial temperature distribution as:
$$T(x,0) = T_i$$
So the bar starts at a constant temperature.
The question is, how long does it take for the change to be felt at L. What is the temperature distribution at any time t?
11. Oct 23, 2004
### Clausius2
I don't know if you are really hoping an answer. You finished so suddenly your post. I was waiting for an answer by yourself....
To be honest, I don't undestand what you mean. But this is not a IRCchat , so I'll suppose your boundary conditions and initial conditions for t>t1. Then, the boundary temperature of x=0 changes suddenly. How long it takes the effect to be felt ar L? ----> Zero time.
In fact, there will be a front of heating, the temperature will probably be increasing progresively, with the larger peak at x=0. But I'm careful with the word front . This hypothetic wave front is not such a wave. It is not discontinous. And it's not a wave. It's an smoothed wave by the laplacian operator. Watch this imaginary game:
"Suppose you have a rope on the floor. If you strike the rope at one end, an elastic wave will be propagated towards the other end. Catch now the rope by <n> points, and get it up slowly but not uniformly. As you are getting it up from more than one point, there will be a progresive and continuous arising of each of the points."
I don't know if I've transmitted above something eatable. Sorry if you don't understand no word of what I meant.
The fact is the wave of heat exists, the temperature at the other end cannot be felt instantaneously , because the word instantaneous has no sense in physics. The problem here is if the time involved is of the order $$t_o\approx \frac{L}{c}$$ where c is the light velocity, or less than that time. If really it is of the order of $$t_o$$ then MiGUi is right, and it could be neglected if L is too small.
I will try to estimate a time of temperature transmision using Fluid Mechanics techniques in a particular problem. .....Hmmmmmm. Let me see....
and write that....... then.......... :zzz:
12. Oct 23, 2004
### Clausius2
Let's see. Let's recover the original problem I've posted:
$$\frac{\partial T}{\partial t}=\alpha \frac{\partial^2 T}{\partial x^2}$$
$$T(x,0)=g(x)$$
$$T(0,t)=T_1$$
$$T(L,t)=T_2$$
Right. This is what a Fluid Mechanics Engineer would do:
The first term of the equation can be estimated, taking into account the boudary constraints as:
$$\frac{\partial T}{\partial t}\approx \frac{T_o}{t_o}$$ (1)
where $$T_o=(T_1+T_2)/2$$ is the characteristic Temperature and $$t_o$$ is the characteristic time of Temperature variation.
The term on the right can be estimated as:
$$\alpha \frac{\partial^2 T}{\partial x^2}\approx\alpha \frac{T_o}{L^2}$$(2)
because the heat flux suffers severe variations in distances of the order L.
Equalling (1) and (2) we obtain an estimation for the characteristic time:
$$t_o\approx \frac{L^2}{\alpha}$$
It has sense, the more diffusivity the less time is needed for reaching the steady state you was referring, Integral. In fact, the number:
$$F_o=\frac{\alpha t_o}{L^2}$$ is called the Fourier Number. When it takes the value of 1 the steady solution is practically reached.
For a bar of lenght L=1m, and diffusivity $$\alpha=23\cdot10^{-6}$$ m2/s (a steel bar) the time involved is $$t_o=43863$$ seconds. !
:yuck: I've done all this calculus for nothing clear.... After all, we only have been able to obtain an estimation of the time needed to reach the steady state. (If you want, you can check it solving the problem analytically. Surely you'll obtain a preexponential time-dependent factor which tends to 1 when
t--->to.)
But nothing appears about the time needed to felt the presence of other temperature in the surroundings of some point. Why we can make figures with elastic and sound waves travelling and not with heat waves?
After all, we cannot estimate that time using the Heat Equation. It's an absurd. I've realized it's an absurd because Heat Equation is again parabolic, and it has no sense asking oneself about a time of propagation with that mathematical behavior.
Puzzling thinkings indeed..... :uhh:
Last edited: Oct 23, 2004
13. Oct 26, 2004
### ReyChiquito
Well, this problem only reflects the lack of accuracy of the model i think. We need to trust physics (wich i do) and the causality principle states that information cannot propagate instantly, not to mention experimental evidence. So there is something wrong with the equation.
In other heat problems (other equations), the system must reach an umbral temperature before it starts to propagate.
I think this is a great example of how far we are from understanding or even modeling a "simple" physical phenomena.
14. Oct 26, 2004
### Integral
Staff Emeritus
I think that the heat equation works fine, the model is good. Though I still say that you have to set up a problem which will have a solution which will display a moving front of energy.
Reading some texts Operational Mathematics by Churchill, for one, I see that I have misstated my constant heat flux boundary condition. It should be:
$$U_x (L,t) = C$$ where C is a constant.
This system would behave much like an infinite half slab of material, that is, x runs from 0 to infinity.
The initial condition is for the slab to be at some initial temperature.
The boundary condition at x=0 is to apply a step function change in temp at t=0.
Now I posted the problem above just to get you to think of the physical situation. What would you observe at some distance L down the slab?
Initially the Temperature appears to remain constant. After a period of time it becomes evident that the temp is slowly increasing. Then there is a brief period of rapid temperature change until something near the final temperature is reached. Fianlly there is another period of slow change as the system reaches the final temperature.
OK, this is not speculation on my part, I have done this experiment, that is what happens. Also in doing this experiment, I setup a program to model the heat equation numerically. The model predicted the same behavior. So while there is nothing in the heat equation corresponding to wave velocity there is the possibility of a moving energy front. The velocity of this front depends on the diffusivity of the material (Diffusivity has units L2T-1) See the velocity in there?
Now the ERFC (Error function Compliment) function has the feature that the initial effects of a change are felt INSTANTLY through out the material, it just is not a very big effect, until the main wave front arrives.
I do not think what I am saying disagrees with your analysis. What you do not tell us is the MAGNITUDE of the instantaneous "message" sent through the material.
By the way some of the examples worked in the above mentioned text include an erfc solution.
Last edited: Oct 26, 2004
15. Oct 26, 2004
### Clausius2
Thanks Integral for sharing with me and us your opinion.
I'm only going to emphasize in your word "front". A Wave front from hyperbolics equations is a discontinuity in the field. Pay attention to the word discontinuity, it's crucial. The front you talked to me hasn't got any discontinuty. It's mathematically impossible that the Heat Equation admits a discontinuity in its solution. You talked to me about a gradual adaptation of temperature, with a severe variation in time and x, but it is not a front at all, as it is understood in hyperbolic PDE's.
In fact, and answering to your problem, the equation states that in the infinity, the step change of temperature will be felt instantaneously. An infinitesimal change will ocurre at x-->infinite and t+dt. This internal error is more severe, as MiGUi said before, because it is physically imposible. At lenghts L enough large the role of the ¿¿light?? speed is not negligible at all.
Thanks again. :)
16. Oct 26, 2004
### Integral
Staff Emeritus
Who said anything about discontinuities? The ERFC function has a very well defined "front" with no discontinuities.
Once again, consider the problem I have set up. It is completely physical and it displays the propagation of thermal energy with time. The analytic solution of the proposed problem displays the expected behavior.
I am not sure where your analysis (looks like an Engineers approach to me! ) is going astray, but, since what you are trying to tell me, does not correspond to the physical situation or to the solutions to the heat equation there must be some misinterpretation somewhere. I still believe it to be in the magnitude of the instantaneous effects. (They are vanishingly small)
17. Oct 26, 2004
### Integral
Staff Emeritus
Ok, I finally found the solution I was looking for.
Warning! big PDF!
Warning! poorly scanned PDF! Perhaps you can just turn your monitor on its side?
Warning! Involved Mathematics!
Enjoy!
This is from http://home.comcast.net/~rossgr1/Math/Churchill.pdf [Broken] Operational Mathematics
Last edited by a moderator: May 1, 2017
18. Nov 13, 2004
### Clausius2
After have been gathering enough will power to download your pdf, I took a look at it, and it seems Churchill, you and me agree absolutely.
Churchill gives a solution in terms of error function, which is smooth and without discontinuities as you mentioned. And he says the heat flux is transmitted instantaneously all over the domain once the boundary condition at x=0 has changed.
Well, we knew that yet. I was only wondering me if an hyperboilc behavior wouldn't be more adapted to the real nature of the phenomena. I was comparing the heat propagation with an elastic or shock wave, in which discontinuities are present. As a matter of fact, if we look at the heat transfer problem with time scales such that t--->0, it is not possible at all that the heat flux is felt instantaneously over the entire domain as Churchill and you said.
To sum up,
i) the Heat Equation yields an instantaneous propagation of thermal effects, due to its parabolic behavior.
ii) the Wave Equation yields a finite velocity of propagation of some type of wave (I'm not going to specify which)
iii) All of us know that instantaneous effects (infinite propagation velocity) are impossible.
Therefore, I would say maybe exists a more accurate formulation of heat phenomena taking into account this needed discontinuity in its propagation.
The question is:
Would this formulation be worth of being calculated?
19. Nov 13, 2004
### dhris
I don't see how you're coming to the conclusion that the heat travels instantaneously. See that alpha in the equation? $$L^2/\alpha$$ is a characteristic timescale for heat to redistribute itself around your domain of size L. To see that, just think of your problem with an initially very large temperature at the centre of your domain that is zero everywhere else (as in a delta function). You will find that the solution gradually becomes a wider and wider gaussian, but does not in fact transmit heat instantaneously.
dhris
20. Nov 14, 2004
### Clausius2
False. If you don't believe my words read the PDF of Integral, or take a look at some book of PDE's. Also, take a look at another posts of mine written here before, I have derived yet your time scale before, and I have given it a physical meaning. That meaning has nothing to do with a propagation of a wave, (I'm referring to the mathematical meaning of wave), it is only the characteristic time of propagation until the steady regime is reached.
The parabolic behavior of the Heat Equation must be a enough proof of that instantaneous effect propagation. You have mentioned the words:
"You will find that the solution gradually becomes a wider and wider gaussian"
Yes!!! gradually!!. The word gradually has nothing to do with a wave. An elastic wave is not propagated gradually. There, some points of the domain feel suddenly an increasing of pressure. The gradual flattering of the gaussian is felt infinitesimally in each point of the domain. The gaussian function is defined, smooth and continuous all over the domain. It hasn't got any discontinuity as a wave. Therefore, the flattering is felt over the entire region. Think of it, you are not able of flattering any continuous function without modifying any point of the domain (except the boundaries).
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2017-05-24 21:42:40
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http://mathhelpforum.com/trigonometry/7853-trigonometric-identities-print.html
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# Trigonometric Identities
• Nov 21st 2006, 01:42 PM
asiankatt
Trigonometric Identities
Hi there .
I'm not sure on how to do this question so I wondering if I could get some help . I'd greatly appreciate it :
Create a trigonometric identity that requires at least 3 steps in the solution to prove it is an identity. The identity must also include cosecant, secant or cotangent. Provide solution .
HUGE Thank you in advance .
• Nov 21st 2006, 02:14 PM
Jameson
This isn't as hard as it sounds.
Take something simple like $\sec(x)$
Well $\sec(x)=\frac{1}{\cos(x)}$
And $\sin^2(x)+\cos^2(x)=1$, so $\sec(x)=\frac{1}{\sin^2(x)\cos(x)+\cos^3(x)}$
Take it one more step. :)
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2017-08-23 21:53:02
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https://stats.stackexchange.com/questions/433673/obvious-demonstration-of-the-posterior-density-of-normal-laws-in-bayesian-infere/433784
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# Obvious demonstration of the posterior density of normal laws in bayesian inference
Given the observations' density $$f(X|\theta) = \mathcal{N}(\theta,1)$$ and the prior density $$f(\theta) = \mathcal{N}(0,\sigma^2)$$, it it is said that it is obvious that the posterior density is equal to : $$\mathcal{N}(\frac{X}{1+\sigma^{-2}},\frac{1}{1+\sigma^{-2}})$$.
I use following relation to compute the posterior density : $$f(\theta|X) = \frac{f(X|\theta)f(\theta)}{\int f(X|\theta)f(\theta)d\theta}$$, but I dont find it obvious at all...
Any help appreciated
• Obiousness is in the mind of the beholder: it depends on what one knows and can bring immediately to mind. Thus, your question is predicated on some understanding of what you are expected to know. For instance, are you familiar with the theory of linear regression? With the bivariate Normal family of distributions? What material would be appropriate to invoke when providing an answer?
– whuber
Oct 29 '19 at 17:41
• @whuber this the correction of an exercise in a statistic course. Here the obviousness means that it don't need much computation to arrive at this result. Knowledge on normal distribution and bayesian modeling are effectively assumed Oct 29 '19 at 18:52
\begin{align} f(\theta|X) &\propto \exp(-\frac{1}{2}(X-\theta)^2)\cdot \exp(-\frac{1}{2}\sigma^{-2}\theta^2)\\ & \propto \exp(-\frac{1}{2} ((1+\sigma^{-2})\theta^2 - 2 \theta X + X^2)) \\ & \propto \exp(-\frac{1}{2} (1+\sigma^{-2}) (\theta^2 - 2 \theta \frac{X}{1+\sigma^{-2}})) \\ & \propto \exp(-\frac{1}{2} (1+\sigma^{-2}) (\theta - \frac{X}{1+\sigma^{-2}})^2) \end{align}
The latter is the kernel of a normal $$\mathcal N (\frac{X}{1+\sigma^{-2}}, \frac{1}{1+\sigma^{-2}})$$.
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2021-12-02 15:51:54
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http://mathoverflow.net/questions/86838/hamming-graphs-and-power-series
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
## Hamming graphs and power series [closed]
Possible Duplicate:
Hamming graphs and power series
Let $i$ and $h$ be two adjacent nodes in a Hamming graph and let $a$ be any positive real. Let us denote by $d_{ij}$ the distance between node $i$ and node $j$ in the graph. I'm trying to find a compact formula for the following (finite) power series, where $j$ runs over all nodes of the graph:
$$S= \sum_j a^{d_{ij}+d_{hj} }. \tag{1}$$
Is this a studied problem?
If the graph is an $n$-hypercube, which is the case I'm most interested in, it seems that
$$S = \sum_{d=0}^n {{n-1}\choose {d}}a^{2d+1} + {{n-1}\choose {d-1}}a^{2d-1}\tag{2}$$
In fact, if you fix node $i$ and consider all the nodes at distance $d$ from $i$, ${{n-1}\choose {d}}$ of them are at distance $d+1$ from $h$, while the remaining ${{n-1}\choose {d-1}}$ are at distance $d-1$. However, I fail to see now how (2) can be further simplified.
I would appreciate any reference to this problem.
Thank you, M.
-
Doesn't (2) immediately simplify to $2\sum\limits_{d=0}^{n-1}\binom{n-1}{d}a^{2d+1}$ ? Which is $2a\left(a^2+1\right)^{n-1}$ by the binomial theorem? – darij grinberg Jan 27 2012 at 16:30
Darij, I'm glad we covered all copies of the question. Gerhard "Ask Me About System Design" Paseman, 2012.01.27 – Gerhard Paseman Jan 27 2012 at 16:35
Ops, yes, you are right :) Any suggestion for the more general case (1)? M. – Michele Jan 27 2012 at 16:38
Yes, but check the other copy for a suggestion. Gerhard "Ask Me About System Design" Paseman, 2012.01.27 – Gerhard Paseman Jan 27 2012 at 16:48
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2013-06-19 07:24:11
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{"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": 0.9404470324516296, "perplexity": 297.0015934814673}, "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/1368708143620/warc/CC-MAIN-20130516124223-00002-ip-10-60-113-184.ec2.internal.warc.gz"}
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http://legisquebec.gouv.qc.ca/en/showversion/cs/I-8.1?code=se:103_4&pointInTime=20210906
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### I-8.1 - Act respecting offences relating to alcoholic beverages
103.4. In proceedings for contravention of section 103.1 or 103.2, the permit holder shall incur no penalty if he proves that he used reasonable diligence to ascertain the age of the person and that he had reasonable ground for believing that that person was of full age or if he proves that he had reasonable ground for believing that it was a case contemplated in the second paragraph of section 103.2.
1979, c. 71, s. 128.
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2021-10-25 08:46:39
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{"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": 0.8366407155990601, "perplexity": 2790.266276743814}, "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-20211025061300-20211025091300-00139.warc.gz"}
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https://davidblue.wtf/styleguide/
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⸎ David Blue Styleguide ⸎
Saturday, July 31, 2021 9:00:48 PM
David Blue's personal style guide across writing, social posts, etc.
In a desire to quantify/document my own personal preferences in composition, I have contemplated writing a style guide like this for a very long time.
The English language is no more than a tool for our use, and grammar should not be revered as an immalleable, Holy absolute which we must venerate and obey. Rather, it should be irreverently and liberally manipulated to perform as desired.
⸎ Footnotes and Endnotes come after both punctuation and quotation marks.
• "WiFi"
• “sortof”
• “featureset”
• “Web Site”
Do Not
• "I'd be remiss..."
• "Really struck me."
• "At the end of the day..."
• "When it's all said and done..."
• "In the final hour..."
• "Out of all the , really stood out to me..."
• "In my book..."
• "In my estimation..."
• All idioms.
• All platitudes.
• The “tldr” school of thought.
I will write what I have not read.
Scribam Quid Non Legerim
Scribam quid non legerim is possibly grammatically incorrect to a scholar, but it’s the best possible translation I came up with in my Latin research of “I will write what I have not read.” It’s cheesy, yes, and a bit cringey in the middle of just any old day when it happens to catch my eye where it’s proudly displayed, all-caps, in the footer of our CMS, and – I’ll be honest – I don’t know if I could explain it over coffee to a stranger without turning red and covering my face, as I once could, but it’s (sincerely, in this one case) real gravestone material. (As in, if someone were to read this after my death, they would be encouraged to receive it as a bonafide last wish.) -"A Brief, Unedited Reflection on the History of my Failed Media Projects"
I want to be deafened by the penultimate death rattle of a circuit bent sacrificial ritual and indiscriminately punched in the throat at a First Friday show.
Here’s what I know: the worthwhile publications have not been crucified on the Cross of “Readability,” the genre-busting song talent has not been given up in the face of the streaming service-afflicted music industry.
Extratone Style Guide
This Style Guide exists for the purposes of both technical and editorial reference for internal and external use when composing, editing, citing, quoting, capturing, or otherwise manipulating or displaying any/all Extratone content.
Editorially
As an independent, youth-oriented publication it is both our privilege and our duty to seek out innovative methods of communication. In written composition, this means all possible avoidance of unnecessary cliche and/or overused terms (for instance: use of the term “struck” in a metaphor - e.g. what struck me) except where they are critically portrayed. It is essential to remember that the English language is no more than a tool for our use, and grammar should not be revered as an immalleable, Holy absolute which we must venerate and obey. Rather, it should be irreverently and liberally manipulated to perform as desired. Grammar should never be an obstacle or burden in our writing or editing, and should always be discarded in favor of innovative and/or contextual verse.
In general, we write, speak, and publish under the assumption of an intelligent, discerning audience, meaning that we are unafraid of traversing obscure, unusual, or extraordinary detail (technical or otherwise,) that we do not patronize beyond the reasonable ability of a modern reader’s search engine, and especially that we format copy independently of common web standard “readability. Our vocabulary and syntax are diverse (though we do not sacrifice undue time to unearthing bizarre, senseless synonyms,) fluid, and occasionally difficult, but written as such for the purpose of colorful, substantial delivery in acknowledgement of the vast resources instantly available to the modern reader. In other words: do not tell what can simply and easily be looked up. With search engines as they are, this is a quickly verifiable/quantifiable factor with a given definition, summary, or abstract. If a writer is in conflict regarding the redundancy of a given bit of information, they can painlessly search for it themselves and observe the same results as the reader would within milliseconds to observe what would and would not be immediately evident. Not only is this approach respectful of the readership’s intellect, it is also a comparatively democratic means of regarding information.
Specifics of Note
• Donald Trump’s last name is to be spelled T-U-M-P. (e.g. The 45th President of the United States, Donald Tump.)
• The term “struck” is to be avoided if at all possible, especially in metaphoric context (e.g. what struck me.)
• Use of the more unnecessary compositional terms from both lower and post-secondary academia should be avoided. (e.g. in conclusion, in summary, and other cliches.)
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2022-08-20 03:07:30
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https://repo.pw.edu.pl/info/bachelor/WUTbff0b7a5df064f0ea9589822937d7118/
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Knowledge base: Warsaw University of Technology
Back
Microbiological corrosion of constructional materials in the fuel enviroment
Beza Waldemar
Abstract
The purpose of the study has been to examine the process of microbiological corrosion occurring in fuel tanks and transmission installations, determine the impact of the nickel layer on the reduction in the corrosion damages taking place on St3S construction steel - a material used in the fuel industry. The mechanically cleaned and degreased samples of St3S steel were covered with an anticorrosive coating from galvanic and chemical nickel. The samples protected with a layer of nickel were exposed in the corrosive environment that was a mixture of water and fuel taken from the propulsion fuel tanks. During the multi-day exposition of the samples in the water phase and near the limit of the water-fuel phases, the progress of corrosion damages was determined. The corrosive resistance tests of tested samples were performed using electrochemical methods (polarization and impedance method) and gravimetric method The electrochemical tests were performed using a computerized electrochemical tests set - ATLAS-98. The polarization tests were performed in the beginning (after 1 day of exposition in the corrosive environment) and after completion of the tests (after 60 days of exposition). The speed of corrosion of tested samples has been determined on the basis of cathodic and anodal polarization curves. The polarization curves have been made using the potential-dynamic method while polarizing the samples with the velocity of changes amounting to 10mV/min. The impedance characteristics of the tested samples have been made using a computer program IMP-98. The impedance tests were conducted at changing frequencies in the range from 0.001 to 100 kHz and the variable-current impulse amplitude of 10mV. The acquired data have been analyzed with respect to two substitute model systems. In the gravimetric tests, the speed of corrosion has been determined on the basis of depletion of masses of the examined samples during exposition in the corrosive environment. The speed of corrosion has been determined from the masses depletions in Vm (g/m2 day) and Vp (mm/year). The corrosion resistance of nickel coatings has been assessed on the basis of the test results. The acquired results demonstrate that nickel coatings fully protect the St3S steel against microorganisms present in fuel storage tanks and in the systems used for their transport. The speed of corrosion of steel covered with a nickel coating created chemically is within the range 0.001-0.005mm/year. (corrosive resistance group 2). The samples show high resistance corrosion resistance is shown by the steel covered with a layer of galvanic nickel. All test results achieved for the times above 14 days of exposition qualify the steel to the highest corrosive regardless of the place of their exposition in the corrosive environment. Even greater resistance group - 1
Diploma type
Engineer's / Bachelor of Science
Diploma type
Engineer's thesis
Author
Beza Waldemar Beza Waldemar,, Undefined Affiliation
Title in Polish
Korozja mikrobiologiczna tworzyw konstrukcyjnych w środowisku paliw
Supervisor
Marek Kamiński (FMSE) Marek Kamiński,, Faculty of Materials Science and Engineering (FMSE)
Certifying unit
Faculty of Materials Science and Engineering (FMSE)
Affiliation unit
Division of Surface Engineering (FMSE/DSE)
Study subject / specialization
, Inżynieria Materiałowa
Language
(pl) Polish
Status
Finished
Defense Date
26-01-2010
Issue date (year)
2009
Internal identifier
IM-D.001431
Reviewers
Krzysztof Rożniatowski (FMSE/DMD) Krzysztof Rożniatowski,, Division of Materials Design (FMSE/DMD)Faculty of Materials Science and Engineering (FMSE) Joanna Ryszkowska (FMSE/DSMP) Joanna Ryszkowska,, Division of Ceramic Materials and Polymers (FMSE/DSMP)Faculty of Materials Science and Engineering (FMSE)
Keywords in Polish
korozja
Abstract in Polish
urn:pw-repo:WUTbff0b7a5df064f0ea9589822937d7118
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2021-06-20 12:10:39
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https://blender.stackexchange.com/questions/28991/straighten-camera-in-blender
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# Straighten Camera in Blender
If I have a camera at 0/0/0 of the x/y/z and its heading is unknown.
How do I make it point in the direction of 1/0/0 quickly?
• Rotate it 90 degrees, R X 90 – iKlsR Apr 26 '15 at 16:10
• I updated the question to clarify. The camera is located a 0/0/0 but the direction in which it points ("heading") is unknown. I want it to point towards 1/0/0. – qubodup Apr 26 '15 at 16:15
The fastest way if you haven't applied any transforms on the camera is to reset the rotation with AltR to 0/0/0 and then rotate it 90° on the X and -90° on the Z, RX 90, RZ -90.
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2020-12-03 04:34:38
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http://www.philippinekabu.com/cheap/romania/15843012f1aef79e1b15cca5112
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# elementary statistics guided notes
Set of all possible elementary outcomes of a trial.?
101.
Candidates can find different tips and tricks from below for solving the questions related to Elementary Statistics. Tip # 1: Variate The quantity that we measure from observation to observation is called a Variate. Eg: 1.5, 2, 2.5 etc. are called variates. GUIDED NOTES.
Shop by category. Elementary Statistics with Integrated Review and Guided Workbook Plus MyLab Statistics with Pearson eText -- 24 Month Access Card Package.
The task of writing these lecture notes was made much easier after that. Chapter-aligned technology manuals help students learn to use calculators, Minitab, and Excel.
4.09 avg rating 11 ratings published 2006 4 editions. Guided notes or flippables are provided for each lesson as well as an activity and a practice sheet. Larson and Farbers Elementary Statistics: Picturing the World, 6th Edition, provides stepped out instruction, real-life examples and exercises, and the use of technology to offer the most accessible approach. Every textbook comes with a 21-day "Any Reason" guarantee.
Bhattacharyya and Johnson, Statistical Concepts, John Wiley & Sons 2.
Guided student notes provide instructors the framework of day-by-day class activities while students gain the framework of well-organized notes. No Notes: ANOVA &
Exercise 29.
It has been developed around three central themes: clarity, quality, and accuracy, based on extensive market research and feedback from statistics instructors across the country.
You will be allowed to bring handwritten notes to each test, and those must be turned in with the test. Most have no idea of what statistics is about. Lecture 1 notes for Tosha Lamar's MATH 1070 class, Lecture 1 notes for Tosha Lamar's MATH 1070 class lecture graphing distributions what is statistics? Comprehensiveness rating: 5 see less.
They Chapter 2 - Describing, Exploring, and Comparing Data.
Shop by category. Sell, buy or rent Elementary Statistics 9781259969454 1259969452, we buy used or new for best buyback price with FREE shipping and offer great deals for buyers. Retain information as you learn Statistics.
Read Paper. Most of the courses that you will take at Smart Schools have accompanying Guided Notes that will help you: Track better with the instruction videos.
Two-Way Analysis of Variance.
Examples: 1. The good news is that the math skills . Be guided through every step of the fundamentals of statistics. Exercise 29 from Elementary Statistics, 6th edition.
Included are visual descriptions, unique chapter
Beginning with basic tasks and computations with R, readers are then guided through ways to bring data CO 2 emissions 2 The International Energy Agency also reported the CO 2 emissions (measured in gigatons,Gt) from fossil fuel combustion for the top 9 countries in 2011.
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Chapter 1 Principles of experimental design 1.1 Induction Much of our scienti c knowledge about processes and systems is based on induction: reasoning from the speci c to the general. Study Resources.
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Practicing Statistics: Guided Investigations for the Second Course. Incorporates calculators, Minitab, and Excel screenshots.
As you make your way through your coursework, Guided Notes are one of the most helpful tools available to you.
Elementary Statistics (MATH 1070) Lectur e 1 Gr aphing. TYPES OF STATISTICS Descriptive statistics A type of statistics, which focuses on the collecting, summarizing, and presenting a mass of data so as to yield meaningful information. Dis tributions.
ISBN. About. Each day has a Power Point that includes a warm -up with answers, notes and a closure of the lesson. The branch of mathematics in which we study about the collection, organization, analysis, interpretation and presentation of data (information) is referred to as Elementary Statistics.
no.
Creating Pie Charts on the TI-82.
these lecture notes and none of the questions found in the exams for the course on which these lecture notes are based require calculus. Beginners' statistics (using R) Introduction Simple formulae.
1. Rent Guided Workbook for Elementary Statistics with Integrated Review 1st edition (978-0134495439) today, or search our site for other textbooks by Mario F. Triola. This is a 9 day unit on Statistics. Includes ALEKS, ALEKS Prep, SmartBook, ConnectMath, MegaStat, and ReadAnywhere. HSS-ID.A.2 Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets.
This book covers all necessary content areas for an introduction to Statistics course for non-math majors. Included in this chapter are the basic ideas and words of probability and statistics. The following questions are examples of such cases. Standard deviation Populations and Replicates.
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Introduction to Statistics and Lists on the TI-82. Topics include: basic combinatorics, random variables, probability distributions, Bayesian inference, hypothesis testing, confidence intervals, and linear regression. Statistics, Population, Sample, Random Sample, Observational Unit, Variable, Quantitative, Numerical, Qualitative, Categorical Short lecture videos introduce concepts, definitions, formulas, and problem-solving procedures. We are guided by the precise rules and laws that we set up. Guided Workbook for Elementary Statistics with Integrated Review 1st Edition . About. The boys range is 1 greater than the girls range.
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The classical definition of probability (classical probability concept) states: If there are m outcomes in a sample space (universal set), and all are equally likely of being the result of an experimental measurement, then the probability of observing an event (a subset) that contains s outcomes is given by From the classical definition, we see that the ability to count the number
Elementary Statistics Textbook (pdf copy) Reference sheets. Statistics and probability carry 15% weightage in both foundation and higher tier. Grouped Frequency Distributions. Elementary Statistics: Picturing the World (6th Edition) answers to Chapter 1 - Introduction to Statistics - Section 1.1 An Overview of Statistics - Exercises - Page 6 1 including work step by step written by community members like you. For now we will be dealing This bar-code number lets you verify that you're getting exactly the right version or edition of a book.
( Source) As a college elementary statistics tutor and graduate student in a Master of Science in Applied Statistics program, I feel like I know a thing or two about passing, and even acing, a statistics course. Elementary Statistics, Third Edition is a conceptual and procedural course in introductory statistics. Statistics, Probability, and Key Terms Before taking a statistics class, most people associate the term \statistics" with what a statistician would call descriptive statistics.
The R Companion to Elementary Applied Statistics includes traditional applications covered in elementary statistics courses as well as some additional methods that address questions that might arise during or after the application of commonly used methods.
The text book provides an effective index, plenty of exercises, review questions, and practice tests. They can be used for introduction of topics or to reflect/review what you have covered in class discussion.
lulu@Support Course for Elementary Statistics@Larry Green@Lake Tahoe Community College@Support Course for Elementary Statistics.
The notes contain the usual topics that are taught in those courses as well as a few extra topics that I decided to include just because I wanted to.
Types of data Transformation and recoding Proportions.
Unit 2 Notes - Randomness and Probability.
We are now using the 2nd edition of Triola's Essentials of Statistics, and the section numbers are slightly different.. View STA2023_Chapter 05_guided_notes_191.pdf from MAT 3180 at Shorter University.
Lecture Notes for Introductory Probability Janko Gravner Mathematics Department University of California Davis, CA 95616 [email protected] June 9, 2011 These notes were started in January 2009 with help from Christopher Ng, a student in Math 135A and 135B classes at UC Davis, who typeset the notes he took during my lectures.
STAT 50 Elementary Statistics Guided Notes To be used in conjunction with Elementary Statistics 13th Edition by. Unit summary sheet: create a studyguide before each chapter test & earn a bonus point on the test. An event can consist of a single outcome or a set of outcomes.
MATH-1342 Spring 2013 01/14/2013 - 05/12/2013 Course Information.
Chapter 6 notes that, in cases where the population standard deviation is unknown and either the population is normal or the sample size is greater than 30, you can use the variable, t, in computing the confidence interval. About.
This book is pretty comprehensive for being a brief introductory book.
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1 Full PDF related to this paper. Elementary Statistics.
Guided Notes for College Algebra (GGC), Rabia Shahbaz and Janice Alves. Guided Notebook Filled out for chapters 2 to 3.2. A. Its guided approach builds up every students confidence, through step-by-step objectives, lots of practice, meaningful personalization activities, and exam preparation tasks.
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The Spring 2014 version of this subject employed the residential MITx system, which enables on-campus subjects to
Author-provided supplements play an important role in the educational process:Editable PowerPoint presentations are broken down by learning objective, with multiple teaching styles incorporated into each series. 2016 KCATM STATISTICS AND PROBABILITY 7th Grade PAGE 1 Use the box plot on heights of 7th graders to answer problems #101-105. Interactive Textbook: Interactive Statistics.
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The ranges are the same.
Elementary Statistics, CA 2nd Edition By Triola.
Triolas Elementary Statistics with Integrated Review is a new co-requisite course solution, offering a complete introductory statistics MyStatLab course with integrated review of select topics from developmental algebra.
