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http://slideplayer.com/slide/255351/ | # Diffie-Hellman Diffie-Hellman is a public key distribution scheme First public-key type scheme, proposed in 1976.
## Presentation on theme: "Diffie-Hellman Diffie-Hellman is a public key distribution scheme First public-key type scheme, proposed in 1976."— Presentation transcript:
Diffie-Hellman Diffie-Hellman is a public key distribution scheme First public-key type scheme, proposed in 1976.
Diffie-Hellman Public-key distribution scheme Cannot be used to exchange an arbitrary message Exchange only a key, whose value depends on the participants (and their private and public key information) The algorithm is based on exponentiation in a finite field, either over integers modulo a prime, or a polynomial field
Diffie-Hellman The algorithm –Alice and Bob agree on two large prime num, p and q. –Alice then chooses another large random number x and calculate A such that A=q ^ x mod p. and send to bob –Bob also chooses a another large num y and calculate B such that B=q ^ y mod p. and send to Alice –Both Alice and Bob can calculate the key as K 1 = B ^ x mod p K2=A ^ y mod p K1 = K2 –The key may then be used in a private-key cipher to secure communications between A and B
Diffie-Hellman Let p = 11 and q = 7 Alice chooses another num x = 3 then we have A = q^ x mod p =7 ^ 3 mod 11 = 2 Alice Sends the number A = 2 to Bob Bob chooses another num y = 6 then we have B =q^ y mod p = 7 ^ 6 mod 11 = 4 Bob sends the number B = 4 to Alice Now Alice generate Secret key, K1 =B ^ x mod p = 4 ^ 3 mod 11 =9 Then Bob generate Secret key, K2 =A ^ y mod p = 2 ^ 6 mod 11 = 9
Key Exchange: Diffie-Hellman Alice Bob A = g ^ x mod n A K1 = B ^ x mod nK2 = A ^ y mod n B B = g ^ y mod n
Mathematical Theory Behind Algorithm First Alice find key K1 = B ^ x mod n but what is B ? B = g ^ y mod n, therefore if we Substitute this value of B in K1 then K1=(g ^ y)^x mod n = g ^ yx mod n Then Bob find key K2 = A ^ y mod n but what is A ? A = g ^ x mod n, therefore if we substitute this value of A in K2 then K2 = (g ^ x)^y mod n = g ^xy mod n Now Basic Maths says that: K^ yx = K^ xy
Similar presentations | {"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": 0.8470885157585144, "perplexity": 1456.722676558486}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039743714.57/warc/CC-MAIN-20181117144031-20181117170031-00409.warc.gz"} |
https://physics.stackexchange.com/questions/390949/if-f-ma-and-w-mg-why-is-using-f-ma-to-calculate-weight-wrong | # If F = $m*a$ and $W = m*g$, why is using $F = m*a$ to calculate weight wrong?
On a test I used $F=m*a$ to calculate the weight of an object on Earth, using $9.8$ $m/s^2$ for $a$. I got the right answer, but the question was marked as wrong because I used the wrong formula and I should have used $W=m*g$.
Aren't the two formulas the same, only that one is "specialized" to calculate weight? What am I missing?
The question was similar to "If an item has mass x, what is its weight on Earth?"
• Someone was picky to you. Just tell him: I decided to use F instead of W and a instead of g. Why? Because I prefer to use F for Weight and a for gravitational acceleration near the surface of earth, that’s why. – J. Manuel Mar 9 '18 at 9:11
You are interpreting $F = m * a$ incorrectly. First of all, it should be $\Sigma F = m * a$. The equation says that all forces applied to a mass $m$ will cause the mass to accelerate by $a$. The left hand side should include all relevant forces per the specific problem. The right hand side is "fixed" as $m*a$.
In your case $W=m*g$ is a force, caused by gravitation. Not to be confused with $m*a$.
Your weight is equal to the force exerted on you by gravity, which (assuming that you're near the surface of the Earth) you can calculate with $W=mg$, where $g=9.8 \ m/s^2$. This tells you nothing about your actual acceleration $a$, because your weight is the same whether
1. You are sitting still, in which case $a=0$, or
2. You are in an upward-accelerating elevator, in which case $a>0$, or
3. You are falling off of your roof, in which case $a<0$
(where I have defined $a>0$ to be upward).
Assuming it has not already been prescribed, to calculate your acceleration you need more information - you need to know the other forces acting on the object. It's a totally different question from calculating the object's weight, despite the apparent similarity in the formulae.
I make the case that there is insufficient difference between $F = ma$, and $W = mg$ to have deprived you of marks in that test.
Firstly, imagine a rocket with a flat front, firing away, but pinned against a sideways wall. Or imagine one of the recent electric drones, that has flown into a ceiling and is stuck there. In both cases there will be no motion. Acceleration is $a = F/m$, where F is due to thrust.
In classical mechanics, how is this fundamentally different from acceleration due to gravity? For all its differences from a rocket engine or a drone propeller, the earth is simply thrusting us towards its centre, is it not?
Secondly, there is the moon happily pulling upwards a part of the earth's ocean that it happens to be above, accelerating it by gravity but also causing motion. The moon generally varies the gravitational pull on earth. In terms of Newtonian mechanics, nothing sacred, just another force.
Thirdly, we have the kilogram-force: "The kilogram-force...is is equal to the magnitude of the force exerted by one kilogram of mass in a 9.80665 $m/s^{2}$ gravitational field" (Ref). Kilogram-force is no different from weight in magnitude. However, it can act in any direction. According to the ref, while newtons $N$ is standard, the kilogram-force is still used in China and by the European Space Agency.
$W = mg$ is merely a special case of $F = ma,\quad$.
There is no conceptual difference between the two.
I grant you permission to use my explanation to confront your teacher(s).
I think, it is a question of being very precise and following conventional notation. What was the exact context?
If the question was: "What is the weight of an object of mass $m$ on Earth?", and you answered: "The object's weight $F$ is $F =m \cdot a$ where $a = 9.81 m/s²$" you would be correct.
I think being precise is an important aspect in physics but this looks exaggerated to me...
• $F = ma$ is conventionally used for net force and net acceleration. Gravitational force has an equivalent acceleration on the surface; but using $F = ma$ implies that the object actually accelerates at $g$. – JMac Mar 8 '18 at 15:41
• As I said, this seems to be about convention. To my eyes, it all depends on the scope of the question but I would not dismiss the answer as absolutely wrong. – lmr Mar 8 '18 at 17:21
They do seem similar and I disagree that you should have lost points, but perhaps when using $F = ma$, it's implied that you're calculating a net force from a net acceleration or vice versa. When calculating weight maybe it's understood better by your teacher when you're using the formula. It does come from newtons law of universal gravitation after all ($F=K*(m_1*m_2)/r^2$ where you're only interested in the attraction between two objects and not other forces. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8552164435386658, "perplexity": 371.1385533671463}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998376.42/warc/CC-MAIN-20190617043021-20190617065021-00258.warc.gz"} |
https://forum.arduino.cc/t/is-this-a-short-circuit/481931 | # is this a short circuit????
Hi, i got come leds from Adafruit and their web tutorial recommend placing a 1000uF between positive and negative.
isnt it a short circuit?
i attach a photo from the tutorial.
thanks
Yes, it is a short for AC voltages. It is an open circuit for DC voltages because a capacitor contains an insulator between the two metallic plates. So, in your case, the capacitor is an insulator for DC, but would be a short or partial short circuit for AC or DC that varies rapidly in voltage value.
Paul
Paul_KD7HB:
Electrolytic capacitor
Paul_KD7HB:
but would be a short or partial short circuit for ... DC that varies rapidly in voltage value.
That doesn't sound right to me. Why would the impedance (resistance, in fact) be any less for rapidly changing DC than for constant or slowly changing DC? It's still DC: no polarity change and thus no return of the stored charge on opposite cycles.
Xc the captivate reactance = 1 / (2 pi f C)
Even if the DC varies rapidly, since the polarity never changes, f=0 and thus Xc is still infinite.
kenwood120s:
That doesn't sound right to me. Why would the impedance (resistance, in fact) be any less for rapidly changing DC than for constant or slowly changing DC? It's still DC: no polarity change and thus no return of the stored charge on opposite cycles.
Xc the captivate reactance = 1 / (2 pi f C)
Even if the DC varies rapidly, since the polarity never changes, f=0 and thus Xc is still infinite.
We are dealing with someone who doesn't know what a capacitor is or how it works an is unwilling to use Google, so I tried my best.
Paul
I'm not a huge fan of analogies, since people often push them too far and they break down but I like the analogy of the capacitor as a diaphragm in a pipe.
By that analogy, it's easy to see that with the current in one direction there's current for a moment while the diaphragm flexes and bulges, that's the capacitor charging. But once that stops, that's it. The pressure's still there, that's the voltage, but no current.
(Unless it's AC in which case the diaphragm returns the water in the bulge (the charge in the cap) whizzing back whence it came, rinse and repeat.)
So, if I put a DC offset on an AC signal so that it never crosses zero, it should not be affected by the capacitor?
KeithRB:
So, if I put a DC offset on an AC signal so that it never crosses zero, it should not be affected by the capacitor?
That's what I'm picturing yes, since f=0, but I'm not stating it as a fact, rather just my understanding. Certainly if f=0 then Xc if infinite from the equation. So I'll restate my understanding as a question: is f=0 for rapidly changing voltage that never crosses 0?
But that all said-maybe we should halt this discussion for OP's sake, and the answer to his question is "No, it's not a short circuit, since for the DC case in the pic, the capacitor charges up until the charge on the plates balances the applied voltage after which nothing moves. A cap is open circuit to DC."
(But I'd like to know if my thinking's flawed and maybe I'll open a thread on that.)
Edit: started this thread to discuss the varying DC stuff.
Any AC, without any regard to DC level, is affected by the capacitor. That is how power supply filters work.
Oh, and "rapidly changing" is not "f=0".
kenwood120s:
That's what I'm picturing yes, since f=0, but I'm not stating it as a fact, rather just my understanding. Certainly if f=0 then Xc if infinite from the equation. So I'll restate my understanding as a question: is f=0 for rapidly changing voltage that never crosses 0?
It appears that you don't know about Fourier analysis. Electrical signals do not possess just one single frequency, they are a sum of sinusoids with different amplitudes, frequencies, and/or phases.
A lot of people use "AC" not to refer to a signal that crosses 0, but a signal with non-zero frequency. I know it's not technically correct by a literal meaning of "alternating current", but that's just how it is.
kenwood120s:
That doesn't sound right to me. Why would the impedance (resistance, in fact) be any less for rapidly changing DC than for constant or slowly changing DC? It's still DC: no polarity change and thus no return of the stored charge on opposite cycles.
Xc the captivate reactance = 1 / (2 pi f C)
Even if the DC varies rapidly, since the polarity never changes, f=0 and thus Xc is still infinite.
I think DC means constant value. Physically hypothetically unchanging with time. If hypothetically or theoretically or physically changing with time in a sinusoidal fashion.... at some particular frequency......... then that's AC.
Also....theoretically.... if mathematically considering a time changing signal that is not purely sinusoidal in behaviour.... then it could '"mathematically" be considered as a collection of sinusoidal signals plus a constant unchanging signal if there is one ...... aka Fourier theory.
The capacitive reactance Xc expression has a negative sign. Eg. -1/(2Πf.C). At relatively high frequency, the capacitor impedance becomes relatively small......and at high enough frequency.... the ideal capacitor can be approximately a short circuit. At theoretical infinite frequency..... which we can never reach... the capacitor approaches a short circuit.
kenwood120s:
That doesn't sound right to me. Why would the impedance (resistance, in fact) be any less for rapidly changing DC than for constant or slowly changing DC?
DC is constant by definition. If something is changing it has a AC component. The impedance
depends on frequency (DC is just frequency zero). The AC component can get through the capacitor.
Hi,
As I pointed out in another thread.
AC CURRENT affects IMPEDANCE, not voltage.
If you vary the voltage on a capacitor form 2V to 5V to 2V to 5V etc, you have AC CURRENT, even though the applied voltage does not change sign.
2V to 5V, CHARGING the cap, current goes into the cap at its postive terminal.
then
5V to 2V DISCHARGING the cap, current goes out of the cap at its positive terminal.
Hence AC current.
Tom…
That is how an analog decoupling cap works in an amplifier.
(Sorry I don’t think they teach analog electronics in school/UNI anymore.) | {"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": 0.8743535876274109, "perplexity": 1535.680936035732}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103347800.25/warc/CC-MAIN-20220628020322-20220628050322-00736.warc.gz"} |
https://nikhildoomra.wordpress.com/category/sharepoint/sharepoint-2010/sharepoint-2010-management-shell/ | # SharePoint : Delete “Scheduling Start Date” and “Scheduling End Date” columns using PowerShell
In general, we can delete the columns using UI from SharePoint pretty easily. But some columns are not allowed to be deleted using UI but are allowed to be deleted using PowerShell. But there are some columns which are not deleted from PowerShell as well without some tweaks.
# Extract WSP Contents using PowerShell
Recently, I encountered a requirement to extract the wsps from SharePoint Content Database and extract its contents. Extraction of WSP is already mentioned in many blogs. So, I would just guide how to extract the WSP contents.
Following is the command which is used to extract the WSP contents and saves its contents to a particular path:
expand <FullPathToWSP> <DirectoryWhereFileNeedsTobeExtracted> -F:*
The above command is a built-in Windows utility which extracts the content of WSP. Following is the example of the above mentioned command.
expand C:\mySolution.wsp C:\SolutionExtract\mySolution -F:* | {"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": 0.915736734867096, "perplexity": 3956.0649484621717}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886103910.54/warc/CC-MAIN-20170817185948-20170817205948-00564.warc.gz"} |
http://mathhelpforum.com/differential-geometry/227028-simple-topology-problem-involving-continuity-print.html | # Simple topology problem involving continuity
• March 20th 2014, 03:17 AM
Bernhard
1 Attachment(s)
Simple topology problem involving continuity
Example 1 in James Munkres' book, Topology (2nd Edition) reads as follows:
Attachment 30471
Munkres states that the map p is 'readily seen' to be surjective, continuous and closed.
My problem is with showing (rigorously) that it is indeed true that the map p is continuous and closed.
Regarding the continuity of a function Munkres says the following on page 102:
--------------------------------------------------------------------------
Let $X$ and $Y$ be topological spaces. A function $f \ : \ X \to Y$ is said to be continuous if for each open subset V of Y, the set $f^{-1} (V)$ is an open subset of X.
---------------------------------------------------------------------------
Yes ... fine ... but how do we use such a definition to prove or demonstrate the continuity of p in the example?
Can someone show me how we use the definition (or some theorems) in practice to demonstrate/ensure continuity?
I have a similar issue with showing p to be a closed map.
On page 137 Munkres writes the following:
--------------------------------------------------------------------------
A map $p$ "is said to be a closed map if for each closed set $A$ of $X$ the set $p(A)$ is closed in $Y$"
---------------------------------------------------------------------------
Again, I understand the definition, I think, but how do we use it to indeed demonstrate/prove the closed nature of the particular map p in Munkres example?
Hope someone can help clarify the above issues?
Peter
NOTE: as an aside, if anyone can tell me how to get images to size to the size of the panel and hence be readable, I would appreciate it.
• March 20th 2014, 06:45 AM
SlipEternal
Re: Simple topology problem involving continuity
Let $U \subset [0,2]$ be open. Let $x \in p^{-1}(U)$ be any point. Since $U$ is open in $[0,2]$, there exists $r>0$ such that $B_r(p(x)) \subset U$. Thus, $B_r(x) \subset p^{-1}(U)$ (If you really want, this can be proven more rigorously by sending $B_r(x)$ through $p$ and showing the image is a subset of $U$). Hence, $p$ is continuous.
To show that it is closed, let $F \subset [0,1]\cup [2,3]$ be a closed set. Let $y \in [0,2]\setminus \left(p(F)\cup \{1\}\right)$ (I am excluding 1 to avoid multiple preimages, as that is the only point with multiple preimages. I will handle that case next). Since $F$ is closed in $[0,1]\cup [2,3]$, there exists $r_0>0$ such that $B_r(p^{-1}(y)) \subset [0,1]\cup [2,3] \setminus F$. Let $r = \min\left\{r_0,\dfrac{|1-y|}{2}\right\}$. Thus, $B_r(y) \subset [0,2]\setminus p(F)$. (Again, if you really want, you can send $B_r(p^{-1}(y))$ through $p$ and show that you get $B_r(y)$).
If $1 \in [0,2]\setminus p(F)$, then you need to handle that separately. Since $p^{-1}(1) = \{1,2\}$, you need to find two radii. Choose $r_1>0$ such that $B_{r_1}(1) \subset [0,1]\cup [2,3] \setminus F$ and choose $r_2>0$ such that $B_{r_2}(2) \subset [0,1]\cup [2,3]\setminus F$. Let $r = \min\{r_1,r_2\}$. Now, $B_r(y) \subset [0,2]\setminus p(F)$. (To show this more rigorously, you should send $B_r(1) \cup B_r(2)$ through $p$ to show you get $B_r(y)$).
This shows $[0,2] \setminus p(F)$ is open, so $p(F)$ is closed. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "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": 0.9777868986129761, "perplexity": 199.16349430751052}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783395546.12/warc/CC-MAIN-20160624154955-00103-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://arxiv.org/abs/1109.4680 | cs.SI
# Title:The Push Algorithm for Spectral Ranking
Abstract: The push algorithm was proposed first by Jeh and Widom in the context of personalized PageRank computations (albeit the name "push algorithm" was actually used by Andersen, Chung and Lang in a subsequent paper). In this note we describe the algorithm at a level of generality that make the computation of the spectral ranking of any nonnegative matrix possible. Actually, the main contribution of this note is that the description is very simple (almost trivial), and it requires only a few elementary linear-algebra computations. Along the way, we give new precise ways of estimating the convergence of the algorithm, and describe some of the contribution of the existing literature, which again turn out to be immediate when recast in our framework.
Subjects: Social and Information Networks (cs.SI); Data Structures and Algorithms (cs.DS); Physics and Society (physics.soc-ph) Cite as: arXiv:1109.4680 [cs.SI] (or arXiv:1109.4680v1 [cs.SI] for this version)
## Submission history
From: Sebastiano Vigna [view email]
[v1] Thu, 22 Sep 2011 00:42:48 UTC (9 KB) | {"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": 0.8220624923706055, "perplexity": 1660.4318560870092}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027330750.45/warc/CC-MAIN-20190825151521-20190825173521-00429.warc.gz"} |
https://www.physicsforums.com/threads/volume-of-solid-of-revolution-y-axis.723283/ | # Volume of solid of revolution - y axis.
1. Nov 17, 2013
### Mutaja
1. The problem statement, all variables and given/known data
Find the volume of the solid of revolution when we rotate the area limited by the x axis and the function f(x) = 1 - cosx where x e [0, 2∏] once around the y-axis?
3. The attempt at a solution
In my notes I have the following equation:
V = ∫ 2∏x f(x) dx
If I put in my limits (upper limit 2∏, lower limit 0) and my function I get the following:
V = 2∏ ∫x(1-cos(x)) dx
V = 2∏ ∫x - xcos(x) dx
V = 2∏[$\frac{x^2}{2}$ - (xsin(x)+cos(x))]
V = 2∏ [$\frac{x^2}{2}$ - (∏sin(∏) + cos(∏)] - 2∏ [$\frac{0^2}{2}$ - (0sin(0) + cos(0)]
Since ∏ sin(∏) = 0, cos(∏) = -1 , 0sin(0) = 0 and cos(0) = 1 I get the following:
V = 2∏ ($\frac{∏^2}{2}$) - 2∏ + 1
V = $2∏^3$ - 4∏ + 2
Is this correct? Am I using the correct formulas/equations?
Please let me know if there is something I need to explain better. Any help and guiding is massively appreciated. Thanks.
Last edited: Nov 17, 2013
2. Nov 17, 2013
### PeroK
Check your integration by parts: should be -cos(x).
3. Nov 17, 2013
### Mutaja
Sloppy mistake by me, there should of course be parenthesis around that equation.
V = 2∏[$\frac{x^2}{2}$ - (xsin(x)+cos(x))]
That slipped past me when I was writing off of my notes - therefore unless I again have overlooked something (which I've double checked I haven't), that mistake was a one off - I've solved the rest of the problem as if there were parenthesis around (xsin(x)+cos(x)). Therefore I get - (-1) -> +1 in my answer which I later multiply by 2.
Does it look ok besides that error?
Draft saved Draft deleted
Similar Discussions: Volume of solid of revolution - y axis. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.9023486375808716, "perplexity": 2033.1860803971301}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886106754.4/warc/CC-MAIN-20170820131332-20170820151332-00244.warc.gz"} |
https://www.physicsforums.com/threads/smallest-unit-of-time.208729/ | # Smallest unit of time
1. ### Mephisto
93
i read somewhere that the smallest unit of time that makes sense is the Planck time, ~10^-43 seconds. I'd like to know how physicists arrived at this number. Also, why is it so impossible to think of a time step lower than that? What equation exactly is in conflict, and what kind of a conflict is it? For example, do you get a division by 0 if you try to use a lower time, somehow? or what happens?
thanks
2. ### nicksauce
1,275
From wikipedia:
"The following thought experiment illuminates this fact. The task is to measure an object's position by bouncing electromagnetic radiation, namely photons, off it. The shorter the wavelength of the photons, and hence the higher their energy, the more accurate the measurement. If the photons are sufficiently energetic to make possible a measurement more precise than a Planck length, their collision with the object would, in principle, create a minuscule black hole. This black hole would "swallow" the photon and thereby make it impossible to obtain a measurement. A simple calculation using dimensional analysis suggests that this problem arises if we attempt to measure an object's position with a precision to within a Planck length."
The Planck time is the time it takes for a particle traveling at c to cross the Planck length.
Also, dimensional analysis shows that the Planck length is the only length that can be obtained from some mix of the 3 important physical constants G, h, c (up to a scale factor), so this shows that it might (must?) have some physical importance. If space and time are discrete it would seem that the Planck length and time would be the natural units for discreteness.
Last edited: Jan 13, 2008
3. ### pam
455
The "Planck time" is just a convenient unit of time for some applications.
It is not a "smallest unit of time" and there is no reason that shorter time intervals
don't exist.
4. ### lazypast
82
that text from wikipedia is so clear. i always thought plank time was the time at which after the big bang no one can describe.
like from t=0s to t=1^-43s physics breaksdown and doesnt describe this period
is this all wrong in which case ive plucked from absolute nothing??
5. ### yogi3939
11
To my knowledge there is no "shortest unit of time". There is only an inability to measure time to determine if one actually exists. So far conventional thinking is divided between the uniform flow camp and the discrete unit camp and no one has devised a way to prove or disprove either one. | {"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": 0.8244112133979797, "perplexity": 582.6552548163025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246633972.52/warc/CC-MAIN-20150417045713-00145-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://ehsmanager.blogspot.com/2011/04/roads-as-solar-and-piezo-electric.html | ## Apr 24, 2011
### Roads as Solar and Piezo-Electric Generators
Another use of a free, wasted byproduct to generate electricity is piezo-electric energy. "Piezo" means pressure. Anything that produces pressure can produce energy.
For example, a train station in Japan installed piezo-electric equipment in the ground, so that the foot traffic of those walking through the train station generates electricity (turnstiles at train, subway and ferry stations, ballparks and amusement parks can also generate electricity).
Similarly, all exercise machines at the gym or at home can be hooked up to produce electricity.
But perhaps the greatest untapped sources of piezo-electric energy are freeways and busy roads. If piezo-electric mats were installed under the busiest sections [a little ways under the surface], the thousands of tons of vehicles passing over each day would generate massive amounts of electricity for the city's use.
A couple of readers thought that sounded nuts.
But as TreeHugger notes today:
Denmark will have to come up with something big to match the latest plan from the Netherlands - the installation of solar panels in roads, starting with bike lanes.
Talk about the efficient use of space: if you're going to have roads (and hopefully you'll have bike lanes), why not put that space to work producing energy? Called the Solaroad, the project is the brainchild of Dutch research firm TNO. The idea is pretty straightforward: a layer of concrete forms the road itself. A centimeter thick layer of crystalline silicon solar cells is laid on top, and covered by a layer of toughened glass. The energy potential: 50kWh per square meter per year, which can then be used to power street lighting, traffic systems and households.
...Scheduled for installation next year, the first Solaroad will hopefully allow its developers better implement many more throughout the country.
Now why not put a piezo-electric mat under the crystalline silicon solar cells, under the layer of toughened glass?
We'd get two different forms of energy generation at once...
The U.S. Wastes More Energy Than it Uses - Partly Because of the Centralization of Power
As shown by the following graphic from Lawrence Berkeley National Laboratory, the U.S. wastes a lot more energy than it uses:
(click for giant graphic.)
America uses 39.97 quads of energy, while it wastes 54.64 quads (i.e. "rejected energy").
As CNET noted in 2007:
Sixty-two percent of the energy consumed in America today is lost through transmission and general inefficiency. In other words, it doesn't go toward running your car or keeping your lights on.
Put another way:
• We waste 650% more energy than all of our nuclear power plants produce
• We waste 280% more energy than we produce by coal
• We waste 235% more energy than we produce by natural gas (using deadly fracking)
• We waste 150% more energy than we generate with other petroleum products
The Department of Energy notes:
Only about 15% of the energy from the fuel you put in your tank gets used to move your car down the road or run useful accessories, such as air conditioning. The rest of the energy is lost to engine and driveline inefficiencies and idling. Therefore, the potential to improve fuel efficiency with advanced technologies is enormous.
According to the DOE, California lost 6.8% of the total amount of electricity used in the state in 2008 through transmission line inefficiencies and losses.
By the time energy is delivered to us in a usable form, it has typically undergone several conversions. Every time energy changes forms, some portion is “lost.” It doesn't disappear, of course. In nature, energy is always conserved. That is, there is exactly as much of it around after something happens as there was before. But with each change, some amount of the original energy turns into forms we don't want or can't use, typically as so-called waste heat that is so diffuse it can't be captured.
Reducing the amount lost – also known as increasing efficiency – is as important to our energy future as finding new sources because gigantic amounts of energy are lost every minute of every day in conversions. Electricity is a good example. By the time the energy content of electric power reaches the end user, it has taken many forms. Most commonly, the process begins when coal is burned in a power station. The chemical energy stored in the coal is liberated in combustion, generating heat that is used to produce steam. The steam turns a turbine, and that mechanical energy is used to turn a generator to produce the electricity.
In the process, the original energy has taken on a series of four different identities and experienced four conversion losses. A typical coal-fired electrical plant might be 38% efficient, so a little more than one-third of the chemical energy content of the fuel is ultimately converted to usable electricity. In other words, as much as 62% of the original energy fails to find its way to the electrical grid. Once electricity leaves the plant, further losses occur during delivery. Finally, it reaches an incandescent lightbulb where it heats a thin wire filament until the metal glows, wasting still more energy as heat. The resulting light contains only about 2% of the energy content of the coal used to produce it. Swap that bulb for a compact fluorescent and the efficiency rises to around 5% – better, but still a small fraction of the original.
Example of energy lost during conversion and transmission. Imagine that the coal needed to illuminate an incandescent light bulb contains 100 units of energy when it enters the power plant. Only two units of that energy eventually light the bulb. The remaining 98 units are lost along the way, primarily as heat.
Read more from "It's Not Just Alternative Energy Versus Fossil Fuels or Nuclear - Energy Has to Become DECENTRALIZED" | {"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": 0.8049251437187195, "perplexity": 1654.964767322575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645367702.92/warc/CC-MAIN-20150827031607-00154-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://theanalysisofdata.com/probability/1_7.html | ## Probability
### The Analysis of Data, volume 1
Probability and Measure Theory
## 1.7. Probability and Measure Theory*
Definition 1.2.1 appears to be formal, and yet is not completely rigorous. It states that a probability function $\P$ assigns real values to events $E\subset \Omega$ in a manner consistent with the three axioms. The problem is that the domain of the probability function $\P$ is not clearly specified. In other words, if $\P$ is a function $\P:\mathcal{F}\to\R$ from a set $\mathcal{F}$ of subsets of $\Omega$ to $\R$, the set $\mathcal{F}$ is not specified. The importance of this issue stems from the fact that the three axioms need to hold for all sets in $\mathcal{F}$.
At first glance this appears to be a minor issue that can be solved by choosing $\mathcal{F}$ to be the power set of $\Omega$: $2^{\Omega}$. This works nicely whenever $\Omega$ is finite or countably infinite. But selecting $\mathcal{F}=2^{\Omega}$ does not work well for uncountably infinite $\Omega$ such as continuous spaces. It is hard to come up with useful functions $\P:2^{\Omega}\to\R$ that satisfy the three axioms for all subsets of $\Omega$.
A satisfactory solution that works for uncountably infinite $\Omega$ is to define $\mathcal{F}$ to be a $\sigma$-algebra of subsets of $\Omega$ that is smaller than $2^{\Omega}$. In particular, when $\Omega\subset\R^d$, the Borel $\sigma$-algebra is sufficiently large to include the "interesting" subsets of $\Omega$ and yet is small enough to not restrict $\P$ too much (see Section E.1 for a definition of a $\sigma$-algebra and the Borel $\sigma$-algebra).
We also note that a probability function $\P$ is nothing but a measure $\mu$ on a measurable space $(\Omega,\mathcal{F})$ satisfying $\mu(\Omega)=1$. In other words, the triplet $(\Omega,\mathcal{F},\P)$ is a measure space where $\mathcal{F}$ is the $\sigma$-algebra of measurable sets and $\P$ is a measure satisfying $\P(\Omega)=1$. Thus, the wide array of mathematical results from measure theory (Chapter E) and Lebesgue integration (Chapter F) are directly applicable to probability theory. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9797101616859436, "perplexity": 85.79019221020029}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189092.35/warc/CC-MAIN-20170322212949-00356-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://braindump.jethro.dev/posts/topic_modelling/ | # Topic Modeling
tags
Machine Learning
## LDA
#### Dirichlet Distribution
https://www2.ee.washington.edu/techsite/papers/documents/UWEETR-2010-0006.pdf
Dirichlet distribution is a family of continuous multivariate probability distributions parameterized by a vector α of positive reals.
$$\theta \sim Dir(\alpha)$$
$$p(\theta) = \frac{1}{\beta(\alpha)} \prod_{i=1}^n \theta_i^{\alpha_i-1} I(\theta \in S)$$
Where $$\theta = (\theta_1, \theta_2, \dots, \theta_n), \alpha = (\alpha_1, \alpha_2, \dots, \alpha_n), \alpha_i > 0$$ and
$$S = \left\{x \in \mathbb{R}^n : x_i \ge 0, \sum_{i=1}^{n} x_i = 1 \right\}$$
and $$\frac{1}{\beta(\alpha)} = \frac{\Gamma(\alpha_0)}{\Gamma(\alpha_1)\Gamma(\alpha_2)\dots\Gamma(\alpha_n)}$$
The infinite-dimensional generalization of the Dirichlet distribution is the Dirichlet process.
The Dirichlet distribution is the conjugate prior distribution of the categorical distribution (a generic discrete probability distribution with a given number of possible outcomes) and multinomial distribution (the distribution over observed counts of each possible category in a set of categorically distributed observations). This means that if a data point has either a categorical or multinomial distribution, and the prior distribution of the distribution’s parameter (the vector of probabilities that generates the data point) is distributed as a Dirichlet, then the posterior distribution of the parameter is also a Dirichlet.
#### Exploring a Corpus with the posterior distribution
Quantities needed for exploring a corpus are the posterior expectations of hidden variables. Each of these quantities are conditioned on the observed corpus.
Visualizing a topic is done by visualizing the posterior topics through their per-topic probabilities $$\hat{\beta}$$.
Visualizing a document uses the posterior topic proportions $$\hat{\theta}_{d,k}$$ and the posterior topic assignments $$\hat{z}_{d,k}$$.
Finding similar documents can be done through the Hellinger distance:
\begin{align*} D_{d,k} = \sum_{k=1}^K \left( \sqrt{\hat{\theta}_{d,k}} - \sqrt{\hat{\theta}_{f,k}}\right)^2 \end{align*}
#### Posterior Inference
• Mean Field Variational Inference
Approximate intractable posterior distribution with a simpler distribution containing free variational parameters. These parameters are fit to approximate the true posterior.
In contrast to the true posterior, the mean field variational distribution for LDA is one where the variables are independent of each other, with and each governed by a different variational parameter.
We fit the variational parameters to minimise the KL-divergence to the true posterior.
The general approach to mean-field variational methods - update each variational parameter with the parameter given by the expectation of the true posterior under the variational distribution - is applicable when the conditional distribution of each variable is the exponential family.
#### Markov Chains
http://setosa.io/ev/markov-chains/
#### Shortcomings
• strong, potentially invalid statistical assumptions:
• topics have no correlation to one another (dirichlet assumes nearly independent)
• solution: CTM: use a logistic normal distribution
• assumes order of documents don’t matter
• solution: DTM: use logistic normal distribution to model topics evolving over time
## TopicRNN
http://www.columbia.edu/~jwp2128/Papers/DiengWangetal2017.pdf
In TopicRNN, latent topic models are used to capture global semantic dependencies so that the RNN can focus its modeling capacity on the local dynamics of the sequences | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 43, "equation": 410, "x-ck12": 0, "texerror": 0, "math_score": 0.9736708402633667, "perplexity": 1364.0018701019371}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00484.warc.gz"} |
http://mathhelpforum.com/pre-calculus/118436-problem-about-vectors.html | # Math Help - Problem about vectors:
1. ## Problem about vectors:
"Find the coordinates of the point which divides the line segment joining A(2,3,5) to B(3, -1, 1) externally in the ratio 2 : 5. "
How does one solve this problem? And what does "externally" mean?
2. Hello karldiesen
Originally Posted by karldiesen
"Find the coordinates of the point which divides the line segment joining A(2,3,5) to B(3, -1, 1) externally in the ratio 2 : 5. "
How does one solve this problem? And what does "externally" mean?
Look at the attached diagram. $P_1$ divides $AB$ internally in the ratio $2:5$ if $AP_1:P_1B = 2:5$; and $P_2$ divides $AB$ externally in the ratio $2:5$ if $P_2A:P_2B = 2:5$; or (taking account of the directions) $AP_2:P_2B = -2:5$.
Now if points $A$ and $B$ have position vectors $\textbf{a}$ and $\textbf{b}$ respectively, then the point $P$ that divides the line segment $AB$ in the ratio $\lambda:\mu$ has position vector
$\textbf{p}=\frac{\mu\textbf{a}+\lambda\textbf{b}}{ \lambda + \mu}$
(It's quite straightforward to prove this: begin with $\textbf{AB} = \textbf{b}-\textbf{a}$, and then say $\textbf{p}= \textbf{a}+\frac{\lambda}{\lambda+\mu}\textbf{AB}$ ... etc)
So you can simply plug the position vectors of $A$ and $B$ into this formula with $\lambda = -2$ and $\mu = 5$. If my arithmetic is correct, I reckon the answer is $(\tfrac43, \tfrac{17}{3},\tfrac{23}{3})$.
Grandad
Attached Thumbnails | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "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": 0.820510745048523, "perplexity": 291.51222299930726}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510271648.4/warc/CC-MAIN-20140728011751-00247-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://cstheory.stackexchange.com/questions/14987/learning-mixture-of-univariate-gaussians | # Learning Mixture of Univariate Gaussians
There are many papers on learning mixtures of multivariate Gaussians, which exploit various separation/projection techniques. What about one-dimensional (univariate) Gaussians -- any formal guarantees for learning these?
UPDATE: I should clarify: I am particularly interested in polynomial-time algorithms, in the number of mixture components.
-
Sorry, but there is lots of evidence that not only is this problem a pain in the univariate case, but moreover that having few dimensions actually makes things worse.
For evidence of the difficulty, please see Proposition 15 in Section 6 ("Exponential dependence on $k$ is inevitable") and the related Figure 1 of the following paper, the follow-up to Ankur Moitra and Greg Valiant's original joint work with Adam Kalai:
Ankur Moitra and Greg Valiant, "Settling the Polynomial Learnability of Mixtures of Gaussians", FOCS 2010.
Proposition 15 states that there exist two mixtures of $k^2$ Gaussians on the line whose $L^1$ distance is exponentially small in $k$. Since the construction is careful and the distribution parameters are well-behaved (polynomially-sized), this dependence on $k$ is real, and thus any algorithm whose performance depends on statistical distance will run into trouble.
For evidence that few dimensions actually forms a bad regime, please see the following very recent work:
Daniel Hsu and Sham Kakade, "Learning mixtures of spherical Gaussians: moment methods and spectral decompositions", ITCS 2013.
This work presents a spectral method for recovering spherical mixtures of Gaussians. So far, so good: univariate Gaussians are spherical. However, this algorithm requires the $k$ individual means to span a rank-$k$ subspace (and the performance depends on the condition number of these points, meaning you can't just perturb your univariate instance in a $k$-dimensional space and then closely recover your original Gaussians). As such, you could view their circumvention of the Moitra-Valiant lower bound as coming from an increase in dimension (and of course by maintaining the speherical nature you find from 1-d).
The above two papers have lots of excellent discussion, techniques, nice references. This is not my field and I can only offer a very surface intuition for what is going on. In the univariate case, things are really crowded, so the contribution of different Gaussians is confused and it's hard to disentangle them. In the high-dimensional case, basically everything is far apart and low density; it's hard to get crowded, and easy to pick out points of high density (which now must reside near means).
I only have one other thing to point out, and that's the connection to $k$-means. $k$-means has a DP in the 1-d case; google tells me it's from the 40s, and while I haven't seen that reference, we can cook up a DP fairly easily: make your table have size $n\times k$, where $n$ is the number of points, and use "optimal $k$-means cost using $j\leq k$ means and the first $m\leq n$ data points" as your subproblem, noting that each mean owns a contiguous subset of the data. As I tried just now to adjust this to the Gaussians case, it was interesting to see how the differences between these two scenarios became acute. First, the hard vs. soft clustering aspect means we can't just knock out contiguous subsets and claim they are owned by some Gaussian (furthermore, for univariate Gaussians of differing variances, the regions where one or the other dominates aren't just two intervals..). Second, we have no relationship between the variances, the univariate assumption really doesn't help us there (it would seem we need more; and of course, $k$-means is like knowing a fixed (common) variance up front).
So, oh well.
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Thanks for the detailed and informative answer, matus! – Aryeh Jan 11 '13 at 7:27
However, let me follow this up with another question. What about adding some assumption on the separation -- not in the statistical distance, but in actual parameter space? – Aryeh Jan 11 '13 at 7:29
@Aryeh: There's a table in Belkin&Sinha's second paper (referenced by siuman) which summarizes the separation assumptions needed in prior work: see Table 1 on page 2 of arxiv.org/abs/1004.4864 ("Polynomial learning of distribution families", also FOCS 2010). The best ones seem to say you need the means to be separated by (standard deviation) * k for the algorithm to work (i.e., performance doesn't simply degrade when separation is smaller). I think that table is up to date; some nice, recent, overview discussion is in the intro to the Hsu&Kakade paper I mentioned in my answer. – matus Jan 11 '13 at 8:08
Thanks again, matus! – Aryeh Jan 11 '13 at 8:12
Since univariate Gaussians are a special case of multivariate Gaussians, in terms of formal guarantees (e.g. provable bounds on run time, confidence, etc.), all papers on learning mixtures of multivariate Guassians also give results on univariate Gaussians. (And often the papers start with the simpler case of learning univariate Gaussians, then reduce the general case of learning multivariate Gaussians to the special case by some tricks, e.g. projections).
Two recent papers on learning mixtures of Gaussians, based on the method of moments:
1. Disentangling Gaussians by Adam Tauman Kalai, Ankur Moitra, and Gregory Valiant. (CACM 2012 and manuscript).
2. Polynomial Learning of Distribution Families by Mikhail Belkin, Kaushik Sinha. (FOCS 2010 and arXiv). See also Table 1 for further references.
They give algorithms for learning mixtures of Gaussians with time bounds polynomial in (the inverse of) precision $\epsilon$, confidence error $\delta$, and some "condition number" measuring the separation in parameters (for each fixed number of Gaussians in the mixture).
-
Right, but none of these is polynomial in the number of components. I'll update the question. – Aryeh Jan 8 '13 at 6:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.867355465888977, "perplexity": 1080.4573411729195}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637900080.19/warc/CC-MAIN-20141030025820-00014-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://science.sciencemag.org/content/198/4322/1153 | Reports
# Water Vapor Maser "Turn-On" in the HII Region W3 (OH)
See allHide authors and affiliations
Science 16 Dec 1977:
Vol. 198, Issue 4322, pp. 1153-1155
DOI: 10.1126/science.198.4322.1153
## Abstract
A line in the water vapor maser spectrum of the HII region W3(OH) was observed to increase in brightness over an 8-day period and then decline to its original intensity over the following 4 weeks. The intensity variation can be explained by a simple maser model, with an impulse of energy suddenly applied. The observed time scale and energy output are consistent with a maser on the outskirts of a dust cocoon surrounding an O5 star, with a momentary "leak," lasting a day or. two, supplying the necessary energy. | {"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": 0.9682385325431824, "perplexity": 4649.768344972515}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376829997.74/warc/CC-MAIN-20181218225003-20181219011003-00522.warc.gz"} |
https://www.easychair.org/publications/preprint/Wh5Z | Maple Program for MC 2021 Paper: a Machine Proof of an Inequality for the Sum of Distances between Four Points on the Unit Hemisphere using Maple Software
EasyChair Preprint no. 6349
26 pagesDate: August 23, 2021
Abstract
In this document, we present the Maple program for our
Maple Conference 2021 paper, where we proved a geometrical inequality
which states that for any four points on the unit hemisphere, the largest
sum of distances between the points is $4+4\sqrt{2}$ using Maple computation.
In our proof, we have constructed a rectangular neighborhood of the local
maximum point in the feasible set, which size is explicitly determined,
and proved that (1): the objective function is bounded by a quadratic
polynomial which takes the local maximum point as the unique critical
point in the neighbor- hood, and (2): the rest part of the feasible set can
be partitioned into a finite union of a large number of very small cubes
so that on each small cube the conjecture can be verified by estimating
the objective function with exact numerical computation. The attched
Maple program is for the second part. The work on first part, i.e., the
construction of the critical neighborhood, has been published recently in
the ADG 2021 (the Thirteenth International Conference on Automated
Deduction in Geometry), where we have proved that the sum of distances
between points contained in the constructed neighborhoods is not larger
than $4+4\sqrt{2}$, also using Maple computatin.
Keyphrases: Branch and Bound, computational geometry, global search algorithm, inequality, Maple program | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9758040904998779, "perplexity": 750.9506848209134}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943750.71/warc/CC-MAIN-20230322051607-20230322081607-00688.warc.gz"} |
https://appliedprobability.blog/category/probability/ | ## Continuous Probability Distributions
We consider distributions that have a continuous range of values. Discrete probability distributions where defined by a probability mass function. Analogously continuous probability distributions are defined by a probability density function.
(This is a section in the notes here.)
## Discrete Probability Distributions
There are some probability distributions that occur frequently. This is because they either have a particularly natural or simple construction. Or they arise as the limit of some simpler distribution. Here we cover
• Bernoulli random variables
• Binomial distribution
• Geometric distribution
• Poisson distribution.
(This is a section in the notes here.)
## Counting Principles
(This is a section in the notes here.)
Counting in Probability. If each outcome is equally likely, i.e. $\mathbb P( \omega ) = p$ for all $\omega \in \Omega$, then since
(where $|\Omega|$ is the number of outcomes in the set $\Omega$ ) it must be that
## Exponential Families
The exponential family of distributions are a particularly tractable, yet broad, class of probability distributions. They are tractable because of a particularly nice [Fenchel] duality relationship between natural parameters and moment parameters. Moment parameters can be estimated by taking the empirical mean of sufficient statistics and the duality relationship can then recover an estimate of the distributions natural parameters. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 6, "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": 0.9836527109146118, "perplexity": 696.7044348591185}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334515.14/warc/CC-MAIN-20220925070216-20220925100216-00548.warc.gz"} |
http://math.libretexts.org/TextMaps/Precalculus_Textmaps/Map%3A_Elementary_Trigonometry_(Corral)/5%3A_Graphing_and_Inverse_Functions/5.1%3A_Graphing_the_Trigonometric_Functions | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 5.1: Graphing the Trigonometric Functions
The first function we will graph is the sine function. We will describe a geometrical way to create the graph, using the unit circle. This is the circle of radius $$1$$ in the $$xy$$-plane consisting of all points $$(x,y)$$ which satisfy the equation $$x^2 + y^2 = 1$$.
Figure 5.1.1
We see in Figure 5.1.1 that any point on the unit circle has coordinates $$(x,y)=(\cos\;\theta,\sin\;\theta)$$, where $$\theta$$ is the angle that the line segment from the
origin to $$(x,y)$$ makes with the positive $$x$$-axis (by definition of sine and cosine). So as the point $$(x,y)$$ goes around the circle, its $$y$$-coordinate is $$\sin\;\theta$$.
We thus get a correspondence between the $$y$$-coordinates of points on the unit circle and the values $$f(\theta)=\sin\;\theta$$, as shown by the horizontal lines from the unit circle to the graph of $$f(\theta)=\sin\;\theta$$ in Figure 5.1.2 for the angles $$\theta = 0$$, $$\tfrac{\pi}{6}$$, $$\tfrac{\pi}{3}$$, $$\tfrac{\pi}{2}$$.
Figure 5.1.2 Graph of sine function based on $$y$$-coordinate of points on unit circle
We can extend the above picture to include angles from $$0$$ to $$2\pi$$ radians, as in Figure 5.1.3. This illustrates what is sometimes called the unit circle definition of the sine function.
Figure 5.1.3 Unit circle definition of the sine function
Since the trigonometric functions repeat every $$2\pi$$ radians ($$360^\circ$$), we get, for example, the following graph of the function $$y=\sin\;x$$ for $$x$$ in the interval $$[-2\pi , 2\pi]$$:
Figure 5.1.4
Graph of $$y = \sin x$$
To graph the cosine function, we could again use the unit circle idea (using the $$x$$-coordinate of a point that moves around the circle), but there is an easier way. Recall from Section 1.5 that $$\cos\;x = \sin\;(x+90^\circ)$$ for all $$x$$. So $$\cos\;0^\circ$$ has the same value as $$\sin\;90^\circ$$, $$\cos\;90^\circ$$ has the same value as $$\sin\;180^\circ$$, $$\cos\;180^\circ$$ has the same value as $$\sin\;270^\circ$$, and so on. In other words, the graph of the cosine function is just the graph of the sine function shifted to the left by $$90^\circ = \pi/2$$ radians, as in Figure 5.1.5:
Figure 5.1.5 Graph of $$y = \cos x$$
To graph the tangent function, use $$\tan\;x = \frac{\sin\;x}{\cos\;x}$$ to get the following graph:
Figure 5.1.6 Graph of $$y = \tan x$$
Recall that the tangent is positive for angles in QI and QIII, and is negative in QII and QIV, and that is indeed what the graph in Figure 5.1.6 shows. We know that $$\tan\;x$$ is not defined when $$\cos\;x = 0$$, i.e. at odd multiples of $$\frac{\pi}{2}$$: $$x=\pm\,\frac{\pi}{2}$$, $$\pm\,\frac{3\pi}{2}$$, $$\pm\,\frac{5\pi}{2}$$, etc. We can figure out what happens near those angles by looking at the sine and cosine functions. For example, for $$x$$ in QI near $$\frac{\pi}{2}$$, $$\sin\;x$$ and $$\cos\;x$$ are both positive, with $$\sin\;x$$ very close to $$1$$ and $$\cos\;x$$ very close to $$0$$, so the quotient $$\tan\;x = \frac{\sin\;x}{\cos\;x}$$ is a positive number that is very large. And the closer $$x$$ gets to $$\frac{\pi}{2}$$, the larger $$\tan\;x$$ gets. Thus, $$x=\frac{\pi}{2}$$ is a vertical asymptote of the graph of $$y=\tan\;x$$.
Likewise, for $$x$$ in QII very close to $$\frac{\pi}{2}$$, $$\sin\;x$$ is very close to $$1$$ and $$\cos\;x$$ is negative and very close to $$0$$, so the quotient $$\tan\;x = \frac{\sin\;x}{\cos\;x}$$ is a negative number that is very large, and it gets larger in the negative direction the closer $$x$$ gets to $$\frac{\pi}{2}$$. The graph shows this. Similarly, we get vertical asymptotes at $$x=-\frac{\pi}{2}$$, $$x=\frac{3\pi}{2}$$, and $$x=-\frac{3\pi}{2}$$, as in Figure 5.1.6. Notice that the graph of the tangent function repeats every $$\pi$$ radians, i.e. two times faster than the graphs of sine and cosine repeat.
The graphs of the remaining trigonometric functions can be determined by looking at the graphs of their reciprocal functions. For example, using $$\csc\;x = \frac{1}{\sin\;x}$$ we can just look at the graph of $$y=\sin\;x$$ and invert the values. We will get vertical asymptotes when $$\sin\;x=0$$, namely at multiples of $$\pi$$: $$x=0$$, $$\pm\,\pi$$, $$\pm\,2\pi$$, etc. Figure 5.1.7 shows the graph of $$y=\csc\;x$$, with the graph of $$y=\sin\;x$$ (the dashed curve) for reference.
Figure 5.1.7 Graph of $$y = \csc x$$
Likewise, Figure 5.1.8 shows the graph of $$y=\sec\;x$$, with the graph of $$y=\cos\;x$$ (the dashed curve) for reference. Note the vertical asymptotes at $$x=\pm\,\frac{\pi}{2}$$, $$\pm\,\frac{3\pi}{2}$$. Notice also that the graph is just the graph of the cosecant function shifted to the left by $$\frac{\pi}{2}$$ radians.
Figure 5.1.8 Graph of $$y = \sec x$$
The graph of $$y=\cot\;x$$ can also be determined by using $$\cot\;x = \frac{1}{\tan\;x}$$. Alternatively, we can use the relation $$\cot\;x = -\tan\;(x+90^\circ)$$ from Section 1.5, so that the graph of the cotangent function is just the graph of the tangent function shifted to the left by $$\frac{\pi}{2}$$ radians and then reflected about the $$x$$-axis, as in Figure 5.1.9:
Figure 5.1.9 Graph of $$y = \cot x$$
Example 5.1
Draw the graph of $$y=-\sin\;x$$ for $$0 \le x \le 2\pi$$.
Solution:
Multiplying a function by $$-1$$ just reflects its graph around the $$x$$-axis. So reflecting the graph of $$y=\sin\;x$$ around the $$x$$-axis gives us the graph of $$y=-\sin\;x$$:
\noindent Note that this graph is the same as the graphs of $$y=\sin\;(x \pm \pi)$$ and $$y=\cos\;(x+\frac{\pi}{2})$$.
It is worthwhile to remember the general shapes of the graphs of the six trigonometric functions, especially for sine, cosine, and tangent. In particular, the graphs of the sine and cosine functions are called sinusoidal curves. Many phenomena in nature exhibit sinusoidal behavior, so recognizing the general shape is important.
Example 5.2
Draw the graph of $$y=1+\cos\;x$$ for $$0 \le x \le 2\pi$$.
Solution:
Adding a constant to a function just moves its graph up or down by that amount, depending on whether the constant is positive or negative, respectively. So adding $$1$$ to $$\cos\;x$$ moves the graph of $$y=\cos\;x$$ upward by $$1$$, giving us the graph of $$y=1+\cos\;x$$: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9536945223808289, "perplexity": 123.07138598263641}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560284352.26/warc/CC-MAIN-20170116095124-00108-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/geometry/geometry-common-core-15th-edition/chapter-2-reasoning-and-proof-2-2-conditional-statements-apply-what-you-ve-learned-page-95/b | ## Geometry: Common Core (15th Edition)
I is the converse of II because the hypothesis and conclusion are swapped. Inverse: If a, b, c and d do not form a 2-by-2 calendar square, then $a+d\ne b+c$. Contrapositive: If $a+d\ne b+c$, then a, b, c and d do not form a 2-by-2 calendar square. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9562876224517822, "perplexity": 699.5076694183564}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583514437.69/warc/CC-MAIN-20181021224001-20181022005501-00120.warc.gz"} |
https://physics.stackexchange.com/questions/625779/fringing-field-effect-on-2d-capacitor | # Fringing Field Effect on 2D Capacitor
I am running a relatively «simple» (or so I thought) 2D electrostatic simulation of a parallel plate capacitor where I would like to see the effects of the fringing fields.
Case 1: air exclusively between two plates
In my initial simulation, I limited the air exclusively to the region between the two plates, and the results matched very nicely with the ideal equations that I extracted from my finite element method software for capacitance, electrostatic energy, electrostatic force, etc.
Case 2: extended the air domain to completely encompass the two capacitor plates
As you can see in Case 2, I extended the air domain to completely encompass the two capacitor plates and ran a parametric sweep on the dimensions of this «air block» to see the effects of the fringing fields. I expected the capacitance and electrostatic energy to both increase slightly and asymptotically reach a steady-state value because, at some point, the additional contributions of the fringing fields become negligible.
I confirmed this to be the case for the capacitance and electrostatic energy, but for the electrostatic force, I had some difficulty interpreting my simulation results.
As you can see in my schematic for Case 2 and the corresponding plots of electrostatic force v. air block dimensions, the force increases at the beginning, then drops sharply, and somewhat stabilizes.
What I am having difficulty understanding is that the final value is less than the case where we don’t take into account the fringing fields. From what I have seen in literature and what I have seemed to remember in my courses is that when we take fringing fields into account, the electrostatic force should increase, not decrease.
Case 3: limited the air block only to the edges of the bottom plate
With that in mind, I decided to simulate what I thought would be a more «realistic» scenario, i.e., Case 3, where I have limited the air block only to the edges of the bottom plate.
My rationale was that very rarely do we see two suspended plates like in Case 2, just floating in the air. Usually, the bottom plate is large (in comparison to the bottom) or rests atop some other material so that the electric field doesn’t have as easy access to the bottom side. When I simulate, the results I obtain for the electrostatic force seem to make more sense in that the final force is indeed bigger than the ideal case with no fringing field.
I realize that this is artificially constraining the field lines and that this behavior is expected, but I wanted to know if this result/behavior in Case 2 and my interpretation is reasonable or whether there is something fundamentally wrong with my simulation or approach.
Thank you in advance for all of your help and I look forward to discussing with you! | {"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": 0.872714102268219, "perplexity": 438.23507048114413}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320302622.39/warc/CC-MAIN-20220120190514-20220120220514-00088.warc.gz"} |
https://ijnaa.semnan.ac.ir/article_5511.html | ### The causal nexus between biomass energy consumption and gross domestic production -ARDL approach
Document Type : Research Paper
Author
Department of Food Science and Technology, College of Agriculture, University of Zabol
Abstract
One of the most important economic goals of countries is fostering more economic growth; this involves the increasing usage of energy sources. Because of the limitations in non-renewable energy sources and environmental pollution caused by burning these sources, renewable energy sources have become a priority. Biomass energy is one of the new and renewable varieties of energy. This energy is more compatible with nature and the environment, and its production and supply cause little environmental pollution; also, since such energies are renewable, there is no near end for their exhaustion. Therefore, biomass energy constitutes a remarkable part of the world energy supply. Because of the importance of energy in economic growth, this study analyzed the relationship between biomass energy consumption and gross domestic production (GDP) during the 1967-2019 time period, using the autoregressive distributed lag modeling approach (ARDL).
Keywords | {"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": 0.8915336728096008, "perplexity": 3610.6759092777806}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104683708.93/warc/CC-MAIN-20220707063442-20220707093442-00441.warc.gz"} |
https://www.physicsforums.com/threads/bounded-lebesgue-integrals.546141/ | # Bounded Lebesgue integrals
1. Nov 1, 2011
### tjkubo
Say f is a non-negative, integrable function over a measurable set E. Suppose
$$\int_{E_k} f\; dm \leq \epsilon$$
for each positive integer $k$, where
$$E_k = E \cap [-k,k]$$
Then, why is it true that
$$\int_E f\; dm \leq \epsilon \quad ?$$
I know that
$$\bigcup_k E_k = E$$
and intuitively it seems reasonable, but I don't know how to prove it.
Last edited: Nov 1, 2011
2. Nov 1, 2011
### Eynstone
Proceed by indirect method : if the integral over E is strictly > e ,then the integral over E_k should also exceed e for sufficiently large k.
3. Nov 1, 2011
### tjkubo
Why is that so? Are you using some property of integrable functions I'm not seeing?
4. Nov 1, 2011
### spamiam
I think you could also prove it this way. Given $\epsilon > 0$, for each $k \in \mathbb{Z}^+$ $\int_{E_k} f \; dm \leq \epsilon/2^k$. Then
$$\int_E f \; dm = \int_{\bigcup_k E_k} f \; dm \leq \sum_{k=1}^\infty \int_{E_k} f \; dm \leq \sum_{k=1}^\infty \frac{\epsilon}{2^k} = \epsilon \frac{1}{2} \frac{1}{1 - \frac{1}{2}} = \epsilon \; .$$
The only question I have is if it's legal to have an epsilon that depends on k, but I think it is since I think your given information held for all $\epsilon > 0$ and all positive integers k.
5. Nov 1, 2011
### tjkubo
Actually, the epsilon is a fixed number. That is why it's stumbling me. Sorry I wasn't clear.
6. Nov 1, 2011
### aesir
define $f_k(x)=\textbf{1}_{E_k}(x)f(x)$ where $\textbf{1}$ is the characteristic function.
From the monotone convergence theorem you have
$$\epsilon \ge \lim_{k\to\infty} \int_{E_k} f \; dm = \lim_{k\to\infty} \int_E f_k \; dm = \int_E f \; dm$$ | {"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": 0.9780451059341431, "perplexity": 810.7552516880222}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247515149.92/warc/CC-MAIN-20190222094419-20190222120419-00420.warc.gz"} |
http://mathhelpforum.com/geometry/202617-nonlinear-system-manually-computing-imaginary-solutions.html | # Math Help - Nonlinear system, manually computing imaginary solutions
1. ## Nonlinear system, manually computing imaginary solutions
$x^2+y^2=4$
$16x^2+9y^2=144$
The answer is not as important as how you arrived there, wolfram alpha can compute these. I would like to know how to solve manually.
I keep getting the equation $25x^2-36x-108$ from substitution(solved for formula 1, subbed into 2),
elimination yields nothing because there are no real roots, the graph is a circle within an ellipse.
Using the quadratic formula with the above equation gives me much different answers than wolfram.
