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# m25 q22
Author Message
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Joined: 17 Mar 2010
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13 May 2010, 11:46
If function f(x) satisfiesf(x) = f(x^2) for all x , which of the following must be true?
* $$f(4) = f(2)f(2)$$
* $$f(16) - f(-2) = 0$$
* $$f(-2) + f(4) = 0$$
* $$f(3) = 3f(3)$$
* $$f(0) = 0$$
$$f(-2) = f(2) = f(4) = f(16)$$ . The other choices are not necessarily true. Consider $$f(x) = 3$$ for all $$x$$ .
I don't really understand why you are doing f(-2) = f(2) = f(4) = f(16). Where does it say in the problem that you can use the function on a previous function ?????
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14 May 2010, 06:41
Very interesting question, I will wait for the specialists.
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18 Sep 2012, 07:08
So the f(x)=f(x^2), so f(2)=f(2^2)=f(4), and f(4)=f(4^2)=f(16).... therefore f(2)=f(4)=f(16).... Cool?
So then f(-2)=f(-2^2)=f(4), and since f(4)=f(2) you can say f(-2)=f(2)=f(4)....... Cool?
Cool.
B.
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18 Sep 2012, 07:21
leilak wrote:
If function f(x) satisfiesf(x) = f(x^2) for all x , which of the following must be true?
* $$f(4) = f(2)f(2)$$
* $$f(16) - f(-2) = 0$$
* $$f(-2) + f(4) = 0$$
* $$f(3) = 3f(3)$$
* $$f(0) = 0$$
$$f(-2) = f(2) = f(4) = f(16)$$ . The other choices are not necessarily true. Consider $$f(x) = 3$$ for all $$x$$ .
I don't really understand why you are doing f(-2) = f(2) = f(4) = f(16). Where does it say in the problem that you can use the function on a previous function ?????
OPEN DISCUSSION OF THIS QUESTION IS HERE: if-function-f-x-satisfies-f-x-f-x-2-for-all-x-107351.html?hilit=function%20satisfies
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Re: m25 q22 [#permalink] 18 Sep 2012, 07:21
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# m25 q22
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,180 | 3,361 | {"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} | 4.03125 | 4 | CC-MAIN-2017-09 | longest | en | 0.843862 |
https://oeis.org/A118276 | 1,679,843,185,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00657.warc.gz | 504,610,317 | 4,199 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A118276 Signature sequence of Phi^2 = 2.618033989... (A104457), where Phi is the golden ratio A001622. 6
1, 2, 3, 1, 4, 2, 5, 3, 6, 1, 4, 7, 2, 5, 8, 3, 6, 1, 9, 4, 7, 2, 10, 5, 8, 3, 11, 6, 1, 9, 4, 12, 7, 2, 10, 5, 13, 8, 3, 11, 6, 14, 1, 9, 4, 12, 7, 15, 2, 10, 5, 13, 8, 16, 3, 11, 6, 14, 1, 9, 17, 4, 12, 7, 15, 2, 10, 18, 5, 13, 8, 16, 3, 11, 19, 6, 14, 1, 9, 17, 4, 12, 20, 7, 15, 2, 10, 18, 5, 13 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Equals A023119 in the first 98 terms, then the sequences differ. [From R. J. Mathar, Aug 08 2008] REFERENCES C. Kimberling, "Fractal Sequences and Interspersions", Ars Combinatoria, vol. 45 p 157 1997. LINKS T. D. Noe, Table of n, a(n) for n=1..1000 Eric Weisstein's World of Mathematics, Signature Sequence MATHEMATICA terms = 90; m = Ceiling[Sqrt[terms]]; s0 = {}; While[s = (Table[i + j*GoldenRatio^2, {i, 1, m}, {j, 1, m}] // Flatten // SortBy[#, N] &)[[1 ;; terms]] /. GoldenRatio -> 0; s != s0, s0 = s; m = 2 m]; s (* Jean-François Alcover, Jan 08 2017 *) CROSSREFS Cf. A084531, A084532, A167970. Sequence in context: A125159 A179548 A023119 * A023123 A023131 A356625 Adjacent sequences: A118273 A118274 A118275 * A118277 A118278 A118279 KEYWORD nonn AUTHOR Casey Mongoven, Apr 21 2006 STATUS approved
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https://www.geeksforgeeks.org/multiply-the-given-number-by-2-such-that-it-is-divisible-by-10/ | 1,611,722,773,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704820894.84/warc/CC-MAIN-20210127024104-20210127054104-00210.warc.gz | 775,683,465 | 25,120 | Related Articles
Multiply the given number by 2 such that it is divisible by 10
• Difficulty Level : Basic
• Last Updated : 07 Dec, 2018
Given a number, the only operation allowed is to multiply the number by 2. Calculate the minimum number of operations to make the number divisible by 10.
NOTE: If it is not possible to convert then print -1.
Examples:
Input: 10
Output: 0
As the given number is itself divisible by 10,
Input: 1
Output: -1
As by multiplying with 2, given no. can’t be
converted into a number that is divisible by 10,
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach: Any given number is divisible by 10 only if the last digit of the number is 0. For this problem, extract the last digit of the input number and check it in the following ways :
1) If the last digit is 0 then it is already divisible by 10 , so the minimum number of steps is 0.
2) If the last digit is 5 then multiplying it by 2 one time will make it divisible by 10, so the minimum number of steps is 1.
3) If the last digit is an even or odd number (apart from 0 and 5) then multiplying it by 2 any number of times will only produce even number so we can never make it divisible by 10. Therefore the number of steps is -1.
## C++
`// C++ code for finding ` `// number of operations ` `#include ` `using` `namespace` `std; ` ` ` `int` `multiplyBy2(``int` `n) ` `{ ` ` ``int` `rem, value; ` ` ` ` ``// Find the last digit or remainder ` ` ``rem = n % 10; ` ` ``switch` `(rem) { ` ` ` ` ``// If the last digit is 0 ` ` ``case` `0: ` ` ``value = 0; ` ` ``break``; ` ` ` ` ``// If the last digit is 5 ` ` ``case` `5: ` ` ``value = 1; ` ` ``break``; ` ` ` ` ``// If last digit is other ` ` ``// than 0 and 5. ` ` ``default``: ` ` ``value = -1; ` ` ``} ` ` ` ` ``return` `value; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` ` ``int` `n = 28; ` ` ``cout << multiplyBy2(n) << endl; ` ` ` ` ``n = 255; ` ` ``cout << multiplyBy2(n) << endl; ` ` ` ` ``return` `0; ` `} `
## Java
`// JAVA code for finding ` `// number of operations ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ``static` `int` `multiplyBy2(``int` `n) ` `{ ` ` ``int` `rem, value; ` ` ` ` ``// Find the last digit ` ` ``// or remainder ` ` ``rem = n % ``10``; ` ` ``switch` `(rem) ` ` ``{ ` ` ` ` ``// If the last digit is 0 ` ` ``case` `0``: ` ` ``value = ``0``; ` ` ``break``; ` ` ` ` ``// If the last digit is 5 ` ` ``case` `5``: ` ` ``value = ``1``; ` ` ``break``; ` ` ` ` ``// If last digit is other ` ` ``// than 0 and 5. ` ` ``default``: ` ` ``value = -``1``; ` ` ``} ` ` ` ` ``return` `value; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ``int` `n = ``28``; ` ` ``System.out.println(multiplyBy2(n)); ` ` ` ` ``n = ``255``; ` ` ``System.out.println(multiplyBy2(n)); ` `} ` `} ` ` ` `// This code is contributed ` `// by shiv_bhakt. `
## Python3
`# Python3 code for finding number ` `# of operations ` `def` `dig(argu): ` ` ``switcher ``=` `{ ` ` ``0``: ``0``, ` ` ``5``: ``1``, ` ` ``} ` ` ``return` `switcher.get(argu, ``-``1``) ` ` ` `def` `multiplyBy2(n): ` ` ` ` ``# Find the last digit or remainder ` ` ``rem ``=` `n ``%` `10``; ` ` ``return` `dig(rem); ` ` ` `# Driver code ` `n ``=` `28``; ` `print``(multiplyBy2(n)); ` ` ` `n ``=` `255``; ` `print``(multiplyBy2(n)); ` ` ` `# This code is contributed by mits `
## C#
`// C# code for finding ` `// number of operations ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ``static` `int` `multiplyBy2(``int` `n) ` ` ``{ ` ` ``int` `rem, value; ` ` ` ` ``// Find the last digit ` ` ``// or remainder ` ` ``rem = n % 10; ` ` ``switch` `(rem) ` ` ``{ ` ` ` ` ``// If the last ` ` ``// digit is 0 ` ` ``case` `0: ` ` ``value = 0; ` ` ``break``; ` ` ` ` ``// If the last ` ` ``// digit is 5 ` ` ``case` `5: ` ` ``value = 1; ` ` ``break``; ` ` ` ` ``// If last digit is ` ` ``// other than 0 and 5. ` ` ``default``: ` ` ``value = -1; ` ` ``break``; ` ` ``} ` ` ` ` ``return` `value; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main () ` `{ ` ` ``int` `n = 28; ` ` ``Console.WriteLine(multiplyBy2(n)); ` ` ` ` ``n = 255; ` ` ``Console.WriteLine(multiplyBy2(n)); ` `} ` `} ` ` ` `// This code is contributed ` `// by shiv_bhakt. `
## PHP
` `
Output:
```-1
1
```
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Show that over any field $F$, the polynomial $x^3-3x+1$ is either irreducible or splits into linear factors. Edited: This is my attempt: Let $f(x)=x^3-3x+1$. Let $a_1,a_2,a_3$ be the roots of $f$. ... | 1,283 | 4,601 | {"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} | 2.859375 | 3 | CC-MAIN-2019-35 | latest | en | 0.867871 |
http://forum.allaboutcircuits.com/threads/i-have-a-circuit-with-an-inductor-a-diode-and-a-neon-light-a-switch-and-10v-i-need-to-explain-it.126150/ | 1,485,271,309,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284429.99/warc/CC-MAIN-20170116095124-00048-ip-10-171-10-70.ec2.internal.warc.gz | 116,961,522 | 18,994 | I have a circuit with an inductor, a diode and a neon light, a switch and 10V. I need to explain it.
Discussion in 'Homework Help' started by Antonella Puricelli, Jul 28, 2016.
1. Antonella Puricelli Thread Starter New Member
Apr 24, 2016
6
0
Hello, good morning/afternoon/night/everythingelese,
I have the following circuit and some questions to answer about its functioning but I'm not exactly sure I'm right about what I thought.
What we had to do, is connect the circuit accordingly and see what happens.
We've seen that the neon light lit up when we opened the switch (but didn't when we closed it). But I don'n know is why this happens.
My theory (after reserching) is that when we close the switch, the inductor generates a tension that goes against the diode and that's why it doesn't light up. Then, when we open the switch, the direction of the tension goes the other way allowing the neon to light up.
However, then, we had to reverse the direction of the diode (to reverse-biased), we could observe that the neon worked the same way, only that its light was brighter. I have no clue as to why this happens. My guess is that it's related to the functioning of the neon light (the thing is that I haven't completely understood how it works).
I know this is a lot, but any kind of help or explaining could help, even providing some links for me to research would be great.
I want to clarify my native language is Spanish so I apologice in advanced for any mistakes and I'm new to the page so I hope I put this in the right place.
Thank you all.