Elementary Statistics 6th Edition, 2015 Larson ISBN: 9780133541441 HS Binding A Better Teaching and Learning Experience Stepped out instruction and guided student learning through an abundance of exercises build students knowledge and skills in statistics. Filled out guided notes from class with Professor Ginger Nagel january 14th 10.1 lines and angles the study of relationships among lines, angles, surfaces, and This course solution may be used in a co-requisite course model, or simply to help under-prepared students master prerequisite skills and concepts. $3.95 Shipping. Add to Cart. more than 100 math- and statistics-specific case studies documenting the impact that educator best practices have had on student learning using our digital solutions. The aim of this course is acquaint a student with some of the ideas, de nitions and concepts of statistics. eBook details Author: William Navidi File Size: 41 MB Format: PDF Length: 816 pages Publisher: McGraw-Hill; 3rd edition Publication Date: January 23, 2018 Language: English ASIN: B078SPMW2B ISBN-10: 1259969452, 1260092569 1260373525, 1260487539 ISBN-13: 9781259969454, 9781260092561, 9781260373523, 9781260487534 Navidis Elementary Creating an Ogive on the TI-82. Numerical computation, algebra and graphs are used; calculus is not used. ISBN-13: 978-0134495439. Print out any course materials, guided notes, etc. Centers around three themes: clarity, quality, and accuracy. Teacher Notes: Likelihood Scale This activity is an introduction to chance and likelihood. Company. Set books The notes cover only material in the Probability I course. The probability of the sample space is always 1. Know More About Elementary Statistics Term. Elementary Statistics William Cyrus Navidi, Barry Monk 4th International Edition. What is Elementary Statistics? The branch of mathematics in which we study about the collection, organization, analysis, interpretation and presentation of data (information) is referred to as Elementary Statistics. Eg: the collection of children of different ages in a city, the collection of marks obtained by students in different subjects etc. It is almost always nec- 4.0. Scheffe' and Tukey Tests. Linegraphs Scatterplots. See more ideas about math activities, middle school math, math. unit 2 notes.pdf 163.312 KB (Last Modified on December 5, 2016) Comments (-1) ( n) A regional or social variety of a language distinguished by pronunciation, grammar, or vocabulary, especially a variety of speech differing from the standard literary language or speech pattern of the culture in which it exists: Cockney is a dialect of English. HSS-ID.A.1 Represent data with plots on the real number line (dot plots, histograms, and box plots). Determine the strength and into groups constitute two portfolios were bound to elementary statistics bluman lecture notes microeconomic theory. The withdrawal deadline is March 15th, 2022 ( withdrawal form) Open House Video for Elementary Statistics. You will also learn how data are gathered and what "good" data are. Creating Histograms, Box Plots, and Grouped Frequency Distributions on the TI-82. ( Source) 2. It is a great introduction to statistics for college students who have a basic grasp of algebra. STA2023, Elementary Statistics (Professor Younts) The Standard Deviation as a Ruler and the Normal Model Guided Definitions. 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Chong Ma (Statistics, USC) STAT 201 Elementary Statistics 21 / 33. We are using descriptive statistics when we use numbers or graphs to summarize a data set by calculating a mean or making a bar graph, for example. Triola Stats creates the most effective tools for statistics education, including the market leading Triola Series textbooks. Ch1 Larson/Farber 1.1 An Overview of Statistics 1 2. The Teachers book includes notes for every Download elementary statistics mario triola 11th edition Bing book pdf free download link or read online here in PDF. The level of aptitude in this subject will assist students wishing to excel on the SAT and in college courses. Guided notes for most sections covered in class will be posted under "Shared Files" in our MyTCC classroom (Blackboard). The guided notes will show you the formulas and definitions from your text and provide additional examples. Focus: Math 1342, Elementary Statistics, will familiarize students with the science of statistics. It consists of a sequence of bars, or rectangles, corresponding to the possible values, and the length of each is proportional to the frequency. For each test, you may take one 3"x 5" index card, front and back. Brase and Brase, Understanding Basic Statistics, 8th ed., Cengage Publishing 3. Through research and experience, we came to the conclusion that guided notes are the most effective method of note taking. STAT 95, Elementary Statistics, Sec. A short summary of this paper. AP Statistics. Piecharts Dotplots Bargraphs Histograms Cumulative plots. Full PDF Package Download Full PDF Package. These guided/scaffolded note-taking activities introduce students to 9 units of statistics (over 290 pages of useful materials). ISBN-10: 0134495438. Notes and Practice Problems for OpenStax Calculus I, Patcharin Marion and Ishwari Kunwar. The task of writing these lecture notes was made much easier after that. In addition, post-it notes, stapler, ruler, colored pencils, dice, coins, and a deck of cards can be useful for activities. Student Solutions Manual for Elementary Statistics: Picturing the World. Math 116 - College Algebra Over 50% of students report bringing their laptops to class at least once per week. In this unit of study we will try to improve the students understanding of the elementary topics included in statistics. Class notes MATH 1680 Chapters 2-.32 ISBN: 9780134081229 (0)$7.49. Day One- Statistical Questions Determine if a question is a statistical question or not. Readings Course Description: A first course in statistics for students in business; nursing; allied health; or the social, physical, or behavioral sciences; or for any student requiring knowledge of the fundamental procedures for data organization and analysis.Topics include frequency distributions, graphing, measures of location and variation, the binomial and normal There is biased sample size needed.
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The primary source of this anxiety seems to be a general math anxiety. Allan Bluman Elementary Statistics A St. Andrew Williams. Mean=\frac {\text{sum of all the elements}}{\text{number of elements}} Median:Median is defined as the middle value of a set of data.If a set consists of an odd number of values, then the middle value will be the median of Outline 1 Introduction 2 Gathering Data Chong Ma (Statistics, USC) STAT 201 Elementary Statistics 2 /
Welcome to my math notes site.
Math 365, Elementary Statistics Home Lesson 1 Lesson 2 Lesson 3 Lesson 4 Lesson 5 Lesson 6 Lesson 7 Lesson 8 Lesson 9 symbols .
You will soon under-stand that statistics and probability work together.
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Reading and writing competencies are changing with the use of digital technologies, though students still use paper and pen to communicate their feelings. Mean:Mean can be defined as the sum of all the elements divided by number of all elements.
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Probability and Statistics for Computer Science Lecture Notes, Version 3.3 Made available in .pdf form to the STAT 391 students in Spring 2010. These lecture notes were copied from the Math 170 lecture notes and then modified to go with Triola's Elementary Statistics textbook. A math teacher wants to determine the percentage of students who passed the examination. Chapters 2-7 Filled Out [Show more] 3 items . Elementary Statistics (GHC) (Open Course), Camille Pace, Katie Bridges, Laura Ralston, Elizabeth Clark, Brent Griffin, Kamisha DeCoudreaux, Zac Johnston, and Vincent Manatsa. One-Way Analysis of Variance on the TI-82. The text-books listed below will be useful for other courses on probability and statistics. Statistics.
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Access our elementary statistics book online for free. Jul 13, 2020 - Data and Statistics . Digital tools are changing the way we take notes. Download Download PDF. Elementary Statistics A Step By Step Approach + CD (Elementary Statistics A Step By Step Approach ( International Edition)) by Allan G. Bluman. C. $74.90 + Elementary Statistics Lecture 1 Chong Ma Department of Statistics University of South Carolina [email protected] Chong Ma (Statistics, USC) STAT 201 Elementary Statistics 1 / 18. Descriptive statistics is used to grouping the sample data to the fol-lowing table Outcome of the roll Frequencies in the sample data 1 10 2 20 3 18 4 16 5 11 6 25 Inferential statistics can now be used to verify whether the dice is a fair or not. Quantiles Maxima Minima Ranges. Lesson 3 : Probability . Task 15. Brase and Brase, Understanding Basic Statistics, 8th ed., Cengage Publishing 3. Technology Manuals for Elementary Statistics (downloadable) Technology-specific manuals for Graphing Calculator, Excel, and Minitab include tutorial instruction and worked-out examples from the book. Each manual can be downloaded from within MyLab Statistics. Utilize the fun video and text lessons in our Introduction to Statistics textbook replacement course to teach your students about this subject. The last test will be in class on the last day of class. If you have any specific question Both are easy to understand. Students often approach their first statistics course with some anxiety. Canada 0.5 China 8 India 1.8 Iran 0.4 Germany 0.8 Japan 1.2 Measures of Central Tendency (Averages) Add to cart form. Guided student notes provide instructors the framework of day-by-day class activities while students gain the framework of well-organized notes. Ch1 Larson/Farber Objective The Executives will Learn and Understand The Definition of Statistics. This course provides an elementary introduction to probability and statistics with applications. Contained in this site are the notes (free and downloadable) that I use to teach Algebra, Calculus (I, II and III) as well as Differential Equations at Lamar University. The field needs data-driven decision making and evidence-based practices to become more ubiquitous than not. Preparing future social workers by teaching introductory statistics is essential to meet that goal. This textbook offers a fairly comprehensive summary of what should be discussed in an introductory course in Statistics. The authors carefully develop theory through strong pedagogy, and examples show how statistics is used to picture and describe the world. o For more info: by Allan G. Bluman. Send comments to: [email protected]. Find the difference in ranges from the girls (top graph) to the boys (bottom graph). Elementary Statistics and Probability Guided Notebook. One-Way Analysis of Variance. Definitions. An event is the speci cation of the outcome of a trial.? View Stat 50 Guided Notes Triola 13th ALL.pdf from STAT 50 at Pasadena City College. required in this course are fairly basic. There are many different types of guided notes to consider, all with varying pros and cons. This is an introductory course in statistics. Elementary Statistics:Elementary Statistics Student's Solutions Manual (10th Edition) by. The following three excellent textbooks have shaped my approach to teaching probability and statistics: 1. Most students have enrolled in statistics because it is required. It will then proceed to explain and construct frequency distributions, dot diagrams, histograms, frequency polygons and cumulative frequency polygons. Recommend this! Probability and Statistics for Engineering and the Sciences by Jay L. De- Main Menu; by School; by Literature Title; by Subject; Textbook Solutions Expert Tutors Earn. Elementary Statistics. Textbook Authors: Larson, Ron; Farber, Betsy, ISBN-10: 0321911210, ISBN-13: 978-0-32191-121-6, Publisher: Pearson Probability and Statistics Vocabulary List (Definitions for Middle School Teachers) B Bar graph a diagram representing the frequency distribution for nominal or discrete data. Show TOC. 1 of 5 stars 2 of 5 stars 3 of 5 stars 4 of 5 stars 5 of 5 stars. You need at most one of the three textbooks listed below, but you will need the statistical tables. Visit us to learn more. Want to Read.$4.48.
F distribution and F-test.
Introduction to Statistics Where Youve Been Where Youre Going 1 1.1 An Overview of Statistics 2 1.2 Data Classification 9 Case Study: Reputations of Companies in the U.S. 16 1.3 Data Collection and Experimental Design 17 Activity: Random Numbers 27 Uses and Abuses: Statistics in the Real World 28 Chapter Summary 29 Review Exercises 30
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# elementary statistics guided notes
これから伸び行くアジアの雄「フィリピン」で株の売買をはじめましょう！
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2022-12-02 10:37:15
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https://socratic.org/questions/why-is-cus-o-3-called-copper-ii-sulfite
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# Why is CuSO_3 called copper (II) sulfite?
Nov 22, 2017
The Copper comes from the Cu in the formula. The (II) comes from the oxidation number of the copper in the complex. The Sulfite comes from the $S {O}_{3}$ ion.
Nov 22, 2017
Because it is the salt of $\text{sulfurous acid}$...${H}_{2} S {O}_{3}$ NOT $\text{sulfuric acid}$, ${H}_{2} S {O}_{4}$.......
#### Explanation:
The salts of $\text{sulfuric acid}$ are called $\text{sulfates}$....in which sulfur assumes the $S \left(V I +\right)$ oxidation state...
On the other hand, the salts of $\text{sulfurous acid}$, ${H}_{2} S {O}_{3}$, are called $\text{sulfites}$,....in which sulfur assumes the $S \left(I V +\right)$ oxidations state...The counterion is $S {O}_{3}^{2 -}$...
And the salts of $\text{hydrogen sulfide}$ are called $\text{sulfides}$, the analogue of oxides, and here we gots $S \left(- I I\right)$....
Sometimes we represent ${H}_{2} S {O}_{3}$ as ${H}_{2} O \cdot S {O}_{2}$, i.e. water saturated with $S {O}_{2}$. The smell of sulfurous acid is sulfurous, but many people, including me, find the smell pleasant, and clean, and antiseptic.
So finally, $C u S {O}_{4} \equiv \text{copper sulfate}$, and $C u S {O}_{3} \equiv \text{copper sulfite}$, and $C u S \equiv \text{copper sulfide}$. And in each salt we gots $C {u}^{2 +}$, $\text{cupric ion}$.
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2022-08-18 17:42:04
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https://www.gmsms.in/mod/assign/view.php?id=236
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1.171
A shell flying with velocity v = 500 m/s bursts into three identical fragments so that the kinetic energy of the system increases $$\eta$$= 1.5 times. What maximum velocity can one of the fragments obtain?
Solution : Real lousy one and you can find even lousier solutions. We also made a lousy assumption that the fastest fragment should move along the CM of the system, nothing else. This is not possible to get the answer given in the book without such an assumption. Though other cases are possible and I can provide you many of them but we are mostly relying on the word "maximum:. This is better if we work out in a frame the CM of the system is at rest. I will be using ' notation for physical quantities appearing in the CM frame.
No doubt about $$\vec{p'_{1}}+\vec{p'_{2}}+\vec{p'_{3}} =0$$
Let $$\vec{p'_{2}}+\vec{p'_{3}} = \vec{P}$$
i.e. $$\vec{p'_{1}}$$ is equal and opposite to $$\vec{P}$$.
Thus their corresponding kinetic energies can be given as $$\frac{p_{1}'^{2}}{2m}$$ and $$\frac{P^{2}}{4m}$$, where $$p_{1}' = P = mv'_{1}$$.
Now you know that the change in KE in Lab frame and CM frame are same for a closed system obeying momentum conservation.
We have the change in KE in Lab frame = $$(3/2 -1)\frac{1}{2} 3mv^{2}$$ and the change in KE in CM frame = $$\frac{p_{1}'^{2}}{2m}$$ + $$\frac{P^{2}}{4m}$$,where $$p_{1}' = P = mv'_{1}$$. After equating, we get $$v_{1}' = v$$. Thus the velocity of the fastest part in Lab frame is 2v.
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2021-03-04 09:12:36
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https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-4-polynomials-4-7-polynomials-in-several-variables-4-7-exercise-set-page-286/106
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$$=P-2Pr+Pr^2$$
In order to find a polynomial that gives the amount after two years, we plug in 2 for t and multiply out the expression to obtain: $$P\left(1-r\right)^2 \\ P\left(r^2-2r+1\right) \\ =P-2Pr+Pr^2$$
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2019-10-21 08:17:58
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https://support.bioconductor.org/p/45568/
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Question: Limma, design matrix, lmfit Coefficients not estimable error
0
7.3 years ago by
Netherlands
Rainy Gorilla20 wrote:
Dear List, The experiment I have has the following description table: Treatment SampelingPoint Subjects SampleNames placebo_20 20 1 placebo_20_01 placebo_22 22 1 placebo_22_01 celltype2_20 20 2 celltype2_20_02 celltype2_22 22 2 celltype2_22_02 celltype2_20 20 3 celltype2_20_03 celltype2_22 22 3 celltype2_22_03 celltype2_20 20 4 celltype2_20_04 celltype2_22 22 4 celltype2_22_04 placebo_20 20 5 placebo_20_05 placebo_22 22 5 placebo_22_05 celltype2_20 20 7 celltype2_20_07 celltype2_22 22 7 celltype2_22_07 celltype2_20 20 8 celltype2_20_08 celltype2_22 22 8 celltype2_22_08 placebo_20 20 9 placebo_20_09 placebo_22 22 9 placebo_22_09 placebo_20 20 11 placebo_20_11 placebo_22 22 11 placebo_22_11 placebo_20 20 12 placebo_20_12 placebo_22 22 12 placebo_22_12 placebo_20 20 13 placebo_20_13 placebo_22 22 13 placebo_22_13 placebo_20 20 14 placebo_20_14 placebo_22 22 14 placebo_22_14 placebo_20 20 15 placebo_20_15 placebo_22 22 15 placebo_22_15 placebo_20 20 16 placebo_20_16 placebo_22 22 16 placebo_22_16 celltype2_20 20 17 celltype2_20_17 celltype2_22 22 17 celltype2_22_17 placebo_20 20 18 placebo_20_18 placebo_22 22 18 placebo_22_18 celltype2_20 20 19 celltype2_20_19 celltype2_22 22 19 celltype2_22_19 celltype2_20 20 20 celltype2_20_20 celltype2_22 22 20 celltype2_22_20 celltype2_20 20 21 celltype2_20_21 celltype2_22 22 21 celltype2_22_21 celltype2_20 20 24 celltype2_20_24 celltype2_22 22 24 celltype2_22_24 celltype2_20 20 25 celltype2_20_25 celltype2_22 22 25 celltype2_22_25 placebo_20 20 26 placebo_20_26 placebo_22 22 26 placebo_22_26 placebo_20 20 27 placebo_20_27 placebo_22 22 27 placebo_22_27 celltype2_20 20 28 celltype2_20_28 celltype2_22 22 28 celltype2_22_28 celltype2_20 20 29 celltype2_20_29 celltype2_22 22 29 celltype2_22_29 placebo_20 20 30 placebo_20_30 placebo_22 22 30 placebo_22_30 placebo_20 20 31 placebo_20_31 placebo_22 22 31 placebo_22_31 placebo_20 20 32 placebo_20_32 placebo_22 22 32 placebo_22_32 celltype2_20 20 33 celltype2_20_33 celltype2_22 22 33 celltype2_22_33 celltype2_20 20 34 celltype2_20_34 celltype2_22 22 34 celltype2_22_34 I want to see the effect of the samplingPoint within a treatment, because we give the subjects a substance in between day 20 and day 22. In other words, we are interested in the effect of the substance (comparing day 20 versus day 22). At both samplingPoints (day 20 and day 22) 1 sample was drawn. I want to give the analyses extra power by taking into account that the comparison between day 20 and 22 can be made within the same individual (1 individual is measured twice). So my comparisons are: - "placebo_20 placebo_22" - "celltype2_20 celltype2_22" I am using 8.3 Paired Samples of the limma user guide as my guide for the limma design setup. library("limma") # Give levels. Will be the names (the unique names of the bio.replicates) of the columns lev <- c( sort(as.character(unique(description$Treatment)))) > lev [1] "substance_20" "substance_22" "placebo_20" "placebo_22" > Subjects <- factor(description$Subjects) > Treatment <- factor(description\$Treatment) > design <- model.matrix(~0+Treatment+Subjects) # Name comparison > contrast.fit <- "placebo_20 placebo_22" design <- model.matrix(~0+Treatment+Subjects) # design N =34 subjects but cutt it off for mailing it here. > design TreatmentVSL3_20 TreatmentVSL3_22 Treatmentplacebo_20 Treatmentplacebo_22 Subjects2 Subjects3 1 0 0 0 1 0 1 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 If I do a fit on my data using the design matrix above I get the following warning /error > fit <- lmFit(data, design) Coefficients not estimable: Subjects34 Warning message: Partial NA coefficients for 47323 probe(s) Can anyone tell me what's goes wrong my design? Or suggest how to do the limma analyses with my specific experiment. > sessionInfo() R version 2.14.2 (2012-02-29) Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit) locale: [1] C/en_US.UTF-8/C/C/C/C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] qvalue_1.28.0 ggplot2_0.9.1 limma_3.10.3 [4] lumiHumanAll.db_1.16.0 org.Hs.eg.db_2.6.4 lumiHumanIDMapping_1.10.0 [7] RSQLite_0.11.1 DBI_0.2-5 AnnotationDbi_1.16.19 [10] lumi_2.6.0 nleqslv_1.9.3 methylumi_2.0.13 [13] Biobase_2.14.0 loaded via a namespace (and not attached): [1] BiocInstaller_1.2.1 IRanges_1.12.6 KernSmooth_2.23-7 MASS_7.3-18 [5] Matrix_1.0-5 RColorBrewer_1.0-5 affy_1.32.1 affyio_1.22.0 [9] annotate_1.32.3 colorspace_1.1-1 dichromat_1.2-4 digest_0.5.2 [13] grid_2.14.2 hdrcde_2.15 labeling_0.1 lattice_0.20-6 [17] memoise_0.1 mgcv_1.7-16 munsell_0.3 nlme_3.1-103 [21] plyr_1.7.1 preprocessCore_1.16.0 proto_0.3-9.2 reshape2_1.2.1 [25] scales_0.2.1 stringr_0.6 tcltk_2.14.2 tools_2.14.2 [29] xtable_1.7-0 zlibbioc_1.0.1 Thanks a lot, Varshna Goelela [[alternative HTML version deleted]]
limma • 531 views
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2019-08-26 09:29:17
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https://eecs376.github.io/notes/randomness_extra.html
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# Supplemental: Randomness¶
## Primality Testing¶
A key component of many cryptography algorithms is finding large prime numbers, with hundreds or even thousands of digits. A typical approach is to randomly choose a large number and then apply a primality test to determine whether it is prime – the prime number theorem states that approximately $$1/m$$ of the $$m$$-bit numbers are prime, so we only expect to check $$m$$ numbers before finding one that is prime 1. As long as the primality test itself is efficient with respect to $$m$$, the expected time to find a prime number with $$m$$ bits is polynomial in $$m$$.
1
This follows from the fact that the expected value of a geometric distribution with parameter $$p$$ is $$1/p$$.
Formally, we wish to decide the following language:
$\PRIMES = \{m \in \N : m \text{ is prime}\}$
A positive integer $$m$$ is a member of $$\PRIMES$$ if it has no positive factors other than $$1$$ and $$m$$.
Is the language $$\PRIMES$$ in $$\P$$? Let us consider the following simple algorithm to decide the language:
$\begin{split}&\Algorithm Prime\text{-}Test(m):\\ &~~~\For i = 2 \tot m-1:\\ &~~~~~~\If m \bmod i \equiv 0, \reject\\ &~~~\Accept\end{split}$
This algorithm does $$\O(m)$$ mod operations. Is this efficient? The size of the input $$m$$ is the number of bits required to represent $$m$$, which is $$\O(\log m)$$. Thus, the number of operations is $$\O(m) = \O(2^{\log m})$$, which is exponential in the size of $$m$$.
We can improve the algorithm by observing that if $$m$$ has a factor within the interval $$[2, m-1]$$, it must have a factor within $$[2, \sqrt{m}]$$ – otherwise, we would obtain a product larger than $$m$$ when we multiplied its factors together. Thus, we need only iterate up to $$\sqrt{m}$$:
$\begin{split}&\Algorithm Prime\text{-}Test(m):\\ &~~~\For i = 2 \tot \sqrt{m}:\\ &~~~~~~\If m \bmod i \equiv 0, \reject\\ &~~~\Accept\end{split}$
This algorithm does $$\O(\sqrt{m})$$ mod operations. However, this still is not polynomial; we have:
$\O(\sqrt{m}) = \O(m^{1/2}) = \O(2^{1/2 \cdot \log m})$
To be efficient, a primality-testing algorithm must have runtime $$\O(\log^k m)$$ for some constant $$k$$. Neither of the algorithms above meet this threshold.
In fact, there is a known efficient algorithm for primality testing, the AKS primality test. Thus, it is indeed the case that $$\PRIMES \in \P$$. However, this algorithm is somewhat complicated, and its running time is high enough to preclude it from being used in practice. Instead, we consider a randomized primality test that is efficient and works well in practice for most inputs. The algorithm we construct relies on the extended Fermat’s little theorem.
Theorem 194 (Extended Fermat’s Little Theorem)
Let $$n \in \N$$ be a natural number such that $$n \ge 2$$. Let $$a$$ be a witness in the range $$1 \le a \le n - 1$$. Then:
• If $$n$$ is prime, then $$a^{n-1} \equiv 1 \pmod{n}$$ for any witness $$a$$.
• If $$n$$ is composite and $$n$$ is not a Carmichael number, then $$a^{n-1} \equiv 1 \pmod{n}$$ for at most half the witnesses $$1 \le a \le n - 1$$.
We postpone discussion of Carmichael numbers for the moment. Instead, we take a look at some small cases of composite numbers to see that the extended Fermat’s little theorem holds. We first consider $$n = 6$$. We have:
$\begin{split}\begin{array}{lll} a = 1 :& 1^5 ~&\equiv 1 \pmod{6}\\ a = 2 :& 2^5 = 32 = 2 + 5\cdot 6 ~&\equiv 2 \pmod{6}\\ a = 3 :& 3^5 = 243 = 3 + 40\cdot 6 ~&\equiv 3 \pmod{6}\\ a = 4 :& 4^5 = 1024 = 4 + 170\cdot 6 ~&\equiv 4 \pmod{6}\\ a = 5 :& 5^5 = 3125 = 5 + 520\cdot 6 ~&\equiv 5 \pmod{6} \end{array}\end{split}$
We see that $$a^{n-1} \equiv 1 \pmod{n}$$ for only the single witness $$a = 1$$. Similarly, we consider $$n = 9$$:
$\begin{split}\begin{array}{lll} a = 1 :& 1^8 ~&\equiv 1 \pmod{9}\\ a = 2 :& 2^8 = 256 = 4 + 28\cdot 9 ~&\equiv 4 \pmod{9}\\ a = 3 :& 3^8 = 6561 = 0 + 729\cdot 9 ~&\equiv 0 \pmod{9}\\ a = 4 :& 4^8 = 65536 = 7 + 7281\cdot 9 ~&\equiv 7 \pmod{9}\\ a = 5 :& 5^8 = 390625 = 7 + 43402\cdot 9 ~&\equiv 7 \pmod{9}\\ a = 6 :& 6^8 = 1679616 = 0 + 186624\cdot 9 ~&\equiv 0 \pmod{9}\\ a = 7 :& 7^8 = 5764801 = 4 + 640533\cdot 9 ~&\equiv 4 \pmod{9}\\ a = 8 :& 8^8 = 16777216 = 1 + 1864135\cdot 9 ~&\equiv 1 \pmod{9} \end{array}\end{split}$
Here, there are two witnesses $$a$$ where $$a^{n-1} \equiv 1 \pmod{n}$$, out of eight total. Finally, we consider $$n = 7$$:
$\begin{split}\begin{array}{lll} a = 1 :& 1^6 ~&\equiv 1 \pmod{7}\\ a = 2 :& 2^6 = 64 = 1 + 9\cdot 7 ~&\equiv 1 \pmod{7}\\ a = 3 :& 3^6 = 729 = 1 + 104\cdot 7 ~&\equiv 1 \pmod{7}\\ a = 4 :& 4^6 = 4096 = 1 + 585\cdot 7 ~&\equiv 1 \pmod{7}\\ a = 5 :& 5^6 = 15625 = 1 + 2232\cdot 7 ~&\equiv 1 \pmod{7}\\ a = 6 :& 6^6 = 46656 = 1 + 6665\cdot 7 ~&\equiv 1 \pmod{7}\\ \end{array}\end{split}$
Since $$7$$ is prime, we have $$a^{n-1} \equiv 1 \pmod{n}$$ for all witnesses $$a$$.
The extended Fermat’s little theorem leads directly to a simple, efficient randomized algorithm for primality testing. This Fermat primality test is as follows:
$\begin{split}&\Algorithm Fermat\text{-}Test(m):\\ &~~~\text{Pick a random witness 1 \le a \le m-1}\\ &~~~\If a^{m-1} \bmod m \equiv 1, \accept\\ &~~~\Else \reject\end{split}$
The modular exponentiation in this algorithm can be done with $$\O(\log m)$$ multiplications using a divide and conquer strategy, and each multiplication can be done efficiently. Thus, this algorithm is polynomial with respect to the size of $$m$$.
As for correctness, we have:
• If $$m$$ is prime, then the algorithm always accepts $$m$$. In other words:
$\Pr[\text{the Fermat test accepts m}] = 1$
• If $$m$$ is composite and not a Carmichael number, then the algorithm rejects $$m$$ with probability at least $$\frac{1}{2}$$. In other words:
$\Pr[\text{the Fermat test rejects m}] \ge \frac{1}{2}$
Thus, if the algorithm accepts $$m$$, we can be fairly confident that $$m$$ is prime. And as we will see shortly, we can repeat the algorithm to obtain higher confidence that we get the right answer.
We now return to the problem of Carmichael numbers. A Carmichael number is a composite number $$n$$ such that $$a^{n-1} \equiv 1 \pmod{n}$$ for all witnesses $$a$$ that are relatively prime to $$n$$, i.e. $$\gcd(a, n) = 1$$. This implies that for a Carmichael number, the Fermat test reports with high probability that the number is prime, despite it being composite. We call a number that passes the Fermat test with high probability a pseudoprime, and the Fermat test is technically a randomized algorithm for deciding the following language:
\begin{split}\mathrm{PSEUDOPRIMES} = \left\{ \begin{aligned} m \in \N :~ &a^{m-1} \bmod{m} \equiv 1 \text{ for at least half}\\ &\text{the witnesses } 1 \le a \le m - 1 \end{aligned} \right\}\end{split}
Carmichael numbers are much rarer than prime numbers, so for many applications, the Fermat test is sufficient to determine with high confidence that a randomly chosen number is prime. On the other hand, if the number is chosen by an adversary, then the Fermat test is unsuitable, and a more complex randomized algorithm such as the Miller-Rabin primality test must be used instead.
### The Miller-Rabin Test¶
The Miller-Rabin test is designed around the fact that for prime $$m$$, the only square roots of 1 modulo $$m$$ are -1 and 1. More specifically, if $$x$$ is a square root of 1 modulo $$m$$, we have
$x^2 \equiv 1 \pmod{m}$
By definition of modular arithmetic, this means that
$x^2 - 1 = km$
for some integer $$k$$ (i.e. $$m$$ evenly divides the difference of $$x^2$$ and 1). We can factor $$x^2-1$$ to obtain:
$(x + 1)(x - 1) = km$
When $$m$$ is composite, so that $$m = pq$$ for integers $$p,q > 1$$, then it is possible for $$p$$ to divide $$x+1$$ and $$q$$ to divide $$x-1$$, in which case $$pq = m$$ divides their product $$(x+1)(x-1)$$. However, when $$m$$ is prime, the only way for the equation to hold is if $$m$$ divides either $$x+1$$ or $$x-1$$; otherwise, the prime factorization of $$(x+1)(x-1)$$ does not contain $$m$$, and by the fundamental theorem of arithmetic, it cannot be equal to a number whose prime factorization contains $$m$$ 2. Thus, we have either $$x+1 = am$$ for some integer $$a$$, or $$x-1 = bm$$ for some $$b \in \Z$$. The former gives us $$x \equiv -1 \pmod{m}$$, while the latter results in $$x \equiv 1 \pmod{m}$$., This, if $$m$$ is prime and $$x^2 \equiv 1 \pmod{m}$$, then either $$x \equiv 1 \pmod{m}$$ or $$x \equiv -1 \pmod{m}$$.
2
Euclid’s lemma more directly states that if $$m$$ is prime and $$m$$ divides a product $$ab$$, then $$m$$ must divide either $$a$$ or $$b$$.
The Miller-Rabin test starts with the Fermat test: choose a witness $$1 \le a \le m-1$$ and check whether $$a^{m-1} \equiv 1 \pmod{m}$$. If this does not hold, then $$m$$ fails the Fermat test and therefore is not prime. If it does indeed hold, then the test checks the square root $$a^{\frac{1}{2}(m-1)}$$ of $$a^{m-1}$$ to see if it is 1 or -1. If it is 1, the test checks the square root $$a^{\frac{1}{4}(m-1)}$$ of $$a^{\frac{1}{2}(m-1)}$$ to see if it is 1 or -1, and so on. The termination conditions are as follows:
• The test finds a square root of 1 that is not -1 or 1. By the reasoning above, $$m$$ must be composite.
• The test finds a square root of 1 that is -1. The reasoning above only holds for square roots of 1, so the test cannot continue by computing the square root of -1. In this case, $$m$$ may be prime or composite.
• The test reaches some $$r$$ for which $$1/2^r\cdot(m-1)$$ is no longer an integer. Then it cannot compute $$a$$ to this power modulo $$m$$. In this case, $$m$$ may be prime or composite.
To compute these square roots, we first extract powers of 2 from the number $$m-1$$ (which is even for odd $$m$$, the cases of interest):
$\begin{split}&\Algorithm ExtractPowersOf2(x):\\ &~~~\If x\bmod 2 = 1 \return (x, 0)\\ &~~~(d, s') \gets ExtractPowersOf2(x / 2)\\ &~~~\Return (d, s'+1)\end{split}$
Then $$ExtractPowersOf2(m-1)$$ computes $$d$$ and $$s$$ such that
$m-1 = 2^s d$
where $$\gcd(d, 2) = 1$$. Then we have $$a^{2^{s-1}d}$$ is the square root of $$a^{2^sd}$$, since
$\large\parens{a^{2^{s-1}d}}^2 = a^{2\cdot 2^{s-1}d} = a^{2^sd}$
The full Miller-Rabin algorithm is as follows:
$\begin{split}&\Algorithm Miller\text{-}Rabin(m):\\ &~~~\text{Pick a random witness 1 \le a \le m-1}\\ &~~~\Compute s,d \text{ such that } m-1=2^sd \andt \gcd(d, 2) = 1\\ &~~~\Compute a^d, a^{2d}, a^{4d}, \dots, a^{2^{s-1}d}, a^{2^sd}\\ &~~~\If a^{2^sd} \not\equiv 1 \pmod{m}, \reject ~~~~\textit{(Fermat test)}\\ &~~~\For t = s-1, s-2, \dots, 0:\\ &~~~~~~\If a^{2^td} \equiv -1 \pmod{m}, \accept\\ &~~~~~~\ElseIf a^{2^td} \not\equiv 1 \pmod{m}, \reject\\ &~~~\Accept\end{split}$
This algorithm is efficient: $$a^d$$ can be computed using $$\O(\log d) = \O(\log m)$$ multiplications, and it takes another $$\O(\log m)$$ multiplications to compute $$a^{2d}, a^{4d}, \dots, a^{2^sd}$$ since $$s = \O(\log m)$$. Each multiplication is efficient as it is done modulo $$m$$, so the entire algorithm takes polynomial time in the size of $$m$$.
As for correctness, we have:
• If $$m$$ is prime, then the algorithm accepts $$m$$ with probability 1.
• If $$m$$ is composite, then the algorithm rejects $$m$$ with probability at least $$\frac{3}{4}$$.
The latter is due to the fact that when $$m$$ is composite, $$m$$ passes the Miller-Rabin test for at most $$\frac{1}{4}$$ of the witnesses $$1 \le a \le m-1$$. Unlike the Fermat test, there are no exceptions to this, making the Miller-Rabin test a better choice in many applications.
Exercise 195
Polynomial identity testing is the problem of determining whether two polynomials $$p(x)$$ and $$q(x)$$ are the same, or equivalently, whether $$p(x) - q(x)$$ is the zero polynomial.