I get $\frac{36\pm \sqrt{12096}}{50}$ Which does not make much sense.
2. ## Re: Nonlinear system, manually computing imaginary solutions
Another system I cannot solve manually is:
$5y^2-x^2=1$
$xy=2$
Due to the same reasons, I think I am forcing myself into extraneous solutions somehow :S
3. ## Re: Nonlinear system, manually computing imaginary solutions
Hello, Greymalkin!
$\begin{array}{cccc}x^2+y^2&=&4 & [1] \\ 16x^2+9y^2 &=& 144 & [2] \end{array}$
$\begin{array}{ccccccc}\text{Multiply -9}\times[1]: & \text{-}9x^2 - 9y^2 &=& \text{-}36 \\ \text{Add [2]:} & 16x^2 + 9y^2 &=& 144 \end{array}$
We have: . $7x^2 \:=\:108 \quad\Rightarrow\quad x^2 \:=\:\tfrac{108}{7} \quad\Rightarrow\quad \boxed{x \:=\:\tfrac{6}{7}\sqrt{21}}$
Substitute into [1]: . $\tfrac{108}{7} + y^2 \:=\:4 \quad\Rightarrow\quad y^2 \:=\:\text{-}\tfrac{80}{7}$
. . . . . . . . . . . . . . . $y \:=\:\pm\sqrt{\text{-}\tfrac{80}{7}} \quad\Rightarrow\quad \boxed{y\:=\:\pm\tfrac{4}{7}i\sqrt{35}}$
4. ## Re: Nonlinear system, manually computing imaginary solutions
Hello agaibn, Greymalkin!
$\begin{array}{cc}5y^2-x^2\:=\:1 & [1] \\ xy\:=\:2 & [2] \end{array}$
From [2]: . $y \,=\,\tfrac{2}{x}$
Substitute into [1]: . $5\left(\tfrac{2}{x}\right)^2 - x^2 \:=\:1 \quad\Rightarrow\quad \frac{20}{x^2} - x^2 \:=\:1$
. . $20 - x^4 \:=\:x^2 \quad\Rightarrow\quad x^4 + x^2 - 20 \:=\:0 \quad\Rightarrow\quad (x^2-4)(x^2+5) \:=\:0$
. . $\begin{Bmatrix}x^2-4 \:=\:0 & \Rightarrow & x^2 \:=\:4 & \Rightarrow & x \:=\:\pm2 \\ x^2+5 \:=\:0 & \Rightarrow & x^2 \:=\:\text{-}5 & \Rightarrow & x \:=\:\pm i\sqrt{5} \end{Bmatrix}$
If $x = \pm2$, [2] becomes: . $(\pm2)y \:=\:2 \quad\Rightarrow\quad y \:=\:\pm1$
If $x = \pm i\sqrt{5$, [2] becomes: . $(\pm i\sqrt{5})}y \:=\:2 \quad\Rightarrow\quad y \:=\:\mp\tfrac{2i}{5}\sqrt{5}$
We have four solutions: . $(\pm2,\,\pm1),\;\left(\pm i\sqrt{5},\:\mp\tfrac{2i}{5}\sqrt{5}\right)$
5. ## Re: Nonlinear system, manually computing imaginary solutions
Thanks again Soroban!
6. ## Re: Nonlinear system, manually computing imaginary solutions
First system has a circle & ellipse that do not interset:
Second system has two real and two complex roots:
Attached Thumbnails | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "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": 0.9749496579170227, "perplexity": 2758.575978539707}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430455241471.82/warc/CC-MAIN-20150501044041-00052-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://images.planetmath.org/idealinvertinginpruferring | ideal inverting in Prüfer ring
Let $\mathfrak{a}_{1}$, …, $\mathfrak{a}_{n}$ be invertible fractional ideals of a Prüfer ring. Then also their sum and intersection are invertible, and the inverse ideals of these are obtained by the formulae resembling de Morgan’s laws:
$(\mathfrak{a}_{1}+\cdots+\mathfrak{a}_{n})^{-1}\;=\;\mathfrak{a}_{1}^{-1}\cap% \cdots\cap\mathfrak{a}_{n}^{-1}$
$(\mathfrak{a}_{1}\cap\cdots\cap\mathfrak{a}_{n})^{-1}\;=\;\mathfrak{a}_{1}^{-1% }+\cdots+\mathfrak{a}_{n}^{-1}$
This is due to the fact, that the sum of any ideals is the smallest ideal containing these ideals and the intersection of the ideals is the largest ideal contained in each of these ideals. Cf. sum of ideals, quotient of ideals.
References
• 1 J. Pahikkala: “Some formulae for multiplying and inverting ideals”. $-$ Annales universitatis turkuensis 183. Turun yliopisto (University of Turku) 1982.
Title ideal inverting in Prüfer ring IdealInvertingInPruferRing 2015-05-06 14:34:48 2015-05-06 14:34:48 pahio (2872) pahio (2872) 15 pahio (2872) Theorem msc 13C13 DualityInMathematics DualityOfGudermannianAndItsInverseFunction | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 5, "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": 0.9487818479537964, "perplexity": 3788.2997552870743}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647545.84/warc/CC-MAIN-20180320224824-20180321004824-00480.warc.gz"} |
https://www.value-at-risk.net/risk-measures/ | # 1.2 Risk Measures
In the context of risk measurement, we distinguish between:
• a risk measure, which is the operation that assigns a value to a risk, and
• a risk metric, which is the attribute of risk that is being measured.
Just as duration and size are attributes of a meeting that might be measured, volatility and credit exposure are attributes of bond risk that might be measured. Volatility and credit exposure are risk metrics. Other examples of risk metrics are delta, beta and duration. Any procedure for calculating these is a risk measure. For any risk metric, there may be multiple risk measures. There are, for example, different ways that the delta of a portfolio might be calculated. Each represents a different risk measure for the single risk metric called delta.
According to Holton (2004), risk has two components:
1. exposure, and
2. uncertainty.
If we swim in shark-infested waters, we are exposed to bodily injury or death from a shark attack. We are uncertain because we don’t know if we will be attacked. Being both exposed and uncertain, we face risk.
Risk metrics typically take one of three forms:
• those that quantify exposure;
• those that quantify uncertainty;
• those that quantify exposure and uncertainty in some combined manner.
Probability of rain is a risk metric that only quantifies uncertainty. It does not address our exposure to rain, which depends upon whether or not we have outdoor plans.
Credit exposure is a risk metric that only quantifies exposure. It indicates how much money we might lose if a counterparty were to default. It says nothing about our uncertainty as to whether or not the counterparty will default.
Risk metrics that quantify uncertainty—either alone or in combination with exposure—are usually probabilistic. Many summarize risk with a parameter of some probability distribution. Standard deviation of tomorrow’s spot price of copper is a risk metric that quantifies uncertainty. It does so with a standard deviation. Average highway deaths per passenger-mile is a risk metric that quantifies uncertainty and exposure. We may interpret it as reflecting the mean of a probability distribution.
###### Exercises
1.2
Give an example of a situation that entails uncertainty but not exposure, and hence no risk.
1.3
Give an example of a situation that entails exposure but not uncertainty, and hence no risk.
1.4
In our example of the deaths per passenger-mile risk metric, for what random variable’s probability distribution may we interpret it as reflecting a mean?
1.5
Give three examples of risk metrics that quantify financial risks:
1. one that quantifies exposure;
2. one that quantifies uncertainty; and
3. one that quantifies uncertainty combined with exposure. | {"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": 0.8486610054969788, "perplexity": 1926.8659803657804}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496667767.6/warc/CC-MAIN-20191114002636-20191114030636-00022.warc.gz"} |
https://chemistry.stackexchange.com/questions/74064/alkoxide-as-a-leaving-group | # Alkoxide as a leaving group
In the reduction of esters to alcohols, the RO group is pushed off. However, the negative charge on the O atom is destabilized by the electron donating R group. Why is the RO group able to function as a leaving group given that the resultant alkoxide is unstable? | {"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": 0.8741969466209412, "perplexity": 3525.25736133221}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875142323.84/warc/CC-MAIN-20200217115308-20200217145308-00432.warc.gz"} |
http://phys.libretexts.org/Under_Construction/Map%3A_Electricity_and_Magnetism_(Tatum)/2%3A_Electrostatic_Potential/2.6%3A_Two_Semicylindrical_Electrodes | $$\require{cancel}$$
# 2.6: Two Semicylindrical Electrodes
This section requires that the reader should be familiar with functions of a complex variable and conformal transformations. For readers not familiar with these, this section can be skipped without prejudice to understanding following chapters. For readers who are familiar, this is a nice example of conformal transformations to solve a physical problem.
FIGURE II.5
We have two semicylindrical electrodes as shown in figure II.5. The potential of the upper one is 0 and the potential of the lower one is $$V_0$$. We'll suppose the radius of the circle is 1; or, what amounts to the same thing, we'll express coordinates x and y in units of the radius. Let us represent the position of any point whose coordinates are (x , y) by a complex number $$z = x + iy$$.
Now let $$w = u + iv$$ be a complex number related to $$z$$ by $$w=i\left (\frac{1-z}{1+z}\right )$$; that is, $$z=\frac{1+ iw}{1- iw}$$. Substitute $$w = u + iv \text{ and }z = x + iy$$ in each of these equations, and equate real and imaginary parts, to obtain
\begin{align}\label{2.6.1}u&=\frac{2y}{(1+x)^2+y^2};\quad\quad &&v=\frac{1-x^2+y^2}{(1+x)^2+y^2};\\ x&=\frac{1-u^2-v^2}{u^2+(1+v)^2}; &&y=\frac{2u}{u^2+(1+v)^2}.\label{2.6.2}\end{align}
In that case, the upper semicircle $$(V = 0)$$ in the xy-plane maps on to the positive u-axis in the uv-plane, and the lower semicircle $$(V = V_0)$$ in the xy-plane maps on to the negative u-axis in the uv-plane. (Figure II.6.) Points inside the circle bounded by the electrodes in the xy-plane map on to points above the u-axis in the uv-plane.
FIGURE II.6
In the uv-plane, the lines of force are semicircles, such as the one shown. The potential goes from 0 at one end of the semicircle to $$V_0$$ at the other, and so equation to the semicircular line of force is
$\label{2.6.3}\frac{V}{V_0}=\frac{\text{arg}\,w}{\pi}$
or
$\label{2.6.4}V=\frac{V_0}{\pi}\tan^{-1}(v/u).$
The equipotentials (V = constant) are straight lines in the uv-plane of the form
$\label{2.6.5}v=fu.$
(You would prefer me to use the symbol m for the slope of the equipotentials, but in a moment you will be glad that I chose the symbol $$f$$.)
If we now transform back to the xy-plane, we see that the equation to the lines of force is
$\label{2.6.6}V=\frac{V_0}{\pi}\tan^{-1} \left (\frac{1-x^2-y^2}{2y}\right ).$
and the equation to the equipotentials is
$\label{2.6.7}1-x^2-y^2=2fy,$
or
$\label{2.6.8}x^2+y^2+2fy-1=0$
Now aren't you glad that I chose $$f$$ ? Those who are handy with conic sections (see Chapter 2 of Celestial Mechanics) will understand that the equipotentials in the xy-plane are circles of radii $$\sqrt{f^2 + 1}$$, whose centres are at $$(0 , \pm f )$$, and which all pass through the points $$(\pm 1 , 0)$$. They are drawn as blue lines in figure II.7. The lines of force are the orthogonal trajectories to these, and are of the form
$\label{2.6.9}x^2+y^2+2gy+1=0$
These are circles of radii $$\sqrt{g^2 −1}$$ and have their centres at $$(0 , \pm g)$$. They are shown as dashed red lines in figure II.7.
FIGURE II.7 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9676727056503296, "perplexity": 385.67643979063433}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281332.92/warc/CC-MAIN-20170116095121-00074-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://ftp.srware.com/manual/xicmanual/node535.html | The math and logical operators are overloaded for zoidlists as follows:
+ , | union - and-not * , & intersection ^ exclusive or ! inverse
The result of the operation is a new zoidlist, with neither of the operands affected.
To test for an empty zoidlist, the == and != comparisons to the value 0 can be applied. Note that ``if (!zlist)'' is an incorrect test for an empty zoidlist; it will invert the list and return true if the inverted list is not empty.
There is a current ``reference'' zoidlist which represents the ``background''. If not explicitly set (with the SetZref function), this is taken as the boundary of the current cell. The reference is used in operations such as inversion and exclusive-or where the size of the background must be assumed. Note that this background can be an arbitrary shape. In binary operators with zoidlists, if one of the operands is a scalar, 0 represents an empty list, and nonzero represents the reference list. | {"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": 0.8424472212791443, "perplexity": 1019.2749633992405}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144027.33/warc/CC-MAIN-20200219030731-20200219060731-00209.warc.gz"} |
https://www.global-sci.org/intro/article_detail/jcm/8864.html | Volume 22, Issue 5
A QP Free Feasible Method
DOI:
J. Comp. Math., 22 (2004), pp. 651-660
Published online: 2004-10
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• Abstract
In [12], a QP free feasible method was proposed for the minimization of a smooth func- tion subject to smooth inequality constraints. This method is based on the solutions of linear systems of equations, the reformulation of the KKT optimality conditions by using the Fischer- Burmeister NCP function. This method ensures the feasibility of all iterations. In this paper, we modify the method in [12] slightly to obtain the local convergence under some weaker conditions. In particular, this method is implementable and globally con- vergent without assuming the linear independence of the gradients of active constrained functions and the uniformly positive definiteness of the submatrix obtained by the Newton or Quasi Newton methods. We also prove that the method has superlinear convergence rate under some mild conditions. Some preliminary numerical results indicate that this new QP free feasible method is quite promising.
• Keywords
Constrained optimization KKT point Multiplier Nonlinear complementarity Convergence | {"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": 0.9141267538070679, "perplexity": 769.9100894280077}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364932.30/warc/CC-MAIN-20210302221633-20210303011633-00529.warc.gz"} |
http://mathhelpforum.com/number-theory/225521-divisor-0-a.html | # Math Help - divisor of 0
1. ## divisor of 0
find all the divisors of 0 in the ring Z6 of integers of modulo 6.
in this system 2x3=0. so divisors of zero are 2 and 3. Again 4x3=0. So 4 is also a divisor of 0. ......
Is my approach right?
2. ## Re: divisor of 0
Yes, the "zero divisors" in $Z_6$ are 2, 3, and 4. More generally, the zero divisors of $Z_n$ are the prime divisors of n and their multiples. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8801977634429932, "perplexity": 873.3631979087766}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246634257.45/warc/CC-MAIN-20150417045714-00163-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://paperity.org/search/?q=authors%3A%22Yanyan+Yu%22 | Search: authors:"Yanyan Yu"
5 papers found.
Use AND, OR, NOT, +word, -word, "long phrase", (parentheses) to fine-tune your search.
Three-dimensional simulations of strong ground motion in the Sichuan basin during the Wenchuan earthquake
A three-dimensional model of the Sichuan basin was established by incorporating the Quaternary Chengdu plain, the near surface sediment, the crystalline basement, the 1D crustal velocity structure, and the realistic surface topography of the basin. Based on this model, and by using the spectral-element method and a parallel computing technique, the low frequency (0.05–0.5 Hz) wave ...
On a Stević-Sharma operator from Hardy spaces to the logarithmic Bloch spaces
Let H ( D ) denote the space of all analytic functions on the unit disc D of the complex plane ℂ, ψ 1 , ψ 2 ∈ H ( D ) , and φ be an analytic self-map of D . In this paper, we characterize the boundedness and compactness of a Stević-Sharma operator T ψ 1 , ψ 2 , φ from Hardy spaces H p (with 1 ≤ p < ∞ ) to the logarithmic Bloch spaces B log . MSC: 47B38, 47B33, 46E15, 30H10.
Fitting Characteristics of N95 Filtering-Facepiece Respirators Used Widely in China
Background Millions of people rely on N95 filtering facepiece respirators to reduce the risk of airborne particles and prevent them from respiratory infections. However, there are no respirator fit testing and training regulations in China. Meanwhile, no study has been conducted to investigate the fit of various respirators. The objective of this study was to investigate whether ...
On a Li-Stević Integral-Type Operators between Different Weighted Bloch-Type Spaces
Let be an analytic self-map of the unit disk , let be the space of analytic functions on , and let . Recently, Li and Stević defined the following operator: , on . The boundedness and compactness of the operator between two weighted Bloch-type spaces are investigated in this paper.
Digital 3-D Headforms Representative of Chinese Workers
Headforms are useful for designing and testing various types of personal protective equipment used to protect millions of workers from occupational hazards in China. Although the Chinese national standard of head-and-face dimensions for adults was first published in 1981, headforms based on those dimensions were never developed. In 2006, an anthropometric survey of 3000 Chinese ... | {"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": 0.892389178276062, "perplexity": 2307.3783874263163}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323870.46/warc/CC-MAIN-20170629051817-20170629071817-00510.warc.gz"} |
https://www.physicsforums.com/threads/probality-questions.102462/ | # Probality questions
1. Dec 2, 2005
thank you.
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2. Dec 2, 2005
### HallsofIvy
Staff Emeritus
What have YOU done on these problems? Where are you stuck?
They look pretty much like applying basic definitions and a little arithmetic.
3. Dec 3, 2005
### rpatel
i tried these questions but i get odd answers.
like for the first bit i got 2. which i dnt think is right?
4. Dec 3, 2005
### HallsofIvy
Staff Emeritus
For the "first bit" you got 2"?? You are aware that a probability must be between 0 and 1, aren't you?
Once again, what have you done on these problems? Not just what answer you got- what have you tried? It will be a lot easier to point out mistakes or give hints if we can see what you are doing.
5. Dec 3, 2005
### rpatel
ok i have redone my calculations . for part a) i calculated the possibilities for outcomes.
spinner 1 probality of it landing on A is 2/5 and the probality of it landing on B is 2/5 also.
spinner 2 probality of it landing on A is 3/5 and the probality of it landing on B is 1/5.
it is impossible for both of them to land on C or D because not both spinner have these two letters.
i thereafter multiply the outcomes of A and multiplied the outcomes of B. i then added these two values together.
A(2/5*3/5) = 6/25
B(2/5*1/5) = 2/25
then i added 6/25+2/25= 8/25 which equals to 0.32.
therefore my answer is 0.32. is this correct?
part b) i firstly calculated the total amount of money which is 0.40*200= £80.
then i calculated the amount of people that should win which is 200*0.32= 64. therefore the profit amy should expect is £80-£64= £16
is this correct?
thanks for your help so far.
regards
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https://langenbergadvies.nl/index.php/detection-by-spooky-photon/ | # Detection by ‘spooky’ photon.
An interesting phenomenon can be evoked in an actual practically feasible experiment with a Mach-Zehnder interferometer as proposed by Elitzur and Valtmann in 1993. With a Mach-Zehnder interferometer it can be investigated whether an object, for example a light-sensitive bomb as Elitzur and Valtmann suggested, is placed in the path of the interferometer without having to use any light to touch the object. For a more detailed description of a Mach-Zehnder interferometer, I refer to another page on this website. If necessary, read the description there first before continuing.
Shown above is a standard configured Mach-Zehnder interferometer. If no extra phase difference is introduced, using the ΔΦ block, and if the path lengths 1-2-4 (upper) and 1-3-4 (lower) are equal, then D1 receives all the light and D2 receives nothing. When we consider light as an EM wave, that phenomenon can be explained by constructive (D1) and destructive (D2) interference at beam splitter 4 by the waves traveling along the upper path and the lower path.
However, when we try to use the interpretation of light as traveling photons, little particles with a fixed amount of energy, their never reaching D2 becomes much more difficult to explain, because how can a photon, being a particle, travel simultaneously two paths in order to achieve interference?
But when we use the non-physical state function – the quantum wave – then the destructive interference at D2 means only that the chance that photons will be detected there is zero and that therefore all photons will manifest on D1. So far this is something I have already explained elsewhere – and more extensively.
Now it becomes interesting if we set up a blockade (for example an extremely light-sensitive bomb that goes off when hit by a single photon) in one of the two paths. See the figure below. Now assume the following circumstances:
• The experimenter knows that the paths in the interferometer are of equal length and that the configuration is closed.
• The experimenter cannot take a look in the interferometer whether or not a blockage has been placed there (the light-sensitive bomb definitely would explode).
• The experimenter only sees the output of D1 and D2,
• The experimenter does not know if and when photons are sent by the laser.
Does the experimenter now have a chance to find out if one of the two paths is blocked? Without setting of the eventual light-sensitive bomb? That’s indeed possible. He has a success chance of 1:4. Kind of reverse Russian roulette.
Explanation: a photon that chooses the upper path (50/50) will now hit the blockade. Neither D1 or D2 will then go off. The experimenter does not know whether photons have been fired and therefore does not yet know if a blockade exists. If the blockade is a ligth sensitive bomb it will explode upon being hit by the photon. However, if a photon takes the lower path, not setting off the bomb, either D1 or D2 goes off. If D1 goes off, the experimenter still knows nothing because that would be the normal expected behavior of an unblocked interferometer. However, if D2 is hit, the experimenter then knows with certainty that there is a blockage. But that photon that hit D2 has never been anywhere near the blockade!
The probability that a photon will reach D2, due to the two consecutive beam splitters, is now 50% x 50% = 25%. So after just a few laser shots the patient experimenter will find out by D2 beeping that there is a blockage and stay alive (or the bomb goes off ending the experiment). If he / she is lucky D2 will beep right at the first laser shot, chance 1:4. But this now raises the spooky behavior question: how does the photon that arrives at beam splitter 4 know that it now has a 50/50 chance of moving on to D2? It arrived there traveling the lower path and should therefore not be aware of any blockade in the upper path. Yet it is. In other words: for a photon traveling the lower path, there should be no difference between blocking the upper path or not, however, when that blockade is removed, the behavior of an identical photon traveling the unaffected lower path suddenly changes its behavior to 100% reflection at beamsplitter 4 towards D1.
Incidentally, the mystery is solved as soon as you use either wave model, EM wave or quantum state wave. The blockage eliminates the possibility of interference of the wave with itself at beam splitter 4. However, the quantum interpretation with physical photons that travel leads to undeniable contradictions.
The only other quantum interpretation known to me that also assumes that the photon is not traveling physically and that can explain this “spooky” phenomenon is John G. Cramer’s Transactional Interpretation that assumes advanced quantum waves that travel back in time from the detector to the laser. But there are serious objections to his interpretation. Among other things, the quantum state wave itself also contributes to the transferred energy in his interpretation and is therefore physical. For further reading about his idea I do recommend Cramers didactically excellent book “The Quantum Handshake.”
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http://www.newton.ac.uk/programmes/BSM/seminars/2012062510101.html | # BSM
## Seminar
### Generic String/M theory Predictions for Particle Physics and Dark Matter
Acharya, B (Abdus Salam International Centre for Theoretical Physics)
Monday 25 June 2012, 10:10-10:45
Seminar Room 1, Newton Institute
#### Abstract
Solutions of string/M theory with low energy supersymmetry have moduli fields with masses which are generically of order the mass of the gravitino. This leads to a spectrum of superpartners with squarks and sleptons just beyond the LHC energy scale. Gluinos could however be produced at the LHC and decay with top and bottom rich final states. The mass of the lightest Higgs boson is predicted to be between 122 and 129 GeV. The moduli fields dominate the pre-BBN Universe leading to a non-thermal history which significantly impacts our picture of dark matter, which is predicted to have two components: axions and W-ino like WIMPS. The tentative high energy photon signal in the Fermi-LAT data can be successfully interpreted as the annihilation of these WIMPs into the Z-boson and a photon. Moreover, we demonstrate that this signal is also consistent with an approximately 50% axion component. The axion component implies that experiments like Planck will NOT observe tensor modes in the CMB.
#### Video
The video for this talk should appear here if JavaScript is enabled.
If it doesn't, something may have gone wrong with our embedded player.
We'll get it fixed as soon as possible. | {"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": 0.8813205361366272, "perplexity": 1664.6031601913676}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997883425.83/warc/CC-MAIN-20140722025803-00012-ip-10-33-131-23.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/49297/the-only-two-rational-values-for-cosine-and-their-connection-to-the-kummer-rings | # The only two rational values for cosine and their connection to the Kummer Rings
I am trying to learn about Kummer Rings, and in particular what makes $n=3,4,6$ so special. (That is the Gaussian and Eisenstein integers)
The only $\theta\in [0,\frac{\pi}{2}]$ which are rational multiples of $\pi$ for which $\cos(\theta)\in \mathbb{Q}$ are $\theta=\frac{\pi}{2},\frac{\pi}{3}$ which corresponds exactly to $n=4,6$ in $\frac{2\pi}{n}$.
Can someone give me an explanation for why $\cos(\theta)$ is rational only in these cases? Also, can we go the other way, and use some nice property of the Kummer Rings to show that $\cos(2\pi/n)$ is rational if and only if $n=1,2,3,4,6$?
Thanks,
Edit: As pointed out by Qiaochu, what I previously wrote above was certainly not the norm.
-
That is not the norm in $\mathbb{Z}[\zeta_n]$. – Qiaochu Yuan Jul 3 '11 at 22:32
@Qiaochu: What would the norm be in the general Kummer Rings? – Eric Naslund Jul 3 '11 at 22:39
I wish the cited wikipedia article gave a source for calling these rings "Kummer rings". I acknowledge that Kummer worked on them (as did others before and since), but every number theorist I know calls them cyclotomic integer rings or something close to that. – Pete L. Clark Jul 3 '11 at 23:31
In a number field $K$, the norm of an element $N_{K/\mathbb{Q}}(a) = N(a)$ can be given various equivalent definitions, one of which is that it is the determinant of the linear map $x \mapsto ax$ acting on $K$ regarded as a vector space over $\mathbb{Q}$. If $\sigma_i : K \to \mathbb{C}$ denote the complex embeddings of $K$, then we also have
$$N(a) = \prod_i \sigma_i(a).$$
The norm is always rational.
If $K$ has degree $n$, then it has $n$ complex embeddings (for example by the primitive element theorem); in particular, fixing a basis of $K$ and expressing $a$ in it, the norm is a homogeneous polynomial of degree $n$. It is a quadratic form if and only if $n = 2$.
Now the degree of $\mathbb{Q}(\zeta_m)$ is equal to $\varphi(m)$, which is equal to $2$ if and only if $m = 3, 4, 6$. In other words, these are the only cyclotomic fields which give quadratic extensions. This is related to the crystallographic restriction theorem.
Yes, you can use cyclotomic fields to prove that $\cos \frac{2\pi}{m}$ is rational only when $m = 1, 2, 3, 4, 6$. Once you know that $\mathbb{Q}(\zeta_m)$ has degree $\varphi(m)$ (but this is not trivial), you can show that that $\mathbb{Q}(\zeta_m + \zeta_m^{-1})$ is a subfield of index $2$, hence is $\mathbb{Q}$ if and only if $\varphi(m) \le 2$.
-
Does this "crystallographic Restriction" have to do with the fact that $\mathbb{Z}[\zeta_n]$ has finitely many units if and only if $n=1,2,3,4,6$ by Dirichlet's unit theorem? – Eric Naslund Jul 3 '11 at 22:43
Well, yes, but that's not how I would think about it. The only number fields $K$ such that $\mathcal{O}_K$ has finitely many units satisfy $r + s = 1$ where $r$ is the number of real embeddings and $s$ the number of pairs of complex embeddings, hence either $r = 1, s = 0$ and $K = \mathbb{Q}$ or $r = 0, s = 1$ and $K$ is an imaginary quadratic number field. This is the important property, as it implies that $\mathcal{O}_K$ naturally embeds into $\mathbb{C}$ as a discrete subgroup, so in particular the units of $\mathcal{O}_K$ act on a lattice. – Qiaochu Yuan Jul 3 '11 at 22:50
Crystallographic restriction is, to my mind, really a theorem about elements of finite order in $\text{GL}_n(\mathbb{Z})$. The point is that an element of order exactly $m$ has the property that its characteristic polynomial is a product of cyclotomic polynomials, at least one of which must be $\Phi_m$, which has degree $\varphi(m)$, hence $\varphi(m) \le n$. Thus in $3$ dimensions we can still only have $m = 1, 2, 3, 4, 6$ but in $4$ dimensions we can have $m = 1, 2, 3, 4, 5, 6, 8$, and so forth. – Qiaochu Yuan Jul 3 '11 at 22:53
One more question: I just want to understand this example: What is the norm in $\mathbb{Z}[\zeta_5]$? So we can write each element as $a+b\zeta_5+c\zeta_5^2+d\zeta_5^3+e\zeta_5^4$. What will the polynomial for the norm be? How does it relate to the cyclotomic polynomial $\Phi_5(X)=X^4+X^3+X^2+X^1+1$? Thanks again for your help! This answer has been very useful! – Eric Naslund Jul 4 '11 at 16:26
@Eric: if $K$ is a Galois extension with Galois group $G$, the norm can be written $N(a) = \prod_{g \in G} ga$. Here the Galois group is cyclic of order $4$ generated by $\zeta_5 \mapsto \zeta_5^2$ (a primitive root modulo $5$). This statement about the Galois group turns out to be equivalent to the statement that the corresponding cyclotomic polynomial is irreducible (together with the existence of primitive roots). By the way, you only need to go up to $\zeta_5^3$. This kind of material should be covered in any good book on algebraic number theory / Galois theory. – Qiaochu Yuan Jul 4 '11 at 16:44 | {"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": 0.9500312805175781, "perplexity": 121.509305261249}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637898141.14/warc/CC-MAIN-20141030025818-00239-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://research.caluniv.ac.in/publication/the-k-l-k-s-mass-difference-in-left-right-symmetric-models | X
the K L -K S mass difference in left-right symmetric models and the right-handed Kobayashi-Maskawa matrix
P BASAK, A DATTA, A RAYCHAUDHURI
Published in Springer-Verlag
1983
Volume: 20
Issue: 4
Pages: 305 - 312
Abstract
KL-KS mass difference is computed in a left-right symmetric model with three generations of quarks. Under the assumption of "manifest" left-right symmetry the mass of the right-handed charged gauge boson consistent with KL-KS constraint appears to be rather high. A low mass right-handed gauge boson can be accommodated in models with broken "manifest" left-right symmetry. The latter scenario has many non-trivial predictions for the decay of heavy (c, t, b) quarks. © 1983 Springer-Veralg. | {"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": 0.9464148879051208, "perplexity": 4441.727242926613}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00575.warc.gz"} |
http://mathhelpforum.com/calculus/164655-volume-problems-2.html | # Math Help - Volume Problems
1. Thank you, everything makes sense now.
2. So for part a I get the volume to be 157.6 Inches cubed, and for part b I still need to figure out the volume of the football using the equation of the parabola, is there a formula I need in order to relate the two equations ?
3. You can integrate to find volume.
4. Would my limits be -5.5 and 5.5 ?
5. You could do 0, 5.5 and multiple by 2. Also, use the volume integral formulas. Regular integration solves area.
6. volume integral formulas ? where do i find them ? or the right one ?
7. $\displaystyle V=\pi\int_{a}^{b}[f(x)]^2dx$
8. Thank you.
9. Ok so when I evaluate my Volume I get a really huge number 3000 something which I know isn't right. This is what I evaluate. 2pi* the integral from 0 to 5.5 of ((-21)/2(5.5)^2pi (x^2)+ 21/2pi))^2dx. When I put that in my calculator and integrate I get a huge number. A friend of mine suggested I try a quad regression but I don't know how to do that on the calculator. Please tell me what I did wrong.
10. I obtained 77 when I integrated it. Show your steps.
11. I did show you my steps. ok I get for the equation Y = (21/2pi) - (21/2pi(5.5)^2) x^2. and then when I put that into the Volume formula I just move 2pi out front and square f(x) I put the whole function into my calculator and I use my calulator to integrate from 0 to 5.5 I get 3000 something
12. Never mind good sir. I was plugging it in to the calculator wrong. I get 77 too now. Thank you
13. $\displaystyle 2\pi\int_{0}^{5.5}\left(\frac{21}{2\pi}-\frac{21}{2\pi *5.5^2}*x^2\right)dx=21\left[x]^{5.5}_0-\frac{21}{5.5^2}\left[\frac{x^3}{3}\right] _0^{5.5}$
$\displaystyle = 21*5.5-\frac{21}{5.5^2}\left[\frac{5.5^3}{3}\right]=77$
14. Yes yes thank you once again.
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https://socratic.org/questions/how-do-you-evaluate-2-5-3-10 | Algebra
Topics
# How do you evaluate 2/5 + 3/10?
Jun 28, 2016
Convert the fractions so they have a common denominator; then add the numerators to get $\frac{7}{10}$
#### Explanation:
Given
$\textcolor{w h i t e}{\text{XXX}} \frac{2}{5} + \frac{3}{10}$
To add fractions we need to write them with the same denominator:
and since $\frac{2}{5} = \frac{4}{10}$
$\textcolor{w h i t e}{\text{XXX}} \frac{2}{5} + \frac{3}{10} = \frac{4}{10} + \frac{3}{10}$
When fractions have the same denominator we can add the numerators and write the sum over that common denominator:
$\textcolor{w h i t e}{\text{XXX}} \frac{4}{10} + \frac{3}{10} = \frac{4 + 3}{10} = \frac{7}{10}$
##### Impact of this question
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http://math.stackexchange.com/questions/76660/whats-the-smallest-area-a-square-can-have-that-a-cube-can-still-be-wrapped-with | # What's the smallest area a square can have that a cube can still be wrapped with it?
My task is to wrap a unit cube with the smallest square sheet of paper possible. The paper is assumed to be infinitely thin of course and no cutting or stretching is permitted.
I must be able to justify any claims I make.
This is what I have so far:
Let $A$ denote this smallest area I seek. (How do I know there is a smallest area? Wouldn't it be the greatest lower bound of a monotonically decreasing sequence of $A$'s with lower bound 6?)
A well-known fact from geometry says that a cube has 11 nets. Since none of the nets are square, my sheet of paper can't be folded like any of them, so $A>6$.
A square $2\sqrt{2}$ on a side can wrap a unit cube. Here's how: make four folds each perpendicular to a diagonal and 3/2 from the paper's corner. Place the cube in the resulting square created by the folds. The corners of the paper will meet at the center of the side on top. Hence, $A\le (2\sqrt{2})^2=8$.
I am stuck at $6<A\le 8$. I can't find anything smaller nor can I prove the square of area 8 is the smallest.
Any suggestions will be much appreciated.
Update: $A=8$. This was proved by Michael L. Catalano-Johnson, Daniel Loeb and John Beebee in "Problem 10716: A cubical gift," American Mathematical Monthly, volume 108, number 1, January 2001, pages 81-82 (posed in volume 106, 1999, page 167).
So what do I do now as far as accepting an answer? Would it be frowned upon if I summarized their solution down below? (That way, I might get someone to comment on it, as I don't completely follow.)
-
Thinking aloud here... each of the 8 corners must be covered at least once by the sheet. And for each point on the sheet that covers a corner, a certain minimal amount of paper must be wasted on folds, and this amount seems to be related to how much of the adjoining edges the local cover reaches. Perhaps a lower bound could be derived from the fact that all 12 edges must be fully covered? But it's probably not that easy -- for convex (but not necessarily square) sheets that will cover the cube there's an easy upper bound of 7.5 (which can be lowered to $8-\sqrt{1/2}$ with a bit of care). – Henning Makholm Oct 28 '11 at 13:22
– Joseph Malkevitch Oct 28 '11 at 15:55
It's perhaps of note that it takes 2 points to enclose a unit segment, and 4 unit segments to enclose a unit square. I would not be entirely surprised if the trend of $2^n$ continued. – A Walker Oct 28 '11 at 19:36
@JosephMalkevitch From the site you linked to, I was able to chase down a few references, one of which pointed to a full solution. That reference is: Michael L. Catalano-Johnson and Daniel Loeb, "Problem 10716: A cubical gift," American Mathematical Monthly, volume 108, number 1, January 2001, pages 81-82 (posed in volume 106, 1999, page 167). Thanks! – sasha Oct 28 '11 at 19:53
@sasha: No, it would not be frowned upon at all! Look at this blog post from Jeff Atwood himself. – Zev Chonoles Oct 28 '11 at 22:31
I am going to summarize the trio's solution here since it's essentially an answer to my question, but I'm hoping that the first paragraph in particular might be clarified by one of you as I don't completely follow. The problem, titled "A Cubical Gift" [10716], was posed in the 1999 volume of the Monthly and phrased differently ("What is the largest cubical present that can be completely wrapped (without cutting) by a unit square of wrapping paper?") Clearly, an answer to this question will solve mine and vice-versa.
Two arbitrary points in the square are chosen and their distance is considered before and after wrapping the cube with the square. They argue that the surface distance between the two points after wrapping is no larger than what it was before wrapping, since the paper was neither cut nor streched. [I'm not convinced. Perhaps the surface distance notion is throwing me off. Is it the distance along the paper? Wouldn't that then be an obvious statement, a tautology even? Otherwise, are we talking about the distance along the cube? Isn't there a lot of choices to call the distance in that case? And how does this paragraph relate to the next?]
Consider an arbitrary point on the cube. There exists another at least twice an edge length away [surface distance, again]. This implies that for any point on the square, another point can be found at distance at least twice an edge length. So, this is also true for the center of the square, implying that the diagonal is at least 4 times the length of an edge. Therefore, the side length of the paper square is at least $2\sqrt{2}$, giving the area of 8. Finally they show that the inequality is tight by demonstrating the folding pattern I described in the OP.
-
By surface distance between two points they mean two things depending on whether the two points are on the sheet or on the cube. On the sheet it is the usual distance between two points. On the cube it is the length of the shortest path along the surface of the cube. If the straight line segment from point $A$ to $B$ on the sheet of paper has length $L$, then, after wrapping, tracing the image of that line on the cube gives a path of length $L$ (no stretching). That path may not be the shortest route connecting the image points $A'$ and $B'$ on the cube, hence you only get inequality. – Jyrki Lahtonen Oct 29 '11 at 6:27
Consider three faces surrounding a corner of the cube. If we suppose that the edge of the square piece of paper is nowhere cutting across these three faces, then in the folding process we inevitably `double cover' near the corner. This waste of paper will, by concavity, amount to at least $1/2$ of a unit of area. If we suppose that only one face, $F$, has a paper edge crossing it, orient $F$ to the top. Then the four other faces surrounding the bottom face each satisfy the above criterion, and we waste at least $2$ units of paper (corresponding to the four bottom corners). This implies that one must use at least 8 units of paper to cover the cube, if one face is crossed by a paper edge.
If no faces are disturbed in this way, then a similar bound holds. Either your bound is optimal or there exists a more efficient one in which between 2 and 6 faces meet the paper's edge.
Hope this helps as a technique. (It seems likely that 8 is optimal.)
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https://papers.nips.cc/paper/2020/hash/ddd808772c035aed516d42ad3559be5f-Abstract.html | #### Authors
Evgenii Chzhen, Christophe Denis, Mohamed Hebiri, Luca Oneto, Massimiliano Pontil
#### Abstract
<p>We study the problem of learning an optimal regression function subject to a fairness constraint. It requires that, conditionally on the sensitive feature, the distribution of the function output remains the same. This constraint naturally extends the notion of demographic parity, often used in classification, to the regression setting. We tackle this problem by leveraging on a proxy-discretized version, for which we derive an explicit expression of the optimal fair predictor. This result naturally suggests a two stage approach, in which we first estimate the (unconstrained) regression function from a set of labeled data and then we recalibrate it with another set of unlabeled data. The recalibration step can be efficiently performed via a smooth optimization. We derive rates of convergence of the proposed estimator to the optimal fair predictor both in terms of the risk and fairness constraint. Finally, we present numerical experiments illustrating that the proposed method is often superior or competitive with state-of-the-art methods.</p> | {"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": 0.8681676387786865, "perplexity": 699.5688570320922}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178361776.13/warc/CC-MAIN-20210228205741-20210228235741-00401.warc.gz"} |
https://www.studypug.com/sg/sg-gce-n(a)-level-a-maths/radian-measure-and-arc-length | # Radian measure and arc length
### Radian measure and arc length
#### Lessons
$arc\;length= r \cdot \theta\;radian$
• 1.
Use the information in each diagram to determine the value of the variable
a)
b)
c)
d)
• 2.
A 20 cm long pendulum swings through an arc length of 60 cm. Through
what angle, in both degrees and radians, does the pendulum swing? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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": 0.9779301881790161, "perplexity": 3049.1305053655115}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057622.15/warc/CC-MAIN-20210925112158-20210925142158-00684.warc.gz"} |
http://ftp.cs.rochester.edu/users/faculty/nelson/courses/csc_160/2011_spring/lectures/signal_proc/signal_proc_slides.html | # Signal Processing -- The Frequency Domain
## Topics and Methods
Read, or have read by now, the "Informal Introduction".
• Classic Comic Book Intro to Frequency Domain
• "Passive" Math (no math. manipulations).
• Very Little Math Justification...have faith.
• Dirac Delta, Sines, Inner Product
• Basis sets of functions.
• Notion of Linear transforms (systems).
• (Invertible) Transform Domains, especially the frequency domain.
• Fourier transformation; properties and use.
• Practical transform-domain signal-processing techniques.
• Web resources
## A Primary Goal
1. See how to transform a signal into a different representation (specifically here, a representation of frequencies in the signal).
2. See why such a transform is a useful practical and abstract thing to do.
## Dirac Delta Function
• Also called the unit impulse function.
• Not technically a "function"
• Has zero width, unit integral (as a vector [...000000100000...].)
• Integral of &delta(x-c)f(x) "samples" f(x)... returns f(c).
• Dirac comb: [...1000100010001..
## Sinusoids
• Amplitude (height)
• Frequency (Hertz Hz) or 1/time or 1/space or C/wavelength
• Wavelength (time, space, or 1/Frequency)
• Phase (left-to right shift)
• Euler Formula:
ei &theta = cos(&theta) + i sin(&theta).
• 2-D sinusoids are "gratings."
Some math leads to: cos(ux +vy) and sin(ux +vy) are 2-D sinusoids as in the figure. Their ridges and troughs fall along the parallel lines ux+vy = k &pi for integer k, and their wavelength is 2 &pi / &radic (u2 + v2). So we can write a 2-D wave as ei(ux + vy).
## Inner Product
Dot Product from high school vectors:
Generalize to n-vectors:
x · y = &Sigma n x(i)y(i).
Generalize to continuous functions:
f · g = &int f(x) g(x) dx
• Dot product measures similarity of unit-magnitude vectors (= cosine of their angular difference).
• Dot product projects one vector onto another. Again, can think of similarity or matching: same-direction, projection is large. Orthogonal vectors, projection is zero.
## Basis Functions
A weighted sum of basis functions is equal to some other function of interest. Very similar to expressing a vector's position in another coordinate system.
Family of something like shifted Dirac Deltas (but not quite): "obviously" true.
Family of Sinusoids: not obviously true (unless your name's Fourier). But easier to formalize.
In 2-D, implies you can build any image out of gratings(!!)
## Basis Functions Cont.
Orthogonal Basis functions have zero inner product with any other function in the family, but positive i.p. with selves (Dirac, Sinusoids both are orthogonal sets.)
Lots of useful basis functions: Bessel (vibrational modes of drumheads), Legendre, Laplace (solving differential equations, e.g... damped sinusoids), spherical harmonics (vib. modes in spheres):
## Linear and Linear Shift-Invariant Systems: Definition
Here, systems of Linear operators, not systems of linear equations (related, but...). Generally a linear system can be represented as a matrix (or some generalization involving continuous functions) operating on an input vector. y = Mx.
Let f1(t) and f2(t) be input functions and g1(t) and g2(t) their corresponding output functions, and &alpha and &beta be scalar weights.
Then for input function &alpha f1(t) + &beta f2(t) the output is &alpha g1(t) + &beta g2(t).
This is the Superposition Principle.
In a linear, shift-invariant (LSI) system, input f(t-h) produces g(t-h).
## LSI Systems as Matrix Multiplication
LSI systems are a specialized subcase of linear systems: they are characterized by a (shiftable) vector or function rather than a full matrix. An example: a linear imaging system---
If x is a one-dimensional scene'' and y its image, and M models a camera,
0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0
1 0 0 1 0 0 0 0 1
2 = 0 0 0 1 0 0 0 * 2
4 0 0 0 0 1 0 0 4
0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0
Can imagine these as infinite objects. Each row is a shifted version of the Dirac delta.
An out-of-focus camera C2 might be modeled like this:
1 1 0 0 0 ...
0 1 1 0 0 ...
0 0 1 1 0 ...
0 0 0 1 1 ...
0 0 0 0 1 ...
....
A camera C3 that behaves nicely in the middle of its field of view but goes blurry around the edges could be:
0 1/2 1/4 0 0 0 0 0 0
1/4 1/4 1/2 1/4 0 0 0 0 0
1 0 0 1 0 0 0 0 1
2 = 0 0 0 1 0 0 0 * 2
4 0 0 0 0 1 0 0 4
1 0 0 0 0 1/4 1/2 1/4 0
0 0 0 0 0 0 1/4 1/2 0
## Observations
• LSIs are even nicer than linear systems because a shifted version of the same input is always the same output, shifted. Their matrices have a simple form: just one row, repeated and shifted.
• The output vector can be considered to be indexed by the amount the single matrix row has been shifted.
• For LSIs, the matrix is a wasteful notation. Each output element is a dot product of the input with a differently-shifted copy of the one row'' in the matrix. The operation that creates the shifted dot products, and indexes the output by the shift, is the discrete convolution operation.
• All extends to continuous domain: a continuous function as input, operated on by a shifting continuous function instead of a matrix row. Continuous addition is integration, so we obtain the (continuous) colvolution integral.
• LSI systems all compute a convolution, and if a system computes a convolution it's LSI!
## Sinusoids as Eigenfunctions
Sinusoid input to ANY linear, shift-invariant system (any h(t)) yields sinusoid output of same frequency (wavelength), possibly shifted (in phase) and changed in amplitude (amplified or attenuated).
This beautiful fact seems surprising, but arises from the definition of convolution and the integral properties of eikt (sinusoids).
This is essentially what it means to be an eigenvector (or eigenfunction) of a linear operator. An eigenvector of a matrix remains in the same "direction" when multiplied by the matrix: though it may be lengthened or shortened (technically it may be multiplied by a scalar). A sinusoid is still a sinusoid of the same frequency when operated on by a LSI. The possible phase shift is a consequence of the vectors actually being complex-valued.
## Linear Systems: Examples
Our LSI systems will operate both in the time-domain (e.g. sound), and in the 2-D spatial domain (e.g. images). We draw them using f(t) as the input, h(t) as the impulse response or point spread function, and g(t) as the output. The box performs a convolution.
The differential d/dt is an LSI operator.
Physical systems can often be approximated by linear systems.
## Linear Systems and Reality
An ideal camera's film frame extends infinitely and exactly records all levels of input, from full dark to infinitely bright.
A ideal camera's point spread function is a Dirac Delta (impulse).
A real PSF is always more complicated!
Linear output response to a linearly increasing input is desired, but...
## Convolution and Correlation
A linear system convolves its impulse response (or PSF) with the input.
g(t) = f(t) &otimes h(t), or g(t) = f(t) * h(t).
Formal def: (f &otimes g )(t) = &int - &infin &infin f(&tau)g(t - &tau)) d &tau .
Note one input is reversed in time (or space) (see camera fig above).
• The convolution is commutative: f &otimes g = g &otimes f
• Array convolution: N-vector &otimes M-vector = (M+N-1)-vector.
• In Matlab, conv([1 2 3 4], [2 4 6]) gives [2 8 20 32 34 24]
• and in 2-D
1 1 * 1 1 = 1 2 1
1 1 1 1 2 4 2
1 2 1
## Correlation
Correlation is the same as convolution, only one of the inputs is not flipped backwards. It is often notated just the same way, with * or &otimes .
g(t) = f(t) &otimes h(t), or g(t) = f(t) * h(t).
Formal def: (f &otimes g )(t) = &int - &infin &infin f(&tau)g( &tau - t)) d &tau .
• Correlation is commutative: f &otimes g = g &otimes f
• Array correlation: N-vector &otimes M-vector = (M+N-1)-vector.
• In Matlab, conv(fliplr([ 1 2 3 4]), [2,4,6]) gives
[8 22 40 28 16 6]
• The correlation also looks like a shifting dot-product, and without the reversal it's easy to consider it as a matching function being tried out at all shifts. Where the arguments best match, the correlation is highest. The autocorrelation of a function with itself is highest at 0 offset.
## In-class Examples
The correlation of the delta function with itself &delta(t) &otimes &delta(t) -- its autocorrelation
The autocorrelation of the delta-comb function.
The correlation of the delta function with an arbitrary function &delta(t) &otimes h(t) -- sometimes correlation of two different functions is called their cross-correlation when it could get mixed up with the autocorrelation in the surrounding prose.
Note the last exercise shows that to discover the impulse-response (PSF) of an unknown system, just give it an impulse: the output is the PSF! Like 'kicking the tyres' but more informative.
The correlation of [1 2 3 2 1] with [1 1 1].
## Fourier Transform
The Fourier transform is a linear mathematical operation that takes time- or space-domain input (sound wave, voltage waveform, image,...) and outputs an equivalent (spatial) frequency-domain representation. The operation is lossless and invertible.
Essentially, it decomposes the input into a number of sinusoids of varying magnitude and phase (and in two dimensions, directions). The inverse transform reverses the process.
The formal definition is
F(&nu) =&int f(t) e - 2 &pi i &nu t dt ,
where t is time or space and &nu is frequency. The inverse is simply related: (We have left out some normalization constants).
f(t) =&int F(&nu) e 2 &pi i &nu t dt .
The Fourier transform (FT) is an inner-product integral that answers the question: How much of the particular sine wave e- 2 &pi i &nu t is in this input function f(t). In mathematical terms, it is projecting f(t) into the transform basis space of sinusoids.
The result is generally it is a complex function (real and imaginary parts).
## FT Properties
• FT is linear, i.e., the FT of a weighted sum of functions is the weighted sum of their FTs.
• FT of any symmetric (even) function is real and even: (the function is a sum of cosines).
• FT of any antisymmetric (odd) function is imaginary and odd: (function is a sum of sines).
• FT of a real function is Hermitian: F(t) = conj(F(-t)).
• In the FT of a shifted function, the magnitude of all the components (the complex numbers) stays the same, but they rotate (their phase changes) as a function of the shift and their frequency.
• Scaling or similarity: FT[f(ax)] = (1/a) F(&nu/a)
• Thus a time-reversal property: FT[f(-x)] = F(- &nu)
• FT of a Gaussian is a Gaussian.
• FT of a Dirac comb is a Dirac comb.
• Scaling examples: FT of a narrower (wider) Gaussian is a wider (narrower) one, FT of a higher (lower) frequency Dirac comb is a lower (higher) frequency one, FT of a (single) Dirac function is flat, with waves of all frequencies and the same magnitude.
• FT has an equivalent for vector (discrete) inputs. It has a very clever implementation called the Fast Fourier Transform, or FFT, which we'll be using.
## (Fast) Fourier Transform Issues
• FFT takes N log(N) time for an N- vector (matrix multiplication, the "Slow FT"), takes N 2 time.
• FFT works on each dimension separately, so the FFT of multi-dimensional arrays can be computed by repeatedly using the 1-D algorithm. The total time required scales exponentially in number of dimensions (as does the number of data points), but retains its N log(N) time with respect to the total number of data points.
• FFT works fastest for arrays whose dimensions are all some power of two: 2D i, but can be adjusted to work for any size..
• Matlab FFT and FFT -1 commands: fft, ifft, fft2, ifft2
• fft(X,N) creates a complex N -long output vector. Input is truncated if longer than N, else padded with 0's.
• Periodic and aperiodic functions. For aperiodic for N-vector one could use fft(X, 2*N), which effectively embeds the function in a larger space of zeros. There may still be high-frequency artifacts arising from discontinuities at the embedding boundaries
• Generally FT takes complex input. That happens especially in intermediate steps, but most real-world input is real. So for N-vector of doubles in, get N-vector of complex out. Twice the numbers but only need N since output is symmetrical for real input.
• The FT (of anything but symmetric real input function ) is complex. This is difficult to visualize. The power spectrum provides a real signal representing quantities of engineering interest that is easy to visualize (i.e. by plotting it).
• Takes a little experience, so experiment.
## Power Spectrum
Analyzing time series: one issue is dominant frequencies. Power spectrum is magnitude of FT: (F .* conj(F)). Below, the PS is displayed with its 0-frequency origin in the middle of the X-axis (at 32).
PS tells how much power the signal contains at a given frequency.
Sampling Issues: bandlimited input, sample at least twice the maximum signal frequency.
## More Power Spectrum
60, 150, 350 Hz. sines plus 0-mean Gaussian noise.
Power Spectrum.
function thePS = PowSpec1D(X,n)
Y = fft(X,n);
thePS = (Y .* conj(Y)) / n;
end
... % and in the calling script
plot(xaxis, fftshift(thePS));
## Complex Numbers
Complex number a + bi is a 2-vector (a,b) living in the complex plane, which has a real axis and an imaginary axis.
Or, a complex number is an ordered pair (a,b) that is a mathematical object having rather funny rules of operation.
The conjugate of a complex number negates the imaginary part:
conj(a + bi) = (a - bi).
You can work out that (a + bi) · conj(a+bi) is a2 + b2, the squared magnitude (length) of the (a,b) vector in the complex plane.
Complex numbers are added, subtracted, multiplied, and divided by formally applying the associative, commutative and distributive laws of algebra, together with the equation i2 = -1 :
• Addition: (a + bi) + (c+di) = (a+c) + (b+d)i
• Subtraction: (a + bi) - (c+di) = (a-c) + (b-d)i
• Multiplication: (a + bi) (c+di) = ac+bci +adi +bdi2 = (ac-bd) + (bc +ad)i
• Division: (a+bi)/(c+di) = (ac+bd)/(c2+d2) + (bc-ad)/(c2+ d2, where c and d are not both zero. Derive this by multiplying both the numerator and the denominator by the conjugate of the denominator c + di, which is (c - di).
## Phasors
A phasor is a complex number, considered as a vector, and considered to rotate around the origin without changing its length. This alters the magnitude of its real and imaginary parts, and is said to change its phase. FT entries are complex. Considered as phasors, each corresponds to a sinusoid at a frequency given by its coordinates in the FT, amplitude equal to its length, and phase given by its the angle to the real axis.
Looking at individual phasors: there are two of them for a sine wave, 180 degrees apart One frequency but two symmetrical conjugate elements in FT. Can see phasor rotate as phase changes. In Image below, for sine (red, 0 phase angle), we get the imaginary (0.0000 +-32.0000i). Shifted sine (blue) gives (29.5641 +-12.2459i).
## The Convolution Theorem
Another FT symmetry property: Convolution in the time (space) domain is dual to elementwise multiplication in the frequency domain.
Let FT denote the Fourier transform operation. Then
FT{f &otimes g} = FT{f} · FT{g}
where · denotes point-wise multiplication.
Also vice-versa:
FT{f · g} = FT{f} &otimes FT{g},
And applying the inverse Fourier transform FT -1 to the first equation , we get:
f &otimes g= FT -1 {FT{f} · FT{g}}
As convolution and correlation are important, The Convolution Theorem is a key idea and technique. We'll see some applications later.
## Linear Systems and the Frequency Domain
Use convolution theorem to make a frequency-domain version of the linear system block diagram:
Inputs and outputs are FT{f(t)} and FT{g(t)}, The function in the box is FT{h(t)}, and the operation of the box is elementwise multiplication (Matlab's .*).
Box is a "graphic equalizer": sound (say) comes in as a pressure wave, the sum of sinusoids of many frequencies. Equalizer amplifies or attenuates (and changes phase) according to H(&nu), the box's function.
FT{h(t)} =H(\nu) is called the Modulation Transfer Function (MTF).
So graphic equalizer is just an MTF, basically. It's a linear system that's easier to think about in the frequency domain than in the temporal domain.
## The Sampling Theorem
Basic Question: Can we exactly reproduce a continuous signal from a finite number of discrete samples?
Yes (!!), if...
• There is a limit to the signal frequency: that is, it's a band-limited function.
• The sampling is done often enough to reconstruct the highest-frequency sinusoid in the signal.
The sampling rate must clearly be at least twice the highest signal frequency. It is called the Nyquist frequency. Turns out any rate above that works (given enough signal).
If sample at high and low peaks of red sine, can get its amplitude, phase, and frequency. But higher-frequency blue sine is undetected, being 0 at all sampling points.
## The Sampling Theorem: Frequency Domain
• The FT of the Dirac comb is another Dirac comb
• Scaling Property: the closer the time (space) domain peaks are together, the farther apart the frequency domain peaks are.
• Multiplying a function f(t) by the Dirac comb effectively gives a sampled version of f(t).
• Convolution Theorem: FT{f · g} = FT{f} &otimes FT{g},
Sampling at too low a rate means copies of FT{signal} overlap (causing 'aliasing') and our strategy fails due to addition and confusion between copies. Faster sampling separates the FT copies in freq. space, and we may imagine snipping one out and retrieving
f(t) = FT -1 {F(&nu)}.
# Three Frequency-domain Operations
## Matching
• Matching known signal in Gaussian noise: optimal is correlation detection or matched filter detection.
• Autocorrelation (f(t) &otimes f(t)) is highest when f(t) lines up with itself.
• Crosscorrelation (f(t) &otimes g(t)) has peak if f(t) lines up with a version of itself itself in g(t). So peaks in cross-correlation can signal positive detections.
• Works for any shape anywhere in an image (like airplanes seen from above in aerial photo): shift invariant. BUT: scale and rotation variant :-{.
• For best peaks, autocorrelation of 'shape' should be as close as possible to an impulse. Leads to design of such functions, and to finding them to assure good results when matching points in images (e.g. for automatic panaroma construction from multiple images).
A 'chirp' and the autocorrelation of a random vector of 1 and -1.
## Deconvolution
Produce good image from one made with an exotic PSF (as in some of today's cameras (see 'coded aperture' and 'computational camera'. Or fix image made with undesireable PSF (bad optics (e.g Hubble), bad 'seeing conditions' (Optical astronomy, atmospheric instability), motion or focus blur,...)
How? Convolution theorem:
FT{f &otimes h} = FT{f} · FT{h}
Where LHS is the image from a camera with point-spread function h and scene data f. If h is an ideal Dirac delta, the output is the scene. Camera motion: h image is spread out along 2-D path. Focus blur: PSF becomes a disk (sensor intersects cone of focused rays NOT at its point).