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2. wayneh Expert
Sep 9, 2010
12,405
3,256
With the switch closed, the neon bulb can never see a voltage greater than the battery voltage less the diode drop. That's not enough to light it.
After current has been flowing in the inductor, it contains the stored energy of the magnetic field. Opening the switch causes nearly immediate collapse of the field. Voltage across the inductor can reach very high levels, plenty to light the bulb.
Do you know the breakdown voltage of the diode?
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3. Antonella Puricelli Thread Starter New Member
Apr 24, 2016
6
0
It's a 1N4007, the datasheet (Vishay) says its breakdown voltage is 1000V.
4. Antonella Puricelli Thread Starter New Member
Apr 24, 2016
6
0
It's a 1N4007, the datasheet (Vishay) says its breakdown voltage is 1000V.
5. wayneh Expert
Sep 9, 2010
12,405
3,256
I think the diode is in the circuit to demonstrate the polarity of the spike coming from the inductor. The spike can light the bulb with the diode in either orientation because the spike is over 1000V, but the light is brighter with the diode in the proper orientation. There's only ~0.7V drop in voltage in that direction, and a drop of about 1000V in the "wrong" direction.
6. Papabravo Expert
Feb 24, 2006
10,340
1,850
What is a vuelta? Is it a unit of inductance in some other language?
Willen likes this.
7. RichardO Well-Known Member
May 4, 2013
1,350
432
L1 = 1200 turns
8. Antonella Puricelli Thread Starter New Member
Apr 24, 2016
6
0
I get what you're saying. However, the following step in the project was to see what happened when we inverted the polarity of the diode. In this case, what happened was that the light was actually much brighter.
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9. wayneh Expert
Sep 9, 2010
12,405
3,256
As predictable. Think about the polarity of the spike. Note the evidence in the photo as well. The bulb has two electrodes, and only one is lit.
I think this is called a relaxation oscillator?
Last edited: Jul 28, 2016
10. Antonella Puricelli Thread Starter New Member
Apr 24, 2016
6
0
Right, okey, I think I got it.
Thank you so much
11. Papabravo Expert
Feb 24, 2006
10,340
1,850
Thanks. Good to know.
12. Antonella Puricelli Thread Starter New Member
Apr 24, 2016
6
0
I don't know about the name, but I think that's because we're using DC instead of AC; I've read something that when using DC only the negative terminal is lit up as opposed to AC in which both are lit up.
13. crutschow Expert
Mar 14, 2008
13,523
3,392
But it's not oscillating.
14. crutschow Expert
Mar 14, 2008
13,523
3,392
When the switch is closed, current starts to flow through the inductor which stores energy in its magnetic field equal to ½LI².
When the switch is opened this stored energy in the field tries to keep the current flowing in the same direction until the energy is dissipated in the external load (the inductor now acting as a current source).
The only path for this current is the neon bulb and diode so a negative voltage (note the current direction to see why it's negative) will build up until the current starts flowing through the bulb and diode.
The current only flows for a short time until all the inductive energy is dissipated in the bulb and diode voltage drop, so just a short flash of light is generated.
If the diode is reverse biased than much more energy is dissipated in the diode compared to the bulb.
If the diode is forward biased than much more energy is dissipated in the bulb.
15. wayneh Expert
Sep 9, 2010
12,405
3,256
I wasn't sure. If the inductor were a capacitor, it would. So says the wiki about neon bulbs.
16. Bordodynov Active Member
May 20, 2015
673
194
See
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17. Tonyr1084 Active Member
Sep 24, 2015
637
112
Inductors and capacitors are similar in nature in that they store electrical energy then give it back. The difference with an inductor is that it can store 10 volts and give it back as a much higher voltage whereas a capacitor can never give more than it has stored.
The capacitor will charge to 10 volts. When the switch is opened the capacitor will supply its voltage (10 volts) until it discharges completely. Depending on the resistance across the capacitor the rate of decay will vary.
The inductor will generate a magnetic field. Its rise time will depend on how much current is supplied. It may take several milliseconds to charge (depending on the inductor). But when the switch is opened the magnetic field is at liberty to collapse completely in just a few microseconds. The changing magnetic field is what generates a voltage. The faster that changes the higher the voltage it generates. So when you open the switch the magnetic field collapses and generates a much higher voltage - one capable of lighting the Neon lamp.
Neon typically requires 70 volts to ignite. (unless things have changed since I messed with them) 10 volts into a coil (inductor) can easily light the neon when its current is interrupted. And as electrons flow from one electrode to the other one will glow. With AC, current is flowing one way then the other. So it appears as if the neon bulb is lit on both electrodes at the same time. We can't see events happening that fast, so our brain assumes both electrodes are glowing at the same time. By reversing the diode you change the direction of flow. One electrode will glow in one orientation, the other electrode will glow when you reverse the system. And the voltage drop across the diode is insignificant to the voltages your coil is generating.
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18. crutschow Expert
Mar 14, 2008
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In this circuit, reversing the diode does not change the direction of current flow.
That's determined by the inductor current direction which does not change.
If the diode is connected with anode to (-) source, the inductor current will flow in the forward direction through the diode when the switch is opened, with low voltage drop.
If the diode is connected with the cathode to (- ) source, the inductor current will flow through the diode in the reverse direction when the switch is opened, causing reverse avalanche breakdown of the diode at high voltage.
19. crutschow Expert
Mar 14, 2008
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Yes.
Replacing the inductor with a capacitor and adding a high value resistor in series with a high voltage source will create a relaxation oscillator. | 1,997 | 7,980 | {"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} | 2.78125 | 3 | CC-MAIN-2017-04 | latest | en | 0.966614 |
https://math.stackexchange.com/questions/1100653/marginal-density-from-joint-distribution-of-3-random-variables | 1,571,357,155,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00133.warc.gz | 603,878,667 | 30,279 | # Marginal density from joint distribution of 3 random variables
Suppose random variables X, Y, Z have joint probability density function given as $$f_{XYZ}(x,y,z) = \frac{2x}{(A_2-A_1)(R_2^{2} - R_1^{2})y^{2}z} e^{-z^{2}/2}$$ where A1,A2,R1,R2 are constant. How do I find the marginal probability density function of X, if the support is \begin{cases}\ R_1yz\leq x \leq R_2yz \\ A_1 \leq y \leq A_2 \\ 0 \leq z < \infty & \end{cases} I know that I need to integrate with respect to Y (the will be two regions, I guess) and Z to find marginal P.D.F. of X. But I don't know what the boundary of the integrals should be, especially I cant understand how to deal with the range of Z ($\infty$ is causing the main problem). What will be the final range of PDF of X. Thanks in advance.
• What is the joint density? What are $A_1$ and $A_2$? By $\inf$ do you mean $\infty$? – Math1000 Jan 12 '15 at 0:25
• @ Math 1000, I have edited the question. Please have a look. – PD007 Jan 12 '15 at 6:08 | 328 | 989 | {"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": 1, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-43 | latest | en | 0.851433 |
https://www.scienceforums.net/topic/67064-why-can-light-push-but-not-be-pushed/page/2/#comment-684296 | 1,720,887,199,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00188.warc.gz | 810,121,318 | 45,212 | # Why can light "push" but not be "pushed?"
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rigney: I suggest you look for some more reputable source, either an expert in the field or a respected science writer. You have found the ramblings of some random nut job who doesn't know what he's talking about. The internet is full of net jobs, crackpots, and charlatan, and you are very good at finding them.
You're right in one respect. I'm gullable enough to believe there may be a bit of truth in just about everything I read or hear. Tell me, is this guy on the level or just pounding salt? Since we both know that I'm not, may I ask if you are working at, or above his level of expertise?
http://www.vanderbilt.edu/AnS/physics/perakis/L9_00/
Edited by rigney
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My "guess" must have been wrong about flying into the projected beam absorbing the ships velocity... as a guess why there is no cumulative velocity. A case of "fuzzy" thinking. You are right. Once light is projected ahead, the ship will never catch up with it, let alone travel into it.
Which leaves the mystery as stated, i.e., why is the beam not boosted to 1&1/2 'c'? Clearly light can not be "pushed" faster than 'c', so what happens to the ship's velocity? I still don't know. But I'm working on it.
The "velocity relative to what?" question (in lieu of no "aether") can be answered by adding some props along the direction of travel. Say some space buoys had been placed along the way, brought to rest (zero velocity) relative to the approaching ship. Say they are placed every 186,000 miles, to make it easy. The ship's light beam passes one every second, while the ship, at 1/2 'c', passes one every two seconds.
After a minute the ship will have passed 30 buoys (having gone 5,580,000 miles) and its beam will have passed 60 of them (having gone 11,160,000 miles.) Then the distance between the ship and the far end of the beam is 5,580,000 miles, yet the light traveled 11,160,000 miles.
What am I missing here?
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Tell me, is this guy on the level or just pounding salt?
Too cutesy!!! And not historically correct. Read "rumor has it" as "I just made this story up".
What Einstein did do was try to envision what it would be like to "ride a beam of light" when he was 16. He didn't get all that far at this young age because he didn't have the necessary math or physics background. Ten years later he had the necessary tools and he had time to spare. He used this time to revisit his thoughts from a decade before. He found a novel and very simple way around one of the most vexing problems of the time, the inherent incompatibility of Newtonian mechanics and Maxwell's electrodynamics. Maxwell's equations strongly imply that the speed of light is the same to all observers. Most physicists were looking for ways around this implication (and around the conflict). Einstein simply took Maxwell's equations to heart. He accepted that the speed of light is the same to all observers and then determined the fall out from this assumption. This provided the answer to his thoughts from a decade before: You cannot ride a beam of light. That beam of light is speeding away from you at the speed of light. It doesn't matter how much faster you go. That beam of light is still going to go away from you at c. Always.
An object with non-zero mass must always move at subluminal speeds. Your mirror cannot go at the speed of light.
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The "velocity relative to what?" question (in lieu of no "aether") can be answered by adding some props along the direction of travel. Say some space buoys had been placed along the way, brought to rest (zero velocity) relative to the approaching ship. Say they are placed every 186,000 miles, to make it easy. The ship's light beam passes one every second, while the ship, at 1/2 'c', passes one every two seconds.
After a minute the ship will have passed 30 buoys (having gone 5,580,000 miles) and its beam will have passed 60 of them (having gone 11,160,000 miles.) Then the distance between the ship and the far end of the beam is 5,580,000 miles, yet the light traveled 11,160,000 miles.
What am I missing here?
If the buoys are at rest with respect to the ship, it doesn't pass by any of them.
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If the buoys are at rest with respect to the ship, it doesn't pass by any of them.
Right. I misspoke, which should have been obvious by the context of the ship and its light beam passing the buoys.
The buoys were meant to represent 'mileposts' on the 'road through space' traveled by the ship.
So, to be accurate, the ship which dropped the buoys, if it were going at half 'c' also would have programmed their retro-rockets to "brake", decreasing velocity relative to the ship until they each had lost 1/2 'c' of velocity.
OK? Now, again, "what am I missing" given that there are only 5,580,000 miles between the ship and the front end of the light beam after a minute of travel?
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Too cutesy!!! And not historically correct. Read "rumor has it" as "I just made this story up".