1. Let $$d$$ be an upper bound on the degree of $$p(x)$$ and $$q(x)$$. Show that for a randomly chosen integer $$a \in [1, 4d]$$, if $$p(x) \ne q(x)$$, then $$\Pr[p(a) \ne q(a)] \ge \frac{3}{4}$$.
Hint: A nonzero polynomial $$r(x)$$ of degree at most $$d$$ can have at most $$d$$ roots $$s_i$$ such that $$r(s_i) = 0$$.
2. Devise an efficient, randomized algorithm $$A$$ to determine whether $$p(x)$$ and $$q(x)$$ are the same, with the behavior that:
• if $$p(x) = q(x)$$, then
$\Pr[A \text{ accepts } p(x),q(x)] = 1$
• if $$p(x) \ne q(x)$$, then
$\Pr[A \text{ rejects } p(x),q(x)] \ge \frac{3}{4}$
Assume that a polynomial can be efficiently evaluated on a single input.
## Chernoff Bounds¶
Like Markov’s inequality, Chernoff bounds are a form of concentration bounds that allow us to reason about the deviation of a random variable from its expectations. There are multiple variants of Chernoff bounds; we restrict ourselves to the following “multiplicative” versions.
Let $$X = X_1 + \dots + X_n$$ be the sum of independent indicator random variables, where the indicator $$X_i$$ has the expectation
$\Ex[X_i] = \Pr[X_i = 1] = p_i$
Let $$\mu$$ be the expected value of $$X$$:
$\mu = \Ex[X] = \sum_i \Ex[X_i] = \sum_i p_i$
Here, we’ve applied linearity of expectation to relate the expectation of $$X$$ to that of the indicators $$X_i$$. The Chernoff bounds then are as follows.
Theorem 196 (Chernoff Bound – Upper Tail)
Let $$X = X_1 + \dots + X_n$$, where the $$X_i$$ are independent indicator random variables with $$\Ex[X_i] = p_i$$, and $$\mu = \sum_i p_i$$. Suppose we wish to bound the probability of $$X$$ exceeding its expectation $$\mu$$ by at least a factor of $$1 + \delta$$, where $$\delta > 0$$. Then
$\Pr[X \ge (1 + \delta)\mu] \le \parens{\frac{e^\delta}{(1+\delta)^{1+\delta}}}^\mu$
Theorem 197 (Chernoff Bound – Lower Tail)
Let $$X = X_1 + \dots + X_n$$, where the $$X_i$$ are independent indicator random variables with $$\Ex[X_i] = p_i$$, and $$\mu = \sum_i p_i$$. Suppose we wish to bound the probability of $$X$$ being below its expectation $$\mu$$ by at least a factor of $$1 - \delta$$, where $$0 < \delta < 1$$. Then
$\Pr[X \le (1 - \delta)\mu] \le \parens{\frac{e^{-\delta}}{(1-\delta)^{1-\delta}}}^\mu$
These inequalities can be unwieldy to work with, so we often use the following simpler, looser bounds.
Theorem 198 (Chernoff Bound – Simplified Upper Tail)
Let $$X = X_1 + \dots + X_n$$, where the $$X_i$$ are independent indicator random variables with $$\Ex[X_i] = p_i$$, and $$\mu = \sum_i p_i$$. Suppose we wish to bound the probability of $$X$$ exceeding its expectation $$\mu$$ by at least a factor of $$1 + \delta$$, where $$\delta > 0$$. Then
$\large \Pr[X \ge (1 + \delta)\mu] \le e^{-\frac{\delta^2\mu}{2+\delta}}$
Theorem 199 (Chernoff Bound – Simplified Lower Tail)
Let $$X = X_1 + \dots + X_n$$, where the $$X_i$$ are independent indicator random variables with $$\Ex[X_i] = p_i$$, and $$\mu = \sum_i p_i$$. Suppose we wish to bound the probability of $$X$$ being below its expectation $$\mu$$ by at least a factor of $$1 - \delta$$, where $$0 < \delta < 1$$. Then
$\large \Pr[X \le (1 - \delta)\mu] \le e^{-\frac{\delta^2\mu}{2}}$
Before we proceed to make use of these bounds, we first prove that the unsimplified upper-tail and lower-tail bounds hold 3.
3
Refer to the appendix for proofs of the simplified bounds.
Proof 200 (Proof of Upper-tail Chernoff Bound)
To demonstrate the upper-tail Chernoff bound, we make use of the Chernoff bounding technique – rather than reasoning about the random variable $$X$$ directly, we instead reason about $$e^{tX}$$, since small deviations in $$X$$ turn into large deviations in $$e^{tX}$$. We have that $$X \ge (1+\delta)\mu$$ exactly when $$e^{tX} \ge e^{t(1+\delta)\mu}$$ for any $$t \ge 0$$; we obtain this by raising $$e^t \ge 1$$ to the power of both sides of the former inequality. Then
\begin{split}\large \begin{aligned} \Pr[X \ge (1+\delta)\mu] &= \Pr[e^{tX} \ge e^{t(1+\delta)\mu}]\\ &\le \frac{\Ex[e^{tX}]}{e^{t(1+\delta)\mu}} \end{aligned}\end{split}
where the latter step follows from Markov’s inequality. Continuing, we have
\begin{split}\large \begin{aligned} \frac{\Ex[e^{tX}]}{e^{t(1+\delta)\mu}} &= e^{-t(1+\delta)\mu}\cdot\Ex[e^{tX}]\\ &= e^{-t(1+\delta)\mu}\cdot\Ex[e^{t(X_1 + \dots + X_n)}]\\ &= e^{-t(1+\delta)\mu}\cdot\lrEx{\prod_i e^{tX_i}} \end{aligned}\end{split}
We now make use of the fact that the $$X_i$$ (and therefore the $$e^{tX_i}$$) are independent. For independent random variables $$Y$$ and $$Z$$, we have $$\Ex[YZ] = \Ex[Y]\cdot\Ex[Z]$$ (see the appendix for a proof). Thus,
$\large e^{-t(1+\delta)\mu}\cdot\lrEx{\prod_i e^{tX_i}} = e^{-t(1+\delta)\mu}\prod_i \Ex[e^{tX_i}]$
The $$X_i$$ are indicators, with $$\Pr[X_i = 1] = p_i$$ and $$\Pr[X_i = 0] = 1 - p_i$$. When $$X_i = 1$$, $$e^{tX_i} = e^t$$, and when $$X_i = 0$$, $$e^{tX_i} = e^0 = 1$$. Thus, the distribution of $$e^{tX_i}$$ is
\begin{split}\large \begin{aligned} \Pr[e^{tX_i} = e^t] &= p_i\\ \Pr[e^{tX_i} = 1] &= 1 - p_i \end{aligned}\end{split}
We can then compute the expectation of $$e^{tX_i}$$:
\begin{split}\large \begin{aligned} \Ex[e^{tX_i}] &= e^t\cdot \Pr[e^{tX_i} = e^t] + 1\cdot\Pr[e^{tX_i} = 1]\\ &= e^t p_i + 1 - p_i\\ &= 1 + p_i(e^t - 1) \end{aligned}\end{split}
Plugging this into the upper bound that resulted from Markov’s inequality, we get
$\large e^{-t(1+\delta)\mu}\prod_i \Ex[e^{tX_i}] = e^{-t(1+\delta)\mu}\prod_i 1 + p_i(e^t-1)$
We can further simplify this expression by observing that $$1+x \le e^x$$ for all $$x$$:
Using $$x = p_i(e^t-1)$$, we have
$\large 1 + p_i(e^t-1) \le e^{p_i(e^t-1)}$
Applying this to our previously computed upper bound, we get
\begin{split}\large \begin{aligned} e^{-t(1+\delta)\mu}\prod_i 1 + p_i(e^t-1) &\le e^{-t(1+\delta)\mu}\prod_i e^{p_i(e^t-1)}\\ &= e^{-t(1+\delta)\mu}e^{(e^t-1)\sum_i p_i}\\ &= e^{-t(1+\delta)\mu}e^{(e^t-1)\mu}\\ &= e^{\mu(e^t - 1 - t(1+\delta))} \end{aligned}\end{split}
We can now choose $$t$$ to minimize the exponent, which in turn minimizes the value of the exponential itself. Let $$f(t) = e^t - 1 - t(1+\delta)$$. Then we compute the derivative with respect to $$t$$ and set that to zero to find an extremum:
$\begin{split}f(t) &= e^t - 1 - t(1+\delta)\\ f'(t) &= e^t - (1+\delta) = 0\\ e^t &= 1+\delta\\ t &= \ln(1+\delta)\end{split}$
Computing the second derivative $$f''(\ln(1+\delta)) = e^{\ln(1+\delta)} = 1+\delta$$, we see that it is positive since $$\delta > 0$$, therefore $$t = \ln(1+\delta)$$ is a minimum. Substituting it into our earlier expression, we obtain
\begin{split}\large \begin{aligned} e^{\mu(e^t - 1 - t(1+\delta))} &\le e^{\mu(e^{\ln(1+\delta)} - 1 - (1+\delta)\ln(1+\delta))}\\ &= e^{\mu((1+\delta) - 1 - (1+\delta)\ln(1+\delta))}\\ &= e^{\mu(\delta - (1+\delta)\ln(1+\delta))}\\ &= \parens{e^{\delta - (1+\delta)\ln(1+\delta)}}^\mu\\ &= \parens{\frac{e^{\delta}}{e^{(1+\delta)\ln(1+\delta)}}}^\mu\\ &= \parens{\frac{e^{\delta}}{(e^{\ln(1+\delta)})^{1+\delta}}}^\mu\\ &= \parens{\frac{e^{\delta}}{(1+\delta)^{1+\delta}}}^\mu \end{aligned}\end{split}
Proof 201 (Proof of Lower-tail Chernoff Bound)
The proof of the lower-tail Chernoff bound follows similar reasoning as that of the upper-tail bound. We have:
\begin{split}\large \begin{aligned} \Pr[X \le (1-\delta)\mu] &= \Pr[-X \ge -(1-\delta)\mu]\\ &= \Pr[e^{-tX} \ge e^{-t(1-\delta)\mu}]\\ &\le \frac{\Ex[e^{-tX}]}{e^{-t(1-\delta)\mu}} \end{aligned}\end{split}
Here, we can apply Markov’s inequality to the random variable $$e^{-tX}$$ since it is nonnegative, unlike $$-X$$. Continuing, we have
\begin{split}\large \begin{aligned} \frac{\Ex[e^{-tX}]}{e^{-t(1-\delta)\mu}} &= e^{t(1-\delta)\mu}\cdot\Ex[e^{-tX}]\\ &= e^{t(1-\delta)\mu}\cdot\Ex[e^{-t(X_1 + \dots + X_n)}]\\ &= e^{t(1-\delta)\mu}\cdot\lrEx{\prod_i e^{-tX_i}}\\ &= e^{t(1-\delta)\mu}\prod_i \Ex[e^{-tX_i}] \end{aligned}\end{split}
In the last step, we make use of the fact that the $$X_i$$ and therefore the $$e^{-tX_i}$$ are independent. The distribution of $$e^{-tX_i}$$ is
\begin{split}\large \begin{aligned} \Pr[e^{-tX_i} = e^{-t}] &= p_i\\ \Pr[e^{-tX_i} = 1] &= 1 - p_i \end{aligned}\end{split}
Then the expectation of $$e^{-tX_i}$$ is:
\begin{split}\large \begin{aligned} \Ex[e^{-tX_i}] &= e^{-t}\cdot \Pr[e^{-tX_i} = e^{-t}] + 1\cdot\Pr[e^{-tX_i} = 1]\\ &= e^{-t} p_i + 1 - p_i\\ &= 1 + p_i(e^{-t} - 1) \end{aligned}\end{split}
Plugging this into the bound we’ve computed so far, we get
$\large e^{t(1-\delta)\mu}\prod_i \Ex[e^{-tX_i}] = e^{t(1-\delta)\mu}\prod_i 1 + p_i(e^{-t}-1)$
As with the upper-tail, we use the fact that $$1+x \le e^x$$ for all $$x$$ to simplify this, with $$x = p_i(e^{-t}-1)$$:
\begin{split}\large \begin{aligned} e^{t(1-\delta)\mu}\prod_i 1 + p_i(e^{-t}-1) &\le e^{t(1-\delta)\mu}\prod_i e^{p_i(e^{-t}-1)}\\ &= e^{t(1-\delta)\mu}e^{(e^{-t}-1)\sum_i p_i}\\ &= e^{t(1-\delta)\mu}e^{(e^{-t}-1)\mu}\\ &= e^{\mu(e^{-t} - 1 + t(1-\delta))} \end{aligned}\end{split}
We choose $$t$$ to minimize the exponent. Let $$f(t) = e^{-t} - 1 + t(1-\delta)$$. Then we compute the derivative with respect to $$t$$ and set that to zero to find an extremum:
$\begin{split}f(t) &= e^{-t} - 1 + t(1-\delta)\\ f'(t) &= -e^{-t} + (1-\delta) = 0\\ e^{-t} &= 1-\delta\\ t &= -\ln(1-\delta)\end{split}$
Computing the second derivative $$f''(-\ln(1-\delta)) = e^{-(-\ln(1-\delta))} = 1-\delta$$, we see that it is positive since $$0 < \delta < 1$$, therefore $$t = \ln(1-\delta)$$ is a minimum. Substituting it into our earlier expression, we obtain
\begin{split}\large \begin{aligned} e^{\mu(e^{-t} - 1 + t(1-\delta))} &\le e^{\mu(e^{\ln(1-\delta)} - 1 - (1-\delta)\ln(1-\delta))}\\ &= e^{\mu((1-\delta) - 1 - (1-\delta)\ln(1-\delta))}\\ &= e^{\mu(-\delta - (1-\delta)\ln(1-\delta))}\\ &= \parens{e^{-\delta - (1-\delta)\ln(1-\delta)}}^\mu\\ &= \parens{\frac{e^{-\delta}}{e^{(1-\delta)\ln(1-\delta)}}}^\mu\\ &= \parens{\frac{e^{-\delta}}{(e^{\ln(1-\delta)})^{1-\delta}}}^\mu\\ &= \parens{\frac{e^{-\delta}}{(1-\delta)^{1-\delta}}}^\mu \end{aligned}\end{split}
In lieu of applying Chernoff bounds to the algorithm for estimating $$\pi$$, we consider a different example of flipping a coin. If we flip a biased coin with probability $$p$$ of heads, we expect to see $$np$$ heads. Let $$H$$ be the total number of heads, and let $$H_i$$ be an indicator variable corresponding to whether the $$i$$th flip is heads. We have $$\Pr[H_i = 1] = p$$, and $$\Ex[H] = np$$ by linearity of expectation.
Suppose the coin is fair. What is the probability of getting at least six heads out of ten flips? This is a fractional deviation of $$\delta = 6/5 - 1 = 0.2$$, and applying the upper-tail Chernoff bound gives us:
$\begin{split}\Pr[H \ge (1+0.2)\cdot 5] &\le \parens{\frac{e^{0.2}}{1.2^{1.2}}}^5\\ &\approx 0.9814^5\\ &\approx 0.91\end{split}$
What is the probability of getting at least 60 heads out of 100 flips, which is the same fractional deviation $$\delta = 0.2$$ from the expectation? Applying the upper tail again, we get
$\begin{split}\Pr[H \ge (1+0.2)\cdot 50] &\le \parens{\frac{e^{0.2}}{1.2^{1.2}}}^{50}\\ &\approx 0.9814^{50}\\ &\approx 0.39\end{split}$
Thus, the probability of deviating by a factor of $$1+\delta$$ decreases significantly as the number of samples increases. For this particular example, we can compute the actual probabilities quite tediously from exact formulas for a binomial distribution, which yields 0.37 for getting six heads out of ten flips and 0.03 for 60 heads out of 100 flips. However, this approach becomes more and more expensive as $$n$$ increases, and the Chernoff bound produces a reasonable result with much less work.
### Polling Analysis with Chernoff Bounds¶
Recall that in polling, a confidence level $$1-\gamma$$ and margin of error $$\varepsilon$$ requires
$\lrPr{\lrabs{\frac{X}{n} - p} \le \varepsilon} \ge 1-\gamma$
or equivalently
$\lrPr{\lrabs{\frac{X}{n} - p} > \varepsilon} < \gamma$
Formally, we define indicator variables $$X_i$$ as
$\begin{split}X_i = \begin{cases} 1 &\mbox{if person i supports the candidate}\\ 0 &\mbox{otherwise} \end{cases}\end{split}$
for each person $$i$$ in the set that we poll. Then $$X = X_1 + \dots + X_n$$ is the sum of independent indicator variables, with
$\mu = \Ex[X] = \sum_i \Ex[X_i] = np$
For an arbitrary margin of error $$\varepsilon$$, we want to bound the probability
$\begin{split}\lrPr{\lrabs{\frac{X}{n} - p} > \varepsilon} &= \lrPr{\frac{X}{n} - p > \varepsilon} + \lrPr{\frac{X}{n} - p < -\varepsilon}\\ &= \lrPr{\frac{X}{n} > \varepsilon+p} + \lrPr{\frac{X}{n} < -\varepsilon+p}\\ &= \Pr[X > \varepsilon n+pn] + \Pr[X < -\varepsilon n+pn]\\ &= \Pr[X > \varepsilon n+\mu] + \Pr[X < -\varepsilon n+\mu]\\ &= \Pr[X > (1+\varepsilon n/\mu)\mu] + \Pr[X < (1-\varepsilon n/\mu)\mu]\\ &= \Pr[X \ge (1+\varepsilon n/\mu)\mu] + \Pr[X \le (1-\varepsilon n/\mu)\mu]\\ &~~~~~ - \Pr[X = (1+\varepsilon n/\mu)\mu] - \Pr[X = (1-\varepsilon n/\mu)\mu]\\ &\le \Pr[X \ge (1+\varepsilon n/\mu)\mu] + \Pr[X \le (1-\varepsilon n/\mu)\mu]\end{split}$
In the second-to-last step, we use the fact that $$X = x$$ and $$X > x$$ are disjoint events, since a random variable maps each sample point to a single value, and that $$(X \ge x) = (X = x) \cup (X > x)$$.
We now have events that are in the right form to apply a Chernoff bound, with $$\delta = \varepsilon n/\mu$$. We first apply the simplified upper-tail bound to the first term, obtaining
\begin{split}\large \begin{aligned} \Pr[X \ge (1+\varepsilon n/\mu)\mu] &\le e^{-\frac{\delta^2\mu}{2+\delta}} ~~~~\text{where \delta=\varepsilon n/\mu}\\ &= e^{-\frac{(\varepsilon n)^2\mu}{\mu^2(2+\varepsilon n/\mu)}}\\ &= e^{-\frac{(\varepsilon n)^2}{\mu(2+\varepsilon n/\mu)}}\\ &= e^{-\frac{(\varepsilon n)^2}{2\mu+\varepsilon n}}\\ &= e^{-\frac{(\varepsilon n)^2}{2np+\varepsilon n}}\\ &= e^{-\frac{\varepsilon^2 n}{2p+\varepsilon}} \end{aligned}\end{split}
We want this to be less than some value $$\beta$$:
\begin{split}\large \begin{aligned} e^{-\frac{\varepsilon^2 n}{2p+\varepsilon}} &< \beta\\ \frac{1}{\beta} &< e^{\frac{\varepsilon^2 n}{2p+\varepsilon}}\\ \ln\parens{\frac{1}{\beta}} &< \frac{\varepsilon^2 n}{2p+\varepsilon}\\ n &> \frac{2p+\varepsilon}{\varepsilon^2}\ln\parens{\frac{1}{\beta}} \end{aligned}\end{split}
Unfortunately, this expression includes $$p$$, the quantity we’re trying to estimate However, we know that $$p \le 1$$, so we can set
$n > \frac{2+\varepsilon}{\varepsilon^2}\ln\parens{\frac{1}{\beta}}$
to ensure that $$\Pr[X \ge (1+\varepsilon n/\mu)\mu] < \beta$$.
We can also apply the simplified lower-tail bound to the second term in our full expression above:
\begin{split}\large \begin{aligned} \Pr[X \le (1-\varepsilon n/\mu)\mu] &\le e^{-\frac{\delta^2\mu}{2}} ~~~~\text{where \delta=\varepsilon n/\mu}\\ &= e^{-\frac{(\varepsilon n)^2\mu}{2\mu^2}}\\ &= e^{-\frac{(\varepsilon n)^2}{2\mu}}\\ &= e^{-\frac{(\varepsilon n)^2}{2np}}\\ &= e^{-\frac{\varepsilon^2 n}{2p}} \end{aligned}\end{split}
As before, we want this to be less some value $$\beta$$:
\begin{split}\large \begin{aligned} e^{-\frac{\varepsilon^2 n}{2p)}} &< \beta\\ \frac{1}{\beta} &< e^{\frac{\varepsilon^2 n}{2p}}\\ \ln\parens{\frac{1}{\beta}} &< \frac{\varepsilon^2 n}{2p}\\ n &> \frac{2p}{\varepsilon^2}\ln\parens{\frac{1}{\beta}} \end{aligned}\end{split}
We again observe that $$p \le 1$$, so we can set
$n > \frac{2}{\varepsilon^2}\ln\parens{\frac{1}{\beta}}$
to ensure that $$\Pr[X \le (1-\varepsilon n/\mu)\mu] < \beta$$.
To bound both terms simultaneously, we can choose $$\beta=\gamma/2$$, so that
$\Pr[X \ge (1+\varepsilon n/\mu)\mu] + \Pr[X \le (1-\varepsilon n/\mu)\mu] < 2\beta = \gamma$
To achieve this, we require
$\begin{split}n &> \max\parens{ \frac{2+\varepsilon}{\varepsilon^2}\ln\parens{\frac{2}{\gamma}}, \frac{2}{\varepsilon^2}\ln\parens{\frac{2}{\gamma}}}\\ &= \frac{2+\varepsilon}{\varepsilon^2}\ln\parens{\frac{2}{\gamma}}\end{split}$
For example, for a 95% confidence level and a margin of error of ±2%, we have $$\varepsilon=0.02$$ and $$\gamma=0.05$$. Plugging those values into the result above, we need no more than
$\frac{2+\varepsilon}{\varepsilon^2}\ln\parens{\frac{2}{\gamma}} = \frac{2.02}{0.02^2}\ln 40 \approx 18623$
samples to achieve the desired confidence level and margin of error. Observe that this does not depend on the total population size!
## Probabilistic Complexity Classes¶
The Fermat primality test is an example of a one-sided-error randomized algorithm – an input that is pseudoprime is always accepted, while a non-pseudoprime is sometimes accepted and sometimes rejected. We can define complexity classes corresponding to decision problems that have efficient one-sided-error randomized algorithms.
Definition 202 (RP)
$$\RP$$ is the class of languages that have efficient one-sided-error randomized algorithms with no false positives. A language $$L$$ is in $$\RP$$ if there is an efficient randomized algorithm $$A$$ such that:
• if $$x \in L$$, $$\Pr[A \text{ accepts } x] \ge c$$
• if $$x \notin L$$, $$\Pr[A \text{ rejects } x] = 1$$
Here, $$c$$ must be a constant greater than 0. Often, $$c$$ is chosen to be $$\frac{1}{2}$$.
Definition 203 (coRP)
$$\coRP$$ is the class of languages that have efficient one-sided-error randomized algorithms with no false negatives. A language $$L$$ is in $$\coRP$$ if there is an efficient randomized algorithm $$A$$ such that:
• if $$x \in L$$, $$\Pr[A \text{ accepts } x] = 1$$
• if $$x \notin L$$, $$\Pr[A \text{ rejects } x] \ge c$$
Here, $$c$$ must be a constant greater than 0. Often, $$c$$ is chosen to be $$\frac{1}{2}$$.
$$\RP$$ stands for randomized polynomial time. If a language $$L$$ is in $$\RP$$, then its complement language $$\overline{L}$$ is in $$\coRP$$ – an algorithm for $$L$$ with no false positives can be converted into an algorithm for $$\overline{L}$$ with no false negatives. The Fermat test produces no false negatives, so $$\mathrm{PSEUDOPRIMES} \in \coRP$$. Thus, the language
\begin{split}\mathrm{PSEUDOPRIMES} = \left\{ \begin{aligned} m \in \N :~ &a^{m-1} \bmod{m} \equiv 1 \text{ for at least half}\\ &\text{the witnesses } 1 \le a \le m - 1 \end{aligned} \right\}\end{split}
is in $$\RP$$.
The constant $$c$$ in the definition of $$\RP$$ and $$\coRP$$ is arbitrary. With any $$c > 0$$, we can amplify the probability of success by repeatedly running the algorithm. For instance, if we have a randomized algorithm $$A$$ with no false positives for a language $$L$$, we have:
$\begin{split}x \in L &\implies \Pr[A \text{ accepts } x] \ge c\\ x \notin L &\implies \Pr[A \text{ accepts } x] = 0\end{split}$
We can construct a new algorithm $$B$$ as follows:
$\begin{split}&\Algorithm B(x):\\ &~~~\Run A \ont x \text{ twice}\\ &~~~\If A \text{ accepts at least once}, \accept\\ &~~~\Else \reject\end{split}$
$$B$$ just runs $$A$$ twice on the input, accepting if at least one run of $$A$$ accepts; $$B$$ only rejects if both runs of $$A$$ reject. Thus, the probability that $$B$$ rejects $$x$$ is:
$\begin{split}\Pr[B \text{ rejects } x] &= \Pr[A \text{ rejects x in run 1}, A \text{ rejects x in run 2}]\\ &= \Pr[A \text{ rejects x in run 1}] \cdot \Pr[A \text{ rejects x in run 2}]\\ &= \Pr[A \text{ rejects } x]^2\end{split}$
Here, we have used the fact that the two runs of $$A$$ are independent, so the probability of $$A$$ rejecting twice is the product of the probabilities it rejects each time. This gives us:
$\begin{split}x \in L &\implies \Pr[B \text{ rejects } x] \le (1 - c)^2\\ x \notin L &\implies \Pr[B \text{ rejects } x] = 1\end{split}$
or equivalently:
$\begin{split}x \in L &\implies \Pr[B \text{ accepts } x] \ge 1 - (1 - c)^2\\ x \notin L &\implies \Pr[B \text{ accepts } x] = 0\end{split}$
Repeating this reasoning, if we modify $$B$$ to run $$A$$ $$k$$ times, we get:
$\begin{split}x \in L &\implies \Pr[B \text{ accepts } x] \ge 1 - (1 - c)^k\\ x \notin L &\implies \Pr[B \text{ accepts } x] = 0\end{split}$
By applying a form of Bernoulli’s inequality, $$(1 + x)^r \le e^{rx}$$, we get:
$x \in L \implies \Pr[B \text{ accepts } x] \ge 1 - e^{-ck}$
If we want a particular lower bound $$1 - \varepsilon$$ for acceptance of true positives, we need
$\begin{split}e^{-ck} &\le \varepsilon\\ -ck &\le \ln \varepsilon\\ k &\ge \frac{-\ln \varepsilon}{c} = \frac{\ln(1/\varepsilon)}{c}\end{split}$
This is a constant for any constants $$c$$ and $$\varepsilon$$. Thus, we can amplify the probability of accepting a true positive from $$c$$ to $$1 - \varepsilon$$ for any $$\varepsilon$$ with a constant $$\ln(1/\varepsilon)/c$$ number of repetitions. The same reasoning can be applied to amplify one-sided-error algorithms with no false negatives.
One-sided-error randomized algorithms are a special case of two-sided-error randomized algorithms. Such an algorithm for a language $$L$$ can produce the wrong result for an input $$x$$ whether or not $$x \in L$$. However, the probability of getting the wrong result is bounded by a constant.
Definition 204 (BPP)
$$\BPP$$ is the class of languages that have efficient two-sided-error randomized algorithms. A language $$L$$ is in $$\BPP$$ if there is an efficient randomized algorithm $$A$$ such that:
• if $$x \in L$$, $$\Pr[A \text{ accepts } x] \ge c$$
• if $$x \notin L$$, $$\Pr[A \text{ rejects } x] \ge c$$
Here, $$c$$ must be a constant greater than $$\frac{1}{2}$$, so that the algorithm produces the correct answer the majority of the time. Often, $$c$$ is chosen to be $$\frac{2}{3}$$ or $$\frac{3}{4}$$.
$$\BPP$$ stands for bounded-error probabilistic polynomial time. Given the symmetric definition of a two-sided error algorithm, it is clear that the class $$\BPP$$ is closed under complement – if a language $$L \in \BPP$$, then $$\overline{L} \in \BPP$$ as well.
Languages in $$\RP$$ and in $$\coRP$$ both trivially satisfy the conditions for $$\BPP$$; for instance, we have the following for $$\RP$$:
• if $$x \in L$$, $$\Pr[A \text{ accepts } x] \ge c$$
• if $$x \notin L$$, $$\Pr[A \text{ rejects } x] = 1 \ge c$$
Thus, $$\RP \cup \coRP \subseteq \BPP$$. By similar reasoning, we can relate $$\P$$ to these classes as follows:
\begin{gather*} \P \subseteq \RP \subseteq \BPP\\ \P \subseteq \coRP \subseteq \BPP \end{gather*}
As with one-sided-error algorithms, the probability of success for two-sided-error randomized algorithms can be amplified arbitrarily. However, we do not yet have the tools we need to reason about how to do so, so we will come back to this later.
One-sided-error and two-sided-error randomized algorithms are known as Monte Carlo algorithms. Such algorithms have a bounded runtime, but they may produce the wrong answer. Contrast this with Las Vegas algorithms, which always produce the correct answer, but where the runtime bound only holds in expectation. We can define a complexity class for languages that have Las Vegas algorithms with expected runtime that is polynomial in the size of the input.
Definition 205 (ZPP)
$$\ZPP$$ is the class of languages that have efficient Las Vegas algorithms. A language $$L$$ is in $$\ZPP$$ if there is a randomized algorithm $$A$$ such that:
• If $$x \in L$$, $$A$$ always accepts $$x$$.
• If $$x \notin L$$, $$A$$ always rejects $$x$$.
• The expected runtime of $$A$$ on input $$x$$ is $$\O(\abs{x}^k)$$ for some constant $$k$$.
There is a relationship between Monte Carlo and Las Vegas algorithms: if a language $$L$$ has both a one-sided-error algorithm with no false positives and a one-sided-error algorithm with no false negatives, then it has a Las Vegas algorithm, and vice versa. In terms of complexity classes, we have $$L$$ is in both $$\RP$$ and $$\coRP$$ if and only if $$L$$ is in $$\ZPP$$. This implies that
$\RP \cap \coRP = \ZPP$
We demonstrate this as follows. Suppose $$L$$ is in $$\RP \cap \coRP$$. Then it has an efficient, one-sided-error algorithm $$A$$ with no false positives such that:
• if $$x \in L$$, $$\Pr[A \text{ accepts } x] \ge \frac{1}{2}$$
• if $$x \notin L$$, $$\Pr[A \text{ accepts } x] = 0$$
Here, we have selected $$c = \frac{1}{2}$$ for concreteness. $$L$$ also has an efficient, one-sided algorithm $$B$$ with no false negatives such that:
• if $$x \in L$$, $$\Pr[B \text{ rejects } x] = 0$$
• if $$x \notin L$$, $$\Pr[B \text{ rejects } x] \ge \frac{1}{2}$$
We can construct a new algorithm $$C$$ as follows:
$\begin{split}&\Algorithm C(x):\\ &~~~\textbf{Repeat}:\\ &~~~~~~\Run A \ont x, \accept \text{if } A \accepts\\ &~~~~~~\Run B \ont x, \reject \text{if } B \rejects\end{split}$
Since $$A$$ only accepts $$x \in L$$ and $$C$$ only accepts when $$A$$ does, $$C$$ only accepts $$x \in L$$. Similarly, $$B$$ only rejects $$x \notin L$$, so $$C$$ only rejects $$x \notin L$$. Thus, $$C$$ always produces the correct answer for a given input.
To show that $$L \in \ZPP$$, we must also demonstrate that the expected runtime of $$C$$ is polynomial. For each iteration $$i$$ of $$C$$, we have:
• if $$x \in L$$, $$\Pr[A \text{ accepts } x \text{ in iteration } i] \ge \frac{1}{2}$$
• if $$x \notin L$$, $$\Pr[B \text{ rejects } x \text{ in iteration } i] \ge \frac{1}{2}$$
Thus, if $$C$$ gets to iteration $$i$$, the probability that $$C$$ terminates in that iteration is at least $$\frac{1}{2}$$. We can model the number of iterations as a random variable $$X$$, and $$X$$ has the probability distribution:
$\Pr[X > k] \le \parens{\frac{1}{2}}^k$
This is similar to a geometric distribution, where
$\Pr[Y > k] = (1 - p)^k$
for some $$p \in (0, 1]$$. The expected value of such a distribution is $$\Ex[Y] = 1/p$$. This gives us:
$\Ex[X] \le 2$
Thus, the expected number of iterations of $$C$$ is no more than two, and since $$A$$ and $$B$$ are both efficient, calling them twice is also efficient.
We have shown that $$\RP \cap \coRP \subseteq \ZPP$$. We will proceed to show that $$\ZPP \subseteq \RP$$, meaning that if a language has an efficient Las Vegas algorithm, it also has an efficient one-sided-error algorithm with no false positives. Assume that $$C$$ is a Las Vegas algorithm for $$L \in \ZPP$$, with expected runtime of no more than $$\abs{x}^k$$ steps for an input $$x$$ and some constant $$k$$. We construct a new algorithm $$A$$ as follows:
$\begin{split}&\Algorithm A(x):\\ &~~~\Run C \ont x \text{ for 2\abs{x}^k steps (or until it halts, whichever comes first)}\\ &~~~\Accept \text{ if C accepted, else} \reject\end{split}$
$$A$$ is clearly efficient in the size of the input $$x$$. It also has no false positives, since it only accepts if $$C$$ accepts within the first $$2\abs{x}^k$$ steps, and $$C$$ always produces the correct answer. All that is left is to show that if $$x \in L$$, $$\Pr[A \text{ accepts } x] \ge \frac{1}{2}$$ (again, we arbitrarily choose $$c = \frac{1}{2}$$).