General treatment: Assume we know or can guess h. In the equation above, divide (elementwise) by FT{h}.
FT{f &otimes h}/FT{h} = FT{f}, so
f = FT -1 [ FT{f &otimes h} / FT{h}]
We now have FT of what we want (the input function) using things we know (PSF and degraded image). Inverse-transforming both sides recovers the original input. It almost works.
Consider boxcar (or disk) blur function, which formalizes either (in 2-D) defocus blur or (in 1-D) straight-line motion blur. Here's FT{h}: note it crosses zero often, so it is near zero often.
Multiplying by 1/FT{h} where the FT is near zero amplifies frequencies there by huge amounts (or Inf if there is a divide by zero error).
Noise at these amplified frequencies can overwhelm the signal. Some care (thresholding before multiplying, say) is needed.
The Gaussian is a user-friendly blur function since its FT is Gaussian, so always positive. Dividing by it amplifies high frequencies.
## Time-Domain Signal Processing
• Probably most SP done in original (time, space) domain: Photoshop, guitar effects...
• e.g. Spatial derivative of image in x-direction ('vertical edge finder'): correlate with [-1 1]: easy. OR can take FT{d/dt}, get an NxN filter, FT the image, multiply the two, inverse FT the result... whew. Too much computation.
• Data analysis usually involves fitting mathematical models (curves) to experimental data, smoothing or interpolating data, creating statistics and visualizations: all spatial or time-domain operations. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.846875786781311, "perplexity": 2876.9834604927373}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823839.40/warc/CC-MAIN-20171020063725-20171020083725-00406.warc.gz"} |
http://scholar.cnki.net/result.aspx?q=%E6%9D%A5%E6%BA%90%3A(General+Relativity+and+Gravitation) | 高级检索
作者：Alfred Molina , Naresh Dadhich , Avas Khugaev 来源：[J].General Relativity and Gravitation(IF 1.902), 2017, Vol.49 (7)Springer 摘要：In the paper (Khugaev et al. in Phys Rev D94:064065. arXiv: 1603.07118 , 2016 ), we have shown that for perfect fluid spheres the pressure isotropy equation for Buchdahl–Vaidya–Tikekar metric ansatz continues to have the same Gauss form in higher dimensions, and hence higher...
作者：Wei Xu , De-Cheng Zou 来源：[J].General Relativity and Gravitation(IF 1.902), 2017, Vol.49 (6)Springer 摘要：In \((2+1)\) -dimensional AdS spacetime, we obtain new exact black hole solutions, including two different models (power parameter \(k=1\) and \(k\ne 1\) ), in the Einstein–Power–Maxwell (EPM) theory with nonminimally coupled scalar field. For the charged hairy black hole wi...
作者：Jingbo Wang 来源：[J].General Relativity and Gravitation(IF 1.902), 2017, Vol.49 (12)Springer 摘要：Recently, the idea that gravity is emergent has attract many people’s attention. The “Emergent Gravity Paradigm” is a program that develop this idea from the thermodynamical point of view. It expresses the Einstein equation in the language of thermodynamics. A key equat...
作者：Károly Z. Csukás 来源：[J].General Relativity and Gravitation(IF 1.902), 2017, Vol.49 (7)Springer 摘要：In geometric inequalities ADM mass plays more fundamental role than the concept of quasi-local mass. This paper is to demonstrate that using the quasi-local mass some new insights can be acquired. In spherically symmetric spacetimes the Misner–Sharp mass and the concept of t...
作者：Hui-Ling Li , Shuai-Ru Chen 来源：[J].General Relativity and Gravitation(IF 1.902), 2017, Vol.49 (10)Springer 摘要：In this paper, we use the generalized uncertainty principle (GUP) and quantum tunneling method to research the formation of the remnant from a Schwarzschild black hole with global monopole. Based on the corrected Hamilton–Jacobi equation, the corrections to the Hawking tempe...
作者：Xavier Jaén , Alfred Molina 来源：[J].General Relativity and Gravitation(IF 1.902), 2017, Vol.49 (8)Springer 摘要：The group of rigid motions is considered to guide the search for a natural system of space-time coordinates in General Relativity. This search leads us to a natural extension of the space-times that support Painlevé–Gullstrand synchronization. As an interesting example, here...
作者：Eugenio Bianchi , Hal M. Haggard , Carlo Rovelli 来源：[J].General Relativity and Gravitation(IF 1.902), 2017, Vol.49 (8)Springer 摘要：We show that in Oeckl’s boundary formalism the boundary vectors that do not have a tensor form represent, in a precise sense, statistical states. Therefore the formalism incorporates quantum statistical mechanics naturally. We formulate general-covariant quantum statistical ...
作者：H. F. Mota , K. Bakke 来源：[J].General Relativity and Gravitation(IF 1.902), 2017, Vol.49 (8)Springer 摘要：We investigate noninertial effects on the scattering problem of a nonrelativistic particle in the cosmic string spacetime. By considering the nonrelativistic limit of the Dirac equation we are able to show, in the regime of small rotational frequencies, that the phase shift has t...
作者：Bartolomé Coll , Joan Josep Ferrando , Juan Antonio Sáez 来源：[J].General Relativity and Gravitation(IF 1.902), 2017, Vol.49 (5)Springer 摘要：Every evolution of a fluid is uniquely described by an energy tensor. But the converse is not true: an energy tensor may describe the evolution of different fluids. The problem of determining them is called here the inverse problem . This problem may admit unphysical or non-...
作者：Zi-Hua Weng 来源：[J].General Relativity and Gravitation(IF 1.902), 2017, Vol.49 (7)Springer 摘要：The paper aims to apply the complex octonion to explore the influence of the energy gradient on the Eötvös experiment, impacting the gravitational mass in the ultra-strong magnetic fields. Until now the Eötvös experiment has never been validated under the ultra-strong magnet... | {"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": 0.9199651479721069, "perplexity": 1747.0001579500456}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875141460.64/warc/CC-MAIN-20200217000519-20200217030519-00516.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=470 | # a little trig
by mercury
Tags: trig
P: n/a this is a question that was part of an mcq test. and i did'nt have any clue as to how to begin! finally i just tried plugging in values for A,B,C to see if they worked. for ex. i put A=B=C=60 (degrees)- equilateral and so on..but i'd like to know the logical way to do it... if A,B,C are the vertices of a triangle, and if cosBcosC + sinAsinBsinC = 1 is the triangle a) equilateral b) isosceles c) right angled isosceles d) right angled but not isosceles e) none of the above.
P: 115 Right, I'm assuming you know sin^2 +cos^2 = 1. Well you can see this is nearly similar, and almost exactly the same if B = C. So there's two angles the same. So now cos^2 + sinA sin^2 = 1 Which means sinA must be 1. Therefore A can only take value of 90 degrees - right angle (450...etc useless here) 180 - 90..other two angles combine as total 90 degrees, so they must be 45 each since you've already found they're the same. Which leaves you a right angled isosceles.
Related Discussions Precalculus Mathematics Homework 7 General Math 4 Calculus & Beyond Homework 1 Precalculus Mathematics Homework 5 Introductory Physics Homework 2 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9175727367401123, "perplexity": 1095.9049319274002}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164554256/warc/CC-MAIN-20131204134234-00097-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://fr.maplesoft.com/support/help/maple/view.aspx?path=updates/Maple2015/compatibility&L=F | Compatibility - Maple Help
Compatibility Issues in Maple 2015
The following is a brief description of the compatibility issues that affect users upgrading from Maple 18 to Maple 2015.
Automatic Evaluation of Pi in Float Expressions
• In Maple 2015, the symbol Pi is automatically converted to a float when it is adjacent to a float in a sum, product, or power.
> 1.0*Pi;
${3.141592654}$ (1)
> Pi^0.5;
${1.772453851}$ (2)
> 2*Pi*1.0;
${6.283185308}$ (3)
> Pi+1.0;
${4.141592654}$ (4)
• If your expression has no floats (that is, it contains only integers and rationals), Pi is not converted to a float and remains a symbol (as in previous versions of Maple).
> 1*Pi;
${\mathrm{\pi }}$ (5)
> Pi^(1/2);
$\sqrt{{\mathrm{\pi }}}$ (6)
> 2*Pi*1;
${2}{}{\mathrm{\pi }}$ (7)
> Pi+1;
${\mathrm{\pi }}{+}{1}$ (8)
• Note that the evaluation and conversion happen only when a float operates on Pi. That is, Maple does not evaluate and convert Pi in cases where Pi could be a float. For example, evalb does not perform the comparison in the following statement because evalb cannot compare a float (3.0) to a symbol (Pi).
> evalb(3.0 < Pi);
${3.0}{<}{\mathrm{\pi }}$ (9)
However, if you perform an arithmetic operation involving Pi and a float, Maple converts Pi to a float, evalb can perform the comparison, and the expected result is returned.
> evalb(3 < 1.0*Pi);
${\mathrm{true}}$ (10)
• In the following example, Pi is added to a product containing a float (sin(1.0)) and a symbol (x). Although Pi and the float are adjacent to each other, Pi is not converted to a float because the term it is being added to cannot be evaluated to a float.
> Pi + sin(1.0)*x;
${\mathrm{\pi }}{+}{0.8414709848}{}{x}$ (11)
Once you assign x to a float, the second term evaluates to a float, and Pi is then converted to a float.
> x := 0.0;
${x}{≔}{0.}$ (12)
> Pi + sin(1.0)*x;
${3.141592654}$ (13)
• Maple 2015 will no longer read .m files located in libname by default. In order to get Maple to read .m files, set readdotm kernel option to true:
• In Maple 2015, the name epsilon will be displayed as the lunate epsilon $\mathrm{\epsilon }$ instead of as $\mathrm{ϵ}$. The lunate epsilon will parse as the name epsilon, while the "inverted 3" epsilon will parse as the name varepsilon.
• Additionally, the names varkappa, varphi, varpi, varrho, varsigma and vartheta will be displayed as $\mathrm{\varkappa }$, $\mathrm{\varphi }$, $\mathrm{\varpi }$, $\mathrm{\varrho }$, $\mathrm{\varsigma }$ and $\mathrm{\vartheta }$ respectively. Conversely, these 2-D math symbols will parse to their corresponding Maple names. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 21, "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": 0.9860977530479431, "perplexity": 2991.609180367298}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571150.88/warc/CC-MAIN-20220810070501-20220810100501-00127.warc.gz"} |
http://math.stackexchange.com/questions/9863/equivalence-relations-on-s2-times-s2 | # Equivalence relations on $S^2 \times S^2$
I am trying to understand (simplify) the following space
$Z = S^2 \times S^2 / E$
Let $(a_1 , a_2 , a_3) \in S^2$ and $(n_1 , n_2 , n_3) \in S^2$. Here ${a_1}^2 + {a_2}^2 + {a_3}^2 = 1$ and ${n_1}^2 + {n_2}^2 + {n_3}^2 = 1$.
The equivalence relations $E$ are the following
1) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(-a_1, -a_2, -a_3);(n_1, n_2, n_3)]$
2) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(a_1, a_2, a_3); (\pi; a_1,a_2,a_3)*(-n_1, -n_2, -n_3)]$
where $(\pi; a_1,a_2,a_3)$ is a rotation of angle $\pi$ along the axis $(a_1, a_2, a_3)$. So the point $(\pi; a_1,a_2,a_3)*(-n_1, -n_2, -n_3)$ is the mirror image of $(n_1, n_2, n_3)$ in the plane whose normal is $(a_1, a_2, a_3)$.
The objective is to simplify the equivalence relation (2) by a homeomorphism of $S^2 \times S^2$. For example, the approach I am trying is to rotate the second $S^2$ such that the mirror plane becomes horizontal (XY plane). Since the normal of the mirror plane is $(a_1, a_2, a_3)$, the amount by which the second $S^2$ has to be rotated depends on the $(a_1, a_2, a_3)$ it corresponds to. And for this transformation to be a homeomorphism the rotation operations should vary continuously. The transformation should not complicate the first equivalence relation; one way this condition could be imposed is by ensuring that the rotation operation corresponding to $(a_1, a_2, a_3)$ and $(-a_1, -a_2, -a_3)$ are be the same. I am not able to come up with such a transformation. All my attempts to simplify (2) end up complicating (1).
To be more specific, I want to get a homeomorphism between the space $Z$ and the following space:
$Z^' = S^2 \times S^2 / E^'$
The equivalence relations $E^'$ are the following
1) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(-a_1, -a_2, -a_3);(n_1, n_2, n_3)]$
2) $[(a_1, a_2, a_3);(n_1, n_2, n_3)] \sim [(a_1, a_2, a_3); (n_1, n_2, -n_3)]$
Any ideas are appreciated. Please let me know if you have any questions.
-
Is this a Homework? – Djaian Nov 11 '10 at 15:38
No, I am graduate student in Materials Science department and it is a part of my project. Just to give a brief background: The space belongs to a product space of rotations and normal vectors $(SO(3) \times S^2)$. The space $S^2 \times S^2$ I mentioned above is a subspace of this space and my final objective is to be able to do statistical analysis on this space. – Srikanth Nov 11 '10 at 15:58
This isn't quite what you asked for, but notice that (1) simply gives us $\mathbb{RP}^2 \times S^2$. I'm not sure I understand your notation for (2); it seems like it doesn't depend at all on $\pi$. If it's true that you're just reflecting $\vec{n}$ through the plane perpendicular to $\vec{a}$, then you're just getting a copy of $D^2$ over every point of $\mathbb{RP}^2$. This is a fiber bundle, and if you think through it I think you'll find that it's the disk bundle of the tangent bundle $T\mathbb{RP}^2$.
You are right. It is just reflecting $\vec{n}$ through the plane perpendicular to $\vec{a}$. When you rotate a vector along an axis by $\pi$ and invert it, you end with a reflection through a plane that is perpendicular to that axis of rotation. I should have just said it is a reflection but I put it that way because that is how my equivalence relation looks after certain simplifications that I performed and I wanted to present the complete problem. – Srikanth Nov 11 '10 at 20:24
Anyways, I understand what you are saying but I am do not have enough experience with algebraic topology (am self-learning). I understand that (1) implies $Z$ is $\mathbb{RP}^2 \times S^2$. And yes, there is just a copy of $D^2$ over every point of $\mathbb{RP}^2$ and hence a fiber bundle. But what is the difference between this space and $Z^'$ that I mentioned above ? – Srikanth Nov 11 '10 at 20:25
$Z$ isn't $\mathbb{RP}^2\times S^2$, that's just what you get from $S^2\times S^2$ and apply only the first equivalence relation. However, now that I'm looking at it closely, the second is definitely not equivalent to the first. There's a fundamental difference between moving your reflecting plane and not. $Z'$ will be a trivial disk bundle, because over each point (= equivalence class $\{\pm \vec{a}\}$) in the first factor you'll have "the same" disk -- i.e. $Z'=\mathbb{RP}^2\times D^2$. This is different from the disk bundle of $T\mathbb{RP}^2$! Does that make sense? – Aaron Mazel-Gee Nov 12 '10 at 5:22
Thanks, Aaron. I totally get $Z^' = \mathbb{RP}^2 \times D^2$. I need to read up on tangent bundles and disk bundles to understand it better. Can you please guide me to some introductory books that introduce these concepts. Thanks, your help is much appreciated. – Srikanth Nov 12 '10 at 15:19
Sure. First of all, disk bundles are really pretty similar to vector bundles -- you can throw away the sphere bundle (i.e. in each fiber throw away the boundary of the disk) and recover something homeomorphic to the original vector bundle. The proof that $T\mathbb{RP}^2$ is nontrivial (i.e. $T\mathbb{RP}^2\not= \mathbb{RP}^2 \times \mathbb{R}^2$) uses something called Stiefel-Whitney classes, the best resource for which is probably Milnor & Stasheff. However, this requires some cohomology theory, so you should make sure that's not totally foreign to you. On the other hand, there's a very – Aaron Mazel-Gee Nov 12 '10 at 17:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8429262638092041, "perplexity": 153.39026794556918}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701147841.50/warc/CC-MAIN-20160205193907-00247-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://cran.cc.uoc.gr/mirrors/CRAN/web/packages/matrixNormal/vignettes/introduction-to-matrixnormal-package.html | # Introduction to Matrix Normal Package
#### 2021-04-02
The matrixNormal package in the Comprehensive R Archive Network (CRAN) [1] consists of two different types of functions: distribution and matrix ones. First, one can compute densities, probabilities and quantiles of the Matrix Normal Distribution. Second, one can perform useful but simple matrix operations like creating identity matrices and matrices of ones, calculating the trace of a matrix, and implementing the matrix operator vec().
## Installation
You can install the released version of matrixNormal from CRAN[1] with install.packages("matrixNormal") and can load the package by:
> library(matrixNormal)
Attaching package: 'matrixNormal'
The following object is masked from 'package:base':
I
## Distribution functions
The matrix normal distribution is a generalization of the multivariate normal distribution to matrix-valued random values. The parameters consist of a n x p mean matrix M and two positive-definite covariance matrices, one for rows U and another for columns V. Suppose that any n x p matrix $$A \sim MatNorm_{n,p}(M, U, V)$$. The mean of A is M, and the variance of vec(A) is the Kronecker product of U and V.
The matrix normal distribution is a conjugate prior of the coefficients used in multivariate regression. Suppose there are p predictors for k dependent variables. In univariate regression, only one dependent variable exists, so the conjugate prior for the k regression coefficients is the multivariate normal distribution. For multivariate regression, the extension of the conjugate prior of the coefficient matrix is the matrix normal distribution. This package also contains the PDF, CDF, and random number generation for the matrix normal distribution (*matnorm). See matrixNormal_Distribution file for more information.
For instance, the USArrests dataset in the dataset package examines statistics in arrests per 100,000 residents for assault, murder, and rape in each of 50 states since 1973. Suppose that a researcher wants to determine whether the outcomes of assault and murder rates are associated with urban population and rape. The researcher then decides to use a Bayesian multiple linear regression and makes the assumption that the states are independent. However, the covariance between assault and murder is nonzero and needs to be taken into account. In fact, it has a correlation of 0.411 as given below.
>
> library(datasets)
> data(USArrests)
> X <- cbind(USArrests$Assault, USArrests$Murder)
> Y <- cbind(USArrests$UrbanPop, USArrests$Rape)
> cor(Y)
[,1] [,2]
[1,] 1.0000000 0.4113412
[2,] 0.4113412 1.0000000
We can assume that the outcome Y follows a matrix normal distribution with mean matrix M, which is the matrix product of the coefficient matrix, $$\Psi$$, times X. The covariance across states is assumed to be independent, and the covariance across the predictors be $$\Sigma$$. Or succinctly, with $$k=2$$ outcomes, $Y \sim MatNorm_{nx2}( M = X\Psi, U = I(n), V = \Sigma_2)$
of coefficient matrix times X, and covariance across the predictors $$\Phi$$. The overall covariance matrix of Y, $$\Sigma$$, is a block diagonal matrix of $$\Phi$$. For instance, suppose that Y has the following distribution, and if we know the parameters, we can calculate its density using dmatnorm().
> # Y is n = 50 x p = 2 that follows a matrix normal with mean matrix M, which is product of
> M <- (100 * toeplitz(50:1))[, 1:2]
> dim(M)
[1] 50 2
[,1] [,2]
[1,] 5000 4900
[2,] 4900 5000
[3,] 4800 4900
[4,] 4700 4800
[5,] 4600 4700
[6,] 4500 4600
> U <- I(50) # Covariance across states: Assumed to be independent
> U[1:5, 1:5]
1 2 3 4 5
1 1 0 0 0 0
2 0 1 0 0 0
3 0 0 1 0 0
4 0 0 0 1 0
5 0 0 0 0 1
> V <- cov(X) # Covariance across predictors
> V
[,1] [,2]
[1,] 6945.1657 291.06237
[2,] 291.0624 18.97047
>
> # Find the density if Y has the density with these arguments.
> matrixNormal::dmatnorm(Y, M, U, V)
[1] -30446783
The coefficient matrix, $$Psi$$, for urban population and rape has dimensions 2 x 3 for the k = 2 outcomes and the p = 3 predictors. A semi-conjugate prior can be constructed on $$\Psi$$ to be a Matrix Normal distribution with mean $$\Psi_0$$ as a matrix of ones, covariance across the predictors as $$X'X$$, and with no covariance across states (due to independence). The conjugate prior for the covariance between the outcomes $$\Sigma$$ is the inverse-Wishart distribution with mean covariance $$\Sigma_0$$ and degrees of freedom $$\nu$$. After defining the parameters, one random matrix is generated from this prior.
> # Generate a random matrix from this prior. The prior mean of regression matrix
> J(2, 3)
1 2 3
1 1 1 1
2 1 1 1
> # The prior variance between rape and population
> t(X) %*% X
[,1] [,2]
[1,] 1798262 80756.0
[2,] 80756 3962.2
> # The prior variance between regression parameters
> I(3)
1 2 3
1 1 0 0
2 0 1 0
3 0 0 1
>
> # Random draw for prior would have these values
> A <- matrixNormal::rmatnorm(M = J(2, 3), U = t(X) %*% X, V = I(3))
> A
1 2 3
[1,] 304.6660 -635.3648 -38.41940
[2,] -883.9216 57.3710 -37.94706
>
> # Predicted Counts for y can be given as:
> ceiling(rowSums(X %*% A))
[1] -98523 -105723 -115523 -77740 -109657 -82129 -43455 -92950 -136967
[10] -92926 -21561 -46541 -100901 -47934 -22572 -47635 -48619 -105223
[19] -32452 -120504 -58802 -104585 -28910 -109520 -73483 -45420 -41367
[28] -103564 -22855 -65087 -115053 -103351 -135631 -17301 -50605 -61442
[37] -62925 -44572 -67165 -115432 -35029 -80805 -85171 -47060 -19619
[46] -64930 -56980 -34826 -21810 -65306
However, the predicted counts do not have much meaning because the prior is uninformed from the data. We should use the posterior distribution to predict Y. Iranmanesh 2010 [2] shows the posterior distributions to be $\Psi | \Sigma, Y, X \sim MatNorm_{kxp}( \frac{\hat{\Psi} +\Psi_0}{2}, (2X^TX)^{-1}, \Sigma)$ and $\Sigma | Y, X \sim W^{-1}_p(S*, 2\nu+n)$ where:
* $$\hat{\Psi}$$ is the maximum likelihood estimate (MLE) for the coefficient $$\Psi$$ matrix. * S is the MLE for the covariance matrix $$\Sigma$$: $$(Y-X\cdot\hat{\Psi})^T*(Y-X\cdot\hat{\Psi})$$ * $$S^{*}$$ is the data-adjusted matrix of inverse Wishart, $S^{*} = 2\Sigma_0 + S + (\hat{\Psi}-\Psi_0)'(2X^TX)^{-1}(\hat{\Psi}-\Psi_0)$ * c is the dimension of $$\Sigma$$ * n is the sample size, the number of rows in Y and X.
Additional details can be found in Iranmanesh et al. 2010 [2]. At any rate, the matrixNormal package can be used in Bayesian Multivariate Linear Regression.
## Matrix functions
In the matrixNormal package, useful but simple matrix operations have been coded:
Filename Description
is.symmetric.matrix Is a matrix square? symmetric? positive definite? or positive semi-definite? A tolerance is included here.
Special_matrices Creates the identity matrix I and matrix of 1’s J.
tr Calculates the usual trace of a matrix
vec Stacks a matrix using matrix operator vec() and has option to keep names.
vech Stacks elements of a numeric symmetric matrix A in lower triangular only (using half vectorization, vech()).
For example, these functions can be applied to the following matrices.
> # Make a 3 x 3 Identity matrix
> I(3)
1 2 3
1 1 0 0
2 0 1 0
3 0 0 1
> # Make a 3 x 4 J matrix
> J(3, 4)
1 2 3 4
1 1 1 1 1
2 1 1 1 1
3 1 1 1 1
> # Make a 3 x 3 J matrix
> J(3, 3)
1 2 3
1 1 1 1
2 1 1 1
3 1 1 1
>
> # Calculate the trace of a J matrix
> tr(J(3, 3)) # Should be 3
[1] 3
>
> # Stack a matrix (used in distribution functions)
> A <- matrix(c(1:4), nrow = 2, dimnames = list(NULL, c("A", "B")))
> A
A B
[1,] 1 3
[2,] 2 4
> vec(A)
1:A 2:A 1:B 2:B
1 2 3 4
>
> # Test if matrix is symmetric (used in distribution function)
> is.symmetric.matrix(A)
[1] "A is not symmetric. Top of the matrix: "
A B
[1,] 1 3
[2,] 2 4
[1] FALSE
## Conclusion
Although other packages on CRAN have some matrixNormal functionality, this package provides a general approach in randomly sampling a matrix normal random variate. The MBSP::matrix.normal , matrixsampling::rmatrixnormal, and LaplacesDemon::rmatrixnorm functions also randomly sample from the matrix normal distribution [35]. The function in [MBSP] (https://CRAN.R-project.org/package=MBSP/) uses Cholesky decomposition of individual matrices U and V [3]. Similarly, the function in [matrixsampling] (https://CRAN.R-project.org/package=matrixsampling/) uses the Spectral decomposition of the individual matrices U and V [4]. Comparatively, the new function, rmatnorm(), in matrixNormal package is flexible in the decomposition of the covariance matrix, which is Kronecker product of U and V. While you can simulate many samples using the rmatrixnormal() function from matrixsampling package, the new rmatnorm() function only generates one variate, but can generate many samples by placing rmatnorm() function inside a for-loop. The [LaplacesDemon] (https://cran.r-project.org/package=LaplacesDemon) package also has a density function, dmatrixnorm(), which calculates the log determinant of Cholesky decomposition of a positive definite matrix [5]. However, the random matrix A that follows a matrix normal distribution does not need to be positive definite [2]. There is no such restriction on the random matrix in matrixNormal package.
In conclusion, the matrixNormal package collects all forms of the Matrix Normal Distribution in one place: calculating the PDF and CDF of the Matrix Normal distribution and simulating a random variate from this distribution. The package allows the users to be flexible in finding the random variate. Its main application in using this package is the Bayesian multivariate regression.
## Computational Details
This vignette is successfully processed using the following.
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collate C
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tz America/New_York
date 2021-04-02
-- Packages -------------------------------------------------------
package * version date lib source
LaplacesDemon 16.1.4 2020-02-06 [3] CRAN (R 4.0.2)
matrixcalc 1.0-3 2012-09-15 [3] CRAN (R 4.0.2)
matrixsampling 2.0.0 2019-08-24 [3] CRAN (R 4.0.2)
MBSP 1.0 2018-04-09 [3] CRAN (R 4.0.2)
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[3] /Library/Frameworks/R.framework/Versions/4.0/Resources/library
## References
1. R Core Team R: A Language and Environment for Statistical Computing; R Foundation for Statistical Computing: Vienna, Austria, 2018;
2. Iranmanesh, A.; Arashi, M.; Tabatabaey, S.M.M. On Conditional Applications of Matrix Variate Normal Distribution. IJMSI 2010, 5, 33–43.
3. Bai, R.; Ghosh, M. MBSP: Multivariate Bayesian Model with Shrinkage Priors; 2018;
4. Laurent, S. Matrixsampling: Simulations of Matrix Variate Distributions; 2018;
5. Statisticat; LLC. LaplacesDemon: Complete Environment for Bayesian Inference; Bayesian-Inference.com, 2018; | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8358182907104492, "perplexity": 1889.322225465246}, "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-2021-31/segments/1627046154457.66/warc/CC-MAIN-20210803092648-20210803122648-00174.warc.gz"} |
http://fluencyuniversity.com/m9o4am/b437h.php?cc98e9=inverse-of-ab | It's called the inverse of A, as I've said three times already. trailer ii Right inverse B with AB I m Left and Right Inverse Let A be a m nmatrix i from MA 1101R at National University of Singapore 0000002429 00000 n But theproductAB has an inverse, if and only if the two factors A and B are separately invertible (and the same size). Also in above statement about square matrices, does C=A? 0000002332 00000 n That is, C-A = B- for some B- from B{1 }. The inverse of a product AB is (AB)−1= B−1A−1. 0000010004 00000 n Let A be an m×n matrix and B be an n×lmatrix. The inverse of A is A-1 only when A × A-1 = A-1 × A = I. While the most common case is that of matrices over the real or complex numbers, all these definitions can be given for matrices over any ring. Hence, if we will get the inverse of a 4 x 4 matrix ‘A’ to be ‘B’, then for checking our work, we will have to simply multiply ‘AB’ and ‘BA’. Inverse of Square Matrices A square matrix A of order n is invertible if there exists a square matrix B of order n such that AB = I n = BA. This man made \$2.8 million swing trading stocks from home. We can calculate the inverse of the matrix in the following steps-1st Step - Calculate a minor matrix. View Answer Answer: adj A⁄|A| 15 Two matrices A and B are multiplied to get AB if A both are rectangular. 0000010688 00000 n If y = f (x) = ab x, then we may solve for x in terms of y using logarithms: x = f –1 (y) = log b (y/a) We see that the inverse of an exponential with base b is a logarithm with base b. But the product ab = −9 does have an inverse, which is 1 3 times − 3. 95 views. Thoroughly talk about the services that you need with potential payroll providers. Let $$A=\begin{bmatrix} a &b \\ c & d \end{bmatrix}$$ be the 2 x 2 matrix. Since there is at most one inverse of AB, all we have to show is that B 1A has the prop-erty required to be an inverse of AB, name, that (AB)(B 1A 1) = (B 1A 1)(AB) = I. The inverse … The inverse of two invertible matrices is the reverse of their individual matrices inverted. The reverse of. In order for a matrix B to be an inverse of A, both equations AB = I and BA = I must be true. 0000021301 00000 n Theorem. He owes his success to 1 strategy. 0000007121 00000 n Matrices, when multiplied by its inverse will give a resultant identity matrix. The inverse of a product AB is.AB/ 1 D B 1A 1: (4) To see why the order is reversed, multiply AB times B 1A 1. The Inverse of a Product AB For two nonzero numbers a and b, the sum a + b might or might not be invertible. Inverse of a matrix A is the reverse of it, represented as A-1. AB = BA = I 2 and therefore A and B are inverse of each other. Here are a few. Additive inverse is a number which on getting added to the original number results in zero. True. 0000002554 00000 n Watch Queue Queue. (4) To see why the order is reversed, multiply AB times B− 1A−1. How about this: 24-24? answered 09/28/14, Math Major, Pursing PhD in Math, with 10+ Years of Teaching Experience. Remember "shoes and socks." The inverse of a matrix is given by. Answer: The rank of a matrix is the extreme number of linearly self-determining column vectors in the matrix. Then AA- = PRA), 7. Inverse definition, reversed in position, order, direction, or tendency. Show transcribed image text. %%EOF > What is tan inverse of (A+B)? More in-depth information read at these rules. Below are four properties of inverses. 0000019057 00000 n 0000012947 00000 n Matrices are array of numbers or values represented in rows and columns. In such a case matrix B is known as the inverse of matrix A. Inverse of matrix A is symbolically represented by 'A-1 '. Then prove the followings. For Free. For two matrices A and B, the situation is similar. We can calculate the inverse of the matrix in the following steps- 0000012063 00000 n It is also common Like 'AB' = 'BA' = 'I.' %PDF-1.6 %���� 0000010572 00000 n I If A = a b c d QED. Note : Let A be square matrix of order n. Then, A −1 exists if and only if A is non-singular. By using this website, you agree to our Cookie Policy. Theorem 6 Chapter 3 (Determinant of product it product of determinants) If A and B are n × n, then det(AB) = (det A)(det B). There may be an easier way to do this, but here goes: We can check that we calculated A correctly by determining A. 529* [-19 -27] -9 -28. If y = f(x) = ab x, then we may solve for x in terms of y using logarithms: . Let us find the inverse of a matrix by working through the following example: Note : Let A be square matrix of order n. Then, A −1 exists if and only if A is non-singular. No packages or subscriptions, pay only for the time you need. The example of finding the inverse of the matrix is given in detail. And if you think about it, if both of these things are true, then actually not only is A inverse the inverse of A, but A is also the inverse of A inverse. 0000026780 00000 n Select the matrix size: 2×2 3×3 4×4 5×5 6×6 7×7. In such a case matrix B is known as the inverse of matrix A. Inverse of matrix A is symbolically represented by 'A-1'. More generally, if A 1, ..., A k are invertible n-by-n matrices, then (A 1 A 2 ⋅⋅⋅A k−1 A k) −1 = A −1 k A −1 k−1 ⋯A −1 2 A −1 1; det A −1 = (det A) −1. Desmos supports an assortment of functions. We moved parentheses to multiply BB−1first. A matrix 'A' of dimension n x n is called invertible only under the condition, if there exists another matrix B of the same dimension, such that AB = BA = I, where I is the identity matrix of the same order. Inverse Calculator Reviews & Tips Inverse Calculator Ideas . Since 7Z(C) = 1Z(AB) 7Z(A), there exists a T E IZ,(A) such that T c V. Choose any {T}-inverse A(T1 of A, and set A- .= A171. Examine why solving a linear system by inverting the matrix using inv(A)*b is inferior to solving it directly using the backslash operator, x = A\b.. By employing this internet matrix inverse calculator, students will come across much time to receive idea of solving the word issues. x�bf������������b�,Gb/�Tnľ�n�������\R�:/X6����ٜk�0b�jM]������D�����T>�� AB = I n, where A and B are inverse of each other. The inverse of a square matrix A is denoted as A-1 and is unique. That is, if B is the left inverse of A, then B is the inverse matrix of A. 0000012776 00000 n But also the determinant cannot be zero (or we end up dividing by zero). Watch Queue Queue The rows of the inverse matrix V of a matrix U are orthonormal to the columns of U (and vice 0000004052 00000 n Answer: $\ \tan^{-1}A+\tan^{-1}B=\tan^{-1}\frac{A+B}{1-AB}$. (We say B is an inverse of A.) Then find the inverse matrix of A. Inside that is BB−= I: Inverse of AB (AB)(B−1A−1) = AIA−1= AA−1= I. So, let's actually use that method in this video right here. To A invertible ⇔ there exists B … 0000001396 00000 n 0000022059 00000 n 0000006020 00000 n 0000013221 00000 n And if we get the inverse of the 4 x 4 matrix 'A' to be 'B,' then we'll only have to multiply 'AB' and 'BA' to test our work. Sponsored by Raging Bull, LLC. This video is unavailable. Start here or give us a call: (312) 646-6365. This illustrates a basic rule of mathematics: Inverses come in reverse order. Cara Marie M. <]>> 3x3 identity matrices involves 3 rows and 3 columns. And if we get the inverse of the 4 x 4 matrix 'A' to be 'B,' then we'll only have to multiply 'AB' and 'BA' to test our work. Inverse of a matrix A is the reverse of it, represented as A-1.Matrices, when multiplied by its inverse will give a resultant identity matrix. 0000019947 00000 n FALSE AB 1 = B 1A 1. Example 1 : ... Unitary method inverse variation. The following properties hold: If B and C are inverses of A then B=C.Thus we can speak about the inverse of a matrix A, A-1. 0000003284 00000 n This problem has been solved! So first let's think about what the determinant of this matrix is. This is one of midterm 1 exam problems at the Ohio State University Spring 2018. You can input only integer numbers or fractions in this online calculator. For any invertible n-by-n matrices A and B, (AB) −1 = B −1 A −1. Question 5: Define the rank of a matrix. 0000011852 00000 n 0000007930 00000 n If A = [a b] and ab - cd does. A link to the app was sent to your phone. Remark 1. Here is the theorem that we are proving. However there are numerous cases where this isn't the scenario, and this is the point where the student faces more of a challenge. 0000024297 00000 n Show transcribed image text. I'm going to use the same matrix that we started off with in the last video. Remark Not all square matrices are invertible. And it turns out there is such a matrix. To find the inverse of a 2x2 matrix: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). The Inverse of Adding is Subtracting. Best Answer 100% (12 ratings) Previous question Next question Transcribed Image Text from this Question. So the inverse of a 2 by 2 matrix is going to be equal to 1 over the determinant of the matrix times the adjugate of the matrix, which sounds like a very fancy word. Free functions inverse calculator - find functions inverse step-by-step This website uses cookies to ensure you get the best experience. If A is nonsingular, then so is A-1 and (A-1) -1 = A ; If A and B are nonsingular matrices, then AB is nonsingular and (AB)-1 = B-1 A-1-1; If A is nonsingular then (A T)-1 = (A-1) T; If A and B are matrices with AB = I … CHALLENGE: Can you nd an inverse for any non-square matrix. Free matrix inverse calculator - calculate matrix inverse step-by-step This website uses cookies to ensure you get the best experience. 0000033026 00000 n 0000012216 00000 n Find the inverse of AB if A^-1 = [5 -3 0 2] and B^-1 = [4 3 4 -4] (AB)^-1 = [_____] Get more help from Chegg. That is, if B is the left inverse of A, then B is the inverse matrix of A. The inverse of a matrix is often used to solve matrix equations. To find the inverse of A using column operations, write A = IA and apply column operations sequentially till I = AB is obtained, where B is the inverse matrix of A. Inverse of a Matrix Formula. 0000002742 00000 n Formula to find inverse of a matrix. _\square 0000025273 00000 n 0000018398 00000 n 0000009968 00000 n P1.5 WHEN IS B-A- A GENERALIZED INVERSE OF AB? C |A| D A. Finding Inverse of 2 x 2 Matrix. Example 1 : ... Unitary method inverse variation. It is hard to say much about the invertibility of A C B. Question: Find The Inverse Of AB If And . False. Finding Inverse of 2 x 2 Matrix. We prove that if AB=I for square matrices A, B, then we have BA=I. Best Answer 100% (12 ratings) Previous question Next question Transcribed Image Text from this Question. Therefore if the operation is associative ABA = AAB and is also an identity element. 0000005349 00000 n Related Topics: Matrices, Determinant of a 2×2 Matrix, Inverse of a 3×3 Matrix. With no prior experience, Kyle Dennis decided to invest in stocks. 0000011305 00000 n Similarly, any other right inverse equals b, b, b, and hence c. c. c. So there is exactly one left inverse and exactly one right inverse, and they coincide, so there is exactly one two-sided inverse. That equals 0, and 1/0 is undefined. Additive number of positive numbers is negative and vice versa. Similarly, we can also say A is the inverse of B written as B-1. But we'll see for by a 2 by 2 matrix, it's not too involved. Free functions inverse calculator - find functions inverse step-by-step This website uses cookies to ensure you get the best experience. If A is a square matrix where n>0, then (A-1) n =A-n; Where A-n = (A n)-1. By using this website, you agree to our Cookie Policy. If A and B are n x n and invertible, then A^-1B^-1 is the inverse of AB. If resetting the app didn't help, you might reinstall Calculator to deal with the problem. Add to solve later Sponsored Links b If A and B are n n invertible matrices then so is AB and the inverse of AB is B if a and b are n n invertible matrices then so is School Colorado School of Mines Otherwise, you might find some weird outcomes. Theorem A.71 Let A: n×n be symmetric, a be an n-vector, and α>0 be any scalar. (b) If the matrix B is nonsingular, then rank(AB)=rank(A). I If A and B are n n and invertible, then A 1B 1 is the inverse of AB. Example: 20 + 9 = 29 can be reversed by 29 − 9 = 20 (back to where we started) such that AB = I and BA = I. But the problem of calculating the inverse of the sum is more difficult. For two matrices A and B, the situation is similar. The numbers a = 3 and b = −3 have inverses 1 3 and − 1 3. Calculadora gratuita para la inversa de una matriz – calcular la inversa de una matriz paso por paso 0000008813 00000 n Adding moves us one way, subtracting moves us the opposite way. [)D*5�oL;�(x*T�c�ʄ4Va��͍�x�*~�(�+�h*����v�Ʀ��I�0���42 [����/���G���h��jq��-*3��������Yڦ�bc+��� -�'���N뺪�����{�Nˋ�q (J�ުq! 0000018772 00000 n The sign of the number is changed and added to get zero. 119 0 obj <>stream 0000011111 00000 n You perform matrix multiplication to get AB: © 2005 - 2020 Wyzant, Inc. - All Rights Reserved, a Question x = f –1 (y) = log b (y/a). D no of rows of A is equal to no of columns of B. Find the inverse of AB if and . 0000010236 00000 n We see that the inverse of an exponential with base b is a logarithm with base b .. Recall that the logarithm is defined only for positive inputs. For binary operation* : A × A → Awithidentity elementeFor element a in A,there is an element b in Asuch thata * b = e = b * aThen, b is called inverse of aAddition+ :R×R→RFor element a in A,there is an element b in Asuch thata * b = e = b * aThen, b is called inverse … Find The Inverse Of AB If And . If A is invertible, then its inverse is unique. 0000023652 00000 n 529 *[ 3 4] * [-1 -5 ] -5 1 -4 -3. But that follows from associativity of matrix multiplication and the facts that AA 1 = A 1A = I and BB 1 = B 1B = I. q.e.d. 0000026910 00000 n C no of columns of A is equal to columns of B. Go through it and learn the problems using the properties of matrices inverse. 0000025677 00000 n 0000006368 00000 n This problem has been solved! Question: Find The Inverse Of AB If And . How Are We Going To Measure The Inverse? > What is tan inverse of (A+B)? By using this website, you agree to our Cookie Policy. * Hans Joachim Werner Institute for Econometrics and Operations Research Econometrics Unit University of Bonn Adenauerallee 24-42 D-53113 Bonn, Germany Submitted by George P H. Styan ABSTRACT In practice factorizations of a generalized inverse often arise from factorizations of the matrix which is to be inverted. Please enter the matrice: A =. If the operation is not commutative or not associative then AB <> BA and it doesn’t work. 14 Additive inverse of a matrix A is A adj A⁄|A| B A². Theorem 3 Chapter 3 (Inverse/Determinant Relationship) A square matrix A is invertible if and only if det A 6 = 0. Math is about vocabulary. That is, AB is almost never equal to BA. B is the inverse of AA means that AAB is an identity element. Find more examples at BYJU’S. AB = BA = I n. then the matrix B is called an inverse of A. Formula to find inverse of a matrix. (a) rank(AB)≤rank(A). Create a random matrix A of order 500 that is constructed so that its condition number, cond(A), is 1e10, and its norm, norm(A), is 1.The exact solution x is a random vector of length 500, and the right side is b = A*x. On rearranging, we can write (AB)A=1, which shows that AB is the inverse of A. for square matrices AB=I then there is some C such that BC=I. 0000026052 00000 n 10. ; If A is invertible and k is a non-zero scalar then kA is invertible and (kA)-1 =1/k A-1. Where the detA = (a*d - b*c) If we go backwards from A-1 to get A, then A = 23* [ 1 -4 ] 5 3. detA = (1*3-(-4*5)) = (3+20) = 23. 0000025021 00000 n Like 'AB' = 'BA' = 'I.' 0000017999 00000 n How Are We Going To Measure The Inverse? It is a general idea in mathematics and has many meanings. In the last video, we stumbled upon a way to figure out the inverse for an invertible matrix. Recall that the logarithm is defined only for positive inputs. It seems like a fairly good matrix. Inside that is BB 1 D I: Inverse of AB .AB/.B 1A 1/ D AIA 1 D AA 1 D I: We movedparentheses to multiplyBB 1 first. Answer: $\ \tan^{-1}A+\tan^{-1}B=\tan^{-1}\frac{A+B}{1-AB}$. AB = BA = I n. then the matrix B is called an inverse of A. See more. 261 further tells us that in this case N(C-) = V and C-C = 1. Get a free answer to a quick problem. In the below Inverse Matrix calculator, enter the values for Matrix (A) and click calculate and calculator will provide you the Adjoint (adj A), Determinant (|A|) and Inverse of a 3x3 Matrix. You can easily nd the inverse of a 2 2 matrix. Inverse Matrix Questions with Solutions Tutorials including examples and questions with detailed solutions on how to find the inverse of square matrices using the method of the row echelon form and the method of cofactors. Get more help from Chegg. We cannot go any further! If so nd one, if not explain why. 0 Inverse means the opposite in effect. For example, if matrix A and B satisfy this condition AB=BA=I, then we can say B is the inverse of A written as A-1 =B. {9��,���ŋ��Z��zKp�L��&fSچ@͋*����HΡs�P%����e. We can check … 0000004891 00000 n A.12 Generalized Inverse 511 Theorem A.70 Let A: n × n be symmetric, a ∈R(A), b ∈R(A),and assume 1+b A+a =0.Then (A+ab)+ = A+ −A +ab A 1+b A+a Proof: Straightforward, using Theorems A.68 and A.69. How to Determine if a Matrix is Invertible The same argument shows that any other left inverse b ′ b' b ′ must equal c, c, c, and hence b. b. b. This Matrix has no Inverse. 0000030372 00000 n startxref Now to get (AB)-1, you take the inverse of matrix AB: (AB)-1= 529* (1/det(AB)) *[-28 27 ] 9 -19. 0000022882 00000 n First of all, to have an inverse the matrix must be "square" (same number of rows and columns). 0000012594 00000 n 0000027678 00000 n Additional features of the inverse matrix calculator. Calculadora gratuita de inversa de una función - Encontrar la inversa de una función paso por paso 0000009110 00000 n 0000000016 00000 n Inverses of 2 2 matrices. The inverse of the 'n' x 'n' matrix 'A' is the 'n' x 'n' matrix 'B.' The inverse of A, written as "A –1" and pronounced "A inverse", would allow you to cancel off the A from the matrix equation and then ... multiplication is not commutative. Inverse of a 2×2 Matrix. Then the following statements are equivalent: (i) αA−aa ≥ 0. The Inverse May Not Exist. 3x3 identity matrices involves 3 rows and 3 columns. 529 * [3*-1+4*-4 3*-5+4*-3] -5*-1+1*-4 5*-5+1*-3. Please Help. These lessons and videos help Algebra students find the inverse of a 2×2 matrix. Their sum a +b = 0 has no inverse. 65 55 The Ugly Side of Inverse Calculator . 65 0 obj <> endobj If A has rank m (m ≤ n), then it has a right inverse, an n -by- m matrix B such that AB = Im. Most questions answered within 4 hours. Please Help. We prove that if AB=I for square matrices A, B, then we have BA=I. A-1. Find the Inverse of a Square Matrix Using the Row Reduction Method This method is also called Gauss–Jordan elimination method. So they're each other's inverses. But the product ab D 9 does have an inverse, which is1 3. times1 3. 0000012140 00000 n SimilarlyB 1A 1 times AB equals I. 0000020721 00000 n Then BC-AB = BC-C = B, thus showing that C-A E B{1}. 0000002987 00000 n When is B-A- a Generalized Inverse of AB? Sometimes there is no inverse at all. A-1 = 1/detA [ d -b ] -c a. Properties of Inverses. Theorem. See the answer. Find the inverse of AB if and . Choose an expert and meet online. Find a nonsingular matrix A such that 3A=A^2+AB, where B is a given matrix. xref 2 B both have same order. Everybody knows that if you consider a product of two square matrices GH, the inverse matrix is given by H-1 G-1. That's all I meant to say. B is the inverse of A^2 implies A^B=1. 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https://encyclopediaofmath.org/wiki/Idempotent_semi-ring | # Idempotent semi-ring
dioid
A semi-ring with idempotent addition. So, a set $A$ equipped with binary operations $\oplus$ (addition) and $\odot$ (multiplication) and neutral elements $0$ and $1$ is called an idempotent semi-ring if the following basic properties are valid for all elements $a,b,c \in A$:
i) $a \oplus a = a$ (idempotent addition);
ii) $a \oplus (b \oplus c) = (a \oplus b) \oplus c$;
iii) $a \oplus b = b \oplus a$;
iv) $0 \oplus a = a \oplus 0 = a$;
v) $0 \odot a = a \odot 0 = 0$;
vi) $1 \odot a = a \odot 1 = a$;
vii) $a \odot (c \oplus c) = (a \odot b) \oplus (a \odot c)$;
viii) $(b \oplus c) \odot a = (b \odot a) \oplus (c \odot a)$.
An idempotent semi-ring $A$ is commutative if $a \odot b = b \odot a$ for all $a,b \in A$. Different versions of this axiomatics are used, see e.g. [a1], [a2], [a3], [a4], [a5], [a6], [a7], [a8], [a9], [a10]. Idempotent semi-rings are often called dioids, see e.g. [a3], [a9]. The concept of an idempotent semi-ring is a basic concept in idempotent analysis. This concept has many applications in different optimization problems (including dynamic programming), computer science, automata and formal language theory, numerical methods, parallel programming, etc. (cf. also Idempotent algorithm and [a1], [a2], [a3], [a4], [a5], [a6], [a7], [a8], [a9], [a10], [a11]).
## Contents
### Example 1.
Let $\mathbf{R}_{\max}$ be the set $A = \mathbf{R} \cup \{-\infty\}$ (where $\mathbf{R}$ is the field of real numbers) equipped with the operations $\oplus = {\max}$ and $\odot = {+}$ (usual addition); set $\mathbf{0} = -\infty$, $\mathbf{1} = 0$. Similarly, let $\mathbf{R}_{\min}$ be the set $\mathbf {R} \cup \{+\infty\}$ equipped with the operations $\oplus = {\min}$ and $\odot = {+}$; in this case $\mathbf{0} = +\infty$ and $\mathbf{1} = 0$. It is easy to check that $\mathbf{R}_{\max}$ and $\mathbf{R}_{\min}$ are (isomorphic) commutative idempotent semi-rings.
### Example 2.
Let $A$ be the set $\mathbf{R}_{+}$ of all non-negative real numbers endowed with the operations $\oplus={\max}$ and $\odot = {\times}$ (usual multiplication); $\mathbf{0} = 0$, $\mathbf{1} = 1$. This idempotent semi-ring is isomorphic to $\mathbf{R}_{\max}$. The isomorphism is given by $x \mapsto \log x$.
### Example 3.
Let $A = [a,b] = \{x \in \mathbf{R} : a \le x \le b \}$ with the operations $\oplus = {\max}$, $\odot = {\min}$ and neutral elements $\mathbf{0} = a$, $\mathbf{1} = b$ (the cases $a = -\infty$, $b = +\infty$ are possible).
### Example 4.
Let $\mathrm{Mat}_n(A)$ be the set of $n \times n$-matrices with entries belonging to an idempotent semi-ring $A$. Then $\mathrm{Mat}_n(A)$ is a non-commutative idempotent semi-ring with respect to matrix addition and matrix multiplication.
The Boolean algebra $\mathcal{B}_2 = \{\mathbf{0},\mathbf{1}\}$ is an example of a finite idempotent semi-ring. There are many other interesting examples of idempotent semi-rings, see e.g. [a1], [a2], [a3], [a4], [a5], [a6], [a7], [a8], [a9], [a10], [a11].
There is a natural partial order on any idempotent semi-ring (as well as on any idempotent semi-group; cf. also Idempotent semi-ring). By definition, $a \le b$ if and only if $a \oplus b = b$. For this relation, reflexivity is equivalent to idempotency of the (generalized) addition, whereas transitivity, respectively anti-symmetry, follow from associativity, respectively commutativity, of this operation. On $\mathbf{R}_{\max}$ (and also on the semi-rings described in the Examples 2 and 3), this ordering relation coincides with the natural one; on $\mathbf{R}_{\min}$ it is the opposite of the natural ordering relation on the real axis. Every element $a$ in an idempotent semi-ring $A$ is "non-negative" : $0 \le a$. Indeed, $0 \oplus a = a$. Similarly, for all $a,b,c \in A$ one has $a \oplus c \le b \oplus c$, and $a \odot c \le b \odot c$ if $a \le b$. Using this standard partial order it is possible to define in the usual way the notions of upper and lower bounds, bounded sets, $\sup M$ (and $\inf N$) for upper- (lower-) bounded sets $M$ (respectively, $N$), etc.
If the multiplication in a semi-ring $A$ is invertible on $A \setminus \{0\}$, then $A$ is called a semi-field. For example, $\mathbf{R}_{\max}$ is a semi-field. Idempotent semi-fields and semi-rings with idempotent multiplication are especially interesting.
#### References
[a1] A.V. Aho, J.E. Hopcroft, J.D. Ullman, "The design and analysis of computer algorithms" , Addison-Wesley (1976) [a2] B.A. Carré, "Graphs and networks" , Clarendon Press and Oxford Univ. Press (1979) [a3] M. Gondran, M. Minoux, "Graphes et algorithms" , Ed. Eyrolles (1979; 1988) [a4] R.A. Cuninghame-Green, "Minimax algebra" , Lecture Notes in Economics and Mathematical Systems , 166 , Springer (1979) ISBN 3-540-09113-0 Zbl 0399.90052 [a5] U. Zimmermann, "Linear and combinatorial optimization in ordered algebraic structures" Ann. Discrete Math. , 10 (1981) pp. 1–380 [a6] "Mathematical aspects of computer engineering" V.P. Maslov (ed.) K.A. Volosov (ed.) , MIR (1988) (In Russian) [a7] "Idempotent analysis" V.P. Maslov (ed.) S.N. Samborskii (ed.) , Amer. Math. Soc. (1992) (In Russian) [a8] V.N. Kolokoltsov, V.P. Maslov, "Idempotent analysis and applications" , Kluwer Acad. Publ. (1996) (In Russian) [a9] F.L. Baccelli, G. Cohen, G.J. Olsder, J.-P. Quadrat, "Synchronization and linearity: an algebra for discrete event systems" , Wiley (1992) [a10] J.S. Golan, "The theory of semirings with applications in mathematics and theoretical computer science" , Pitman monographs and surveys in pure and applied mathematics , 54 , Longman (1992) [a11] "Idempotency" J. Gunawardena (ed.) , Publ. Isaac Newton Institute 11, Cambridge Univ. Press (1998) ISBN 0-521-55344-X Zbl 0882.00035
How to Cite This Entry:
Idempotent semi-ring. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Idempotent_semi-ring&oldid=51546
This article was adapted from an original article by G.L. Litvinov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9343970417976379, "perplexity": 2044.412523396754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00574.warc.gz"} |
http://mathhelpforum.com/calculus/75555-differntial-caculus-salt-problem.html | # Thread: Differntial Caculus Salt Problem
1. ## Differntial Caculus Salt Problem
A resevoir contains 700 liters of pure water. Brine contaning 0.02 kg/L of salt enters at a rate of 5 L/min. Another source of brine contanining 0.05 kg/L enters at a rate of 2L/min. The resevoir is well mixed and drains at a rate of 7L/min.How much salt is thier in the resevoir (i) after t min (ii) after 10 min?
2. In our problem, the level of water remains the same, as we have $5\frac{\mbox{L}}{\mbox{min}}$ and $2\frac{\mbox{L}}{\mbox{min}}$ flowing in and $7\frac{\mbox{L}}{\mbox{min}}$ flowing out.
If we call the salt level $S(t)$, then we have
$S(0)=0\,\mbox{kg}.$
To find $\frac{dS}{dt}$, we subtract the rate of salt flowing out from the rate flowing in:
$\frac{dS}{dt}=\frac{dS_{\mbox{\scriptsize{in}}}}{d t}-\frac{dS_{\mbox{\scriptsize{out}}}}{dt}.$
The rate flowing in is
\begin{aligned}
\frac{dS_{\scriptsize{\mbox{in}}}}{dt}&=0.02\frac{ \mbox{kg}}{\mbox{L}}\cdot 5\frac{\mbox{L}}{\mbox{min}}+0.05\frac{\mbox{kg}}{ \mbox{L}}\cdot 2\frac{\mbox{L}}{\mbox{min}} \\
&=0.1\frac{\mbox{kg}}{\mbox{min}}+0.1\frac{\mbox{k g}}{\mbox{min}}\\
&=0.2\frac{\mbox{kg}}{\mbox{min}}.
\end{aligned}
Since the fluid is well-mixed, the rate flowing out will be in proportion to $7\,\mbox{L}$ as the current salt level is to the entire resevoir. Therefore,
\begin{aligned}
\frac{dS_{\scriptsize{\mbox{out}}}}{dt}&=\frac{S}{ 700\,\mbox{L}}\cdot 7\frac{\mbox{L}}{\mbox{min}} \\
&=\frac{S}{100}\,\mbox{min}^{-1}.\\
\end{aligned}
For $\frac{dS}{dt}$, we obtain
\begin{aligned}
\frac{dS}{dt}&=\frac{dS_{\mbox{\scriptsize{in}}}}{ dt}-\frac{dS_{\mbox{\scriptsize{out}}}}{dt}\\
&=0.2\frac{\mbox{kg}}{\mbox{min}}-\frac{S}{100}\,\mbox{min}^{-1}.
\end{aligned}
Now, all that remains is to solve the differential equation using the initial conditions given. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8079133033752441, "perplexity": 2161.4719685692207}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281574.78/warc/CC-MAIN-20170116095121-00511-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://repository.uantwerpen.be/link/irua/92427 | Publication
Title
Crystal fields, disorder, and antiferromagnetic short-range order in $Yb_{0.24}Sn_{0.76}Ru$
Author
Abstract
We report extensive measurements on a new compound (Yb0.24Sn0.76)Ru that crystallizes in the cubic CsCl structure. Valence-band photoemission (PES) and L3 x-ray absorption show no divalent component in the 4f configuration of Yb. Inelastic neutron scattering (INS) indicates that the eight-fold degenerate J-multiplet of Yb3+ is split by the crystalline electric field (CEF) into a Γ7-doublet ground state and a Γ8 quartet at an excitation energy 20 meV. The magnetic susceptibility can be fit very well by this CEF scheme under the assumption that a Γ6-excited state resides at 32 meV; however, the Γ8/Γ6 transition expected at 12 meV was not observed in the INS. The resistivity follows a Bloch-Grüneisen law shunted by a parallel resistor, as is typical of systems subject to phonon scattering with no apparent magnetic scattering. All of these properties can be understood as representing simple local moment behavior of the trivalent Yb ion. At 1 K there is a peak in specific heat that is too broad to represent a magnetic-phase transition, consistent with absence of magnetic reflections in neutron diffraction. On the other hand this peak also is too narrow to represent the Kondo effect in the Γ7-doublet ground state. On the basis of the field dependence of the specific heat, we argue that antiferromagnetic (AF) short-range order (SRO) (possibly coexisting with Kondo physics) occurs at low temperatures. The long-range magnetic order is suppressed because the Yb site occupancy is below the percolation threshold for this disordered compound.