What Einstein did do was try to envision what it would be like to "ride a beam of light" when he was 16. He didn't get all that far at this young age because he didn't have the necessary math or physics background. Ten years later he had the necessary tools and he had time to spare. He used this time to revisit his thoughts from a decade before. He found a novel and very simple way around one of the most vexing problems of the time, the inherent incompatibility of Newtonian mechanics and Maxwell's electrodynamics. Maxwell's equations strongly imply that the speed of light is the same to all observers. Most physicists were looking for ways around this implication (and around the conflict). Einstein simply took Maxwell's equations to heart. He accepted that the speed of light is the same to all observers and then determined the fall out from this assumption. This provided the answer to his thoughts from a decade before: You cannot ride a beam of light. That beam of light is speeding away from you at the speed of light. It doesn't matter how much faster you go. That beam of light is still going to go away from you at c. Always.
An object with non-zero mass must always move at subluminal speeds. Your mirror cannot go at the speed of light.
As ignorant as I am concerning both math and physics, I only invisioned the lazer traveling at c. and shooting out a beam also at c It stands to reason that a massless particle such as a photon would likely be the only thing that could travel at c. But now that a (particle/wave?) what ever, with a sub-mass (tactons) are staring us in the face, how do we address the issue? I know it's only theory that they exist, but what if? Edited by rigney
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What if? As DH said above... Basically you are asking "what do the laws of physics say would happen if we violated the laws of physics?"
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What if? As DH said above... Basically you are asking "what do the laws of physics say would happen if we violated the laws of physics?"
What will happen if they are violated?
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What will happen if they are violated?
Nothing.
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Nothing.
Right! We just pick up sticks and go on with it.
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What will happen if they are violated?
What happens in a game of chess when someone moves a rook along a diagonal?
To be a bit less flippant, there are two possible answers to your question.
1.
2. It can't happen, so stop asking.
This is equivalent to asking what happens when a rook moves along a diagonal. It's an illegal move. It can't happen. There is however a problem with this answer: It assumes the laws of physics perfectly describe the universe. They don't. The rules of chess are made by humankind, so we know exactly what they are. The rules of the universe are fumblingly ferreted out by humankind. Our job in doing so isn't perfect or complete. Our laws of physics are best guesses based on evidence, logic, and math. Not perfect. So what if they're wrong? That leads to answer #2:
3. We haven't the foggiest idea what would happen, so stop asking.
Physicists have occasionally performed experiments where the outcome was "Whoa! That can't happen!" These experiments, if confirmed, have the effect of turning physics upside down. The Michelson-Morley experiment is one of the most famous of such experiments. Isaac Asimov: The most exciting phrase to hear in science, the one that heralds the most discoveries, is not "Eureka!" (I found it!) but "That's funny..."
In the case of your mirror, we have very, very good reason to think that your question falls into category #1: "It can't happen, so stop asking." Suppose that it could happen. This means everything we know is wrong. Very, very wrong. This makes your question fall into category #2: "We haven't the foggiest idea what would happen, so stop asking."
Edited by D H
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So now my OP questions are ignored and the thread has gone into off-topic chatter. I would call it "thread hijacking" but I don't like the label, having been accused of it so much myself.
Thanks DH for making it personal and then derailing it onto a sidetrack.
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So now my OP questions are ignored and the thread has gone into off-topic chatter. I would call it "thread hijacking" but I don't like the label, having been accused of it so much myself.
Thanks DH for making it personal and then derailing it onto a sidetrack.
I apologise owl, it was my fault. Your questions were fine and on subject. Mine were simply unmindful questions on the issues. DHs answers were offensive, but hopefully he will eventually get over his arrogance. I wonder, does he even play chess? Edited by rigney
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Right! We just pick up sticks and go on with it.
Nothing that happens will violate physical laws. If something does, then they weren't laws.
Better?
What would happen if physical laws were violated?
If there's a law that says your answer isn't so or is impossible, just consider the law violated.
Though, a more reasonable answer comes from considering the limits of what is physically possible, for example not considering something moving at c but rather what happens in the limit as v approaches c. This is what Einstein did in some of the examples.
But to get back on track... owl, what is missing is a basic understanding of special relativity. I apologize, as having an incomplete understanding of things pointed out by someone seems to be considered offensive around here. Why not start with a simple example of SR found elsewhere on the web and ask about what doesn't make sense to you, rather than devising your own complicated thought experiment that doesn't help you see how it works?
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OK? Now, again, "what am I missing" given that there are only 5,580,000 miles between the ship and the front end of the light beam after a minute of travel?
You are once again assuming that the universe is Newtonian is by thinking that time and distance are immutable. They are not. What you are missing is that time is subject to time dilation, length to length contraction.
In the frame of the ship, those buoys are not 186,000 miles apart. They are only 161080.725 miles apart. At the time the ship emits the beam, the ships captain will compute that the beam will travel 4,832,421.75 miles. (Note well: It's not 60*161080.725 miles = 9,664,843.5 miles. It's half that. In the captain's frame, that last buoy is moving toward the captain at 1/2 c.) The clocks on the ship and the clocks in the space harbor master's frame (the frame in which the buoys are at rest) aren't running at the same rates, either. At the point that the ship has passed 30 buoys it's clock will not show 30 seconds. It will instead show 25.9807 seconds. From the captain's perspective, the beam of light is moving at 4,832,421.75 miles / 25.9807 seconds = 186,000 miles/second. (I've used your value for the speed of light here.)
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Right. I misspoke, which should have been obvious by the context of the ship and its light beam passing the buoys.
The buoys were meant to represent 'mileposts' on the 'road through space' traveled by the ship.
So, to be accurate, the ship which dropped the buoys, if it were going at half 'c' also would have programmed their retro-rockets to "brake", decreasing velocity relative to the ship until they each had lost 1/2 'c' of velocity.
OK? Now, again, "what am I missing" given that there are only 5,580,000 miles between the ship and the front end of the light beam after a minute of travel?
What frame is used to measure the distance between the buoys? Relative to the rest frame of the buoys the distance will be contracted in the spaceship frame by a factor of 0.866 from traveling at c/2. And that minute of travel — is it a minute as measured in the buoy frame? From your setup it's not clear.
(edit: xpost with D H)
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What frame is used to measure the distance between the buoys? Relative to the rest frame of the buoys the distance will be contracted in the spaceship frame by a factor of 0.866 from traveling at c/2. And that minute of travel — is it a minute as measured in the buoy frame? From your setup it's not clear.
(edit: xpost with D H)
I was obviously using the earth standard mile as derived from earth's frame. That is what "a mile" means. So the ship that dropped the buoys would have used that standard for the distances between buoys, once 'fixed' in place.
Of course the ship would have compensated for its 1/2 'c' velocity and the resulting .866 apparent contraction of distance between drops.
I was striving to demonstrate how the whole example would 'pan out' in old fashioned earth miles marked out over the course of the travel through space in question.
Another part of the photon "pushing" issue raised in the OP was raised again in post 4.
In all possible cases of light pushing or being pushed, the question for me still remains, "How is it that mass-less light acts like mass? Or, "how does mass-less momentum "push" at all with no 'substance' (so to speak) to impact whatever it pushes on?"
I redirect attention to the above and ask again.
Edited by owl
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You seem to be asking about a change in the rest mass of light, even though light can never be at rest by definition. Perhaps that's the source of your confusion?
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In all possible cases of light pushing or being pushed, the question for me still remains, "How is it that mass-less light acts like mass? Or, "how does mass-less momentum "push" at all with no 'substance' (so to speak) to impact whatever it pushes on?"
Because a massless particle, i.e. in this case photon, still has momentum. Classical physics is not completely correct and its failure is obvious in some circumstances but these are situations not encountered in everyday experience. The effects from moving fast, strong gravity and/or having massless particles are not explained by pre-1900 physics. The fact that EM radiation would have momentum, though, is part of classical theory. If you have a transfer of energy you are going to transfer momentum as well.
I was obviously using the earth standard mile as derived from earth's frame. That is what "a mile" means. So the ship that dropped the buoys would have used that standard for the distances between buoys, once 'fixed' in place.
Of course the ship would have compensated for its 1/2 'c' velocity and the resulting .866 apparent contraction of distance between drops.
I was striving to demonstrate how the whole example would 'pan out' in old fashioned earth miles marked out over the course of the travel through space in question.
The definition of the mile is not the issue, it was the frame in which the mile was measured. But this is now moot, as D H has explained the details by choosing one of the frames.
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Which leaves the mystery as stated, i.e., why is the beam not boosted to 1&1/2 'c'? Clearly light can not be "pushed" faster than 'c', so what happens to the ship's velocity? I still don't know. But I'm working on it.
The "velocity relative to what?" question (in lieu of no "aether") can be answered by adding some props along the direction of travel. Say some space buoys had been placed along the way, brought to rest (zero velocity) relative to the approaching ship. Say they are placed every 186,000 miles, to make it easy. The ship's light beam passes one every second, while the ship, at 1/2 'c', passes one every two seconds.
After a minute the ship will have passed 30 buoys (having gone 5,580,000 miles) and its beam will have passed 60 of them (having gone 11,160,000 miles.) Then the distance between the ship and the far end of the beam is 5,580,000 miles, yet the light traveled 11,160,000 miles.
What am I missing here?
Start view from Earth:
> Spaceship
- Lightray
o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o bouys
Finish view from Earth:
________________________ 30 buoys __________________________> Spaceship
______________________________________________________________________________________ 60 buoys ________________________- Lightray
o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o_o bouys
From an Earth view nothing happens to the ships velocity, it is moving in the trail behind the photons it emitts, getting further and further behind the front peak of the lightray. The mystery is how the crew on the ship can measure the lightray to move away from the ship with c when a measurement from Earth shows a speed difference of 1/2 c between the ship and the lightray. The answer from relativity is that the crew is in *distorted* space and time compared to observers on Earth, rulers and clocks on the spaceship are no longer calibrated to an equal scale as if they where on Earth.
(Of course from the crews point of view it is the observers on Earth that are in *distorted* space and time.)
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You seem to be asking about a change in the rest mass of light, even though light can never be at rest by definition. Perhaps that's the source of your confusion?
I know that light does not "rest" to be measured as resting mass and that its momentum acts like mass. My question is, how does the energy/momentum of photons "impact" or apply force to objects without a "substantial" agent or carrier of that force, i.e., without mass.
I really don't understand how momentum sans mass acts as a force "pushing" on things as in the OP examples.
Because a massless particle, i.e. in this case photon, still has momentum.
I know and have been referring to that fact.
Classical physics is not completely correct and its failure is obvious in some circumstances but these are situations not encountered in everyday experience. The effects from moving fast, strong gravity and/or having massless particles are not explained by pre-1900 physics. The fact that EM radiation would have momentum, though, is part of classical theory. If you have a transfer of energy you are going to transfer momentum as well.
Check. Also already known. I am not arguing for classical physics. See above reply to iNow for my continuing inquiry about how momentum transfers energy and force ("push") as in my examples.
I also already know that clocks slow down in rate of "timekeeping" as they move faster or experience stronger gravity and that lengths/distances appear contracted from higher velocity frames.
We need not go over all that again.
The definition of the mile is not the issue, it was the frame in which the mile was measured. But this is now moot, as D H has explained the details by choosing one of the frames.
It's my inquiry and I'll tell you what my issue is.
I am not interested in how the string of buoys *appears* to the moving ship. The ship would have used its SR manual and computers to adjust for its clock running slower than earth clocks and for the apparent length contraction (to .866/1 if you figure is correct) it experiences because of that while dropping the buoys.