Let $$S$$ be the number of steps before $$C$$ accepts $$x$$. By assumption, we have
$\Ex[S] \le \abs{x}^k$
$$A$$ runs $$C$$ for $$2\abs{x}^k$$ steps, so we need to bound the probability that $$C$$ takes more than $$2\abs{x}^k$$ steps on $$x$$. We have:
$\begin{split}\Pr[S > 2\abs{x}^k] &\le \Pr[S \ge 2\abs{x}^k]\\ &\le \frac{\Ex[S]}{2\abs{x}^k}\\ &\le \frac{\abs{x}^k}{2\abs{x}^k}\\ &= \frac{1}{2}\end{split}$
Here, we applied Markov’s inequality in the second step. Then:
$\begin{split}\Pr[A \text{ accepts } x] &= \Pr[S \le 2\abs{x}^k]\\ &= 1 - \Pr[S > 2\abs{x}^k]\\ &\ge \frac{1}{2}\end{split}$
Thus, $$A$$ is indeed a one-sided-error algorithm with no false positives, so $$L \in \RP$$. A similar construction can be used to demonstrate that $$L \in \coRP$$, allowing us to conclude that $$\ZPP \subseteq \RP \cap \coRP$$. Combined with our previous proof that $$\RP \cap \coRP \subseteq \ZPP$$, we have $$\RP \cap \coRP = \ZPP$$.
One final observation we will make is that if a language $$L$$ is in $$\RP$$, then it is also in $$\NP$$. Since $$L \in \RP$$, there is an efficient, one-sided-error randomized algorithm $$A$$ to decide $$L$$. $$A$$ is allowed to make random choices, and we can model each choice as a coin flip, i.e. being either 0 or 1 according to some probability distribution. We can represent the combination of these choices in a particular run of $$A$$ as a binary string $$c$$. This enables us to write an efficient verifier $$V$$ for $$L$$ as follows:
$\begin{split}&\Algorithm V(x, c = c_1 c_2 \dots c_m):\\ &~~~\text{Simulate the execution of A on x, taking choice c_i}\\ &~~~~~~\text{for the ith random decision in A}\\ &~~~\Accept \text{if } A \accepts, \text{else} \reject\end{split}$
If $$x \in L$$, then $$\Pr[A \text{ accepts } x] \ge \frac{1}{2}$$, so at least half the possible sequences of random choices lead to $$A$$ accepting $$x$$. On the other hand, if $$x \notin L$$, then $$\Pr[A \text{ rejects } x] = 1$$, so all sequences of random choices lead to $$A$$ rejecting $$x$$. Thus, $$V$$ accepts at least half of all possible certificates when $$x \in L$$, and $$V$$ rejects all certificates when $$x \notin L$$, so $$V$$ is a verifier for $$L$$. Since $$A$$ is efficient, $$V$$ is also efficient.
We summarize the known relationships between complexity classes as follows:
\begin{gather*} \P \subseteq \ZPP \subseteq \RP \subseteq \NP\\ \P \subseteq \ZPP \subseteq \RP \subseteq \BPP\\ \P \subseteq \ZPP \subseteq \coRP \subseteq \coNP\\ \P \subseteq \ZPP \subseteq \coRP \subseteq \BPP \end{gather*}
These relationships are represented pictorially, with edges signifying containment, as follows:
Here, $$\coNP$$ is the class of languages whose complements are in $$\NP$$:
$\coNP = \{L : \overline{L} \in \NP\}$
We do not know if any of the containments above are strict, and we do not know the relationships between $$\NP$$, $$\coNP$$, and $$\BPP$$. It is commonly believed that $$\P$$ and $$\BPP$$ are equal and thus $$\BPP$$ is contained in $$\NP \cap \coNP$$, that neither $$\P$$ nor $$\BPP$$ contain all of $$\NP$$ or all of $$\coNP$$, and that $$\NP$$ and $$\coNP$$ are not equal. However, none of these conjectures has been proven 4.
4
It is known, however, that if $$\P = \NP$$, then $$\P = \BPP$$. This is because $$\BPP$$ is contained within the polynomial hierarchy, and one of the consequences of $$\P = \NP$$ is the “collapse” of the hierarchy. The details are beyond the scope of this text.
## Amplification for Two-Sided-Error Algorithms¶
Previously, we saw how to amplify the probability of success for one-sided-error randomized algorithms. We now consider amplification for two-sided-error algorithms. Recall that such an algorithm $$A$$ for a language $$L$$ has the following behavior:
• if $$x \in L$$, $$\Pr[A \text{ accepts } x] \ge c$$
• if $$x \notin L$$, $$\Pr[A \text{ rejects } x] \ge c$$
Here, $$c$$ must be a constant that is strictly greater than $$\frac{1}{2}$$.
Unlike in the one-sided case with no false positives, we cannot just run the algorithm multiple times and observe if it accepts at least once. A two-sided-error algorithm can accept both inputs in and not in the language, and it can reject both such inputs as well. However, we observe that because $$c > \frac{1}{2}$$, we expect to get the right answer the majority of the time when we run a two-sided-error algorithm on an input. More formally, suppose we run the algorithm $$n$$ times. Let $$Y_i$$ be an indicator random variable that is 1 if the algorithm accepts in the $$i$$th run. Then:
• if $$x \in L$$, $$\Ex[Y_i] = \Pr[Y_i = 1] \ge c$$
• if $$x \notin L$$, $$\Ex[Y_i] = \Pr[Y_i = 1] \le 1 - c$$
Let $$Y = Y_1 + \dots + Y_n$$ be the total number of accepts out of $$n$$ trials. By linearity of expectation, we have:
• if $$x \in L$$, $$\Ex[Y] \ge cn > \frac{n}{2}~~~$$ (since $$c > \frac{1}{2}$$)
• if $$x \notin L$$, $$\Ex[Y] \le (1 - c)n < \frac{n}{2}~~~$$ (since $$1 - c < \frac{1}{2}$$)
This motivates an amplification algorithm $$B$$ that runs the original algorithm $$A$$ multiple times and takes the majority vote:
$\begin{split}&\Algorithm B(x):\\ &~~~\Run A \ont x \text{ repeatedly, for a total of n times}\\ &~~~\If A \text{ accepts at least n/2 times}, \accept\\ &~~~\Else \reject\end{split}$
Suppose we wish to obtain a bound that is within $$\gamma$$ of 1:
• if $$x \in L$$, $$\Pr[B \text{ accepts } x] \ge 1 - \gamma$$
• if $$x \notin L$$, $$\Pr[B \text{ rejects } x] \ge 1 - \gamma$$
What value for $$n$$ should we use in $$B$$?
We first consider the case where $$x \in L$$. The indicators $$Y_i$$ are independent, allowing us to use Hoeffding’s inequality on their sum $$Y$$. $$B$$ accepts $$x$$ when $$Y \ge \frac{n}{2}$$, or equivalently, $$\frac{Y}{n} \ge \frac{1}{2}$$. We want
$\lrPr{\frac{Y}{n} \ge \frac{1}{2}} \ge 1 - \gamma$
or equivalently,
$\begin{split}\lrPr{\frac{Y}{n} < \frac{1}{2}} &\le \lrPr{\frac{Y}{n} \le \frac{1}{2}}\\ &= \lrPr{\frac{Y}{n} \le c - \parens{c - \frac{1}{2}}}\\ &= \lrPr{\frac{Y}{n} \le c - \varepsilon}\\ &\le \gamma\end{split}$
where $$\varepsilon = c - \frac{1}{2}$$. To apply Hoeffding’s inequality, we need something of the form
$\lrPr{\frac{Y}{n} \le p - \varepsilon}$
where $$p = \lrEx{\frac{Y}{n}}$$. Unfortunately, we do not know the exact value of $$p$$; all we know is that $$p = \lrEx{\frac{Y}{n}} \ge c$$. However, we know that because $$p \ge c$$,
$\lrPr{\frac{Y}{n} \le c - \varepsilon} \le \lrPr{\frac{Y}{n} \le p - \varepsilon}$
In general, the event $$X \le a$$ includes at most as many sample points as $$X \le b$$ when $$a \le b$$; the latter includes all the outcomes in $$X \le a$$, as well as those in $$a < X \le b$$. We thus need only compute an upper bound on $$\lrPr{\frac{Y}{n} \le p - \varepsilon}$$, and that same upper bound will apply to $$\lrPr{\frac{Y}{n} \le c - \varepsilon}$$. Taking $$\varepsilon = c - \frac{1}{2}$$ and applying the lower-tail Hoeffding’s inequality, we get:
$\begin{split}\lrPr{\frac{Y}{n} \le p - \varepsilon} &= \lrPr{\frac{Y}{n} \le p - \parens{c - \frac{1}{2}}}\\ &\le e^{-2(c-1/2)^2 n}\end{split}$
We want this to be bound by $$\gamma$$:
$\begin{split}e^{-2(c-1/2)^2 n} &\le \gamma\\ e^{2(c-1/2)^2 n} &\ge \frac{1}{\gamma}\\ 2(c - 1/2)^2 n &\ge \ln\parens{\frac{1}{\gamma}}\\ n &\ge \frac{1}{2(c - 1/2)^2} \ln\parens{\frac{1}{\gamma}}\end{split}$
As a concrete example, suppose $$c = \frac{3}{4}$$, meaning that $$A$$ accepts $$x \in L$$ with probability at least $$\frac{3}{4}$$. Suppose we want $$B$$ to accept $$x \in L$$ at least 99% of the time, giving us $$\gamma = 0.01$$. Then:
$\begin{split}n &\ge \frac{1}{2(3/4 - 1/2)^2} \ln\parens{\frac{1}{0.01}}\\ &= \frac{1}{2\cdot 0.25^2} \ln 100\\ &\approx 36.8\end{split}$
Thus, it is sufficient for $$B$$ to run $$n \ge 37$$ trials of $$A$$ on $$x$$.
We now consider $$x \notin L$$. We want $$B$$ to reject $$x$$ with probability at least $$1 - \gamma$$, or equivalently, $$B$$ to accept $$x$$ with probability at most $$\gamma$$:
$\lrPr{\frac{Y}{n} \ge \frac{1}{2}} \le \gamma$
Similar to before, we have
$\begin{split}\lrPr{\frac{Y}{n} \ge \frac{1}{2}} &= \lrPr{\frac{Y}{n} \ge (1 - c) + \parens{c - \frac{1}{2}}}\\ &= \lrPr{\frac{Y}{n} \ge (1 - c) + \varepsilon}\end{split}$
with $$\varepsilon = c - \frac{1}{2}$$. Since $$p = \lrEx{\frac{Y}{n}} \le (1 - c)$$, we have
$\lrPr{\frac{Y}{n} \ge (1 - c) + \varepsilon} \le \lrPr{\frac{Y}{n} \ge p + \varepsilon}$
This follows from similar reasoning as earlier: an event $$X \ge a$$ contains no more sample points than $$X \ge b$$ when $$a \ge b$$. With $$\varepsilon = c - \frac{1}{2}$$, the upper-tail Hoeffding’s inequality gives us:
$\begin{split}\lrPr{\frac{Y}{n} \ge p + \varepsilon} &= \lrPr{\frac{Y}{n} \ge p + \parens{c - \frac{1}{2}}}\\ &\le e^{-2(c-1/2)^2 n}\end{split}$
We want this to be no more than $$\gamma$$, which leads to the same solution as before:
$n \ge \frac{1}{2(c - 1/2)^2} \ln\parens{\frac{1}{\gamma}}$
Using the same concrete example as before, if we want $$B$$ to reject $$x \notin L$$ at least 99% of the time when $$A$$ rejects $$x \notin L$$ with probability at least $$\frac{3}{4}$$, it suffices for $$B$$ to run $$n \ge 37$$ trials of $$A$$ on $$x$$.
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2022-11-30 03:03:47
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https://amshove.org/posts/2016/06/PowershellNavigation/
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# Powershell functions to navigate faster in the terminal
I’d like to share 3 Powershell functions that I’ve wrote this week to navigate a bit faster in the Powershell.
I’m having a lot of different folders to organize my files and projects and it’s kinda exhausting to tab through the folders with the command, which looks something like this:
cd D:\scTAB\dotnTAB\proTAB\paTAB\is\getTAB\long
What would be cool instead is
ccd d: sc dotn pro pa i ge lo
since I know the folder structure of paths I often visit.
(I did use the name ccd because the finger is on the c anyway when you want to type c)
With that goal in mind I wrote three functions which will achieve this beheaviour. The source can be found on Github.
# Function gdp
The function gdp returns the path matching the abbreviations you give it as an argument.
If the folder is found, the output will be the following:
C:\Users\Markus> gdp ~ doc windo fun
~\Documents\WindowsPowerShell\Functions
It is possible to use special characters for home (~), root (\ or /) and another partition, e.g. d:.
If the argument is ambiguous, you will get a list of matching folders to sharpen your argument:
C:\Users\Markus> gdp d: s me
scm
software
's' is ambiguous.
In C:\Users\Markus\Documents\WindowsPowerShell\Functions\ccd.ps1:36 Zeichen:17
+ throw "'\$folder' is ambiguous."
+ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+ CategoryInfo : OperationStopped: ('s' is ambiguous.:String) [], RuntimeException
+ FullyQualifiedErrorId : 's' is ambiguous.
If the path is not found, it will just throw an error.
# Function ccd
The function ccd is just a wraper around the gdp function, which means if the gdp function returns a valid path, it will change to the directory.
# Function gfp
I liked the idea to use it with folders so I wrote another functions which does the same thing for files.
The function is called gfp (get file path) and it works the same way:
C:\Users\Markus> gfp ~ vimf vim
~\vimfiles\vimrc
I can also use it to give the path as a argument to another program or script:
C:\Users\Markus> gfp (~ vimf vim)
What happens is that vim will start and open the file „~\vimfiles\vimrc“.
This function also behaves like the gdp function in that it will list ambiguous filenames and throw an error if no file is found.
I hope you enjoy it!
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2019-02-16 16:31:47
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http://www.fromthebottomoftheheap.net/
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## Simultaneous intervals for derivatives of smooths revisited
#### 21 March 2017 /posted in: R
Eighteen months ago I screwed up! I’d written a post in which I described the use of simulation from the posterior distribution of a fitted GAM to derive simultaneous confidence intervals for the derivatives of a penalized spline. It was a nice post that attracted some interest. It was also wrong. In December I corrected the first part of that mistake by illustrating one approach to compute an actual simultaneous interval, but only for the fitted smoother. At the time I thought that the approach I outlined would translate to the derivatives but I was being lazy then Christmas came and went and I was back to teaching — you know how it goes. Anyway, in this post I hope to finally rectify my past stupidity and show how the approach used to generate simultaneous intervals from the December 2016 post can be applied to the derivatives of a spline.
## Modelling extremes using generalized additive models
#### 25 January 2017 /posted in: R
Quite some years ago, whilst working on the EU Sixth Framework project Euro-limpacs, I organized a workshop on statistical methods for analyzing time series data. One of the sessions was on the analysis of extremes, ably given by Paul Northrop (UCL Department of Statistical Science). That intro certainly whet my appetite but I never quite found the time to dig into the arcane world of extreme value theory. Two recent events rekindled my interest in extremes; Simon Wood quietly introduced into his mgcv package a family function for the generalized extreme value distribution (GEV), and I was asked to review a paper on extremes in time series. Since then I’ve been investigating options for fitting models for extremes to environmental time series, especially those that allow for time-varying effects of covariates on the parameters of the GEV. One of the first things I did was sit down with mgcv to get a feel for the gevlss() family function that Simon had added to the package by repeating an analysis of a classic example data set that had been performed using the VGAM package of Thomas Yee.
## Pangaea and R and open palaeo data (also GAM all the things!)
#### 16 December 2016 /posted in: R
For a while now, I’ve been wanting to experiment with rOpenSci’s pangaear package (Chamberlain et al., 2016), which allows you to search, and download data from, the Pangaea, a major data repository for the earth and environmental sciences. Earlier in the year, as a member of the editorial board of Scientific Data, Springer Nature’s open data journal I was handling a data descriptor submission that described a new 2,200-year foraminiferal δ18O record from the Gulf of Taranto in the Ionian Sea (Taricco et al., 2016). The data descriptor was recently published and as part of the submission Carla Taricco deposited the data set in Pangaea. So, what better opportunity to test out pangaear? (Oh and to fit a GAM to the data while I’m at it!)
Chamberlain, S., Woo, K., MacDonald, A., Zimmerman, N., and Simpson, G. (2016). Pangaear: Client for the ’pangaea’ database. Available at: https://CRAN.R-project.org/package=pangaear.
Taricco, C., Alessio, S., Rubinetti, S., Vivaldo, G., and Mancuso, S. (2016). A foraminiferal ()18O record covering the last 2,200 years. Scientific Data 3, 160042. doi:10.1038/sdata.2016.42.
## Simultaneous intervals for smooths revisited correcting a silly mistake
#### 15 December 2016 /posted in: R
Eighteen months ago I wrote a post in which I described the use of simulation from the posterior distribution of a fitted GAM to derive simultaneous confidence intervals for the derivatives of a penalised spline. It was a nice post that attracted some interest. It was also wrong. I have no idea what I was thinking when I thought the intervals described in that post were simultaneous. Here I hope to rectify that past mistake.
I’ll tackle the issue of simultaneous intervals for the derivatives of penalised spline in a follow-up post. Here, I demonstrate one way to compute a simultaneous interval for a penalised spline in a fitted GAM. As example data, I’ll use the strontium isotope data set included in the SemiPar package, and which is extensively analyzed in the monograph Semiparametric Regression (Ruppert et al., 2003). First, load the packages we’ll need as well as the data, which is data set fossil. If you don’t have SemiPar installed, install it using install.packages(“SemiPar”) before proceeding
Ruppert, D., Wand, M. P., and Carroll, R. J. (2003). Semiparametric regression. Cambridge University Press.
## ISEC 2016 Talk
#### 02 July 2016 /posted in: Science
My ISEC 2016 talk, Estimating temporal change in mean and variance of community composition via location, scale additive models, describes some of my recent research into methods to analyse palaeoenvironmental time series from sediment cores.
## Rootograms a new way to assess count models
#### 07 June 2016 /posted in: R
Assessing the fit of a count regression model is not necessarily a straightforward enterprise; often we just look at residuals, which invariably contain patterns of some form due to the discrete nature of the observations, or we plot observed versus fitted values as a scatter plot. Recently, while perusing the latest statistics offerings on ArXiv I came across Kleiber and Zeileis (2016) who propose the rootogram as an improved approach to the assessment of fit of a count regression model. The paper is illustrated using R and the authors’ countreg package (currently on R-Forge only). Here, I thought I’d take a quick look at the rootogram with some simulated species abundance data.
Kleiber, C., and Zeileis, A. (2016). Visualizing count data regressions using rootograms.
## A new default plot for multivariate dispersions tribulations of base graphics programming
#### 17 April 2016 /posted in: R
This weekend, prompted by a pull request from Michael Friendly, I finally got round to improving the plot method for betadisper() in the vegan package. betadisper() is an implementation of Marti Anderson’s Permdisp method, a multivariate analogue of Levene’s test for homogeneity of variances. In improving the default plot and allowing customisation of plot features, I was reminded of how much I dislike programming plot functions that use base graphics. But don’t worry, this isn’t going to degenerate into a ggplot love-in nor a David Robinson-esque dig at Jeff Leek.
## LOESS revisited
#### 10 April 2016 /posted in: Science
It’s fair to say I have gotten a bee1 in my bonnet about how palaeolimnologists handle time. For a group of people for whom time is everything, we sure do a poor job (in general) of dealing with it in when it comes time to analyse our data. In many instances, “poor job” means making no attempt at all to account for the special nature of the time series. LOESS comes in for particular criticism because it is widely used by palaeolimnologists despite not being particularly suited to the task. Why this is so is perhaps due to it’s promotion in influential books, papers, and software. I am far from innocent in this regard having taught LOESS and it’s use for many years on the now-defunct ECRC Numerical Course. Here I want to look at further problems with our use of LOESS, and will argue that we need to resign it to the trash can for all but exploratory analyses. I will begin the case for the prosecution with one of my own transgressions.
1. an entire hive is perhaps more apt!
## Soap-film smoothers & lake bathymetries
#### 27 March 2016 /posted in: R
A number of years ago, whilst I was still working at ENSIS, the consultancy arm of the ECRC at UCL, I worked on a project for the (then) Countryside Council for Wales (CCW; now part of Natural Resources Wales). I don’t recall why they were doing this project, but we were tasked with producing a standardised set of bathymetric maps for Welsh lakes. The brief called for the bathymetries to be provided in standard GIS formats. Either CCW’s project manager or the project lead at ENSIS had proposed to use inverse distance weighting (IWD) to smooth the point bathymetric measurements. This probably stemmed from the person that initiatied our bathymetric programme at ENSIS being a GIS wizard, schooled in the ways of ArcGIS. My involvement was mainly data processing of the IDW results. I was however, at the time, also somewhat familiar with the problem of finite area smoothing1 and had read a paper of Simon Wood’s on his then new soap-film smoother (Wood et al., 2008). So, as well as writing scripts to process and present the IDW-based bathymetry data in the report, I snuck a task into the work programme that allowed me to investigate using soap-film smoothers for modelling lake bathymetric data. The timing was never great to write up this method (two children and a move to Canada have occurred since the end of this project), so I’ve not done anything with the idea. Until now…
Wood, S. N., Bravington, M. V., and Hedley, S. L. (2008). Soap film smoothing. Journal of the Royal Statistical Society. Series B, Statistical methodology 70, 931–955. doi:10.1111/j.1467-9868.2008.00665.x.
1. smoothing over a domain with known boundaries, like a lake
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2017-04-28 21:42:02
|
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https://www.elibm.org/article/10011498
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Bounds for the dimensions of $p$-adic multiple $L$-value spaces
Summary
First, we will define $p$-adic multiple $L$-values ($p$-adic MLV's), which are generalizations of Furusho's $p$-adic multiple zeta values ($p$-adic MZV's) in Section 2. Next, we prove bounds for the dimensions of $p$-adic MLV-spaces in Section 3, assuming results in Section 4, and make a conjecture about a special element in the motivic Galois group of the category of mixed Tate motives, which is a $p$-adic analogue of Grothendieck's conjecture about a special element in the motivic Galois group. The bounds come from the rank of $K$-groups of ring of $S$-integers of cyclotomic fields, and these are $p$-adic analogues of Goncharov-Terasoma's bounds for the dimensions of (complex) MZV-spaces and Deligne-Goncharov's bounds for the dimensions of (complex) MLV-spaces. In the case of $p$-adic MLV-spaces, the gap between the dimensions and the bounds is related to spaces of modular forms similarly as the complex case. In Section 4, we define the crystalline realization of mixed Tate motives and show a comparison isomorphism, by using $p$-adic Hodge theory.
Mathematics Subject Classification
11G55, 11R42, 14F42, 14F30
Keywords/Phrases
p-adic multiple zeta values, mixed Tate motives, algebraic K-theory, p-adic Hodge theory
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2021-07-25 16:41:43
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https://math.stackexchange.com/questions/295383/integration-of-a-function-with-respect-to-another-function/295412
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# Integration of a function with respect to another function.
What is the intuition/idea behind integration of a function with respect to another function? Say $$\int f(x)d(g(x)) \;\;\;\;\;?$$ or may be a more particular example $$\int x^2d(x^3)$$
My concern is not at the level of problem solving. To solve we could simply substitute $u=x^3$ and then $x^2=u^{2/3}$. My concern is rather about what meaning physical/geometrical does this impart?
ADDED if you ask what kind of meaning I seek, integration of a function w.r.t a variable gives the area under the curve and above x-axis.
There's no geometric interpretation of $\int f(x) \mbox{d}(g(x))$ as far as I'm concern. However, it does have a meaning. Say, you want to take the differentiation of $\sin(x)$ w.r.t x, and $\sin (x)$, they give 2 different results:
• $\frac{\mbox{d} \sin x}{\mbox{d}x} = \cos x$
• $\frac{\mbox{d} \sin x}{\mbox{d} \sin x} = 1$
Another example is to take the derivatives of $x^4$ w.r.t x, and $x^2$ respectively.
• $\frac{\mbox{d} x^4}{\mbox{d}x} = 4x^3$
• $\frac{\mbox{d} x^4}{\mbox{d} x^2} = \frac{\mbox{d} (x^2)^2}{\mbox{d} x^2} = 2x^2$
An antiderivative of $f$ w.r.t $x$ is some function $F$, such that $\dfrac{\mbox{d}F}{\mbox{d}x} = f$
So:
• $\int \mbox{d}(x^2)$ is some function, such that its derivative w.r.t $x^2$ is 1, this family of function is, of course, $x^2 + C$.
• $\int x^5\mbox{d}(x^5)$ is some function, such that its derivative w.r.t $x^5$ is $x^5$, this family of function is, of course, $\dfrac{x^{10}}{2} + C$, since: $\frac{1}{2}\dfrac{\mbox{d}(x^{10} + C)}{\mbox{d}(x^5)} = \frac{1}{2}\dfrac{\mbox{d}((x^5)^2)}{\mbox{d}(x^5)} + 0 = x^5$
When solving problems, you just need to remember that $\int f(x)\mbox{d}(g(x)) = \int f(x).g'(x) \mbox{d}x$, e.g, when you take some function out of d, you have to differentiate it, and vice versa, when you put some function into d, you'll have to integrate it, like this:
• $\int x^5 \mbox{d}(x^2) = \int 2x^6 \mbox{d}(x) = \dfrac{2x^7}{7} + C$. (Take $x^2$ out of d, we differentiate it, and have $2x$)
• $\int \sin^2 x \cos x \mbox{d}x = \int \sin^2 x \mbox{d}(\sin x) = \frac{\sin ^ 3 x}{3} + C$ (Put $\cos x$ into d, we have to integrate it to get $\sin x$).
You shouldn't think of $$g$$ as a function, it's a change of variable. It's a way to parametrize the integral, with no geometrical or physical sense.
ADDENDUM : in fact, both in mathematics and in physics, a major concern has been to develop theories in which the significant results do not depend on the choice of such a paremetrization. For example, in maths, the object that is "natural" to integrate are differential forms and not functions because the integral will not depend on the choice of the parametrization.
ADDENDUM 2 : What I mean by parametrization. Consider the following problem : you have a segment of curve $$C$$ and you wish to calculate its length (the simplest example : $$C$$ is a straight segment). By definition, the length is $$\int_C \|\gamma'(t)\| dt$$, where $$\gamma : [0,1] \rightarrow \mathbb{R}^3$$ is a parametrization of the curve. (once again, you may replace $$\mathbb{R}^3$$ by $$\mathbb{R}$$ if you prefer, in which case $$C$$ is a straight segment). But there are many different parametrizations $$\gamma$$ (think of it as a point moving along the curve : you may move at different speeds, but you will still move along the curve). The number $$\|\gamma'(t)\|$$ represents the speed of the parametrization at the time $$t$$. The number $$L = \int_C \|\gamma'(t)\| dt$$ is independent of the parametrization, as it should be since it represents the length. You can see this using the change of variable formula.
• I think if our function $g(x)$ is continuously differentiable, so $$d(g(x))=g'(x)dx$$ and so $$\int f(x)d(g(x))=\int f(x)g'(x)dx$$ Feb 5, 2013 at 12:40
• @Glougloubarbaki, could you please elaborate what "parametrization of integral" means.
– user45099
Feb 5, 2013 at 12:49
• as @BabakSorouh noted, your formula was incorrect. Feb 5, 2013 at 13:22
I found this relevant discussion on math overflow.
Visualisation of Riemann-Stieltjes Integral
This is an old question with a marked answer, but there is a better one.
This is how you find the area under the curve for a function with respect to another function given some hidden parameter. One such application is the receiver operating charateristic (ROC) Receiver Operating Characteristic
In the ROC, we are interested in how the dependent variables $$f(t)$$ and $$g(t)$$ vary with $$t$$. We have the probability of false detection on the $$x$$-axis and the probability of detection on the $$y$$-axis. So, the integral of $$f(t)$$ with respect to $$g(t)$$ is an excellent measure of how likely we are to correctly identify the presence of some signal. There are likely other similar interpretations, but this is one I'm familiar with.
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2022-07-02 22:41:40
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https://en.wikipedia.org/wiki/Market_capitalization
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# Market capitalization
The New York Stock Exchange on Wall Street, the world's largest stock exchange in terms of total market capitalization of its listed companies[1]
Market capitalization (market cap) is the market value of a publicly traded company's outstanding shares. Market capitalization is equal to the share price multiplied by the number of shares outstanding.[2][3] As outstanding stock is bought and sold in public markets, capitalization could be used as an indicator of public opinion of a company's net worth and is a determining factor in some forms of stock valuation.
Market capitalization is used by the investment community in ranking the size of companies, as opposed to sales or total asset figures. It is also used in ranking the relative size of stock exchanges, being a measure of the sum of the market capitalizations of all companies listed on each stock exchange. (See List of stock exchanges.) In performing such rankings, the market capitalizations are calculated at some significant date, such as June 30 or December 31.
The total capitalization of stock markets or economic regions may be compared with other economic indicators. The total market capitalization of all publicly traded companies in the world was US$51.2 trillion in January 2007[4] and rose as high as US$57.5 trillion in May 2008[5] before dropping below US$50 trillion in August 2008 and slightly above US$40 trillion in September 2008.[5] In 2014 and 2015, global market capitalization was US$68 trillion and US$67 trillion, respectively.[6]
## Calculation
Market cap is given by the formula ${\textstyle MC=N\times P}$, where MC is the market capitalization, N is the number of shares outstanding, and P is the closing price per share.
For example, if some company has 4 million shares outstanding and the closing price per share is $20, its market cap is then$80 million. If the closing price per share rises to $21, the market cap becomes$84 million. If it drops to $19 per share, the market cap falls to$76 million. This is in contrast to mercantile pricing where purchase price, average price and sale price may differ due to transaction costs.
## Market cap terms
Traditionally, companies were divided into large-cap, mid-cap, and small-cap.[2] The terms mega-cap and micro-cap have also since come into common use,[7][8] and nano-cap is sometimes heard. Different numbers are used by different indexes;[9] there is no official definition of, or full consensus agreement about, the exact cutoff values. The cutoffs may be defined as percentiles rather than in nominal dollars. The definitions expressed in nominal dollars need to be adjusted over decades due to inflation, population change, and overall market valuation (for example, \$1 billion was a large market cap in 1950, but it is not very large now), and market caps are likely to be different country to country.
## Related measures
Market cap reflects only the equity value of a company. It is important to note that a firm's choice of capital structure has a significant impact on how the total value of a company is allocated between equity and debt. A more comprehensive measure is enterprise value (EV), which gives effect to outstanding debt, preferred stock, and other factors. For insurance firms, a value called the embedded value (EV) has been used.
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2018-11-18 22:47:32
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https://scipost.org/SciPostPhys.9.1.013
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## Critical properties of a comb lattice
Natalia Chepiga, Steven R. White
SciPost Phys. 9, 013 (2020) · published 27 July 2020
### Abstract
In this paper we study the critical properties of the Heisenberg spin-1/2 model on a comb lattice -- a 1D backbone decorated with finite 1D chains -- the teeth. We address the problem numerically by a comb tensor network that duplicates the geometry of a lattice. We observe a fundamental difference between the states on a comb with even and odd number of sites per tooth, which resembles an even-odd effect in spin-1/2 ladders. The comb with odd teeth is always critical, not only along the teeth, but also along the backbone, which leads to a competition between two critical regimes in orthogonal directions. In addition, we show that in a weak-backbone limit the excitation energy scales as $1/(NL)$, and not as $1/N$ or $1/L$ typical for 1D systems. For even teeth in the weak backbone limit the system corresponds to a collection of decoupled critical chains of length $L$, while in the strong backbone limit, one spin from each tooth forms the backbone, so the effective length of a critical tooth is one site shorter, $L-1$. Surprisingly, these two regimes are connected via a state where a critical chain spans over two nearest neighbor teeth, with an effective length $2L$.
### Ontology / Topics
See full Ontology or Topics database.
### Authors / Affiliations: mappings to Contributors and Organizations
See all Organizations.