Language
English
Source (journal)
Physical review : B : condensed matter and materials physics. - Lancaster, Pa, 1998 - 2015
Publication
Lancaster, Pa : 2011
ISSN
1098-0121 [print]
1550-235X [online]
Volume/pages
84:7(2011), p. 075152,1-075152,8
Article Reference
075152
ISI
000293830800003
Medium
E-only publicatie
Full text (Publisher's DOI)
Full text (open access)
UAntwerpen
Faculty/Department Research group Publication type Subject Affiliation Publications with a UAntwerp address | {"extraction_info": {"found_math": true, "script_math_tex": 1, "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": 0.8634181022644043, "perplexity": 4346.782626352894}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934808972.93/warc/CC-MAIN-20171124214510-20171124234510-00098.warc.gz"} |
https://www.physicsforums.com/members/vedicguru.135289/recent-content | # Recent content by vedicguru
1. ### Groups and Subgroups!
ohh yeah I am sorry about that...I was in hurry or something to write it down! but yeah I believe everything else is fine and I would not forget to mention that <a> is not equal to {e}..it is G! Thanks!
2. ### Groups and Subgroups!
oh yeah sure! I meant how would say, after constructing it! I forgot it is equal to {e} it is equal to G...this is what I get! We could say that, let a E G - {e} then we construct <a> = a, a^2, a^3... a^n=1 <- cyclic subgroup generated by element a. Since <a> is subgroup, it must coincide with G...
3. ### Groups and Subgroups!
how do I construct <a> ??
4. ### Groups and Subgroups!
Homework Statement Prove that if a group G has no subgroup other than G and {e}, then G is cyclic.... Homework Equations The Attempt at a Solution we could say that, let a E G - {e} then we construct <a>... | {"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": 0.9239959716796875, "perplexity": 2408.6931689770263}, "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-2021-31/segments/1627046154042.23/warc/CC-MAIN-20210731011529-20210731041529-00680.warc.gz"} |
http://mathhelpforum.com/math-topics/59622-urgent-physics-question-please.html | 1. ## URGENT physics question please.
A diffraction grating (with 10,000 lines per cm) is held 2 meters from a Hydrogen lamp (lamp which emits
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2. Originally Posted by InsPiRation
A diffraction grating (with 10,000 lines per cm) is held 2 meters from a Hydrogen lamp (lamp which emits
Hydrogen’s spectrum). As you look through the diffraction grating a blue line is observed to lie 1.11 meters on either
side of the lamp.
a) At what distance from the lamp (central maximum) will the second occurrence of this blue line be observed at?
Bohr’s model of the hydrogen atom states that the energy levels are given by the formula E = -13.6 eV/n2, where n labels
the energy state of the hydrogen atom.
b) What transition (from initial n to final n) does this blue line correspond to?
c) What series does this line belong to?
You will find the relevant theory in any relevant textbook.
a) $1.11 = \frac{(1)(\lambda)(2)}{10^{-6}} \Rightarrow \lambda = 555 \, \text{nm}$. Therefore $X_2 = \frac{(2)(555 \times 10^{-9})(2)}{10^{-6}}$ m. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9503999352455139, "perplexity": 2564.1240300291265}, "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-2013-48/segments/1387345758214/warc/CC-MAIN-20131218054918-00004-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://eprints.kfupm.edu.sa/id/eprint/140064/ | # t-Reductions of Ideals in Integral Domains
t-Reductions of Ideals in Integral Domains. PhD thesis, King Fahd University of Petroleum and Minerals.
Preview
PDF
Dissertation.pdf
This Ph.D. thesis traverses two chapters which contribute to the study of multiplicative ideal theoretic properties of integral domains. Let $R$ be an integral domain and $I$ a nonzero ideal of $R$. An ideal $J\subseteq I$ is a $t$-reduction of $I$ if $(JI^{n})_{t}=(I^{n+1})_{t}$ for some integer $n\geq0$. An element $x\in R$ is $t$-integral over $I$ if there is an equation $x^{n}+a_{1}x^{n-1}+...+a_{n-1}x+a_{n}=0$ with $a_{i}\in (I^{i})_{t}$ for $i=1,...,n$. The set of all elements that are $t$-integral over $I$ is called the $t$-integral closure of $I$. The first chapter investigates the $t$-reductions and $t$-integral closure of ideals. Our objective is to establish satisfactory $t$-analogues of well-known results, in the literature, on the integral closure of ideals and its correlation with reductions. Namely, Section 1.2 identifies basic properties of $t$-reductions of ideals and features explicit examples discriminating between the notions of reduction and $t$-reduction. Section 1.3 investigates the concept of $t$-integral closure of ideals, including its correlation with $t$-reductions. Section 1.4 studies the persistence and contraction of $t$-integral closure of ideals under ring homomorphisms. All along the chapter, the main results are illustrated with original examples. An ideal $I$ is $t$-basic if it has no $t$-reduction other than the trivial ones. The second chapter investigates $t$-reductions of ideals in pullback constructions. Section 2.2 examines the correlation between the notions of reduction and $t$-reduction in pseudo-valuation domains. Section 2.3 solves an open problem on whether the finite $t$-basic and $v$-basic ideal properties are distinct. We prove that these two notions coincide in any arbitrary domain. Section 2.4 features the main result, which establishes the transfer of the finite $t$-basic ideal property to pullbacks in line with Fontana-Gabelli's result on Pr\"ufer $v$-Multiplication Domains (P$v$MDs) and Gabelli-Houston's result on $v$-domains. This allows us to enrich the literature with new families of examples, which put the class of domains subject to the finite $t$-basic ideal property strictly between the two classes of $v$-domains and integrally closed domains. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8655351400375366, "perplexity": 424.6168681116995}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735885.72/warc/CC-MAIN-20200804220455-20200805010455-00396.warc.gz"} |
http://mathhelpforum.com/calculus/279854-even-odd-parts-piecewise-defined-function.html | # Thread: Even and odd parts of a piecewise defined function
1. ## Even and odd parts of a piecewise defined function
Hello,
I'm supposed to find the even and odd parts of the following $$f(x)=\begin{cases}x^2 & x<0 \\ e^{-x} & x>0.\end{cases}$$
We have a formula, $f(x)=f_e(x)+f_o(x)=1/2 (f(x)+f(-x))+1/2(f(x)-f(-x))$.
So for even/odd to mean anything $x$ should live in an interval which is symmetric with respect to 0. For otherwise there would exist some $x_0$ such that either $f(x)\neq f(-x)$ or $f(-x) \neq -f(x)$ simply because the function isn't defined on appropriate values.
I tried just naively shoving function values into the formula. So $f_e(x)=1/2 (f(x)+f(-x))=(e^{-x}+x^2)$ but since $f(-x)=1/2(e^{-(-x)}+(x)^2)$ this function is not even. So this can't be correct.
Is there some mistake?
2. ## Re: Even and odd parts of a piecewise defined function
use the formula more carefully
$f_e(x) = \begin{cases}\dfrac{x^2 + e^x}{2} &x < 0 \\\dfrac{e^{-x}+x^2}{2} &0 < x \end{cases}$
3. ## Re: Even and odd parts of a piecewise defined function
Originally Posted by bkbowser
I'm supposed to find the even and odd parts of the following $$f(x)=\begin{cases}x^2 & x<0 \\ e^{-x} & x>0.\end{cases}$$
We have a formula, $f(x)=f_e(x)+f_o(x)=1/2 (f(x)+f(-x))+1/2(f(x)-f(-x))$.
What is the exact formulation of $f(-x)$ you are using.
i.e. $f(-x)=\begin{cases}? & x<0 \\ ?? & x>0.\end{cases}$
FYI This is a very important idea: Any function is sum of an even function and an odd function.
4. ## Re: Even and odd parts of a piecewise defined function
Originally Posted by bkbowser
Hello,
I'm supposed to find the even and odd parts of the following $$f(x)=\begin{cases}x^2 & x<0 \\ e^{-x} & x>0.\end{cases}$$
We have a formula, $f(x)=f_e(x)+f_o(x)=1/2 (f(x)+f(-x))+1/2(f(x)-f(-x))$.
So for even/odd to mean anything $x$ should live in an interval which is symmetric with respect to 0. For otherwise there would exist some $x_0$ such that either $f(x)\neq f(-x)$ or $f(-x) \neq -f(x)$ simply because the function isn't defined on appropriate values.
I tried just naively shoving function values into the formula. So $f_e(x)=1/2 (f(x)+f(-x))=(e^{-x}+x^2)$ but since $f(-x)=1/2(e^{-(-x)}+(x)^2)$ this function is not even. So this can't be correct.
Is there some mistake?
The "even part" of a function is $\frac{f(x)+ f(-x)}{2}$.
If x< 0 then -x> 0 so for x< 0, $f(x)= x^2$ and $f(-x)= e^{-(-x)}= e^x$. If x< 0, $\frac{f(x)+ f(-x)}{2}= \frac{x^2+ e^x}{2}$. If x> 0, x< 0 for $f(x)= e^{-x}$ and $f(-x)= (-x)^2= x^2$ so $\frac{f(x)+ f(-x)}{2}= \frac{e^{-x}+ x^2}{2}. 5. ## Re: Even and odd parts of a piecewise defined function Originally Posted by HallsofIvy The "even part" of a function is$\frac{f(x)+ f(-x)}{2}$. If x< 0 then -x> 0 so for x< 0,$f(x)= x^2$and$f(-x)= e^{-(-x)}= e^x$. If x< 0,$\frac{f(x)+ f(-x)}{2}= \frac{x^2+ e^x}{2}$. If x> 0, x< 0 for$f(x)= e^{-x}$and$f(-x)= (-x)^2= x^2$so$\frac{f(x)+ f(-x)}{2}= \frac{e^{-x}+ x^2}{2}.
Sorry, it may just me but I cannot read the above.
I still ask how this is defined: $f(-x)=\begin{cases}? & x<0 \\ ?? & x>0.\end{cases}$ | {"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": 0.947843074798584, "perplexity": 455.6957452885821}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814249.56/warc/CC-MAIN-20180222180516-20180222200516-00286.warc.gz"} |
http://www.researchgate.net/researcher/50039926_S_M_Hansen | # S. Hansen
University of California Observatories, Santa Cruz, California, United States
Are you S. Hansen?
## Publications (191)505.76 Total impact
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##### Article: First SN Discoveries from the Dark Energy Survey
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ABSTRACT: Distributed as an Instant Email Notice Supernovae Credential Certification: Masao Sako ([email protected]) Subjects: Optical, Supernovae Referred to by ATel #: 4725, 4741, 4800, 4826 First SN Discoveries from the Dark Energy Survey The Dark Energy Survey (DES) report the discovery of the first set of supernovae (SN) from the project. Images were observed as part of the DES Science Verification phase using the newly-installed 570-Megapixel Dark Energy Camera on the CTIO Blanco 4-m telescope by observers J. Annis, E. Buckley-Geer, and H. Lin. SN observations are planned throughout the observing campaign on a regular cadence of 4-6 days in each of the ten 3-deg2 fields in the DES griz filters. The SN candidates are named according to the season and field in which they were discovered. We adopt the convention -- DES{season}{field}{index} -- where {season} is the year pertaining to the beginning of each observing season, {field} denotes one of the ten SN search fields (E1,E2,S1,S2,X1,X2,X3,C1,C2,C3) in Elais-S1 (E), Stripe 82 (S), XMM-LSS (X) and CDF-S (C), and {index} is one or more lower-case letters starting from a-z, then aa-az, and so on. The DES SN Survey strategy is described in Bernstein et al. (2012, ApJ, 753, 152). Spectroscopic classifications were performed by the OzDES collaboration from spectra (350-900 nm) obtained at the Anglo-Australian Telescope with AAOmega-2dF observed by C. Lidman, R. Sharp, and S. A. Uddin. Classifications were performed using Superfit (Howell et al 2002, BAAS, 34, 1256) or SNID (Blondin & Tonry, 2007, ApJ, 666, 1024). Redshifts measured from narrow galaxy lines are quoted to 3 significant figures. Those measured from broad SN features are quoted to 2 significant figures. SN phases are based on both the optical spectra and multi-band light curves at the time of the spectroscopic measurements. Name | RA(J2000) | Dec(J2000) | Discovery date (UT) | Discovery r mag| Spectrum date (UT) | redshift | type | phase DES12C1a | 03:38:54.5 | -27:32:28.2 | 2012 Dec 07 | 22.0 | 2012 Dec 13 | 0.303 | Ia | near max DES12C1b | 03:35:05.8 | -26:45:53.9 | 2012 Dec 07 | 20.9 | 2012 Dec 13 | 0.243 | Ia | near max DES12C2a | 03:41:13.1 | -28:59:37.9 | 2012 Dec 04 | 21.5 | 2012 Dec 14 | 0.21 | Ia | near max
The Astronomer's Telegram. 12/2012;
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##### Article: Hydrostatic Level Sensors as High Precision Ground Motion Instrumentation for Tevatron and Other Energy Frontier Accelerators
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ABSTRACT: Particle accelerators pushed the limits of our knowledge in search of the answers to most fundamental questions about micro-world and our Universe. In these pursuits, accelerators progressed to higher and higher energies and particle beam intensities as well as increasingly smaller and smaller beam sizes. As the result, modern existing and planned energy frontier accelerators demand very tight tolerances on alignment and stability of their elements: magnets, accelerating cavities, vacuum chambers, etc. In this article we describe the instruments developed for and used in such accelerators as Fermilab's Tevatron (FNAL, Batavia, IL USA) and for the studies toward an International Linear Collider (ILC). The instrumentation includes Hydrostatic Level Sensors (HLS) for very low frequency measurements. We present design features of the sensors, outline their technical parameters, describe test and calibration procedures and discuss different regimes of operation. Experimental results of the ground motion measurements with these detectors will be presented in subsequent paper.
Journal of Instrumentation 05/2012; 7(1). · 1.66 Impact Factor
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##### Article: The Blanco Cosmology Survey: Data Acquisition, Processing, Calibration, Quality Diagnostics and Data Release
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ABSTRACT: The Blanco Cosmology Survey (BCS) is a 60 night imaging survey of $\sim$80 deg$^2$ of the southern sky located in two fields: ($\alpha$,$\delta$)= (5 hr, $-55^{\circ}$) and (23 hr, $-55^{\circ}$). The survey was carried out between 2005 and 2008 in $griz$ bands with the Mosaic2 imager on the Blanco 4m telescope. The primary aim of the BCS survey is to provide the data required to optically confirm and measure photometric redshifts for Sunyaev-Zel'dovich effect selected galaxy clusters from the South Pole Telescope and the Atacama Cosmology Telescope. We process and calibrate the BCS data, carrying out PSF corrected model fitting photometry for all detected objects. The median 10$\sigma$ galaxy (point source) depths over the survey in $griz$ are approximately 23.3 (23.9), 23.4 (24.0), 23.0 (23.6) and 21.3 (22.1), respectively. The astrometric accuracy relative to the USNO-B survey is $\sim45$ milli-arcsec. We calibrate our absolute photometry using the stellar locus in $grizJ$ bands, and thus our absolute photometric scale derives from 2MASS which has $\sim2$% accuracy. The scatter of stars about the stellar locus indicates a systematics floor in the relative stellar photometric scatter in $griz$ that is $\sim$1.9%, $\sim$2.2%, $\sim$2.7% and$\sim$2.7%, respectively. A simple cut in the AstrOmatic star-galaxy classifier {\tt spread\_model} produces a star sample with good spatial uniformity. We use the resulting photometric catalogs to calibrate photometric redshifts for the survey and demonstrate scatter $\delta z/(1+z)=0.054$ with an outlier fraction $\eta<5$% to $z\sim1$. We highlight some selected science results to date and provide a full description of the released data products.
The Astrophysical Journal 04/2012; 757(1). · 6.73 Impact Factor
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##### Article: The Serendipitous Observation of a Gravitationally Lensed Galaxy at z = 0.9057 from the Blanco Cosmology Survey: The Elliot Arc
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ABSTRACT: We report on the serendipitous discovery in the Blanco Cosmology Survey (BCS) imaging data of a z = 0.9057 galaxy that is being strongly lensed by a massive galaxy cluster at a redshift of z = 0.3838. The lens (BCS J2352–5452) was discovered while examining i- and z-band images being acquired in 2006 October during a BCS observing run. Follow-up spectroscopic observations with the Gemini Multi-Object Spectrograph instrument on the Gemini-South 8 m telescope confirmed the lensing nature of this system. Using weak-plus-strong lensing, velocity dispersion, cluster richness N 200, and fitting to a Navarro-Frenk-White (NFW) cluster mass density profile, we have made three independent estimates of the mass M 200 which are all very consistent with each other. The combination of the results from the three methods gives M 200 = (5.1 ± 1.3) × 1014 M ☉, which is fully consistent with the individual measurements. The final NFW concentration c 200 from the combined fit is c 200 = 5.4+1.4 – 1.1. We have compared our measurements of M 200 and c 200 with predictions for (1) clusters from ΛCDM simulations, (2) lensing-selected clusters from simulations, and (3) a real sample of cluster lenses. We find that we are most compatible with the predictions for ΛCDM simulations for lensing clusters, and we see no evidence based on this one system for an increased concentration compared to ΛCDM. Finally, using the flux measured from the [O II]3727 line we have determined the star formation rate of the source galaxy and find it to be rather modest given the assumed lens magnification.
The Astrophysical Journal 11/2011; 742(1):48. · 6.73 Impact Factor
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##### Article: The XMM-BCS galaxy cluster survey: I. The X-ray selected cluster catalog from the initial 6 deg$^2$
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ABSTRACT: The XMM-Newton - Blanco Cosmology Survey project (XMM-BCS) is a coordinated X-ray, optical and mid-infrared cluster survey in a field also covered by Sunyaev-Zel'dovich effect surveys by the South Pole Telescope and the Atacama Cosmology Telescope. The aim of the project is to study the cluster population in a 14 deg$^2$ field. In this work, we present a catalog of 46 X-ray selected clusters from the initial 6 deg$^2$ survey core. We describe the XMM-BCS source detection pipeline and derive physical properties of the clusters. We provide photometric redshift estimates derived from the BCS imaging data and spectroscopic redshift measurements for a low redshift subset of the clusters. We derive the cluster log N - log S relation using an approximation to the survey selection function and find it in good agreement with previous studies. We carry out an initial comparison between X-ray luminosity derived masses and masses from optical estimators from the Southern Cosmology Survey for a subset of the cluster sample. Optical masses based on cluster richness and total optical luminosity are found to be significantly higher than the X-ray values. (abridged)
Astronomy and Astrophysics 11/2011; · 5.08 Impact Factor
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##### Article: Longitudinal bunch monitoring at the Fermilab Tevatron and Main Injector synchrotrons
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ABSTRACT: The measurement of the longitudinal behavior of the accelerated particle beams at Fermilab is crucial to the optimization and control of the beam and the maximizing of the integrated luminosity for the particle physics experiments. Longitudinal measurements in the Tevatron and Main Injector synchrotrons are based on the analysis of signals from resistive wall current monitors. This article describes the signal processing performed by a 2 GHz-bandwidth oscilloscope together with a computer running a LabVIEW program which calculates the longitudinal beam parameters.
Journal of Instrumentation 10/2011; 6(10):T10004. · 1.66 Impact Factor
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##### Article: Robust Optical Richness Estimation with Reduced Scatter
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ABSTRACT: Reducing the scatter between cluster mass and optical richness is a key goal for cluster cosmology from photometric catalogs. We consider various modifications to the red-sequence matched filter richness estimator of Rozo et al. (2009), and evaluate their impact on the scatter in X-ray luminosity at fixed richness. Most significantly, we find that deeper luminosity cuts can reduce the recovered scatter, finding that sigma_lnLX|lambda=0.63+/-0.02 for clusters with M_500c >~ 1.6e14 h_70^-1 M_sun. The corresponding scatter in mass at fixed richness is sigma_lnM|lambda ~ 0.2-0.3 depending on the richness, comparable to that for total X-ray luminosity. We find that including blue galaxies in the richness estimate increases the scatter, as does weighting galaxies by their optical luminosity. We further demonstrate that our richness estimator is very robust. Specifically, the filter employed when estimating richness can be calibrated directly from the data, without requiring a-priori calibrations of the red-sequence. We also demonstrate that the recovered richness is robust to up to 50% uncertainties in the galaxy background, as well as to the choice of photometric filter employed, so long as the filters span the 4000 A break of red-sequence galaxies. Consequently, our richness estimator can be used to compare richness estimates of different clusters, even if they do not share the same photometric data. Appendix 1 includes "easy-bake" instructions for implementing our optimal richness estimator, and we are releasing an implementation of the code that works with SDSS data, as well as an augmented maxBCG catalog with the lambda richness measured for each cluster.
The Astrophysical Journal 04/2011; 746(2). · 6.73 Impact Factor
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##### Article: A Multiband Study of the Galaxy Populations of the First Four Sunyaev--Zeldovich Effect selected Galaxy Clusters
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ABSTRACT: We present first results of an examination of the optical properties of the galaxy populations in SZE selected galaxy clusters. Using clusters selected by the South Pole Telescope survey and deep multiband optical data from the Blanco Cosmology Survey, we measure the radial profile, the luminosity function, the blue fraction and the halo occupation number of the galaxy populations of these four clusters with redshifts ranging from 0.3 to 1. Our goal is to understand whether there are differences among the galaxy populations of these SZE selected clusters and previously studied clusters selected in the optical and the X-ray. The radial distributions of galaxies in the four systems are consistent with NFW profiles with a galaxy concentration of 3 to 6. We show that the characteristic luminosities in $griz$ bands are consistent with passively evolving populations emerging from a single burst at redshift $z=3$. The faint end power law slope of the luminosity function is found to be on average $\alpha \approx -1.2$ in griz. Halo occupation numbers (to $m^*+2$) for these systems appear to be consistent with those based on X-ray selected clusters. The blue fraction estimated to $0.36L^*$, for the three lower redshift systems, suggests an increase with redshift, although with the current sample the uncertainties are still large. Overall, this pilot study of the first four clusters provides no evidence that the galaxy populations in these systems differ significantly from those in previously studied cluster populations selected in the X-ray or the optical.
The Astrophysical Journal 03/2011; 734(1). · 6.73 Impact Factor
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##### Conference Paper: The Chicagoland Observatory Underground for Particle Physics cosmic ray veto system
[Hide abstract]
ABSTRACT: A photomultiplier (PMT) readout system has been designed for use by the cosmic ray veto systems of two warm liquid bubble chambers built at Fermilab by the Chicagoland Observatory Underground for Particle Physics (COUPP) collaboration. The systems are designed to minimize the infrastructure necessary for installation. Up to five PMTs can be daisy-chained on a single data link using standard Category 5 network cable. The cables is also serve distribute to low voltage power. High voltage is generated locally on each PMT base. Analog and digital signal processing is also performed locally. The PMT base and system controller design and performance measurements are presented.
Nuclear Science Symposium Conference Record (NSS/MIC), 2010 IEEE; 12/2010
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##### Article: Optical Redshift and Richness Estimates for Galaxy Clusters Selected with the Sunyaev-Zel'dovich Effect from 2008 South Pole Telescope Observations
[Hide abstract]
ABSTRACT: We present redshifts and optical richness properties of 21 galaxy clusters uniformly selected by their Sunyaev-Zel'dovich (SZ) signature. These clusters, plus an additional, unconfirmed candidate, were detected in a 178 deg2 area surveyed by the South Pole Telescope (SPT) in 2008. Using griz imaging from the Blanco Cosmology Survey and from pointed Magellan telescope observations, as well as spectroscopy using Magellan facilities, we confirm the existence of clustered red-sequence galaxies, report red-sequence photometric redshifts, present spectroscopic redshifts for a subsample, and derive R 200 radii and M 200 masses from optical richness. The clusters span redshifts from 0.15 to greater than 1, with a median redshift of 0.74; three clusters are estimated to be at z>1. Redshifts inferred from mean red-sequence colors exhibit 2% rms scatter in σ z /(1 + z) with respect to the spectroscopic subsample for z < 1. We show that the M 200 cluster masses derived from optical richness correlate with masses derived from SPT data and agree with previously derived scaling relations to within the uncertainties. Optical and infrared imaging is an efficient means of cluster identification and redshift estimation in large SZ surveys, and exploiting the same data for richness measurements, as we have done, will be useful for constraining cluster masses and radii for large samples in cosmological analysis.
The Astrophysical Journal 10/2010; 723(2):1736. · 6.73 Impact Factor
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##### Article: Cosmological Constraints from the Sloan Digital Sky Survey maxBCG Cluster Catalog
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ABSTRACT: We use the abundance and weak-lensing mass measurements of the Sloan Digital Sky Survey maxBCG cluster catalog to simultaneously constrain cosmology and the richness-mass relation of the clusters. Assuming a flat ΛCDM cosmology, we find σ8(Ω m /0.25)0.41 = 0.832 ± 0.033 after marginalization over all systematics. In common with previous studies, our error budget is dominated by systematic uncertainties, the primary two being the absolute mass scale of the weak-lensing masses of the maxBCG clusters, and uncertainty in the scatter of the richness-mass relation. Our constraints are fully consistent with the WMAP five-year data, and in a joint analysis we find σ8 = 0.807 ± 0.020 and Ω m = 0.265 ± 0.016, an improvement of nearly a factor of 2 relative to WMAP5 alone. Our results are also in excellent agreement with and comparable in precision to the latest cosmological constraints from X-ray cluster abundances. The remarkable consistency among these results demonstrates that cluster abundance constraints are not only tight but also robust, and highlight the power of optically selected cluster samples to produce precision constraints on cosmological parameters.
The Astrophysical Journal 12/2009; 708(1):645. · 6.73 Impact Factor
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##### Article: Cross-correlation Weak Lensing of SDSS Galaxy Clusters. III. Mass-to-Light Ratios
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ABSTRACT: We present measurements of the excess mass-to-light ratio (M/L) measured around MaxBCG galaxy clusters observed in the Sloan Digital Sky Survey. This red-sequence cluster sample includes objects from small groups with M 200 ~ 5 × 1012 h –1 M ☉ to clusters with M 200 ~ 1015 h –1 M ☉. Using cross-correlation weak lensing, we measure the excess mass density profile above the universal mean for clusters in bins of richness and optical luminosity. We also measure the excess luminosity density measured in the z = 0.25 i band. For both mass and light, we de-project the profiles to produce three-dimensional mass and light profiles over scales from 25 h –1 kpc to 22 h –1 Mpc. From these profiles we calculate the cumulative excess mass ΔM(r) and excess light ΔL(r) as a function of separation from the BCG. On small scales, where , the integrated mass-to-light profile (ΔM/ΔL)(r) may be interpreted as the cluster M/L. We find the (ΔM/ΔL)200, the M/L within r 200, scales with cluster mass as a power law with index 0.33 ± 0.02. On large scales, where , the ΔM/ΔL approaches an asymptotic value independent of cluster richness. For small groups, the mean (ΔM/ΔL)200 is much smaller than the asymptotic value, while for large clusters (ΔM/ΔL)200 is consistent with the asymptotic value. This asymptotic value should be proportional to the mean M/L of the universe M/L. We find M/Lb–2 M/L = 362 ± 54h (statistical). There is additional uncertainty in the overall calibration at the ~10% level. The parameter b 2 M/L is primarily a function of the bias of the L L * galaxies used as light tracers, and should be of order unity. Multiplying by the luminosity density in the same bandpass we find Ω m b–2 M/L = 0.20 ± 0.03, independent of the Hubble parameter.
The Astrophysical Journal 09/2009; 703(2):2232. · 6.73 Impact Factor
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##### Article: Charged Kaon Mass Measurement using the Cherenkov Effect
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ABSTRACT: The two most recent and precise measurements of the charged kaon mass use X-rays from kaonic atoms and report uncertainties of 14 ppm and 22 ppm yet differ from each other by 122 ppm. We describe the possibility of an independent mass measurement using the measurement of Cherenkov light from a narrow-band beam of kaons, pions, and protons. This technique was demonstrated using data taken opportunistically by the Main Injector Particle Production experiment at Fermi National Accelerator Laboratory which recorded beams of protons, kaons, and pions ranging in momentum from +37 GeV/c to +63 GeV/c. The measured value is 491.3 +/- 1.7 MeV/c^2, which is within 1.4 sigma of the world average. An improvement of two orders of magnitude in precision would make this technique useful for resolving the ambiguity in the X-ray data and may be achievable in a dedicated experiment. Comment: 6 pages, 7 figures. Submitted to Nuclear Instruments and Methods A
Nuclear Instruments and Methods in Physics Research Section A Accelerators Spectrometers Detectors and Associated Equipment 09/2009; · 1.14 Impact Factor
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##### Article: Improvement of the Richness Estimates of maxBCG Clusters
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ABSTRACT: Minimizing the scatter between cluster mass and accessible observables is an important goal for cluster cosmology. In this work, we introduce a new matched filter richness estimator, and test its performance using the maxBCG cluster catalog. Our new estimator significantly reduces the variance in the LX -richness relation, from to . Relative to the maxBCG richness estimate, it also removes the strong redshift dependence of the LX -richness scaling relations, and is significantly more robust to photometric and redshift errors. These improvements are largely due to the better treatment of galaxy color data. We also demonstrate the scatter in the LX -richness relation depends on the aperture used to estimate cluster richness, and introduce a novel approach for optimizing said aperture which can easily be generalized to other mass tracers.
The Astrophysical Journal 08/2009; 703(1):601. · 6.73 Impact Factor
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##### Article: Constraining the Scatter in the Mass-richness Relation of maxBCG Clusters with Weak Lensing and X-ray Data
[Hide abstract]
ABSTRACT: We measure the logarithmic scatter in mass at fixed richness for clusters in the maxBCG cluster catalog, an optically selected cluster sample drawn from Sloan Digital Sky Survey imaging data. Our measurement is achieved by demanding consistency between available weak-lensing and X-ray measurements of the maxBCG clusters, and the X-ray luminosity-mass relation inferred from the 400 days X-ray cluster survey, a flux-limited X-ray cluster survey. We find (95% CL) at N 200 40, where N 200 is the number of red sequence galaxies in a cluster. As a byproduct of our analysis, we also obtain a constraint on the correlation coefficient between ln LX and ln M at fixed richness, which is best expressed as a lower limit, r L,M|N ≥ 0.85(95% CL). This is the first observational constraint placed on a correlation coefficient involving two different cluster mass tracers. We use our results to produce a state-of-the-art estimate of the halo mass function at z = 0.23—the median redshift of the maxBCG cluster sample—and find that it is consistent with the WMAP5 cosmology. Both the mass function data and its covariance matrix are presented.
The Astrophysical Journal 06/2009; 699(1):768. · 6.73 Impact Factor
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##### Article: A Merged Catalog of Clusters of Galaxies from Early Sloan Digital Sky Survey Data
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ABSTRACT: We present a catalog of 799 clusters of galaxies in the redshift range zest = 0.05-0.3 selected from ~400 deg2 of early Sloan Digital Sky Survey (SDSS) commissioning data along the celestial equator. The catalog is based on merging two independent selection methods—a color-magnitude red-sequence maxBCG technique (B), and a hybrid matched filter method (H). The BH catalog includes clusters with richness Λ ≥ 40 (matched filter) and Ngal ≥ 13 (maxBCG), corresponding to typical velocity dispersion of σv 400 km s-1 and mass (within 0.6 h-1 Mpc radius) 5 × 1013 h-1 M. This threshold is below Abell richness class 0 clusters. The average space density of these clusters is 2 × 10-5 h3 Mpc-3. All NORAS X-ray clusters and 53 of the 58 Abell clusters in the survey region are detected in the catalog; the five additional Abell clusters are detected below the BH catalog cuts. The cluster richness function is determined and found to exhibit a steeply decreasing cluster abundance with increasing richness. We derive observational scaling relations between cluster richness and observed cluster luminosity and cluster velocity dispersion; these scaling relations provide important physical calibrations for the clusters. The catalog can be used for studies of individual clusters, for comparisons with other sources such as X-ray clusters and active galactic nuclei, and, with proper correction for the relevant selection functions, also for statistical analyses of clusters.
The Astrophysical Journal Supplement Series 12/2008; 148(2):243. · 16.24 Impact Factor
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##### Article: The Cluster Mass Function from Early Sloan Digital Sky Survey Data: Cosmological Implications
[Hide abstract]
ABSTRACT: The mass function of clusters of galaxies is determined from 400 deg2 of early commissioning imaging data of the Sloan Digital Sky Survey using ~300 clusters in the redshift range z = 0.1-0.2. Clusters are selected using two independent selection methods: a matched filter and a red-sequence color-magnitude technique. The two methods yield consistent results. The cluster mass function is compared with large-scale cosmological simulations. We find a best-fit cluster normalization relation of σ8Ω = 0.33 ± 0.03 (for 0.1 Ωm 0.4) or, equivalently, σ8 = (0.16/Ωm)0.6. The amplitude of this relation is significantly lower than the previous canonical value, implying that either Ωm is lower than previously expected (Ωm = 0.16 if σ8 = 1) or σ8 is lower than expected (σ8 = 0.7 if Ωm = 0.3). The shape of the cluster mass function partially breaks this classic degeneracy. We find best-fit parameters of Ωm = 0.19 ± and σ8 = 0.9 ±. High values of Ωm (0.4) and low σ8 (0.6) are excluded at 2 σ.
The Astrophysical Journal 12/2008; 585(1):182. · 6.73 Impact Factor
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##### Article: An Improved Cluster Richness Estimator
[Hide abstract]
ABSTRACT: Minimizing the scatter between cluster mass and accessible observables is an important goal for cluster cosmology. In this work, we introduce a new matched filter richness estimator, and test its performance using the maxBCG cluster catalog. Our new estimator significantly reduces the variance in the L_X-richness relation, from \sigma_{\ln L_X}^2=(0.86\pm0.02)^2 to \sigma_{\ln L_X}^2=(0.69\pm0.02)^2. Relative to the maxBCG richness estimate, it also removes the strong redshift dependence of the richness scaling relations, and is significantly more robust to photometric and redshift errors. These improvements are largely due to our more sophisticated treatment of galaxy color data. We also demonstrate the scatter in the L_X-richness relation depends on the aperture used to estimate cluster richness, and introduce a novel approach for optimizing said aperture which can be easily generalized to other mass tracers.
10/2008;
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##### Article: The Galaxy Content of SDSS Clusters and Groups
[Hide abstract]
ABSTRACT: Imaging data from the Sloan Digital Sky Survey are used to characterize the population of galaxies in groups and clusters detected with the MaxBCG algorithm. We investigate the dependence of Brightest Cluster Galaxy (BCG) luminosity, and the distributions of satellite galaxy luminosity and satellite color, on cluster properties over the redshift range 0.1 < z < 0.3. The size of the dataset allows us to make measurements in many bins of cluster richness, radius and redshift. We find that, within r_200 of clusters with mass above 3e13 h-1 M_sun, the luminosity function of both red and blue satellites is only weakly dependent on richness. We further find that the shape of the satellite luminosity function does not depend on cluster-centric distance for magnitudes brighter than ^{0.25}M_i - 5log(h) < -19. However, the mix of faint red and blue galaxies changes dramatically. The satellite red fraction is dependent on cluster-centric distance, galaxy luminosity and cluster mass, and also increases by ~5% between redshifts 0.28 and 0.2, independent of richness. We find that BCG luminosity is tightly correlated with cluster richness, scaling as L_{BCG} ~ M_{200}^{0.3}, and has a Gaussian distribution at fixed richness, with sigma_{log L} ~ 0.17 for massive clusters. The ratios of BCG luminosity to total cluster luminosity and characteristic satellite luminosity scale strongly with cluster richness: in richer systems, BCGs contribute a smaller fraction of the total light, but are brighter compared to typical satellites. This study demonstrates the power of cross-correlation techniques for measuring galaxy populations in purely photometric data. Comment: 22 pages, 14 figures, submitted to ApJ
The Astrophysical Journal 10/2007; · 6.73 Impact Factor
#### Publication Stats
1k Citations 505.76 Total Impact Points
#### Institutions
• ###### University of California Observatories
Santa Cruz, California, United States
• ###### Fermi National Accelerator Laboratory (Fermilab)
Batavia, Illinois, United States
• ###### University of California, Santa Cruz
• Department of Astronomy and Astrophysics
Santa Cruz, CA, United States
• ###### University of Chicago
• Department of Astronomy and Astrophysics
Chicago, IL, United States
• ###### University of Michigan
• Department of Physics
Ann Arbor, MI, United States
• ###### Joint Institute for Nuclear Research
Dubno, Moskovskaya, Russia
• ###### University of Oklahoma
Norman, Oklahoma, United States
• ###### CUNY Graduate Center
New York City, New York, United States
• ###### Northeastern University
Boston, Massachusetts, United States
• ###### University of Buenos Aires
Buenos Aires, Buenos Aires F.D., Argentina
• ###### Institute of High Energy Physics
Protvino, Moskovskaya, Russia
• ###### Los Andes University (Colombia)
Μπογκοτά, Bogota D.C., Colombia | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8616040349006653, "perplexity": 3997.6219056231917}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447562878.33/warc/CC-MAIN-20141224185922-00005-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/95770/positive-forms-and-dot-products-in-some-basis | # Positive forms and dot products in some basis
For a real $n \times n$ matrix $A$ the following are equivalent:
• $A$ is symmetric and positive definite
• There is an invertible matrix $P$ such that $A = P^{t} P$
This makes geometric sense when viewing $A$ as representing the dot product in some basis and the standard proof is based around this.
But is there a way to prove this viewing $A$ as a linear operator. What exactly is the geometric meaning of decomposing a symmetric positive definite operator into $P^{t}P$? What is the linear operator $P$ and $P^t$ doing that results in $P^t P =A$?
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https://math.stackexchange.com/questions/2472720/what-is-the-meaning-of-operatornamehom-a-m-operatornamehom-b-n-p | # What is the meaning of $\operatorname{Hom}_A (M, \operatorname{Hom}_{B} (N, P ))$?
Let $A$ and $B$ be rings, $M$ an $A$-module, $P$ an $B$-module, $N$ an $(A, B)$-bimodule. Then what is the meaning of $\operatorname{Hom}_A (M, \operatorname{Hom}_{B} (N, P ))$?
If $\operatorname{Hom}_A (M, \operatorname{Hom}_{B} (N, P ))$ is the $A$-linear maps from $M$ to $\operatorname{Hom}_{B} (N, P )$, doesn't $\operatorname{Hom}_{B} (N, P )$ have to be an $A$-module? Should it be made into one in some sense?
It can indeed be made an $A$-module, in the following way: \begin{align} A\times \operatorname{Hom}_B(N,P)&\longrightarrow \operatorname{Hom}_B(N,P)\\ (a,f)&\longmapsto \bigl(af\colon n\mapsto f(an)\bigr) \end{align} You can check the axioms for the structure of $A$-module from the compatibility conditions for the structure of $(A,B)$-bimodule of $N$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9276506900787354, "perplexity": 240.81808146699464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573258.74/warc/CC-MAIN-20190918065330-20190918091330-00286.warc.gz"} |
https://www.physicsforums.com/threads/polynomials-and-isomorphism.674976/ | # Polynomials and isomorphism
1. Feb 27, 2013
### trap101
Ok for the longest while I've been at war with polynomials and isomorphisms in linear algebra, for the death of me I always have a brain freeze when dealing with them. With that said here is my question:
Is this pair of vector spaces isomorphic? If so, find an isomorphism T: V-->W.
V= R4 , W = {p$\in$P4(R) | p(0) = 0}
Here is the issue. What kind of polynomial am I examining?
It says it is the set of polynomials of degree 4 s.t p(0) = 0.
What does the p(0) = 0 mean? For example if I used the set of standard basis vectors of P4(R), what would the set of p(0) = 0 look like? All I could picture is
P(1) = 1 , p(x) = 0 , p(x2) = 0,.....p(x4) = 0.
Is that the right way to look at it?
There was no specified transformation given either.
Thanks
2. Feb 27, 2013
### jbunniii
A general polynomial of degree 4 looks like $p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x^1 + a_0$. So plug in $x = 0$ to see what $p(0) = 0$ means.
By the way, are you sure $P_4(\mathbb{R})$ means the set of polynomials of degree 4, and not the set of polynomials with degree less than or equal to 4?
3. Feb 27, 2013
### trap101
It would be the set of degree less than or equal to 4. Ok but doing that would then mean all the polynomials get reduced down to their constant term so the dimension is 1 which is not the same as the dimension of R4 i.e not isomorphic.
4. Feb 27, 2013
### jbunniii
No, that's not true. What do you get when plug $x = 0$ into $p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x^1 + a_0$? Which coefficient(s) does the condition $p(0) = 0$ force to zero?
5. Feb 27, 2013
### trap101
it forces all of the coefficients except for a0 to be 0's. So it would be the set of polynomials represented only by their constants. But how can I show its dimension in order to prove it's an isomorphism.
6. Feb 27, 2013
### jbunniii
No, that's not right. If $p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x^1 + a_0$, then $p(0) = a_0$. So $p(0) = 0$ forces $a_0 = 0$. All the other coefficients can be anything.
7. Feb 28, 2013
### trap101
unfortunately you had to give me the answer to finally see it, but I get your logic behind it now. So the set of polynomials where p(0)=0 implies that the constant is going to be 0. Now since any of the other terms can be anything, which in this case reduces the dimension of the polynomial by 1 making it the same dimension as R4 therefore it is isomorphic.
Now I have to try and find an isomprphism........which is the other thing I suck at. So I'm going to have the general form of the polynomial:
$p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x^1 + a_0$ but a0 = 0 so it is now of the form $p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x^1$
now for it to be an isomorphism I have to get this polynomial to turn into a vector of 4 elements by a transformation. How do I create that transformation?
8. Feb 28, 2013
### pasmith
What are the first four derivatives of $p(x)$ at $x = 0$?
9. Feb 28, 2013
### trap101
hmmm..... I suppose each one of those derivatives could be considered a component of the vector then. Interesting.
10. Feb 28, 2013
### jbunniii
Well, you have four coefficients, not counting the one that was forced to zero. The natural thing to try is to group those four coefficients into a vector, $(a_1, a_2, a_3, a_4)$ [or in whichever order you prefer]. Then check whether this is indeed an isomoprhism.
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http://mathonline.wikidot.com/the-product-of-two-series-of-real-numbers | The Product of Two Series of Real Numbers
# The Product of Two Series of Real Numbers
Consider two series of real numbers, $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$. Then we can consider the product of these two series:
(1)
\begin{align} \quad \left ( \sum_{n=0}^{\infty} a_n \right ) \left ( \sum_{n=0}^{\infty} b_n \right ) = (a_0 + a_1 + a_2 + ...)(b_0 + b_1 + b_2 + ... ) \end{align}
We can organize the terms of the expanding the product on the righthand side above in the following array:
(2)
\begin{align} \quad \begin{matrix} a_0b_0 & a_0b_1 & a_0b_2 & \cdots & a_0b_n & \cdots \\ a_1b_0 & a_1b_1 & a_1b_2 & \cdots & a_1b_n & \cdots \\ a_2b_0 & a_2b_1 & a_2b_2 & \cdots & a_2b_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \\ a_nb_0 & a_nb_1 & a_nb_2 & \cdots & a_nb_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \end{matrix} \end{align}
It is not clear how exactly we should sum up the terms in this array since the original multiplication involved multiplication of infinite sums. One such way is defined (generically) below.
Definition: Let $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ be two series of real numbers. The Product of these two series denoted $\displaystyle{\left ( \sum_{n=0}^{\infty} a_n \right ) \left ( \sum_{n=0}^{\infty} b_n \right )}$ is given by the partial sum sequence $\displaystyle{S_n = \sum_{i=0}^{n} \left ( \sum_{j=0}^{n} a_ib_j \right ) = \left ( \sum_{i=0}^{n} a_i \right ) \left ( \sum_{j=0}^{n} b_j \right )}$.
This type of series multiplication tells us exactly how the add the terms in the array above. $S_0$ is given by summing up the red terms below:
(3)
\begin{align} \quad \begin{matrix} \color{red}{a_0b_0} & a_0b_1 & a_0b_2 & \cdots & a_0b_n & \cdots \\ a_1b_0 & a_1b_1 & a_1b_2 & \cdots & a_1b_n & \cdots \\ a_2b_0 & a_2b_1 & a_2b_2 & \cdots & a_2b_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \\ a_nb_0 & a_nb_1 & a_nb_2 & \cdots & a_nb_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \end{matrix} \end{align}
$S_1$ is given by summing up the orange terms below:
(4)
\begin{align} \quad \begin{matrix} \color{orange}{a_0b_0} & \color{orange}{a_0b_1} & a_0b_2 & \cdots & a_0b_n & \cdots \\ \color{orange}{a_1b_0} & \color{orange}{a_1b_1} & a_1b_2 & \cdots & a_1b_n & \cdots \\ a_2b_0 & a_2b_1 & a_2b_2 & \cdots & a_2b_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \\ a_nb_0 & a_nb_1 & a_nb_2 & \cdots & a_nb_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \end{matrix} \end{align}
$S_2$ is given by summing up the yellow terms below:
(5)
\begin{align} \quad \begin{matrix} \color{gold}{a_0b_0} & \color{gold}{a_0b_1} & \color{gold}{a_0b_2} & \cdots & a_0b_n & \cdots \\ \color{gold}{a_1b_0} & \color{gold}{a_1b_1} & \color{gold}{a_1b_2} & \cdots & a_1b_n & \cdots \\ \color{gold}{a_2b_0} & \color{gold}{a_2b_1} & \color{gold}{a_2b_2} & \cdots & a_2b_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \\ a_nb_0 & a_nb_1 & a_nb_2 & \cdots & a_nb_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \end{matrix} \end{align}
In general, $S_n$ is given by summing up all terms in the $n + 1$ by $n+1$ top-left subarray above.
We will now look at a nice theorem which tells us if the two series in the product converge to $A$ and $B$, then the product of the two series will also converge to the product, $AB$.
Theorem 1: If $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ converge to $A$ and $B$ respectively, then the product $\displaystyle{\left ( \sum_{n=0}^{\infty} a_n \right ) \left ( \sum_{n=0}^{\infty} b_n \right )}$ converges to $AB$.
• Proof: Let $(A_n)_{n=0}^{\infty}$ and $(B_n)_{n=0}^{\infty}$ be the sequences of partial sums for $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ respectively. If $S_n$ is the sequence of partial sums for the product of these two series, then:
(6)
\begin{align} \quad S_n = \left ( \sum_{i=0}^{n} a_i \right ) \left ( \sum_{j=0}^{n} b_j \right ) = A_nB_n \end{align}
• Since the sequences $A_n$ and $B_n$ converge to $A$ and $B$ as $n \to \infty$, we have that $S_n$ converges to $AB$ as $n \to \infty$. Therefore:
(7)
\begin{align} \quad \left ( \sum_{n=0}^{\infty} a_n \right ) \left ( \sum_{n=0}^{\infty} b_n \right ) = AB \quad \blacksquare \end{align} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 7, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000097751617432, "perplexity": 206.35614185130342}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304515.74/warc/CC-MAIN-20220124054039-20220124084039-00275.warc.gz"} |
https://tw.answers.yahoo.com/question/index?qid=20131207000010KK02196 | Ian Chen 發問時間： 科學數學 · 6 年前
# 我有數學的問題~急x)
1. let f and g be polynomials over a field what is the degree of the polynomial f(g(x))?
2. does the set of all nonzero polynomials over a field form a group with respect to multiplication?
### 1 個解答
• 6 年前
最佳解答
1. We denote by F the field considered. We suppose that the degree of f is p and that of g is q, with (p,q)∈N^2. (N={0,1,2,....})
Let f(x)=a0+a1*x+...+ap*x^p with (a0,...,ap)∈F^p and ap≠0
g(x)=b0+b1*x+...+bq*x^q with (b0,...,bq)∈F^q and bq≠0
then f(g(x))=ap(b0+b1*x+...+bq*x^q)^p+.....+a1(b0+b1*x+...+bq*x^q)+a0
=ap*bq^p*x^pq+h(x) where h is a polynomial whose degree is inferior to pq.
Because ap*bq^p is nonzero (since ap≠0 and bq≠0), we can conclude that the degree of f(g(x)) is pq, which is also deg(f)*deg(g).
Therefore, deg(f o g)=deg(f)*deg(g) .
2.Let's take the polynomial f(x)=x for example. (It's a nonzero polynomial)
There is no polynomial g such that fg=1, where 1 is the identity element for the group.
So, the set of all nonzero polynomials over a field DOESN'T form a multipicative group.
參考資料： Myself I study maths in France | {"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": 0.9874854683876038, "perplexity": 1427.7329569678936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670559.66/warc/CC-MAIN-20191120134617-20191120162617-00206.warc.gz"} |
https://lysario.de/2014/02 | In problem [1, p. 94 exercise 4.1] we will show that the periodogram is an inconsistent estimator by examining the estimator at $f=0$, or
$\hat{P}_{PER}(0)=\frac{1}{N}\left(\sum\limits_{n=0}^{N-1}x[n]\right)^{2}.$ (1)
If $x[n]$ is real white Gaussian noise process with PSD
$P_{xx}(f)=\sigma_{x}^{2}$ (2)
we are asked to find the mean an variance of $hat{P}_{PER}(0)$. We are asked if the variance converge to zero as $N \rightarrow \infty$. The hint provided within the exercise is to note that
$\hat{P}_{PER}(0)= \sigma_{x}^{2} \left(\sum\limits_{n=0}^{N-1}\frac{x[n]}{\sigma_{x}\sqrt{N}}\right)^{2}$ (3)
where the quantiiy inside the parenthesis is $\sim N(0,1)$ read the conclusion > | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "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": 0.9494830965995789, "perplexity": 290.717146597784}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257259.71/warc/CC-MAIN-20190523143923-20190523165923-00403.warc.gz"} |
http://docs.oracle.com/cd/E28280_01/bi.1111/b32121/pbr_uxprt004.htm | # 10.4 Printer-Related Files
This section explains the different printing related files. It gives an overview of these files and also provides information for editing these files for common printing needs.
## 10.4.1 Overview of Files
Table 10-2 lists files used by Oracle Reports for printing on UNIX.
Table 10-2 Printer-Related Files Overview
File Name/Extension Description
`.ppd`
PostScript Printer Definition file
`.hpd`
HP glue file
`.afm`
Adobe font metrics file
`.tfm`
PCL font metrics file
`uifont.ali`
font aliasing file
`uiprint.txt`
printer configuration file
## 10.4.2 PPD Files
PostScript is Adobe's page description programming language. PPD files define what capabilities a printer has for applications like Oracle Reports. For example, a PPD file might define which paper tray to use, what paper sizes are available, what is the physical dimension of the paper, and what font is available. Currently, Oracle Reports reads the paper sizes and fonts available on the printer as well as its default resolution from this file. In the future, more information may be used, such as memory for proper image partitioning.
The only reason to modify the PPD file is to allow Oracle Reports to recognize newly added fonts or memory. You can also change the `DefaultPageSize` to your preferred page size.
Note:
Page sizes, like all PPD entries, are case sensitive. Other entries in the PPD file should generally be left undisturbed.
When you select a printer that is not listed in `uiprint.txt` or change the type of printer to a PostScript type in the Choose Printer dialog box, you are prompted for the PPD file for the printer. You must choose the PPD file for a printer that most closely resembles the printer being used. PPD file names typically bear some resemblance to the printer model name.
In `uiprint.txt`, a PPD file must be specified for each printer. If an invalid PPD file is specified for the current printer (for example, no PPD file is found or the PPD file format is wrong), Oracle Reports will use `default.ppd` for that printer. You should make `default.ppd` a copy of another PPD file that better reflects the most likely default, local printer.
Oracle Reports includes a common set of PPD files, but sometimes you may need to get specific PPD files for your printers from the vendor. Table 10-3 shows some examples of PPD files that are shipped with Oracle Reports:
Table 10-3 Common PPD Files Shipped with Oracle Reports
PPD File Name Corresponding Printer
`appl230.ppd`
Apple LaserWriter v23.0
`datap462.ppd`
Dataproducts LZR-2665
`declps32.ppd`
Digital PrintServer 40
`default.ppd`
Default Level 1 PostScript Printer
`hpljet41.ppd`
HP LaserJet 4/4M PostScript 600DPI
`lwntx470.ppd`
Apple LaserWriter II NTX
`nccps801.ppd`
NEC Colormate PS/80
`tkphzr33.ppd`
Tektronix Phaser III PXi v2011.108
`l530_523.ppd`
Linotronic 530
`screenprinter.ppd`
Default PPD file to be used when a printer is not available on UNIX.
If you need a PPD file that is not among those shipped with Oracle Reports, you must do one of the following (in order of preference):
• Ask the printer vendor for the PPD file.
• Copy an existing PPD file and edit it.
• Ask Adobe for the PPD specs and write the PPD file.
The PostScript file only has the font information not the font metrics. Oracle Reports refers to the AFM file installed for the font metrics information. The font vendors provide these AFM files. Oracle Reports ships AFM files for some of the most commonly used fonts. The printer must have the required font installed in order to correctly print the PostScript file generated by Oracle Reports.
### 10.4.2.1 Local Customization of PPD files
A PPD file is a static representation of the features of a printer. It contains default factory settings. Once a printer is installed, features such as additional memory, paper trays, and fonts may be added to the device. The task of managing a device is a dynamic issue that requires keeping track of fonts downloaded to disk, error handlers, RAM-based fonts and procedure sets, default device setup, and so forth. This kind of device management is beyond the scope of PPD files. However, there are some provisions for customizing the information contained in PPD files to adapt them to local instances of printers or to specific applications when necessary.
Instead of modifying the original PPD file, another approach would be having a new file having the local customization of certain parameters and refer to the primary file for the remaining information. The local customization file must contain a reference to the primary PPD file in this format:
```*Include: "filename"
```
where `filename` is the name of the primary PPD file. This referencing allows a system administrator to later replace the primary PPD file without forcing users to edit their local customization files. A file referenced by the `*Include` keyword is treated as though it were in the including (local customization) file.
For example, suppose that the `default.ppd` file is defined as:
```*PPD-Adobe: "4.0"
*Include: "datap462.ppd"
*% Page definitions
*DefaultPageSize: Letter
…………………………
*DefaultPageRegion: Letter
```
The primary PPD file is `datap462.ppd`.
Administrators should change the name of the included file to conform to their site's default printer type.
When a local customization file includes a primary PPD file, there might be several instances of the same keyword in the composite file. Hence, the location of the primary file in the customization file (beginning or end) is important and effects the changes made by the customization file.
## 10.4.3 HPD Files
HPD files provide functionality for PCL printers that is similar to what PPD files provide for PostScript printers. HPD or HP glue files provide information on what fonts are available for a PCL printer. The HPD file format can be found in the HP PCL5 Developer's Guide.
Just as PostScript has AFM files, every HP font must have an associated TFM file. The font vendor should provide TFM files and new fonts should be added to the HPD file for your printer when installed. For a new font, you should specify the following fields in the HPD file:
```FONT={fontname}
/tfm={tfm-filename}
```
where
`fontname` is a descriptive name for the font.
`tfm-filename` is the base file name for the TFM file.
If the TFM file isn't specific enough, you can also specify the following after the `FONT` field:
```/ptsize={size {size ...}}
```
If the specified font is a bit mapped font but is listed in the TFM file as a scalable font, you can limit the point sizes used by listing the acceptable sizes as follows:
```/symset={symset {symset ...}}
```
This field limits the supported symbol sets to those listed. See the HP PCL documentation for a list of recognized symbol sets.
Oracle Reports also supports the `defaultpaper` field for printing to PCL format. This field can be used to set the `defaultpaper` to be used by the Toolkit. The format of this field is:
```<defaultpaper={papername}>
```
For example, the following sets the paper name to A4:
```<defaultpaper=A4>
```
The paper name is case insensitive. If you specify `defaultpaper` in more than one place, then the last instance of `defaultpaper` is used. If you specify a paper name that is not supported by the printer, `defaultpaper` is ignored and `LETTER` is used as the paper name instead. Similarly, if the paper name is incorrect, then `LETTER` is used.
## 10.4.4 Font Metrics Files
Oracle Reports supports two kinds of font metrics files:
### 10.4.4.1 AFM files
Each AFM files contains the font-related metrics for a single font. The metrics include various font attributes such as style, weight, width, and character set. AFM files and a description of the AFM file format are typically available from the font or printer vendors.
To install the AFM file, just copy it to the AFM file location, which is listed in Section 10.2.2, "Verifying the Printer Setup for Oracle Reports". The name of the file must match name of the font without the`.afm` extension. For example, if the font name is `CodedreineunBold`, the file name must be `CodedreineunBold`.
To verify the font name, you can look for the fontname string in the AFM file. Please note that the AFM files are not font files, they are metrics files, which give information on how to properly format the characters for the printer. If you have an AFM file for a font, but the font is not present on the printer, Oracle Reports cannot generate the correct output on the printer because of the font metrics mismatch. You must ensure that the font used to design the report is also available on the printer.
### 10.4.4.2 TFM files
PCL uses HPD and TFM files. The HPD file contains the list of available fonts for the printer and each font refers to a TFM file. TFM files serve the same purpose as Adobe's AFM files, with each file listing information about a single font. The HPD file is an ASCII file, which can be edited, but the TFM file is a binary file, which cannot be edited.
To use a new font in Oracle Reports and have it appear correctly in PCL output, you need the HPD and TFM files for the printer. You can copy an HPD file from an existing one, after you ensure it is suitable for your printer. The fonts specified in the HPD file must be available on the printer.
Oracle Reports includes a common set of TFM files. If you need other font metrics files for your printer, you should obtain them from your font or printer vendor. To install the TFM file, just copy it to the TFM file location, which is listed in Section 10.2.2, "Verifying the Printer Setup for Oracle Reports".
## 10.4.5 uifont.ali
The `uifont.ali` file defines the font aliases used by Oracle Reports. It is an extremely useful tool for cross-platform development because it enables you to define which fonts to substitute when a particular font is unavailable. `uifont.ali` is located in:
On Windows: `ORACLE_INSTANCE\config\FRComponent\frcommon\tools\common`
On UNIX: `ORACLE_INSTANCE/config/FRComponent/frcommon/guicommon/tk/admin`
To alias a font, use the following syntax:
```source_font = destination_font
```
For each font, you may also specify the following attributes:
```face.size.style.weight.width.character_set
```
Styles may also be combined using a plus sign + to delimit the styles. For example:
```Arial.Italic+Overstrike = Helvetica.12.Italic.Bold
```
This entry maps any Arial font that has both Italic and Overstrike styles to a 12-point, bold, and italic Helvetica font. Font faces can be case sensitive depending on the platform and the surface; that is, printer or system.
Chapter 9, "Managing Fonts in Oracle Reports" for more font-related information.
## 10.4.6 uiprint.txt
`uiprint.txt` provides a convenient way for you to provide details about the printer queue, such as the type of printer driver and the printer description. You should edit `uiprint.txt` for each instance of Oracle Reports.