But its mission was to place the buoys 186,000 earth-based miles apart... not length-contracted "miles" as "seen" from the ship's high velocity.
I have no argument with what DH said about that as per "in the frame of the ship.":
In the frame of the ship, those buoys are not 186,000 miles apart. They are only 161080.725 miles apart.
Also no argument with the rest of that post and all the math.
My thought experiment specified earth miles as the units between buoys. The question then remains regarding the distance between the ship and the end of the light beam (5,580,000 earth-frame miles) after a minute of travel.
I agree with Spyman here:
The answer from relativity is that the crew is in *distorted* space and time compared to observers on Earth, rulers and clocks on the spaceship are no longer calibrated to an equal scale as if they where on Earth.
(Of course from the crews point of view it is the observers on Earth that are in *distorted* space and time.)
My setup was meant to illustrate the example as seen from earth's frame of reference, assuming that the whole operation (both ships) were earth-based at launch and that their 1/2 'c' velocity was relative to earth... not about creating an alternative reality "as seen from the ships' frame of reference."... again, not disputing the observations as seen from the ships.
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I know that light does not "rest" to be measured as resting mass and that its momentum acts like mass. My question is, how does the energy/momentum of photons "impact" or apply force to objects without a "substantial" agent or carrier of that force, i.e., without mass.
I really don't understand how momentum sans mass acts as a force "pushing" on things as in the OP examples.
I know and have been referring to that fact.
Check. Also already known. I am not arguing for classical physics. See above reply to iNow for my continuing inquiry about how momentum transfers energy and force ("push") as in my examples.
A change in momentum is the same as a push. You can't have one without the other. The amount of push additionally depends on the time interval over which the change in momentum happened. If this is from a reflection, then the "how" can be viewed as that the photon is absorbed and re-emitted in the opposite direction. There's an E&M discussion as well, which involves applying the boundary conditions of how the fields have to behave when you go from one medium to the next. In either case, you have redirected the light, so there is a change in momentum. Thus, a force.
It's my inquiry and I'll tell you what my issue is.
I am not interested in how the string of buoys *appears* to the moving ship. The ship would have used its SR manual and computers to adjust for its clock running slower than earth clocks and for the apparent length contraction (to .866/1 if you figure is correct) it experiences because of that while dropping the buoys.
But its mission was to place the buoys 186,000 earth-based miles apart... not length-contracted "miles" as "seen" from the ship's high velocity.
I have no argument with what DH said about that as per "in the frame of the ship.":
Also no argument with the rest of that post and all the math.
My thought experiment specified earth miles as the units between buoys. The question then remains regarding the distance between the ship and the end of the light beam (5,580,000 earth-frame miles) after a minute of travel.
The point I was making is that any time you reference a distance, you must say in what frame it has been measured. You have specified the earth frame, so you see a difference of 5,580,000 miles between the light and the ship, because one moved at c and one moved at c/2.
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I have no argument with what DH said about that as per "in the frame of the ship.":
Also no argument with the rest of that post and all the math.
My thought experiment specified earth miles as the units between buoys. The question then remains regarding the distance between the ship and the end of the light beam (5,580,000 earth-frame miles) after a minute of travel.
What question? I laid it all out. What part did you not understand?
My setup was meant to illustrate the example as seen from earth's frame of reference, assuming that the whole operation (both ships) were earth-based at launch and that their 1/2 'c' velocity was relative to earth... not about creating an alternative reality "as seen from the ships' frame of reference."... again, not disputing the observations as seen from the ships.
Alternative reality? The spaceship POV is just as real as the Earth-based POV, and it is part and parcel of the same reality. It just isn't your Newtonian reality, which you apparently just cannot give up. You don't even know that you are making assumptions of a Newtonian universe in every post you make. You say you accept the math, but it's obvious that deep down, you don't.
You will not be able to understand these concepts until you can see that (1) you are assuming a Newtonian universe, and (2) you are willing to give up that point of view. It's not easy. Relativity is weird and can be very counterintuitive to our intuition.
Our physical intuition is at best Newtonian, and more often than not its even more primitive than that. Our physical intuition is based on what we see around us. For example, the natural state of a rock is obviously at rest. A rock might roll down a hill if someone or something gives it a shove, but it inevitably comes to a stop. This makes it hard for many to accept Newtonian mechanics, let alone something as foreign as relativity.
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The point I was making is that any time you reference a distance, you must say in what frame it has been measured. You have specified the earth frame, so you see a difference of 5,580,000 miles between the light and the ship, because one moved at c and one moved at c/2.
First, on the mechanics of light pushing on things...
say on a solar sail just to keep it simple: How does the energy of its momentum transfer force to the sail with no mass to push against the matter of the sail? The sail absorbs the light, no doubt, but how does the "push" manifest?
As to "distance":
I said that the distance between buoys must be 186,000 in earth miles, not contracted "miles" as measured from the speeding ship... requiring the adjustment to compensate as already discussed. Also I already specified that there is, as you said "a difference of 5,580,000 miles between the light and the ship, because one moved at c and one moved at c/2." I laid that out already. Do you have a point in repeating it all?
The question remains, why are there only 5,580,000 earth-frame miles between the ship and the far end of the light beam if the light traveled at 'c' ahead relative to the ship as well as relative to the buoys during that minute?
Edited by owl
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First, on the mechanics of light pushing on things...
say on a solar sail just to keep it simple: How does the energy of its momentum transfer force to the sail with no mass to push against the matter of the sail? The sail absorbs the light, no doubt, but how does the "push" manifest?
Newtonian mechanics implies that massless objects can't have momentum. You are once again showing your Newtonian biases. You need to drop this. If you cannot do that you will not understand.
The answer is simple: Massless objects can and do have momentum. Even though photons are massless, they do have momentum and energy.
As to "distance":
I said that the distance between buoys must be 186,000 in earth miles, not contracted "miles" as measured from the speeding ship... requiring the adjustment to compensate as already discussed. Also I already specified that there is, as you said "a difference of 5,580,000 miles between the light and the ship, because one moved at c and one moved at c/2." I laid that out already. Do you have a point in repeating it all?
The question remains, why are there only 5,580,000 earth-frame miles between the ship and the far end of the light beam if the light traveled at 'c' ahead relative to the ship as well as relative to the buoys during that minute?
I. Give. Up.
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## Syntax Questions
Hello All,
Sorry for previous posts did not tag a answer, I am a newbie, will not happen again haha. For this post I am trying to read through some previous mathcad solutions and am unsure what this syntax means, does anyone have any idea? Is it saying if mod(j,2)=0 AND mod(j,4)=2,-1,1) then return 0 (screenshot 1)? I have a older version of mathcad so I am also unsure how to write this in my version (screenshot 2). I have also attached the problem document where I found this (PDF).
Best Regards.
1 ACCEPTED SOLUTION
Accepted Solutions
24-Ruby IV
(To:UK_9916373)
Additionally to what Alan already explained I'd like to make clear to you that "if" exists in two flavours.
The first one is the if-function which is used in your original document an the syntax was already explained by Alan. Its similar to the if-function in a spreadsheet like Excel or Calc and if nested it can be confusing and quite hard to read. The advantage in Prime is that the if function is not a premium function and therefore can also be used in the free Prime Express. In Mathcad 15 (and below you don't need to use the programming toolbar to use the if-function, just type it in.
Additionally there is also the if construct in a program and this is what you had tried, given the picture you sent. IMHO using a multiline program may take up more space but is significantly easier to read (and debug, if necessary).
Here are a few way to achieve the very same result, done in Mathcad 15, which may be the version you are using. I save the file in the format of Mathcad 11 to further increase the likelihood that you can actually read the file.
I hope this helps clarify your question.
4 REPLIES 4
15-Moonstone
(To:UK_9916373)
if(a,b,c) means if a is true return b, if a is false return c.
This can be nested, so if(a, if(x,y,z), c) means if a is true then if x is true return y, if x is false return z, if a is false return c
So, with j as an integer, if(mod(j,2)=0, if(mod(j,4)=2, -1,1), 0) means
if j is even (ie mod(j,2)=0) then: if the remainder when dividing j by 4 is 2, return -1, if the remainder when dividing j by 4 isn't 2, return 1,
if j is odd then return 0.
Alan
5-Regular Member
(To:AlanStevens)
Thank you very much Alan!
24-Ruby IV
(To:UK_9916373)
Additionally to what Alan already explained I'd like to make clear to you that "if" exists in two flavours.
The first one is the if-function which is used in your original document an the syntax was already explained by Alan. Its similar to the if-function in a spreadsheet like Excel or Calc and if nested it can be confusing and quite hard to read. The advantage in Prime is that the if function is not a premium function and therefore can also be used in the free Prime Express. In Mathcad 15 (and below you don't need to use the programming toolbar to use the if-function, just type it in.
Additionally there is also the if construct in a program and this is what you had tried, given the picture you sent. IMHO using a multiline program may take up more space but is significantly easier to read (and debug, if necessary).
Here are a few way to achieve the very same result, done in Mathcad 15, which may be the version you are using. I save the file in the format of Mathcad 11 to further increase the likelihood that you can actually read the file.
I hope this helps clarify your question.
5-Regular Member
(To:Werner_E)
This is fantastic. Thank you Alan and Werner!
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https://www.printablemultiplication.com/worksheets/Multiplication-As-Repeated-Addition-Worksheets-3rd-Grade | 1,618,545,371,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00365.warc.gz | 833,141,791 | 76,485 | Multiplication Chart X30
…discovered up coming, multiplication or department? Multiplication is shorthand for addition. At this point, children use a firm understand of addition. As a result, multiplication is the next logical type…
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…counting, addition, subtraction, multiplication, lastly section. This document brings about the query why find out arithmetic with this series? Moreover, why learn multiplication after counting, addition, and subtraction before department?…
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Understanding multiplication right after counting, addition, and subtraction is good. Youngsters understand arithmetic using a all-natural progression. This advancement of understanding arithmetic is usually the subsequent: counting, addition, subtraction, multiplication,… | 504 | 2,728 | {"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} | 3.453125 | 3 | CC-MAIN-2021-17 | latest | en | 0.874216 |
http://math.stackexchange.com/questions/309147/system-with-infinite-number-of-axioms | 1,469,682,482,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00312-ip-10-185-27-174.ec2.internal.warc.gz | 159,644,117 | 18,276 | # System with infinite number of axioms
Assume we have a set of axioms $A_0$. There exists a statement that can be formulated with these axioms that cannot be proven to be true with this system. Assume we give such a statement axiomatic status and add it to $A_0$ and construct a new set of axioms $A_1$. Continue this process indefinitely. Assume we end up with a set of axioms $A_{\infty}$, that contains infinite number of axioms. Is this set effectively generated? Is this system complete? Is it consistent?
I don't know anything about these things, apart from what can be read in Wikipedia about Gödel's incompleteness theorems. Just a thought I had.
-
If you are generating the $A_i$ by a process, then you can't get completeness - there are statements that are unprovable that can't be resolved by that process. If the $A_i$ are not enumerated by a process, then they are not an effective axiom system. – Thomas Andrews Feb 20 '13 at 15:06
Second, if your original theory $A_0$ is effectively generated, then your $A_\infty$ will be too -- there's a mechanical, deterministic process that will eventually print every axiom in $A_\infty$.