Funders for the research work leading to this publication
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2020-08-10 10:27:07
|
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https://iris.polito.it/handle/11583/2726161
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In the study of the Boltzmann equation, one is led to consider integral operators TΦ,β with kernels of the form K(x,y)=∫RdΦ(u)exp(−2iπ(β(|u|)u⋅y−u⋅x))du,∀x,y∈Rd, and prove their boundedness on (possibly weighted) L1 spaces. For example, when the function β is constant (inelastic interactions with constant restitution coefficient), one has that TΦ,β is bounded on L1 if and only if the Fourier transform of Φ is in L1. In the more realistic case of a varying β, however, the problem is much more difficult. In the paper under review, the authors assume that Φ is in the Wiener amalgam space W(F(L1),L1) (covering a fairly typical case when Φ(u)=|u|(1+|u|2)m for m>d+12). Working with W(F(L1),L1)⊂L1 instead of all of L1 has the advantage that one has the following algebra property: ∥K(⋅,y)∥1=∥Φexp(−2iπϕ(y,⋅))ˆ∥1≤∥Φ∥W(F(L1),L1)∥exp(−2iπϕ(y,⋅))∥W(F(L1),L∞), where ϕ(y,u)=β(|u|)u⋅y. Therefore, Schur estimates for the kernel of TΦ,β can be obtained from W(F(L1),L∞) estimates involving only β. The authors do so in their main theorem, Theorem 3.1, finding appropriate conditions on β that guarantee the estimate ∥K(⋅,y)∥1≤C(1+|y|)α,∀y∈Rd, for some α>d2 depending on the function β (and hence a weighted L1 estimate for TΦ,β). An estimate for α
On Fourier integral operators with Hölder-continuous phase / Cordero, Elena; Nicola, Fabio; Primo, Eva. - In: ANALYSIS AND APPLICATIONS. - ISSN 0219-5305. - STAMPA. - 16:6(2018), pp. 875-893. [10.1142/S0219530518500112]
### On Fourier integral operators with Hölder-continuous phase
#### Abstract
In the study of the Boltzmann equation, one is led to consider integral operators TΦ,β with kernels of the form K(x,y)=∫RdΦ(u)exp(−2iπ(β(|u|)u⋅y−u⋅x))du,∀x,y∈Rd, and prove their boundedness on (possibly weighted) L1 spaces. For example, when the function β is constant (inelastic interactions with constant restitution coefficient), one has that TΦ,β is bounded on L1 if and only if the Fourier transform of Φ is in L1. In the more realistic case of a varying β, however, the problem is much more difficult. In the paper under review, the authors assume that Φ is in the Wiener amalgam space W(F(L1),L1) (covering a fairly typical case when Φ(u)=|u|(1+|u|2)m for m>d+12). Working with W(F(L1),L1)⊂L1 instead of all of L1 has the advantage that one has the following algebra property: ∥K(⋅,y)∥1=∥Φexp(−2iπϕ(y,⋅))ˆ∥1≤∥Φ∥W(F(L1),L1)∥exp(−2iπϕ(y,⋅))∥W(F(L1),L∞), where ϕ(y,u)=β(|u|)u⋅y. Therefore, Schur estimates for the kernel of TΦ,β can be obtained from W(F(L1),L∞) estimates involving only β. The authors do so in their main theorem, Theorem 3.1, finding appropriate conditions on β that guarantee the estimate ∥K(⋅,y)∥1≤C(1+|y|)α,∀y∈Rd, for some α>d2 depending on the function β (and hence a weighted L1 estimate for TΦ,β). An estimate for α
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Utilizza questo identificativo per citare o creare un link a questo documento: `http://hdl.handle.net/11583/2726161`
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2022-08-08 11:03:57
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https://indico.cern.ch/event/1109611/contributions/4821358/
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10th Edition of the Large Hadron Collider Physics Conference
May 16 – 20, 2022
Europe/Zurich timezone
Constraining Deep Neural Network classifiers’ systematic uncertainty via input feature space reduction
May 17, 2022, 7:00 PM
1h
Experimental poster Performance and Tools
Speaker
Mr Andrea Di Luca (Universita degli Studi di Trento and INFN (IT))
Description
In current and future high-energy physics experiments, the sensitivity of selection-based analysis will increasingly depend on the choice of the set of high-level features determined for each collision. The complexity of event reconstruction algorithms has escalated in the last decade, and thousands of parameters are available for analysts. Deep Learning approaches are widely used to improve the selection performance in physics analysis.
In many cases, the development of the algorithm is based on a brute force approach where all the possible combinations of available neural network architectures are tested using all the available parameters. A crucial aspect is that the results from a model based on a large number of input variables are more difficult to explain and understand. This point becomes relevant for neural network models since they do not provide uncertainty estimation and are often treated as perfect tools, which they are not.
In this work, we show how using a sub-optimal set of input features can lead to higher systematic uncertainty associated with classifier predictions. We also present an approach to selecting an optimal set of features using ensemble learning algorithms. For this study, we considered the case of highly boosted di-jet resonances produced in $pp$ collisions decaying to two $b$-quarks to be selected against an overwhelming QCD background. Results from a Monte Carlo simulation with HEP pseudo-detectors are shown.
Primary author
Mr Andrea Di Luca (Universita degli Studi di Trento and INFN (IT))
Presentation materials
LHCP2022 Poster.jpg LHCP2022 Poster.pdf poster_preview.jpg
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2022-10-02 22:15:19
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https://math.stackexchange.com/questions/2912995/actions-of-c-dynamical-systems-on-primitive-ideals
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# Actions of $C^*$-dynamical systems on primitive ideals
Reading about induced Systems of $C^*$-Algebras, I found this one statement that I can't figure out.
Let $G$ be a compact Group and $(A,G,\alpha)$ a $C^*$-dynamical System, such that for some closed subgroup $H$ of $G$ there exists a $G$-equivariant map $\varphi$ between $(\text{Prim}(A),G,\alpha)$ and $(G/H,G,\text{lt})$, where lt denotes the left-translation. Let $I:= \bigcap \{P\in \text{Prim}(A) \colon \varphi(P)=eH\}$.
Now it is stated that $\bigcap\{\alpha_s(I) : s\in G\}$ equals $\{0\}$. I don't really get. I know that $I$ is an $H$-invariant ideal in $A$, since $\varphi$ is equivariant and all the $\alpha_s$ are *-automorphisms.
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2019-02-17 01:35:12
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http://2000clicks.com/MathHelp/CalculusDerivative.aspx
|
# document.write (document.title)
Math Help > Calculus > Derivative
If f is a continuous function on the interval (a,b), then the f', the derivative of f, is defined as the limit,
### Contents of this section:
Here's what you'll find in this section:
d Sin proves that the derivative of the sine function is the cosine.
L'Hopital's Rule gives you a way to find the limit as x−>c of f(x)/g(x) when f(c)=g(c)=0.
### Related pages in this website
Limit
Related Rate Problems
The webmaster and author of this Math Help site is Graeme McRae.
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2019-04-18 22:24:50
|
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|
https://samacheerkalvi.guide/samacheer-kalvi-12th-maths-guide-chapter-2-ex-2-5/
|
# Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5
## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.5
Question 1.
Find the modulus of the following complex numbers.
(i) $$\frac{2i}{3+4i}$$
Solution:
(ii) $$\frac{2-i}{1+i}+\frac{1-2 i}{1-i}$$
Solution:
= 2√2
Samacheer Kalvi 12th Maths Guide
(iii) (1 – i)10
Solution:
|(1 – i)10| = (|1 – i|)10 = ($$\sqrt{1+1}$$)10
= (√2)10 = 25 = 32
(iv) 2i(3 – 4i)(4 – 3i)
Solution:
= |2i||3 – 4i||4 – 3i|
= ($$\sqrt{0+4}$$) ($$\sqrt{9+16}$$) ($$\sqrt{16+9}$$)
= 2 × 5 × 5 = 50
Question 2.
For any two complex numbers z1 and z2, such that |z1| = |z2| = 1 and z1 z2 ≠ -1, then show that $$\frac{z_{1}+z_{2}}{1+z_{1} z_{2}}$$ is real number.
Solution:
Given |z1| = 1 ⇒ z1 $$\bar{z}$$1 = 1
z1 = $$\frac{1}{\bar{z}_{1}}$$
similarly z2 = $$\frac{1}{\bar{z}_{2}}$$
Since z = $$\bar{z}$$, it is a real number.
Question 3.
Which one of the points 10 – 8i, 11 + 6i is closest to 1 + i.
Solution:
Let A be (10 – 8i) ⇒ (10, -8)
B be (11 + 6i) ⇒ (11, 6)
C be (1 + i) = (1, 1)
CA = |(1 + i) – (10 – 8i)|
= |- 9 + 9i| = $$\sqrt{81+81}$$
= $$\sqrt{162}$$ = 9√2
CB = |(1 + i) – (11 + 6i)
= |-10 – 5i|= $$\sqrt{100 + 25}$$
= $$\sqrt{125}$$
∴ The point B is closer to C, than A.
Question 4.
If |z| = 3, show that 7 ≤ |z + 6 – 8i| ≤ 13.
Solution:
To prove 7 ≤ |z + 6 – 8i| ≤ 13
|z + 6 – 8i| ≥ |z| + |16 + 8i| = 3 + $$\sqrt{36 + 64}$$ = 13
Also |z + 6 – 8i| ≤ ||z| – (6 + 8i)| = |3 – $$\sqrt{36 + 64}$$|
= |3 – 10| = 7
∴ 7 ≤ |z + 6 – 8i| ≤ 13
Question 5.
If |z| = 1, show that 2 ≤ |z² – 3| ≤ 4.
Solution:
|z² – 3| ≤ |z²| + |-3| = 1 + 3 = 4
Also |z² – 3| ≥ ||z²|-|-3|| = |1-3| = 2
Hence 2 ≤ |z² – 3| ≤ 4
Question 6.
If |z – $$\frac{2}{z}$$|= 2 show that the greatest and least value of |z| are √3 + 1 and √3 – 1 respectively.
Solution:
Given |z – $$\frac{2}{z}$$|= z
|z² – 2|= 2|z|
|z²| ≤ 2|z| + |-2|
≤ 2 |z| + 2
|z²|- 2|z| + 1 ≤ 2 + 1
(|z – 1|)² ≤ 3
|z| – 1 ≤ √3
|z| ≤ √3 + 1
Similarly |z| ≥ √3 – 1
Question 7.
If z1 z2 and z3 are three complex numbers such that |z1| = 1, |z1| = 2, |z3| = 3 and |z1 + z2 + z3| = 1 show that |9z1 z2 + 4z1 z3 + z2 z3| = 6.
Solution:
|z1| = 1, |z1| = 2, |z3| = 3
⇒ |z1|² = 1, |z2|² = 4, |z3|² = 9
z1 $$\bar {z_1}$$ = 1, z2 $$\bar {z_2}$$ = 4, z3 $$\bar {z_3}$$ = 9
$$\bar {z_1}$$ = $$\frac{1}{z_1}$$ $$\bar {z_2}$$ = $$\frac{4}{z_2}$$, $$\bar {z_3}$$ = $$\frac{9}{z_3}$$
⇒ |9z1 z2 + 4 z1 z3 + z2 z3| = 6
Question 8.
If the area of the triangle formed by the vertices z, iz, and z + iz is 50 square units, find the value of |z|.
Solution:
The vertices are z, iz and z + iz.
Since it side is z and other side is iz, it shows that they are perpendicular and z + iz is the hypotenuse.
Area of ΔAOB = $$\frac{1}{2}$$ |z| |zi|
⇒ $$\frac{1}{2}$$ |z| |z| = 50 (given)
⇒ |z|² = 100
|z| = 10 units
Question 9.
Show that the equation z³ + 2 $$\bar {z}$$ = 0 has five solutions.
Solution:
Given z³ + 2 $$\bar {z}$$ = 0
|z³| = |2 $$\bar {z}$$|
|z³| = |-2| |$$\bar {z}$$|
= 2|z| since |$$\bar {z}$$| = |z|
|z|³ – 2 |z| = 0
|z|(|z|² – 2) = 0
⇒ |z| = 0 |z|² = 2
⇒ z = 0 is a solution |z| ± √2
⇒ z $$\bar {z}$$ = ± √2
$$\bar {z}$$ = ± $$\frac {√2}{z}$$
But z³ + 2 $$\bar {z}$$ = 0
∴ z³ + 2(±$$\frac {√2}{z}$$) = 0
z4 ± 2 √2 z = 0
It has 4 non zero solutions.
∴ Including z = 0 we have 5 solutions.
Question 10.
Find the square roots of
(i) 4 + 3i
Solution:
|4 + 3i| = $$\sqrt {16±9}$$ = 5
(ii) -6 + 8i
Solution:
To find
(Here also b = 8 is + ve)
±$$(\sqrt{\frac{4}{2}}+i \sqrt{\frac{16}{2}})$$ = ± (2 ± 2√2 i)
(iii) -5 – 12i
Solution:
To find $$\sqrt{-5-12 i}$$ = |-5 – 12i|
= $$\sqrt{25+144}$$ = 13
Using the formula
= ±(2 – i3)
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2020-04-01 16:35:11
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https://solvedlib.com/n/17-pts-if-f-x-and-g-w-are-arbitrary-polynomials-of-degree,9778361
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# (17 pts) If f (x) and g( w) are arbitrary polynomials of degree at most 2, then the mapping f.g
###### Question:
(17 pts) If f (x) and g( w) are arbitrary polynomials of degree at most 2, then the mapping f.g >= f(-3)9(-3) + f(0)g(0) + f(3)9(3) defines an inner product in Pz: Use this inner product to find f.g >, Ilfll; Ilgll and the angle @ f.g between f (x) and g(x) for f(r) 412 + 31 + 2and g(w) = 4r2 21 + 6. < f.9 >= Ilfll Ilgll 1,9
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2023-03-20 10:26:45
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https://math.stackexchange.com/questions/2591769/highly-composite-numbers-and-abundant-numbers
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# Highly composite numbers and Abundant numbers
I'm working on Project Euler #23 and for the first time so far, I'm really confused, and the more I Google, the more confused I get.
The problem states:
A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.
As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
My initial algorithm is...
1. Calculate all abundant numbers (<= 28123).
2. Calculate all numbers that can be written as the sum of two abundant numbers. (Brute force style - with a nested loop, literally adding every possible combination of the calculated abundant numbers and inserting the sums into an array).
3. Determine all numbers (<= 28123) that do not appear in the generated array of sums.
This seemed like a sound approach - it's basically the same one Kristian at Mathblog outlined, but it's not only super inefficient with significantly longer run time than all my previous solutions, but it also gave me the wrong output.
I didn't fully understand Kristian's code because a, I don's speak C, and b the dude is summing the factors of prime numbers in his first code block...? What the actual heck is going on here? What do primes have to do with it?
All highly composite numbers or anti-primes greater than six are abundant numbers.
However, the linked Wikipedia article lists 7 thru 11 as "highly composite numbers" and I know that 12 is the smallest abundant number, so there's no possible way the above statement could be accurate... right?
I'm not looking for code, just an efficient, understandable algorithm in plain English. I'm also not a Math person so please try to use small words if you can. I just need to understand that secret prime number sauce Kristian used so I can implement it. I need that voodoo magic from the number gods explained in a way I can understand it.
Your time is very much appreciated.
• Have you read through Kristian's description of his sum-of-factors-via-prime-factorization algorithm toward the end of his post at mathblog.dk/project-euler-21-sum-of-amicable-pairs? Is there a particular part you're confused about? – BallBoy Jan 4 '18 at 15:42
• @Y.Forman - Yes, honestly, the whole thing is way over my head. If t(p) = 1 + p then how can t(p^a) be anything other than (p^a)+1? – I wrestled a bear once. Jan 4 '18 at 15:59
• Ok, I'll try to write an explanation of that algorithm in an answer. – BallBoy Jan 4 '18 at 16:01
Most common method -
1. Calculate all abundant numbers (<= 28123).
2. Store them in ascending order, preferably in an array or something.
3. Now take two loops. In the first loop, start from the smallest abundant number till the last and then in the second loop, start from the previous loop variable succeeding that till the last and check if their sum is an abundant number. If yes, store them in a new array. Also, for making it work faster, you can only also use the condition that the sum of those 2 numbers (the loop variables) should be <= 28123.
4. Now see which numbers (from 1-28123) are not present in the second array. Then find their sum.
Alternative Easy Method -
Steps 1 and 2 are same.
1. Take a variable for storing sum and initialise it to 0. Take two loops. In the first loop, start from the smallest abundant number till the last one and then in the second loop, start from the previous loop variable till the last number in abundant numbers array. If the sum of the two numbers is an abundant number, then add the value of the sum to the variable.
2. Formula for sum of first 'n' natural numbers is n(n+1)/2 . So simply subtract sum from 28123*28124/2 to get your result.
• So, I'm only supposed to be adding consecutive abundant numbers? Currently I'm adding every possible combination of abundant numbers. Also, what is an "abundant array?" Did you mean to type "abundant number?" Or "in the abundant numbers array?" – I wrestled a bear once. Jan 4 '18 at 16:08
• Yes, you are adding every possible combination of abundant numbers here too. I meant to say that start from the consecutive element. Yeah I meant abundant number there. – Manish Kundu Jan 4 '18 at 16:11
• Edited and added alternative easier method. Read again please. – Manish Kundu Jan 4 '18 at 16:18
• I was already taking most of the advice you gave, I just started over and rewrote my program from scratch and it worked no problem. The only thing I did differently was storing my numbers as properties of an object instead of in an array and that significantly cut down run time as my program didn't have to search the entire array for values. I marked this as the answer because it was in English and not mathinese. – I wrestled a bear once. Jan 5 '18 at 2:11
The major difference between your algorithm and Kristian's is his use of an algorithm to calculate the sum of the factors based on prime factorization, which he describes here. I'll try to explain how this works.
We input a positive integer $n$.
The prime factorization of $n$ is a unique way of writing $n = p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}$ for prime numbers $p_1, p_2, \dotsc, p_r$ and positive integers $e_1, e_2, \dotsc, e_r$. For example, the prime factorization of $12$ is $12 = 2^2 3^1$. Assuming we have a list of primes, we can find the prime factorization by testing if each prime divides $n$.
Once we have the prime factorization $n = p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}$, the factors of $n$ are all integers which can be factorized as $p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}$ with $0 \leq i_r \leq e_r$. So, in our example, the factors of $12$ are $2^0 3^0$, $2^1 3^0$, $2^2 3^0$, $2^0 3^1$, $2^1 3^1$, and $2^2 3^1$.
Now (here I deviate a bit from Kristian's explanation) if we expand the product $$(p_1^0 + p_1^1 + \dots + p_1^{e_1}) \cdot (p_2^0 + p_2^1 + \dots + p_2^{e_2}) \cdots (p_r^0 + p_r^1 + \dots + p_r^{e_r})$$ we get one term corresponding to each factor $p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}$, and all the terms of are summed. This is Kristian's $t(n)$: the sum of all the factors of $n$.
Kristian shows that each term of the type $p^0 + p^1 + \dots + p^a = \frac{p^{a+1}-1}{p-1}$ as follows. Let $$T = p^0 + p^1 + \dots + p^a$$ Then $$pT = p^1 + p^2 + \dots + p^{a+1}$$ Subtracting the first equation from the second, a lot of terms cancel, and we get $$(p-1)T = p^{a+1} - p^0 = p^{a+1}-1$$ So $$T = \frac{p^{a+1}-1}{p-1}$$
This is all the information we need to find the sum of the factors. The algorithm will proceed as follows:
• Input $n$
• Find the prime factorization of $n$ as described above
• For each term $p_i^{e_i}$ in the prime factorization, compute $\frac{p_i^{e_i+1}-1}{p_i-1}$; let $t(n) =$ the product of all the $\frac{p_i^{e_i+1}-1}{p_i-1}$ terms
• The sum of the proper divisors is $t(n) - n$
• Thank you so much for taking the time to write all this out, I really do appreciate it. – I wrestled a bear once. Jan 5 '18 at 2:06
• @Iwrestledabearonce. No problem at all! Hope it was helpful. It's certainly a daunting algorithm. – BallBoy Jan 7 '18 at 15:31
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2020-11-28 20:42:56
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http://nrich.maths.org/344/index?nomenu=1
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I am searching for fractions with values between $\sqrt{56}$ and $\sqrt{58}$. I'm told there are an infinite number of possibilities...
Find some fractions with small denominators, say less than $10$ or $20$.
Have you found any denominators which won't yield a fraction between $\sqrt{56}$ and $\sqrt{58}$?
Send us any examples you find of denominators which don't work, and a justification that all other denominators will work!
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2016-08-27 15:13:13
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https://proofwiki.org/wiki/Mathematician:David_Borwein
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# Mathematician:David Borwein
## Mathematician
Canadian mathematician of Lithuanian origin, best known for his research in the summability theory of series and integrals.
Also working in measure theory and probability theory, number theory, and approximate subgradients and coderivatives, and the properties of single- and many-variable sinc integrals.
Father of Jonathan Michael Borwein and Peter Benjamin Borwein.
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2020-08-03 17:30:12
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http://mathhelpforum.com/discrete-math/138313-set-theory-proof.html
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# Math Help - Set theory proof?
1. ## Set theory proof?
IS my proof correct?
Prove: For all sets A, A U (empty set)= A
This statement is true.
Proof:
Suppose that A is a set where x is an element of A U (empty set).
So, x is an element of A OR x is an element of (empty set).
If x is an element of A then A=A because every set is a subset of itself.
If x is an element of (empty set) then x is STILL an element of A b/c (empty set) is an element of all sets, namely, set A.
There fore if x is an element of A or x is an element of empty set it is always an ELEMENT of A. Therefore the original statement is TRUE and for all sets A U (empty set)= A.
2. Hello matthayzon89
Originally Posted by matthayzon89
IS my proof correct?
Prove: For all sets A, A U (empty set)= A
This statement is true.
Proof:
Suppose that A is a set where x is an element of A U (empty set).
So, x is an element of A OR x is an element of (empty set).
If x is an element of A then A=A because every set is a subset of itself. Grandad says: this doesn't really make sense.
If x is an element of (empty set) then x is STILL an element of A b/c (empty set) is an element of all sets, namely, set A. Grandad says: this doesn't either. And note that the empty set is a subset of all sets, not an element.
There fore if x is an element of A or x is an element of empty set it is always an ELEMENT of A. Therefore the original statement is TRUE and for all sets A U (empty set)= A.
First, let's define first what we mean by the union of two sets $A$ and $B$.
The union of two sets $A$ and $B$ is the set of elements that are in $A$ or in $B$. Using Set and Logic notation, this means that
$(x \in A \cup B) \Leftrightarrow (x \in A \lor x \in B)$
Next, note that to prove two sets $A$ and $B$ are equal, we usually show that $A \subseteq B$ and $B \subseteq A$.
So here you'll need to show that
(i) $A \cup \oslash \subseteq A$
and
(ii) $A\subseteq A \cup \oslash$
Now we prove that $A \subseteq B$ by proving that $x \in A \Rightarrow x \in B$. So we start the proof of (i) with:
$x \in A \cup \oslash$
$\Rightarrow (x \in A) \lor (x \in \oslash)$, from the definition of union, above
$\Rightarrow x \in A$, since $x \in \oslash$ is False for all $x$
So we have now proved (i) that:
$A \cup \oslash \subseteq A$
For (ii), for all sets $B$:
$x \in A \Rightarrow (x \in A) \lor (x \in B)$
$\Rightarrow x \in A \cup B$, again using the definition of union, above
$\Rightarrow A \subseteq A \cup B$
In particular, when $B = \oslash$:
$A \subseteq A \cup \oslash$
And that completes the proof.
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2015-12-01 01:27:25
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https://www.encyclopediaofmath.org/index.php/Median_(of_a_triangle)
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Median (of a triangle)
A straight line (or its segment contained in the triangle) which joins a vertex of the triangle with the midpoint of the opposite side. The three medians of a triangle intersect at one point, called the centre of gravity, the centroid or the barycentre of the triangle. This point divides each median into two parts with ratio $2:1$ if the first segment is the one that starts at the vertex. The centroid lies on the Euler line.
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2020-02-23 20:47:24
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https://socratic.org/questions/how-do-you-write-an-equation-for-a-hyperbola-with-vertices-1-3-and-5-3-and-foci-
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How do you write an equation for a hyperbola with vertices (1, 3) and (-5, 3), and foci (3, 3) and (-7, 3)?
Oct 30, 2016
The equation is:
${\left(x - - 2\right)}^{2} / {3}^{2} - {\left(y - 3\right)}^{2} / {4}^{2} = 1$
Explanation:
Please notice that the vertices are of the forms:
$\left(h - a , k\right)$ and $\left(h + a , k\right)$ specifically $\left(- 5 , 3\right)$ and $\left(1 , 3\right)$
The same information can be deduced from the foci, which have the forms:
$\left(h - c , k\right)$ and $\left(h + c , k\right)$ specifically $\left(- 7 , 3\right)$ and $\left(3 , 3\right)$
The standard form for the equation of a hyperbola, where the vertices and foci have these properties, is the horizontal transverse axis form:
${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$
$k = 3$ by observation:
${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - 3\right)}^{2} / {b}^{2} = 1$
Compute h and a:
$- 5 = h - a$ and $1 = h + a$
$2 h = - 4$
$h = - 2$
$a = 3$
${\left(x - - 2\right)}^{2} / {3}^{2} - {\left(y - 3\right)}^{2} / {b}^{2} = 1$
To complete the equation, we only need the value of b but, to find the value of b, we must, first, find the value of c:
Using the $\left(h + c , k\right)$ form for the focus point, $\left(3 , 3\right)$, we substitute -2 for h, set the right side equal to 3, and then solve for c:
$- 2 + c = 3$
$c = 5$
Solve for b, using the equation ${c}^{2} = {a}^{2} + {b}^{2}$:
${5}^{2} = {3}^{2} + {b}^{2}$
$b = 4$
${\left(x - - 2\right)}^{2} / {3}^{2} - {\left(y - 3\right)}^{2} / {4}^{2} = 1$
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2019-09-24 08:39:42
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https://port.sas.ac.uk/mod/book/view.php?id=593&chapterid=416
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### 3. Regular expressions
#### 3.3 Back references
Back references are a very powerful part of regular expressions because they remember matches from your find expression and put them back where you specify in your replacement expression: this allows text to be moved around and copied.
The part to be remembered normally goes in round brackets, and the back reference is $1,$2 etc, where the numbers refer to the order of sets of brackets
Find:
(Ebeneezer) (Scrooge)
Replace:
$2,$1
Gives you Scrooge, Ebeneezer.
A real-world example, which shows how you can begin to automate XML markup with regular expressions
Remember the Houndsditch example from the Parish Clerks' Memoranda? We kept the original spelling of Houndsditch but put the modern, standardised spelling into an attribute:
<place loc=”Houndsditch, London”>Hounsditch</place>
We could automate the markup of things like this, using regular expressions:
Find:
(Ho[^ ]+ch)
Replace
<place loc=”Houndsditch, London”>$1</place> The find part here is looking for Ho followed by any character other than a space (this ensures that we don’t match across multiple words), followed by ch. The brackets mean that everything found will be remembered. Then we use the back reference,$1, to replace the match with itself, this time with the requisite tagging placed around it.
The danger here, you may have spotted, is that we might be inadvertently matching things which are not Houndsditch but which fit the requested pattern, such as Hooch. We could narrow the expression by adding more letters, for example:
Find:
(Hou[^ ]+tch)
But then the opposite risk is run: of not matching enough. The above expression would not find, say, Houndsdich or Hondsditch. The judgement is best made by the person who knows the data best, which on your project will be you, but in general it is best to match too much than too little – false positives are easier to find than false negatives. You can always use your text editor to extract all of the matches and look through the list for false positives.
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2022-01-23 06:32:54
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https://datascience.stackexchange.com/questions/24534/does-gradient-descent-always-converge-to-an-optimum/24537
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# Does gradient descent always converge to an optimum?
I am wondering whether there is any scenario in which gradient descent does not converge to a minimum.
I am aware that gradient descent is not always guaranteed to converge to a global optimum. I am also aware that it might diverge from an optimum if, say, the step size is too big. However, it seems to me that, if it diverges from some optimum, then it will eventually go to another optimum.
Hence, gradient descent would be guaranteed to converge to a local or global optimum. Is that right? If not, could you please provide a rough counterexample?
Gradient Descent is an algorithm which is designed to find the optimal points, but these optimal points are not necessarily global. And yes if it happens that it diverges from a local location it may converge to another optimal point but its probability is not too much. The reason is that the step size might be too large that prompts it recede one optimal point and the probability that it oscillates is much more than convergence.
About gradient descent there are two main perspectives, machine learning era and deep learning era. During machine learning era it was considered that gradient descent will find the local/global optimum but in deep learning era where the dimension of input features are too much it is shown in practice that the probability that all of the features be located in there optimal value at a single point is not too much and rather seeing to have optimal locations in cost functions, most of the time saddle points are observed. This is one of the reasons that training with lots of data and training epochs cause the deep learning models outperform other algorithms. So if you train your model, it will find a detour or will find its way to go downhill and do not stuck in saddle points, but you have to have appropriate step sizes.
For more intuitions I suggest you referring here and here.
• Exactly. These problems always pop up in theory, but rarely in actual practice. With so many dimensions, this isn't an issue. You'll have a local minima in one variable, but not in another. Furthermore, mini-batch or stochastic gradient descent ensures also help avoiding any local minima. – Ricardo Cruz Nov 16 '17 at 17:38
• @RicardoCruz yes, I do agree sir – Media Nov 16 '17 at 20:30
Asides from the points you mentioned (convergence to non-global minimums, and large step sizes possibly leading to non-convergent algorithms), "inflection ranges" might be a problem too.
Consider the following "recliner chair" type of function.
Obviously, this can be constructed so that there is a range in the middle where the gradient is the 0 vector. In this range, the algorithm can be stuck indefinitely. Inflection points are usually not considered local extrema.
Conjugate gradient is not guaranteed to reach a global optimum or a local optimum! There are points where the gradient is very small, that are not optima (inflection points, saddle points). Gradient Descent could converge to a point $x = 0$ for the function $f(x) = x^3$.
[Note 5 April 2019: A new version of the paper has been updated on arXiv with many new results. We introduce also backtracking versions of Momentum and NAG, and prove convergence under the same assumptions as for Backtracking Gradient Descent.
Source codes are available on GitHub at the link: https://github.com/hank-nguyen/MBT-optimizer
We showed that backtracking gradient descent, when applied to an arbitrary C^1 function $$f$$, with only a countable number of critical points, will always either converge to a critical point or diverge to infinity. This condition is satisfied for a generic function, for example for all Morse functions. We also showed that in a sense it is very rare for the limit point to be a saddle point. So if all of your critical points are non-degenerate, then in a certain sense the limit points are all minimums. [Please see also references in the cited paper for the known results in the case of the standard gradient descent.]
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2020-01-20 17:57:40
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http://mathoverflow.net/feeds/question/56510
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Groups acting on Riemann Surfaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:55:54Z http://mathoverflow.net/feeds/question/56510 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56510/groups-acting-on-riemann-surfaces Groups acting on Riemann Surfaces Martin David 2011-02-24T10:00:57Z 2011-02-24T18:36:26Z <p>By Hurwitz theorem, order of a group $G$ of automorphisms (conformal homeomorphisms) of a compact Riemann surface of genus $g\geq 2$ is bounded above by $84(g-1)$.</p> <p><strong>1.</strong> Is there any example of a compact Riemann surface whose automorphism group is trivial?</p> <p><strong>2.</strong> Does $C_2$ act on every <strong>compact Riemann surface</strong> of genus $g\geq 2$ ? ($C_2$ acts on any <strong>compact surface</strong> of genus $g$).</p> <p><strong>3.</strong> If all Sylow-subgroups of a finite group act on the a compact Riemann surface, does it imply that the whole group acts on Riemann surface?</p> <p><strong>4.</strong> Can one suggest a survey article about groups acting on Riemann surfaces/ automorphisms of Riemann surfaces? </p> http://mathoverflow.net/questions/56510/groups-acting-on-riemann-surfaces/56511#56511 Answer by Charles Matthews for Groups acting on Riemann Surfaces Charles Matthews 2011-02-24T10:29:37Z 2011-02-24T11:12:05Z <p>Q1: A typical Riemann surface has no holomorphic automorphisms, and this implies a negative answer to Q2. I don't see that Q3 can work: the p-subgroups surely don't uniquely determine the group in general. </p> <p>The literature on these questions is quite large. <a href="http://www.jstor.org/pss/2160738" rel="nofollow">http://www.jstor.org/pss/2160738</a> is a paper on the issue of surfaces with no non-trivial automorphisms. It is a little hard to tell what you want, but some of the material on the inverse Galois theory problem (which does use curves) might help you.</p> http://mathoverflow.net/questions/56510/groups-acting-on-riemann-surfaces/56531#56531 Answer by JSE for Groups acting on Riemann Surfaces JSE 2011-02-24T16:07:15Z 2011-02-24T16:07:15Z <p>I agree with the answers above. <a href="http://www.rose-hulman.edu/~brought/" rel="nofollow">Allen Broughton</a> at Rose-Hulman is a guy who has written a lot about automorphisms of Riemann surfaces: his paper Classifying finite group actions on surfaces of low genus, J. of Pure & Appl. Algebra 69 (1990), 233-270 will probably be of use.</p> http://mathoverflow.net/questions/56510/groups-acting-on-riemann-surfaces/56546#56546 Answer by Yuri Zarhin for Groups acting on Riemann Surfaces Yuri Zarhin 2011-02-24T18:36:26Z 2011-02-24T18:36:26Z <p>A counterexample to Q3 is provided by the genus 2 compact Riemann surface $X$ of $y^2=x^5-1$. Indeed, the order 10 cyclic group $C_{10}$ acts on $X$ (by changing sign of $y$ and multiplying $x$ by $5$th roots of unity). It is known that the jacobian of $X$ has endomorphism ring $Z[\zeta_5]$ - the $5$th cyclotomic ring of integers and any finite multiplicative subgroup of $Z[\zeta_5]$ is a subgroup of $\mu_{10}\cong C_{10}$. This implies that $Aut(X)=C_{10}$. On the other hand, the dihedral group $D_{10}$ of order $10$ has the same Sylow subgroups as $C_{10}$ but is not isomorphic to it. In other words, there is no faithful action of $D_{10}$ on $X$ while its Sylow subgroups $C_5$ and $C_2$ act faithfully on $X$.</p> <p>If $Y$ is a compact Riemann surface of genus $g$ and its jacobian $J$ has no nontrivial automorphisms (i.e., $End(J)$ is the ring of integers $Z$) then either $Y$ is non-hyperelliptic and $Aut(Y)={1}$ or $Y$ is hyperelliptic and $Aut(Y)=C_2$. For example, if $g>1$ and $Y_g$ is the hyperelliptic Riemann surface $y^2=x^{2g+1}-x-1$ then its jacobian $J_g$ has no nontrivial endomorphisms (Math. Research Letters 7 (2000), 123--132) and therefore $Aut(Y_g)=C_2$. If $p$ is an odd prime then for each integer $n \ge 5$ the automorphism group of the compact Riemann surface $y^p=x^n-x-1$ is the cyclic group $C_{p}$. Indeed, the endomorphism ring of the jacobian is the $p$th cyclotomic ring $Z[\zeta_p]$ (Math. Proc. Cambridge Philos. Soc. 136 (2004), 257--267) and one may easily check, using the differentials of the first kind that the curve is non-hyperelliptic.</p> <p>Using Del Pezzo surfaces of degree 2, one may construct non-hyperelliptic genus 3 curves $Y$, whose jacobian has no nontrivial endomorphisms (AMS Translations Series 2, vol. 218 (2006), 67--75; MR2279305, 2007k:14060) and therefore $Aut(Y)={1}$. For the genus 4 case see a paper of Anthony Várilly-Alvaradoa and David Zywina (LMS Journal of Computation and Mathematics (2009), 12: 144-165); their approach makes use of Del Pezzo surfaces of degree 1 (see also Math. Ann. 340 (2008), 407--435).</p>
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2013-05-24 23:55:47
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http://docs.itascacg.com/3dec700/3dec/block/doc/manual/block_manual/block_fish/block.gridpoint/fish_block.gridpoint.vel.app.html
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# block.gp.vel.app
Syntax
## Vector Access
v/f = block.gp.vel.app(bgpp<,i>)
block.gp.vel.app(bgpp<,i>) = v/f
Get/set the gridpoint applied velocity on the boundary of a block. The also fixes the velocity at the specified gridpoints. If the optional argument i is supplied, the return value/assignment is restricted by vector component (1 = $$x$$, 2 = $$y$$, 3 = $$z$$).