Section 10.3.1, "Editing uiprint.txt File" for more information about `uiprint.txt`.
## 10.4.7 Editing the Printer-Related Files
This section describes how to edit the various print-related files:
### 10.4.7.1 Editing PPD files
In some cases, you may need to change certain attributes in your PPD file. The sections that follow describe some of the attributes that you would commonly want to change:
#### 10.4.7.1.1 Changing the default paper size
Suppose that you need the page size to be A4 for some of your reports. On UNIX platforms, the printer driver is specified in `uiprint.txt` and the default page size is not necessarily set to A4. For example, `hpljet41.ppd` has LETTER as the default page size. Note that the default page size setting for each printer queue is taken from the corresponding PPD file.
To set A4 as the default page size, you would do the following:
1. Edit `uiprint.txt` to include a PostScript Printer Description file (extension is .ppd) that supports the A4 page size. For example, you might include `hpljet41.ppd`.
2. As a backup, make a copy of `hpljet41.ppd`.
3. Add an entry to `uiprint.txt`:
```Printer_name: PostScript:1: the printer on floor1:hpljet41.ppd
```
4. Edit `hpljet41.ppd` and change these settings as follows:
```DefaultPageSize: A4
DefaultPageRegion: A4
DefaultImageableArea: A4
DefaultPaperDimension: A4
```
#### 10.4.7.1.2 Changing the printer margin settings
To change the margins, you must modify the ImageableArea section in the PPD file. ImageableArea provides the bounding box of the area in which the printer may print for the page size named mediaOption. There will be one statement for each named page size supported by the device. *DefaultImageableArea provides the mediaOption name of the default imageable area. Since there can be only one default page size, this value should be the same as the value of *DefaultPageSize, *DefaultPageRegion, and *DefaultPaperDimension.
The syntax for defining imageable area is as follows:
```*ImageableArea mediaOption: "llx lly urx ury "
*DefaultImageableArea: mediaOption | Unknown
```
`ll` stands for lower left corner and `ur` for upper right corner. The bounding box value of *ImageableArea is given as four real numbers, representing the x and y coordinates of the lower left and upper right corners of the region, respectively, in the PostScript language default user space coordinate system. The x and y axes of a given page size correspond to the x and y axes of that page size in the *PaperDimension entry.
The imageable area is defined as the part of the page where the printer may actually make marks. On some printers, the imageable area of a given page size varies as a result of the current resolution, amount of memory, the direction of paper feed, and other factors. In PPD files where the imageable area of a given page size can vary, the imageable area recorded for that page size will be the intersection of all possible imageable areas for that page size. This formula ensures that the available imageable area is never smaller than that shown in the PPD file and all marks made within the imageable area will be visible. It does, however, also mean that the imageable area in the current configuration might actually be larger than the imageable area shown in the PPD file.
The following table contains the option keywords currently registered for `mediaOption`, which designates a given page size on a device:
Table 10-4 `mediaOption` Keywords
mediaOption (Paper Size) Size (pts) Size (mm) Size (inches)
Letter
612 * 792
215.9 * 279.4
8.5 * 11
Legal
612 * 1008
215.9 * 355.6
8.5 * 14
Ledger
1224 * 792
431.8 * 279.4
17 * 11
Tabloid
792 * 1224
279.4 * 431.8
11 * 17
A3
842 * 1191
297 * 420
11.69 * 16.54
A4
595 * 842
210 * 297
8.27 * 11.69
A5
420 * 595
148 * 210
5.83 * 8.27
B4
729 * 1032
257 * 364
10.12 * 14.33
B5
516 * 729
182 * 257
7.17 * 10.12
Example
To change the margins of an A4 page in the `default.ppd`, you would perform the following steps:
1. Modify the default page from `Letter` to `A4` in the following sections:
```*% Page definitions
*DefaultPageSize: A4
*PageSize A4: " "
*% These entries set up the frame buffer. Usually used with manual feed.
*DefaultPageRegion: A4
*PageRegion A4: "A4"
*% These provide the physical dimensions of the paper (by keyword)
*DefaultPaperDimension: A4
*PaperDimension A4: "595 842"
```
2. Add the margin definition in the following sections:
```*% Imageable (writable) areas for each page size, in pixels
*DefaultImageableArea: A4
*ImageableArea A4: "2 2 593 840 "
```
Note:
All PPD entries are case sensitive.
#### 10.4.7.1.3 Adding a new font entry to PPD files
On PostScript printers, Oracle Reports only enables you to use fonts known to be available on the printer. Since printers are rarely available for personal requests on multiprocess operating systems, Oracle Reports gets a complete list of fonts from the PPD file.
When a new font is installed on the printer, a corresponding font entry needs to be added to the printer's PPD file. The format for a font entry is:
```*Font {fontname}: {encoding} "({version})" {charset}
```
where
`{fontname}` is the Adobe font face name as specified in PostScript.
`{encoding}` is the PostScript encoding name.
`{version}` is the FontInfo version number.
`{charset}` is the Adobe character set.
The `encoding` value has slightly different meanings depending on the font type. If the encoding cannot be determined, the value of encoding may be set to unknown. Fonts are usually re-encoded by applications to provide other encodings; the `charset` value for each font indicates which encodings are possible for that font. For more information, please refer to the PPD specification from Adobe.
When new fonts are added to the printer, the matching AFM files must also be added to the font metrics directory. Oracle Reports requires the AFM files to get the actual font attributes and properly place text on the printed page.
Example
Suppose you add a new font, CodedreineunBold, and want to edit the PPD file to include the new font.
1. In the PPD file, search for:
```*% Font Information
```
2. For the new font, append the following at the end of the paragraph:
```*Font CodedreineunBold: Standard "(00.1001)" Standard ROM
```
#### 10.4.7.1.4 Overriding the printer tray setting
The PostScript output generated by Oracle Reports has the tray information embedded into it. The PPD file defines the default tray to be used and is followed by the definitions of valid trays for the printer. To print to a different tray, the `DefaultInputSlot` entry in the PPD file must be updated.
In the PPD file, you should find a section that lists the default tray and the valid input slots. The section typically starts with a line like this one:
```*OpenUI *InputSlot: <PickOne>
```
The default tray entry looks like the following:
```*DefaultInputSlot: Lower
```
The defined slots typically follow the default entry and look like the following:
```*InputSlot Upper/Multipurpose Tray: "
...
*InputSlot Lower/Paper Cassette: "
```
The section ends with a line like the following:
```*CloseUI: *InputSlot
```
You can set `DefaultInputSlot` to be any of the values in the list of defined slots.
### 10.4.7.2 Editing HPD files for PCL printing
In some cases, you may need to change certain attributes in you HPD file. The sections that follow describe some of the attributes that you would commonly want to change:
#### 10.4.7.2.1 Changing the paper size
For example, to change the papersize to A4, add the following to the HPD file used:
``` <defaultpaper=A4>
```
#### 10.4.7.2.2 Adding a new font entry
As with PostScript's AFM files, every HP font must have a TFM file in order for Oracle Reports to use it. The font vendor should provide TFM files. You should add new fonts to the HPD file when you install them.
You must specify the following settings in the HPD file for any new font:
```FONT={fontname} # {fontname} is a descriptive name for the font
/tfm={tfm-filename} # {tfm-filename} is the base filename for TFM file
```
Note:
The font name entries in HPD files must be unique. | {"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": 0.8395427465438843, "perplexity": 3456.5497219936947}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447555227.45/warc/CC-MAIN-20141224185915-00039-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/253581/reverse-triangle-inequality-with-exponent-p | # Reverse triangle inequality with exponent $p$
I am trying to prove that for any $p > 1$ and for any real numbers $a,b > 0$, the following inequality holds: $$| a^{\frac{1}{p}} - b^{\frac{1}{p}} |^p \leq 2^p|a-b|$$
In the case where $p$ is a positive integer, it seems like I could construct an even tighter bound than the one given: $$| a^{\frac{1}{p}} - b^{\frac{1}{p}} | + \min(a^{\frac{1}{p}},b^{\frac{1}{p}}) = \max(a^{\frac{1}{p}},b^{\frac{1}{p}})$$ thus raising both sides to the $p$-th power, applying the binomial theorem and noting that all the cross-terms in the binomial expansion are nonnegative, we have $$| a^{\frac{1}{p}} - b^{\frac{1}{p}} |^p + \min(a^{\frac{1}{p}},b^{\frac{1}{p}})^p \leq \max(a^{\frac{1}{p}},b^{\frac{1}{p}})^p \\ \implies | a^{\frac{1}{p}} - b^{\frac{1}{p}} |^p \leq \max(a,b) - \min(a,b) = |a-b| \leq 2^p |a-b|$$
However, I am not seeing how to generalize this argument to non-integer $p$.
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We can assume that $a \geq b$. For $x \in [0,1]$ the inequalities $$0 \leq 1-x^{1/p} \leq 1-x \leq (1 - x)^{1/p}$$ hold since $p>1$. Substituting $x \leftarrow b/a$ and raising to the power $p$ results in $$\left(1- \left(\frac{b}{a}\right)^{\frac{1}{p}}\right)^p \leq 1 - \frac{b}{a}$$ and after multiplication by $a$ $$\left(a^{\frac{1}{p}}- b^{\frac{1}{p}}\right)^p \leq a - b$$ | {"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": 0.9906911253929138, "perplexity": 53.193860328334765}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049275181.24/warc/CC-MAIN-20160524002115-00122-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/11659/number-of-nodes-on-diagonal-planes-of-a-cube | Number of nodes on diagonal planes of a cube
How many elements are in the set:
$\{(x,y,z) \mid x+y+z = k \text{ and } x,y,z \in \{0,\dotsc,n\}\}$
I can see that when $k \le n$, it's $\binom{k+2}{2}$. And, by symmetry, the value for $k$ is the same as that for $3n-k$. So, I've figured out the two pyramids, and I'm just missing the octahedron in the middle.
-
In the cases you've figured out, the points form an equilateral triangle. When $n < k < 2n$, you can think of the arrangement as an truncated equilateral triangle; that should let you count the number of points. – Rahul Nov 24 '10 at 6:21
@Rahul: Right, I see that now. So, the solution will be of the form (A choose 2) - 3 (B choose 2)? – Neil G Nov 24 '10 at 6:24
... or maybe $\binom{A}{2} - 3\binom{B}{2} - 3\binom{C}{2}$ – Neil G Nov 24 '10 at 6:31
Oh, I didn't see your response when I was writing my answer. Well, there it is for you now. – Rahul Nov 24 '10 at 6:42
Thanks......... – Neil G Nov 24 '10 at 6:52
When $n < k < 2n$, the number of points for which $x,y,z \in \mathbb N$ is $\binom{k+2}{2}$. Of these, $\binom{k-n+1}{2}$ have $x > n$ (as you can see by letting $x = x'+n+1$), and similarly for $y$ and $z$. These are the only exceptions, because when $k < 2n$, more than one of the variables cannot exceed $n$. So the number of points with $x,y,z \le n$ is $\binom{k+2}{2} - 3\binom{k-n+1}{2}$. This looks somewhat asymmetrical, but it's actually equal to $(k-n)(2n-k) + \binom{n+2}{2}$. I like writing it in this form because then it's obvious that it satisfies the "boundary conditions". | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9315322041511536, "perplexity": 182.08076894033348}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257826916.34/warc/CC-MAIN-20160723071026-00009-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://higgs.ph.ed.ac.uk/colloquia/the-structure-of-amplitudes-in-gauge-theory/ | The main page content begins here.
# The structure of amplitudes in gauge theory
Speaker:
## Abstract
Observing new physics at hadron colliders requires a detailed description of multi-particle scattering events. It is not practical to evaluate Feynman diagrams directly for all significant processes. Moreover, adding all diagrams reveals many cancellations: scattering amplitudes in gauge theories such as QCD take remarkably simple forms. This simplicity is a clue that the perturbative theory is best understood without reference to Feynman diagrams. In fact, it has recently become possible to explain some of this simplicity and derive amplitudes efficiently and elegantly. Symmetry can be preserved by working recursively on the mass shell: amplitudes are expressed in terms of other amplitudes, based on analysis of their singularities. I will describe these insights and some applications. | {"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": 0.8689185380935669, "perplexity": 710.781813490302}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00557.warc.gz"} |
http://physics.stackexchange.com/questions/31402/does-standard-model-confirm-that-mass-assigned-by-higgs-mechanism-creates-gravit/31438 | # Does Standard Model confirm that mass assigned by Higgs Mechanism creates gravitational field?
I am not comparing passive gravitational mass with rest inertial mass. Is there an evidence in Standard Model which says that active gravitational mass is essentially mass assigned by Higgs mechanism.
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I've been working on a similar question. Let me refer to this video that started my thinking: youtube.com/watch?v=9Uh5mTxRQcg, which says "the standard model misses out on gravity". I should qualify that I have no idea what that means, but clearly, the Higgs mechanism accomplishes something along the lines of allowing mass values to exist within QFT, which the theory might have otherwise precluded. Now, exactly how well does the standard model allow for these values to give rise to inertia and spacetime curvature (GR topics)? Good question! – Alan Rominger Jul 5 '12 at 18:29
Look at table 1 and you will see that in the microcosm of quarks and leptons the gravitational interaction is so weak that it is completely irrelevant and certainly its effect on the values used of the standard model cannot be measured with our present experimental accuracies.
Is there an evidence in Standard Model which says that active gravitational mass is essentially mass assigned by Higgs mechanism.
The standard model is mainly descriptive, it is a method to mathematically tie together a very large number of experimental data, economically, for the three stronger forces, strong, electromagnetic and weak. Because the strength of the coupling of the gravitational interaction is very much smaller than the coupling of these three forces ( Table 1 in link) there is no measurable predictable effect. In any case, as the quark and lepton masses are parameters in the SM, not predicted values,any tiny effect will be absorbed in the definitions.
The Stndard Model is not a theory of everything, but must be embedded in a theory of everything because it is really a shorthand for all the data we have up to now on quarks and leptons. A theory of everything will of course incorporate the gravitational force, and theorists are working on it, with prime candidate the string theories.
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But, why are you answering here? – SS-3.1415926535897932384626433 Jul 6 '12 at 6:45
because the question you deleted is a reasonable one, and the answer simple: coupling strengths. I will edit my answer to make answer directly this question also. – anna v Jul 6 '12 at 6:57
Anyone over 10k reputation can see my deleted question at physics.stackexchange.com/questions/31434/… – SS-3.1415926535897932384626433 Jul 6 '12 at 7:07
Can anyone undelete it please? – SS-3.1415926535897932384626433 Jul 6 '12 at 7:08
You could repeat it I guess – anna v Jul 6 '12 at 7:16
The mass of a particle is the energy that it has when it is at rest. The Higgs only makes it that particles oscillate between different helicities, so that you can make them be at rest, and their energy at rest is equal to the rate of oscillation between the two helicities. This energy gravitates like any other energy.
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Sounds good around Higgs ju-ju, but how is it an answer? – SS-3.1415926535897932384626433 Jul 5 '12 at 19:57
@Sachin: please re-read Ron's answer: This energy gravitates like any other energy. – Christoph Jul 5 '12 at 20:10
@Christoph I read that. Comparing with a general energy attribute can't be an answer. – SS-3.1415926535897932384626433 Jul 6 '12 at 3:35
@SachinShekhar: I am not comapring anything--- the energy of a particle at rest is it's mass. That's what mass means. The Higgs gives a fermionic particle at rest an energy (from helicity swapping) a mass proportional to the coupling to the higgs, or the swapping rate (in amplitude). – Ron Maimon Jul 6 '12 at 4:43
@SachinShekhar: The gravity is independent of whether it's a Higgs or a gluon making the energy. The Higgs gravitates, the massless things gravitate, and the massive things made from the Higgs plus the massless things also gravitate, all according to the energy and this has nothing to do with producing mass. Gravity is sourced by energy-momentum and stress, not by rest-mass. – Ron Maimon Jul 6 '12 at 7:27
The standard model itself says nothing about gravity at all, so no.
In any case, the thing that couples to gravity is energy, not just mass. This is different from the mass that the Higgs field gives to particles, which is actually just (rest) mass, and does not count energy.
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The energy you are talking about has to do with passive gravitational mass (in sense of General Theory of Relativity). I am talking about active one. – SS-3.1415926535897932384626433 Jul 5 '12 at 18:47
The active one isn't doing work against field to involve with energy.. – SS-3.1415926535897932384626433 Jul 5 '12 at 18:51
Curious - if the Higgs mechanism had something to do with active mass (or matter-energy in general), then I would expect photons (among others) to interact with the Higgs field. My guess is that they don't, and I think this graph confirms that: en.wikipedia.org/wiki/File:Elementary_particle_interactions.svg – Alan Rominger Jul 5 '12 at 19:05
Both kinds of gravitational mass are actually energy. @AlanSE, you're right that photons don't directly interact with the Higgs field, and yet they do couple to gravity, which is one way you can tell that the Higgs is not the sole "producer" of gravitational mass. – David Z Jul 5 '12 at 19:31
-1: Come on! This is the second time you encourage the persistent wrong thing that rest mas does not count as energy. Of course it counts as energy! All mass is energy. – Ron Maimon Jul 5 '12 at 19:32 | {"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": 0.8609433174133301, "perplexity": 772.8872337326326}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398455246.70/warc/CC-MAIN-20151124205415-00065-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://proofwiki.org/wiki/Category:Definitions/Abelian_Groups | Category:Definitions/Abelian Groups
This category contains definitions related to Abelian Groups.
Related results can be found in Category:Abelian Groups.
An abelian group is a group $G$ where:
$\forall a, b \in G: a b = b a$
That is, every element in $G$ commutes with every other element in $G$.
Subcategories
This category has the following 3 subcategories, out of 3 total.
Pages in category "Definitions/Abelian Groups"
The following 14 pages are in this category, out of 14 total. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9207485914230347, "perplexity": 1639.4148825977823}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348509972.80/warc/CC-MAIN-20200606031557-20200606061557-00222.warc.gz"} |
http://cameragossip.blogspot.com/2013/ | ## Friday, 6 December 2013
### Nikon's new 58mm F1.4g is a disappointment
If Nikon's aim was to better Sigma's 50mm f1.4 and Canon's 50mm f1.2 with the new 58mm f1.4g then I think Nikon has probably succeeded. Tests indicate that the new Nikon beats the Sigma as far as Coma correction is concerned. See the links below for comparative tests carried out by Thomas Rubach - whose lens tests are published by CameraLabs.
The new 58mm is also superior to Nikon's 50mm f1.4g, which exhibits significant coma at f1.4 as evidenced below, again thanks to testing done by Thomas.
Clearly the 58mm is an improvement. My disappointment though stems from the fact that Nikon has claimed that Sagittal coma flare is effectively minimised across the entire frame with the result that point light sources such as city lights are reproduced as fine rounded points, even at the periphery of the image, enabling unparalleled nightscapes. Yet clearly as demonstrated by the test above, this is not the case. Even Nikon's own sample below exhibits coma at the edges were point light sources are distorted.
If the Zeiss Otus 55mm f1.4 had not been present, we might have consoled ourselves with the thought that perhaps it was not physically possible to do better than what the 58mm f1.4 achieves in coma correction. Except that the Zeiss Otus is better corrected for coma than the Nikon as can be seen in example below (again from Thomas Rubach's testing):
Not only is the Zeiss better corrected for coma, it is also much sharper.
The question is in what way can the 58mm f1.4g be justified, given its price tag. If one wants the best performance then clearly the Zeiss Otus is the undisputed choice. If better bokeh is desired then perhaps the 85mm is a better focal length anyway. | {"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": 0.8618782162666321, "perplexity": 4201.588291789694}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105961.34/warc/CC-MAIN-20170820015021-20170820035021-00048.warc.gz"} |
http://mathoverflow.net/questions/71629/function-theory-of-a-hyperbolic-variable | # Function theory of a hyperbolic variable
I've found quite a number of articles on the basics of function theory in one hyperbolic (split-complex, dual, duplex, motro,..) variable, perhaps the most notable being http://arxiv.org/PS_cache/math-ph/pdf/0507/0507053v2.pdf. What is covered in this and the other articles are more or less only the counterparts of most elementary topics in complex analysis. Are there really no deeper results in this theory or are they just so hard to find? If so, some links or summaries of results would be dearly appriciated.
Particularly http://clifford-algebras.org/v8/81/MOTTER81.pdf aksed for a hyperbolic equvalence of Cauchy integral formula. Thus two rather different answers were provided in http://arxiv.org/PS_cache/arxiv/pdf/0712/0712.0375v1.pdf and http://www.springerlink.com/content/kp44rl074g7187n2 . Both articles mention immense applications that follow directly from their formulas, but I havn't been unable to find a single article discussing them.
And it is quite apparent that hyperbolic Cauchy-like formulas don't yield the percise same results as in complex analysis since it is easy to show not only that hyperbolic holomorphic functions are not allways analytic, but that they need not be even $C^2$! So could somebody explain what those mentioned direct implications of hyperbolic Cauchy formulas and also explain why are there two formulas to begin with? Thank you!
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The problem is that "adjoining" a "number" $j$ to $\mathbb R$ such that $j^2=+1$ rather than $i^2=-1$ gives, by Sun-Ze's theorem, $$\mathbb R[j] \approx \mathbb R[x]/\langle x^2-1\rangle \approx \mathbb R[x]/\langle x-1\rangle \oplus \mathbb R[x]/\langle x+1\rangle \approx \mathbb R\oplus \mathbb R$$ In particular, there are $0$-divisors, such as $(0,1)\cdot (1,0)=(0,0)$. A corresponding change of coordinates gives a analogue of the Cauchy-Riemann operator: just $\frac{\partial}{\partial x}\pm \frac{\partial}{\partial y}$.
Notably, these two operators are the factors of the one (spatial) dimensional wave equation, whose analytic features/failings caused so much consternation/interest pre-1800, namely, any function of the form $F(x,y)=f(x-y)$ is apparently annihilated by $\frac{\partial}{\partial x}+\frac{\partial}{\partial y}$, even when $f$ is not as smooth as one might think it ought to be.
This is in extreme contrast to the $i^2=-1$ situation, where the Cauchy-Riemann operator's vanishing gives a definite constraint (a.k.a. "ellipticity").
Edit: [thx for helpful edit-corrections...] If one uses the "usual" norm/length in the denominator, in the definition of "derivative", this does avoid zero-divisors, but changes the thing to being something else entirely, I think.
The fact that $\mathbb R$ with $j$ such that $j^2=+1$ adjoined is not a field is inescapable, and quite unlike the case of complex numbers. On another hand, it is certainly true that Clifford analysis is useful, if not quite in this fashion. Yes, "Dirac operators" have many roles, as factoring second-order differential operators into first-order. Yes, the Laplacian in $\mathbb R^2$ factors into Cauchy-Riemann and its conjugate, and the one-dimensional wave equation (as noted above) factors into two "real" linear operators. This is the goal, actually. For higher dimensions, these operators cannot factor into scalar differential operators, but Clifford-algebra-valued ones. Nevertheless, I don't think it's quite that the underlying "scalars" are made more exotic, but, rather, are enhanced by allowing operator-valued functions.
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The structure $\mathbb{R}[x]/(x-1)$ corresponds to $z=x+jy$ for $x,y$ real. In this notation zero divisors are percisely those numbers with $x^2-y^2=(x+y)(x-y)=0$. Than you can construct a basis alternative to $\{1,j\}$ from zero divisors $\{(1+j)/2,(1-j)/2\}$ so that $$z=x+jy=\frac{1+j}{2}(x+y)+\frac{x-y}{2}(x-y)$$ Now in this form, the CR operator indeed gets the form $\frac{1+j}{2}\frac{\partial}{\partial (x-y)}+\frac{1+j}{2}\frac{\partial}{\partial (x-y)}$ which can easily be proven equivalent to the form $\frac{1}{2}\Big (\frac{\partial}{\partial x}-j{\partial}{\partial y}$ – HeWhoHungers Jul 30 '11 at 14:55
...to the form $\frac{1}{2}(\frac{\partial}{\partial x}-j\frac{partial}{\partial y})$. Your account is incorrect because of the following fact: consider the operator $$\frac{1+j}{2}\frac{\partial}{\partial (x+y)}+\frac{1-j}{2}\frac{\partial}{\partial (x-y)}$$ The function annihilated by it are the hyperbolic equivalent of antiholomorphic. – HeWhoHungers Jul 30 '11 at 14:57
Also if one defines differentiability for functions in $\mathbb{R}[j]$ as the existance of such $f'(z)$ for every direction $$\lim_{h\to 0}\frac{f(z+h)-f(z)-f'(z)h}{\lVert h\rVert}=0$$ Here $\lVert h\rVert$ is the standard euclidean norm on $\mathbb{R}^2$, using which prevents the problem of zero devisors with the definition of derivatives. Notice that the condition that $f'(z)$ exists is the same as requiering CR operator to annihilate the function at $z$, simmilarly as in complex analisys. – HeWhoHungers Jul 30 '11 at 14:58
It does however not follow that such functions need be analytic or even $C^{2}$. Therefore the hyperbolic CR operator perhaps doesn not impose as much structure on functions it annihilates as does its complex counterpart, but it gives some structure none the less. **please forgive me for consistantly failing to write this down correctly: what I wanted to say was that the CR operator is equal to $\frac{1}{2}(\frac{\partial}{\partial x}+j\frac{\partial}{\partial y})$. – HeWhoHungers Jul 30 '11 at 15:01 | {"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": 0.9750640392303467, "perplexity": 384.3151394150699}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1404776440593.58/warc/CC-MAIN-20140707234040-00012-ip-10-180-212-248.ec2.internal.warc.gz"} |
https://en.m.wikibooks.org/wiki/FHSST_Physics/Work_and_Energy/Work | # FHSST Physics/Work and Energy/Work
Work and Energy The Free High School Science Texts: A Textbook for High School Students Studying Physics Main Page - << Previous Chapter (Momentum) - Next Chapter (Collisions and Explosions) >> Definition - Work - Energy - Mechanical Energy and Energy Conservation - Important Quantities, Equations, and Concepts - Sources
# WorkEdit
To do work on an object, one must move the object by applying a force with at least a component in the direction of motion. The work done is given by
${\displaystyle W=F_{\|}s}$ W : work done (N.m or J) F| : component of applied force parallel to motion (N) s : displacement of the object (m)
It is very important to note that for work to be done there must be a component of the applied force in the direction of motion. Forces perpendicular to the direction of motion do no work.
As with all physical quantities, work must have units. As follows from the definition, work is measured in N.m. The name given to this combination of S.I. units is the joule (J).
Definition: 1 joule is the work done when an object is moved 1m under the application of a force of 1N in the direction of motion.
The work done by an object can be positive or negative. Since force (F|) and displacement (s) are both vectors, the result of the above equation depends on their directions:
• If F| acts in the same direction as the motion then positive work is being done. In this case the object on which the force is applied gains energy.
• If the direction of motion and F| are opposite, then negative work is being done. This means that energy is transferred in the opposite direction. For example, if you try to push a car uphill by applying a force up the slope and instead the car rolls down the hill you are doing negative work on the car. Alternatively, the car is doing positive work on you!
## Worked Example 38 Calculating Work Done IEdit
Question: If you push a box 20m forward by applying a force of 15N in the forward direction, what is the work you have done on the box?
Step 1 : Analyse the question to determine what information is provided
• The force applied is F = 15N.
• The distance moved is s = 20m.
• The applied force and distance moved are in the same direction. Therefore, ${\displaystyle F_{\|}=15N}$
These quantities are all in the correct units, so no unit conversions are required.
Step 2 : Analyse the question to determine what is being asked
• We are asked to find the work done on the box. We know from the definition that work done is W = F|s
Step 3 : Next we substitute the values and calculate the work done
${\displaystyle {\begin{matrix}W&=&F_{\|}s\\&=&(15N)(20m)\\&=&300\ N\cdot m\\&=&300\ J\end{matrix}}}$
Remember that the answer must be positive as the applied force and the motion are in the same direction (forwards). In this case, you (the pusher) lose energy, while the box gains energy.
## Worked Example 39 Calculating Work Done IIEdit
Question: What is the work done by you on a car, if you try to push the car up a hill by applying a force of 40N directed up the slope, but it slides downhill 30cm?
Step 1 : Analyse the question to determine what information is provided
• The force applied is F = 40N
• The applied force and distance moved are in opposite directions. Therefore, if we take s = 0.3m, then ${\displaystyle F_{\|}=-40N}$ .
Step 2 : Analyse the question to determine what is being asked
• We are asked to find the work done on the car by you. We know that work done is W = F|s
Step 3 : Substitute the values and calculate the work done
Again we have the applied force and the distance moved so we can proceed with calculating the work done:
${\displaystyle {\begin{matrix}W&=&F_{\|}s\\&=&(-40N)(0.3m)\\&=&-12N\cdot m\\&=&-12\ J\end{matrix}}}$
Note that the answer must be negative as the applied force and the motion are in opposite directions. In this case the car does work on the person trying to push.
What happens when the applied force and the motion are not parallel? If there is an angle between the direction of motion and the applied force then to determine the work done we have to calculate the component of the applied force parallel to the direction of motion. Note that this means a force perpendicular to the direction of motion can do no work.
## Worked Example 40 Calculating Work Done IIIEdit
Question: Calculate the work done on a box, if it is pulled 5m along the ground by applying a force of F = 10N at an angle of 60o to the horizontal.
Step 1 :
Analyse the question to determine what information is provided
• The force applied is F = 10N
• The distance moved is s = 5m along the ground
• The angle between the applied force and the motion is 60o
These quantities are in the correct units so we do not need to perform any unit conversions.
Step 2 : Analyse the question to determine what is being asked
• We are asked to find the work done on the box.
Step 3 : Calculate the component of the applied force in the direction of motion
Since the force and the motion are not in the same direction, we must first calculate the component of the force in the direction of the motion.
From the force diagram we see that the component of the applied force parallel to the ground is
${\displaystyle {\begin{matrix}F_{||}&=&F\cdot \cos(60^{o})\\&=&10N\cdot \cos(60^{o})\\&=&5\ N\end{matrix}}}$
Step 4 : Substitute and calculate the work done
Now we can calculate the work done on the box:
${\displaystyle {\begin{matrix}W&=&F_{\|}s\\&=&(5N)(5m)\\&=&25\ J\end{matrix}}}$
Note that the answer is positive as the component of the force F| is in the same direction as the motion.
We will now discuss energy in greater detail. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "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": 0.9453588724136353, "perplexity": 278.63662386394464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125937780.9/warc/CC-MAIN-20180420120351-20180420140351-00321.warc.gz"} |
http://mathhelpforum.com/discrete-math/199986-set-open.html | # Thread: Is this set open?
1. ## Is this set open?
A is a subset of B, both sets are open.
Is the set B/A open?
My answer would be no, because the complement of an open set is always closed. Is this the correct definition in this case?
2. ## Re: Is this set open?
But a set can be both open and closed.
You can look at A=(0,1), B=(0,2) as subset of the real line with the usual topology.
3. ## Re: Is this set open?
girdav: In your example, A\B is [1,2) which is not open.
However, let B be the union of two arbitrary disjoint open subsets A and C. B is open.
Then B\A = C is open by construction.
4. ## Re: Is this set open?
But why is the set open? If it's the complement of A and A is open, shouldn't the complement of an open set always be closed?
5. ## Re: Is this set open?
Originally Posted by infernalmich
But why is the set open? If it's the complement of A and A is open, shouldn't the complement of an open set always be closed?
You must understand that $B\setminus A=B\cap A^c$ and is known as a relative complement.
That is the complement of A relative to B.
You see $A$ is an open set in the underlying space which may not be $B$.
In the real numbers $((0,2)\setminus (0,1)=[1,2)$ which is not open.
6. ## Re: Is this set open?
Originally Posted by infernalmich
But why is the set open? If it's the complement of A and A is open, shouldn't the complement of an open set always be closed?
Yes, that's true. But "B\A" is NOT the "complement" of A, which is, by definition, the set all points not in A, whether they are in B or not. If $B= (0, 1)\cup (2, 3)$ and $A= (0, 1)$, then $B\A= (2, 3)$ an open set. The [b]complement of A is $(-\infty, 0]\cup [1, \infty$, a closed set. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "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": 0.8969346880912781, "perplexity": 494.0911027814441}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825141.95/warc/CC-MAIN-20171022041437-20171022061437-00233.warc.gz"} |
http://mathhelpforum.com/geometry/215157-parametric-equations-finding-dy-dx.html | # Math Help - Parametric Equations. Finding dy/dx?
1. ## Parametric Equations. Finding dy/dx?
I'm unsure on the following C4 Coordinate Geometry Question
A curve has the following parametric equations
$x= \frac{t}{2-t}$ $y= \frac{1}{1+t}$
a) show that
$\frac{dy}{dx}= -\frac{1}{2}(\frac{2-t}{1+t})^2$, would this be done using dy/dx = (dy/dt)/(dx/dt)?
b) find an equation for the normal to the curve at point where t = 1
this would just be substituting t=1 into the x, y and dy/dx equations and using (y-y1) = m(x-x1), yeah?
c) show that the cartesian equation of the curve can be written in the form
$y= \frac{1+x}{1+3x}$
unsure on this
2. ## Re: Parametric Equations. Finding dy/dx?
I'm unsure on the following C4 Coordinate Geometry Question
A curve has the following parametric equations
$x= \frac{t}{2-t}$ $y= \frac{1}{1+t}$
a) show that
$\frac{dy}{dx}= -\frac{1}{2}(\frac{2-t}{1+t})^2$, would this be done using dy/dx = (dy/dt)/(dx/dt)?
Yes, that's the idea.
b) find an equation for the normal to the curve at point where t = 1
this would just be substituting t=1 into the x, y and dy/dx equations and using (y-y1) = m(x-x1), yeah?
No, that would give you the tangent to the curve. The normal is perpendicular to the tangent line.
c) show that the cartesian equation of the curve can be written in the form
$y= \frac{1+x}{1+3x}$
unsure on this
Your parametric equations are $x= \frac{t}{2- t}$ and $y= \frac{1}{1+ t}$.
Solve each of those equations for t, set them equal, and solve for y.
3. ## Re: Parametric Equations. Finding dy/dx?
I now understand the methods but seem to be getting different answers to those that the answer wants. I must be making mistakes :/ Could you go through them...pretty please :P
4. ## Re: Parametric Equations. Finding dy/dx?
If you have gotten some answers, please post what you have done so we can check them and give you some guidance where you have made your mistakes... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "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": 0.9167859554290771, "perplexity": 835.7387408454467}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500829955.75/warc/CC-MAIN-20140820021349-00080-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://mathspace.co/textbooks/syllabuses/Syllabus-410/topics/Topic-7293/subtopics/Subtopic-97357/?activeTab=theory | New Zealand
Level 7 - NCEA Level 2
Graph sine or cosine curves (deg)
Lesson
Previously we looked at key features and transformations of sine and cosine functions. We use these key concepts now to discuss graphing.
The general form of the equation of a sine curve is
$f\left(x\right)=a\sin\left(bx-c\right)+d$f(x)=asin(bxc)+d
Recall that:
• The amplitude of the wave shape is given by the constant $a$a
• The period is determined from the coefficient $b$b by dividing $360^\circ$360° (or $2\pi$2π radians) by $b$b
• After writing the function as $f\left(x\right)=a\sin\left(b(x-\frac{c}{b})\right)+d$f(x)=asin(b(xcb))+d we see that the phase shift is $\frac{c}{b}$cb
• The vertical shift of the central line (and hence the entire curve) is given by the constant $d$d.
To sketch the graphs of sine curves, a number of approaches can be taken. Which one you choose may depend on your preference or the question you are given.
Here are two approaches.
APPROACH 1
Walk through the transformations and change the stem graph $y=\sin x$y=sinx accordingly.
Example 1
Graph $y=2\sin\left(\frac{x}{2}\right)+3$y=2sin(x2)+3
Start with a sketch of $y=\sin x$y=sinx
Apply the vertical translation - move the graph up $3$3 units, this in now the graph of $y=\sin x+3$y=sinx+3
Increase the amplitude - the amplitude of this graph is $2$2 units, so we dilate the graph. Move the maximum and minimum out an extra unit. This is now the graph of $y=2\sin x+3$y=2sinx+3
The period of the function has been changed from $360^\circ$360° to $\frac{360^\circ}{\frac{1}{2}}=720^\circ$360°12=720°.This is a horizontal dilation. Stretch out the graph, keeping the starting point the same. This is now the graph of $y=2\sin\left(\frac{x}{2}\right)+3$y=2sin(x2)+3. (Because the scale has been kept the same, only a half period is shown in the diagram.)
APPROACH 2
Step through the important components and create a dot-to-dot style map of the function.
Example 2
Graph $y=2\sin\left(\frac{x}{2}\right)+3$y=2sin(x2)+3
Step 1 - identify the transformations from the graph by identifying the following
• amplitude: $a=2$a=2
• reflection: no reflection as $a$a is positive
• period: $\frac{360^\circ}{b}=\frac{360^\circ}{\frac{1}{2}}=720^\circ$360°b=360°12=720°
• phase shift: $\frac{c}{b}=0$cb=0 as $c=0$c=0 in this case
• vertical translation: $d=3$d=3
Step 2 - draw the central line (indicated by the vertical translation)
Step 3 - mark on the maximum and minimum by measuring the amplitude above and below the central line
Step 4 - check for a phase shift (this would shift the initial starting position)
There is no phase shift for this function as $c=0$c=0
Step 5 - mark out the distance of the full period. At this stage mark out half way and quarter way marks, this will help us sketch the curve.
Step 6 - check for a reflection (this would change the initial starting direction
There is no reflection for this function as $a>0$a>0
Step 7 - Create some dots on the starting and ending positions of the cycle (on the central line), also mark out the maximum and minimum points (on the quarter lines).
Step 8 - sketch the curve lightly, joining the preparatory dots. Developing the skills for smooth curve drawing takes practice so don't get disheartened.
Some people prefer step-by-step constructions, some prefer the fluid changes of transformations, others develop their own order and approach to sketching sine functions. Regardless of your approach, it will need to use the specific features of the sine curve.
Worked Examples
Question 1
Consider the function $y=\sin x-2$y=sinx2.
1. Determine the period of the function, giving your answer in degrees.
2. Determine the amplitude of the function.
3. Determine the maximum value of the function.
4. Determine the minimum value of the function.
5. Graph the function.
Question 2
Consider the function $y=\sin\left(\frac{2}{3}x\right)$y=sin(23x).
1. Identify the amplitude of the function.
2. Identify the period of the function, giving your answer in degrees.
3. Graph the function.
Question 3
Consider the function $y=3\sin2x-2$y=3sin2x2.
1. Determine the period of the function, giving your answer in degrees.
2. Determine the amplitude of the function.
3. Determine the maximum value of the function.
4. Determine the minimum value of the function.
5. Graph the function.
Outcomes
M7-2
Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs
91257
Apply graphical methods in solving problems | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8084917068481445, "perplexity": 1376.9035864820178}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300616.11/warc/CC-MAIN-20220117182124-20220117212124-00677.warc.gz"} |
http://physics.stackexchange.com/questions/1968/lunar-twilight-and-sixth-magnitude-stars | # Lunar twilight and sixth magnitude stars
Summary: when the Moon is x degrees below the horizon, it interferes with stargazing the same as astronomical twilight would. What is x (as a function of the Moon's phase)?
We define civil, nautical, and astronomical twilight as when the sun is 0-6, 6-12, and 12-18 degrees below the horizon respectively. This corresponds roughly to what most people call twilight, the ability to distinguish a horizon at sea, and the ability to see 6th magnitude stars at the zenith.
However, even when below the horizon, the Moon shines brightly enough to interfere with stargazing. What are the equivalent twilight angles for the Moon? I realize that even the full Moon overhead isn't bright enough for civil twilight, so my real interest is in astronomical twilight. Of course, this will vary greatly with the Moon's phase, and slightly with Moon's distance.
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Very interesting question. It would be great to see more astronomy questions on this site! – user346 Dec 16 '10 at 4:20
@space_cadet: well I think it would be great to see more astrophysics questions on the site. But this is not really an astronomy site, and there's some point at which astronomy is just astronomy, not really physics. I'm not quite sure about this question. If you answer it by calculating the ratio of light intensities for the Sun vs. the moon, that would be physics, but the way Kostya answered it, that makes it seem more astronomical. – David Z Dec 17 '10 at 11:24
It was really difficult to find any mentioning of the "lunar twilight" in astronomical literature. But I managed to find one in an old book by G.V.Rosenberg called "General picture of twilight phenomena". The book is in russian, but there is not too much about the moon. So I'll just translate all the relevant stuff.
First of all there is this plot for the solar twilight (I've translated it):
Where E is the illuminance of a horizontal surface in luxes. And lg -- is a logarithm base 10.
And then there is a small chapter:
Lunar twilight
Lunar twilight is similar to solar twilight, but it is much more bleak. The illumination from the moon varies from $10^{-9}$ (new moon) to $2\cdot 10^{-8}$ (full moon) from the illumination from the sun at the same point in the sky. Therefore the illumination from the full moon high in the sky corresponds approximately to the middle of nautical twilight. And lunar twilight ends virtually when the moon gets under the horizon.
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What is the x axis here? Distance in degrees from the zenith? If so, is the author saying twilight begins when the sun is 6 degrees ABOVE the horizon? – barrycarter Dec 17 '10 at 14:35
Sorry. Of course these are degrees from the zenith. And yes -- author is saying that "the moment of acceleration of decrease of illumination should be considered as a start of twilight". The book was written in 1963 in Russia, so maybe a different convention was adopted since then. – Kostya Dec 17 '10 at 15:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8573533296585083, "perplexity": 364.87472346492984}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257825365.1/warc/CC-MAIN-20160723071025-00144-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/94838-integration-help.html | # Math Help - Integration Help
1. ## Integration Help
Hey guys, I'm new to integration and these are giving me a really rough time, even just a push in the right direction would be great. Thanks for taking a look! So far, I've only learned the basic integration techniques and u-substitution.
1) $\int\frac{x+1}{\sqrt{x^2+2x-3}}dx$
2) $\int\frac{dx}{x^{\frac{3}{4}}(x^\frac{1}{4}+2)^2}$
I solved 2) and got $16\sqrt[4]{x}-\frac{4}{\sqrt[4]{x}}$. Not sure if this is right, I did expanded completely and brought everything up top before integrating.
3) $\int3x\sin(4-x^2)dx$
4) $\int\frac{dx}{\cos^2x\sqrt{1+\tan{x}}}$
Thanks for any pointers!
2. For the first one, I'd try a $u$-substitution as $u=x^2+2x-3 \Rightarrow du=2x+2 dx$.
Hence the integral is worth $\frac{1}{2} \int \frac{du}{\sqrt u}$. Finish it.
Edit: I also see a u-sub for the third integral : let $u=4-x^2$... can you finish it?
3. Originally Posted by arbolis
For the first one, I'd try a $u$-substitution as $u=x^2+2x-3 \Rightarrow du=2x+2 dx$.
Hence the integral is worth $\frac{1}{2} \int \frac{du}{\sqrt u}$. Finish it.
Edit: I also see a u-sub for the third integral : let $u=4-x^2$... can you finish it?
Wicked I think I worked out a solution for both of them, if you have a sec would you mind checking over my work?
1) $\int\frac{x+1}{\sqrt{x^2+2x-3}}dx$
Let $u=x^2+2x-3$
$du=2x+2dx$
$\frac{1}{2}du=(x+1)dx$
Therefore
$\frac{1}{2}\int\frac{du}{\sqrt{u}}$
$=\frac{1}{2}\int u^{-{\frac{1}{2}}}du$
$=(\frac{1}{2})(2u^{\frac{1}{2}})+c$
$=\sqrt{u}+c$
$\boxed{=\sqrt{x^2+2x-3}+c}$
----------------------------------------------------------
And for number 3) I got the following;
$
\int3x\sin{(4-x^2)}dx$
Let $u=4-x^2$
$du=-2xdx$
$-\frac{3}{2}du=3xdx$
Therefore
$\int(\sin{(u)})(-\frac{3}{2}du)$
$=-\frac{3}{2}\int\sin(u)du$
$
=-\frac{3}{2}(-\cos{u})+C$
$
\boxed{=\frac{3}{2}\cos{(4-x^2)}+C}$
What do you think?
4. Both looks right to me. Very good!
I'll let others helping you for the 2 remaining ones.
5. Cool, thanks! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 30, "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": 0.9950076341629028, "perplexity": 745.0692038780745}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824757.62/warc/CC-MAIN-20160723071024-00209-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-equations/155537-modelling-differential-equation-print.html | # Modelling with differential equation
• Sep 8th 2010, 03:05 AM
acevipa
Modelling with differential equation
A cyclist freewheeling on a level road experiences a negative acceleration which is proportional to the square of his speed. His speed is reduced from $20m/s$ to $10m/s$ in a distance of $100m$. Find the average speed (w.r.t. time) during this period.
Would the equation be
$\dfrac{dv}{dt}=-kv^2$ where $k$ is a constant of proportionality
• Sep 8th 2010, 03:57 AM
Quote:
Originally Posted by acevipa
A cyclist freewheeling on a level road experiences a negative acceleration which is proportional to the square of his speed. His speed is reduced from $20m/s$ to $10m/s$ in a distance of $100m$. Find the average speed (w.r.t. time) during this period.
Would the equation be
$\dfrac{dv}{dt}=-kv^2$ where $k$ is a constant of proportionality
yes.
• Sep 8th 2010, 04:49 AM
acevipa
Quote:
yes.
Thanks. So I'm a little unsure of what to do next.
$\dfrac{dv}{dt}=-kv^2$
$-\dfrac{dv}{v^2}=k \ dt$
$\displaystyle\int -\dfrac{dv}{v^2}=\int k \ dt$
$\dfrac{1}{v}=kt+C$
• Sep 8th 2010, 05:37 AM
Quote:
Originally Posted by acevipa
Thanks. So I'm a little unsure of what to do next.
$\dfrac{dv}{dt}=-kv^2$
$-\dfrac{dv}{v^2}=k \ dt$
$\displaystyle\int -\dfrac{dv}{v^2}=\int k \ dt$
$\dfrac{1}{v}=kt+C$
You have to set up a differential equation which involves the displacement and velocity.
Hint: Notice that a=dv/dx=(dv/dx) x (dx/dt) = v (dv/dx)
• Sep 8th 2010, 05:56 AM
chisigma
Quote:
Originally Posted by acevipa
Thanks. So I'm a little unsure of what to do next.
$\dfrac{dv}{dt}=-kv^2$
$-\dfrac{dv}{v^2}=k \ dt$
$\displaystyle\int -\dfrac{dv}{v^2}=\int k \ dt$
$\dfrac{1}{v}=kt+C$
... all right!... the solution is then...
$\displaystyle v(t)= \frac{1}{k t + c}$ (1)
... where of course is $c= \frac{1}{v(0)}$. Now suppose that $v(0)$ is negative, for example $v(0) = -1$, so that the solution is...
$\displaystyle v(t)= \frac{1}{k t -1}$ (2)
A little question: what does if happend at the time $t=\frac{1}{k}$ ?(Surprised) ...
Kind regards
$\chi$ $\sigma$
• Sep 8th 2010, 06:14 AM
acevipa
Quote:
You have to set up a differential equation which involves the displacement and velocity.
Hint: Notice that a=dv/dx=(dv/dx) x (dx/dt) = v (dv/dx)
Thanks that makes a lot of sense.
$a=-kv^2 \Longrightarrow \dfrac{dv}{dt}=-kv^2$
$v\dfrac{dv}{dx}=-kv^2$
$\dfrac{dv}{dx}=-kv$
$\displaystyle\int\dfrac{dv}{v}=\int -k \ dx$
$\ln v=-kx+C$
When $x=0, v=20 \Longrightarrow C = \ln 20$
$\ln v =-kx+\ln 20$
When $x=100, v=10$
$\ln 10=-100k+\ln 20$
$100k = \ln 20 - \ln 10$
$100k = \ln 2$
$k = \dfrac{\ln 2}{100}$
Have I done everything right? If so, I'm not too sure what to do next.
The answer in the book is $20\ln 2$
• Sep 11th 2010, 09:27 PM
Quote:
Originally Posted by acevipa
Thanks that makes a lot of sense.
$a=-kv^2 \Longrightarrow \dfrac{dv}{dt}=-kv^2$
$v\dfrac{dv}{dx}=-kv^2$
$\dfrac{dv}{dx}=-kv$
$\displaystyle\int\dfrac{dv}{v}=\int -k \ dx$
$\ln v=-kx+C$
When $x=0, v=20 \Longrightarrow C = \ln 20$
$\ln v =-kx+\ln 20$
When $x=100, v=10$
$\ln 10=-100k+\ln 20$
$100k = \ln 20 - \ln 10$
$100k = \ln 2$
$k = \dfrac{\ln 2}{100}$
Have I done everything right? If so, I'm not too sure what to do next.
The answer in the book is $20\ln 2$
That's correct. Then, work from $\dfrac{dv}{dt}=-kv^2$
For velocity, integrate from 20 to 10 and for time, integrate from 0 to t. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 57, "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": 0.9338727593421936, "perplexity": 1004.8282239698642}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812756.57/warc/CC-MAIN-20180219151705-20180219171705-00489.warc.gz"} |
https://bt.gateoverflow.in/general-biotechnology/biochemistry | # Recent questions and answers in Biochemistry
1
If the dissociation constant for solute-adsorbent binding is $K_D$, the retention time of the solute in a chromatography column increases with increasing $K_D$ decreases with increasing $K_D$ passes through minimum with increasing $K_D$ is independent of $K_D$
2
The product(s) resulting from the hydrolysis of maltose is/are a mixture of $\alpha$-D-Glucose and $\beta$-D-Glucose a mixture of D-Glucose and L-Glucose $\alpha$-D-Glucose only $\beta$-D-Glucose only
3
Amino acid residue which is most likely to be found in the interior of water-soluble globular proteins is Threonine Aspartic acid Valine Histidine
4
The reactions leading to the formation of amino acids from the TCA cycle intermediates are carboxylation isomerization transamination decarboxylation
5
The total number of fragments generated by the complete and sequential cleavage of the polypeptide given below by Trypsin followed by $CNBr$ is __________. $Phe-Trp-Met-Gly-Ala-Lys-Leu-Pro-Met-Asp-Gly-Arg-Cys-Ala-Gln$
6
Match the herbicides in $\text{Group I}$ with the target enzymes in $\text{Group II}$ ... $\text{P-4, Q-1, R-2, S-3}$ $\text{P-4, Q-3, R-2, S-1}$ $\text{P-3, Q-2, R-4, S-1}$
7
Match the entries in $\text{Group I}$ with the enzymes in $\text{Group II}$ ... $\text{P-4, Q-1, R-3, S-2}$ $\text{P-3, Q-1, R-4, S-2}$ $\text{P-3, Q-4, R-2, S-1}$
8
The catalytic efficiency for an enzyme is defined as $K_{cat}$ $\frac{V_{max}}{K_{cat}}$ $\frac{k_{cat}}{K_m}$ $\frac{K_{cat}}{V_{max}}$
9
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10
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11
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12
In a chemostat, the feed flow rate and culture volume are $100$ ml/h and $1.0$ L, respectively. With glucose as substrate, the values of $\mu_{\text{max}}$ and $K_s$ are $0.2 \: h^{-1}$ and $1 \: g/L$, respectively. For a glucose concentration of $10 \: g/L$ in the feed, the effluent substrate concentration (in $g/L$) is _______
13
Yeast converts glucose to ethanol and carbon dioxide by glycolysis as per the following reaction: $C_6H_{12}O_6 \to 2C_2H_5OH + 2CO_2$Assuming complete conversion, the amount of ethanol produced (in $g$) from $200 \: g$ of glucose is (up to two decimal places) ____________
14
Consider a simple uni-substrate enzyme that follows Michaelis-Menten kinetics. When the enzyme catalyzed reaction was carried out in the presence of $10$ nM concentration of an inhibitor, there was no change in the maximal velocity. However. the slope of the Lineweaver-Burk plot increased $3$-fold. The dissociation constant for the enzyme-inhibitor complex (in nM) is ____________
A rod shaped bacterium has a length of $2 \: \mu m$, diameter of $1 \mu m$ and density the same as that of water. If proteins constitute $15 \%$ of the cell mass and the average protein has a mass of $50$ kDa, the number of proteins in the cell is __________ ($1$ Da $= 1.6 \times 10^{-24}$g) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8733391761779785, "perplexity": 2823.4977410420156}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178389472.95/warc/CC-MAIN-20210309061538-20210309091538-00465.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/71749-normal-distributions.html | # Thread: Normal Distributions
1. ## Normal Distributions
all are normal distributions
1) Sample size = 6. mean of 10 and variance of 2. Calculate the probability that 2 of the values are less than 9 and the other 4 are greater than nine.
I can find the probability of 1 being greater than or less than 9. That would just be (8-10)/(sqrt 2) right?
2)The sizes of claims are normal random variables. Mean = 1800. SD=400. Find probability that 2 random claims differ by more than 500.
3) n = 80 they are independent. indiv mean = 574 with sd = 186. P[average claim <565]
P[Z<565-574/(Sqrt(186^2/80)). that gave me -.433 as my Z and then i took 1-.6664 (chart value of .433). I got .3336 as my answer.
I got (sqrt(186^2/80) as my variance for each average term. the math was 1/80^2 * var(s)= 80*Var/(80^2). Var=SD^2 so that gave me (sqrt(186^2/80)
4) X = rate of interest. mean=.06 SD=.015 Amount in account after year => Y=10,000 e^x Find E[x] after a year
2. Originally Posted by PensFan10
all are normal distributions
1) Sample size = 6. mean of 10 and variance of 2. Calculate the probability that 2 of the values are less than 9 and the other 4 are greater than nine.
I can find the probability of 1 being greater than or less than 9. That would just be (8-10)/(sqrt 2) right? Mr F says: Can a probabilty be negative, I wonder .....? Surely you mean the z-value, in which case it is actually (9 - 10)/(sqrt 2) = -0.7071.
[snip]
I assume you mean that the sample is taken from a normal population whose mean is 10 and variance is 2 ....
Pr(Z < -0.7071) = 0.23975 (correct to 5 decimal places)
Now let Y be the random variable number of values less than 9.
Y ~ Binomial(n = 6, p = 0.23975).
Calculate Pr(Y = 2).
3. Originally Posted by PensFan10
[snip]
2)The sizes of claims are normal random variables. Mean = 1800. SD=400. Find probability that 2 random claims differ by more than 500.
[snip]
Apply a well known result for the difference of two independent normal random variables:
$U = X_1 - X_2$ ~ $N(\mu = 1800 - 1800 = 0, \sigma^2 = 400^2 + 400^2 = 3200)$.
Calculate $\Pr(U \geq 500)$.
4. Is the probability Y=2 (0.23975)(0.23975) = .05748?
Then for Greater than nine:
probability .500
then for the other 4 (.500)^4 = .0625?
5. Originally Posted by PensFan10
[snip]4) X = rate of interest. mean=.06 SD=.015 Amount in account after year => Y=10,000 e^x Find E[x] after a year
$E(10,000 e^X) = 10,000 E(e^X)$.
$E(e^X) = \int_{-\infty}^{+\infty} e^x \, f(x) \, dx$
where f(x) is the pdf of the given normal distribution (the formula will be in your class notes or textbook).
I assume an approximate answer is sufficient ie. use technology to get a decimal approximation of the integral.
(I cannot understand your question 3. PLease post the whole question, exactly as it's written in your textbook or wherever it is you got it from).
6. ## for number 2
500-0/(3200) = .1563
.1563 = .5636
thats for less than.
greater than would be 1-(.5636)= .4364
7. number 3 =
An insurance portfolio consists of 80 policies. The claim amount for the policies are independent, identically distributed random variables with a normal distribution. The individual mean claim is 574 with a SD of 186. Find the probability that the average claim is below 565
8. Originally Posted by PensFan10
Is the probability Y=2 (0.23975)(0.23975) = .05748?
Then for Greater than nine:
probability .500
then for the other 4 (.500)^4 = .0625?
No. And I cannot follow where this working has come from (nor will your instructor be able to either, I'll bet).
Use the binomial distribution in the way suggested by my post. Have you been taught the binomial distribution?
I get 0.288 (correct to three decimal places).
And please .... quote the posts that you are asking about (use the Quote tool at the bottom right).
9. Originally Posted by PensFan10
number 3 =
An insurance portfolio consists of 80 policies. The claim amount for the policies are independent, identically distributed random variables with a normal distribution. The individual mean claim is 574 with a SD of 186. Find the probability that the average claim is below 565
Use the sampling distribution of the mean: Read this thread: http://www.mathhelpforum.com/math-he...stic-help.html
10. Originally Posted by PensFan10
500-0/(3200) = .1563
.1563 = .5636
thats for less than.
greater than would be 1-(.5636)= .4364
This working makes no sense at all. How can 0.1563 = 0.5636? It doesn't matter what you mean, what matters is what you've written. Statements like 0.1563 = 0.5636 are ridiculous. Please post working that is well set out and that makes sense if you expect people to read and comment on your work.
The question requires you to calculate Pr(U > 0.1563). Solve this in the way you've been taught how to deal with normal probabilities.
11. Originally Posted by mr fantastic
No. And I cannot follow where this working has come from (nor will your instructor be able to either, I'll bet).
Use the binomial distribution in the way suggested by my post. Have you been taught the binomial distribution?
I get 0.288 (correct to three decimal places).
And please .... quote the posts that you are asking about (use the Quote tool at the bottom right).
We went over it in one class. We just did some basic example.
I assume you mean that the sample is taken from a normal population whose mean is 10 and variance is 2 ....
Pr(Z < -0.7071) = 0.23975 (correct to 5 decimal places)
Now let Y be the random variable number of values less than 9.
Y ~ Binomial(n = 6, p = 0.23975).
Calculate Pr(Y = 2).
I figured that .23975 was probability that value is less than nine. Then to get the probability that the figure is greater than nine i did this:
10- 10/2 = which gave me 0. the probability for the z value is .500.
I used 10-10 because 10 is the first number greater than nine and 10 is the mean.
The probability that corresponds to 0 = .500. I raised .500 to the fourth because there are 4 numbers selected to be greater than 9.
(.500)^4 = .0625
(0.23975)^2 = .05748
then add them together to get = .11998
12. Originally Posted by PensFan10
We went over it in one class. We just did some basic example.
I figured that .23975 was probability that value is less than nine. Then to get the probability that the figure is greater than nine i did this:
10- 10/2 = which gave me 0. the probability for the z value is .500.