$A_\infty$ will be consistent too, if $A_0$ is -- by the "compactness" property of formal logic which says that every inconsistent system of axioms has a finite subsystem that's also inconsistent. (If there's a proof of a contradiction in the system, then because proofs are finite by definition, the proof depends on only finitely many of the axioms). Every finite subset of $A_\infty$ will be a subset of one of the $A_n$'s, and all of these are consistent by construction.
Since $A_\infty$ is effectively generated and (because it extends $A_0$) sufficiently rich, we can repeat the Gödel process on it, and get at Gödel sentence for $A_\infty$. Add that to $A_\infty$ to get $A_{\infty+1}$, and proceed ad nauseam. (In this context it is traditional to write $\omega$ instead of $\infty$, and there's a theory of the necessary numbers "beyond infinity" under the name "ordinal numbers"). | 511 | 2,040 | {"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} | 3.25 | 3 | CC-MAIN-2016-30 | latest | en | 0.923177 |
http://myonlinegrades.com/apjava/quiz/quiz22.php | 1,553,097,471,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202433.77/warc/CC-MAIN-20190320150106-20190320172106-00155.warc.gz | 143,384,876 | 2,882 | Loading
notes intro/old submit AP Problems the dump links
Quiz 22 AP Test Knowledge id: 1.True or false: The multiple choice is worth the same as the free response true false 2. True or false: You have 75 minutes each for the mc and the free response true false 3. If you get a question right it counts as +1 point, if you get a question wrong, it counts as (hint it never hurts to guess) -1 point -.2 points -.25 points Doesnt effect your score 4. To get a 3 (of 5 score), the book says you need to do adequately on the free response and get this percentage right on the multiple choice 50-60 60-70 40-50 70-80 5. a b c d e 6. a b c d e | 184 | 655 | {"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} | 2.75 | 3 | CC-MAIN-2019-13 | latest | en | 0.845667 |
https://hubpages.com/community/hubpages-calculator | 1,652,878,995,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00797.warc.gz | 369,912,796 | 30,616 | # A HubPages Calculator Tool for Estimating How Many Hubs You Need to Make Money
Updated on April 2, 2013
One of my friends was trying to make some quick bucks from Hubpages. I told him to not to expect to be a millionaire overnight. One day he came to me to know how many hubs is needed for making \$500 per month. I calculated his projected income and the number of hubs needed to reach his goal. From that time I thought of building a calculator tool for Hubpages. But I thought that I’ll do it later and I forgot about it. Yesterday I this thought came to me again and I opened my laptop to build it.
## HubPages Calculator
This calculator tool is made of Flash. It took my many hours to develop it. Actually, I am not a developer of any sort. It made my job difficult and time consuming. Anyway, I am glad that it is completed. I am glad that this tool might be helpful for my fellow hubbers (and also for me too).
## How to Use the Calculator
This calculator tool will estimate the number of hubs you needed to make a certain amount of money each month. You need to provide input data for your current number of hubs, total views of last 7 days and your last 7 days total income. The output results will show your current CPM, that is how much you are earning now for each 1000 visitors. Note, this is your CPM not Hubpages’s. And also, the results will so total number of hubs needed to achieve your targeted monthly income and how many more hubs you shall have to write.
## An Example
Let us take an example. Suppose a hubber named “ I. M. Not” has a total of 45 hubs. Mr. Not’s last 7days traffic is 6840. Let us assume that he made \$14.25 in the last 7 days. He wanted to make \$1000 per month from Hubpages. We shall calculate how many hubs are required for earning that amount.
So, if we put the inputs in the calculator, what will happen? The results will show that Mr.Not will have to write a total of 737 hubs to achieve his target. His income from each 1000 views is \$2.08 and his per hub income is \$1.35 per month. He needs to add 692 more hubs to his portfolio. Is it clear now?
## Get it from Below
You can get this calculator from here. It is completely free. It is my property and I give you the right of use and distribute to anybody. If you do not have flash in your computer then you need to install flash player plug-in from here to run the application.
If you want to run it on your Android or iPhone, you need to have a flash player app in your device.
## Is it Safe to Use?
It is completely safe and free of any malware or bots. It is made by me and I am not a hacker or such thing. If you I looked like a criminal to you then do not download the tool.
I shall be happy if this calculator helps you to do your calculations properly.
working | 654 | 2,787 | {"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} | 3.65625 | 4 | CC-MAIN-2022-21 | latest | en | 0.966826 |
https://decodedscience.org/math-may-distinguish-flukes-from-flu-epidemic-outbreaks-2/ | 1,618,526,660,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088264.43/warc/CC-MAIN-20210415222106-20210416012106-00245.warc.gz | 284,385,412 | 16,253 | # Math May Distinguish Flukes from Flu Epidemic Outbreaks
Home / Math May Distinguish Flukes from Flu Epidemic Outbreaks
One sick person can transmit a disease far and wide. Image by SCA Svenska Cellulosa Aktiebolaget
## Implications for Public Health in Recognizing a Flu Epidemic
This paper has two obvious implications for public health and clinical research organizations.
• First, when routine sentinel surveillance can reliably determine that an outbreak is highly unlikely to lead to an epidemic, a full outbreak investigation would not be initiated. Instead, resources would be directed to isolating the reservoir and protecting those most likely to become the head of a chain.
• Second, when an outbreak investigation is required, this method can determine the threat of an influenza epidemic much more quickly and economically by estimating ‘R through determining ‘F’, rather than by exhaustively reviewing each patient’s source of infection./li>
## Lead Author Dr. Simon Cauchemez Responds to Questions
Decoded Science had the opportunity to ask Dr. Simon Cauchemez, lead author of the research paper, some questions to clarify this report.
Decoded Science: Under routine sentinel surveillance, R=1-G and is a trustworthy estimate if G is large enough (close to 1). Would a smaller value of G trigger the outbreak investigation, or cast doubt on the accuracy of the estimate? If so, what value of G would be recommended to trigger further investigation?
Dr. Cauchemez: “If the sample size is large and G<=5%, I’d strongly suspect that R may be above 1.” (further investigation would be indicated in this situation.)
Decoded Science: Both ‘G‘ and ‘F‘ must be between zero and one, so the estimate for ‘R‘ cannot be greater than one. But can ‘R‘ actually be greater than one?
Dr. Cauchemez: “The reproduction number R in an outbreak can be larger than 1. Interpretation of statistics G and F works as follows:
1. If 1-G<1, this gives evidence that R is smaller than 1 and we can estimate R as R=1-G;
2. If 1-G~1, this means that R is above or larger than 1 but we can’t provide a precise estimate of R.
…Our method is designed to estimate R in the context of subcritical outbreaks, i.e., R<1.
Decoded Science: Is it worth determining a value for R>1 by the old-fashioned and labour intensive process in the outbreak investigation?
Dr. Cauchemez: “Indeed, we’re not claiming that this approach should replace outbreak investigations. It’s just adding another method to the toolbox of epidemiologists, allowing simple interpretation of standard surveillance data. It’s also filling a gap for situations where it’s very hard to track cases. Outbreak investigations remain needed to get a precise estimate of R when R>1 and there are many other reasons why we should keep on doing outbreak investigations.
Decoded Science: What would be a threshold value for F in an outbreak investigation that would trigger the more intense approach?
Dr. Cauchemez: “The main concern for public health is if R becomes larger than 1 so (F close to zero). But we’re worried that a virus adapts to increase its ability to transmit between humans. So, if we could document an increase in transmissibility (e.g. from R=0 to R=0.7) – that would be source of concern, even in R remains <1.
## Source
Cauchemez S, Epperson S, Biggerstaff M, Swerdlow D, Finelli L, et al. Using Routine Surveillance Data to Estimate the Epidemic Potential of Emerging Zoonoses: Application to the Emergence of US Swine Origin Influenza A H3N2v Virus. (2013). PLoS Med 10(3): e1001399. doi:10.1371/journal.pmed.1001399. Accessed March 5, 2013.
Categories Uncategorized | 846 | 3,652 | {"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} | 3.578125 | 4 | CC-MAIN-2021-17 | latest | en | 0.933026 |
http://57.bigm-event.com/worksheet-on-application-of-quadratic-equations/ | 1,601,233,733,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401578485.67/warc/CC-MAIN-20200927183616-20200927213616-00398.warc.gz | 3,078,906 | 31,493 | # Worksheet On Application Of Quadratic Equations
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A037409 Numbers n such that the set of base-2 digits of n equals the set of base-4 digits of n. 1
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OFFSET 1,2 LINKS John Cerkan, Table of n, a(n) for n = 1..10000 MATHEMATICA Select[Range[1500], Union[IntegerDigits[#, 2]]==Union[IntegerDigits[#, 4]]&] (* Harvey P. Dale, Sep 14 2014 *) CROSSREFS Subsequence of A000695. Sequence in context: A140299 A119007 A043052 * A178364 A034121 A281685 Adjacent sequences: A037406 A037407 A037408 * A037410 A037411 A037412 KEYWORD nonn,base AUTHOR STATUS approved
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https://wikimili.com/en/Event_(philosophy) | 1,585,431,196,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493120.15/warc/CC-MAIN-20200328194743-20200328224743-00344.warc.gz | 786,711,010 | 23,254 | Event (philosophy)
Last updated
In philosophy, events are objects in time or instantiations of properties in objects.
Kim’s property-exemplification
Jaegwon Kim theorized that events are structured.
They are composed of three things:
1. Object(s) ${\displaystyle [x]}$,
2. a property ${\displaystyle [P]}$ and
3. time or a temporal interval ${\displaystyle [t]}$.
Events are defined using the operation ${\displaystyle [x,P,t]}$.
A unique event is defined by two principles:
a) the existence condition and
b) the identity condition.
The existence condition states “${\displaystyle [x,P,t]}$ exists if and only if object ${\displaystyle x}$ exemplifies the ${\displaystyle n}$-adic ${\displaystyle P}$ at time ${\displaystyle t}$.” This means a unique event exists if the above is met. The identity condition states “${\displaystyle [x,P,t]}$ is ${\displaystyle [y,Q,t']}$ if and only if ${\displaystyle x=y}$, ${\displaystyle P=Q}$ and ${\displaystyle t=t'}$.”
Kim uses these to define events under five conditions:
1. One, they are unrepeatable, unchangeable particulars that include changes and the states and conditions of that event.
2. Two, they have a semi-temporal location.
3. Three, only their constructive property creates distinct events.
4. Four, holding a constructive property as a generic event creates a type-token relationship between events, and events are not limited to their three requirements (i.e. ${\displaystyle [x,P,t]}$). Critics of this theory such as Myles Brand have suggested that the theory be modified so that an event had a spatiotemporal region; consider the event of a flash of lightning. The idea is that an event must include both the span of time of the flash of lightning and the area in which it occurred.
Other problems exist within Kim's theory, as he never specified what properties were (e.g. universals, tropes, natural classes, etc.). In addition, it is not specified if properties are few or abundant. The following is Kim's response to the above.
. . . [T]he basic generic events may be best picked out relative to a scientific theory, whether the theory is a common-sense theory of the behavior of middle-sized objects or a highly sophisticated physical theory. They are among the important properties, relative to the theory, in terms of which lawful regularities can be discovered, described, and explained. The basic parameters in terms of which the laws of the theory are formulated would, on this view, give us our basic generic events, and the usual logical, mathematical, and perhaps other types of operations on them would yield complex, defined generic events. We commonly recognize such properties as motion, colors, temperatures, weights, pushing, and breaking, as generic events and states, but we must view this against the background of our common-sense explanatory and predictive scheme of the world around us. I think it highly likely that we cannot pick out generic events completely a priori. [1]
There is also a major debate about the essentiality of a constitutive object. There are two major questions involved in this: If one event occurs, could it have occurred in the same manner if it were another person, and could it occur in the same manner if it would have occurred at a different time? Kim holds that neither are true and that different conditions (i.e. a different person or time) would lead to a separate event. However, some consider it natural to assume the opposite.