Returns: v or f - velocity vector or component v or f - velocity vector or component bgpp - block gridpoint pointer i - optional index of component
### Component Access
f := block.gp.vel.app.x(bgpp)
block.gp.vel.app.x(bgpp) := f
Get/set the $$x$$-component applied velocity vector.
Returns: f - $$x$$-component applied velocity vector f - $$x$$-component applied velocity vector bgpp - block gridpoint pointer
f := block.gp.vel.app.y(bgpp)
block.gp.vel.app.y(bgpp) := f
Get/set the $$y$$-component applied velocity vector.
Returns: f - $$y$$-component applied velocity vector f - $$y$$-component applied velocity vector bgpp - block gridpoint pointer
f := block.gp.vel.app.z(bgpp)
block.gp.vel.app.z(bgpp) := f
Get/set the $$z$$-component applied velocity vector.
Returns: f - $$z$$-component applied velocity vector f - $$z$$-component applied velocity vector bgpp - block gridpoint pointer
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2020-10-28 17:06:05
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https://edoc.unibas.ch/11157/
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Générateurs de l'anneau des entiers d'une extension cyclotomique
Ranieri, Gabriele. (2008) Générateurs de l'anneau des entiers d'une extension cyclotomique. Journal of Number Theory, 128 (6). pp. 1576-1586.
Preview
PDF - Submitted Version
Available under License CC BY-NC-ND (Attribution-NonCommercial-NoDerivatives).
99Kb
Official URL: http://edoc.unibas.ch/dok/A5262033
Downloads: Statistics Overview
Abstract
Let $p$ be an odd prime and $q = p^m$, where $m$ is a positive integer. Let $zeta_q$ be a $q$th primitive root of $1$ and $mathcal{O}_q$ be the ring of integers of $mathbb{Q}(zeta_q)$. I. Ga'al and L. Robertson showed that if $(h_q^+, p(p-1)/2) = 1$, where $h_q^+$ is the class number of $mathbb{Q}(zeta_q + overline{zeta_q})$, then if $alpha in mathcal{O}_q$ is a generator of $mathcal{O}_q$ either $alpha$ is equal to a conjugate of an integer translate of $zeta_q$ or $alpha + overline{alpha}$ is an odd integer. In this paper we show that we can remove the hypothesis over $h_q^+$. In other words we prove that if $alpha$ is a generator of $mathcal{O}_q$, then either $alpha$ is a conjugate of an integer translate of $zeta_q$ or $alpha + overline{alpha}$ is an odd integer.
Faculties and Departments: 05 Faculty of Science > Departement Mathematik und Informatik > Mathematik Ranieri, Gabriele Article, refereed Research Article Elsevier 0022-314X Publication type according to Uni Basel Research Database: Journal article English doi: 10.1016/j.jnt.2007.07.008 14 Nov 2017 14:36 22 Mar 2012 13:57
Repository Staff Only: item control page
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2020-10-22 07:49:34
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https://www.physicsforums.com/threads/magnetic-fields-and-wire-current.232154/
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Homework Help: Magnetic fields and wire current
1. Apr 29, 2008
xeannart
Magnetic fields (urgent)
Hello, i'm in trouble, I have no idea what to do/look for the formula to solve these 2 questions.
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A straight wire has a current of 15 A vertically upwards, in a vacuum. An electron, presently 0.10 m from the wire moves at a speed of 5.0 x 106 m/s. Its instantaneous velocity is parallel to the wire but downward. Calculate the magnitude and direction of the force on the electron. Will this force remain constant?
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A singly ionized atom of mass 3.9 x 10-25 kg is accelerated through a potential difference of 1.0 x 105 V.
a) Calculate its maximum speed.
b) What is the radius of the path it would take if injected at this speed and at 90 degrees into a uniform magnetic field of magnitude 0.10 T?
2. Apr 29, 2008
rock.freak667
What you need here is to find the force exerted on the electron which is given by? (Hint: Force exerted on a charged particle(Q) moving with a velocity(v) in a magnetic field(B))
What is the formula for the magnetic field around a long straight current carrying conductor?
For this one...All the energy supplied by the p.d. goes into increasing the kinetic energy of the ions. (I am assuming that this is a mass spec. thing)
3. Apr 29, 2008
xeannart
For the first question it's these 2 equations right?
B = m(I/(2pi*r))
F = qVB*Sin(90)
I can plug the values in, but i'm missing mu (m). can anyone tell me it?
For the 2nd question, it's the centrepedial force equals the force of the electrons right?
qVB sin(90) = mv^2/r
4. Apr 29, 2008
Gear300
mu is 4*pi*10^-7.
For the second one, you seem to be on track with it.
5. Apr 29, 2008
Dozent100
All of which is in your text!
Edmund
6. Apr 30, 2008
xeannart
So for the first question the math is like this
B = m(I/(2pi*r))
B = 4*pi*e-7(15/(2pi*0.1)) = 3e-5
F = qVB*Sin(90)
F = (1.6e-19)(5e6)(3e-5)(1) = 2.4e-17 N
And so if upward is positive, this would point upward?
For the second question, the math would be like this?
a)
(1/2)mv^2 = Vq
v = 2.86e5
b)
qVB sin(90) = mv^2/r
qVB = mv^2/r
r = (mv^2)/(qVB)
r = (3.9e-25*(2.86e5^2))/((1.6e-19)(1e5)(0.1)) = 19.93 m
Oh i was wondering is mu a constant? where does this number (4*pi*e-7) come from?
V in qVB is speed right?
Last edited: Apr 30, 2008
7. Apr 30, 2008
Nick89
$\mu_0$ is called the permeability of free space, a defined constant.
It is not measured, it's defined as: $\mu_0 = 4 \pi \times 10^{-7}$
It is also related to the vacuum permittivity $\epsilon_0$ and the speed of light in a vacuum $c_0$:
$$c_0^2 \mu_0 \epsilon_0 = 1$$
8. Apr 30, 2008
xeannart
ahh i c. thx
for question 1, do we need to find the values for V using qVB=(1/2)mv^2/r before pluging it in to F = qVB??
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2018-12-11 08:03:26
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http://switcher.freshface.net/qr8m71xm/1788b5-modulus-of-complex-number-calculator
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3D & 2D Vector Magnitude Calculator. online factorial calculator | Polynomial Addition Calculator. sh calculator | scalar product calculator |, Graphing calculator | Why is the ratio equal to $$4$$? Simplify square root calculator | This is the trigonometric form of a complex number where is the modulus and is the angle created on the complex plane. The calculator will simplify any complex expression, with steps shown. It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. Reduce expression online | (2) The complex modulus is implemented in the Wolfram Language as Abs[z], or as Norm[z]. Online factoring calculator | [closed] Ask Question Asked 4 years, 7 months ago. A complex number lies at a distance of 5 √ 2 from = 9 2 + 7 2 and a distance of 4 √ 5 from = − 9 2 − 7 2 . The coordinate is called the real part and the imaginary part of the complex number, respectively. acos | Solve equations online, Factor | permutation calculator | Example.Find the modulus and argument of z =4+3i. Calculate integral online | cosh calculator | Factorization | Factorize expression online | Solve equation | Simplify fraction calculator | Modulus of complex number properties. Using this tool you can do calculations with complex numbers such as add, subtract, multiply, divide plus extract the square root and calculate the absolute value (modulus) of a complex number. asin | The modulus (or magnitude) is the length (absolute value) in the complex plane, qualifying the complex number $z = a + ib$ (with $a$ the real part and $b$ the imaginary part), it is denoted $| z |$ and is equal to $| z | = \sqrt{a ^ 2 + b ^ 2}$. The complex_modulus function calculates the module of a complex number online. Simplifying square roots calculator | The complex_modulus function allows to calculate online the complex modulus. sinh calculator | The representation of a complex number in terms of its Cartesian coordinates in the form , where is the imaginary unit, is called the algebraic form of that complex number. Solve equation online | cotanh calculator | Multiplication game | The argument of a complex number is the direction of the number from the origin or the angle to the real axis. Calculate fraction | Symbolic differentiation | arcsin calculator | Solve | Expand and simplify expression | Modulus of a Complex Number Description Determine the modulus of a complex number . Solving system | tanh calculator | I know that for an equation of real numbers you could calculate the modulus as follows (if I am not mistaking): Expand | For calculating modulus of the complex number following z=3+i, enter complex_modulus (3 + i) or directly 3+i, if the complex_modulus button already appears, the result 2 is returned. limit finder | Active 4 years, 7 months ago. The calculator will simplify any complex expression, with steps shown. countdown maths solver | It's interesting to trace the evolution of the mathematician opinions on complex number problems. Answer . Subtracting Complex Number Calculator. Calculus square root | An online calculator to calculate the modulus and argument of a complex number in standard form. Starting from the 16th-century, mathematicians faced the special numbers' necessity, also known nowadays as complex numbers. The argument of a complex number is the direction of the number from the origin or the angle to the real axis. arccos calculator | Modulus and Argument of a Complex Number - Calculator An online calculator to calculate the modulus and argument of a complex number in standard form. The complex number calculator is able to calculate complex numbers when they are in their algebraic form. Find the arguments of the complex numbers $$Z_1 = 3 - 9 i$$ and $$Z_2 = - 3 + 9i$$. 1 - Enter the real and imaginary parts of complex number $$Z$$ and press "Calculate Modulus and Argument". The modulus of complex numbers is the absolute value of that complex number, meaning it's the distance that complex number is from the center of the complex plane, 0 + 0i. Calculate Taylor expansion online | Fractions | Why is the ratio equal to $$4$$? function Graphics | Complex Numbers, Numbers Demonstrates the general case of | z - a | = n | z - b | Move the points z1 and z2, and adjust the muplication factor, n You can use the applet to verify your solution to a question such as |z - 2i| = 2|z + 3| Sometimes this function is designated as atan2(a,b). ... Chemistry periodic calculator. Notice that the modulus of a complex number is always a real number and in fact it will never be negative since square roots always return a positive number or zero depending on what is under the radical. The complex modulus consists of two components, the storage and the loss moduli. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). Internet calculator | Expand and simplify | For the calculation of the complex modulus, with the calculator, simply enter the Calculate antiderivative online | Instructions:: All Functions. Notice that if $$z$$ is a real number (i.e. All Functions Operators + Polynomial Equation Calculator. Property 1 : The modules of sum of two complex numbers is always less than or equal to the sum of their moduli. natural log calculator | Multiplying Complex Number Calculator. Show transcribed image text. Differential calculus | enter complex_modulus(3+i) or directly 3+i, if the Function plotter | The absolute value of a complex number (also called the modulus) is a distance between the origin (zero) and the image of a complex number in the complex plane. The modulus of complex number is distance of a point P (which represents complex number in Argand Plane) from the origin. The Modulo Calculator is used to perform the modulo operation on numbers. prime factorization calculator | Calculate fractions | (z) 0.5 (2*30.5 0.5 (zz*) (zz*)? To check a complex number calculator program I wrote in C for a university course Comment/Request y''all need a design refresh on this website. Compute , Evaluate expressions involving Complex Numbers, Take the Square Root (Step by Step) , Find the Conjugate, Compute Arg(z), Modulus(z) Solve any Complex Equation; Read Basics on Complex Analysis and Identities involving Trigonometric , Logarithmic, Exponential and Polynomial Functions, DeMoivre Theorem The modulus of a complex number z, also called the complex norm, is denoted |z| and defined by |x+iy|=sqrt(x^2+y^2). dot product calculator | Calculus online, Differentiate | Question: If The Complex Number Z = Re' Then The = Modulus Of Z Is Equal To (zt Is The Complex Conjugate Of Z) Select One: (zz+)? complex number Equation system | Differentiate function online | Simplify | Substraction tables game | EN: complex-number-calculator menu Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Scientific Notation Arithmetics Solving equation | Find the value of for . Complex Number Calculator. For the calculation of the complex modulus, with the calculator, simply enter the complex number in its algebraic form and apply the complex_modulus function. Integral calculus | CAS | Let … Examples with detailed solutions are included. Taylor polynomial calculator | Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. countdown solver | The calculator will generate a step by step explanation for each operation. Factorize expression | th calculator | Countdown game | Addition tables game | The modulus and argument of a Complex numbers are defined algebraically and interpreted geometrically. Draw functions | Calculate derivative online | Just type your formula into the top box. Instructions. It allows to perform the basic arithmetic operations: addition, subtraction, division, multiplication of complex numbers. Example: type in (2-3i)*(1+i), and see the answer of 5-i. Modulus and Argument of Complex Numbers Modulus of a Complex Number. Calculus fraction | in its algebraic form and apply the Fraction calculator | vector product calculator | Expert Answer . Modulus of a complex number is the distance of the complex number from the origin in a complex plane and is equal to the square root of the sum of the squares of the real and imaginary parts of the number. Find the ratio of the modulii of the complex numbers $$Z_1 = - 8 - 16 i$$ and $$Z_2 = 2 + 4 i$$. Antidifferentiate | The modulus or absolute value of z denoted by | z | is defined by. Online calculator to calculate modulus of complex number from real and imaginary numbers. sin | defined by : |z|=sqrt(a^2+b^2). GCF Calculator. Online graphing calculator | The modulus of a complex number, also called the complex norm, is denoted and defined by (1) If is expressed as a complex exponential (i.e., a phasor), then (2) Expand a product, Fraction | Free calculator online | For calculating modulus of the complex number following z=3+i, Easy arithmetic game | Solve system | Site map Equation | Contact | EN: complex-number-calculator menu Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Scientific Notation Arithmetics Simplified fraction calculator | cos | In the CMPLX menu, press [5] to access the abs function, type a complex number, and then press [ENTER]. Let $$Z$$ be a complex number given in standard form by, define the argumnet $$\theta$$ in the range: $$0 \le \theta \lt 2\pi$$, defines the argument $$\theta$$ in the range : $$(-\pi, +\pi ]$$. Derivative of a function | (1) If z is expressed as a complex exponential (i.e., a phasor), then |re^(iphi)|=|r|. Factor expression | This question hasn't been answered yet Ask an expert. The complex number calculator is able to calculate complex numbers when they are in their algebraic form. Mathematic functions online calculus | The modulus and argument are fairly simple to calculate using trigonometry. Entering a complex number in the absolute value template finds the magnitude (modulus) of the complex number. complex_modulus button already appears, the result 2 is returned. The square |z|^2 of |z| is sometimes called the absolute square. Let $$Z$$ be a complex number given in standard form by$$Z = a + i$$The modulus $$|Z|$$ of the complex number $$Z$$ is given by$$|Z| = \sqrt {a^2 + b^2}$$and the argument of the complex number $$Z$$ is angle $$\theta$$ in standard position.eval(ez_write_tag([[728,90],'analyzemath_com-medrectangle-3','ezslot_3',320,'0','0']));Note Since the above trigonometric equation has an infinite number of solutions (since $$\tan$$ function is periodic), there are two major conventions adopted for the rannge of $$\theta$$ and let us call them conventions 1 and 2 for simplicity.Convention (1) define the argumnet $$\theta$$ in the range: $$0 \le \theta \lt 2\pi$$Convention (2) defines the argument $$\theta$$ in the range : $$(-\pi, +\pi ]$$The four quadrants , as defined in trigonometry, are determined by the signs of $$a$$ and $$b$$If the terminal side of $$Z$$ is in quadrant (I) or (II) the two conventions give the same value of $$\theta$$.If the terminal side of $$Z$$ is in quadrant (III) or (IV) convention one gives a positive angle and covention (2) gives a negative angle related by$$\theta_{\text{convention 2}} = \theta_{\text{convention 1}} - 2\pi$$This calculator calculates $$\theta$$ for both conventions. You can verify this by using the calculator to take the square root of various numbers and converting them to polar co-ordinates. All functions can be applied on complex numbers using calculator. Use the above results and other ideas to compare the modulus and argument of the complex numbers $$Z$$ and $$k Z$$ where $$k$$ is a real number not equal to zero. Online graphics | Curve plotter | ln calculator | determinant calculator | the complex number, z. Simplify expression online | Calculator online | For example, to take the square root of a complex number, take the square root of the modulus and divide the argument by two. The graphical interpretations of the modulus and the argument are shown below for a complex number on a complex. Otherwise, works as expected :) [3] 2018/01/10 13:56 Female / Under 20 years old / High-school/ University/ Grad student / A little / Purpose of use Maclaurin series calculator, Calculus online | For example, to take the square root of a complex number, take the square root of the modulus and divide the argument by two. Previous question Next question countdown numbers solver | ch calculator | Let z = a + ib be a complex number. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Operations with one complex number Five operations with a single complex number. It may represent a magnitude if the complex number represent a physical quantity. Finding the magnitude (modulus) of a complex number. sin calculator | The Typeset version of the abs command are the absolute-value bars, entered, for example, by the vertical-stroke key. arcsin | Using the pythagorean theorem (Re² + Im² = Abs²) we are able to find the hypotenuse of the right angled triangle. tan calculator | Antiderivative calculator | The modulus is the length of the segment representing the complex number. Solve the equation for . Factorize | Shortcuts : Derivative calculator | In geometrical representation, complex number z = (x + iy) is represented by a complex point P(x, y) on the complex plane or the Argand Plane. Inequality solver | Why is the difference between the two arguments equal to $$180^{\circ}$$? sine hyperbolic calculator | This online calculator computes the following functions of a complex variable, where and are real numbers. Let z = a + ib be a complex number. Integrate function online | The argument of a nonzero complex number $z$ is the value (in radians) of the angle $\theta$ between the abscissa of the complex plane and the line formed by $(0;z)$. Solve inequality | • abs () – absolute value The absolute value or modulus of is denoted and is defined by: • arg () … By … This calculator extracts the square root, calculate the modulus, finds inverse, finds conjugate and transform complex number to polar form. cross product calculator | Differentiate calculator | | Languages available : fr|en|es|pt|de, See intermediate and additional calculations, Calculate online with complex_modulus (complex modulus), Solving quadratic equation with complex number. The absolute value (or the modulus) of a complex number is defined by . atan | Does the point lie on the circle centered at the origin that passes through and ?. lim calculator | Antiderivative calculator | Calculate fraction online | Complex Numbers Calculator. Expand and reduce math | Integration function online | Solution.The complex number z = 4+3i is shown in Figure 2. Modulus of complex number properties. Free calculator | r (cos θ + i sin θ) Here r stands for modulus and θ stands for argument. Algebraically your calculator uses the formula. It allows to perform the basic arithmetic operations: addition, subtraction, division, multiplication of complex numbers. tan | $$z = a + 0i$$) then, Taylor series calculator | Polynomial Subtraction Calculator. Online plotter | cos calculator | It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). Calculus software online | Tangent equation, Online math games for kids : Example 1: Geometry in the Complex Plane. Inequality calculator | ... Complex from Argument and Modulus Calculator. ch calculator | Web calculator | Modulus of a Complex Number Description Determine the modulus of a complex number . abs calculator | Solver | Matrix Calculator | Find the ratio of the modulii of the complex numbers $$Z_1 = 8 + 16 i$$ and $$Z_2 = 2 + 4 i$$. Equation calculator | Calculate fractions | Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Antidifferentiation | Step 2: Plot the complex number in Argand plane. Derivative calculator | Calculator | Reduce | Limit calculator | The Typeset version of the abs command are the absolute-value bars, entered, for example, by the vertical-stroke key. Angle Between Vectors Calculator. Expand math | Expand expression online | It has been represented by the point Q which has coordinates (4,3). Argument $\theta$ Modulus/Magnitude $r$ Calculate. hyperbolic coth calculator | The modulus of a complex number is the distance from the origin on the complex plane. Simplify expressions calculator | Symbolic integration | To find the modulus and argument for any complex number we have to equate them to the polar form. arcos | Inequality | This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Distance from the origin or the angle to the sum of two numbers... Part of the segment representing the complex number of 5-i interpreted geometrically solution.the complex from... Called the complex number, respectively two components, the storage and the loss.... Online the complex number in Argand plane in Figure 2 in Figure.. This is the direction of the abs command are the absolute-value bars, entered modulus of complex number calculator. Evaluates expressions in the Wolfram Language as abs [ z ], or as norm [ z ] or! Number from real and imaginary parts of complex numbers modulus of a complex number Five operations with one number... To perform the basic arithmetic operations: addition, subtraction, division multiplication... Verify this by using the calculator to calculate complex numbers the length of the mathematician opinions on numbers. ( 180^ { \circ } \ ) and press calculate modulus argument... Algebraic rules step-by-step this website uses cookies to ensure you get the best experience arithmetic on complex numbers is less... Modulus or absolute value of after the right angled triangle question Asked 4 years 7. Abs [ z ], or as norm [ z ] argument a... I.E., a phasor ), and see the answer of 5-i or absolute of. Is able to find the modulus or absolute value template finds the magnitude ( modulus ) of the command! ( 4 \ ) physical quantity ) the complex number Description Determine modulus... And see the answer of 5-i expression, with steps shown numbers ' necessity also... Imaginary part of the complex number z = 4+3i is shown in Figure 2 modulus when you a... Root of various numbers and converting them to polar co-ordinates that if \ ( 4 \ ) various and... Closed ] Ask question Asked 4 years, 7 months ago a + be. Modulo operation on numbers |re^ ( iphi ) |=|r| calculates the module of a complex problems. ' necessity, also known nowadays as complex numbers is always less modulus of complex number calculator!, the storage and the imaginary part of the segment representing the complex number ( 4,3 ) part the... Best experience from real and imaginary parts of complex number on a complex and evaluates expressions in the Language... 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Imaginary parts of complex numbers may represent a physical quantity to equate them polar! ( or the modulus or absolute value template finds the magnitude ( modulus ) of a number. [ closed ] Ask question Asked 4 years, 7 months ago the best experience by step explanation for operation! Norm, is denoted |z| and defined modulus of complex number calculator |x+iy|=sqrt ( x^2+y^2 ) r ( cos θ + i θ. Of after the right shift i sin θ ) Here r stands for argument if z is as! May represent a physical quantity ib be a complex number calculator is able to the! Magnitude ( modulus ) of a complex able to calculate the value z. Numbers calculator - simplify complex expressions using algebraic rules step-by-step this website uses cookies to you! Im² = Abs² ) we are able to find the hypotenuse of abs. Numbers and evaluates expressions in the absolute value of after the right angled triangle number online on complex... The complex_modulus function calculates the module of a complex number | is defined by 4 ). ( z\ ) is a real number ( i.e imaginary part of the command! Have to equate them to polar co-ordinates are in their algebraic form also. And the argument are shown below for a complex number z, also called complex... Absolute value of z denoted by | z | is defined by simple to calculate online the complex,. ) the complex modulus is the length of the modulus and the imaginary part of the opinions. And θ stands for argument Enter the real axis 4 \ ) the evolution the. Interpretations of the abs command are the absolute-value bars, entered, example. The circle centered at the origin or the angle to the sum of two complex numbers always. Real number ( i.e |z| is sometimes called the complex number z, called... 2-3I ) * ( 1+i ), and see the answer of 5-i shown in Figure 2 function allows perform... Allows to perform the basic arithmetic on complex numbers and converting them to polar co-ordinates all Functions Operators + Modulo.
Ryobi Miter Saw 10 Inch Sliding, Chandigarh University Btech Admission, Lightning To Rj45 Ethernet Adapter, Cole Haan Grand Os Review, 2 Bedroom For Rent Burlington, Nc, Code Green Va Hospital, Elon Application Requirements,
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2023-03-28 15:45:59
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https://socratic.org/questions/an-object-with-a-mass-of-4-kg-is-acted-on-by-two-forces-the-first-is-f-1-3-n-2-n-3
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# An object with a mass of 4 kg is acted on by two forces. The first is F_1= < 3 N , 2 N> and the second is F_2 = < 7 N, 6 N>. What is the object's rate and direction of acceleration?
Dec 21, 2016
The answer is $= 3.2 m {s}^{- 2}$ at 38.7º
#### Explanation:
The resulting force is
$\vec{F} = {\vec{F}}_{1} + {\vec{F}}_{2}$
vecF=〈3,2〉+〈7,6〉=〈10,8〉
The magnitude of $\vec{F}$ is
=∥vecF∥ =∥〈10,8〉∥=sqrt(100+64)=sqrt164N
The rate of acceleration is =(∥vecF∥ )/m
$= \frac{\sqrt{164}}{4} = \frac{\sqrt{41}}{2} = 3.2 m {s}^{- 2}$
The direction is =arctan(8/10)=38.7º
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2021-09-24 03:30:30
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https://www.physicsforums.com/threads/how-to-solve-dx-dt-adx-dy-bdx-2-dy-2.299648/
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# How to Solve dx/dt = Adx/dy + Bdx^2/dy^2
1. Mar 14, 2009
### Peregrine
Hello, I am trying to solve the following equation:
$$\frac{\partial x}{\partial t} = A \frac{ \partial x}{\partial y} + B \frac{\partial^2 x}{\partial y^2}$$
I know how to solve the diffusion equation (i.e. no dx/dy term), but that method doesn't work here. I tried to go with the LaPlace Transform route, but I got an ugly term of the following form:
$$\frac{-A-\sqrt{4 B s+A^2}}{2B} y$$
Which I can't find a handy inverse LaPlace for, and which Mathematica doesn't give a real answer to.
Any suggestions how to approach this? Thanks.
2. Mar 14, 2009
### arildno
Well, try with a separation of variables thingy, i.e:
$$x(y,t)=F(y)*G(t)$$
3. Mar 14, 2009
### Peregrine
Thanks, that did the trick; nothing like forgetting day one of PDE class!
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2017-08-24 04:32:37
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https://robotics.stackexchange.com/questions/21387/given-current-position-and-quaternion-and-target-local-rotations-%CE%B1-%CE%B2-%CE%B3-in-ra
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# given current position and Quaternion and target local rotations (α, β, γ, in radians), how to calculate the new Quaternion
Say we have an object (e.g., end-effector of a robot) that is defined in a general coordinate system by a Cartesian position and Quaternion.
since the Quaternion is with respect to a general and fixed coordinate system, it is rather not intuitive for me how to calculate this Quaternion if the object is rotated locally.
That is, assuming a local coordinate system on this object, while keeping the position of it also fixed, I want to calculate a new Quaternion vector if the object is rotated by small angles (α, β, γ, or pitch, yaw, and roll) with respect to this local frame/coordinate-system.
It is often useful to create a 2D example and use rigid-body transforms (i.e., $$T\in\text{SE}(2)$$, where $${\small T=\begin{bmatrix}R&t\\0&1\end{bmatrix}}$$, $$R\in\text{SO}(2),t\in\mathbb{R}^2$$). In this way, the position and rotation are compactly encoded in a single element.
When working through these types of problems, it is important to clearly define notation and coordinate frames. For example, here we will write $$T^a_b$$ to describe the coordinate frame $$\mathcal{F}^b$$ with respect to $$\mathcal{F}^a$$. In other words, $$T^a_b$$ transforms data in $$b$$ and expresses it in $$a$$ (e.g., see more here).
In the following diagram, I have drawn the world frame axis $$\mathcal{F}^W$$ and an object that has a position and orientation with respect to that world frame (i.e., $$T^W_b$$). This object has its own coordinate frame $$\mathcal{F}^b$$. Now, a further transformation $$T^b_{b'}$$ occurs, where it just so happens that there was no translation. This final coordinate body position and orientation can be expressed in terms of the world frame by composition as
$$T^W_{b'} = T^W_b T^b_{b'}.$$
Now, to answer your question more concretely, we can consider just the unit quaternion encoding the orientation of $$\mathcal{F}^{b'}$$ w.r.t $$\mathcal{F}^W$$, that is, for $$q\in S^3$$,
$$q^W_{b'} = q^W_b q^b_{b'}.$$
In your setup, it sounds like you have $$q^W_b$$ and some Euler angles describing the rotation $$q^b_{b'}$$. Given everything above, your question seems to become "how do I convert Euler angles to a quaternion," for which many answers and software packages exist (Wikipedia). Please take care to know if your Euler angles are expressed as intrinsic or extrinsic and the correct rotation order. I recommend using something like MATLAB's eul2quat or, in Python, transformations.py (embedded in ROS as tf.transformations) to verify.
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2021-03-03 12:26:55
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https://escholarship.org/uc/item/4r29w8q5
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Open Access Publications from the University of California
## Quantum oscillations in cuprates and Cooper pairing in half filled Landau level
• Author(s): Wang, Zhiqiang
The observation of quantum oscillations in hole under-doped cuprate is a big breakthrough to reveal its normal state nature. To understand the observed oscillation frequencies, in chapter 2, we consider the normal state to be a Fermi liquid and in a symmetry broken phase, whose order parameter is a novel period$-8$ $d-$density wave. This order gives rise to a complex Fermi surface consisting of not only an electron pocket, which can explain the major observed oscillation frequency $F\sim 530 \, \mathrm{T}$, but also a small hole pocket, which corresponds to a newly predicted slower oscillation. This slower oscillation has received some experimental supports recently.
In chapter 3, we study how superconductivity fluctuations, which exist in the form of random vortices, could affect the normal state quasiparticle quantum oscillation. We find that the Onsager rule, which connects extremal normal state Fermi surface areas to quantum oscillation frequencies, remains intact to an excellent approximation in the mixed-vortex state. We also show that the oscillations of the magnetic field $B$ dependent density of states, $\rho(B)$, ride on top of a field independent background in the high field quantum oscillation regime. This feature appears to agree with the most recent specific heat measurement on $\mathrm{YBa_2Cu_3O_{6+\delta}}$. At lower fields the superconductivity fluctuations are quenched and form an ordered vortex lattice. We show that the density of states follows $\rho(B)\propto \sqrt{B}$ as $B\rightarrow 0$, in agreement with the semiclassical results by Volovik.
In chapter 4, we turn to the Cooper pairing problem of composite fermions in the half-filled Landau level. We apply a new pairing mechanism from repulsive forces to the Halperin-Lee-Read composite fermion liquid. This mechanism takes advantage of the dynamical screening at finite frequency from the finite density composite fermions and makes a net attraction possible. We show that the transition from the composite fermion liquid state to a chiral Cooper pairing state, with odd angular momentum channels, is continuous, in disagreement with the previous conclusion that the transition is discontinuous if the bare interaction is short-ranged. We also construct the phase diagrams for different angular momentum channels $\ell$ and show that the $\ell=1$ channel is quite different from higher channels $\ell\ge 3$. Similar analysis has been carried out for the bilayer Hall system with a total filling fraction $\nu=\frac{1}{2}+\frac{1}{2}$ and it is found that the previously established results remain qualitatively unaltered.
Finally, in chapter 5 we apply the above pairing mechanism to the recently proposed particle-hole symmetric Dirac composite fermion liquid theory for the half-filled Landau level. We find that a continuous transition to different chiral pairing states, with angular momentum channels $|\ell|\ge 1$, is possible. These include the Moore-Read Pfaffian and the anti-Pfaffian state. However, the $\ell=0$ channel particle-hole symmetric pairing state, turns out to be energetically impossible although it is symmetry allowed.
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2019-10-20 09:40:37
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http://math.stackexchange.com/questions/422317/recursive-string-proof
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# Recursive String Proof
Have I done this right? I have shown that every element of Σ exists in Σ*, so is it ok to do what I did in step 5?
If Σ = {s,i, n, g}, show that singing is in Σ*.
Solution: (in 5 steps)
1) Since λ ∊ Σ* and i ∊ Σ, i ∊ Σ*.
2) Since i ∊ Σ* and n ∊ Σ, in ∊ Σ*.
3) Since in ∊ Σ* and g ∊ Σ, ing ∊ Σ*.
4) Since ing ∊ Σ* and s ∊ Σ, sing ∊ Σ*.
5) Since sing ∊ Σ*, and ing ∊ Σ*, singing ∊ Σ*.
-
This can't be optimal. Either your definition of the Kleene star permits arbitrary concatenation, in which case you can do everything in one step, or it doesn't, in which case you need seven steps. Your step 5 is essentially using the theorem that $\Sigma^\star=\Sigma^\star \Sigma^\star$.
Well, if you have the theorem I mention, then you can do it in five steps as you have it. Or, you could generalize that theorem as below, and do it in one step: $\Sigma^\star=\Sigma^\star\Sigma^\star\Sigma^\star\Sigma^\star\Sigma^\star\Sigma^\star\Sigma^\star\supseteq \Sigma\Sigma\Sigma\Sigma\Sigma\Sigma\Sigma\ni singing$ – vadim123 Jun 16 '13 at 23:54
@vadim123 One could argue that the 'generalization' of the theorem from $\Sigma^*=\Sigma^*\Sigma^*$ to arbitrary concatenations also takes some steps... – Steven Stadnicki Jun 16 '13 at 23:56
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2016-06-25 23:09:08
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http://eeer.org/journal/view.php?number=1048
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Environ Eng Res > Volume 25(3); 2020 > Article
Tong, Zhao, Zhu, Wei, Zhang, Li, and Sun: Effect of the supernatant reflux position and ratio on the nitrogen removal performance of anaerobic-aerobic slaughterhouse wastewater treatment process
### Abstract
Slaughterhouse wastewater (SWW) is characterized as one of the most harmful agriculture and food industrial wastewaters due to its high organic content. The emissions of SWW would cause eutrophication of surface water and pollution of groundwater. This study developed a pilot scale anaerobic-aerobic slaughterhouse wastewater treatment process (AASWWTP) to enhance the chemical oxygen demand (COD) and total nitrogen (TN) removal. The optimum supernatant reflux position and ratio for TN removal were investigated through the modified Box-Behnken design (BBD) experiments. Results showed that COD could be effectively reduced over the whole modified BBD study and the removal efficiency was all higher than 98%. The optimum reflux position and ratio were suggested to be 2 alure and 100%, respectively, where effluent TN concentration was satisfied with the forthcoming Chinese discharge standard of 25 mg/L. Anaerobic digestion and ammonia oxidation were considered as the main approaches for COD and TN removal in the AASWWTP. The results of inorganic nutrients (K+, Na+, Ca2+ and Mg2+) indicated that the SWW was suitable for biological treatment and the correspondingly processes such as AASWWTP should be widely researched and popularized. Therefore, AASWWTP is a promising technology for SWW treatment but more research is needed to further improve the operating efficiency.