I used 10-10 because 10 is the first number greater than nine and 10 is the mean.
The probability that corresponds to 0 = .500. I raised .500 to the fourth because there are 4 numbers selected to be greater than 9.
(.500)^4 = .0625
(0.23975)^2 = .05748
then add them together to get = .11998
Do it as I outlined in post #2.
13. Originally Posted by mr fantastic
Apply a well known result for the difference of two independent normal random variables:
$U = X_1 - X_2$ ~ $N(\mu = 1800 - 1800 = 0, \sigma^2 = 400^2 + 400^2 = 3200)$.
Calculate $\Pr(U \geq 500)$.
Mean = 0 with the variance = 3200
What i did was take the X-Mean/(variance)
500-0/3200 = .15625 as my z value
From this i got 500-0/3200
P[Z(greater than or equal to)500] = since the z value of .16 is .5636
since the z value gives numbers less than i subtracted .5636 from 1 which gave me:
1-.5636 = .4364
14. Originally Posted by mr fantastic
I assume you mean that the sample is taken from a normal population whose mean is 10 and variance is 2 ....
Pr(Z < -0.7071) = 0.23975 (correct to 5 decimal places)
Now let Y be the random variable number of values less than 9.
Y ~ Binomial(n = 6, p = 0.23975).
Calculate Pr(Y = 2).
Okay, I see how you got 0.23975.
Using the Binomial theorem I get
(6C2)(.23975)^2(.76025)^4 = .2880
15. Originally Posted by PensFan10
Mean = 0 with the variance = 3200
What i did was take the X-Mean/(variance)
500-0/3200 = .15625 as my z value Mr F says: 3200 is NOT the sd .... It's the variance. Your z-value is therefore vey wrong.
From this i got 500-0/3200
P[Z(greater than or equal to)500] = since the z value of .16 is .5636
since the z value gives numbers less than i subtracted .5636 from 1 which gave me:
1-.5636 = .4364
Please take greater care.
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https://www.physicsforums.com/threads/divisors-and-riemann-roch-intuition.756344/ | # Divisors and Riemann-Roch Intuition
1. Jun 2, 2014
### bolbteppa
Could anybody explain what divisors and the Riemann-Roch theorem are intuitively, motivating them, without any jargon or vagueries (i.e. using actual math), and preferably offering a nice example necessitating this stuff?
I'm sure there is a nice way to explain it in an absolutely natural way, that explains why it applies to classical algebraic curves, differential forms, homology etc... without ever having to use a definition like abelian group, or vector space, i.e. only using definitions Riemann had handy. Thank you.
2. Jun 9, 2014
### mathwonk
Riemann tried to describe math by the most intrinsic properties. take a polynomial. if you know the roots then you know the polynomial up to a constant multiple since if the roots are a,b,c,...d, then the polynomial is c.(x-a)(x-b)(x-c)...(x-d). for some c.
ion the same way Riemann tried tom describe meromorphic functions by their zeroes and "poles" i.e. points where the value was not zero but infinity. His ideas was that any meromorphic function on a compact surface is described up to a constant multipkle by knowing its zeroes and poles.
"divisor" is a fancy name for the zeroes and poles of a function. i.e. if a meromorphic functions has zeroes of order ri at pi and poles of order si at qi, then the "divisor" of zeroes and poles is
r1p1+...rnpn +s1q1+...+smqm.
the riemann roch theorem tells you the dimension of the space of meromorphic functions with given zeroes and poles in terms of the number of zeroes and poles, i.e. in terms of the degree of the divisor of those zeroes and poles, plus the actual location of those points.
i.e. if we specofy zeroes at p1,...,pn of orders r1,...,rn and polkes at q1,...,qm of orders s1,..,sm,
then the dimension oif meromorphic functions with zeroes at least of those orders at those points and poles at most of those orders atn those points,
is Equal to 1-g + d + i, where g = the topological genus of the surface,d = r1+...+rn -s1-...-sm, and i = the dimension of the vector space of holomorphic differential forms vanishing on the given points pi and with pokles at most on the given qi, of the given orders,
e.g. on an elliptic curve, i.e. a curve of genus 1, no differential forms have zeroes, so if we have a positive divisor of degree d, the space of meromorphic functions with poles at most at the points of our divisor has dimension d. E.g. given 2 points p1+p2, there are two independent meromorphic functions with poles at most at p1 +p2. One such is constant so we have one other defining a map S-->P^1, of degree 2, from our elliptic curve S to the Riemann sphere P^1.
probably the simplest example is the Riemann sphere, where we have all meromorphic functions are rational, and g=0, and there are no holomorphic differentials, so given any set of point p1,...,pn, there are exactly n
=1 independent rational functions with poles at those points, presumably the linear combinations of 1, 1/(x-p1),....,1/(x-pn).
does this help?
3. Jun 9, 2014
### bolbteppa
That's exactly what I was hoping for, thank you very much.
4. Jun 10, 2014
### mathwonk
If you want much more, consult:
http://www.math.uga.edu/%7Eroy/rrt.pdf [Broken]
Last edited by a moderator: May 6, 2017
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https://math.stackexchange.com/questions/2102008/adding-of-two-elliptic-curve-points | Adding of two elliptic curve points
I'm working on elliptic curve points, and I know that some possible points require separate consideration. For example, if I want to add points P and Q, and P is a point in infinity and Q is a point in infinity - I will have the infinity point. If Q = -P, adding P + Q results in the infinity point. But what in the cases:
• if $P = (x,y1)$ and $Q = (-x, y2)$
• if $P = (x, y1)$ and $Q = (-x, y1)$
Points P and Q are not opposite, since the plot is symmetric over axes x, not axes y. Should I treat those cases somehow separately? Are those points even on the same curve?
No, the points may not be on the same curve. Say, we have the elliptic curve $y^2=x^3+17$ over $\mathbb{Q}$. Then the point $(x,y_1)=(8,23)$ lies on the curve, but there is no rational point $(-x,y_2)=(-8,y_2)$ on it. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8616471290588379, "perplexity": 284.79787959493274}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987841291.79/warc/CC-MAIN-20191024040131-20191024063631-00224.warc.gz"} |
http://www.mathisfunforum.com/post.php?tid=2683&qid=26476 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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irspow
2006-02-07 13:43:30
johny, the answer in post #30 is correct given the functions you had and the angle of swing. What we are still questioning is only why the length of blade does not match what you said originally.
A ≈ 1501.3cm²
johny
2006-02-07 07:12:14
Seriously guys thanks alot for giving soo much time to it!!!!! If i find something out then i wil let u know.........and once again thanks
irspow
2006-02-06 07:31:43
Has anyone definitively answered the question? I tried my best in post #30. I thought that I followed mathematical convention exactly, yet it does not match all of the information given in the original problem. For all of the talk and views that this thread has gotten, my humble calculation is the only one that offers the area in question? I would say with the talent in this forum and the lack of a definitive answer would indeed suggest that the changing length of blade has complicated the problem.
In fact, without the blade variation, no integration is even needed. See my edit in 35.
edit
I see your point Ricky. If all of the information was correct from the beginning, yes, it would be a rather simple calculation. The changing length would not make it more difficult if the functions were defined correctly.
Ricky
2006-02-06 07:07:42
Yes, but why are you saying the changing lenght of the blade complicates the problem? Those functions describe that change.
irspow
2006-02-06 06:48:35
I interpreted the functions given like this;
y1 represents the height of the top of the blade
y2 represents the height of the bottom of the blade
Is that what you just said? haha
I believe either y1 or y2 is incorrect.
edit*
Interestingly, using simple geometry and assuming no blade length variations produces a pretty good approximation.
120° represents 1/3 of a circle
1/3 π(40)² - 1/3 π(10)² = 500π ≈ 1570.796327cm²
Ricky
2006-02-06 06:44:04
Well, the way I read it, the functions given by the question are describing the changing length of the blade. It does change, after all, but it is 3cm off from 20. Is the 20-30cm change just supposed to be approximate? If so, then the equations as stated properly represent the area of integration. If not, then there may be some oversight in assigning the functions or 120° by the creator.
Or the function don't represent the changing lenght of the blade, in which case I believe this question is impossible.
I'd rather go with the first option personally.
irspow
2006-02-06 06:38:32
Yeah, so it isn't just me. My numbers say that the blade actually increased in length instead of the opposite, from 30 to >32 cm. Something is fishy. Can you at least see why I have been making all of the noise here now?
It all stemmed from how the intersection points were defined. And the ho-hum, by the way, varying length blade does indeed make the situation more complicated. Like I have said all along, this problem should have been really simple, but something is not quite right.
Ricky
2006-02-06 06:29:00
And shouldn't the y3(right boundary function) be y = 40 - x/:raic3?
Yep. Big mistake on my part. Nice catch.
When I find the length of the wiper blade based on my functions (with the √3 change), I get the blade goes from 23.5307 to 30cm. Not exactly 20-30, but close. Is the question wrong? Is the interpretation of it? Not sure.
John E. Franklin
2006-02-06 04:11:08
I think that the wiper blade is centered on the end of the arm, so when vertical, then 40 minus 30/2 is exposed of the arm, not 40 minus 30. Why would you hold the wiper at the bottom?
irspow
2006-02-06 03:50:14
Look at post #23 and #24 to see where I am coming from. It is the varying length of the wiper blade that is causing my confusion.
And shouldn't the y3(right boundary function) be y = 40 - x/√3 ?
If I follow all of the rules of finding the intersections of the lines and using them as the limits of integrations I get:
y1 = y3(my y3 that is) at x = 1.63899453, we'll call this b
y3 = y2 at x = 32.55915054, we'll call this c
We will use 0 = a for the origin
We will assume the functions given are correct so that;
y1 = x²/50 + 30
y2 = x²/50
y3 = 40 - x/√3
We will multiply the integrals by two because the two halves are symmetrical.
2 {∫y1 - y2 dx]from a to b + ∫y3 - y2 dx]from b to c} which gives;
2(∫30 dx]a to b + ∫40 - x/√3 - x²/50 dx]b to c)
2{ [30x]a,b + [40x - x²/√12 - x³/150]b,c}
If you plug in the values of a, b, and c you will get;
A = 1501.301771cm²
That is strictly by the rules.
The problem is with the points b and c. (y1 at b and y2 at c)
pb = (1.638994538, 30.05372606)
pc = (32.55915054, 21.20196568)
We all know how to find the distance between these two points, and in this case it equals;
blade length = √[(dx)² + (dy)²]
That measurement contradicts the stipulated conditions, but it was done using the equations given for the endpoints of the wiper blade.
This is what I was talking about all along in a formal way. The blade should be 20cm long at the end of its swing and 30cm at the origin. I could not and still can not find where the discrepency lies.
Ricky
2006-02-06 03:34:03
irspow, I'm not understanding your objections at all.
The problem gives two curves and two lines, and want you to find the area inbetween all of them.
What's wrong with this?
irspow
2006-02-06 02:38:08
If you use Ricky's conclusions, then the area would be;
2{[∫y1 - y2 dx] + [∫y3 - y2]}
If y1 = x²/50 + 30, y2 = x²/50, y3 = 40 - x/√3 then the above becomes;
2 {[30x]0 to b + [40x - x²/√(12) - x³/150]b to x max}
That would indeed give you the area that you seek, but you would have to have the correct limits of integration. This determination is where all of the discussion has taken place.
Right now you just have;
-x³/75 -x²/√3 + 80x + b³/75 + b²/√3 - 20b = A
But that is assuming that you can figure out what the proper value of x max and b are to plug into the above formula. If you read the last several posts there are some questions raised as to what exactly they should be given the original conditions of the problem. Are you absolutely sure that you did not leave anything out? Especially check out posts 18 through 24 again to see what I am talking about.
johny
2006-02-05 22:07:34
I dont want to be pain, but just one last question becuase i too am getting a bit confused.....so i need to simplify the equation given on post no. 18???? And that would give me the area????
irspow
2006-02-05 16:39:49
No John, the arm is 40cm long. I know this getting to be a long thread, but there is a good illustration posted somewhere above. When the arm is vertical only 10cm is exposed.
I think that my last post raised serious questions about the conditions stated in the problem. I used equivalent trigonomic functions and found that they would not produce the same values that are stipulated by the functions originally given. I think that we are missing something here.
In other words, this type of problem is quite easy to solve normally, but all of the conditions stated in the original problem can not be satisfied.
John E. Franklin
2006-02-05 16:24:32
Is the wiper blade centered on the end of the wiper arm? | {"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": 0.821029782295227, "perplexity": 1604.5508711350126}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00466-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://math.stackexchange.com/users/22405/brett-frankel?tab=activity | Brett Frankel
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Jan 5 awarded Yearling Jun 1 awarded Necromancer Feb 11 revised How would you go about proving that any complex line is a curve on the Riemann sphere? Removed Riemannian geometry tag Jan 18 asked Semisimplicity of Frobenius action on H^1(X,Q_p) Jan 5 awarded Yearling Nov 30 comment Compositum of all Galois extensions of prime power degree It's an isomorphism. Nov 29 comment Compositum of all Galois extensions of prime power degree @AngeloRendina By $E_i$, do you mean what I called $K_i$? If so, automorphisms of the $E_i$ are not subgroups but quotient groups. So it should be correct as I originally wrote it. Nov 27 revised Compositum of all Galois extensions of prime power degree added 17 characters in body Nov 27 revised Compositum of all Galois extensions of prime power degree added 664 characters in body Nov 27 comment Compositum of all Galois extensions of prime power degree Yes, there's some work to be done here, but the fact that the subextensions are Galois means it will work out. Give me a few minutes and I'll sketch out something more complete. Nov 27 comment Compositum of all Galois extensions of prime power degree Well, what is the compositum? It's the set of elements that can be obtained from elements of these finite extensions. So you just have to show, by induction, that if you take the compositum of two finite Galois extensions of $p$-power order, the compositum is again of $p$-power order (and so is its Galois closure) Nov 27 answered Compositum of all Galois extensions of prime power degree Nov 21 comment Prove that any subfield of $\Bbb R$ contains $\Bbb Q$ Your proof is correct, but could perhaps use some more details (why do you reach each conclusion?) Also, why doesn't your proof work for $\mathbb Z/p$? You must be using some property of $\mathbb R$, but you don't explicitly say what property you are using. Nov 17 comment Why every central simple algebra has a splitting field @Dune The Tensor product of a central simple algebra with a simple algebra is always simple. Nov 17 revised Why every central simple algebra has a splitting field edited tags Nov 9 comment A proposition on Mobius map The only group-theoretic observation, which we can make without using the language of groups, is that given an automorphism $\psi$ with $\psi(a)=0$ for some $a$, we can write $\psi=\tau \circ \varphi_a$, where $\tau$ maps $0$ to $0$. Nov 8 answered A proposition on Mobius map Oct 21 revised product considering the period of index over a cyclotomic extension added 31 characters in body; edited title Oct 7 awarded Nice Answer Sep 30 reviewed Looks OK Thinking about a probability problem in terms of sets. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9061292409896851, "perplexity": 383.7415603609856}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701153585.76/warc/CC-MAIN-20160205193913-00293-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://davegiles.blogspot.ca/2014_04_10_archive.html | ## Thursday, April 10, 2014
### Proof of a Result About the "Adjusted" Coefficient of Determination
In a post last year I discussed the conditions under which the "adjusted" coefficient of determination (RA2) will increase or decrease, when regressors are deleted from (added to) a regression model. Without going over the full discussion again, here is one of the key results:
Adding a group of regressors to the model will increase (decrease) RA2 depending on whether the F-statistic for testing that their coefficients are all zero is greater (less) than one in value. RA2 is unchanged if that F-statistic is exactly equal to one.
A few days ago, "Zeba" reminded me that I had promised to post a simple proof of this result, but I still hadn't done so. Shame on me! A proof is given below. As a bonus, I've given the proof for a more general result - we don't have to be imposing "zero" restrictions on some of the coefficients - any exact linear restrictions will suffice.
Let's take a look at the proof. | {"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": 0.8485188484191895, "perplexity": 618.11565126543}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218203536.73/warc/CC-MAIN-20170322213003-00354-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://infoscience.epfl.ch/record/179364 | ## A converse approach to the calculation of NMR shielding tensors
We introduce an alternative approach to the first-principles calculation of NMR shielding tensors. These are obtained from the derivative of the orbital magnetization with respect to the application of a microscopic, localized magnetic dipole. The approach is simple, general, and can be applied to either isolated or periodic systems. Calculated results for simple hydrocarbons, crystalline diamond, and liquid water show very good agreement with established methods and experimental results. (C) 2009 American Institute of Physics. [doi:10.1063/1.3216028]
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2009
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https://www.educator.com/mathematics/probability/murray/the-central-limit-theorem.php | ×
Start learning today, and be successful in your academic & professional career. Start Today!
William Murray
The Central Limit Theorem
Slide Duration:
Section 1: Probability by Counting
Experiments, Outcomes, Samples, Spaces, Events
59m 30s
Intro
0:00
Terminology
0:19
Experiment
0:26
Outcome
0:56
Sample Space
1:16
Event
1:55
Key Formula
2:47
Formula for Finding the Probability of an Event
2:48
Example: Drawing a Card
3:36
Example I
5:01
Experiment
5:38
Outcomes
5:54
Probability of the Event
8:11
Example II
12:00
Experiment
12:17
Outcomes
12:34
Probability of the Event
13:49
Example III
16:33
Experiment
17:09
Outcomes
17:33
Probability of the Event
18:25
Example IV
21:20
Experiment
21:21
Outcomes
22:00
Probability of the Event
23:22
Example V
31:41
Experiment
32:14
Outcomes
32:35
Probability of the Event
33:27
Alternate Solution
40:16
Example VI
43:33
Experiment
44:08
Outcomes
44:24
Probability of the Event
53:35
1h 2m 47s
Intro
0:00
Unions of Events
0:40
Unions of Events
0:41
Disjoint Events
3:42
Intersections of Events
4:18
Intersections of Events
4:19
Conditional Probability
5:47
Conditional Probability
5:48
Independence
8:20
Independence
8:21
Warning: Independent Does Not Mean Disjoint
9:53
If A and B are Independent
11:20
Example I: Choosing a Number at Random
12:41
Solving by Counting
12:52
Solving by Probability
17:26
Example II: Combination
22:07
Combination Deal at a Restaurant
22:08
Example III: Rolling Two Dice
24:18
Define the Events
24:20
Solving by Counting
27:35
Solving by Probability
29:32
Example IV: Flipping a Coin
35:07
Flipping a Coin Four Times
35:08
Example V: Conditional Probabilities
41:22
Define the Events
42:23
Calculate the Conditional Probabilities
46:21
Example VI: Independent Events
53:42
Define the Events
53:43
Are Events Independent?
55:21
Choices: Combinations & Permutations
56m 3s
Intro
0:00
Choices: With or Without Replacement?
0:12
Choices: With or Without Replacement?
0:13
Example: With Replacement
2:17
Example: Without Replacement
2:55
Choices: Ordered or Unordered?
4:10
Choices: Ordered or Unordered?
4:11
Example: Unordered
4:52
Example: Ordered
6:08
Combinations
9:23
Definition & Equation: Combinations
9:24
Example: Combinations
12:12
Permutations
13:56
Definition & Equation: Permutations
13:57
Example: Permutations
15:00
Key Formulas
17:19
Number of Ways to Pick r Things from n Possibilities
17:20
Example I: Five Different Candy Bars
18:31
Example II: Five Identical Candy Bars
24:53
Example III: Five Identical Candy Bars
31:56
Example IV: Five Different Candy Bars
39:21
Example V: Pizza & Toppings
45:03
Inclusion & Exclusion
43m 40s
Intro
0:00
Inclusion/Exclusion: Two Events
0:09
Inclusion/Exclusion: Two Events
0:10
Inclusion/Exclusion: Three Events
2:30
Inclusion/Exclusion: Three Events
2:31
Example I: Inclusion & Exclusion
6:24
Example II: Inclusion & Exclusion
11:01
Example III: Inclusion & Exclusion
18:41
Example IV: Inclusion & Exclusion
28:24
Example V: Inclusion & Exclusion
39:33
Independence
46m 9s
Intro
0:00
Formula and Intuition
0:12
Definition of Independence
0:19
Intuition
0:49
Common Misinterpretations
1:37
Myth & Truth 1
1:38
Myth & Truth 2
2:23
Combining Independent Events
3:56
Recall: Formula for Conditional Probability
3:58
Combining Independent Events
4:10
Example I: Independence
5:36
Example II: Independence
14:14
Example III: Independence
21:10
Example IV: Independence
32:45
Example V: Independence
41:13
Bayes' Rule
1h 2m 10s
Intro
0:00
When to Use Bayes' Rule
0:08
When to Use Bayes' Rule: Disjoint Union of Events
0:09
Bayes' Rule for Two Choices
2:50
Bayes' Rule for Two Choices
2:51
Bayes' Rule for Multiple Choices
5:03
Bayes' Rule for Multiple Choices
5:04
Example I: What is the Chance that She is Diabetic?
6:55
Example I: Setting up the Events
6:56
Example I: Solution
11:33
Example II: What is the chance that It Belongs to a Woman?
19:28
Example II: Setting up the Events
19:29
Example II: Solution
21:45
Example III: What is the Probability that She is a Democrat?
27:31
Example III: Setting up the Events
27:32
Example III: Solution
32:08
Example IV: What is the chance that the Fruit is an Apple?
39:11
Example IV: Setting up the Events
39:12
Example IV: Solution
43:50
Example V: What is the Probability that the Oldest Child is a Girl?
51:16
Example V: Setting up the Events
51:17
Example V: Solution
53:07
Section 2: Random Variables
Random Variables & Probability Distribution
38m 21s
Intro
0:00
Intuition
0:15
Intuition for Random Variable
0:16
Example: Random Variable
0:44
Intuition, Cont.
2:52
Example: Random Variable as Payoff
2:57
Definition
5:11
Definition of a Random Variable
5:13
Example: Random Variable in Baseball
6:02
Probability Distributions
7:18
Probability Distributions
7:19
Example I: Probability Distribution for the Random Variable
9:29
Example II: Probability Distribution for the Random Variable
14:52
Example III: Probability Distribution for the Random Variable
21:52
Example IV: Probability Distribution for the Random Variable
27:25
Example V: Probability Distribution for the Random Variable
34:12
Expected Value (Mean)
46m 14s
Intro
0:00
Definition of Expected Value
0:20
Expected Value of a (Discrete) Random Variable or Mean
0:21
Indicator Variables
3:03
Indicator Variable
3:04
Linearity of Expectation
4:36
Linearity of Expectation for Random Variables
4:37
Expected Value of a Function
6:03
Expected Value of a Function
6:04
Example I: Expected Value
7:30
Example II: Expected Value
14:14
Example III: Expected Value of Flipping a Coin
21:42
Example III: Part A
21:43
Example III: Part B
30:43
Example IV: Semester Average
36:39
Example V: Expected Value of a Function of a Random Variable
41:28
Variance & Standard Deviation
47m 23s
Intro
0:00
Definition of Variance
0:11
Variance of a Random Variable
0:12
Variance is a Measure of the Variability, or Volatility
1:06
Most Useful Way to Calculate Variance
2:46
Definition of Standard Deviation
3:44
Standard Deviation of a Random Variable
3:45
Example I: Which of the Following Sets of Data Has the Largest Variance?
5:34
Example II: Which of the Following Would be the Least Useful in Understanding a Set of Data?
9:02
Example III: Calculate the Mean, Variance, & Standard Deviation
11:48
Example III: Mean
12:56
Example III: Variance
14:06
Example III: Standard Deviation
15:42
Example IV: Calculate the Mean, Variance, & Standard Deviation
17:54
Example IV: Mean
18:47
Example IV: Variance
20:36
Example IV: Standard Deviation
25:34
Example V: Calculate the Mean, Variance, & Standard Deviation
29:56
Example V: Mean
30:13
Example V: Variance
33:28
Example V: Standard Deviation
34:48
Example VI: Calculate the Mean, Variance, & Standard Deviation
37:29
Example VI: Possible Outcomes
38:09
Example VI: Mean
39:29
Example VI: Variance
41:22
Example VI: Standard Deviation
43:28
Markov's Inequality
26m 45s
Intro
0:00
Markov's Inequality
0:25
Markov's Inequality: Definition & Condition
0:26
Markov's Inequality: Equation
1:15
Markov's Inequality: Reverse Equation
2:48
Example I: Money
4:11
Example II: Rental Car
9:23
Example III: Probability of an Earthquake
12:22
Example IV: Defective Laptops
16:52
Example V: Cans of Tuna
21:06
Tchebysheff's Inequality
42m 11s
Intro
0:00
Tchebysheff's Inequality (Also Known as Chebyshev's Inequality)
0:52
Tchebysheff's Inequality: Definition
0:53
Tchebysheff's Inequality: Equation
1:19
Tchebysheff's Inequality: Intuition
3:21
Tchebysheff's Inequality in Reverse
4:09
Tchebysheff's Inequality in Reverse
4:10
Intuition
5:13
Example I: Money
5:55
Example II: College Units
13:20
Example III: Using Tchebysheff's Inequality to Estimate Proportion
16:40
Example IV: Probability of an Earthquake
25:21
Example V: Using Tchebysheff's Inequality to Estimate Proportion
32:57
Section 3: Discrete Distributions
Binomial Distribution (Bernoulli Trials)
52m 36s
Intro
0:00
Binomial Distribution
0:29
Binomial Distribution (Bernoulli Trials) Overview
0:30
Prototypical Examples: Flipping a Coin n Times
1:36
Process with Two Outcomes: Games Between Teams
2:12
Process with Two Outcomes: Rolling a Die to Get a 6
2:42
Formula for the Binomial Distribution
3:45
Fixed Parameters
3:46
Formula for the Binomial Distribution
6:27
Key Properties of the Binomial Distribution
9:54
Mean
9:55
Variance
10:56
Standard Deviation
11:13
Example I: Games Between Teams
11:36
Example II: Exam Score
17:01
Example III: Expected Grade & Standard Deviation
25:59
Example IV: Pogo-sticking Championship, Part A
33:25
Example IV: Pogo-sticking Championship, Part B
38:24
Example V: Expected Championships Winning & Standard Deviation
45:22
Geometric Distribution
52m 50s
Intro
0:00
Geometric Distribution
0:22
Geometric Distribution: Definition
0:23
Prototypical Example: Flipping a Coin Until We Get a Head
1:08
Geometric Distribution vs. Binomial Distribution.
1:31
Formula for the Geometric Distribution
2:13
Fixed Parameters
2:14
Random Variable
2:49
Formula for the Geometric Distribution
3:16
Key Properties of the Geometric Distribution
6:47
Mean
6:48
Variance
7:10
Standard Deviation
7:25
Geometric Series
7:46
Recall from Calculus II: Sum of Infinite Series
7:47
Application to Geometric Distribution
10:10
Example I: Drawing Cards from a Deck (With Replacement) Until You Get an Ace
13:02
Example I: Question & Solution
13:03
Example II: Mean & Standard Deviation of Winning Pin the Tail on the Donkey
16:32
Example II: Mean
16:33
Example II: Standard Deviation
18:37
Example III: Rolling a Die
22:09
Example III: Setting Up
22:10
Example III: Part A
24:18
Example III: Part B
26:01
Example III: Part C
27:38
Example III: Summary
32:02
Example IV: Job Interview
35:16
Example IV: Setting Up
35:15
Example IV: Part A
37:26
Example IV: Part B
38:33
Example IV: Summary
39:37
Example V: Mean & Standard Deviation of Time to Conduct All the Interviews
41:13
Example V: Setting Up
42:50
Example V: Mean
46:05
Example V: Variance
47:37
Example V: Standard Deviation
48:22
Example V: Summary
49:36
Negative Binomial Distribution
51m 39s
Intro
0:00
Negative Binomial Distribution
0:11
Negative Binomial Distribution: Definition
0:12
Prototypical Example: Flipping a Coin Until We Get r Successes
0:46
Negative Binomial Distribution vs. Binomial Distribution
1:04
Negative Binomial Distribution vs. Geometric Distribution
1:33
Formula for Negative Binomial Distribution
3:39
Fixed Parameters
3:40
Random Variable
4:57
Formula for Negative Binomial Distribution
5:18
Key Properties of Negative Binomial
7:44
Mean
7:47
Variance
8:03
Standard Deviation
8:09
Example I: Drawing Cards from a Deck (With Replacement) Until You Get Four Aces
8:32
Example I: Question & Solution
8:33
Example II: Chinchilla Grooming
12:37
Example II: Mean
12:38
Example II: Variance
15:09
Example II: Standard Deviation
15:51
Example II: Summary
17:10
Example III: Rolling a Die Until You Get Four Sixes
18:27
Example III: Setting Up
19:38
Example III: Mean
19:38
Example III: Variance
20:31
Example III: Standard Deviation
21:21
Example IV: Job Applicants
24:00
Example IV: Setting Up
24:01
Example IV: Part A
26:16
Example IV: Part B
29:53
Example V: Mean & Standard Deviation of Time to Conduct All the Interviews
40:10
Example V: Setting Up
40:11
Example V: Mean
45:24
Example V: Variance
46:22
Example V: Standard Deviation
47:01
Example V: Summary
48:16
Hypergeometric Distribution
36m 27s
Intro
0:00
Hypergeometric Distribution
0:11
Hypergeometric Distribution: Definition
0:12
Random Variable
1:38
Formula for the Hypergeometric Distribution
1:50
Fixed Parameters
1:51
Formula for the Hypergeometric Distribution
2:53
Key Properties of Hypergeometric
6:14
Mean
6:15
Variance
6:42
Standard Deviation
7:16
Example I: Students Committee
7:30
Example II: Expected Number of Women on the Committee in Example I
11:08
Example III: Pairs of Shoes
13:49
Example IV: What is the Expected Number of Left Shoes in Example III?
20:46
Example V: Using Indicator Variables & Linearity of Expectation
25:40
Poisson Distribution
52m 19s
Intro
0:00
Poisson Distribution
0:18
Poisson Distribution: Definition
0:19
Formula for the Poisson Distribution
2:16
Fixed Parameter
2:17
Formula for the Poisson Distribution
2:59
Key Properties of the Poisson Distribution
5:30
Mean
5:34
Variance
6:07
Standard Deviation
6:27
Example I: Forest Fires
6:41
Example II: Call Center, Part A
15:56
Example II: Call Center, Part B
20:50
Example III: Confirming that the Mean of the Poisson Distribution is λ
26:53
Example IV: Find E (Y²) for the Poisson Distribution
35:24
Example V: Earthquakes, Part A
37:57
Example V: Earthquakes, Part B
44:02
Section 4: Continuous Distributions
Density & Cumulative Distribution Functions
57m 17s
Intro
0:00
Density Functions
0:43
Density Functions
0:44
Density Function to Calculate Probabilities
2:41
Cumulative Distribution Functions
4:28
Cumulative Distribution Functions
4:29
Using F to Calculate Probabilities
5:58
Properties of the CDF (Density & Cumulative Distribution Functions)
7:27
F(-∞) = 0
7:34
F(∞) = 1
8:30
F is Increasing
9:14
F'(y) = f(y)
9:21
Example I: Density & Cumulative Distribution Functions, Part A
9:43
Example I: Density & Cumulative Distribution Functions, Part B
14:16
Example II: Density & Cumulative Distribution Functions, Part A
21:41
Example II: Density & Cumulative Distribution Functions, Part B
26:16
Example III: Density & Cumulative Distribution Functions, Part A
32:17
Example III: Density & Cumulative Distribution Functions, Part B
37:08
Example IV: Density & Cumulative Distribution Functions
43:34
Example V: Density & Cumulative Distribution Functions, Part A
51:53
Example V: Density & Cumulative Distribution Functions, Part B
54:19
Mean & Variance for Continuous Distributions
36m 18s
Intro
0:00
Mean
0:32
Mean for a Continuous Random Variable
0:33
Expectation is Linear
2:07
Variance
2:55
Variance for Continuous random Variable
2:56
Easier to Calculate Via the Mean
3:26
Standard Deviation
5:03
Standard Deviation
5:04
Example I: Mean & Variance for Continuous Distributions
5:43
Example II: Mean & Variance for Continuous Distributions
10:09
Example III: Mean & Variance for Continuous Distributions
16:05
Example IV: Mean & Variance for Continuous Distributions
26:40
Example V: Mean & Variance for Continuous Distributions
30:12
Uniform Distribution
32m 49s
Intro
0:00
Uniform Distribution
0:15
Uniform Distribution
0:16
Each Part of the Region is Equally Probable
1:39
Key Properties of the Uniform Distribution
2:45
Mean
2:46
Variance
3:27
Standard Deviation
3:48
Example I: Newspaper Delivery
5:25
Example II: Picking a Real Number from a Uniform Distribution
8:21
Example III: Dinner Date
11:02
Example IV: Proving that a Variable is Uniformly Distributed
18:50
Example V: Ice Cream Serving
27:22
Normal (Gaussian) Distribution
1h 3m 54s
Intro
0:00
Normal (Gaussian) Distribution
0:35
Normal (Gaussian) Distribution & The Bell Curve
0:36
Fixed Parameters
0:55
Formula for the Normal Distribution
1:32
Formula for the Normal Distribution
1:33
Calculating on the Normal Distribution can be Tricky
3:32
Standard Normal Distribution
5:12
Standard Normal Distribution
5:13
Graphing the Standard Normal Distribution
6:13
Standard Normal Distribution, Cont.
8:30
Standard Normal Distribution Chart
8:31
Nonstandard Normal Distribution
14:44
Nonstandard Normal Variable & Associated Standard Normal
14:45
Finding Probabilities for Z
15:39
Example I: Chance that Standard Normal Variable Will Land Between 1 and 2?
16:46
Example I: Setting Up the Equation & Graph
16:47
Example I: Solving for z Using the Standard Normal Chart
19:05
Example II: What Proportion of the Data Lies within Two Standard Deviations of the Mean?
20:41
Example II: Setting Up the Equation & Graph
20:42
Example II: Solving for z Using the Standard Normal Chart
24:38
Example III: Scores on an Exam
27:34
Example III: Setting Up the Equation & Graph, Part A
27:35
Example III: Setting Up the Equation & Graph, Part B
33:48
Example III: Solving for z Using the Standard Normal Chart, Part A
38:23
Example III: Solving for z Using the Standard Normal Chart, Part B
40:49
Example IV: Temperatures
42:54
Example IV: Setting Up the Equation & Graph
42:55
Example IV: Solving for z Using the Standard Normal Chart
47:03
Example V: Scores on an Exam
48:41
Example V: Setting Up the Equation & Graph, Part A
48:42
Example V: Setting Up the Equation & Graph, Part B
53:20
Example V: Solving for z Using the Standard Normal Chart, Part A
57:45
Example V: Solving for z Using the Standard Normal Chart, Part B
59:17
Gamma Distribution (with Exponential & Chi-square)
1h 8m 27s
Intro
0:00
Gamma Function
0:49
The Gamma Function
0:50
Properties of the Gamma Function
2:07
Formula for the Gamma Distribution
3:50
Fixed Parameters
3:51
Density Function for Gamma Distribution
4:07
Key Properties of the Gamma Distribution
7:13
Mean
7:14
Variance
7:25
Standard Deviation
7:30
Exponential Distribution
8:03
Definition of Exponential Distribution
8:04
Density
11:23
Mean
13:26
Variance
13:48
Standard Deviation
13:55
Chi-square Distribution
14:34
Chi-square Distribution: Overview
14:35
Chi-square Distribution: Mean
16:27
Chi-square Distribution: Variance
16:37
Chi-square Distribution: Standard Deviation
16:55
Example I: Graphing Gamma Distribution
17:30
Example I: Graphing Gamma Distribution
17:31
Example I: Describe the Effects of Changing α and β on the Shape of the Graph
23:33
Example II: Exponential Distribution
27:11
Example II: Using the Exponential Distribution
27:12
Example II: Summary
35:34
Example III: Earthquake
37:05
Example III: Estimate Using Markov's Inequality
37:06
Example III: Estimate Using Tchebysheff's Inequality
40:13
Example III: Summary
44:13
Example IV: Finding Exact Probability of Earthquakes
46:45
Example IV: Finding Exact Probability of Earthquakes
46:46
Example IV: Summary
51:44
Example V: Prove and Interpret Why the Exponential Distribution is Called 'Memoryless'
52:51
Example V: Prove
52:52
Example V: Interpretation
57:44
Example V: Summary
1:03:54
Beta Distribution
52m 45s
Intro
0:00
Beta Function
0:29
Fixed parameters
0:30
Defining the Beta Function
1:19
Relationship between the Gamma & Beta Functions
2:02
Beta Distribution
3:31
Density Function for the Beta Distribution
3:32
Key Properties of the Beta Distribution
6:56
Mean
6:57
Variance
7:16
Standard Deviation
7:37
Example I: Calculate B(3,4)
8:10
Example II: Graphing the Density Functions for the Beta Distribution
12:25
Example III: Show that the Uniform Distribution is a Special Case of the Beta Distribution
24:57
Example IV: Show that this Triangular Distribution is a Special Case of the Beta Distribution
31:20
Example V: Morning Commute
37:39
Example V: Identify the Density Function
38:45
Example V: Morning Commute, Part A
42:22
Example V: Morning Commute, Part B
44:19
Example V: Summary
49:13
Moment-Generating Functions
51m 58s
Intro
0:00
Moments
0:30
Definition of Moments
0:31
Moment-Generating Functions (MGFs)
3:53
Moment-Generating Functions
3:54
Using the MGF to Calculate the Moments
5:21
Moment-Generating Functions for the Discrete Distributions
8:22
Moment-Generating Functions for Binomial Distribution
8:36
Moment-Generating Functions for Geometric Distribution
9:06
Moment-Generating Functions for Negative Binomial Distribution
9:28
Moment-Generating Functions for Hypergeometric Distribution
9:43
Moment-Generating Functions for Poisson Distribution
9:57
Moment-Generating Functions for the Continuous Distributions
11:34
Moment-Generating Functions for the Uniform Distributions
11:43
Moment-Generating Functions for the Normal Distributions
12:24
Moment-Generating Functions for the Gamma Distributions
12:36
Moment-Generating Functions for the Exponential Distributions
12:44
Moment-Generating Functions for the Chi-square Distributions
13:11
Moment-Generating Functions for the Beta Distributions
13:48
Useful Formulas with Moment-Generating Functions
15:02
Useful Formulas with Moment-Generating Functions 1
15:03
Useful Formulas with Moment-Generating Functions 2
16:21
Example I: Moment-Generating Function for the Binomial Distribution
17:33
Example II: Use the MGF for the Binomial Distribution to Find the Mean of the Distribution
24:40
Example III: Find the Moment Generating Function for the Poisson Distribution
29:28
Example IV: Use the MGF for Poisson Distribution to Find the Mean and Variance of the Distribution
36:27
Example V: Find the Moment-generating Function for the Uniform Distribution
44:47
Section 5: Multivariate Distributions
Bivariate Density & Distribution Functions
50m 52s
Intro
0:00
Bivariate Density Functions
0:21
Two Variables
0:23
Bivariate Density Function
0:52
Properties of the Density Function
1:57
Properties of the Density Function 1
1:59
Properties of the Density Function 2
2:20
We Can Calculate Probabilities
2:53
If You Have a Discrete Distribution
4:36
Bivariate Distribution Functions
5:25
Bivariate Distribution Functions
5:26
Properties of the Bivariate Distribution Functions 1
7:19
Properties of the Bivariate Distribution Functions 2
7:36
Example I: Bivariate Density & Distribution Functions
8:08
Example II: Bivariate Density & Distribution Functions
14:40
Example III: Bivariate Density & Distribution Functions
24:33
Example IV: Bivariate Density & Distribution Functions
32:04
Example V: Bivariate Density & Distribution Functions
40:26
Marginal Probability
42m 38s
Intro
0:00
Discrete Case
0:48
Marginal Probability Functions
0:49
Continuous Case
3:07
Marginal Density Functions
3:08
Example I: Compute the Marginal Probability Function
5:58
Example II: Compute the Marginal Probability Function
14:07
Example III: Marginal Density Function
24:01
Example IV: Marginal Density Function
30:47
Example V: Marginal Density Function
36:05
Conditional Probability & Conditional Expectation
1h 2m 24s
Intro
0:00
Review of Marginal Probability
0:46
Recall the Marginal Probability Functions & Marginal Density Functions
0:47
Conditional Probability, Discrete Case
3:14
Conditional Probability, Discrete Case
3:15
Conditional Probability, Continuous Case
4:15
Conditional Density of Y₁ given that Y₂ = y₂
4:16
Interpret This as a Density on Y₁ & Calculate Conditional Probability
5:03
Conditional Expectation
6:44
Conditional Expectation: Continuous
6:45
Conditional Expectation: Discrete
8:03
Example I: Conditional Probability
8:29
Example II: Conditional Probability
23:59
Example III: Conditional Probability
34:28
Example IV: Conditional Expectation
43:16
Example V: Conditional Expectation
48:28
Independent Random Variables
51m 39s
Intro
0:00
Intuition
0:55
Experiment with Two Random Variables
0:56
Intuition Formula
2:17
Definition and Formulas
4:43
Definition
4:44
Short Version: Discrete
5:10
Short Version: Continuous
5:48
Theorem
9:33
For Continuous Random Variables, Y₁ & Y₂ are Independent If & Only If: Condition 1
9:34
For Continuous Random Variables, Y₁ & Y₂ are Independent If & Only If: Condition 2
11:22
Example I: Use the Definition to Determine if Y₁ and Y₂ are Independent
12:49
Example II: Use the Definition to Determine if Y₁ and Y₂ are Independent
21:33
Example III: Are Y₁ and Y₂ Independent?
27:01
Example IV: Are Y₁ and Y₂ Independent?
34:51
Example V: Are Y₁ and Y₂ Independent?
43:44
Expected Value of a Function of Random Variables
37m 7s
Intro
0:00
Review of Single Variable Case
0:29
Expected Value of a Single Variable
0:30
Expected Value of a Function g(Y)
1:12
Bivariate Case
2:11
Expected Value of a Function g(Y₁, Y₂)
2:12
Linearity of Expectation
3:24
Linearity of Expectation 1
3:25
Linearity of Expectation 2
3:38
4:03
Example I: Calculate E (Y₁ + Y₂)
4:39
Example II: Calculate E (Y₁Y₂)
14:47
Example III: Calculate E (U₁) and E(U₂)
19:33
Example IV: Calculate E (Y₁) and E(Y₂)
22:50
Example V: Calculate E (2Y₁ + 3Y₂)
33:05
Covariance, Correlation & Linear Functions
59m 50s
Intro
0:00
Definition and Formulas for Covariance
0:38
Definition of Covariance
0:39
Formulas to Calculate Covariance
1:36
Intuition for Covariance
3:54
Covariance is a Measure of Dependence
3:55
Dependence Doesn't Necessarily Mean that the Variables Do the Same Thing
4:12
If Variables Move Together
4:47
If Variables Move Against Each Other
5:04
Both Cases Show Dependence!
5:30
Independence Theorem
8:10
Independence Theorem
8:11
The Converse is Not True
8:32
Correlation Coefficient
9:33
Correlation Coefficient
9:34
Linear Functions of Random Variables
11:57
Linear Functions of Random Variables: Expected Value
11:58
Linear Functions of Random Variables: Variance
12:58
Linear Functions of Random Variables, Cont.
14:30
Linear Functions of Random Variables: Covariance
14:35
Example I: Calculate E (Y₁), E (Y₂), and E (Y₁Y₂)
15:31
Example II: Are Y₁ and Y₂ Independent?
29:16
Example III: Calculate V (U₁) and V (U₂)
36:14
Example IV: Calculate the Covariance Correlation Coefficient
42:12
Example V: Find the Mean and Variance of the Average
52:19
Section 6: Distributions of Functions of Random Variables
Distribution Functions
1h 7m 35s
Intro
0:00
Premise
0:44
Premise
0:45
Goal
1:38
Goal Number 1: Find the Full Distribution Function
1:39
Goal Number 2: Find the Density Function
1:55
Goal Number 3: Calculate Probabilities
2:17
Three Methods
3:05
Method 1: Distribution Functions
3:06
Method 2: Transformations
3:38
Method 3: Moment-generating Functions
3:47
Distribution Functions
4:03
Distribution Functions
4:04
Example I: Find the Density Function
6:41
Step 1: Find the Distribution Function
6:42
Step 2: Find the Density Function
10:20
Summary
11:51
Example II: Find the Density Function
14:36
Step 1: Find the Distribution Function
14:37
Step 2: Find the Density Function
18:19
Summary
19:22
Example III: Find the Cumulative Distribution & Density Functions
20:39
Step 1: Find the Cumulative Distribution
20:40
Step 2: Find the Density Function
28:58
Summary
30:20
Example IV: Find the Density Function
33:01
Step 1: Setting Up the Equation & Graph
33:02
Step 2: If u ≤ 1
38:32
Step 3: If u ≥ 1
41:02
Step 4: Find the Distribution Function
42:40
Step 5: Find the Density Function
43:11
Summary
45:03
Example V: Find the Density Function
48:32
Step 1: Exponential
48:33
Step 2: Independence
50:48
Step 2: Find the Distribution Function
51:47
Step 3: Find the Density Function
1:00:17
Summary
1:02:05
Transformations
1h 16s
Intro
0:00
Premise
0:32
Premise
0:33
Goal
1:37
Goal Number 1: Find the Full Distribution Function
1:38
Goal Number 2: Find the Density Function
1:49
Goal Number 3: Calculate Probabilities
2:04
Three Methods
2:34
Method 1: Distribution Functions
2:35
Method 2: Transformations
2:57
Method 3: Moment-generating Functions
3:05
Requirements for Transformation Method
3:22
The Transformation Method Only Works for Single-variable Situations
3:23
Must be a Strictly Monotonic Function
3:50
Example: Strictly Monotonic Function
4:50
If the Function is Monotonic, Then It is Invertible
5:30
Formula for Transformations
7:09
Formula for Transformations
7:11
Example I: Determine whether the Function is Monotonic, and if so, Find Its Inverse
8:26
Example II: Find the Density Function
12:07
Example III: Determine whether the Function is Monotonic, and if so, Find Its Inverse
17:12
Example IV: Find the Density Function for the Magnitude of the Next Earthquake
21:30
Example V: Find the Expected Magnitude of the Next Earthquake
33:20
Example VI: Find the Density Function, Including the Range of Possible Values for u
47:42
Moment-Generating Functions
1h 18m 52s
Intro
0:00
Premise
0:30
Premise
0:31
Goal
1:40
Goal Number 1: Find the Full Distribution Function
1:41
Goal Number 2: Find the Density Function
1:51
Goal Number 3: Calculate Probabilities
2:01
Three Methods
2:39
Method 1: Distribution Functions
2:40
Method 2: Transformations
2:50
Method 3: Moment-Generating Functions
2:55
Review of Moment-Generating Functions
3:04
Recall: The Moment-Generating Function for a Random Variable Y
3:05
The Moment-Generating Function is a Function of t (Not y)
3:45
Moment-Generating Functions for the Discrete Distributions
4:31
Binomial
4:50
Geometric
5:12
Negative Binomial
5:24
Hypergeometric
5:33
Poisson
5:42
Moment-Generating Functions for the Continuous Distributions
6:08
Uniform
6:09
Normal
6:17
Gamma
6:29
Exponential
6:34
Chi-square
7:05
Beta
7:48
Useful Formulas with the Moment-Generating Functions
8:48
Useful Formula 1
8:49
Useful Formula 2
9:51
How to Use Moment-Generating Functions
10:41
How to Use Moment-Generating Functions
10:42
Example I: Find the Density Function
12:22
Example II: Find the Density Function
30:58
Example III: Find the Probability Function
43:29
Example IV: Find the Probability Function
51:43
Example V: Find the Distribution
1:00:14
Example VI: Find the Density Function
1:12:10
Order Statistics
1h 4m 56s
Intro
0:00
Premise
0:11
Example Question: How Tall Will the Tallest Student in My Next Semester's Probability Class Be?
0:12
Setting
0:56
Definition 1
1:49
Definition 2
2:01
Question: What are the Distributions & Densities?
4:08
Formulas
4:47
Distribution of Max
5:11
Density of Max
6:00
Distribution of Min
7:08
Density of Min
7:18
Example I: Distribution & Density Functions
8:29
Example I: Distribution
8:30
Example I: Density
11:07
Example I: Summary
12:33
Example II: Distribution & Density Functions
14:25
Example II: Distribution
14:26
Example II: Density
17:21
Example II: Summary
19:00
Example III: Mean & Variance
20:32
Example III: Mean
20:33
Example III: Variance
25:48
Example III: Summary
30:57
Example IV: Distribution & Density Functions
35:43
Example IV: Distribution
35:44
Example IV: Density
43:03
Example IV: Summary
46:11
Example V: Find the Expected Time Until the Team's First Injury
51:14
Example V: Solution
51:15
Example V: Summary
1:01:11
Sampling from a Normal Distribution
1h 7s
Intro
0:00
Setting
0:36
Setting
0:37
Assumptions and Notation
2:18
Assumption Forever
2:19
Assumption for this Lecture Only
3:21
Notation
3:49
The Sample Mean
4:15
Statistic We'll Study the Sample Mean
4:16
Theorem
5:40
Standard Normal Distribution
7:03
Standard Normal Distribution
7:04
Converting to Standard Normal
10:11
Recall
10:12
Corollary to Theorem
10:41
Example I: Heights of Students
13:18
Example II: What Happens to This Probability as n → ∞
22:36
Example III: Units at a University
32:24
Example IV: Probability of Sample Mean
40:53
Example V: How Many Samples Should We Take?
48:34
The Central Limit Theorem
1h 9m 55s
Intro
0:00
Setting
0:52
Setting
0:53
Assumptions and Notation
2:53
Our Samples are Independent (Independent Identically Distributed)
2:54
No Longer Assume that the Population is Normally Distributed
3:30
The Central Limit Theorem
4:36
The Central Limit Theorem Overview
4:38
The Central Limit Theorem in Practice
6:24
Standard Normal Distribution
8:09
Standard Normal Distribution
8:13
Converting to Standard Normal
10:13
Recall: If Y is Normal, Then …
10:14
Corollary to Theorem
11:09
Example I: Probability of Finishing Your Homework
12:56
Example I: Solution
12:57
Example I: Summary
18:20
Example I: Confirming with the Standard Normal Distribution Chart
20:18
Example II: Probability of Selling Muffins
21:26
Example II: Solution
21:27
Example II: Summary
29:09
Example II: Confirming with the Standard Normal Distribution Chart
31:09
Example III: Probability that a Soda Dispenser Gives the Correct Amount of Soda
32:41
Example III: Solution
32:42
Example III: Summary
38:03
Example III: Confirming with the Standard Normal Distribution Chart
40:58
Example IV: How Many Samples Should She Take?
42:06
Example IV: Solution
42:07
Example IV: Summary
49:18
Example IV: Confirming with the Standard Normal Distribution Chart
51:57
Example V: Restaurant Revenue
54:41
Example V: Solution
54:42
Example V: Summary
1:04:21
Example V: Confirming with the Standard Normal Distribution Chart
1:06:48
• ## Transcription
1 answerLast reply by: Dr. William MurrayTue Sep 18, 2018 4:32 PMPost by Said Sabir on September 15, 2018I am confused between the sampling distribution and probability distribution, could you please explain the difference, and what is the use case of each?