Davidson
Donald Davidson and John Lemmon proposed a theory of events that had two major conditions, respectively: a causal criterion and a spatiotemporal criterion.
The causal criterion defines an event as two events being the same if and only if they have the same cause and effect.
The spatiotemporal criterion defines an event as two events being the same if and only if they occur in the same space at the same time. Davidson however provided this scenario; if a metal ball becomes warmer during a certain minute, and during the same minute rotates through 35 degrees, must we say that these are the same event? However, one can argue that the warming of the ball and the rotation are possibly temporally separated and are therefore separate events.
Lewis
David Lewis theorized that events are merely spatiotemporal regions and properties (i.e. membership of a class). He defines an event as “e is an event only if it is a class of spatiotemporal regions, both thisworldly (assuming it occurs in the actual world) and otherworldly.” The only problem with this definition is it only tells us what an event could be, but does not define a unique event. This theory entails modal realism, which assumes possible worlds exist; worlds are defined as sets containing all objects that exist as a part of that set. However, this theory is controversial. Some philosophers have attempted to remove possible worlds, and reduce them to other entities. They hold that the world we exist in is the only world that actually exists, and that possible worlds are only possibilities.
Lewis’ theory is composed of four key points. Firstly, the non-duplication principle; it states that x and y are separate events if and only if there is one member of x that is not a member of y (or vice versa). Secondly, there exist regions that are subsets of possible worlds and thirdly, events are not structured by an essential time.
In Being and Event, Alain Badiou writes that the event (événement) is a multiple which basically does not make sense according to the rules of the "situation," in other words existence. Hence, the event "is not," and therefore, in order for there to be an event, there must be an "intervention" which changes the rules of the situation in order to allow that particular event to be ("to be" meaning to be a multiple which belongs to the multiple of the situation — these terms are drawn from or defined in reference to set theory). In his view, there is no "one," and everything that is is a "multiple." "One" happens when the situation "counts," or accounts for, acknowledges, or defines something: it "counts it as one." For the event to be counted as one by the situation, or counted in the one of the situation, an intervention needs to decide its belonging to the situation. This is because his definition of the event violates the prohibition against self-belonging (in other words, it is a set-theoretical definition which violates set theory's rules of consistency), thus does not count as extant on its own. [2]
Deleuze
Gilles Deleuze lectured on the concept of event on March 10, 1987. A sense of the lecture is described by James Williams. [3] Williams also wrote, "From the point of view of the difference between two possible worlds, the event is all important". [4] He also stated, "Every event is revolutionary due to an integration of signs, acts and structures through the whole event. Events are distinguished by the intensity of this revolution, rather than the types of freedom or chance." [5] In 1988 Deleuze published a magazine article "Signes et événements" [6]
In his book Nietszche and Philosophy, he addresses the question "Which one is beautiful?" In the preface to the English translation he wrote:
The one that ... does not refer to an individual, to a person, but rather to an event, that is, to the forces in their various relationships to a proposition or phenomenon, and the genetic relationship that determines these forces (power). [7]
Kirkeby
The Danish philosopher Ole Fogh Kirkeby deserves mentioning, as he has written a comprehensive trilogy about the event, or in Danish "begivenheden". In the first work of the trilogy "Eventum tantum – begivenhedens ethos" [8] (Eventum tantum - the ethos of the event) he distinguishes between three levels of the event, inspired from Nicola Cusanus: Eventum tantum as non aliud, the alma-event and the proto-event.
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References
1. Jaegwon Kim (1993) Supervenience and Mind, page 37, Cambridge University Press
2. Alain Badiou (1988) L'Être et l'Événement
3. Charles J. Stivale (editor) (2011) Gilles Deleuze: Key Concepts, 2nd edition, chapter 6: Event, pp 80–90
4. James Williams (2003) Gilles Deleuze’s Difference and Repetition: A Critical Introduction and Guide, page 78, Edinburgh University Press
5. Williams 2003 p xi
6. Gilles Deleuze (1988) "Signes et événements", Magazine Littéraire, #257, pages 16 to 25
7. Michael Hart (1993) Gilles Deleuze: An apprenticeship in philosophy, page 31, University of Minnesota Press ISBN 0-8166-2160-8
8. Ole Fogh Kirkeby (2005) Eventum tantum : Begivenhedens ethos. København: Samfundslitteratur | 3,954 | 18,115 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "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} | 2.71875 | 3 | CC-MAIN-2020-16 | latest | en | 0.901948 |
https://justaaa.com/finance/153514-consider-an-asset-that-costs-595197-and-is | 1,723,331,276,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00547.warc.gz | 259,863,528 | 9,072 | Question
# Consider an asset that costs \$595197 and is depreciated straight-line to zero over its seven-year tax...
Consider an asset that costs \$595197 and is depreciated straight-line to zero over its seven-year tax life. The asset is to be used in a five-year project; at the end of the project, the asset can be sold for \$152603. If the relevant tax rate is 35 percent, what is the aftertax cash flow from the sale of this asset?
Depreciation per year = Asset price / total life
Depreciation = 595197 / 7 = 85028.14286
Book value of asset = Asset purchase price or cost - Depreciation x total number of usage in years
Book value of asset = 595197 - 85028.14286 x 5
Book value of asset = 170056.2857
Loss on sales of asset = Sales value – Book value
Loss on sales of asset = 152603 - 170056.2857
Loss on sales of asset = - 17453.2857
After tax cash flow = Sales value + |Loss on sale of asset| x tax rate
After tax cash flow = 152603 + 17453.2857 x 35%
After tax cash flow = \$158,711.65
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 290 | 1,066 | {"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} | 3.1875 | 3 | CC-MAIN-2024-33 | latest | en | 0.892893 |
https://rd.springer.com/article/10.1007%2Fs40879-018-00311-6 | 1,547,989,974,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583716358.66/warc/CC-MAIN-20190120123138-20190120145138-00087.warc.gz | 644,672,593 | 16,698 | # A parallel metrization theorem
• Taras Banakh
• Olena Hryniv
Open Access
Research Article
## Abstract
Two non-empty sets AB of a metric space (Xd) are called parallel if $$d(a,B)=d(A,B)=d(A,b)$$ for any points $$a\in A$$ and $$b\in B$$. Answering a question posed on mathoverflow.net, we prove that for a cover $${\mathscr {C}}$$ of a metrizable space X by compact subsets, the following conditions are equivalent: (i) the topology of X is generated by a metric d such that any two sets $$A,B\in {\mathscr {C}}$$ are parallel; (ii) the cover $${\mathscr {C}}$$ is disjoint, lower semicontinuous and upper semicontinuous.
## Keywords
Metrization Parallel sets Metric space
## Mathematics Subject Classification
54E35
In this paper we shall prove a “parallel” metrization theorem answering a question [1] of the Mathoverflow user116515. The question concerns parallel sets in metric spaces.
Two non-empty sets AB in a metric space (Xd) are called parallel if
\begin{aligned} d(a,B)=d(A,B)=d(A,b)\quad \hbox {for any } a\in A \hbox { and } b\in B. \end{aligned}
Here $$d(A,B)=\inf \{ d(a,b):a\in A,\, b\in B\}$$ and for $$x\in X$$. Observe that two closed parallel sets AB in a metric space are either disjoint or coincide.
Let $${\mathscr {C}}$$ be a family of non-empty closed subsets of a topological space X. A metric d on X is defined to be $${\mathscr {C}}$$-parallel if any two sets $$A,B\in {\mathscr {C}}$$ are parallel with respect to the metric d.
A family $${\mathscr {C}}$$ of subsets of X is called a compact cover of X if $$X=\bigcup {\mathscr {C}}$$ and each set $$C\in {\mathscr {C}}$$ is compact.
### Problem
([1]) For which compact covers $${\mathscr {C}}$$ of a topological space X the topology of X is generated by a $${\mathscr {C}}$$-parallel metric?
A metric generating the topology of a given topological space will be called admissible. A necessary condition for the existence of an admissible $${\mathscr {C}}$$-parallel metric is the upper and lower semicontinuity of the cover $${\mathscr {C}}$$.
A family $${\mathscr {C}}$$ of subsets of a topological space X is called
• lower semicontinuous if for any open set $$U\subset X$$ its $${\mathscr {C}}$$-star is open in X;
• upper semicontinuous if for any closed set $$F\subset X$$ its $${\mathscr {C}}$$-star is closed in X;
• continuous if $${\mathscr {C}}$$ is both lower and upper semicontinuous;
• disjoint if any distinct sets $$A,B\in {\mathscr {C}}$$ are disjoint.
The following theorem is the main result of the paper, answering Problem.
### Theorem
For a compact cover $${\mathscr {C}}$$ of a metrizable topological space X the following conditions are equivalent:
1. (i)
the topology of X is generated by a $${\mathscr {C}}$$-parallel metric;
2. (ii)
the family $${\mathscr {C}}$$ is disjoint and continuous.
### Proof
(i) $$\Rightarrow$$ (ii). Assume that d is an admissible $${\mathscr {C}}$$-parallel metric on X. The disjointness of the cover $${\mathscr {C}}$$ follows from the obvious observation that two closed parallel sets in a metric space are either disjoint or coincide.
To see that $${\mathscr {C}}$$ is lower semicontinuous, fix any open set $$U\subset X$$ and consider its $${\mathscr {C}}$$-star . To see that is open, take any point and find a set $$C\in {\mathscr {C}}$$ such that $$s\in C$$ and $$C\cap U\ne \varnothing$$. Fix a point $$u\in U\cap C$$ and find $$\varepsilon >0$$ such that the $$\varepsilon$$-ball $$B(u;\varepsilon )=\{ x\in X:d(x,u)<\varepsilon \}$$ is contained in U. We claim that . Indeed, for any $$x\in B(s;\varepsilon )$$ we can find a set $$C_x\in {\mathscr {C}}$$ containing x and conclude that $$d(C_x,u)=d(C_x,C)\leqslant d(x,s)<\varepsilon$$ and hence $$C_x\cap U\ne \varnothing$$ and .
To see that $${\mathscr {C}}$$ is upper semicontinuous, fix any closed set $$F\subset X$$ and consider its $${\mathscr {C}}$$-star . To see that is closed, take any point and find a set $$C\in {\mathscr {C}}$$ such that $$s\in C$$. It follows from that $$C\cap F=\varnothing$$ and hence by the compactness of C. We claim that . Assuming the opposite, we can find a point and a set $$C_x\in {\mathscr {C}}$$ such that $$x\in C_x$$ and $$C_x\cap F\ne \varnothing$$. Fix a point $$z\in C_x\cap F$$ and observe that $$d(C,F)\leqslant d(C,z)=d(C,C_x)\leqslant d(s,x)<\varepsilon =d(C,F)$$, which is a desired contradiction.
(ii) $$\Rightarrow$$ (i). The proof of this implication is more difficult. Assume that $${\mathscr {C}}$$ is disjoint and continuous. Fix any admissible metric $$\rho \leqslant 1$$ on X.