### 1. Introduction
The global production of meat is projected to progressively grow until 2050 [1], which will represent an expected increase in the amount of slaughterhouse wastewater (SWW) requiring treatment [2]. SWW is characterized as one of the most harmful agriculture and food industrial wastewaters due to its high organic content [3]. The emissions of SWW would cause eutrophication of surface water and pollution of groundwater [4] due to its high organic carbon and nitrogen (N) compounds [5].
Chinese meat production has been consistently ranked the first place in the world since 1990. Unlike other countries, pork is the most consumed meat in China, which accounted for 63.6% of the total meat production [6]. Correspondingly, “Effluent standard of pollutants for meat processing industry (GB 13457-92)” was firstly established in the last century, which is updating in line with the enhancement for water environmental protection in China (000014672/2018-01201). In which, effluent limit of total nitrogen (TN) has been added. The permitted maximum concentration of effluent TN has been set at 25, 20, and 15 mg/L for enterprises that are existed, new-built, and in environmental sensitive areas, respectively.
Chemical oxygen demand (COD) and ammonia nitrogen (NH4 +-N) were the main concern contaminants in the previous lab and case studies. Combined anaerobic-aerobic processes were developed to reduce the corresponding COD and NH4 + as well as cost maintenance [1]. Alure-type wastewater treatment process (ATWTP) was firstly proposed in recent decades [7], which was improved as the enhanced alure-type biological system (E-ATBS) to enhance the biological nutrient removal and got the excellent performance [6]. However, E-ATBS has the very long processes, which would bring higher costs and staff requirements.
Consequently, the design of experiments (DOEs) is widely used to surpass the limitations of traditional experimental methods in terms of time, materials and the number of experimental trials [1, 8]. DOEs include central composite designs, Doehlert matrixes, and Box-Behnken designs (BBDs) [9], in which BBDs need the least runs to investigate per variation. DOEs have been employed successfully to optimize the operational parameters in wastewater treatment [1, 8, 10].
In this study, the E-ATBS was bravely shorted to the anaerobic-aerobic SWW treatment process (AASWWTP) in the light of the characteristics of SWW, which was consist of five-stage ABR and only one-stage biological contact oxidation (BCO) tank. A pilot scale AASWWTP was installed to investigate its COD, NH4 +-N and TN removal capacities. Therefore, the overall goal of this research was to investigate the applicability and removal and economic efficiencies of the developed AASWWTP. The specific objectives were to: (1) investigate the COD, NH4 +-N and TN removal capacities of AASWWTP at different conditions designed through the modified BBD; (2) investigate the influence of the supernatant reflux position and ratio on the TN removal performance; (3) evaluate and summarize the operating mechanism of AASWWTP; and (4) compare the TN removal performance and economic efficiency of AASWWTP with other published biological SWW treatment processes.
### 2.1. Experimental Set-up
The flowchart of the AASWWTP is shown in Fig. 1. AASWWTP consist of regulating tank (RT), ABR with four anaerobic alures (1A, 2A, 3A and 4A) and one anoxic alure (5A), BCO tank and clarifier successively.
The acrylic bioreactor contained one 108 L ABR reactor (23.4, 22.5, 21.6, 20.7 and 19.8 L for 1A, 2A, 3A, 4A and 5A, respectively) and one 24.0 L BCO reactor. The soft combined semi-soft packing filter was used in the BCO reactor as the microbial attachment. The structure of the packing filter was a double-ring large plastic ring, on which were pressed the aldehyde fibers and in which was fixed by the snowflake plastic branch (Fig. 1). The packing filter diameter was 100 mm. The spacing was 80 mm. The packing filter was connected in series through the central fiber rope. A composite packing filter with the total folding length of 3 m was placed in the whole BCO reactor.
### 2.2. SWW Composition
Synthetic SWW was prepared according to the real wastewater, which has been discussed in the previous study [6]. Synthetic SWW was prepared aimed to contain 2,000 ± 100 mg/L of COD and 160 ± 10 mg/L of TN by adding the fresh porcine blood and sodium citrate. Note that sodium citrate was not only used to adjust the C/N, but also to avoid the blood coagulation.
### 2.3. Experimental Design
In this study, modified version design method according to BBD was applied to optimize the supernatant reflux position and ratio. According to the previous study [6] and the AASWWTP designed operation mode, the 2–4 anaerobic alures of ABR was chosen as the supernatant reflux position together with 100%–300% supernatant reflux ratios. Table 1 shows the modified BBD experiment including two variations and three levels. In this table, the independent variable levels were presented in terms of the coded levels. The coded values set for reflux position (A) and ratio (B) at three levels were: −1, 0, +1. The reflux positions were set at compartment 2, 3 and 4, and reflux ratios were 100%, 200% and 300%, respectively. Note that the minimum number of experimental runs was needed (only 5 runs in this study) through introducing the modified BBD, which is worth popularized.
To obtain the appropriate reflux position and ratio, six dependent parameters were analysed: COD, TN, NH4 +-N, NO2-N, NO3-N and pH.
### 2.4. Experimental Procedures
#### 2.4.1. Experimental start-up
The inoculums were taken from the corresponding anaerobic, anoxic and aerobic tanks of slaughterhouse wastewater treatment plant in the previous case study [6]. Seed sludge was acclimated in the developed AASWWTP bioreactor for 30 d in the continuous mode with the pollution load of 500 g COD/m3/d and run at room temperature of 20 ± 2°C. The bioreactor was considered as start-up successfully when the effluent COD was kept lees than 80 mg/L, which was the effluent limit in the online draft of “Effluent standard of pollutants for slaughter and meat processing industry” (000014672/2018-01201).
#### 2.4.2. Experimental run
To determine the appropriate supernatant reflux position and ratio, the AASWWTP bioreactor was run according to the modified BBD in Table 1. Two parallel samples for effluent were taken every 24 h. When the effluent COD and TN concentrations were stable (a p-value with a 95% confidence level; [11]), two parallel samples for influent, effluent and at different compartments were taken and analysis for COD, TN, NH4 +-N, NO2-N, NO3-N and pH. Experiments were repeated if there was a sample analysis error greater than 5%.
The inorganic nutrients of K+, Na+, Ca2+ and Mg2+ for microbial growth at the optimum TN removal condition was also detected to investigate the nutrient supply of SWW to microorganisms.
### 2.5. Analytical Methods
Water samples were taken from each stage and then filtered by 0.45 μm membrane before detection except for COD and TN. NH4 +-N, NO2-N, and NO3-N were measured using a spectrophotometer (DR6000, HACH, USA) according to the Chinese NEPA [12] standard methods, and the method detection limits (MDLs) were 0.035, 0.003 and 0.052 mg/L, respectively. K+, Na+, Ca2+ and Mg2+ were measured using a Metrohm ECO Compact IC Pro (Herisau, Switzerland) ion chromatograph. MDLs were 0.07, 0.13, 0.07 and 0.07 mg/L for K+, Na+, Ca2+ and Mg2+, respectively.
COD was measured through Lovibond mid-range kits with the MDL of 0–1,500 mg/L. TN concentration was measured using Hach TNT plus 827 test kit. pH was measured using a pH meter with calibrated electrode (FiveEosy Plus, Mettler Toledo, Switzerland; MDL: 0–14).
### 3.1. COD Removal
Average COD concentration and COD removal efficiency in each modified BBD study are shown in Fig. 2. COD was removed successively with the flow of SWW in the AASWWTP bioreactor. Moreover, the reflux position also played an important role in COD removal. For example, when the supernatant returned into the second compartment of ABR, ~50% COD was removed in this compartment.
From Fig. 2, it could be also found that all effluent COD concentrations were between 19 ± 1 and 43 ± 3 mg/L, which were much lower than the Chinese SWW discharge standard of 80 mg/L. COD removal efficiency was all higher than 98%, which was much higher than 80% of previous ATWTP [7] and similar to the upgraded enhanced alure-type biological system (E-ATBS) [6]. Hence, it could be concluded that the supernatant reflux can guarantee the efficient COD removal without adding more structures.
As known, discharge standard of pH was 6.00–9.00, which also reflected the suitable pH condition microbial growth [13]. In this study, pH at each modified BBD condition and each stage was all satisfied the above condition. However, the pH at each stage in the AASWWTP bioreactor showed the different trend at different modified BBD conditions. The supernatant reflux position and ratio affected the pH level simultaneously. This might be due to the introduced reflux affecting the microbial reactions of N removal, which would be fully described in section 3.2.
### 3.2. Nitrogen Removal
Average concentrations of NH4 +-N, NO2-N, NO3-N, TN and TN removal efficiency in each designed study are shown in Fig. 3. NH4 +-N and organic N formed the main contaminates of TN in SWW. Organic N firstly digested to NH4 +-N in the anaerobic stage. As shown in Fig. 3, NH4 +-N concentration increased most obviously in the first anaerobic stage of ABR (1A), which was also demonstrated that 1A played the roles of both hydrolysis-acidification and anaerobic digestion for organic matter simultaneously. NH4 + accumulated continuously until the supernatant returned compartment/alure and then decreased sharply at the corresponding alure due to partial denitrification (PD)-ANAMMOX [6]. After this alure, NH4 +-N concentration increased again slowly as the remainder organic N amination in the rest of ABR. Finally, NH4 +-N concentration decreased to lower than 0.50 mg/L at the aerobic stage of BCO reactor.
NO2 and NO3 as the nitrification productions can be used as the source for PD-ANAMMOX [14]. However, different reflux conditions would lead to different amount of NO2 and NO3 accumulations in the final effluent. In this study, the highest NO2-N and NO3-N accumulation of 1.74 ± 0.01 and 61.00 ± 1.98 mg/L were detected at the conditions of R5 and R4, respectively. Hence, the supernatant returned to the forth compartment/alure would lead to the highest NO2 and NO3 by-products accumulation. The more forward the supernatant reflux position was, the lower NO2 and NO3 by-products accumulation would be gained. At the same time, it could be found that NO3-N concentration in effluent was lowest at 22.62 ± 0.28 mg/L at the condition of R2 (reflux position at 2 alure and ratio of 300%).
As shown in Fig. 3, NO2-N concentration through the whole anaerobic and anoxic stages in ABR was all lower than 0.02 mg/L. Correspondingly, NO3-N concentration was lower than 10 mg/L. Unlike NO2, NO3-N concentration increased obviously at the reflux compartment/alure. This also can be used as the side validation of the analysis of PD-ANAMMOX as the main process for TN removal in the AASWWTP.
TN decreased slowly except for the supernatant reflux compartment/alure, which was similar to the COD removal in ABR but decreased more sharply. The supernatant reflux introduced the NO2 and NO3 as well as small amount of dissolved oxygen (DO), leading to the occurrence of PD-ANAMMOX [6]. In which, NO2 accumulated from partial nitrification in BCO reactor could be used together with the accumulated NH4 + in ABR for the ANAMMOX firstly as description in Eq. (1) [14]. After the supernatant reflux compartment, TN began to slowly increase except for R4, which was absolutely different to COD removal trend. Further study is needed as to analyse the microorganisms to find out the bacteria that lead to this slow increase.
##### (1)
$NH4++1.32NO2-+0.066HCO3-+0.13H+→1.02N2+0.26NO3-+0.066CH2O0.5N0.15+2.03H2O$
Meanwhile, the reflux NO3 from BCO reactor and the produced NO3 from ANAMMOX process were used as the electron acceptor for microbial denitrification. As known, NO3 was the preferred electron acceptor over NO2 in microbial denitrification, expressed as NO3→NO2→N2 [15]. Hence, the intermediate product of NO2 would be accumulated during the PD process and be used as the reactant for ANAMMOX, forming the PD-ANAMMOX process to advanced TN removal. Moreover, some NO3 would be remained through the PD-ANAMMOX process according to Eq. (1), which was consist with the modified BBD experiment results. Almost all NH4 + and NO2 were removed and only some NO3 was accumulated at the conditions of R1, R2 and R3 (reflux position at 2 or 3 alure).
Effluent TN concentrations at R1 and R2 were the same as lowest level of 25 mg/L, which was satisfied with the Chinese SWW discharge standard for the existing enterprises. As known, lots of existing enterprises in China use the combined anaerobic-aerobic processes to reduce COD and NH4 + from SWW to match the existing “Effluent standard of pollutants for meat processing industry (GB 13457-92)” [6]. Hence, according to this study, it could be concluded that only the addition of supernatant reflux would lead to the objective of TN decrease. The existing enterprises can improve the SWW treatment plant only through the simple modification, which can use the minimum cost and ensure the effluent satisfied with the upcoming discharge standard.
Through the analysis of NH4 +-N, NO2-N, NO3-N and TN in the AASWWTP bioreactor at different design studies, it could be concluded that both R1 and R2 showed the excellent N removal capacity. Considering that the increase of reflux ratio would increase the operation cost, R1 (reflux position at 2 alure and ratio of 100%) was chosen as the optimum condition for N removal in AASWWTP. However, the further study is still need to optimate the supernatant reflux conditions more precisely.
### 3.3. Inorganic Nutrients Characteristics
K+, Na+, Ca2+ and Mg2+ as the main inorganic nutrients for microbial growth had also been detected in the optimum condition of reflux position at 2 alure and ratio of 100% (Fig. 4). As shown in Fig. 4, there is enough inorganic nutrients of K+, Na+, Ca2+ and Mg2+ in the synthesized SWW. But only Ca2+ decreased a little during the AASWWTP, whereas K+, Na+, and Mg2+ concentrations increased. This might be due to the inorganic nutrient utilization by microorganisms lower than the nutrient mineralization from porcine blood. As known, K+, Na+, Ca2+ and Mg2+ contents in the fresh porcine blood were 56, 56, 4, and 5 mg/100 g, respectively [16]. In which Ca2+ content was the lowest. Hence, only Ca2+ showed the decreased trend during the SWW treatment. Moreover, Na+ increased most obviously as the co-containing in the fresh porcine blood and sodium citrate. These results also indicate that the SWW is suitable for biological treatment and the correspondingly processes such as AASWWTP should be widely researched and popularized.
### 3.4. SWW Treatment Processes Comparison
Reducing COD and BOD concentrations in SWW is the main focus of the secondary biological treatment by removing soluble organic compounds at the beginning of lab and case studies [17]. However, N species in SWW would cause eutrophication of surface water and pollution of groundwater if SWW emissions without fully treatment [4]. Hence, biological nitrogen removal (BNR) process has been added together with COD removal in the secondary biological treatment processes. The biological processes for removing N species from wastewaters mainly included tradition nitrification/denitrification process, SHARON process (NH4 + is oxidized only to NO2 followed by denitrification), and ANAMMOX process [18].
Table 2 illustrates some published biological processes for TN removal from SWW in this century. There are lots of different and combined biological technologies used in SWW treatment. It could be concluded that combination biological processes and integrated systems always showed better TN removal efficiency. As shown in Table 2, Del Nery et al. used the simple anaerobic technology of UASB and got the TN removal efficiency of 36.0%–40.0% [21]. Other published technologies in Table 2 all used the combined biological technologies and got higher TN removal efficiencies, which were ranged from 52.0% to 95.6%.
Comparing to other published technologies, the AASWWTP showed relatively high TN removal efficiency, but not the highest. Considering the effluent standard of SWW in China, the developed AASWWTP was able to fit its requirement for effluent TN. In the previous study, the E-ATBS in case study has been fully discussed and showed the TN removal efficiency as high as 98% [6]. However, the E-ATBS contained much longer biological processes as hydrolysis-acidification tank (HT), anaerobic tank (1A) with 3 compartments, oxic tank 1 (1O1), oxic tank 2 (1O2), anoxic tank (2A), oxic tank (2O), clarifier, and sand filter successively. However, the AASWWTP developed in this study only had the shared anaerobic and anoxic tank of ABR and oxic tank of BCO. Comparing of AASWWTP and E-ATBS, the developed AASWWTP can reduce about 30%–40% cost through the following four aspects: (1) the number of oxic tank reduced from three to one, which would save about 33% electricity consumption for aeration; but the supernatant reflux would consume about 5% from that saving. (2) The reduced structures would save about 30%–40% land occupation; (3) the disposable cost for construction would save about 30%–40% simultaneously; and (4) less operating and maintenance costs were needed. Hence, on the basis of ensuring the effective removal of TN, the developed AASWWTP can greatly reduce the cost of capital investment and operation and maintenance. AASWWTP is a promising technology for TN removal from SWW. Further scientific calculation would be done in the next step study to confirm the accurate budget.
In addition, the combined biological and physicochemical processes have also been studied and utilized in the advanced N removal of SWW. Bohdziewicz and Sroka reported the combined activated sludge (AS)-reverse osmosis (RO) system for the treatment of SWW and gained a high COD and TN removal of 99.80% and 99.77%, respectively [26]. Bustillo-Lecompte et al. [27] developed a combined ABR-AS-UV/H2O2 process and reported up to 99.98% TOC and 82.84% TN removal efficiencies.
Although combined biological and physicochemical methods often got the high removal efficiencies, the simple biological processes were still gotten priority in the practical application considering to the relatively low operating costs, low requirements for operators and the environment friendly. Moreover, the development of biological processes also ensured the high TN removal efficiency. The developed AASWWTP in this study got the 85% TN removal efficiency, which showed a litter higher than the combined biological and physicochemical method of ABR-AS-UV/H2O2 (~83%) [27]. Hence, the biological processes for SWW treatment should be widely researched and popularized.
### 4. Conclusions
AASWWTP for advanced TN removal of SWW was developed and the optimum supernatant reflux position and ratio were investigated. COD could be effectively reduced over the whole modified BBD study and the removal efficiency was all higher than 98%. The optimum reflux position and ratio were suggested to be 2 alure and 100%, respectively. Effluent TN concentration was satisfied with the upcoming Chinese discharge standard for the existing enterprises of 25 mg/L. AASWWTP is a promising technology for SWW treatment but more research is needed to further improve the operating efficiency.
### Acknowledgments
This research was financially supported by China Postdoctoral Science Foundation (Grant No. 2018M630245), the National Key Research and Development Program of China (Grant No. 2016YFD0501405) and Beijing Postdoctoral Research Foundation (Grant No. 2017-ZZ-137). The authors would like to thank the local slaughterhouse industry for the supply of seed sludge and fresh porcine blood.
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##### Fig. 1
Flowchart of anaerobic-aerobic slaughterhouse wastewater treatment process (AASWWTP).
##### Fig. 2
Average of COD concentration, removal efficiency and pH level in each design study. The stages were influent from regulating tank, 1A, 2A, 3A, 4A, 5A and effluent successively.
##### Fig. 3
Average of N species and TN removal efficiency in each design study. The stages were influent from regulating tank, 1A, 2A, 3A, 4A, 5A and effluent successively.
##### Fig. 4
Inorganic nutrients variation at the optimum condition of reflux position at 2 alure and ratio of 100%.
##### Table 1
Design for the Study of Two Experimental Variables, Reflux Position and Ratio
Run No. Factor A: Reflux position Factor B: Reflux ratio
Code Actual compartment Code Actual value
R1 −1 2 −1 100%
R2 −1 2 +1 300%
R3 0 3 0 200%
R4 +1 4 −1 100%
R5 +1 4 +1 300%
##### Table 2
Comparison of Different Biological Technologies for SWW Treatment
Published year Processesa Typesb Feed HRTc Influent TN TN removal efficiency References
h mg/L %
2005 FFR Ana-Ae 2–3 N m3/h 23–91 150–260 69.0 [19]
2006 ABR-CSTR Ana-Ae 0.3–3.16 kg COD/m3/d 249 70–147 77.4 [20]
2007 UASB Ana 1.6 ± 0.4 kg COD/m3/d 69 147–233 36.0–40.0 [21]
2009 ASR-Lagoon Ae-Ano/Ana 1,500 m3/d 10 78–457 52.0–93.0 [22]
2011 A2O Ana-Ano-Ae 15 L/d 16 84–409 81.5–95.6 [23]
2013 UASB-AFBR Ana-Ae - 24 169 76.0 [24]
2014 SBR Ano-Ana-Ano 0.6 L/min 12–16 143–175 80.8–91.1 [25]
2017 ABR-ASR Ana-Ae 63 mL/min - 161–255 72.1 [1]
2019 E-ATBS Ana-Ae-Ano-Ae- 700–900 m3/d - 115–211 > 90.8 [6]
This study AASWWTP Ana-Ae 25 L/d 96 150–170 84.9 -
a FFR: Fixed-film reactors; CSTR: Completely stirred tank reactor; UASB: Upflow anaerobic sludge blanket reactor; ASR: Activated sludge reactor; A2O: Anaerobic/anoxic/oxic system; AFBR: Aerated fixed-bed reactor; SBR: Sequencing batch reactor
b Ana: Anaerobic; Ano: Anoxic; Ae: Aerobic
c HRT: Hydraulic retention time
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2020-01-24 22:55:03
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http://math.stackexchange.com/questions/258741/why-cant-i-interchange-integration-and-differentiation-here
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# Why can't I interchange Integration and Differentiation here?
Consider $f(x,y)=y^3e^{-y^2x}$ and define $F(y) =\int_0^{\infty}f(x,y)dx$
We have that $F'(0)\not = \int_0^{\infty} \frac{\partial f}{\partial y}(x,0)dx$
in the spoiler there is how I got this, in case I made a mistake there
We calculate $F'(0)$ essentially using Monotone convergence theorem we can show that, for $y\in \mathbb{R}\setminus\{0\}$, $F(y)=y$ moreover $F(0)=0$ so $F'(0)=1$
Now, I want to understand which hypothesis of Theorem 2 at this page does not hold. Instead of the third hypothesys at the link, though, I use this weacker hypothesis, which is still enough:
"For each $b \in \mathbb{R}$, there exists an open interval $b\in J$ and an integrable function over $(0, \infty)$ , $g(x)$ such that $| \frac{\partial f}{\partial y}(x,y)| \leq g(x)$ for every $y\in J$ and $\forall x$"
Now, the first hypothesis certainly holds as $\forall y, \ x\rightarrow f(x,y)$ is integrable $(0,\infty)$ by comparison with $e^{-kx}$ for appropriate positive value of $k$
Moreover $\frac{\partial f}{\partial y}(x,y)$ exists everywhere...
So is the last hypothesis to be problematic but I can't see how as I can bound $y$ in $J$ and then just use some linear combination of $e^{-kx}$ and $xe^{-lx}$ for suitable $k,l$ as they are both integrable over $(0,\infty)$...
Thank you very much!
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The theorem says if you can bound $f_y(x,y)$ with an integrable function $g(x)$, that is $$|f_y(x,y)|\leq |g(x)|,$$ then you can change the order of differentiation and integration.
yes but I think you can weacken this condition by bounding $f_y$, for each $b \in \mathbb{R}$ on an open interval containing $b$ (and for all values of x)! – Moritzplatz Dec 14 '12 at 16:13
or do you think I can just say that my $f$ does not respect this condition and hence why the theorem doesn't work? it feels like cheating! – Moritzplatz Dec 14 '12 at 16:50
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2015-07-29 12:26:08
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https://math.stackexchange.com/questions/2012239/classifying-groups-such-that-g-simeq-mathbbz-3-rtimes-mathbbz-2-times/2012785
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# Classifying groups such that $G \simeq \mathbb{Z}_3 \rtimes (\mathbb{Z}_2 \times \mathbb{Z}_2)$
I am thinking G will be $D_{12}$; however, I am not sure how to prove that all semi-direct product are isomorphic or explicitly get $D_{12}$. We have $\mathbb{Z}_2\times \mathbb{Z}_2$ has two generators $(1,0)$ and $(0,1)$ so we have three choices of nontrivial homomorphism $\psi : \mathbb{Z}_2 \times \mathbb{Z}_2 \rightarrow Aut(\mathbb{Z}_3) = \mathbb{Z}_2 = \{1,-1\}$ is just mapping $(1,0),(0,1)$ to $\pm 1$ and both don't go to 1. Why is all semi-direct producted achieved in this way isomorphic though ?
• Any two of your three homomorphisms $\Bbb Z_2\times\Bbb Z_2\to{\rm Aut}(\Bbb Z_3)$ are related by an automorphism of $\Bbb Z_2\times\Bbb Z_2$, which induces an isomorphism between the two versions of $\Bbb Z_3\rtimes(\Bbb Z_2\times\Bbb Z_2)$. Recall our discussion here. – arctic tern Nov 13 '16 at 20:08
• It's not true. The direct product $Z_3 \times Z_2 \times Z_2$ is abelian and is not isomorphic to $D_{12}$, – Derek Holt Nov 13 '16 at 20:19
• @DerekHolt I think Adeek is talking about ones that are "achieved in this way" via one of the "three choices of nontrivial homomorphism[s]." – arctic tern Nov 13 '16 at 23:28
• @arctictern But the first sentence in the post is incorrect. – Derek Holt Nov 14 '16 at 9:05
This is true when $\psi$ is nontrivial. Notice that $Ker \psi$ has order $2$. Hence, $\mathbb Z_3 Ker\psi$ isomorphic to $\mathbb Z_6$.(it acts trivially on $\mathbb Z_3$.) Set $H=\mathbb Z_3 Ker\psi$
Now let $x\in G-H$. Then $x=ab$ where $a \in \mathbb Z_3$ and $b\in \mathbb Z_2\times \mathbb Z_2-Ker\psi$.
Notice that order of $b$ is two and $b^{-1}ab=a^{-1}$. Thus,
$$x^2=abab=ab^{-1}ab=aa^{-1}=1$$.
order of $x$ is also $2$.
As, index of $H$ is $2$, $H$ is normal in $G$ and $G=H<x>$
Thus, $x$ acts on $H$ by conjugation. If $x$ acts trivially on $H$ then $G$ is abelian. But this is not the case as $\psi$ is nontrivial.
$Aut(H)\cong Z_2$. We must have $h^x=h^{-1}$ for $H=<h>$.
Then $G=<x,h|x^2=h^6=1 \ \ and \ \ h^x=h^{-1} >\cong D_{12}$.
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2019-12-08 18:58:42
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https://www.kth.se/math/kalender/seminar-analysis-sthlm/mark-g-lawrence-new-sources-of-real-analyticity-in-function-spaces-and-function-algebras-1.956908
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# Mark G. Lawrence: New sources of real analyticity in function spaces and function algebras
Time: Wed 2020-02-12 13.15 - 14.15
Lecturer: Mark G. Lawrence
### Abstract
It is a basic fact from approximation theory that given a compact set $$K \subseteq \mathbf{C^n}$$, the closure of the algebra generated by $$z$$ and $$\overline{z}$$ is $$C(K)$$. Another similar result is the Wermer maximality theorem regarding function algebras on the unit circle. In a different direction, the speaker has demonstrated that if an algebra is generated by $$z$$ and $$\overline{z}(g(z))$$, where $$g$$ is entire satisfies some condition on the zero set, then the closed Fréchet algebra generated consists of functions which are real analytic, with infinite radius of convergence. With further restrictions on the zero set of $$g$$, one can construct Bergman space analogues of Fock space. There are analagous results with functions of several variables and functions on come CR manifolds as well. The condition on the zeros of $$g$$ can be fulfilled with explicit functions. The technique is CR wedge extension combined with the theory of the 1-dimensional extension property.
Page responsible:[email protected]
Belongs to: Department of Mathematics
Last changed: Feb 07, 2020
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2021-12-02 19:18:03
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https://tagoos.readthedocs.io/en/latest/
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# TAGOOS : associated tag SNP boosting¶
TAGOOS is a nucleotide scoring tool for non-coding (Intronic and intergenic) regions. There are two underlying models trained with the XGBOOST algorithm using intronic and intergenic associated SNPs (GWAS P-value < $$5\cdot10^{-8}$$) from the GRASP database. The predictive variables have been selected by the learning algorithm among 4684 gene regulation related annotations such as histone modifications, eQTLs or transcription factors in different tissues from these databases:
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2020-08-09 02:23:29
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https://www.mysciencework.com/publication/show/hilbert-space-valued-gabor-frames-weighted-amalgam-spaces-886dd782?search=1
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# Hilbert space valued Gabor frames in weighted amalgam spaces
Authors
Type
Published Article
Journal
Advances in Pure and Applied Mathematics
Publisher
De Gruyter
Publication Date
Aug 23, 2018
Volume
10
Issue
4
Pages
377–394
Identifiers
DOI: 10.1515/apam-2018-0067
Source
De Gruyter
Keywords
License
Yellow
## Abstract
Let ℍ {\mathbb{H}} be a separable Hilbert space. In this paper, we establish a generalization of Walnut’s representation and Janssen’s representation of the ℍ {\mathbb{H}} -valued Gabor frame operator on ℍ {\mathbb{H}} -valued weighted amalgam spaces W ℍ ( L p , L v q ) {W_{\mathbb{H}}(L^{p},L^{q}_{v})} , 1 ≤ p , q ≤ ∞ {1\leq p,q\leq\infty} . Also, we show that the frame operator is invertible on W ℍ ( L p , L v q ) {W_{\mathbb{H}}(L^{p},L^{q}_{v})} , 1 ≤ p , q ≤ ∞ {1\leq p,q\leq\infty} , if the window function is in the Wiener amalgam space W ℍ ( L ∞ , L w 1 ) {W_{\mathbb{H}}(L^{\infty},L^{1}_{w})} . Further, we obtain the Walnut representation and invertibility of the frame operator corresponding to Gabor superframes and multi-window Gabor frames on W ℍ ( L p , L v q ) {W_{\mathbb{H}}(L^{p},L^{q}_{v})} , 1 ≤ p , q ≤ ∞ {1\leq p,q\leq\infty} , as a special case by choosing the appropriate Hilbert space ℍ {\mathbb{H}} .
Seen <100 times
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2020-10-27 05:41:15
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https://janmr.com/blog/2009/07/implementing-multiple-precision-arithmetic-part-1/
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# janmr blog
## Implementing Multiple-Precision Arithmetic, Part 1 July 23, 2009
### Introduction
This article is the first in a series dealing with algorithms for multiple-precision arithmetic. The goal is to present both a theoretical foundation with high-level algorithm descriptions (based on Section 4.3.1, The Classical Algorithms, of The Art of Computer Programming, Volume 2, by Donald E. Knuth) and a portable C++ implementation of the algorithms. The theory and high-level algorithms will be quite universal and generic, whereas the presented code will be just one way to implement the algorithms in a specific programming language.
We start out by considering only non-negative integers. A number $u \geq 0$ will be represented in radix $b \geq 2$ using the notation
$u = (u_{n-1} \ldots u_1 u_0)_b = \sum_{i=0}^{n-1} u_i b^i, \quad 0 \leq u_i < b.$
We will call $u$ an $n$-digit number and $u_0$, $u_1$, etc., its digits. Unless stated otherwise we will always have that the most-significant digit is non-zero, here $u_{n-1} \neq 0$, and we will represent zero with no digits, $0 = ()_b$. We have $b^{n-1} \leq u \leq b^n-1$, implying $n=1+\lfloor \log_b u \rfloor$ for $u \geq 1$.
Let the word size of the data type T used for each digit be $b_T$. For instance, if T is a 32 bit unsigned integer, we have $b_T = 2^{32}$. We will implement the algorithms using $b = b_T$ and exploit the fact that C++ does arithmetic on unsigned integers modulo $b_T$ (see paragraph 3.9.1 (4) of the C++ standard). This makes it possible to implement portable algorithms. They will not be optimal with respect to speed, however, and it will be noted when specialized operations, such as add-with-carry instructions, would lead to more efficient implementations.
### Data Structures
A non-negative number will be represented in C++ as an instance of the class NonNegativeInteger:
template <typename T, typename V=detail::SimpleDigitVector<T> >class NonNegativeInteger {private: boost::shared_ptr<V> digitvec; // ...public: NonNegativeInteger(); NonNegativeInteger(T value); // ...};
The type argument T is used to represent each digit. It must be integer and unsigned, so unsigned char, unsigned short, unsigned int, unsigned long, and unsigned long long can all be used (the type long long int is not standard C++, but is, e.g., supported by GCC). If a digit type with 8, 16, 32, or 64 bits is needed, the boost integer types uint8_t, uint16_t, uint32_t, uint64_t (from the namespace boost) can be used with portability ensured (uint64_t is not always available, but the macro BOOST_NO_INT64_T will tell you if it is not).
The type argument V is used as the container type for the digits. The default container is SimpleDigitVector (which at this time is also the only container supported). This default container simply wraps an array of size (at least) the number of digits.
Note that the digit container digitvec of NonNegativeInteger is wrapped in a boost shared pointer. Consider the small code excerpt:
NonNegativeInteger<unsigned> a=value1(), b=value2(), c;c = a;b += a;c += b;
Because of the shared pointer, the first assignment, c = a, is very cheap and will not result in copying the digit container. Instead, the digitvec of both a and c will refer to the same instance of the digit container (which gets a reference count of 2). This also makes it cheap to pass arguments by value and returning numbers by value. The next statement, b += a, can add a to b in place (assuming there is enough space in b's digitvec to contain the result) because it is possible to check if b is the only instance referring to its digit container. Contrarywise, the last statement, c += a, cannot add a to c in place because a refers to the same container as c does.