### The Central Limit Theorem
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Setting 0:52
• Setting
• Assumptions and Notation 2:53
• Our Samples are Independent (Independent Identically Distributed)
• No Longer Assume that the Population is Normally Distributed
• The Central Limit Theorem 4:36
• The Central Limit Theorem Overview
• The Central Limit Theorem in Practice
• Standard Normal Distribution 8:09
• Standard Normal Distribution
• Converting to Standard Normal 10:13
• Recall: If Y is Normal, Then …
• Corollary to Theorem
• Example I: Probability of Finishing Your Homework 12:56
• Example I: Solution
• Example I: Summary
• Example I: Confirming with the Standard Normal Distribution Chart
• Example II: Probability of Selling Muffins 21:26
• Example II: Solution
• Example II: Summary
• Example II: Confirming with the Standard Normal Distribution Chart
• Example III: Probability that a Soda Dispenser Gives the Correct Amount of Soda 32:41
• Example III: Solution
• Example III: Summary
• Example III: Confirming with the Standard Normal Distribution Chart
• Example IV: How Many Samples Should She Take? 42:06
• Example IV: Solution
• Example IV: Summary
• Example IV: Confirming with the Standard Normal Distribution Chart
• Example V: Restaurant Revenue 54:41
• Example V: Solution
• Example V: Summary
• Example V: Confirming with the Standard Normal Distribution Chart
### Transcription: The Central Limit Theorem
Hi, welcome back to the probability lectures here on www.educator.com, my name is Will Murray.0000
This is our very last probability lecture, I’m want to say a special thank you0006
to those of you who stuck with me through all the videos.0010
Today, we are going to talk about the central limit theorem which is one of the crown jewels of probability.0014
I'm very excited to talk about the central limit theorem and show you how it plays a role in sampling.0019
We will be doing a lot of problems, solving questions about samplings.0026
I need to give you the background here.0031
It starts out just like the previous video.0033
If you watched the previous video, sampling from a normal distribution, then the first slide is going to be exactly the same.0036
You can safely skip that and then, I will show you what the difference is when we are using the central limit theorem .0044
Let us jump into that, the setting here, like I said, this is exactly the same as in the previous video, at least for the first slide.0051
The idea is that we have a population of stuff.0061
For example, we could have a whole bunch of students at a university and each student is a different height.0063
We have some distribution of heights at the university.0071
There are some population mean which means the average of all the students at the university.0074
We might or might not know that, that μ might be known or it might not be known.0080
There is some variance σ².0085
In the problems that we are going to solve today, we will need to know what the variance is.0088
That should be given to you in the problems.0092
And then, we are going to take some samples which means we are going to go out in the quad of the university.0094
We will stop some students randomly and survey them on how tall they are.0102
Or if we do not like measuring how tall people are, you can ask them how many units they are carrying0107
or how much student that they have, or what their bank balance is, or other GPA.0113
It does not really matter for the purposes of probability, what quantity we are keeping track of,0119
the important thing is that we are taking random samples.0124
The way we are going to keep track of them is, each student that we talked to counts as 1 random variable.0130
For example, if we are talking about the heights of the students then Y1 is the height of the first student,0137
Y2 is the height of the second student, and so on, until nth student.0142
If we want to survey N students then YN is the height of the last student.0148
Each one of those counts as a random variable and we will calculate the average of those samples.0154
We will talk about probability questions related to whether the average of our sample0161
is really close to the average of the entire population.0166
That is all the same as in the previous lecture.0171
What else is the same as the previous lecture is that our samples are independent.0174
There is this catch phrase that you hear in probability a lot and in statistics,0179
independent identically distributed random variables.0184
They are independent meaning that, if we meet 1 student and that student is very tall,0188
it does not really tell us that the next student is going to be tall or short because they are independent.0194
Identically distributed means they are all coming from the same population.0200
That is a buzz phrase to say independent identically distributed random variables.0206
Here is where this lecture is different from the previous lecture.0212
In the previous lecture, we have to assume that our population was normally distributed,0216
which is not really valid when you are talking about heights of students.0221
Because one thing, a student can never have a negative height so it is not really normally distributed.0225
In this lecture, using the central limit theorem, which I have not gotten to yet,0232
we do not have to assume that the population is normally distributed.0236
The beautiful thing about the central limit theorem is that the population could have any distribution at all.0241
The central limit theorem is very broad and it applies to any distribution at all.0247
In particular, when you are doing sampling, you do not have to know the general parameters of the population at all.0254
You do not need to know that your population is a normal distribution.0261
That is the key feature of the central limit theorem is that,0265
you do not need to know ahead of time what kind of distribution you are working with.0269
Let me actually tell you, what the central limit theorem is.0275
It says that, if you have independent samples from any population with mean μ and variance σ²,0279
these are IID independent identically distributed samples.0287
The conclusion of the central limit theorem says that, Y ̅ is the sample mean.0292
That means you take the samples that you collect and you take their average,0300
just of those samples, and that is a new random variable.0309
That is a function of the random variables you had before.0313
It says that, the distribution of that random variable approaches as N goes to infinity,0316
it gets closer and closer to a normal distribution with mean μ and variance σ²/N.0324
This is really one of the most extraordinary facts in all of mathematics,0335
which is that we did not assume that the original population was normally distributed.0340
But, even without that assumption, the sample mean approaches a normal distribution.0347
This is sort of why the bell curve, the normal distribution is considered the most important distribution in all probability and statistics.0356
you always end up with a normal distribution, as you take samples and you look at the sample mean.0371
That is really quite extraordinary but the math is very powerful and it does work out that way.0379
Let me mention that in practice, it is kind of the rule of thumb that people use in practice0386
when applying the central limit theorem is that, it starts to kick in, remembering it applies as N goes to infinity.0398
It really starts to become useful when N is bigger than about 30.0408
N is the number of samples that you take.0416
If you take more than 30 samples, you can safely assume that your sample mean will follow a normal distribution.0418
We can invoke the central limit theorem and say that the sample mean will have a normal distribution,0430
a normal distribution with mean μ and variance σ²/N.0446
That is kind of how the central limit theorem is used in practice.0451
As long as you take at least of 30 samples then, you can say that your sample mean is going to have a normal distribution.0456
It does not even matter, what the distribution of your original population was.0466
What do we actually do with that, once we know that the sample mean has a normal distribution,0471
what are we supposed to do with that.0476
From then on, it is pretty much the same as in the previous lecture.0478
We can walk you through that, in case you did not just watch the previous video.0483
What you do with a normal distribution is you convert it to a standard normal distribution.0488
A standard normal distribution, I will remind you is a normal distribution with mean 0 and variance 1.0494
It is what it means to be a standard normal distribution as mean 0 and variance 1.0507
The point of the standard normal distribution is that, you can look up probabilities for a standard normal distribution using charts.0513
Of course, there also lots of online applets that you can use,0521
a lot of computer programs will know tell you probabilities for a standard normal distribution.0525
For the videos in this lecture, I'm going to use charts.0531
If you are lucky enough to have access to the some kind of online tool or computer program,0536
that will tell you probabilities for a standard normal distribution then by all means, have at it and use that.0541
This is kind of an archaic method that I'm showing you here.0550
But still, in a lot of classroom settings people still use chart that is why I’m showing it to you.0553
That is how you can look up probabilities for a standard normal distribution.0558
This picture is one that is really useful to keep in mind.0563
This chart, the way it works is, it tells you the probability of being above a certain cutoff.0570
If you want to find the probability of being less than that cutoff, then you have to do something like subtracting from 1.0577
If you want to be between 2 cutoffs, then you have to figure out the probability of being in the tails0583
and then subtract those from 1.0589
That is a kind of computations that you have to do to use these charts.0591
That is all based on a standard normal distribution.0596
In practice, you usually do not get a standard normal distribution.0602
You usually get some kind of random normal distribution.0606
Let me show you how you convert it to a standard normal distribution.0609
I say recall because I did a whole lecture on this, earlier on in these probability lectures.0615
If this is totally new to you, what you might want to do is go back0621
and work through the lecture on the normal distribution, that we studied earlier on in this lecture series.0624
But if you already worked through that lecture, maybe you just need a quick refresher, here you go.0631
What we learned back then was that, if Y is any normal distribution then0635
what you can do is convert it to a standard normal distribution,0642
by subtracting off the mean and dividing by its standard deviation.0645
We call that new variable Z, and what we learned is that Z is a standard normal distribution.0651
The point of getting a standard normal distribution is then,0660
you can look up probabilities in terms of Z and convert them back to find probabilities in terms of Y.0662
What we learned in our central limit theorem was that, Y ̅ is essentially,0670
it approaches a normal distribution with mean μ and variance σ²/N.0679
What that means is that, if we do Y ̅ and we subtract its mean, that is Y ̅ - μ and divide by,0687
its standard deviation is always the square root of its variance.0697
I have to do √ σ ⁺N/N, I’m going to call that Z.0702
Now, you notice if I take the denominator and flip it upside down because of the fraction in the denominator,0710
then I will get σ², √σ² is just σ.0719
√N is going to flip up to the numerator, that is why I get that √N/σ × Y – μ.0726
That is where I’m getting this expression right here, that is where that comes from.0734
The variable that we just created is a standard normal variable.0741
I can use the charts to look up probabilities for that standard normal variable.0745
That is how I’m going to be solving the examples.0751
I’m going to ask you some kind of question about Y ̅,0755
and then what we will do is we will build up the standard normal variable, translate it into a question about Z.0760
And then, we will use the charts to look up probabilities on Z, that is how that is going to work.0768
Let us jump into the exercises and practice that.0775
In example 1, this is a very realistic problem.0778
Homework problems take you an average of 12 minutes each, but there is a lot of variation there,0782
there is a standard deviation of 10 minutes.0787
Maybe, if you get a real quick problem, you can quickly dispense of it in 2 minutes.0790
Or if you get a really tough one, it could take you 22 minutes or possibly even longer.0797
Your assignment is to solve 36 problems, this is going to take awhile.0802
What is the probability that it will take you more than 9 hours?0806
Let us think about that, first of all I have a conversion to solve here.0811
9 hours is 9 hours × 60 minutes per hour, that is 540 minutes.0819
If I'm going to take 540 minutes, I want to convert that into an average .0838
Remember, what I want to use is that Z is √N/σ × Y ̅ – μ.0845
I know that, that will be a standard normal variable.0858
Somehow, I got to buildup those quantities.0861
If my total time spent on the homework is going to be 9 hours, that is 540 minutes.0864
How much time will that be per problem on average?0871
My Y ̅, which is the average time per problem, if I spent 540 minutes total then0875
that would be 540 minutes divided by 36 problems.0888
I rigged up those numbers to work fairly nicely.0894
That is 15 minutes per problem.0898
I want to know the likelihood that I'm going to end up spending more than 15 minutes per problem,0909
over a 36 problem assignment.0915
I want to find the probability that my Y ̅ is bigger than 15.0922
Somehow, I want to build up this standard normal variable so I can use my normal distribution charts to solve this.0935
If Y ̅ is bigger than 15, that mean Y ̅ - μ is bigger than, what was my μ?0942
My μ is the average of all homework problems, 12 minutes on average,0951
is what it takes me to solve a homework problem.0958
15 -12, I will put in a σ and what is my σ?0962
My σ is standard deviation, that is 10.0973
The last ingredient here is √N, √N I'm going to include that.0980
What is my √N, N is the number of problems that I have.0991
N was 36, the √N is 6, this is 3 × 6/10, 18/10 is 1.8.0995
That was my variable Z, I want to find the probability that Z, my standard normal variable is going to be bigger than 1.8.1007
I'm going to look that up on the next page because I have a standard normal chart all set to go, on the next page.1021
From the chart on the next page, and I will show you in a moment where that comes from.1035
The probability what was it, it was 0.0359 is what we are going to find on the next page.1040
If I want to think about in terms of percentages, that is just about 3.6%.1052
That is my probability that I'm going to spend more than 9 hours on this homework assignment.1060
It was about 3.6% chance that I'm going to spend more than 9 hours on this homework assignment.1068
Maybe, I’m worried because I have something I need to do in 9 hours.1074
I’m worried I would not get finished in time.1076
It actually looks pretty good, it looks like there is more than 96% chance that I will finish on time, that is kind of reassuring.1080
That is the answer to the problem, there is 3.6% chance that we will spend more than 9 hours on this homework assignment.1088
Let me just recap the steps there, before I show you that one missing step of looking it up on the chart.1095
We want to, first of all, convert into a standard unit here.1102
I got 9 hours and I got 12 minutes, I decided to convert the hours into minutes.1107
You can also convert the other way, if you wanted, but I think it is a little easier this way.1114
9 hours is 540 minutes, that was a total on our time that I would spend doing all the homework problems.1119
Since, I know that I have a result about averages here, I wanted to convert that into an average amount of time.1128
The average time is 540 minutes divided by 36 problems and that is just 15 minutes per problem.1138
The question is really, how likely am I to spend an average of more than 15 minutes per problem?1148
I want to find the probability that Y ̅ is bigger than 15.1154
For Y ̅ to be bigger than 15, I want to build up this expression Y ̅ - μ/σ × √N.1159
I filled in my μ is 12, μ right there is the average.1168
My σ was 10, there is σ and I got that from the problem here.1176
My √N is 6, that comes from N = 36, the number of problems that we have to solve here.1185
Then, I just simplified the numbers 15 -12 is 3, 3 × 6 is 18, 18 divided by 10 is 1.8.1193
I want to find the probability that my standard normal variable is bigger than 1.8.1199
That is what I’m going to confirm on the next page.1205
I will show you where that comes from, but we will see that it comes out to 0.0359.1207
If you think about that as a percentage, that is just about 3.6%.1214
I just want to confirm the result that we used on the previous page.1225
What we use on the previous page was, we calculated the probability that a standard normal variable,1231
because we had converted a Y ̅ to a standard normal variable, was bigger than 1.8.1236
I’m going to look up 1.8 on this chart, 1.80.1245
There is 1.8, there is 1.80, it is 0.0359 which was the number that I gave you back on the previous slide.1249
This shows you where that come from, it just come from this chart.1260
.0359, that is where I got that answer of 3.6% that we used on the previous slide.1263
That just completes that little gap that we had on the previous side.1274
That totally answers our probability of having to spend more than 9 hours on this horrible homework assignment.1278
In example 2, we have a bakery that is charting how many muffins do they start per day.1288
We have figure out a long-term average of 30 muffins per day but there is a lot of variation in there.1295
Maybe, they sell more muffins on the weekend and fewer on a weekday.1300
They figure out that, there is a standard deviation of 8 muffins.1305
What they are doing is, they are planning out the next month or so, actually 36 days.1309
They are worried about, what is the chance that they will sell more than 1000 muffins?1315
Maybe, they are worried about whether they are going to have to order some more supplies,1320
some more flours, some more eggs, or something like that.1323
Or maybe, they are worry about whether they are going to make enough money.1327
They know they need to sell 1000 muffins in the next 36 days.1330
It was the kind of calculations that a business person would make.1334
We are going to answer them using the central limit theorem.1338
Let me remind you what we have, our mean theorem is that, if we start out with Y ̅ – μ, very important distinction there.1342
Y ̅ - μ × √N/σ is a standard normal variable, that is kind of our main result for this lecture.1357
We want to figure out how we can use that here.1371
I see a Y ̅ there, that Y ̅ is the average of the number of muffins we are going to sell each day.1375
If the total muffins is going to be 1000, then that means the daily average is Y ̅ which will be 1000/36,1384
because there are 36 days.1412
It does simply a bit, I can take a 4 out of top and bottom there, simplify that down to 250/9,1413
still not the nicest fraction in the world.1421
I'm going to try to build up the standard normal variable and get an answer that I can look up easily on the normal charts.1425
Y ̅ - μ is 250/9 - μ is the overall average which we figure out is 30 muffins per day, that is -30.1434
That is a little awkward, let me go ahead and try to combine those fractions.1451
I did do this one in fractions because I rigged it up so the fraction work fairly nicely,1457
something we can work out in our heads.1462
If the fractions did not work nicely, I would probably just be going to a decimal right now.1464
But, this one works nicely.1468
250/9 - 30 is 270/9, we get -20/9, you can convert that into a decimal, if you like.1471
Let me continue to build up the standard normal variable.1486
√N × Y ̅ - μ/σ, I want to be bigger than the values that we have, because we want to sell more than 1000 muffins.1490
This should have been a greater than or equal to.1507
This should be greater than or equal to -20/9.1512
I’m multiplying by √N, the √N is 36, that is because N is 36.1517
√36 is 6, and now I have a σ.1522
Σ is our standard deviation, that is 8.1531
That is because, what we are told in the problem here.1537
I think this does simple fairly well, this is - 20/8 could simplify to 5/2, 6/2 could simplify to 3/1.1542
3/9 could simplify to 1/3, we just get -5/3.1557
Now, I'm going to convert it into a decimal.1563
The point of this was that, this was a standard normal variable.1566
I want that Z to be bigger than or equal to -5/3 which is as a decimal is -1.67.1573
2/3 is about 0.67, most technically that is an approximation.1580
I do not want any pure mathematicians to complain about that.1587
It is -1.67, and now I want to figure out the probability that Z will be bigger than -1.67.1591
Let me draw what I'm going to be looking for, then we will use a chart on the next page to actually calculate that.1602
-1.67 was down here somewhere.1610
I’m looking for the probability that Z is bigger than that.1617
I’m looking for all that probability.1619
The way the chart works is it will tell me probabilities of being bigger than a certain cutoff.1622
What I'm going to do is find the probability that Z is bigger than 1.67 and then subtract that.1630
The probability that Z is bigger than -1.67 is going to be 1 - the probability that Z is bigger than +1.67.1639
This is what we are looking for, this probability that Z is bigger than -1.67.1660
But I can figure it out as 1- this probability, the probability that Z is bigger than 1.67.1670
That is how I'm going to calculate that out.1681
The rest of it is simply a matter of looking it up on the chart, because that is the form that I can look things up on the chart.1684
I have already done this, if I look it up on the chart on the next page.1690
Let me tell you what the answer comes out to be.1696
It is 1- 0.0475 and that comes from the chart on the next page.1699
And then, 1- 0.0475 is 0.9525, and that is approximately 95%.1708
It is in fact very likely that, this bakery is going to sell more than 1000 muffins in the next 36 days.1721
If they are planning on buying supplies, buying flour, eggs for their muffins,1730
then they better go out and buy more supplies because it is very likely that they will sell more than 1000 muffins.1735
If they are worried about revenue then things are looking pretty good,1742
because there was a good chance that they will sell more than 1000 muffins.1745
Let me recap the steps here.1750
There is one missing step which is the chart, which I will fill in on the next slide.1751
In the meantime, the total muffins, we want that to be bigger than 1000,1757
which means the daily average should be bigger than 1000/36, which is more than 250/9.1762
I was just reducing the fractions there.1770
I rigged this one up to give us nice fractions.1772
And then, I kind of built up this standard normal variable.1775
I subtract a μ, the μ was the average of 30 muffins per day, that comes from there.1779
That is where that 30, and subtracted it and I got a negative number.1787
It is significant that it is negative there.1792
We do want to keep track of the negative sign.1795
And then, I multiply by √N which was, there is my N is 36.1798
√N is my 6 right there, divided by σ which is the standard deviation, 8 muffins right there, there is my 8.1804
And then, I just did some simplifying fractions there, got down to -5/3.1813
And I convert that back into a decimal which is -1.67.1820
In order to find the probability of Z being bigger than -1.67, I flipped it around and I calculated that the probability of being in this tail.1827
That is the probability of being in the tail, the probability that Z is bigger than 1.67.1840
For that, I’m going to use the chart on the next page.1847
I hope I have been reading the chart correctly.1849
When we look on the next page, it really will be 0.0475.1852
It will work out then to be 1- that is 95% chance that this bakery will sell more than 1000 muffins.1856
That is just filling that one missing step from the chart.1867
This is the normal distribution chart, this will tell you the probability that in normal variable,1871
we will end up being bigger than a particular cutoff.1878
In this case, our cutoff is 1.67, we are using that to solve the problem on the previous slide.1881
The probability that Z is bigger than 1.67, here is 1.6 and the second decimal place is 7, it is right there.1887
I hope this works out, 1.6 and 0.0475, that is what we use before.1899
It is 0.0475 and that was the answer that we plugged into our calculations on the previous side.1907
Just take this answer, drop into the calculations on the previous side.1918
And then, we did some more work and we got our answer to be 95%.1923
That fills in the one missing step from the previous slide.1932
It is just a matter of taking this 1.67 and matching up 1.6 and 0.07, and finding the right probability.1938
Of course, if you are using electronic tools, you probably do not need to use this chart.1948
You can just ask what is the probability that a standard normal variable will be bigger than 1.67,1953
and it should just spit out the answer for you.1959
In example 3, we have a technician fixing a soda machine.1963
It is supposed to give a certain amount of soda and she wants to check out whether it is dispensing the right amount of soda.1969
She takes 100 samples and the standard deviation in this machine is 2.5 ml.1976
We want to find the chance that her sample mean, that is the Y ̅,1986
that is the mean of her 100 samples is within 0.5 ml of the true average amount.1990
That is the true population mean, that is the μ of soda dispensed.1997
Let me setup that one out for you.2002
The whole point of this lecture is that, we look at Y ̅ - μ and we convert that to a standard normal variable.2005
The way we do that is by multiplying by √N/σ.2016
That turned out to be a standard normal variable meaning it has mean 0 and standard of deviation 1.2022
What we want to is figure out Y ̅ – μ.2032
In this case, we want the sample mean to be within 5 ml of the true mean.2038
Within 5 ml means it could go 0.5 ml either way.2044
I’m going to put absolute values here and set it less than or equal to 0.5 ml.2051
I’m going to try to build up my standard normal variable.2061
I'm going to build up by putting √N here and dividing by σ.2065
Of course, I got to do that on the other side, as well, √N and divided by σ.2071
The point of that is, that gives me a standard normal variable or actually there is a mass of values their,2078
I put the absolute values on Z as well.2084
That means, I want the absolute value of Z less than or equal to, let me fill in what I can here.2087
0.5 √N, N is 100 because we are taking 100 samples here.2093
That is × 10, √100 is 10, I rigged that up to make the numbers easy.2101
What is my σ, σ is 2.5 here.2108
10 divided by 2.5 is 4, this is 0.5 × 4 which is 2.2117
That worked out really nicely, is not it.2125
I want the probability that the absolute value of Z will be less than 2.2127
It is the same as the probability that Z is between -2 and 2.2134
Let me draw a little picture, it is always useful when you are working with these normal distributions2146
to draw a picture and figure out what it is you are actually calculating.2150
I want Z to be between -2 and 2, there is -2 and 2.2155
I want to be in between there.2162
The way my chart is set up, your chart might be different, but the way my chart works is,2164
it will tell you the probability of Z being in the positive tail.2170
It will tell you that area right there.2175
In order to find the probability in the middle, what I'm going to do is2179
calculate that probability in the tail and then subtract off 2 tails, that will give me the probability in the middle.2184
I will do 1-2 × the probability that Z is bigger than 2.2192
That should give me the probability of being between -2 and 2.2199
That is something now that I can look up on my chart,2205
or if you do not like charts and you got access to some kind of electronic tool, you can look that up more quickly.2209
Just drop the number 2 into your tool.2216
I’m going to use the chart and it is on the next page where I set up the charts.2219
Let me go ahead and tell you the numerical answer now, and then let me just confirm that on the next page.2224
I found the probability of Z being bigger than 2 from the chart.2229
It was 0.0228, and that is 1- 2 × 0.0228 is 0.0456.2234
That is 0.9544, and that is approximately 95%, just slightly over 95%.2250
That is my answer, that is the probability that this technician’s sample mean will be within 0.5 ml of the true mean.2271
Let me recap the steps there.2286
There is one missing step which is the step from the chart, which will confirm that on the next slide.2288
Just while we have the slide in front of us, let me recap the steps.2295
I set up my standard normal variable, I know that Y ̅ - μ is always × √N/σ is a standard normal.2299
I wanted Y ̅ - μ to be within 0.5 of each other.2311
Y ̅ μ should be within 0.5 of each other.2317
When we say two things are within a certain distance from each other, that really means that you want to bound their absolute value.2320
The distance from A to B is the absolute value of A – B.2329
We want the absolute value of Y ̅ - μ to be less than 0.5.2333
I just tacked on these other quantities, √N/σ.2338
The point of that was that gave me a standard normal variable, I can call that Z.2343
And then, I wanted to fill in what my √N and σ were.2350
N was 100, √N gave me 10, that is where that came from.2354
The σ was the standard deviation given there, that is my σ 2.5.2360
The numbers simplified nicely, that was me being clever, setting up nice numbers there.2367
It simplified down to 2, but now you have to think a bit more because you want the probability that Z is less than 2.2374
Absolute value, that means Z is between -2 and 2, that is the middle region here.2383
The way my chart works, it will tell you the tail region, that would not you the middle region directly.2391
What I did was, I said you could solve this by doing 1-2 tails, because there is a lower tail and there is an upper tail there.2397
I'm going to look up on the chart on the next page and we will see that the probability of Z being bigger than to is 0.0228.2407
It is 1-2 × that, 2 × that is 0.0456.2416
We finally get 0.9544, get a 95% chance that we will be within 0.5 ml.2421
By the way, example 4 is a follow up to example 3.2430
I want to make sure that you understand example 3.2435
It there is any steps in here that you are a little fuzzy on, just watch the video again and2437
just make sure that you are very clear on all the steps here,2442
because in example 4, we are going to tweak the numbers a little bit.2445
It really will help if example 3 is already very solid for you.2449
One missing step here is where that number comes from.2453
Let us fill in that step, this is still example 3 and we had one missing step2457
which was the probability that Z was bigger than 2, that was from the previous side.2464
In order to solve that, I see that there is 2.0 right there.2471
2.00 gives me 0. 0228 is what I used on the previous side.2477
I use that number on the previous side and you can catch up with all the rest of the arithmetic on the previous side.2492
If you do not remember how that worked out but we did some calculations with that.2501
We came up with an answer of 95% for this soda dispensing machine.2505
That wraps up example 3, we are going to use the same scenario for example 4.2514
I do want to make sure that you understand example 3, before you go ahead and try example 4.2519
In example 4, we have the same technician from example 3.2528
Remember, this technician is taking samples from a soda dispensing machine.2532
She wants to guarantee with probability 95% that her sample mean will be with 0.4ml of the true average.2537
This looks a lot like example 3, the difference, I will just remind you was with examples 3,2547
we had not 0.4 but a 0.5 ml tolerance.2555
Here we are restricting that to 0.4 ml tolerance.2562
We are deciding that 0.5 is not close enough, I want to get a 0.4 ml tolerance.2566
I still want to keep the probability at 95%.2572
That means, I have to change something else.2576
Since, I want a more accurate answer, that means I'm going to need to take more samples than I did before.2580
We are going to try and solve that together.2587
First thing is to figure out with this probability 95%, what is that mean?2591
Let me try to illustrate that graphically.2597
I want to find some cutoff that gives me 95% of the probability in the middle.2602
I will put a value of Z there, that is –Z.2611
I want to get 95% of the probability in between these cutoffs, for the standard normal variable.2614
This is supposed to be 95% here and I want to figure out what value of Z will give me that.2621
The way to find that, since I have a chart that will tell me how much probability is in the tail of the normal distribution,2631
what I can do is say that these tails, I got two tails here,2644
the area in 1 tail should be 1- 95%, 1 - 0.95/2, which is 0.05/2 which is 0.025.2649
I want to find a Z value, since that the probability of Z being bigger than that cutoff z is 0.025.2668
I will check this on the next slide, you will see, we will look it up together.2680
I do not want to break my flow for this slide, I will tell you right now that it comes out to be Z = 1.96.2684
I want to make sure that was the right value that I used.2694
Yes 1.96, that is the cutoff Z value that we are looking for.2696
Let me go back and show you how that factors in with all the other numbers in the problem.2702
I will just remind you that Z is our standard normal variable.2706
The way we get it is we do √N/σ × Y ̅ – μ.2711
Let us work that, I wanted out to have Y ̅ – μ.2720
I wanted those two quantities, Y ̅ is the sample mean, that is Y ̅ right there.2726
The true average of this machine is μ, I do not know what that is, by the way.2733
I want this to be within 0.4 of each other.2739
I want the difference between those two in absolute value to be less than 0.4.2741
When I want to do is build up this standard normal variable.2755
I’m going to multiply on √N/σ here.2760
I will multiply on √N/σ, as well here.2764
The point of that is this gives me my Z value, my standard normal variable.2771
My Z is less than or equal to √N × 0.4.2779
I know what σ is, my σ was given to me in example 3, that was 2.5.2786
That is 2.5 from example 3 was where that was given to us.2793
Let me fill that in, example 3 was where that information came from.2799
The N in example 3, N was 100 because we took 100 samples.2805
We cannot use that anymore because now we are trying to go for more accurate estimation.2810
We are going to have to increase the number of samples.2817
I know it is going to be increased because we have to get a more accurate answer.2820
What we are really going to do is solve this for N.2823
We are going to use this value of Z that we figured out over here, 1.96, and then we will solve for N.2831
1.96 is less than or equal to √N × 0.4/2.5.2841
Now, I’m just going to manipulate the algebra a little bit until I can get a value for N.2852
I do not think this one is going to work out particularly nicely, I did not rigged the numbers for this one very well.2860
2.5, I will multiply by both sides, I will divide both sides by 0.4, that should still be less than √N.2865
N should be greater than or equal to 1.96 × 2.5 divided by 0.4.2876
Since, I changed from √N to N, I'm squaring both sides.2888
We should square that expression.2893
That is not a number you will have work out in your head.2897
I did that on a calculator and I will show you what I got there.2901
I got 150.063, just slightly over 150 there.2905
N is the number of samples that we are going to take.2916
It is got to be a whole number, because you cannot take half of the sample.2919
In order to make this work, I need a whole number bigger than 150.6063.2923
N = 151, I will round that up.2931
You always round up, if you are talking about the number of samples.2936
The samples is enough, I solved that except for one detail of showing you on the chart where that Z value came from.2940
I will go back over the steps here and then we will jump forward to the chart.2952
I will show you where that Z value came from.2957
To go back to be getting here.2960
I start out with this probability of 95%.2964
I’m going for a 95% probability here.2967
In order to get 95% in the middle, that means my values on the tail, there is two tails, they are going to split the left over.2972
The left over probability is 1 - 0.95/2.2982
1-0.95 is 0.05, .05/2 is 0.025.2988
I’m looking for cutoff value of z, that when I find the probability bigger than that, it is 0.025.2994
It will come from the chart on the next slide.3003
L will fill the part in, if you are willing to be a little patient with me, or you can skip ahead and see the chart on the next slide.3007
Then, I would just going to hang onto that Z for a little while.3015
I'm going to go back and I'm going to build up my standard normal variable here, that comes from this formula here.3018
That is supposed to be a standard normal distribution.3023
I started off with Y - μ and their absolute value, that comes from the word within here.3029
I want two things to be within 0.4 ml.3035
I want the absolute value of less than 0.4.3039
And then, I built up my √N/σ.3042
I do not know what the √N is because I have not been told how many samples I want at this point,3046
that is what I'm solving for.3051
The N is the variable but my σ, I was given the standard deviation in example 3.3054
I dropped that in, it is 2.5.3059
I fill in my Z value, that comes from over here.3062
That is where that Z value came from.3066
I just take this equation and I solve it for N.3071
That is just a matter of manipulating the algebra around, squaring both sides because I had a √N.3075
I get N = 150.063, and I need it to be a whole number.3080
To be safe, I round it up.3089
Because 150 samples would not be quite enough, I have to go for 151 samples.3092
And then, I know with probability 95% that my sample mean will be close to the true mean.3098
One missing step here is where that 1.96 came from and it comes from this 0.025.3106
But, I'm using the chart that we will see on the next slide.3114
I want to fill in that one missing step from example 4.3118
We want to find a Z for which the probability of Z being bigger than that Z is 0.025.3122
We work that out from the previous slide, that was what we are looking for.3130
Let me draw a little picture of what we are dealing with.3139
We want to find a Z, says that tail probability there is 0.025 which means I have to look for 0.025 in my chart.3143
I’m going to start here, they are getting smaller 0.7, 0.6, 0.5, 0.4, 0.3, 0.02.3158
It is getting close, 0.027, 0.026, 0.0256, 0.0250, there it is right there.3172
I look at where row and column that happened in, 1.96 and 0.06, that tells me that my Z value is 1.96.3182
You might also have electronic applets, you do not have to do this kind of old fashioned method of looking up on charts.3197
That is totally fine with me, if you are okay using electronic tools in your class.3205
You can jump from 0.025 to 1.96, that is totally fine with me.3211
This 1.96, we went on and use that in our calculations on the previous side.3217
Somehow, that work out on the previous side to tell us that we need N = 151 samples,3228
in order to guarantee a particular accuracy of our sample mean.3237
Just to recap there, most of the work was done on the previous side.3245
I figure out on the previous side that I was looking for cutoff value of Z,3249
such that the probability of being bigger than that was 0.025.3253
We just look through this chart until I found 0.025, found that in the 1.9 row and the 0.06 column.3257
I put those together and I get 1.96.3265
The rest of it goes back to the previous side where we threw that 1.96 in a bunch of calculations3268
and came back to N = 151 samples.3275
In our final example here, we have a restaurant that is worried about how much money it is going to make tonight.3282
It has done some studies and it is found that its customers spend an average of \$30.00 per customer.3294
But, they have a standard deviation of \$10.00 which means,3301
maybe if somebody just has an appetizer and a drink, maybe they will spend \$20.00.3303
Maybe, if they really go for the full menu and have drinks and desserts, and a few different extras,3308
then they are going to end up spending \$40.00 or even more.3317
The restaurant, their average is \$30.00 and they have 25 reservations tonight.3320
I guess the reservation only restaurant, you cannot just walk in here, you have to have a reservation.3326
They are expecting 25 customers tonight.3331
They want to know the chance that their total revenue tonight, we are not worried about profit,3334
we are not worried about what we are spending on supplies, total revenue will be between \$725 and \$800.3338
Let me show you how this turns into a central limit theorem problem.3348
Because it is not totally obvious right now, we are talking about total revenue.3353
Let me show you here, let me remind you of what we are given at the beginning of this lecture.3356
We are given that Y ̅ is sample mean – μ the global mean, divided by σ the standard deviation, and multiplied by √N.3361
N is the size of the sample.3378
The point of that was that would give you a standard normal variable.3381
We are going to call it Z and that is a standard normal variable.3386
In turn, the point of a standard normal variable is it is very easy to calculate probabilities.3392
The way I’m doing it is I'm using charts.3397
You might use charts for your class or you might have more sophisticated electronic tools, and that is okay with me.3400
What is this have to do with this restaurant?3407
They want to make between \$725 and \$800 total.3409
If the total is going to be between 725, I said they want to make between that.3416
Of course, they would be happy if they made more.3425
They are worried about, they want calculate how likely is it that they will make between 725 and 800 total.3428
What we have here is a result that has to do with the mean, the sample mean.3434
How do we convert that into a mean?3440
We just divide by the customers, the number of customers,3442
and convert that into an average amount that each customer would spend.3448
The mean, the average, Y ̅ would have to be between 725 divided by 25 and 800/253453
because that is how much the average customer would have to spend, in order to get the total between 725 and 800.3468
I just did a little arithmetic here, I rigged this up so that the numbers came out fairly nicely.3476
800/25 is 32 and 725/25 is 28, 725/25 is 29.3482
Y ̅ would have to be between 29 and 32.3495
What that means is, all the customers that come in tonight,3501
they would have to spend an average of between \$29.00 and \$32.00.3504
It does not mean they will have to spend between that,3510
but it means you can still have some big spenders to come in and drop 50 bucks on a meal.3512
You can still have some cheapskates to just buy an appetizer and then slid out of there after spending \$10.00.3517
But on average, it has to come out between \$29 and \$32.00 per customer for tonight's customers.3523
What I would like to do is kind of buildup that standard normal variable.3534
Y ̅ - μ would be between, my μ is my global average.3540
That is right there, that is the 30, that is how much a customer spend on the average in the long term.3547
That is 29 -30 and 32 -30, let me go ahead and divide by σ.3554
My σ is the standard deviation, there it is \$10.00 right there.3569
I also need to multiply by √N, I did not really leave myself enough space to do that.3577
I will give myself another line there.3583
Y ̅ - μ/σ × √N, this not absolute value, this is not one of those within problems like the previous one.3584
We have to be careful about what is positive, what is negative, no absolute value here.3595
√N, N is 25, that is the number of customers we are going to be working with tonight.3599
√N is 5, 5 × 29 -30/10 and 5 × 32 -30/10.3604
That simplifies fairly nicely, 5/10 is ½, ½ × -1 is - ½.3619
I will write that as -0.5.3628
And then, the point of this was we are building up a standard normal variable.3631
That is my Z right there, this is between -0.5 and 5/10 is still ½, 32 -30 is 2, 2 × ½ is just 1.0.3635
I have a standard normal variable and I want to find the probability that it is between -1/2 and +1.3651
Let me draw a graph of what I’m looking for.3659
Possibly, if you have the right electronic tool, you can jump to the answer at this point.3663
Just drop these numbers in your electronic tool, but let me show you how you can figure out using your charts.3668
There is -0.5 and there is 1.0.3675
I want to find the probability of being in between those two.3680
What my chart will do is, it will tell me the probability of being in a tail.3685
It will tell me that probability right there.3690
It will also tell me that probability right there, but those are not the same because 0.5 and 1.0 are not symmetric.3693
This is the probability that Z is greater than 1.0.3701
This is the probability that Z is less than -0.5, but it is also the same as Z being bigger than 0.5.3707
What I really want here is, my probability that Z is between -0.5, 0.5, and 1.0.3716
Let me write that a little more clearly, - 0.5 and 1.03730
What I can do is, I can subtract off the two tails to get the probability.3741
That is 1- the probability that Z is bigger than 0.5 - the probability that Z is bigger than 1.0.3746
We have some other problems that is like this where we subtracted off the two tails.3758
Those are symmetric ones, they started out with absolute values.3761
We can just find one tail and then multiply it by 2.3765
But these tails are not symmetric, I would have to do two separate calculations and3769
look up two separate numbers on a chart there.3775
Let me say that I will do the steps on the chart on the next page.3780
I will just tell you what the answers are for now, and then I will prove to you by showing you the chart on the next page.3785
The probability that Z is bigger than 0.5, I look that up on my chart, I got 0.3085.3793
The probability that Z is bigger than 1.0 was 0.1587.3804
Now, it is just 1 - 0.3085 0.1587.3812
I was lazy, I threw that into a calculator and what I got was a 0.5328,3817
could have done that by hand, that would have been that bad.3827
If you want to estimate that, that would be just 53%.3832
That is the probability that this restaurant, their total profit for tonight is between \$725 and \$800 tonight.3840
There is that one missing step from the chart.3853
We will confirm that on the next slide but before I turn the page from the slide, let me show you the steps here.3856
I want to find the probability that my total was between 725 and 800.3862
What I really know is a result about the average that each customer tonight is going to spend.3867
I want to convert that total into an average.3875
I just divided it by the number of customers.3878
The total divided by the number of customers gives me the average.3883
725/25 gives 29, 800/25 gives me 32.63887
And then, I start to build up this formula for a standard normal variable.3894
I subtracted μ from both sides, that μ was the average that all the customers in the world spend at this restaurant.3899
I subtracted 30 from both sides and then I divided by σ, where is my σ, there is my σ right there, the standard deviation.3908
I divided by σ and then I multiply by √N on the next line.3919
N was the number of customers, there is 25 of them, I’m going to multiply both sides by 5.3926
5 is the √25.3934
I simplified the numbers here, they have worked out pretty nicely.3936
They are rigged to work nicely, simplified down to -0.5 and +1.0.3940
We are really looking for the probability between -0.5 and 1.0 for a standard normal variable.3947
The way my chart works and some people's chart work a little differently,3959
but the way my chart works is it will tell you these tail probabilities.3959
It tells you the positive tails but you can work out the negative tails the same way.3964
What I will do is, I will look up the two different tails there and subtract them off from 1.3969
That will give me this probability in the middle, that I'm really looking for.3976
I will confirm those on the next page with the chart, it is 0.3085 and 0.1587.3982
Once I look those up, I can drop them back into the calculation and just reduce it down to 0.5328 which is about 53%.3989
That is my probability that my restaurant is going to make between \$725 and \$800 tonight.3999
The one missing piece of the puzzle from the previous slide,4009
we are still answering example 5 now, is to find those two probabilities.4012
We were finding the probabilities of being less than - 0.5 or being bigger than 1.0.4020
We actually want to find the probability of being between those cutoffs.4029
This chart will tell you the probability of being in the tails.4033
The probability of Z being less than -0.5 is the same as being bigger than 0.5.4039
It should be in this chart somewhere, here it is 0.5 and 0.00 is 0.3085.4048
The probability of Z being bigger than 1.0, here is 1.0, it is 0.1587.4060
I took those numbers and I do not think I'm going to rehash all the calculations that I did on the previous side.4078
I will just say, you plug those numbers in to the appropriate place on the previous side.4084
You can go back and watch it, if you do not remember how it works out.4089
It would be good if I can spell previous though.4094
We work through some calculations and we came up with a 53% probability4101
that this restaurant is going to make between \$725 and \$800 in their nightly revenue.4107
That wraps up example 5, most of the work was done on the previous side.4118
You can go back and check it out, just the missing step on the previous side was4125
where these two numbers came from, the 0.3085 and 0.1587.4129
Where they came from was by looking up 0.5 and 1.0, and then getting these two numbers from my standard normal chart.4136
If you do not like using charts, if you have an electronic way of getting probabilities4146
for a standard normal distribution, then by all means use that.4151
It is definitely something quicker than this slightly archaic method.4156
In some probability classes, they are still using charts so I want to show you that way.4160
That wraps up this lecture on the central limit theorem and4166
that is the last lecture in the probability series here on www.educator.com.4169
My name is Will Murray, I have really enjoyed making these lectures.4174
I hope you have learned something about probability.4178
I hope you have been working through the examples and learning something along with me.4180
I thank you very much for sticking with me through these probability lectures.4186
I hope you are enjoying your probability class and all your math classes, bye now.4190
OR
### Start Learning Now
Our free lessons will get you started (Adobe Flash® required). | {"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": 0.8502492308616638, "perplexity": 1430.775825557022}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669730.38/warc/CC-MAIN-20191118080848-20191118104848-00022.warc.gz"} |
https://thatsmaths.com/2019/10/10/the-wonders-of-complex-analysis/ | ### The Wonders of Complex Analysis
Augustin-Louis Cauchy (1789–1857)
If you love mathematics and have never studied complex function theory, then you are missing something wonderful. It is one of the most beautiful branches of maths, with many amazing results. Don’t be put off by the name: complex does not mean complicated. With elementary calculus and a basic knowledge of imaginary numbers, a whole world of wonder is within your grasp.
In the early nineteenth century, Augustin-Louis Cauchy (1789–1857) constructed the foundations of what became a major new branch of mathematics, the theory of functions of a complex variable.
Complex Analysis
Complex analysis is the branch of mathematics dealing with the theory of functions of a complex variable. A function ${f(z)}$ of a complex variable ${z}$ is differentiable at a point ${z_0}$ in the complex plane if the following limit exists:
$\displaystyle f^\prime(z_0) = \lim_{ z\rightarrow z_0} \frac{f(z)-f(z_0)}{z-z_0}$
This looks like the usual definition for functions of a real variable, but it is a much stronger condition: the value of the limit must be independent of the direction in which the limit is approached; that is, it must be independent of the phase of ${z-z_0}$.
Differentiability of a complex functions has powerful consequences. If the derivative of ${f(z)}$ exists in a region ${\Omega}$, then the function is said to be holomorphic in ${\Omega}$. Holomorphic functions are infinitely differentiable. This contrasts sharply from functions of a real variable, where the existence of an ${n}$-th derivative does not imply existence of the ${(n+1)}$-th derivative.
Every holomorphic function is analytic. That is, it can be represented in its domain by a power series (the Taylor series). Again, the contrast with real functions is sharp. For the real function
$\displaystyle f(x) = \exp(-1/x^2),\ \ \ x\ne0\,, \qquad f(0) = 0$
all derivatives vanish at the origin, so the Taylor series about ${x=0}$ (the Maclaurin series) is identically zero and does not represent the function.
If ${f(z) = u(x,y)+i v(x,y)}$ is analytic in ${\Omega}$, where ${u(x,y)}$ and ${v(x,y)}$ are real functions, then the following equations hold:
$\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \,, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \,.$
These are the Cauchy-Riemann Equations. We can see from these that the real and imaginary parts of ${f(z)}$ are strongly interlinked.
Line integrals
Integrals along curves in the complex plane ${\mathbb{C}}$ are of great importance in compex analysis. If the function ${f(z)}$ of a complex variable ${z}$ is analytic in ${\Omega}$ and on its boundary ${\Gamma}$ then
$\displaystyle \oint_\Gamma f(z)\,\mathrm{d}z = 0 \,.$
This is Cauchy’s Theorem. It has many consequences. One is that, for any two points ${z_1}$ and ${z_2}$ in ${\Omega}$, the integral
$\displaystyle \int_{z_1}^{z_2} f(z)\,\mathrm{d}z = 0$
is independent of the path (in ${\Omega}$) between the points.
If ${f(z)}$ is known on the boundary ${\Gamma}$ of a region ${\Omega}$ and analytic in the region, then the value of ${f(z)}$ at any point ${a}$ within the region is given using Cauchy’s integral formulas. This is somewhat amazing: the values of ${f(z)}$ and its derivatives anywhere within the region follow from a knowledge of the values of ${f(z)}$ on the boundary. These remarkable formulas are
$\displaystyle f(a) = \frac{1}{2\pi i} \oint_\Gamma \frac{f(z)}{z-a} \,\mathrm{d}z = 0$
and
$\displaystyle f^{(n)}(a) = \frac{n!}{2\pi i} \oint_\Gamma \frac{f(z)}{(z-a)^{n+1}} \,\mathrm{d}z = 0 \,.$
Laurent Series
Suppose that ${f(z)}$ is analytic in the annular region between two concentric circles ${\mathbf{C}_1}$ and ${\mathbf{C}_2}$ and ${\gamma}$ is a simple closed curve between them (see Figure). Then ${f(z)}$ can be expanded in a doubly-infinite series
$\displaystyle f(z) = \sum_{n=-\infty}^{+\infty} a_n z^n \,, \qquad\mbox{where}\qquad a_n = \frac{1}{2\pi i}\oint_{\gamma} \frac{f(z)}{z^{n+1}}\,\mathrm{d}z \quad\mbox{for}\quad n \in \mathbb{Z} \,.$
This is Laurent’s expansion. The quantity ${a_{-1} = (1/2\pi i)\oint_{\mathbf{C}} f(z) \,\mathrm{d}z}$ is called the residue (at ${z=0}$) of ${f(z)}$.
The Residue Theorem
For a simple pole at ${z=a}$, the residue is ${\lim_{z\rightarrow a} (z-a)f(z)}$. Suppose that ${f(z)}$ is analytic inside a simple curve ${\mathbf{C}}$ except for simple poles at points ${z_1, z_2, \dots ,z_N}$ within ${\mathbf{C}}$ and the residues at these poles are ${r_1, r_2, \dots ,r_N}$. Then
$\displaystyle \oint_{\mathbf{C}} f(z) \,\mathrm{d}z = 2\pi i (r_1 + r_2 + \dots + r_N) \,.$
The residue theorem is valid under more general conditions than stated here. It is of great value in evaluating definite integrals such as
$\displaystyle \int_{-\infty}^{+\infty} \frac{\sin x}{x}\mathrm{d}x = \pi$
and for summing infinite series.
Conclusion
The above results are just a taste of the wonderful richness of complex analysis. There are many excellent and accessible textbooks on this subject. Drop into the library and have fun.
Sources
${\bullet}$ Wikipedia article Complex Analysis . | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 57, "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": 0.9417566061019897, "perplexity": 132.07211240729615}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662577757.82/warc/CC-MAIN-20220524233716-20220525023716-00397.warc.gz"} |
https://brilliant.org/problems/riding-the-sinusoid/ | # Riding The Sinusoid
There is a point mass $$M$$ on a sinusoid path. The height of the curve is described by the equation $$y(x)=A\cos(x)+B$$ where $$A=B=1\text{ m}$$. The initial position of the point is $${x}_{0}=0.7$$ and starts moving from rest.
Find the period of the motion in seconds to the nearest integer.
Details and Assumptions:
• Take $$g=10 \text{ m/s}^2$$.
There is no friction between the body and the surface.
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9610893726348877, "perplexity": 273.64319448344634}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281421.33/warc/CC-MAIN-20170116095121-00168-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/116237/the-product-of-integrable-random-variables-need-not-be-integrable | # The product of integrable random variables need not be integrable
The product of integrable random variables need not be integrable
@Did gave a great example showing that in general the product of two Lebesgue integrable functions need not be integrable. I thought I could just tweak that example and prove this. However, I found the tweaking'' is not easy. Suppose random variable $X,Y\in L^{1}$. It means that $E(X),E(Y)<+\infty$ , that is, $\int XdP,\int YdP<+\infty$. I want to find an example where $\int XYdP$ is infinite.
The problem with the definition of expectation is that it is not convenient to calculate. The convenient way of computing the expectation is to use the density function : $E(X)=\int_{-\infty}^{\infty}xf(x)dx$ . However, if I use the density function, then unless $X$ and $Y$ are independent, contructing the density function for $XY$ and making $E(XY)$ infinite will be tricky.
Any suggestion? Thank you.
-
As usual: $X=Y$ with density something like $2x^{-3}\cdot[x\gt1]$. (But any integrable independent $X$ and $Y$ yield an integrable product $XY$.) – Did Mar 4 '12 at 8:37
For finding the expected value of a function of $X$ and $Y$, you never want to look at the density of that function of $X$ and $Y$. Instead, use the Law of the Unconscious Statistician. – Robert Israel Mar 4 '12 at 8:58
I have been thinking how @Didier Piau came up with good counterexamples. Then, I realized that in both counterexamples he used the properties of $1/x$: $\log(x)$ approaches infinity both at 0 and at infinity. Also, Robert Israel is right: don't mess with joint density function. Instead, since here $X=Y$ by construction, we only need to deal with $X^2$ and only the density function of $X$. Thank you both. – user16859 Mar 4 '12 at 16:23
1. Taking $X=Y$ does not limit our chances of finding a counterexample, because $XY\le X^2+Y^2$. (So, if $XY$ has infinite expected value, then one of $X^2$ and $Y^2$ does too.)
2. The focus is on the large values of $X$, because this is where squaring hurts.
3. The integral that gives the expected value of $X$ should converge, but not too rapidly (like the Gaussian), more like $\int_1^\infty x^{-p}\,dx$ with $p>1$ to be determined.
4. According to 3, the pdf of $X$ is $c_px^{-p-1}\cdot [x>1]$ where $c_p$ is normalization constant and $[\ \ ]$ the Iverson bracket.
5. Instead of computing the pdf of $X^2$, we can directly calculate $E(X^2)$ using the pdf of $X$: it is $c_p\int_1^\infty x^2x^{-p-1}\,dx=c_p\int_1^\infty x^{-p+1}\,dx$
6. The integral in 5 diverges when $p\le 2$. Thus, any $p$ in the range $1< p\le 2$ could be used. We can just as well pick the integer value $p=2$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9762129187583923, "perplexity": 189.69612585827593}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988860.4/warc/CC-MAIN-20150728002308-00052-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://laborunionreport.com/orztoa11/952131-how-to-find-eigenvalues-of-a-symmetric-matrix | Find all eigenvalues of a matrix using the characteristic polynomial. Also, determine the identity matrix I of the same order. Find the eigenvalues and a set of mutually orthogonal eigenvectors improve our educational resources. Example To find the eigenvalues of the matrix we substitute A into the equation det(A-kI)=0 and solve for k. The matrix A-kI is given by which has determinant k^2-2k-3. of the matrix A. we substitute A into the equation det(A-kI)=0 and solve for k. The matrix will have found the eigenvalues Thus, if you are not sure content located an eigenvector For a matrix A 2 Cn⇥n (potentially real), we want to find 2 C and x 6=0 such that Ax = x. <-2,1> and <3,-2>) one for each eigenvalue. Rensselaer Polytechnic Institute, Bachelor of Science, Ceramic Sciences and Engineering. To find the eigenvectors Find the eigenvalues of the symmetric matrix. eigenvalues here (x and z); thus, eigenvectors for k=-1 must have the form y=-2x-2z which Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. A symmetric matrix A is a square matrix with the property that A_ij=A_ji with eigenvalue k=-1+i. We can thus find two linearly independent eigenvectors (say Example: Find the eigenvalues and eigenvectors of the real symmetric (special case of Hermitian) matrix below. Theorem If A is a real symmetric matrix then there exists an orthonormal matrix P such that (i) P−1AP = D, where D a diagonal matrix. A-kI is given by. We now examine (A+I)v=0 to find the eigenvectors for the eigenvalue k=-1: It is easily seen that this system reduces to the single equation 2x+y+2z=0 roots k=-1+i and k=-1-i. So, of course, we have k=3 or k=-1 . Eigenvalues[m] gives a list of the eigenvalues of the square matrix m. Eigenvalues[{m, a}] gives the generalized eigenvalues of m with respect to a. Eigenvalues[m, k] gives the first k eigenvalues of m. Eigenvalues[{m, a}, k] gives the first k generalized eigenvalues. are symmetric matrices. result is a 3x1 (column) vector. EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4 . All eigenvalues are solutions diagonal. If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe of (A-I)v=0 and are thus of the form . We figured out the eigenvalues for a 2 by 2 matrix, so let's see if we can figure out the eigenvalues for a 3 by 3 matrix. Now we need to substitute into or matrix in order to find the eigenvectors. of equations Beware, however, that row-reducing to row-echelon form and obtaining a triangular matrix does not give you the eigenvalues, as row-reduction changes the eigenvalues of the matrix … St. Louis, MO 63105. Eigenvalues and eigenvectors How hard are they to find? Thus, the characteristic equation is (k-8)(k+1)^2=0 which has roots k=-1, when the eigenvalues are not distinct. In these notes, we will compute the eigenvalues and eigenvectors of A, and then find the real orthogonal matrix that diagonalizes A. Then prove the following statements. Let's say that A is equal to the matrix 1, 2, and 4, 3. Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially identity matrix, Av=v for any vector v, i.e. Most relevant problems: I A symmetric (and large) I A spd (and large) I Astochasticmatrix,i.e.,allentries0 aij 1 are probabilities, and thus only and If A is a real skew-symmetric matrix then its eigenvalue will be equal to zero. (The corresponding eigenvector is $[1~0~0~0~0]^T$.) University. the vector to stretch (or shrink) and/or reverse direction. mututally orthogonal. [Vector Calculus Home] Alternatively, we can say, non-zero eigenvalues of A are non-real. Varsity Tutors. To find the eigenvalues, we need to minus lambda along the main diagonal and then take the determinant, then solve for lambda. link to the specific question (not just the name of the question) that contains the content and a description of then the characteristic equation is . Let's find the eigenvector, v 1, associated with the eigenvalue, λ 1 =-1, first. Send your complaint to our designated agent at: Charles Cohn So, we now have two orthogonal vectors It can also be shown that the eigenvectors for k=8 If . Math 2940: Symmetric matrices have real eigenvalues The Spectral Theorem states that if Ais an n nsymmetric matrix with real entries, then it has northogonal eigenvectors. k is called the eigenvalue (or characteristic value) of the vector v. If Av=kv for scalar(s) k. Rearranging, we have Av-kv=0. Condition for block symmetric real matrix eigenvalues to be real. linearly independent. In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. A matrix P is said to be orthonormal if its columns are unit vectors and P is orthogonal. This is in equation form is , which can be rewritten as . (A-(-1+i)I)v=0 for v: The second equation is a constant multiple of the first equation so the In order to determine the eigenvectors of a matrix, you must first determine the eigenvalues. Understand the geometry of 2 × 2 and 3 × 3 matrices with a complex eigenvalue. I am struggling to find a method in numpy or scipy that does this for me, the ones I have tried give complex valued eigenvectors. The easiest ones to pick are , and . Your Infringement Notice may be forwarded to the party that made the content available or to third parties such Let's verify these facts with some random matrices: Let's verify these facts with some random matrices: which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Here, you already know that the matrix is rank deficient, since one column is zero. First, choose anything, say s=1 and t=0: <1,-2,0>. With the help of the community we can continue to contact us. A vector v for which this equation Now lets use the quadratic equation to solve for . non-zero solution for v if which has determinant k^2-2k-3. Note that we have listed k=-1 twice since it is a double root. Real number λ and vector z are called an eigen pair of matrix A, if Az = λz.For a real matrix A there could be both the problem of finding the eigenvalues and the problem of finding the eigenvalues and eigenvectors.. Let A be an n×n matrix and let λ1,…,λn be its eigenvalues. Since This will be orthogonal to our other vectors, no matter what value of , we pick. Learn to find complex eigenvalues and eigenvectors of a matrix. eigenvectors. If you have questions or comments, don't hestitate to 0. Eigenvalue of Skew Symmetric Matrix. are of the form <2r,r,2r> for any value of r. It is easy to check that Add to solve later Sponsored Links the following: Thus, the situation encountered with the matrix D in the example This equation has a So if lambda is an eigenvalue of A, then this right here tells us that the determinant of lambda times the identity matrix, so it's going to be the identity matrix in R2. Explanation: . For convenience, let's pick , then our eigenvector is. for all i and j. So for example, choosing y=2 yeilds the vector <3,2> which is thus An easy choice here is x=4 and z=-5. Matrix norm the maximum gain max x6=0 kAxk kxk is called the matrix norm or spectral norm of A and is denoted kAk max x6=0 Your name, address, telephone number and email address; and I have a real symmetric matrix with a lot of degenerate eigenvalues, and I would like to find the real valued eigenvectors of this matrix. The eigenvalues are immediately found, and finding eigenvectors for these matrices then becomes much easier. But kv=kIv where I is By definition, if and only if-- I'll write it like this. There are two parameters sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require equations (A-3I)v=0: Since the second equation is a constant multiple of the first, this system So let's do a simple 2 by 2, let's do an R2. The eigenvalue for the 1x1 is 3 = 3 and the normalized eigenvector is (c 11 ) =(1). A description of the nature and exact location of the content that you claim to infringe your copyright, in \ Find the eigenvalues and set of mutually orthogonal. this vector is orthogonal to the other two we have for any choice of r. So, Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. of the form are eigenvectors of A with eigenvalue k=-1-i. In Section 5.1 we discussed how to decide whether a given number λ is an eigenvalue of a matrix, and if Iowa State University, Bachelor of Science, Mathematics. We must choose values a (Enter your answers as a comma-separated list. This system k^2-2k-3=0. Similarly in characteristic different from 2, each diagonal element of a skew-symmetric matrix must be zero, since each is its own negative.. so … We need to take the dot product and set it equal to zero, and pick a value for , and . Massachusetts Institute of Technolog... Emory University, Bachelor of Science, Mathematics/Economics. An identification of the copyright claimed to have been infringed; Diagonalization of a 2× 2 real symmetric matrix Consider the most general real symmetric 2×2 matrix A = a c c b , where a, b and c are arbitrary real numbers. eigenvalues: eigenvectors from distinct eigenvalues are Specifically, we are interested in those vectors v for which Av=kv where 0 ⋮ Vote. © 2007-2020 All Rights Reserved, Eigenvalues And Eigenvectors Of Symmetric Matrices. Let A be a real skew-symmetric matrix, that is, AT=−A. of s and t that yield two orthogonal vectors (the third comes from the eigenvalue For simple matrices, you can often find the eigenvalues and eigenvectors by observation. In symmetric matrices the upper right half and Add to solve later Sponsored Links Now we need to get the last eigenvector for . of the symmetric matrix. The determinant of a triangular matrix is easy to find - it is simply the product of the diagonal elements. Learn to recognize a rotation-scaling matrix, and compute by how much the matrix rotates and scales. Section 5.5 Complex Eigenvalues ¶ permalink Objectives. Step 2: Estimate the matrix A – λ I A – \lambda I A … homogeneous system of n equations [Math There are many an takes a vector, operates on it, and returns a new vector. of A. The 3x3 matrix can be thought of as an operator Now we pick another value for , and so that the result is zero. In order to find the eigenvalues of a nxn matrix A (if any), we solve your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the do not exist two linearly independent eigenvectors for the two eigenvalues The diagonal elements of a triangular matrix are equal to its eigenvalues. <1,-2,0> and <4,2,-5> that correspond to the two instances 3 have the form <2t,3t> where t is any real number. Proposition An orthonormal matrix P has the property that P−1 = PT. of the eigenvalue k=-1. Example: Find Eigenvalues and Eigenvectors of a 2x2 Matrix. A nxn symmetric matrix A not only has a nice structure, but it also satisfies Enter your answers from smallest to largest. a matrix has more than one The matrices. This process is then repeated for each of the remaining eigenvalues. Varsity Tutors LLC means of the most recent email address, if any, provided by such party to Varsity Tutors. In this problem, we will get three eigen values and eigen vectors since it's a symmetric matrix. Eigenvalues of symmetric matrices suppose A ∈ Rn×n is symmetric, i.e., A = AT ... Symmetric matrices, quadratic forms, matrix norm, and SVD 15–19. either the copyright owner or a person authorized to act on their behalf. Lemma 0.1. In vector form it looks like, . If you've found an issue with this question, please let us know. Eigenvalues and eigenvectors of a real symmetric matrix. for each eigenvalue). system reduces to the single equation (2-i)x-y=0 which implies y=(2-i)x. vectors are left "essentially unchanged" by the operation of the matrix. (A-(-1-i)I)v=0 it can also be shown that vectors The or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing λ 1 =-1, λ 2 =-2. eigenvector the associated eigenvalues can be different for the different A has exactly n (not necessarily distinct) eigenvalues. information described below to the designated agent listed below. shown (by solving the system (A+I)v=0) also has non-distinct eigenvalues of 1 and 1. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are 101 S. Hanley Rd, Suite 300 There exists a set of n eigenvectors, one for each eigenvalue, that are misrepresent that a product or activity is infringing your copyrights. The eigenvalues of a symmetric matrix are always real and the eigenvectors are always orthogonal! The first step into solving for eigenvalues, is adding in a along the main diagonal. since the other two equations are twice this one. Hence, in this case there that have quite nice properties concerning eigenvalues and eigenvectors. Thus, by finding the zeros of the polynomial in k determined by the characteristic equation det(A-kI)=0, we will have found the eigenvalues of the matrix A. Show that (1) det(A)=n∏i=1λi (2) tr(A)=n∑i=1λi Here det(A) is the determinant of the matrix A and tr(A) is the trace of the matrix A. Namely, prove that (1) the determinant of A is the product of its eigenvalues, and (2) the trace of A is the sum of the eigenvalues. let's take r=1. has an infinite number of solutions. By examining the system of equations Now we need to get the matrix into reduced echelon form. vector such that. that has eigenvalue k=3. The following examples illustrate that the situation is not so clear cut and there exist n linearly independent eigenvectors (because of orthogonality) Dirk. k=8). and the two eigenvalues are . Now find a if the determinant det(A-kI) is zero. 1 7 1 1 1 7 di = 6,9 For each eigenvalue, find the dimension of the corresponding eigenspace. any values of s and t. There is a very important class of matrices called symmetric matrices There are once again an infinite number of eigenvectors of A of the form (a) Each eigenvalue of the real skew-symmetric matrix A is either 0or a purely imaginary number. The rst step of the proof is to show that all the roots of the characteristic polynomial of A(i.e. So lambda is an eigenvalue of A. of Mathematics, Oregon State and a set of 3 orthogonal (and thus linearly independent) eigenvectors (one And I want to find the eigenvalues of A. SOLUTION: • In such problems, we first find the eigenvalues of the matrix. Vote. Do not list the same eigenvalue multiple times.) In this problem, we will get three eigen values and eigen vectors since it's a symmetric matrix. Notice that this is a block diagonal matrix, consisting of a 2x2 and a 1x1. the lower left half of the matrix are mirror images of each other about the the eigenvalues of A) are real numbers. This leads to the characteristic equation k^2+2k+2=0 which has complex If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. Learn some strategies for finding the zeros of a polynomial. the The row vector is called a left eigenvector of . We must find two eigenvectors for k=-1 and one for k=8. Once you guess an eigenvalue, its easy to find the eigenvector by solving the linear system $(A-\lambda I)x=0$. Find the eigenvalues and corresponding eigenvalues for the matrix. Hence, we are looking for values k satisfying above cannot happen with a symmetric matrix: A symmetric matrix has n eigenvalues Find max/min eigenvalue of a symmetric matrix. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. reduces to the single equation -x+(3/2)y=0 or equivalently x=1.5y. Infringement Notice, it will make a good faith attempt to contact the party that made such content available by even if the eigenvalues are not distinct. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9372198581695557, "perplexity": 394.7800072095815}, "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/1634323584913.24/warc/CC-MAIN-20211016170013-20211016200013-00446.warc.gz"} |
https://homework.cpm.org/category/CON_FOUND/textbook/mc1/chapter/9/lesson/9.2.2/problem/9-71 | ### Home > MC1 > Chapter 9 > Lesson 9.2.2 > Problem9-71
9-71.
In a previous course you may have learned that $8^{3} = 8 · 8 · 8 = 512$. Rewrite the following expressions using multiplication and then calculate the final value for each of them (as shown for $8^{3}$ above).
1. $10^{4}$
Each exponent (the smaller number to the top right of the $10$) represents another $10$ by which to multiply.
$\left(10\right)\left(10\right)\left(10\right)\left(10\right) = 10{,}000$
1. $5^{3}$
What is the exponent in this problem?