Let $${\mathscr {U}}_0(C)=\{X\}$$ for every $$C\in {\mathscr {C}}$$.
### Claim
For every $$n\in {\mathbb {N}}$$ and every $$C\in {\mathscr {C}}$$ there exists a finite cover $${\mathscr {U}}_n(C)$$ of C by open subsets of X such that
1. (a)
each set $$U\in {\mathscr {U}}_n(C)$$ has $$\rho$$-diameter $$\leqslant 1/{2^n}$$;
2. (b)
if a set $$A\in {\mathscr {C}}$$ meets some set $$U\in {\mathscr {U}}_n(C)$$, then $$A\subset \bigcup {\mathscr {U}}_n(C)$$ and A meets each set $$U'\in {\mathscr {U}}_n(C)$$.
### Proof
Using the paracompactness [2, 5.1.3] of the metrizable space X, choose a locally finite open cover $${\mathscr {V}}$$ of X consisting of sets of $$\rho$$-diameter $$<1/{2^n}$$.
For every compact set $$C\in {\mathscr {C}}$$ consider the finite subfamily of the locally finite cover $${\mathscr {V}}$$. Since the cover $${\mathscr {C}}$$ is upper semicontinuous, the set is closed and disjoint with the set C. Since $${\mathscr {C}}$$ is lower semicontinuous, for any open set $$V\in {\mathscr {V}}(C)$$ the set is open and hence is an open neighborhood of C.
Put and observe that $${\mathscr {U}}_n$$ satisfies condition (a).
Let us show that the cover $${\mathscr {U}}_n(C)$$ satisfies condition (b). Assume that a set $$A\in {\mathscr {C}}$$ meets some set $$U\in {\mathscr {U}}_n(C)$$. First we show that $$A\subset \bigcup {\mathscr {U}}_n(C)$$. Find a set $$V\in {\mathscr {V}}(C)$$ such that $$U=W(C)\cap V$$. It follows from $$\varnothing \ne A\cap U\subset A\cap W(C)$$ that the set A meets W(C) and hence is contained in W(C) and is disjoint with $$F_C$$. Hence
Next, take any set $$U'\in {\mathscr {U}}_n(C)$$ and find a set $$V'\in {\mathscr {V}}(C)$$ with $$U'=W(C)\cap V'$$. The relation $$A\cap W(C)\cap V=A\cap U\ne \varnothing$$ and the definition of the set $$W(C)\supset A$$ imply that A intersects the set $$V'\in {\mathscr {V}}(C)$$ and hence intersects the set $$U'=W(C)\cap V'$$. $$\blacksquare$$
Given two points $$x,y\in X$$ let
Adjust the function $$\delta$$ to a pseudometric d letting
where the infimum is taken over all sequences $$x=x_0$$, $$\dots$$, $$x_m=y$$. Condition (a) of Claim implies that $$\rho (x,y)\leqslant \delta (x,y)$$ and hence $$\rho (x,y)\leqslant d(x,y)$$ for any $$x,y\in X$$. So, the pseudometric d is a metric on X such that the identity map $$(X,d)\rightarrow (X,\rho )$$ is continuous. To see that this map is a homeomorphism, take any point $$x\in X$$ and $$\varepsilon >0$$. Find $$n\in {\mathbb {N}}$$ such that $$1/{2^n}<\varepsilon$$ and choose a set $$C\in {\mathscr {C}}$$ with $$x\in C$$ and a set $$U\in {\mathscr {U}}_n(C)$$ with $$x\in U$$. Then for any $$y\in U$$ we get $$d(y,x)\leqslant \delta (x,y)\leqslant 1/{2^n}<\varepsilon$$, which means that the map $$X\rightarrow (X,d)$$ is continuous.
Finally, let us prove that the metric d is $${\mathscr {C}}$$-parallel. Pick any two distinct compact sets $$A,B\in {\mathscr {C}}$$. We need to show that $$d(a,B)=d(A,B)=d(A,b)$$ for any $$a\in A$$, $$b\in B$$. Assuming that this inequality is not true, we conclude that either $$d(a,B)>d(A,B)>0$$ or $$d(A,b)>d(A,B)>0$$ for some $$a\in A$$ and $$b\in B$$.
First assume that $$d(a,B)>d(A,B)$$ for some $$a\in A$$. Choose points $$a'\in A$$, $$b'\in B$$ such that $$d(a',b')=d(A,B)<d(a,B)$$. By the definition of the distance $$d(a',b')<d(a,B)$$, there exists a chain $$a'=x'_0, x'_1,\dots ,x'_m=b'$$ such that $$\sum _{ i=1}^{ m}\delta (x'_{i-1},x'_i)<d(a,B)$$. We can assume that the points $$x'_0,\dots ,x'_m$$ are pairwise distinct, so for every $$i\leqslant m$$ there exist $$n_i\geqslant 0$$ such that $$\delta (x'_{i-1},x'_i)= 1/{2^{n_i}}$$ and hence $$x'_{i-1},x'_i\in U_i'$$ for some $$C_i\in {\mathscr {C}}$$ and $$U_i'\in {\mathscr {U}}_{n_i}(C_i)$$. For every $$i\leqslant m$$ let $$A_i\in {\mathscr {C}}$$ be the unique set with $$x_i'\in A_i$$. Then $$A_0=A$$ and $$A_m=B$$.
Using condition (b), we can inductively construct a sequence of points $$a=x_0,x_1,\dots ,x_m\in B$$ such that for every positive $$i\leqslant m$$ the point $$x_i$$ belongs to $$A_i$$ and the points $$x_{i-1},x_i$$ belong to some set $$U_i\in {\mathscr {U}}_{n_i}(C_i)$$. The chain $$a=x_0,x_1,\dots ,x_m\in A_m=B$$ witnesses that
which is a desired contradiction. By analogy we can prove that the case $$d(A,B)<d(A,b)$$ leads to a contradiction. $$\square$$
## References
1. 1.
user116515: Making compact subsets “parallel”. https://mathoverflow.net/questions/284544/making-compact-subsets-parallel. Accessed 27 Oct 2017
2. 2.
Engelking, R.: General Topology. Sigma Series in Pure Mathematics, vol. 6, 2nd edn. Heldermann, Berlin (1989) | 3,041 | 9,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-04 | latest | en | 0.824328 |
https://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/archimedes_repository/large/fabri_tract_026_la_1646/index.meta&start=11&viewMode=text&pn=454&tocMode=figures | 1,675,840,092,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00285.warc.gz | 235,355,549 | 4,175 | Fabri, Honoré, Tractatus physicus de motu locali, 1646
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1
APPENDIX PRIMA
PHYSICOMATHEMATICA,
De centro percuſsionis.
DE duplici centro hactenus actum eſt,
magnitudinis, ſcilicet, & grauitatis;
præſertim de hoc vltimo:
in quo certè
opere non ſine maxima laude præ
ſtantiſſimi Mathematici deſudarunt,
ſcilicet Archimedes, Commandinus, Lucas Vale
rius, Steuinus, Guldinus, Galileus paucis:
ſed du
plex aliud centrum conſiderari poteſt;
primum di
citur centrum impreſſionis: vtrumque prorſus inta
ctum aliis doctâ paucarum licèt propoſitionum co
ronâ, vel peripheria in hac appendice corona
mus.
DEFINITIO I.
CEntrum grauitatis eſt punctum, quod omnia grauitatis momenta æqua
liter dirimit.
Clara eſt definitio; centrum enim grauitatis eſt illud punctum, ex
quo pendulum corpus per quamlibet lineam ſeruat æquilibrium.
Definitio 2.
Centrum impreſſionis eſt illud, per quod, ſi ducatur planum vtrimque, di
rimit æqualem impetum.
Hæc etiam clara eſt; conſideratur autem impetus non modò ratione | 580 | 1,410 | {"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} | 2.96875 | 3 | CC-MAIN-2023-06 | latest | en | 0.139216 |
https://testpremier.com/book/mercer-graduate-practice-aptitude-test-pack-2022/ | 1,656,156,260,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00260.warc.gz | 632,471,749 | 67,575 | Mercer Graduate Practice Aptitude Test Pack 2022
\$19
Practice Full-length aptitude tests questions for Mercer graduate jobs. Featuring the latest questions of the same style and difficulty of the actual test. Practice now, track your scores and ace it!
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Enhance your practice for Mercer graduate assessment with extensive practice questions from the Mercer Graduate Practice pack, and featuring all the sections on the actual exams. This book provides insight into what to expect, and helps you develop effective study strategies.
Kick off your preparation with our Mercer assessment prep bundle (all-in-one pack):
• Mercer Numerical reasoning test / Mercer Assessment
• Mercer Verbal reasoning test / Mercer Assessment
• Mercer Abstract reasoning tests / Mercer Assessment
• Mercer Logical reasoning tests / Mercer Assessment
With step by step explanations on every questions, and hints on how to solve them faster.
Mercer Graduate Practice pack tests formats; What to expect:
The sections on the assessments may include any of the following test sections, depending on the role that you applied to and also the country you are applying from:
1. Numerical Reasoning
2. Verbal Reasoning
3. Diagramatic
4. Interview
5. Personality Test
Sample Mercer Graduate Practice pack test Questions
Mercer Graduate Practice pack Numerical Reasoning
Question 1
Approximately what proportion of Beauty & Fragrance are Store-Brand Product?
A. 12.5%
B. 15.8%
C. 16.8%
D. 10.8%
E. 19.7%
Question 2
What proportion of Jewelry & Watches are National Product?
A. 13.41%
B. 18.29%
C. 31.71%
D. 26.83%
E. 9.76%
Question 3
What is the approximate ratio of National Product to Imported products?
A. none
B. 3.73:1
C. 1:3.73
D. 1:2.73
E. 2.73:1
Question 4
What proportion of Clothing & Shoes and Home Products are Imported Products.
A. 50%
B. 51%
C. 54%
D. 53%
E. 52%
1. What proportion of beauty and fragrance are stored-brand products:
Beauty and fragrance that are brand = 30
Total stored-brand products = 45 + 30 + 60 +15 = 190
So therefore, proportion of beauty and fragrance that are stored-brand products:
30/190 x 100 = 15.8%. Answer: (B)
2. Jewelry and watches that are natural products = 80
Total natural products = 160 + 150 + 200 + 220 + 80 = 820
So therefore, proportion of Jewelry and watches that are natural products are:
80/820 x 100 = 9.76%. Answer: (E)
3. Total natural products = 820, total imported products = 2240
Ratio of national products to imported products = 820: 2240 =
820/2240 = 41/114 = 41:112 = 1:2.73 (D)
4. What proportion of clothing and shoe and home products are imported products
Clothing and shoes = 680 for imported products
Homes products = 345 for imported products, total = 1025
1025/2240 x 100 = 54%. Answer: (C)
Mercer Graduate Practice pack Verbal Reasoning
Sages Store enjoys an international reputation for quality and style. Nowhere is this more important than in the dress and appearance of its staff. The company sets minimum standards of appearance which are demanded of all shop floor staff, although some departments have specific additional requirements. Hair must be clean, tidy and well cut at all times. With very few exceptions, such as “Designer Corner”, which operates a different staff dress code reflecting their particular style, business dress must be worn…Women should wear tailored suits, with a white or cream blouse. Men should wear dark grey trousers together with a white shirt and navy blazer.