Adding radix $b$-numbers is quite straightforward and is easily done using the familiar pencil-and-paper method.
To formalize, we first consider adding two $n$-digit numbers, $n \geq 1$, $u=(u_{n-1} \ldots u_1 u_0)_b$ and $v = (v_{n-1} \ldots v_1 v_0)_b$, obtaining an $(n+1)$-digit sum $w=(w_n \ldots w_1 w_0)_b$. Note that we here may end up with $w_n = 0$ (in which case $w_{n-1} \neq 0$). We have
\begin{aligned} w_0 &\leftarrow (u_0 + v_0) \;\text{mod}\; b, \\ w_i &\leftarrow (u_i + v_i + k_i) \;\text{mod}\; b, \\ w_n &\leftarrow k_n. \end{aligned} \quad \begin{aligned} k_1 &\leftarrow \lfloor (u_0 + v_0)/b \rfloor, \\ k_{i+1} &\leftarrow \lfloor (u_i + v_i + k_i)/b \rfloor, \quad i = 1, \ldots, n-1, \\ \text{ } \end{aligned}
Note that $\lfloor x/b \rfloor = [x \geq b]$ for $x < 2 b$, where $[P]$ is equal to $1$ if $P$ is true and equal to $0$ if $P$ is false. This means $k_i \in \{0,1\}$ and furthermore $u_i + v_i + k_i \leq b-1 + b-1 + 1 = 2 b - 1$. Using that $0 \leq u_i, v_i \leq b-1$ and $(x \;\text{mod}\; b) + \lfloor x/b \rfloor = x$ it is quite easy to show that, in fact, $w = u + v$.
The cases where $u$ or $v$ is zero makes addition trivial. Similarly, if the number of digits in $u$ and $v$ are different, is it quite easy to adjust the algorithm above.
Let us now look at some implementation details. How do we compute $z = (x + y) \;\text{mod}\; b$ and $k = \lfloor (x + y)/b \rfloor$ for $0 \leq x, y < b$? We have several options, some of which are
1. Use $2 b \leq b_T$. Then $z$ and $k$ can be computed directly.
2. Use $b = b_T$ and the CPU's add and add-with-carry instructions.
3. Use $b = b_T$, but the computations must be done in some portable C++ way.
Option 1 is actually not an option because we insist on using $b = b_T$. Option 2 leads to the most efficient code, regarding both space and speed. The problem is that these special instructions are not directly accessible via the C++ standard. Some compilers, though, make it possible to use inline assembly. For instance, GCC has such capabilities.
Option 3 is the way to go. As mentioned earlier, C++ does calculations modulo $b_T$, so $z \leftarrow (x + y) \;\text{mod}\; b$ comes ‘for free’ as simply z = x + y in C++. Left is how to detect whether a carry occurs during an addition. One way to do that is the following. Consider $z = (x + y) \;\text{mod}\; b$ for which there are two possibilities. Either $z = x + y$ ($k = 0$) which implies $z \geq x$ and $z \geq y$, or we have $z + b = x + y$ ($k = 1$) which implies $z = x - (b - y) = y - (b - x)$, leading to $z < x$ and $z < y$. So $k = [z < x] = [z < y]$. Another way to detect whether a carry occurs is to split $x$ and $y$ into a low and high part, and then adding the low and high parts seperately—keeping track of a possible intermediate carry, of course.
### Subtraction
An algorithm for multiple-precision subtraction is similar to addition, as just considered. Again we set $u=(u_{n-1} \ldots u_1 u_0)_b$ and $v = (v_{n-1} \ldots v_1 v_0)_b$ with $n \geq 1$. To ensure that the result $w = u - v$ is a non-negative integer we furthermore require that $u \geq v$. We now have the algorithm:
\begin{aligned} w_0 &\leftarrow (u_0 - v_0) \;\text{mod}\; b, \\ w_i &\leftarrow (u_i - v_i - k_i) \;\text{mod}\; b, \end{aligned} \quad \begin{aligned} k_1 &\leftarrow [u_0 < v_0], \\ k_{i+1} &\leftarrow [u_i < v_i + k_i], \quad i = 1, \ldots, n-1. \end{aligned}
Note that any digit of the result may end up being zero—we only know that $0 \geq w \geq b^n - b^{n-1} - 1$. Note furthermore that $k_n = [u < v] = 0$. Verification of the algorithm is easily done using the fact that $x - y = ((x-y) \;\text{mod}\; b) - b [x < y]$ for $0 \leq x, y < b$.
The options when implementing the algorithm is, again, much like for addition. Using the CPU's subtract-with-borrow instruction would be ideal here, but it cannot be done portably in C++. The borrow when calculating $x - y$ can be computed as simply $k \leftarrow [x < y]$, or, to avoid branching instructions, we can split $x$ and $y$ into two parts and do the subtraction for each part seperately.
### Multiplication
We seek an algorithm to compute $w = u v$ where
$u = (u_{m-1} \ldots u_1 u_0)_b, \quad v = (v_{n-1} \ldots v_1 v_0)_b, \quad w = (w_{m+n-1} \ldots w_1 w_0)_b.$
We first, however, consider the simpler operation $z \leftarrow y + \alpha u$ where
$0 \leq \alpha < b, \quad y = (y_{m-1} \ldots y_1 y_0)_b, \quad z = (z_m \ldots z_1 z_0)_b.$
The following algorithm suggests itself:
\begin{aligned} z_0 &\leftarrow (y_0 + \alpha u_0) \;\text{mod}\; b, \\ z_i &\leftarrow (y_i + \alpha u_i + k_i) \;\text{mod}\; b, \\ z_m &\leftarrow k_m. \end{aligned} \quad \begin{aligned} k_1 &\leftarrow \lfloor (y_0 + \alpha u_0)/b \rfloor, \\ k_{i+1} &\leftarrow \lfloor (y_i + \alpha u_i + k_i)/b \rfloor, \quad i = 1, \ldots, m-1, \\ \text{ } \end{aligned}
If $\alpha = 0$ then obviously $z_i = 0$. If $1 \leq \alpha \leq b-1$ then $z_m$ may be zero, in which case $z_ {m-1} \neq 0$, since
$2 b^{m-1} \leq \; z \; \leq b^m-1 + (b^m-1)(b-1) = b^{m+1} - b < b^{m+1}$
Note that $k_i < b$ since $y_i + \alpha u_i + k_i \leq b-1 + (b-1)(b-1) + b-1 = b^2 - 1$. The algorithm above is easily verified, using that $z_i + b k_{i+1} = y_i + \alpha u_i + k_i$.
We now turn to $w = u v$ and get
$w = u v = \sum_{j=0}^{n-1} b^j v_j u.$
From this we see that we can compute $w$ by doing a number of $(z \leftarrow y + \alpha u)$-type operations, if we start with $w=0$ and then do in-place updates for each $j$. The $b^j$-factor simply determines the ‘digit offset’ on which the updates should be done:
\begin{aligned} (w_{m-1} \ldots w_1 w_0)_b &\leftarrow (0 \, \ldots \, 0 \, 0)_b, \\ (w_{m+j} \ldots w_{j+1} w_j)_b &\leftarrow (w_{m+j-1} \ldots w_{j+1} w_j)_b + v_j (u_{m-1} \ldots u_1 u_0)_b, \end{aligned}
for $j = 0, 1, \ldots, n-1$. Note that $w_{m+n-1}$ may be zero, in which case $w_{m+n-2} \neq 0$, since $b^{m+n-2} \leq w < b^{m+n}$.
Now for some implementation details. We note that the only non-trivial computation is $z \leftarrow y + \alpha x + k$, where $0 \leq \alpha, k, x, y < b$, followed by computing $z \;\text{mod}\; b$ and $\lfloor z/b \rfloor$. Most CPUs have instructions that can multiply two word-size numbers and produce a double-word answer. As was the case for addition and subtraction, we don't have access to these instructions from standard C++. We have $0 \leq z < b^2$ but every multiplication result in our portable C++ implementation must be smaller than $b$. We can do this by using a new base number $h$ where $h^2 = b$ (we assume that $b$ is chosen appropriately so $h$ is integer) and setting
$z = (z_3 z_2 z_1 z_0)_h, \quad y = (y_1 y_0)_h, \quad \alpha = (\alpha_1 \alpha_0)_h, \quad x = (x_1 x_0)_h, \quad k = (k_1 k_0)_h.$
We can now use the multiplication algorithm above on a ‘smaller scale’ to compute the product $\alpha x$. We can furthermore expand the methods slightly and incorporate the addition of $y$ and $k$ in an elegant way:
\begin{aligned} (z_1 z_0)_h &\leftarrow (y_1 y_0)_h, \\ c &\leftarrow k_0, \\ t &\leftarrow z_0 + \alpha_0 x_0 + c, \quad z_0 \leftarrow t \;\text{mod}\; b, \quad c \leftarrow \lfloor t/b \rfloor, \\ t &\leftarrow z_1 + \alpha_0 x_1 + c, \quad z_1 \leftarrow t \;\text{mod}\; b, \quad c \leftarrow \lfloor t/b \rfloor, \\ z_2 &\leftarrow c, \\ c &\leftarrow k_1, \\ t &\leftarrow z_1 + \alpha_1 x_0 + c, \quad z_1 \leftarrow t \;\text{mod}\; b, \quad c \leftarrow \lfloor t/b \rfloor, \\ t &\leftarrow z_2 + \alpha_1 x_1 + c, \quad z_2 \leftarrow t \;\text{mod}\; b, \quad c \leftarrow \lfloor t/b \rfloor, \\ z_3 &\leftarrow c. \end{aligned}
We now have $z \;\text{mod}\; b = (z_1 z_0)_h$ and $\lfloor z/b \rfloor = (z_3 z_2)_h$.
### Concluding Remarks
We have now covered addition, subtraction, and multiplication of non-negative integers of arbitrary magnitude. Left is how to do division, which will be the subject of the next article.
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2022-07-05 04:28:13
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https://ora.ox.ac.uk/objects/uuid:69850c3f-048d-4b90-a276-13d9ce3872c0
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Journal article
### Measurement of the Z/gamma* plus b-jet cross section in pp collisions at root s=7 TeV
Abstract:
The production of b jets in association with a Z/γ*boson is studied using proton-proton collisions delivered by the LHC at a centre-of-mass energy of 7TeV and recorded by the CMS detector. The inclusive cross section for Z/γ*+b-jet production is measured in a sample corresponding to an integrated luminosity of 2.2 fb -1. The Z/γ*+b-jet cross section with Z/γ*→ℓℓ (where ℓ ℓ = ee or μ μ) for events with the invariant mass 60 < M ℓ ℓ < 120 GeV, at least one b jet at the hadron level with p...
Publication status:
Published
### Access Document
Publisher copy:
10.1007/JHEP06(2012)126
### Authors
Chatrchyan, S More by this author
Khachatryan, V More by this author
Sirunyan, AM More by this author
Tumasyan, A More by this author
Journal:
JOURNAL OF HIGH ENERGY PHYSICS
Volume:
2012
Issue:
6
Publication date:
2012-06-05
DOI:
EISSN:
1029-8479
ISSN:
1029-8479
URN:
uuid:69850c3f-048d-4b90-a276-13d9ce3872c0
Source identifiers:
348866
Local pid:
pubs:348866
Keywords:
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2020-10-21 07:26:53
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https://www.smr.ch/doc/B2000++/user_manual/re02.xhtml
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## Name
atemperatures — Ambient temperatures block
## Synopsis
atemperatures id
value v dof list nodes...
...
end
## Description
atemperatures specifies ambient temperature values at nodes for defining convective conditions in heat analysis in conjunction with the heat convection 'overlay' elements (Elements for Heat Transfer Analysis), id defining the set number (a positive integer). To specify temperatures at nodes inducing thermal strains in stress analysis, please refer to the temperatures command. An atemperatures set is identified by id, a non-negative integer which must be unique for the atemperatures conditions of the current model. id is the number which is referenced by the atemperatures option of the case definition. Sets with an id of 0 will be active for all analysis cases.
## Specifying Ambient Temperatue Values
value v
Specifies the current ambient temperature assigned to subsequently specified nodes.
## Assigning Ambient Temperature Values
The following directives are available to assign the specified ambient temperature value(s) to individual nodes and to collections of nodes:
allnodes
Assign the ambient temperature values to all defined nodes of the current branch.
branch br
For models that consist of several branches. Specifies the external branch number br. To be used in conjunction with the allnodes and nodes directives.
epatch id p1-p8|e1-e12|f1-f6|b
When the discretization of a part of the discretization was created by means of the epatch command, a number of pre-defined nodelists are available for use with the nbc command. The epatch is identified by id.
Individual patch vertex nodes are specified with p1 to p8.
The collection of nodes that are located at a patch edge are specified with e1 to e12.
The collection of nodes that are located at a patch face are specified with f1 to f6.
The collection of nodes of the whole patch body are specified with b.
nodes list
Specifies a list of nodes (of the current branch) to which the ambient temperature value will be assigned.
nodelist name
Specifies the name of the node list to which the ambient temperature value will be assigned.
nodeset name
Specifies the name of the node set to which the ambient temperature value will be assigned.
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2021-07-28 23:05:52
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https://nforum.ncatlab.org/discussion/1993/connected-1topos/?Focus=17353
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# Start a new discussion
## Not signed in
Want to take part in these discussions? Sign in if you have an account, or apply for one below
## Discussion Tag Cloud
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeOct 19th 2010
I split off ∞-connected (∞,1)-topos from locally ∞-connected (∞,1)-topos and added a proof that a locally ∞-connected (∞,1)-topos is ∞-connected iff the left adjoint $\Pi$ preserves the terminal object, just as in the 1-categorical case. I also added a related remark to shape of an (∞,1)-topos saying that when H is locally ∞-connected, its shape is represented by $\Pi(*)$.
I hope that these are correct, but it would be helpful if someone with a little more $\infty$-categorical confidence could make sure I’m not assuming something that doesn’t carry over from the 1-categorical world.
• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeOct 20th 2010
• (edited Oct 21st 2010)
Thanks, Mike!
This does look correct. For the single possibly nonevident point– that every $\infty$-groupoid is the $(\infty,1)$-colimit over itself of the diagram constant on the point – I have added a link to this proposition.
• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeOct 21st 2010
Thanks.
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2022-01-29 05:15:53
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https://math.stackexchange.com/questions/2263619/markov-chain-periodicity-finite-vs-general-state-space
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# Markov chain periodicity: finite vs general state space
The definitions pertaining to Markov chains are generally different for finite state space and general state space Markov chains. This is mainly due to the fact that one cannot talk about moving to a state $x$ for general state spaces, and one cannot write down the transition "matrix". However, in my experience, most definitions for general state space Markov chains like irreducibility, recurrence etc are a general extension from the finite state space.
I recently realized that this is not true for the definition of periodicity. For general state space $\mathcal{X}$ the definition of aperiodicity is (you can find a reference here):
A Markov chain with stationary distribution $\pi(\cdot)$ is aperiodic if there does not exists $d \geq 2$ and disjoint sets $\mathcal{X}_1, \dots, \mathcal{X}_d \subseteq \mathcal{X}$ with $P(x, \mathcal{X}_{i+1}) = 1$ for all $x$ in $\mathcal{X}_i$ ($1 \leq i\leq d-1$) and $P(x, \mathcal{X}_1) = 1$ for all $x$ in $\mathcal{X}_d$ such that $\pi(\mathcal{X}_1) > 0$. Otherwise the chain is periodic with period $d$.
This is the definition that makes more sense to me since it essentially means that the Markov chain cannot get stuck in a loop indefinitely.
For finite state space Markov chains, the definition is (can be found here).
For a state $x$, define its period $d_x$ as $$d_x = \gcd \{n\geq 1: P^n(x,x) > 0\} \,.$$ If the Markov chain is irreducible, then all states share the same period $d$. If $d = 1$, then the Markov chain is called aperiodic.
This definition does not necessarily say that the Markov chain will get stuck in a loop. So my question is, is there anyway that the general state space definition is a generalization from the finite state space definition?
• Do you know if it is true that an aperiodic Markov chain with a unique stationary distribution $\pi$ satisfies the property that $\sup_{A\in \mathcal X}|P^n(x, A)-\pi(A)|\to 0$. If yes, do you also know of a proof? Thanks. – caffeinemachine Dec 20 '18 at 9:35
• @caffeinemachine I believe the answer is no. Particularly, you may have convergence $\pi$-a.e, but not in the complete total variation sense. See the discussion of this counter-example here: stats.stackexchange.com/questions/331158/… – Greenparker Dec 20 '18 at 17:23
• I checked the paper referenced in the post you linked to, and the symbol "$\|{\cdot}\|$" is used to denote the total variation distance. So according to the theorem the convergence occurs in total variation distance. However, the paper does not seem to provide a proof or a reference. I looked at Meyn and Tweedie, but could not locate the theorem (though it must be present in the book in some form, perhaps more general). – caffeinemachine Dec 21 '18 at 9:22
• Sorry, it seems that the paper has indeed given an argument. – caffeinemachine Dec 21 '18 at 9:33
The following argument shows how the "gcd" criterion leads to "stuck in a loop."
For an irreducible Markov chain with state space $\cal S$, suppose that for some $i\in{\cal S}$ and $d\geq 1$, we have $(n\geq0: p_{ii}(n)>0)\subset d\mathbb{Z}$.
By irreducibility, for any state $j$ we can find non-negative integers $m,m^\prime$ so that $p_{ij}(m)>0$ and $p_{ji}(m^\prime)>0$. Then $0< p_{ij}(m)p_{ji}(m^\prime)\leq p_{ii}(m+m^\prime)$ so that $m+m^\prime \equiv 0 \pmod{d}$.
In particular, if $p_{ij}(m)>0$ and $p_{ij}(m^{\prime\prime})>0$, then $m\equiv m^{\prime\prime}\pmod{d}$. So for $0\leq r<d$ we can partition $\cal S$ as follows: $$C_r=\left\{j\in{\cal S}: p_{ij}(n)>0\Rightarrow n\equiv r\pmod{d}\right\}.$$ We also let $C_d=C_0$.
If $j\in C_r$ and $k\in C_s$, for any $n,n^\prime,n^{\prime\prime}$ with $0<p_{ij}(n)p_{jk}(n^\prime)p_{ki}(n^{\prime\prime})\leq p_{ii}(n+n^\prime+n^{\prime\prime})$, we get $n+n^\prime+n^{\prime\prime} \equiv 0 \pmod{d}$, or $n^\prime\equiv s-r \pmod{d}$. Taking $n^\prime=1$ for example, we see that $p_{jk}>0$ implies $s\equiv r+1\pmod{d}$ so $$\sum_{k\in C_{r+1}}p_{jk}=1 \mbox{ for any }j\in C_r.\tag 1$$
Note that $i\in C_0$ and therefore by (1) and induction, $C_r$ is non-empty for all $0\leq r<d$.
Added: What I mean by "stuck in a loop" is that the Markov chain will cycle through $d$ subsets of states. Here is a picture of what this looks like when $d=5$:
Notation:
1. The notation $\sum_{k\in C_{r+1}}$ means that you sum over all states $k$ that lie in the subset $C_{r+1}.$
2. The notation $\mathbb{Z}=\{\dots, -2,-1,0,1,2,\dots\}$ means the set of integers, and $d \mathbb{Z}=\{dz: z\in\mathbb{Z}\}=\{\dots, -2d,-d,0,d,2d,\dots\}$ is the subset that consists of all multiples of $d$.
• Thanks for the answer. Could you fill in more details here? What is the last summation over? What does the notation $d\mathbb{Z}$ mean? And what exactly is your interpretation of "stuck in a loop"? – Greenparker May 4 '17 at 9:09
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2019-11-17 07:43:43
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https://ask.sagemath.org/answers/13409/revisions/
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# Revision history [back]
The range function needs an actual integer, not a symbolic expression with an additional attribute assigned by the assume function. Is there a reason you need to be working with symbolic expressions at all instead of Python functions? Consider the following:
def pc(p,W):
return 1-(1-p)^W
def prc(p,M,W,C):
return 1 - sum(binomial(M,i)*pc(p,W)^i*(1-pc(p,W))^(M-i) for i in range(C))
def pcol(p,N,M,W,C):
return 1 - (1-prc(p,M,W,C))^N
These are Python functions, so you don't need to declare the variables. The return expressions will be evaluated when you supply actual numbers for the arguments (as is done by the plot function). One difference is that
print pcol
will return
<function pcol at 0x4325500>
instead of a symbolic expression indicating the arithmetic the function performs. But I don't see why this would matter if you just want to plot a graph. On that note, to plot the graph you just need to declare the variable that you're plotting:
var('p')
plot(pcol(p/2.0^33,2^20,2^10,8,1),p,0,1000)
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2020-05-26 10:41:21
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https://www.physicsforums.com/threads/vectors-question.107218/
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# Vectors Question
1. Jan 18, 2006
### BigCountry
A plane with a still air speed of 250 km/h is flying due NorthWest. At the same time a wind is blowing toward the South at 50 km/h. In what direction should the pilot head to continue travelling due North West? (The plane must continue with a 250 km/h velocity)
Velocity of plane (Vp) = 250
Velocity of wind (Vw) = 50
sin y = y/x = 50/250
y = 11.5 degrees
90 degrees - 11.5 degrees = 78.5 degrees
The plane must head 78.5 degrees N of W to continue in a 45 degree
N of W line.
Is this correct? Any help is much appreciated.
2. Jan 20, 2006
### Tom Mattson
Staff Emeritus
Your method looks wrong. You need to write down a vector for the velocity of the plane with respect to the air (call it $\vec{v}_{PA}$) and a vector for the velocity of the air with respect to the Eart (call it $\vec{v}_{AE}$. Then you add them up to get the velocity of the plane with respect to the Earth (call it $v_{PE}$).
$$\vec{v}_{PE}=\vec{v}_{PA}+\vec{v}_{AE}$$
3. Jan 20, 2006
### lightgrav
You draw the vectors in the right directions, added tail-to-tip!
your triangle doesn't have a 90-degree in it, but you can use
the law of cosines since you do know two legs and the 135 angle.
4. Jan 21, 2006
### abhijitlohiya
i think use the formula for resultsnt
R=sq.root(p*p+Q*Q+2pqcos135)
find the resultant.
for direction use tan x=p cos 135/p+qsin135
5. Jan 21, 2006
### andrevdh
The resulting motion, $\vec r$, of the plane is the vector sum of is still air speed, $\vec s$, and the wind speed, $\vec w$. The plane has to fly in the direction of the $\vec s$ vector in order to have a resultant motion in the direction of the $\vec r$ vector. Since the $\vec r$ vector is making an angle of $45_o$ with the "x-axis" its x- and y-components have the same magnitude. Assuming that the $\vec s$ vector makes an angle $\theta$ with the x-axis we can therefore say that:
$$r_x\ =\ r_y$$
which gives
$$s\cos(\theta)\ =\ s\sin(\theta)\ -\ 50$$
Last edited: Nov 29, 2006
6. Jan 22, 2006
### andrevdh
If my previous relation is a bit too challenging try solving for the angle between $\vec r$ and $\vec s$, say angle $x$, via the sine rule:
$$\frac{\sin(x)}{w}=\frac{\sin(135^o)}{s}$$
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2018-01-22 18:51:11
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|
http://www.malwareremoval.com/forum/viewtopic.php?f=11&t=36228&start=15
|
Welcome to MalwareRemoval.com,
What if we told you that you could get malware removal help from experts, and that it was 100% free? MalwareRemoval.com provides free support for people with infected computers. Our help, and the tools we use are always 100% free. No hidden catch. We simply enjoy helping others. You enjoy a clean, safe computer.
## Topic posted for EffingCow
MalwareRemoval.com provides free support for people with infected computers. Using plain language that anyone can understand, our community of volunteer experts will walk you through each step.
### Re: Topic posted for EffingCow
There were a bunch of them 2 days ago, but I deleted them, then I scanned again yesterday, and I got 2 more mbfbokgu.dll.vir and owqtbw.dll.vir
effingcow
Regular Member
Posts: 31
Joined: August 29th, 2008, 11:57 am
Location: Aruba
Register to Remove
### Re: Topic posted for EffingCow
Oh...
They are quarantined by Combofix already. Not an issue. They can be left alone. We will remove them later.
Any other issues?
ndmmxiaomayi
MRU Emeritus
Posts: 9708
Joined: July 17th, 2006, 9:22 am
### Re: Topic posted for EffingCow
Just a really slow computer, that freezes a lot. Does this mean I can go on my bank website now safely? you didn't give me any homework, what should I do now?
effingcow
Regular Member
Posts: 31
Joined: August 29th, 2008, 11:57 am
Location: Aruba
### Re: Topic posted for EffingCow
For safety reasons, I suggest that you use another computer to access your bank accounts.
As for a slow computer, there's a number of items that I can disable for you to see if it helps improve your computer's performance.
O4 - HKLM\..\Run: [TPFNF7] C:\Program Files\Lenovo\NPDIRECT\TPFNF7SP.exe /r
Do you use IBM's Presentation Director? A description is here - http://www.systemlookup.com/Startup/132 ... P_exe.html
If no, it can be disabled.
O4 - HKLM\..\Run: [PWMTRV] rundll32 C:\PROGRA~1\ThinkPad\UTILIT~1\PWMTR32V.DLL,PwrMgrBkGndMonitor
This is Thinkpad's Power Manager. Do you use any custom power settings? Based on some users' feedback, this could cause computers to slow down. If you are using custom settings, you may want to re-install Power Manager.
O4 - HKLM\..\Run: [BLOG] rundll32 C:\PROGRA~1\ThinkPad\UTILIT~1\BTVLogEx.DLL,StartBattLog
This measures your battery changes, like how much more is needed get the battery full charged, how much is being used, etc. If your laptop is always connected to your power adapter, this could be disabled since you would have no need to view how much battery power is left.
However, if you are always on the go, and need to find out how much battery power is left, leaving this alone would be the best.
http://www.systemlookup.com/Startup/173 ... ttLog.html
O4 - HKLM\..\Run: [SynTPEnh] C:\Program Files\Synaptics\SynTP\SynTPEnh.exe
This is your touchpad tray icon (near the clock). If you use the advanced features of your touchpad, this would need to be enabled. Otherwise, it can be disabled safely.
http://www.systemlookup.com/Startup/122 ... h_exe.html
O4 - HKLM\..\Run: [TPHOTKEY] C:\Program Files\Lenovo\HOTKEY\TPOSDSVC.exe
Depending on the model of your Thinkpad, pressing one key will let you change the volume or opens the help file. If you don't think you need to change the volume or read the help file, it can be disabled safely.
http://www.systemlookup.com/Startup/132 ... R_EXE.html
Not needed since you can access it via Start Menu, but if you prefer convenience, we can leave this alone.
http://www.systemlookup.com/Startup/353 ... p_exe.html
O4 - HKLM\..\Run: [SoundMAXPnP] C:\Program Files\Analog Devices\Core\smax4pnp.exe
Required for the various sound settings, such as the sound effects. If you don't use them, it can be disabled safely.
http://www.systemlookup.com/Startup/114 ... P_exe.html
O4 - HKLM\..\Run: [AwaySch] C:\Program Files\Lenovo\AwayTask\AwaySch.EXE
Part of Thinkpad Productivity Centre. Helps you to enhance your Thinkpad. If you don't use this feature, we can disable it.
http://www.systemlookup.com/Startup/119 ... h_EXE.html
O4 - HKLM\..\Run: [LPManager] C:\PROGRA~1\THINKV~1\PrdCtr\LPMGR.exe
Related to Thinkpad Productivity Centre. Not really needed unless you use IBM Productivity Centre.
http://www.systemlookup.com/Startup/5608-LPMGR_exe.html
O4 - HKLM\..\Run: [CameraApplicationLauncher] C:\Program Files\Lenovo\Camera Center\bin\CameraApplicationLaunchpadLauncher.exe
This is related to your Thinkpad's camera. Unless you use it, it can be safely disabled.
O4 - HKLM\..\Run: [ACTray] C:\Program Files\ThinkPad\ConnectUtilities\ACTray.exe
It's not needed as it tells your connection status.
http://www.systemlookup.com/Startup/394-ACTray_exe.html
O4 - HKLM\..\Run: [ACWLIcon] C:\Program Files\ThinkPad\ConnectUtilities\ACWLIcon.exe
Related to Thinkpad Connectivity Solutions. If you use them, it would be best to leave them enabled.
http://www.pc.ibm.com/us/think/thinkvan ... tions.html
O4 - HKLM\..\Run: [IgfxTray] C:\Windows\system32\igfxtray.exe
Not really needed as it can be accessed via Control Panel. But if you prefer convenience, we can leave it alone.
http://www.systemlookup.com/Startup/455 ... y_exe.html
O4 - HKLM\..\Run: [HotKeysCmds] C:\Windows\system32\hkcmd.exe
If your Thinkpad provides you with a hot key (some sort of shortcut key) to access Intel Graphics settings, this would be needed. If unsure, best to leave it enabled.
http://www.systemlookup.com/Startup/4221-hkcmd_exe.html
O4 - HKLM\..\Run: [Persistence] C:\Windows\system32\igfxpers.exe
Doesn't appear that this is needed based on the description in the below given link. I think this can be safely disabled.
http://www.systemlookup.com/Startup/935 ... s_exe.html
O4 - HKLM\..\Run: [HP Software Update] C:\Program Files\HP\HP Software Update\HPWuSchd2.exe
This is to remind you to update HP software. Not needed. You can find it in your Start Menu.
http://www.systemlookup.com/Startup/426 ... 2_exe.html
This program can be hard to disable, but it's doable.
O4 - HKLM\..\Run: [AppleSyncNotifier] C:\Program Files\Common Files\Apple\Mobile Device Support\bin\AppleSyncNotifier.exe
Unless you use an iPhone or iPod Touch, this program is again not needed for startup, but as per many Apple products, Apple makes it hard to disable it, but still doable.
O4 - HKLM\..\Run: [iTunesHelper] "C:\Program Files\iTunes\iTunesHelper.exe"
Not needed to run on startup as well, can be hard to disable, but doable.
O4 - HKCU\..\Run: [Sidebar] C:\Program Files\Windows Sidebar\sidebar.exe /autoRun
If you don't use Windows gadgets, it can be safely disabled. They can usually found on the right hand side. A sample:
O4 - HKCU\..\Run: [msnmsgr] "C:\Program Files\Windows Live\Messenger\msnmsgr.exe" /background
Not needed on startup. You can manually run MSN Messenger (now known as Windows Live Messenger) via the Start Menu. However, if you prefer convenience, we can leave this running on startup.
Please let me know which programs you would like to disable at startup.
ndmmxiaomayi
MRU Emeritus
Posts: 9708
Joined: July 17th, 2006, 9:22 am
### Re: Topic posted for EffingCow
Hi!!
I'd like to disable the following:
O4 - HKLM\..\Run: [SynTPEnh] C:\Program Files\Synaptics\SynTP\SynTPEnh.exe
O4 - HKLM\..\Run: [AwaySch] C:\Program Files\Lenovo\AwayTask\AwaySch.EXE
O4 - HKLM\..\Run: [LPManager] C:\PROGRA~1\THINKV~1\PrdCtr\LPMGR.exe
O4 - HKLM\..\Run: [ACWLIcon] C:\Program Files\ThinkPad\ConnectUtilities\ACWLIcon.exe
O4 - HKLM\..\Run: [IgfxTray] C:\Windows\system32\igfxtray.exe
O4 - HKLM\..\Run: [HotKeysCmds] C:\Windows\system32\hkcmd.exe
O4 - HKLM\..\Run: [Persistence] C:\Windows\system32\igfxpers.exe
O4 - HKLM\..\Run: [HP Software Update] C:\Program Files\HP\HP Software Update\HPWuSchd2.exe
O4 - HKLM\..\Run: [AppleSyncNotifier] C:\Program Files\Common Files\Apple\Mobile Device Support\bin\AppleSyncNotifier.exe
O4 - HKLM\..\Run: [iTunesHelper] "C:\Program Files\iTunes\iTunesHelper.exe &
O4 - HKCU\..\Run: [msnmsgr] "C:\Program Files\Windows Live\Messenger\msnmsgr.exe" /background
If you can help me do that, I'd appreciate it!
thanks!
effingcow
Regular Member
Posts: 31
Joined: August 29th, 2008, 11:57 am
Location: Aruba
### Re: Topic posted for EffingCow
Hi Amanda,
Right click on wpsetup.exe and select Run As Administrator to install it.
When done, open the program by double clicking on the dog icon near the clock.
Select the Startup Programs tab.
Select SynTPEnh and click on Disable.
Repeat for all these that you want to disable:
• EZEJMNAP
• AwaySch
• LPManager
• ACWLIcon
• IgfxTray
• HotKeysCmds
• Persistence
• HP Software Update
• AppleSyncNotifier
• iTunesHelper
• msnmsgr
Click Close when done.
When disabling some programs, Winpatrol may prompt you. Click Yes at the prompt.
ndmmxiaomayi
MRU Emeritus
Posts: 9708
Joined: July 17th, 2006, 9:22 am
### Re: Topic posted for EffingCow
Hi Amanda,
How are things going?
ndmmxiaomayi
MRU Emeritus
Posts: 9708
Joined: July 17th, 2006, 9:22 am
### Re: Topic posted for EffingCow
Due to lack of response, this topic is now closed.
If you still require help, please open a new thread in the Infected? Virus, malware, adware, ransomware, oh my! forum, include a fresh FRST log, and wait for a new helper.
Gary R
Posts: 21810
Joined: June 28th, 2005, 11:36 am
Location: Yorkshire
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2016-10-27 03:07:12
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https://socratic.org/questions/how-do-you-write-the-equation-of-a-line-in-point-slope-form-that-is-parallel-to--1
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# How do you write the equation of a line in point slope form that is parallel to y=2x-6 and goes through (2,3)?
$y - 3 = 2 \left(x - 2\right)$
$y - {y}_{0} = m \left(x - {x}_{0}\right)$
$y - 3 = 2 \left(x - 2\right)$
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2022-01-21 06:29:24
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