That will be the number of fives to multiply together.
$\left(5\right)\left(5\right)\left(5\right) = 125$
1. $(\frac { 1 } { 2 })^2$
Remember, when multiplying fractions, you simply multiply the numerators together and denominators together.
1. $\left(–2\right)^{3}$
The product of three negative numbers is a negative number. | {"extraction_info": {"found_math": true, "script_math_tex": 10, "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": 0.9841743111610413, "perplexity": 874.6582138163893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00530.warc.gz"} |
http://eguruchela.com/math/calculator/ordering-fraction | ### Ordering Fractions Calculation
Enter the fraction separated by comma and press the calculate button. The system will show the entered fractions in ascending and descending order. To compare fractions the calculator first finds the least common denominator (LCD), converts the fractions to equivalent fractions using the LCD, then compares the numerators for equality.
Please enter fraction separated by comma , (example: 1/3,4/2,3/2,7/5) Ascending Order (Least to Greatest) Descending Order (Greatest to Least)
There can be three cases as follows:
Case 1 : Order the fractions with like denominators example :
$$\frac{14}{7} , \frac{8}{7}, \frac{17}{7}, \frac{25}{7}, \frac{51}{7}$$ In this case we compare the numerators and arrange the fractions in the order based on numerators as follows:
$$\text {Ascending order : }\frac {8}{7}, \frac {14}{7}, \frac {17}{7}, \frac {25}{7} and \frac {51}{7}$$ $$\text {Descending order: }\frac {51}{7}, \frac {25}{7}, \frac {17}{7}, \frac {14}{7} and \frac {8}{7}$$ Case 2 : Order the fractions with unlike denominators Example
$$\frac {3}{2}, \frac {8}{3}, \frac {10}{9}, \frac {5}{4}$$ In this case We will derive the least common denominator (LCD) to achieve the fractions with same denominator. In other words this will become case 1 (like denominators)
So lets find the LCD for denominators 2, 3, 9 and 4 which is 36.
Now to find the equivalent fractions have denominator as 36 therefore
$$\frac {3}{2} = \frac {(n \times 3)}{36} = \frac {(18 \times 3)}{(18 \times 2)} = \frac {54}{36}$$ $$\frac {8}{3} = \frac {(n \times 8)}{36} = \frac {(12 \times 8)}{(12 \times 3)} = \frac {96}{36}$$ $$\frac {10}{9} = \frac {(n \times 10)}{36} = \frac {(4 \times 10)}{(4 \times9)} = \frac {40}{36}$$ $$\frac {5}{4} = \frac {(n \times 5)}{36} = \frac {(9 \times5)}{(9 \times 4)} = \frac {45}{36}$$ Now compare the derived numerators and arrange the fractions in the order based on derived numerators like :
$$\text {Ascending order : } \frac {10}{9}, \frac {5}{4}, \frac {3}{2} and \frac {8}{3}$$ $$\text {Descending order: } \frac {8}{3}, \frac {3}{2}, \frac {5}{4} and \frac {10}{9}$$ Case 3: Order the fractions with like numerators example:
$$\frac {1}{9}, \frac {1}{5}, \frac {1}{11}, \frac {1}{6}$$ If the fractions having like numerators, we will compare the denominators and arrange the fraction in order based on denominators. The important point is here that fraction with the smaller denominator is the larger fraction In this case we compare the denominators and arrange the fractions in the order based on denominators as follows:
$$\text {Ascending order : } \frac {1}{11}, \frac {1}{9}, \frac {1}{6} and \frac {1}{5}$$ $$\text {Descending order: } \frac {1}{5}, \frac {1}{6}, \frac {1}{9} and \frac {1}{11}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9802488088607788, "perplexity": 893.4976925625024}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256887.36/warc/CC-MAIN-20190522163302-20190522185302-00505.warc.gz"} |
https://tjoresearchnotes.wordpress.com/2009/03/09/solving-symmetric-disordered-systems/ | ## Solving symmetric disordered systems
In this post I want to describe a solution to a special class of symmetric disordered quantum systems. This solution is probably not new (it is pretty hard to come up with any solvable system which hasn’t been discovered before!) but I haven’t been able to find anything quite like it in a preliminary search of the literature. So I thought I’d write it up here; if anyone has seen something like this before then please let me know!
This research is intended to be part of a larger project focussed on the computational complexity of disordered quantum systems: I’m starting by collecting results on solvable models to subsequently utilise in the analysis of algorithms like the density matrix renormalisation group.
1. Disorder and symmetric quantum systems
The system I want to consider today is that of a single scalar particle hopping on a finite graph. Such systems arise, eg., in the study of conduction when one studies electrons moving through some regular lattice of atoms. However, as you’ll see, the model I want to talk about today doesn’t really arise in from any naturally occurring material.
Usually, when one studies the conduction of electrons one can reduce the problem, after several approximations, to understanding how a single electron hops between bound states of atoms in the lattice. Reflecting the approximation that the electron is assumed to be located in a definite orbital of atom ${j}$ we often approximate the electron position with a basis ${|j\rangle}$, where ${j}$ runs over all locations in the lattice. Notice that we have reduced the continuous degree of freedom ${|x\rangle}$ of the electron to a discrete degree of freedom ${|j\rangle}$.
Now, since the orbitals of the atoms in the lattice are not completely localised there is some overlap between them. In a regular lattice there is usually only an appreciable overlap between the orbitals of neighbouring atoms. So there is some possibility for the electron at some lattice location to tunnel to a neighbouring location. Thus the energy of a single electron in the lattice ${L}$ may be approximated by
where ${v_j}$ is the potential felt by the electron at lattice site ${j}$ and ${\sum_{\langle j, k\rangle}}$ represents a sum over all neighbouring lattice sites. The magnitude of the overlap between the neighbouring orbitals is wrapped up in the constant ${\tau}$. Please note that this discussion is purely metaphorical and I’ve made a caricature of the actual details of the actual approximations involved in the correct derivation; I’ve swept a great deal under the carpet. Now, the first term in (1) is the hopping term, and this generates the dynamics of an initially localised electron at position ${|j\rangle}$. In real systems there is a lot of dirt and defects, so the potential ${v_j}$ tends to vary from site to site. In fact, one can often successfully model ${v_j}$ as a random variable. If we do this in the case of an electron hopping along a ${1}$D line then we obtain the Anderson model.
When a system such as (1) is clean, i.e., not in the presence of a disordered potential, one typically finds that the eigenstates of the system are completely delocalised (they are regular eigenstates of momentum). In this case the electron can zoom through the lattice freely and the material becomes an excellent conductor; you should think of the electron not as a particle here, but as a wave, like a beam of light, propagating through some transparent medium. However, something remarkable happens when the potential ${v_j}$ is random: the electron gets scattered by the impurities in the potential. In fact, it gets scattered multiple times: it gets scattered so much that it becomes trapped — localised — at it’s initial location. This is the ubiquitous phenomenon of Anderson localisation.
Anderson localisation is a quantum phenomena: a classical argument based on the kinetic energy alone would lead you to expect that a classical particle would simply zoom over the rugged potential landscape without ever bouncing back. Actually, I should say it is a wave phenomena because, amazingly, Anderson localisation occurs in a wide variety of systems. There is a wonderful example explained by Sir Michael Berry: take a stack of overhead transparency films. This is a stack of transparent films with slightly random widths. At the boundary between the films an incident wave of light is partially reflected and transmitted. Now, this can be modelled as a 1D lattice of random scatterers — just like the Anderson model. One would therefore expect that light is localised in such a medium, i.e., it can’t propagate far into the stack. Thus the light can’t shine through the stack, and is thus reflected: a stack of OHP transparencies looks silver because all incident light is reflected thanks to Anderson localisation!
Disordered systems such as the Anderson model are extremely challenging to solve, to put it mildly. There are very many subtle mathematical problems which arise when trying to solve systems such as (1). The literature is roughly broken into two branches: the physical and the mathematical. The physical literature is far ahead of the mathematical literature at the current time, and I think it is fair to say that we have a pretty solid understanding of the physics of disordered systems such as (1). There are only a couple of controversies left in the theory for systems on regular lattices, centred on the two-dimensional case where some issues are not resolved. The physical literature makes heavy use of quantum-field theoretical methods, (which is what makes it hard to make rigourous) and the premiere tool to study disordered systems is the supersymmetric method pioneered by Efetov.
The mathematical literature is far behind the predictions obtained via physical arguments. The state of the art is summarised in this paper. Some things are known rigourously: it is understood that systems like (1) are localised for strong enough disorder. However, little, if anything, is known about systems which remain delocalised in the presence of small amounts of disorder. Additionally, it is extremely hard to say anything rigourously about observable quantities which are averaged over disorder. Even using physical arguments this is still rather challenging: the supersymmetric method is well-adapted to such questions, but it is restricted (as far as I can tell) to gaussian-distributed disorder.
Now this discussion is meant to be a motivation of sorts for the actual system whose solution I want to describe today. This system is far from realistic, although it might arise as a mean-field limit of a strongly interacting collection of particles with long-range interactions. The system is given by
i.e., it represents a scalar particle which can hop between any of ${n}$ locations. The disorder is taken to be gaussian-distributed with variance ${\gamma}$ (but I think this can be lifted fairly easily). Obviously this is pretty unrealistic in any lattice, but it may represent a decent approximation to complex molecules… This is something I need to think about. I should note that the solution I’ll describe can actually be applied to a more general model, namely anything of the form
where ${\psi_j}$ can be arbitrary. This class defies any immediate physical interpretation, but, who knows?
This whole post rests on a wonderful result about matrices that I learnt while teaching financial mathematics (it arises when studying portfolio selection in the single-index model). It is simple to state and is essentially elementary (I’ll write it out in quantum notation, though, to facilitate the remainder of this discussion).
Proposition 1 Suppose ${M}$ is an ${n\times n}$ matrix of the form
$\displaystyle M = \sum_{j,k=1}^n \psi_j\psi_k|j\rangle\langle k| + \sum_{j=1}^n v_j|j\rangle\langle j|, \ \ \ \ \ (4)$
where ${\psi_j}$ and ${v_j}$ are arbitrary complex numbers. Then, when it exists, the inverse of ${M}$ is given by
$\displaystyle M^{-1} = \sum_{j=1}^n \frac{1}{v_j}|j\rangle\langle j| - \sum_{j,k=1}^n \frac{\psi_j\psi_k}{v_jv_k\Phi}|j\rangle\langle k|, \ \ \ \ \ (5)$
where
$\displaystyle \Phi = 1+\sum_{j=1}^n \frac{\psi_j^2}{v_j}. \ \ \ \ \ (6)$
Proof: Couldn’t be simpler: just write out ${M^{-1}M}$ and notice that it equals ${\mathbb{I}}$! $\Box$
3. Solving our disordered system
Proposition 1 is the result which I’m going to leverage to solve the system (2). By “solve” here I mean that I’ll describe how to work out the locations of the eigenvalues of ${H}$ and learn statistical results about the eigenstates of ${H}$. Before I get to the actual solution (although I’m sure it’s relatively obvious what I’m going to do), I want to digress for a moment and discuss Green’s function.
Definition 2 Let ${H}$ be an ${n\times n}$ hermitian matrix. Then we define the retarded green’s function to be
$\displaystyle G_{-}(z) = \frac{\mathbb{I}}{z-i\delta+H} \ \ \ \ \ (7)$
and the advanced green’s function to be
$\displaystyle G_{+}(z) = \frac{\mathbb{I}}{z+i\delta+H}. \ \ \ \ \ (8)$
Green’s function are useful for a variety of reasons, usually because they are easier to study than the eigenvalues and eigenfunctions directly. We have the following
Lemma 3 Let ${H}$ be an ${n\times n}$ hermitian matrix. Then the eigenvalue density function ${\rho(z) = \sum_{j=1}^n \delta(z-E_j)}$, where ${E_j}$ are the eigenvalues of ${H}$ and ${\delta(z)}$ is the Dirac delta function, is given by
$\displaystyle \rho(z) = \lim_{\delta \rightarrow 0} \frac{1}{\pi} \mbox{Im}\,\mbox{tr}(G_+(z)). \ \ \ \ \ (9)$
(The eigenvalue density function has a delta-function spike at the location of each one of the eigenvalues of ${H}$.)
Proof: Let’s write out he green’s functions in the eigenbasis of ${H}$ and take the trace: we get
$\displaystyle \mbox{tr}(G_{\pm}(z)) = \sum_{j=1}^n \frac{1}{z\pm i\delta-E_j} \ \ \ \ \ (10)$
so that
$\displaystyle \mbox{tr}(G_+(z)-G_-(z)) = \sum_{j=1}^n \frac{2i\delta}{(z-E_j)^2+\delta^2} \ \ \ \ \ (11)$
and thus
$\displaystyle \mbox{Im}\,\mbox{tr}(G_+(z)) = \sum_{j=1}^n \frac{\delta}{(z-E_j)^2+\delta^2}. \ \ \ \ \ (12)$
Taking the limit ${\delta \rightarrow 0}$ and using the fact that ${\lim_{\delta\rightarrow0} \frac{\delta}{(z-\alpha)+\delta^2} = \pi\delta(z-\alpha)}$ gives us
$\displaystyle \rho(z) = \lim_{\delta \rightarrow 0} \frac1\pi\mbox{Im}\,\mbox{tr}(G_+(z)) = \sum_{j=1}^n\delta(z-E_j). \ \ \ \ \ (13)$
$\Box$
We’re now going to use proposition 1 to work out the averaged eigenvalue density function. Before we do we have the following corollary of proposition 1.
Corollary 4 Let ${H}$ be the system (2). Then
$\displaystyle G_{\pm}(z) = \sum_{j=1}^n \frac{1}{z\pm i\delta-v_j}|j\rangle\langle j| - \frac{1}{\Phi}\sum_{j,k=1}^n \frac{1}{(z\pm i\delta-v_j)(z\pm i\delta-v_k)}|j\rangle\langle k|, \ \ \ \ \ (14)$
where
$\displaystyle \Phi = 1 + \sum_{j=1}^n \frac{1}{z\pm i\delta-v_l}. \ \ \ \ \ (15)$
Now we can present the first main result
Proposition 5 Let ${H}$ be the system (2). Then the averaged eigenvalue density (i.e. the eigenvalue density function averaged over the disorder) is given by
$\displaystyle \langle \rho(z) \rangle = \frac{ne^{-\frac{z^2}{\gamma^2}}}{\gamma\sqrt{\pi}}+ f(z), \ \ \ \ \ (16)$
where
$\displaystyle f(z)= - \lim_{\delta\rightarrow 0}\frac{1}{\pi\gamma^n\pi^{n/2}}\int e^{-\frac{\|\mathbf{v}\|}{\gamma^2}}\mbox{Im}\left(\frac{1}{\Phi} \sum_{j=1}^n\frac{1}{(z+i\delta-v_j)^2}\right)dv. \ \ \ \ \ (17)$
Proof: First note that
We write
$\displaystyle l(z) = - \lim_{\delta\rightarrow 0}\frac{1}{\pi}\mbox{Im}\left(\frac{1}{\Phi}\sum_{j=1}^n\frac{1}{(z+i\delta-v_j)^2}\right) \ \ \ \ \ (19)$
To work out the averaged eigenvalue density we take the expectation over the ${v}$s:
$\displaystyle \langle \rho (z)\rangle = \sum_{j=1}^n \mathbb{E}_{v}[\delta(z-v_j)] - \mathbb{E}_{v}[l(z)]. \ \ \ \ \ (20)$
Using the fact that ${\mathbb{E}_{v}[\delta(z-v_j)] = \frac{e^{-\frac{z^2}{\gamma^2}}}{\gamma\sqrt{\pi}}}$ gives us the result, upon writing ${f(z) = -\mathbb{E}_{v}[l(z)]}$. $\Box$
That’s it for today: I spent most of the day making stupid mistakes (eg. working out exact expressions for ${\rho(z)}$ which couldn’t possibly have ever worked.)
In future posts we’ll take a look at the small-${\gamma}$ asymptotics of ${f(z)}$ — we know exactly what ${f(z)}$ looks like at ${\gamma=0}$, i.e. ${f(z; \gamma=0) = -\delta(z) + \delta(z-n)}$, because we know the eigenvalues of ${H}$ when ${v_j=0}$. I’m not sure what the best method is to get at the asymptotics at the moment: possibly fixing ${\delta}$ to be small and doing some contour integrals…?
### 5 Responses to Solving symmetric disordered systems
1. Steve Flammia says:
Hi Tobias,
I’m not sure if you’re aware of two well-known results that are related to Prop. 1. The first is a generalization:
http://en.wikipedia.org/wiki/Woodbury_matrix_identity
and the second is just closely related:
http://en.wikipedia.org/wiki/Matrix_determinant_lemma
Maybe you will find these useful.
2. tobiasosborne says:
Dear Steve,
Many thanks for the references! Embarrassingly I didn’t actually know of them…
T
3. […] and eigenstates of the system: the system will exhibit Anderson localisation (see my post here for a brief description of this phenomena). In two and higher dimensions this is not necessarily […]
4. Ghiret says:
For some reason, equations (1-3) don’t appear. It would help with readability.
5. […] wavefunctions become localised and all diffusion is suppressed. (I refer you to my previous post here for further discussion.) A caricature of this phenomenon is then that static disorder eliminates […] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 72, "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": 0.9089271426200867, "perplexity": 323.1304162908469}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549426951.85/warc/CC-MAIN-20170727022134-20170727042134-00196.warc.gz"} |
https://en.wikipedia.org/wiki/J-invariant | j-invariant
Klein's j-invariant in the complex plane
In mathematics, Klein's j-invariant or j function, regarded as a function of a complex variable τ, is a modular function of weight zero for SL(2, Z) defined on the upper half-plane of complex numbers. It is the unique such function which is holomorphic away from a simple pole at the cusp such that
$j\left(e^{\frac{2}{3}\pi i}\right) = 0, \quad j(i) = 1728.$
Rational functions of j are modular, and in fact give all modular functions. Classically, the j-invariant was studied as a parameterization of elliptic curves over C, but it also has surprising connections to the symmetries of the Monster group (this connection is referred to as monstrous moonshine).
Definition
Real part of the j-invariant as a function of the nome q on the unit disk
Phase of the j-invariant as a function of the nome q on the unit disk
While the j-invariant can be defined purely in terms of certain infinite sums (see g2, g3 below), these can be motivated by considering isomorphism classes of elliptic curves. Every elliptic curve E over C is a complex torus, and thus can be identified with a rank 2 lattice; i.e., two-dimensional lattice of C. This is done by identifying opposite edges of each parallelogram in the lattice. It turns out that multiplying the lattice by complex numbers, which corresponds to rotating and scaling the lattice, preserves the isomorphism class of the elliptic curve, and thus we can consider the lattice generated by 1 and some τ in H (where H is the Upper half-plane). Conversely, if we define
\begin{align} g_2 &= 60\sum_{(m,n) \neq (0,0)} (m + n\tau)^{-4}, \\ g_3 &= 140\sum_{(m,n) \neq (0,0)} (m + n\tau)^{-6}, \end{align}
then this lattice corresponds to the elliptic curve over C defined by y2 = 4x3g2x - g3 via the Weierstrass elliptic functions. Then the j-invariant is defined as
$j(\tau) = 1728 \frac{g_2^3}{\Delta}$
where the modular discriminant Δ is
$\Delta = g_2^3 - 27g_3^2$
It can be shown that Δ is a modular form of weight twelve, and g2 one of weight four, so that its third power is also of weight twelve. Thus their quotient, and therefore j, is a modular function of weight zero, in particular a meromorphic function HC invariant under the action of SL(2, Z). As explained below, j is surjective, which means that it gives a bijection between isomorphism classes of elliptic curves over C and the complex numbers.
The fundamental region
The fundamental domain of the modular group acting on the upper half plane.
The two transformations ττ + 1 and τ → -τ−1 together generate a group called the modular group, which we may identify with the projective special linear group PSL(2, Z). By a suitable choice of transformation belonging to this group,
$\tau \mapsto \frac{a\tau + b}{c\tau +d}, \qquad ad-bc =1,$
we may reduce τ to a value giving the same value for j, and lying in the fundamental region for j, which consists of values for τ satisfying the conditions
\begin{align} |\tau| &\ge 1 \\ -\tfrac{1}{2} &< \mathfrak{R}(\tau) \le \tfrac{1}{2} \\ -\tfrac{1}{2} &< \mathfrak{R}(\tau) < 0 \Rightarrow |\tau| > 1 \end{align}
The function j(τ) when restricted to this region still takes on every value in the complex numbers C exactly once. In other words, for every c in C, there is a unique τ in the fundamental region such that c = j(τ). Thus, j has the property of mapping the fundamental region to the entire complex plane.
As a Riemann surface, the fundamental region has genus 0, and every (level one) modular function is a rational function in j; and, conversely, every rational function in j is a modular function. In other words the field of modular functions is C(j).
Class field theory and j
The j-invariant has many remarkable properties:
• The field extension Q[j(τ), τ]/Q(τ) is abelian, that is, it has an abelian Galois group.
• Let Λ be the lattice in C generated by {1, τ}. It is easy to see that all of the elements of Q(τ) which fix Λ under multiplication form a ring with units, called an order. The other lattices with generators {1, τ′}, associated in like manner to the same order define the algebraic conjugates j(τ′) of j(τ) over Q(τ). Ordered by inclusion, the unique maximal order in Q(τ) is the ring of algebraic integers of Q(τ), and values of τ having it as its associated order lead to unramified extensions of Q(τ).
These classical results are the starting point for the theory of complex multiplication.
Transcendence properties
In 1937 Theodor Schneider proved the aforementioned result that if τ is a quadratic irrational number in the upper half plane then j(τ) is an algebraic integer. In addition he proved that if τ is an algebraic number but not imaginary quadratic then j(τ) is transcendental.
The j function has numerous other transcendental properties. Kurt Mahler conjectured a particular transcendence result that is often referred to as Mahler's conjecture, though it was proved as a corollary of results by Yu. V. Nesterenko and Patrice Phillipon in the 1990s. Mahler's conjecture was that if τ was in the upper half plane then exp(2πiτ) and j(τ) were never both simultaneously algebraic. Stronger results are now known, for example if exp(2πiτ) is algebraic then the following three numbers are algebraically independent, and thus at least two of them transcendental:
$j(\tau), \frac{j^\prime(\tau)}{\pi}, \frac{j^{\prime\prime}(\tau)}{\pi^2}$
The q-expansion and moonshine
Several remarkable properties of j have to do with its q-expansion (Fourier series expansion), written as a Laurent series in terms of q = exp(2πiτ), which begins:
$j(\tau) = {1 \over q} + 744 + 196884 q + 21493760 q^2 + 864299970 q^3 + 20245856256 q^4 + \cdots$
Note that j has a simple pole at the cusp, so its q-expansion has no terms below q−1.
All the Fourier coefficients are integers, which results in several almost integers, notably Ramanujan's constant:
$e^{\pi \sqrt{163}} \approx 640320^3 + 744$.
The asymptotic formula for the coefficient of qn is given by
$\frac{e^{4\pi\sqrt{n}}}{\sqrt{2}n^{3/4}}$,
as can be proved by Hardy–Littlewood circle method.[2][3]
Moonshine
More remarkably, the Fourier coefficients for the positive exponents of q are the dimensions of the graded part of an infinite-dimensional graded algebra representation of the monster group called the moonshine module – specifically, the coefficient of qn is the dimension of grade-n part of the moonshine module, the first example being the Griess algebra, which has dimension 196,884, corresponding to the term 196884q. This startling observation was the starting point for moonshine theory.
The study of the Moonshine conjecture led J.H. Conway and Simon P. Norton to look at the genus-zero modular functions. If they are normalized to have the form
$q^{-1} + {O}(q)$
then Thompson showed that there are only a finite number of such functions (of some finite level), and Cummins later showed that there are exactly 6486 of them, 616 of which have integral coefficients.[4]
Alternate Expressions
We have
$j(\tau) = \frac{256(1-x)^3}{x^2}$
where x = λ(1−λ) and λ is the modular lambda function
$\lambda(\tau) = \frac{\theta_2^4(0,\tau)}{\theta_3^4(0,\tau)} = k^2(\tau)$
a ratio of Jacobi theta functions $\theta_{m}$, and is the square of the elliptic modulus $k(\tau)$.[5] The value of j is unchanged when λ is replaced by any of the six values of the cross-ratio:[6]
$\left\lbrace { \lambda, \frac{1}{1-\lambda}, \frac{\lambda-1}{\lambda}, \frac{1}{\lambda}, \frac{\lambda}{\lambda-1}, 1-\lambda } \right\rbrace$
The branch points of j are at {0, 1, ∞}, so that j is a Belyi function.[7]
Expressions in terms of theta functions
Define the nome $q=e^{\pi i \tau}$ and the Jacobi theta function,
$\vartheta(0; \tau) = \vartheta_{00}(0; \tau) = 1 + 2 \sum_{n=1}^\infty \left(e^{\pi i\tau}\right)^{n^2} = \sum_{n=-\infty}^\infty q^{n^2}$
from which one can derive the auxiliary theta functions. Let,
\begin{align} a &= \theta_{2}(0; q) = \vartheta_{10}(0; \tau) \\ b &= \theta_{3}(0; q) = \vartheta_{00}(0; \tau) \\ c &= \theta_{4}(0; q) = \vartheta_{01}(0; \tau) \end{align}
where $\theta_{m}$ and $\vartheta_{n}$ are alternative notations, and $a^4 - b^4 + c^4 = 0$. Then,
\begin{align} g_2(\tau) &= \tfrac{2}{3}\pi^4 \left(a^8 + b^8 + c^8\right) \\ g_3(\tau) &= \tfrac{4}{27}\pi^6 \sqrt{\frac{(a^8+b^8+c^8)^3-54(abc)^8}{2}} \\ \Delta &= g_2^3-27g_3^2 = (2\pi)^{12} \left(\tfrac{1}{2}a b c\right)^8 = (2\pi)^{12}\eta(\tau)^{24} \end{align}
for Weierstrass invariants g2, g3, and Dedekind eta function η(τ). We can then express j(τ) in a form which can rapidly be computed.
$j(\tau) = 1728\frac{g_2^3}{g_2^3-27g_3^2} = 32 {(a^8 + b^8 + c^8)^3 \over (a b c)^8}$
Algebraic definition
So far we have been considering j as a function of a complex variable. However, as an invariant for isomorphism classes of elliptic curves, it can be defined purely algebraically. Let
$y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$
be a plane elliptic curve over any field. Then we may define
$b_2 = a_1^2 + 4a_2,\quad b_4 = a_1a_3 + 2a_4$
$b_6 = a_3^2 + 4a_6,\quad b_8 = a_1^2a_6 - a_1a_3a_4 + a_2a_3^2 + 4a_2a_6 - a_4^2$
$c_4 = b_2^2 - 24b_4,\quad c_6 = -b_2^3 + 36b_2b_4 - 216b_6$
and
$\Delta = -b_2^2b_8 + 9b_2b_4b_6 - 8b_4^3 - 27b_6^2$
the latter expression is the discriminant of the curve.
The j-invariant for the elliptic curve may now be defined as
$j = {c_4^3 \over \Delta}$
In the case that the field over which the curve is defined has characteristic different from 2 or 3, this definition can also be written as
$j = 1728{c_4^3 \over c_4^3-c_6^2}$
Inverse function
The inverse function of the j-invariant can be expressed in terms of the hypergeometric function 2F1 (see also the article Picard–Fuchs equation). Explicitly, given a number N, to solve the equation j(τ) = N for τ can be done in at least four ways.
Method 1: Solving the sextic in λ,
$j(\tau) = \frac{256(1-\lambda(1-\lambda))^3}{(\lambda(1-\lambda))^2} = \frac{256(1-x)^3}{x^2}$
where x = λ(1−λ), and λ is the modular lambda function so the sextic can be solved as a cubic in x. Then,
$\tau = i \ \frac{{}_2F_1 \left (\tfrac{1}{2},\tfrac{1}{2},1,1 - \lambda \right )}{{}_2F_1 \left (\tfrac{1}{2},\tfrac{1}{2},1,\lambda \right)}$
for any of the six values of λ.
Method 2: Solving the quartic in γ,
$j(\tau) = \frac{27(1 + 8\gamma)^3}{\gamma(1 - \gamma)^3}$
then for any of the four roots,
$\tau = \frac{i}{\sqrt{3}} \frac{{}_2F_1 \left (\tfrac{1}{3},\tfrac{2}{3},1,1-\gamma \right)}{{}_2F_1 \left(\tfrac{1}{3},\tfrac{2}{3},1,\gamma \right )}$
Method 3: Solving the cubic in β,
$j(\tau) = \frac{64(1+3\beta)^3}{\beta(1-\beta)^2}$
then for any of the three roots,
$\tau = \frac{i}{\sqrt{2}} \frac{{}_2F_1 \left (\tfrac{1}{4},\tfrac{3}{4},1,1-\beta \right)}{{}_2F_1 \left(\tfrac{1}{4},\tfrac{3}{4},1,\beta \right )}$
Method 4: Solving the quadratic in α,
$j(\tau)=\frac{1728}{4\alpha(1-\alpha)}$
then,
$\tau = i \ \frac{{}_2F_1 \left (\tfrac{1}{6},\tfrac{5}{6},1,1-\alpha \right)}{{}_2F_1 \left(\tfrac{1}{6},\tfrac{5}{6},1,\alpha \right )}$
One root gives τ, and the other gives 1/τ, but since j(τ) = j(1/τ), then it doesn't make a difference which α is chosen. The latter three methods can be found in Ramanujan's theory of elliptic functions to alternative bases.
The inversion is highly relevant to applications via enabling high-precision calculations of elliptic functions periods even as their ratios become unbounded. A related result is the expressibility via quadratic radicals of the values of j at the points of the imaginary axis whose magnitudes are powers of 2 (thus permitting compass and straightedge constructions). The latter result is hardly evident since the modular equation of level 2 is cubic.
Pi formulas
The Chudnovsky brothers found in 1987,[8]
$\frac{1}{\pi} = \frac{12}{640320^{3/2}} \sum_{k=0}^\infty \frac{(6k)! (163 \cdot 3344418k + 13591409)}{(3k)!(k!)^3 (-640320)^{3k}}$
which uses the fact that $j\big(\tfrac{1+\sqrt{-163}}{2}\big) = -640320^3$. For similar formulas, see the Ramanujan–Sato series.
Special values
The j-invariant vanishes at the "corner" of the fundamental domain at
$\;\;\tfrac{1}{2}\left(1 + i \sqrt{3}\right).$
Here are a few more special values (only the first four of which are well known; in what follows, j means J/1728 throughout):
\begin{align} j(i) &= j \left( \tfrac{1 + i}{2} \right) = 1 \\ j\left(\sqrt{2}i\right) &= \left(\tfrac{5}{3}\right)^3 \\ j(2i) &= \left(\tfrac{11}{2}\right)^3 \\ j\left(2\sqrt{2}i\right) &= \tfrac{125}{216} \left(19 + 13\sqrt{2} \right)^3\\ j(4i) &= \tfrac{1}{64} \left(724 + 513\sqrt{2} \right)^3\\ j\left( \tfrac{1 + 2i}{2} \right) &= \tfrac{1}{64} \left(724 - 513\sqrt{2} \right)^3\\ j\left( \tfrac{1 + 2\sqrt{2}i}{3} \right) &= \tfrac{125}{216} \left(19 - 13\sqrt{2} \right)^3\\ j(3i) &= \tfrac{1}{27} \left(2 + \sqrt{3}\right)^2 \left(21 + 20\sqrt{3}\right )^3 \\ j\left(2\sqrt{3}i\right) &= \tfrac{125}{16} \left(30 + 17\sqrt{3}\right)^3\\ j\left( \tfrac{1 + 7\sqrt{3}i}{2} \right) &= -\tfrac{64000}{7} \left(651 + 142\sqrt{21} \right)^3\\ j\left(\tfrac{1 + 3\sqrt{11}i}{10} \right) &= \tfrac{64}{27} \left(23 - 4\sqrt{33}\right)^2 \left(-77 + 15\sqrt{33} \right)^3\\ j\left(\sqrt{21}i\right) &= \tfrac{1}{32} \left(5 + 3\sqrt{3}\right)^2 \left(3 + \sqrt{7} \right)^2 \left( 65 + 34\sqrt{3} + 26\sqrt{7} + 15\sqrt{21}\right)^3\\ j\left( \tfrac{\sqrt{30}i}{1} \right) &= \tfrac{1}{16} \left(10 + 7\sqrt{2} + 4\sqrt{5} + 3\sqrt{10} \right)^4 \left( 55 + 30\sqrt{2} + 12\sqrt{5} + 10\sqrt{10} \right)^3\\ j\left( \tfrac{\sqrt{30}i}{2} \right) &= \tfrac{1}{16} \left(10 + 7\sqrt{2} - 4\sqrt{5} - 3\sqrt{10} \right)^4 \left( 55 + 30\sqrt{2} - 12\sqrt{5} - 10\sqrt{10} \right)^3\\ j\left( \tfrac{\sqrt{30}i}{5} \right) &= \tfrac{1}{16} \left(10 - 7\sqrt{2} + 4\sqrt{5} - 3\sqrt{10} \right)^4 \left( 55 - 30\sqrt{2} + 12\sqrt{5} - 10\sqrt{10} \right)^3\\ j\left( \tfrac{\sqrt{30}i}{10} \right) &= \tfrac{1}{16} \left(10 - 7\sqrt{2} - 4\sqrt{5} + 3\sqrt{10} \right)^4 \left( 55 - 30\sqrt{2} - 12\sqrt{5} + 10\sqrt{10} \right)^3\\ j\left(\tfrac{1+\sqrt{31}i}{2}\right) &= \left(1-\left(1+\frac{\sqrt{19}}{2}\left(\sqrt{\tfrac{13-\sqrt{93}}{13+\sqrt{93}}} \cdot \sqrt[3]{\tfrac{\sqrt{31}+\sqrt{27}}{\sqrt{31}-\sqrt{27}}}+ \sqrt{\tfrac{13+\sqrt{93}}{13-\sqrt{93}}} \cdot \sqrt[3]{\tfrac{\sqrt{31}-\sqrt{27}}{\sqrt{31}+\sqrt{27}}}\right)\right)^2\right)^3\\ j(\sqrt{70}i) &= \left(1 + \tfrac{9}{4}\left(303 + 220\sqrt{2} + 139\sqrt{5} + 96\sqrt{10}\right)^2 \right)^3 \\ j(7i) &= \left( 1 + \tfrac{9}{4}\sqrt{21+8\sqrt{7}} \left(30 + 11\sqrt{7} + \left (6+\sqrt{7} \right )\sqrt{21+8\sqrt{7}}\right)^2 \right)^3\\ j(8i) &= \left( 1 + \tfrac{9}{4} \sqrt[4]{2} \left (1 + \sqrt{2} \right) \left(123 + 104\sqrt[4]{2} + 88\sqrt{2} + 73\sqrt[4]{8}\right)^2 \right)^3\\ j(10i) &= \left(1 + \tfrac{9}{8}\left(2402 + 1607\sqrt[4]{5} + 1074\sqrt[4]{25} + 719\sqrt[4]{125}\right)^2 \right)^3\\ j \left( \tfrac{5i}{2} \right) &= \left(1 + \tfrac{9}{8}\left(2402 - 1607\sqrt[4]{5} + 1074\sqrt[4]{25} - 719\sqrt[4]{125}\right)^2 \right)^3\\ j(2\sqrt{58}i) &= \left(1+\tfrac{9}{256}\left(1+\sqrt{2}\right)^5\left(5+\sqrt{29}\right)^5\left(793+907\sqrt{2}+237\sqrt{29}+103\sqrt{58}\right)^2\right)^3\\ j\left( \tfrac{1 + \sqrt{1435}i}{2} \right) &= \left( 1 - 9 \left ( 9892538 + 4424079\sqrt{5} + 1544955\sqrt{41} + 690925\sqrt{205} \right )^2 \right)^3\\ j\left( \tfrac{1 + \sqrt{1555}i}{2} \right) &= \left( 1 - 9 \left ( 22297077 + 9971556\sqrt{5} + \left ( 3571365 + 1597163\sqrt{5} \right ) \sqrt{\tfrac{31 + 21\sqrt{5}}{2}} \right)^2 \right)^3\\ \end{align}
Several special values were calculated in 2014:[9]
\begin{align} j \left( \tfrac{5i + 1}{2} \right) &= \left( \frac{2927 - 1323 \sqrt{5}}{2} \right)^3,\\ j(5i) &= \left( \frac{2927 + 1323 \sqrt{5}}{2} \right)^3,\\ \end{align}
and let,
\begin{align} a_1, a_2, a_3, a_4 &= 1190448488, 858585699, 540309076, 374537880 \\ b_1, b_2, b_3, b_4 &= 693172512, 595746414, 407357424, 240819696 \end{align}
\begin{align} j \left( \tfrac{5 i + 2}{4} \right) &= \left ( \frac{\left( 1 + \sqrt{5}\right)^{37}}{2^{39}} \left ( a_1- a_2 \sqrt{2} + a_3 \sqrt{5} - a_4 \sqrt{10} -\sqrt[4]{5} \left(b_1 - b_2 \sqrt{2} + b_3 \sqrt{5} - b_4 \sqrt{10} \right) \right) \right)^3,\\ j \left( \tfrac{10i + 1}{2} \right) &= \left ( \frac{\left( 1 + \sqrt{5} \right)^{37}}{2^{39}} \left (a_1- a_2 \sqrt{2} + a_3 \sqrt{5} - a_4 \sqrt{10} + \sqrt[4]{5} \left( b_1- b_2\sqrt{2} + b_3\sqrt{5} - b_4\sqrt{10} \, \right) \right) \right)^3,\\ j \left( \tfrac{5 i}{4} \right) &= \left( \frac{\left( 1 + \sqrt{5} \right)^{37}}{2^{39}} \left (a_1+ a_2\, \sqrt{2} + a_3 \sqrt{5} + a_4 \sqrt{10} - \sqrt[4]{5} \left( b_1+ b_2 \sqrt{2} + b_3 \sqrt{5} + b_4 \sqrt{10} \right) \right) \right)^3,\\ j(20i) &= \left ( \frac{\left( 1 + \sqrt{5} \right)^{37}}{2^{39}} \left ( a_1+ a_2 \sqrt{2} + a_3 \sqrt{5} + a_4 \sqrt{10} + \sqrt[4]{5} \left( b_1+ b_2 \sqrt{2} + b_3 \sqrt{5} + b_4 \sqrt{10} \right) \right) \right)^3. \end{align}
All preceding values are real. A complex conjugate pair might be inferred exploiting the symmetry described in the reference, along with the values for $j(10i)$ and $j(5 i/2)$, given above:
$j \left( \tfrac{1}{4}(5 i \pm 1) \right) = \left(1 - \tfrac{9}{8}\left((2402 - 1074\sqrt{5}) i \pm (1607 - 719\sqrt{5}) \sqrt[4]{5} \right)^2 \right)^3.$
Four more special values are given as two complex conjugate pairs:[10]
\begin{align} j \left( \tfrac{4}{13} \left( 5 i \pm 1 \right) \right) = \left( \frac{\left( 1 - \sqrt{5} \right)^{37}}{2^{39}} \left( a_1- a_2\sqrt{2} - a_3 \sqrt{5} + a_4 \sqrt{10} \pm i \sqrt[4]{5} \left( b_1 - b_2 \sqrt{2} - b_3 \sqrt{5} + b_4 \sqrt{10} \right) \right) \right)^3,\\ j \left( \tfrac{5}{17} \left( 4i \pm 1 \right) \right) = \left( \frac{\left( 1 - \sqrt{5} \right)^{37}}{2^{39}} \left(a_1+a_2 \sqrt{2} - a_3 \sqrt{5} - a_4\sqrt{10} \pm i \sqrt[4]{5} \left( b_1+ b_2\sqrt{2} - b_3\sqrt{5} - b_4 \sqrt{10} \right) \right) \right)^3 \end{align}
References
1. ^ Silverman, Joseph H. (1986). The Arithmetic of Elliptic Curves. Graduate Texts in Mathematics 106. Springer-Verlag. p. 339. ISBN 0-387-96203-4. Zbl 0585.14026.
2. ^ Petersson, Hans (1932). Über die Entwicklungskoeffizienten der automorphen Formen. Acta Mathematica 58 (1). pp. 169–215. doi:10.1007/BF02547776. MR 1555346.
3. ^ Rademacher, Hans (1938). The Fourier coefficients of the modular invariant j(τ). American Journal of Mathematics 60 (2) (The Johns Hopkins University Press). pp. 501–512. doi:10.2307/2371313. JSTOR 2371313. MR 1507331.
4. ^ Cummins, C.J. (2004). "Congruence subgroups of groups commensurable with PSL(2,Z)\$ of genus 0 and 1". Exp. Math. 13 (3): 361–382. ISSN 1058-6458. Zbl 1099.11022.
5. ^ Chandrasekharan (1985) p.108
6. ^ Chandrasekharan, K. (1985), Elliptic Functions, Grundlehren der mathematischen Wissenschaften 281, Springer-Verlag, p. 110, ISBN 3-540-15295-4, Zbl 0575.33001
7. ^ Girondo, Ernesto; González-Diez, Gabino (2012), Introduction to compact Riemann surfaces and dessins d'enfants, London Mathematical Society Student Texts 79, Cambridge: Cambridge University Press, p. 267, ISBN 978-0-521-74022-7, Zbl 1253.30001
8. ^ Chudnovsky, David V.; Chudnovsky, Gregory V. (1989), "The Computation of Classical Constants", Proceedings of the National Academy of Sciences of the United States of America 86 (21): 8178–8182, doi:10.1073/pnas.86.21.8178, ISSN 0027-8424, JSTOR 34831, PMC 298242, PMID 16594075.
9. ^ Adlaj, Semjon. "Multiplication and division on elliptic curves, torsion points and roots of modular equations" (PDF). Retrieved 17 October 2014.
10. ^ Adlaj, Semjon (2014). "Torsion points on elliptic curves and modular polynomial symmetries" (PDF). The joined MSU-CCRAS Computer Algebra Seminar. Moscow, Russia. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 50, "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": 0.9998279213905334, "perplexity": 1566.3123117118982}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443738004493.88/warc/CC-MAIN-20151001222004-00097-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-7th-edition/chapter-4-exponential-and-logarithmic-functions-section-4-1-exponential-functions-4-1-exercises-page-372/13 | ## College Algebra 7th Edition
To graph the given function, perform the following steps: (1) Create a table of values by assigning values to $x$ then solving for the corresponding $y$ value of each one. (refer to the attached table below) (2) Plot each ordered pair of the table of values then connect them using a smooth curve. (refer to the attached image in the answer part above) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8107068538665771, "perplexity": 414.0472868430163}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156376.8/warc/CC-MAIN-20180920020606-20180920040606-00089.warc.gz"} |
https://math.stackexchange.com/questions/737665/when-to-use-in-or-subseteq | # When to use $\in$ or $\subseteq$?
If I have a family of $n$ sets like this $\mathcal{F}=\{\{S_1\}, \{S_2\}, \dotsc, \{S_n\}\}$.
What is the right notation: for some $i\;\{S_i\}\in\mathcal{F}$ or $\{S_i\}\subseteq\mathcal{F}$?
In general, when to use $\in$ and when to use $\subseteq$?
• They are two very different relations between sets and their members; a soldier is a member of a platoon; a platoon is a subset of a company. A soldier is not a subset of a company. – Mauro ALLEGRANZA Apr 3 '14 at 6:26
Given a set $X$ we write $A\in X$ when $A$ is an element of $X$. We write $A\subseteq X$ when every element of $A$ is an element of $X$. It's slightly more difficult to express $\in$ because that is taken to be the basic relation from which we define everything else. So we can really just say that $A\in X$ when $A$ is an element of $X$ (which is not much to say).
For example, $\{a\}\subseteq\{a,b\}$ because every element of $\{a\}$ is an element of $\{a,b\}$. On the other hand $a\in\{a\}$.
Note that it is perfectly possible for a set to be both an element and a subset of another set. Consider for example $A=\{a\}$ and $X=\{a,A\}=\{a,\{a\}\}$. We have that $A\in X$ and $A\subseteq X$.
In your case, $\{S_i\}$ is an element of $\cal F$. It might be a subset of $\cal F$ in the case where $S_i=\{S_j\}$ for some $i$ and $j$. But your question does not include sufficient information to decide about that.
• Good. It is a new topic that I am learning so I do not have much knowledge. Thank you for your help. – zighalo Apr 3 '14 at 14:26
The second suggested notation is wrong, because $\{S_i\}$ is not a subset of $\mathcal{F}$. To be a subset, each element of $\{S_i\}$ must be an element of $\mathcal{F}$, and that is not true: $S_i$ is not an element of $\mathcal{F}$. What is in $\mathcal{F}$ is $\{S_i\}$, which is the set containing $S_i$.
The first suggested notation is correct.
$\in$ denotes element of, $\subseteq$ denotes subset of.
• I get it. Thank you very much for your help. – zighalo Apr 3 '14 at 14:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.940085768699646, "perplexity": 78.10122475660935}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670559.66/warc/CC-MAIN-20191120134617-20191120162617-00417.warc.gz"} |
https://openstax.org/books/chemistry/pages/14-2-ph-and-poh | Chemistry
# 14.2pH and pOH
Chemistry14.2 pH and pOH
### Learning Objectives
By the end of this section, you will be able to:
• Explain the characterization of aqueous solutions as acidic, basic, or neutral
• Express hydronium and hydroxide ion concentrations on the pH and pOH scales
• Perform calculations relating pH and pOH
As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water (Kw). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.
A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “X” is the quantity of interest and “log” is the base-10 logarithm:
$pX=−log XpX=−log X$
The pH of a solution is therefore defined as shown here, where [H3O+] is the molar concentration of hydronium ion in the solution:
$pH=−log[H3O+]pH=−log[H3O+]$
Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:
$[H3O+]=10−pH[H3O+]=10−pH$
Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH:
$pOH=−log[OH−]pOH=−log[OH−]$
or
$[OH−]=10−pOH[OH−]=10−pOH$
Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the Kw expression:
$Kw=[H3O+][OH−]Kw=[H3O+][OH−]$
$−logKw=−log([H3O+][OH−])=−log[H3O+]+−log[OH−]−logKw=−log([H3O+][OH−])=−log[H3O+]+−log[OH−]$
$pKw=pH+pOHpKw=pH+pOH$
At 25 °C, the value of Kw is 1.0 $××$ 10−14, and so:
$14.00=pH+pOH14.00=pH+pOH$
As was shown in Example 14.1, the hydronium ion molarity in pure water (or any neutral solution) is 1.0 $××$ 10−7 M at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:
$pH=−log[H3O+]=−log(1.0×10−7)=7.00pH=−log[H3O+]=−log(1.0×10−7)=7.00$
$pOH=−log[OH−]=−log(1.0×10−7)=7.00pOH=−log[OH−]=−log(1.0×10−7)=7.00$
And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than 1.0 $××$ 10−7 M and hydroxide ion molarities less than 1.0 $××$ 10−7 M (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than 1.0 $××$ 10−7 M and hydroxide ion molarities greater than 1.0 $××$ 10−7 M (corresponding to pH values greater than 7.00 and pOH values less than 7.00).
Since the autoionization constant Kw is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the “Check Your Learning” exercise accompanying Example 14.1 showed the hydronium molarity of pure water at 80 °C is 4.9 $××$ 10−7 M, which corresponds to pH and pOH values of:
$pH=−log[H3O+]=−log(4.9×10−7)=6.31pH=−log[H3O+]=−log(4.9×10−7)=6.31$
$pOH=−log[OH−]=−log(4.9×10−7)=6.31pOH=−log[OH−]=−log(4.9×10−7)=6.31$
At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 °C) (Table 14.1).
Summary of Relations for Acidic, Basic and Neutral Solutions
Classification Relative Ion Concentrations pH at 25 °C
acidic [H3O+] > [OH] pH < 7
neutral [H3O+] = [OH] pH = 7
basic [H3O+] < [OH] pH > 7
Table 14.1
Figure 14.2 shows the relationships between [H3O+], [OH], pH, and pOH, and gives values for these properties at standard temperatures for some common substances.
Figure 14.2 The pH and pOH scales represent concentrations of [H3O+] and OH, respectively. The pH and pOH values of some common substances at standard temperature (25 °C) are shown in this chart.
### Example 14.4
#### Calculation of pH from [H3O+]
What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of 1.2 $××$ 10−3 M?
#### Solution
$pH=−log[H3O+]pH=−log[H3O+]$
$=−log(1.2×10−3)=−log(1.2×10−3)$
$=−(−2.92)=2.92=−(−2.92)=2.92$
(The use of logarithms is explained in Appendix B. Recall that, as we have done here, when taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)
Water exposed to air contains carbonic acid, H2CO3, due to the reaction between carbon dioxide and water:
$CO2(aq)+H2O(l)⇌H2CO3(aq)CO2(aq)+H2O(l)⇌H2CO3(aq)$
Air-saturated water has a hydronium ion concentration caused by the dissolved CO2 of 2.0 $××$ 10−6 M, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C.
5.70
### Example 14.5
#### Calculation of Hydronium Ion Concentration from pH
Calculate the hydronium ion concentration of blood, the pH of which is 7.3 (slightly alkaline).
#### Solution
$pH=−log[H3O+]=7.3pH=−log[H3O+]=7.3$
$log[H3O+]=−7.3log[H3O+]=−7.3$
$[H3O+]=10−7.3or[H3O+]=antilog of −7.3[H3O+]=10−7.3or[H3O+]=antilog of −7.3$
$[H3O+]=5×10−8M[H3O+]=5×10−8M$
(On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate 10−7.3.)
Calculate the hydronium ion concentration of a solution with a pH of −1.07.
12 M
### How Sciences Interconnect
#### Environmental Science
Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO2 which forms carbonic acid:
$H2O(l)+CO2(g)⟶H2CO3(aq)H2O(l)+CO2(g)⟶H2CO3(aq)$
$H2CO3(aq)⇌H+(aq)+HCO3−(aq)H2CO3(aq)⇌H+(aq)+HCO3−(aq)$
Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:
$H2O(l)+SO3(g)⟶H2SO4(aq)H2O(l)+SO3(g)⟶H2SO4(aq)$
$H2SO4(aq)⟶H+(aq)+HSO4−(aq)H2SO4(aq)⟶H+(aq)+HSO4−(aq)$
Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.
Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure 14.3). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.
For further information on acid rain, visit this website hosted by the US Environmental Protection Agency.
Figure 14.3 (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by “Eden, Janine and Jim”/Flickr)
### Example 14.6
#### Calculation of pOH
What are the pOH and the pH of a 0.0125-M solution of potassium hydroxide, KOH?
#### Solution
Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH] = 0.0125 M:
$pOH=−log[OH−]=−log0.0125pOH=−log[OH−]=−log0.0125$
$=−(−1.903)=1.903=−(−1.903)=1.903$
The pH can be found from the pOH:
$pH+pOH=14.00pH+pOH=14.00$
$pH=14.00−pOH=14.00−1.903=12.10pH=14.00−pOH=14.00−1.903=12.10$
The hydronium ion concentration of vinegar is approximately 4 $××$ 10−3 M. What are the corresponding values of pOH and pH?
pOH = 11.6, pH = 2.4
The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 14.4).
Figure 14.4 (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of ± 0.002 pH units, and may cost in excess of \$1000. (b) A portable pH meter has lower resolution (0.01 pH units), lower accuracy (± 0.2 pH units), and a far lower price tag. (credit b: modification of work by Jacopo Werther)
The pH of a solution may also be visually estimated using colored indicators (Figure 14.5).
Figure 14.5 (a) A universal indicator assumes a different color in solutions of different pH values. Thus, it can be added to a solution to determine the pH of the solution. The eight vials each contain a universal indicator and 0.1-M solutions of progressively weaker acids: HCl (pH = l), CH3CO2H (pH = 3), and NH4Cl (pH = 5), deionized water, a neutral substance (pH = 7); and 0.1-M solutions of the progressively stronger bases: KCl (pH = 7), aniline, C6H5NH2 (pH = 9), NH3 (pH = 11), and NaOH (pH = 13). (b) pH paper contains a mixture of indicators that give different colors in solutions of differing pH values. (credit: modification of work by Sahar Atwa)
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As an Amazon Associate we earn from qualifying purchases. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 39, "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": 0.8010730743408203, "perplexity": 4077.144508362825}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585204.68/warc/CC-MAIN-20211018155442-20211018185442-00289.warc.gz"} |
https://stats.stackexchange.com/questions/408734/is-the-difference-between-the-residual-and-error-term-in-a-regression-just-the-a | # Is the difference between the residual and error term in a regression just the ability to observe it?
According to what I read online, the error term and the residual are often interchangeable. Please let me know if my understanding below is correct:
However, the difference is that the error term is the difference between our predicted value and the ACTUAL population value. So for instance, if we are to measure the relationship between salary level and experience level for a 35 year old male in the US, we wouldn't be able to get the data for ALL of the millions of 35 year old males in the US. However, we can get a sample population of that whole population. So therefore, our regression of Y would include an error term which indicates the value of what we have and the ACTUAL value of the population that we CANNOT OBSERVE. However, the RESIDUAL is the difference between the actual regression line and the actual POINTS of the scatter plot of data that we have that we have actually OBSERVED from going out and collecting data from some source we may have.
Is my understanding correct?
• To see why the concept of observability is not at the root of this, suppose you actually did have a complete and accurate database of salary and experience for 35 y.o. US males. Because these two variables would not exhibit a linear (or even monotonic) relationship, there would still be residuals, there would still be error terms, and the two would not necessarily coincide because (1) the residuals depend on the procedure you use for fitting the model and (2) by taking the view that this dataset is the entire population of interest, the errors no longer are random variables.
– whuber
May 17 '19 at 14:54
• So then in this case, what exactly IS the error term and what is the residual? I've always understood residual is just the "physical distance" between the data points from the best fit line when doing an OLS regression. It reminds me of like high school math (square root of (x2-x1)^2+ (y2-y1)^2)
– John
May 17 '19 at 15:21
• Simply put, the error term is a construct in a model of the population or process and the residual is the difference between an observation and the value assigned to that observation by your regression procedure. As a concrete example, suppose a single value $X$ is randomly selected from a population known to have a Normal distribution with unknown mean $\mu;$ that is, the model is that $X=\mu+\epsilon.$ The error is $\epsilon=X-\mu.$ The OLS estimate of $\mu$ is $X$ itself, with residual $X-X=0.$ In this case the residual and the error almost surely are unequal.
– whuber
May 17 '19 at 15:28
Your wording seems to imply that the error term exists because we deal with samples and the error term captures information about the non-sampled part of the population. That's not correct. The error term in a regression model represents factors other than the observed variables included in the model as $$X$$'s (explanatory/independent variables) that affect the dependent variable $$Y$$. Regression model (e.g., $$y = \beta_{0} + \beta_{1}x + \epsilon$$) begins from assuming what the relationship between $$X$$ and $$Y$$ variables is in the population, so the error term exists even in the population model. The model you end up estimating with sample data allows you to estimate the parameters of that population model.
So the error term is NOT the difference between observed and predicted values of $$Y$$. Repeating myself, it represents unobserved factors affecting $$Y$$.
Once the population regression model is assumed, we proceed to estimating the model with randomly sampled data. The estimation/fitting procedure we use estimates the values of $$\beta$$'s and we can then compute predicted/fitted values of $$Y$$ based on those estimated values of $$\beta$$'s and the observed values of $$X$$'s. The estimated regression equation takes on the form: $$y_{i} = \hat\beta_{0} +\hat\beta_{1}x_{i} + \hat\epsilon_{i}$$, with those hats denoting estimated values. $$\hat\epsilon_{i}$$'s are the residuals in the estimated equation (differences between observed and predicted values of $$Y$$ for each individual in the sample), while $$\epsilon$$'s are the errors in the equation containing population parameters $$\beta_{0}$$ and $$\beta_{1}$$. The errors are not observable, while the residuals are computed from the data.
• Yes, terms are often used interchangeably, but the technical definition is what is in the answer above. If you run a regression and compute a difference observed and predicted values of $Y$ for each observation, they are residuals. May 17 '19 at 19:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9137962460517883, "perplexity": 231.79966336061727}, "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/1634323585199.76/warc/CC-MAIN-20211018062819-20211018092819-00349.warc.gz"} |
http://books.duhnnae.com/2017/jun4/149715351862-Water-Fractions-in-Extrasolar-Planetesimals-M-Jura-S-Xu.php | Water Fractions in Extrasolar Planetesimals
With the goal of using externally-polluted white dwarfs to investigate the water fractions of extrasolar planetesimals, we assemble from the literature a sample that we estimate to be more than 60% complete of DB white dwarfs warmer than 13,000 K, more luminous than 3 ${\times}$ 10$^{-3}$ L${\odot}$ and within 80 pc of the Sun. When considering all the stars together, we find the summed mass accretion rate of heavy atoms exceeds that of hydrogen by over a factor of 1000. If so, this sub-population of extrasolar asteroids treated as an ensemble has little water and is at least a factor of 20 drier than CI chondrites, the most primitive meteorites. In contrast, while an apparent -excess- of oxygen in a single DB can be interpreted as evidence that the accreted material originated in a water-rich parent body, we show that at least in some cases, there can be sufficient uncertainties in the time history of the accretion rate that such an argument may be ambiguous. Regardless of the difficulty associated with interpreting the results from an individual object, our analysis of the population of polluted DBs provides indirect observational support for the theoretical view that a snow line is important in disks where rocky planetesimals form.
Author: M. Jura; S. Xu
Source: https://archive.org/ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8993958830833435, "perplexity": 1178.3807950912521}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824618.72/warc/CC-MAIN-20171021062002-20171021082002-00257.warc.gz"} |
http://mathhelpforum.com/differential-geometry/109827-solved-counterexample-needed.html | Math Help - [SOLVED] counterexample needed
1. [SOLVED] counterexample needed
I have to prove or disprove the following:
f is continuous on (a,b) implies that f is bounded on (a,b).
Im pretty sure this is false, but cant come up with a viable counterexample to disprove it. Any help?
2. Originally Posted by dannyboycurtis
I have to prove or disprove the following:
f is continuous on (a,b) implies that f is bounded on (a,b).
Im pretty sure this is false, but cant come up with a viable counterexample to disprove it. Any help?
$f(x)=\frac{1}{x}$ on $(0,1)$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9548653960227966, "perplexity": 672.8697750290842}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246651471.95/warc/CC-MAIN-20150417045731-00230-ip-10-235-10-82.ec2.internal.warc.gz"} |
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