Question 1
Women in “Designer Corner” are allowed to wear jewelry.
A. True
B. False
C. Cannot say
Question 2
Business dress must be worn by staff in all departments.
A. True
B. False
C. Cannot say
The early chaos of the home computing industry in the USA, where it developed, probably had a more detrimental effect in Europe than it did in the States. All the inovators in the field were companies which were too small to cope with or understand foreign sales. As a result, all US companies sold exclusively through European distributors, some of which were only interested in making maximum profits in a minimum amount of time. Home computing in Europe got off to a slow start because greedy distributors worked through incompetent suppliers, none of which had any real interest in the long term future of the technology.
Question 3
Incompetent suppliers were one of the reasons for the slow development of home computing in Europe.
A. True
B. False
C. Cannot say
Question 4
None of the American innovators in the field were able to deal adequately with foreign sales.
A. True
B. False
C. Cannot say
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A: Testpremier practice packs do not only contain randomly created simulated questions. They infact includes actual tests that appear in the past, which are shared with us by previous candidates of the exams. We guarantee that studying with these materials, you will increase your chance of passing and scoring higher by 99.9% or you get your money back. Please see our terms and conditions for all the details. | 1,302 | 5,531 | {"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} | 3.59375 | 4 | CC-MAIN-2022-27 | latest | en | 0.806346 |
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## Difficulty level
Lecture title:
The probability of a hypothesis, given data.
Difficulty level: Beginner
Duration: 7:57
Speaker: : Barton Poulson
Lecture title:
Why math is useful in data science.
Difficulty level: Beginner
Duration: 1:35
Speaker: : Barton Poulson
Lecture title:
Why statistics are useful for data science.
Difficulty level: Beginner
Duration: 4:01
Speaker: : Barton Poulson
Lecture title:
Statistics is exploring data.
Difficulty level: Beginner
Duration: 2:23
Speaker: : Barton Poulson
Lecture title:
Graphical data exploration
Difficulty level: Beginner
Duration: 8:01
Speaker: : Barton Poulson
Lecture title:
Numerical data exploration
Difficulty level: Beginner
Duration: 5:05
Speaker: : Barton Poulson
Lecture title:
Simple description of statistical data.
Difficulty level: Beginner
Duration: 10:16
Speaker: : Barton Poulson
Lecture title:
Basics of hypothesis testing.
Difficulty level: Beginner
Duration: 06:04
Speaker: : Barton Poulson
In this lecture, the speaker demonstrates Neurokernel's module interfacing feature by using it to integrate independently developed models of olfactory and vision LPUs based upon experimentally obtained connectivity information.
Difficulty level: Intermediate
Duration: 29:56
Speaker: : Aurel A. Lazar
Lecture title:
Introduction to the Mathematics chapter of Datalabcc's "Foundations in Data Science" series.
Difficulty level: Beginner
Duration: 2:53
Speaker: : Barton Poulson
Lecture title:
Primer on elementary algebra
Difficulty level: Beginner
Duration: 3:03
Speaker: : Barton Poulson
Lecture title:
Primer on linear algebra
Difficulty level: Beginner
Duration: 5:38
Speaker: : Barton Poulson
Lecture title:
Primer on systems of linear equations
Difficulty level: Beginner
Duration: 5:24
Speaker: : Barton Poulson
Lecture title:
Primer on calculus
Difficulty level: Beginner
Duration: 4:17
Speaker: : Barton Poulson
Lecture title:
How calculus relates to optimization
Difficulty level: Beginner
Duration: 8:43
Speaker: : Barton Poulson
Lecture title:
Big O notation
Difficulty level: Beginner
Duration: 5:19
Speaker: : Barton Poulson
Lecture title:
Basics of probability.
Difficulty level: Beginner
Duration: 7:33
Speaker: : Barton Poulson
Lecture title:
A basic introduction to clinical presentation of schizophrenia, its etiology, and current treatment options.
Difficulty level: Beginner
Duration: 51:49
Lecture title:
Introduction to the types of glial cells, homeostasis (influence of cerebral blood flow and influence on neurons), insulation and protection of axons (myelin sheath; nodes of Ranvier), microglia and reactions of the CNS to injury.
Difficulty level: Beginner
Duration: 40:32 | 665 | 2,726 | {"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} | 2.8125 | 3 | CC-MAIN-2019-43 | latest | en | 0.674077 |
http://www.howtodothings.com/home-garden/how-to-crack-a-master-lock-combination-lock | 1,556,295,195,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578841544.98/warc/CC-MAIN-20190426153423-20190426175423-00127.warc.gz | 238,392,894 | 7,281 | # How To Crack a Master Lock Combination Lock
Sometimes, you may forget or misplace the combination sequence to open a "Master Lock" combination and lock. Short of physically hacking away at the lock to open it, there are a few ways in which you can crack the combination and open the lock. Some of these ways are listed below for your reference.
## Step 1
Set the dial. To begin, set the dial on the lock to zero. Next, tug on the lock loop and keep twisting the dial until it doesn't move anymore. Make a note of the number is stops at; if the dial has stopped midway between numbers, for example, between 4 and 5, note the number as 4.5.
## Step 2
Release the tension. Take away the pressure you've been applying and begin moving the dial once more. Tug the lock loop again as in step one and twist the dial till it stops. Note down the number the dial stops at, again. Continue repeating this action till the dial reaches 0 every time. Note down all the twelve numbers you get on a piece of paper. You should have 7 fractions and five full numbers. Of the five whole numbers, four will end with the same number. The fifth whole number is the 3rd number in your combination sequence. Mark this as so.
## Step 3
Getting the remaining numbers in the sequence. Take the number you've isolated in step two and divide it by 4. Note down the result, a number falling between 0 and 3.
## Step 4
Selection for first number of the combination sequence. Take the number found in the previous step and keep adding 4 to it until you've got 10 new numbers. These are the probables for the first number of the sequence.
## Step 5
Selection for the second combination number. Take the same number from step 3 and add 2 this time, till you get another set of ten numbers. Any of these numbers could be the second digit in the combination.
## Step 6
Permutations and combinations. Take the single number from step 3 and the two sets of 10 numbers from the next two steps. Try out all the possible combinations with these three sets to arrive at the best possible combination sequence for your master lock.
You will probably spend the most time on the last step; given that there can be eighty possible combinations. However, you may not need to try out all 80 combinations, before you find the correct one. | 518 | 2,306 | {"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} | 3.75 | 4 | CC-MAIN-2019-18 | latest | en | 0.916847 |
https://study.com/academy/answer/the-can-division-of-fruit-products-inc-manufactures-and-sells-tin-cans-externally-for-0-60-per-can-its-unit-variable-costs-and-unit-fixed-costs-are-0-24-and-0-08-respectively-the-packaging-division-wants-to-purchase-50-000-cans-at-0-32-a-can-sell.html | 1,652,805,413,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00104.warc.gz | 625,001,110 | 22,135 | The Can Division of Fruit Products Inc. manufactures and sells tin cans externally for $0.60 per... Question: The Can Division of Fruit Products Inc. manufactures and sells tin cans externally for {eq}\$0.60 {/eq} per can. Its unit variable costs and unit fixed costs are {eq}\$0.24 {/eq} and {eq}\$0.08 {/eq}, respectively. The packaging division wants to purchase 50,000 cans at {eq}\$0.32 {/eq} a can. Selling internally will save {eq}\$0.02 {/eq} a can. Assuming the Can Division has sufficient capacity, what is the minimum transfer price it should accept?
A) {eq}\$0.24 {/eq}. B) {eq}\$0.32 {/eq}.
C) {eq}\$0.22 {/eq}. D) {eq}\$0.30 {/eq}.
Transfer Price:
Transfer price is the price at which one department of a company transfers its goods or services to another department of the same company. The transfer price should be such that it does not result in loss to the transferor department.
The correct option is (C) $0.22 The Can division will recover its fixed costs from its normal capacity. So, it will not include fixed costs in the transfer price. The transfer price would be the effective variable cost of the can division. {eq}\text{Transfer Price} = \text{Variable Cost per unit} - \text{Savings in cost per unit}\\ \text{Transfer Price} = \$0.24 - \$0.02\\ \text{Transfer Price} = \$0.22 {/eq}
Negotiated Transfer Pricing: Definition & Examples
from
Chapter 10 / Lesson 6
8.6K
Negotiated transfer pricing is where company representatives negotiate prices themselves, not basing purely on market prices. Learn several advantages and disadvantages to using negotiated transfer pricing demonstrated through two companies' negotiations. | 424 | 1,658 | {"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} | 2.890625 | 3 | CC-MAIN-2022-21 | latest | en | 0.913685 |
https://hrtanswers.com/what-is-the-mass-of-1-56-x-1021-atoms-of-magnesium-in-grams-solved/ | 1,670,518,102,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00451.warc.gz | 328,885,734 | 13,041 | # What is the mass of 1.56 x 1021 atoms of magnesium in grams [Solved]
(6.02X10^23)X24.3=1.46286X10^25
mass of magnesium is 0.0629g
1 Mole of Mg is 24.3g.
1 Mole of Mg is 24.3g.
1.56X1021/1.46286X10^25=
m=0.00259 x 24.3= 0.0629g of Mg
mass of magnesium is 0.0629g
To work out the mass of 1.56 x 10^21 atoms of magnesium is grams is split into 2 parts first we have to convert the number of atoms into the number of moles and we do that using this formula
n=N/NA
n=number of moles
N=molecules of substance
NA=avogadros number which is 6.02 x 10^23
mass of magnesium is 0.0629g
then to work out the mass you have to multiply the number of mole(n)
by the molar mass(M) of magnesium which is 24.3
m=n x M
m=mass
n=number of moles
M=molar mass
m=0.00259 x 24.3= 0.0629g of Mg
1.5610^21 atoms x (1 mol / 6.02210^23 atoms ) = 2.5910^-3 moles
1.09X10^-22g
1.56X1021/1.46286X10^25=
1.09X10^-22g
(6.02X10^23)X24.3=1.46286X10^25
mass of magnesium is 0.0629g
This post is last updated on hrtanswers.com at Date : 1st of September – 2022
Get Answer for Assuming equal concentrations, rank these aqueous solutions by their freezing point: Na3PO4,SnCl4,LiNO3,K2SO4 [Solved] | 462 | 1,178 | {"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} | 3.78125 | 4 | CC-MAIN-2022-49 | latest | en | 0.774487 |
https://studdy.ai/shared-solution/3a78407a-02ba-49d7-b2be-3a32bf84c6ac | 1,721,143,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00171.warc.gz | 486,830,317 | 8,770 | # Summary Snap
## Suppose you want to calculate how much workit takes to lift a 160 N barbell. Besides themass of the barbell, what other information doyou need to know?a. the shape of the weightsb. how high the barbell is being liftedc. the strength of the person doing the liftingd. None of the above
##### SUMMARY
- The text is discussing what information is needed to calculate the amount of work required to lift a barbell weighing 160 N.
- In addition to knowing the mass of the barbell, the text suggests that the following information could be needed:
- The height to which the barbell is being lifted.
- The shape of the weights on the barbell.
- The strength of the person lifting the barbell.
- The text also offers an option that none of the above information is necessary. | 177 | 788 | {"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} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.934979 |
https://technonguide.com/ways-to-use-currency-data/ | 1,723,445,533,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00859.warc.gz | 430,925,375 | 21,068 | # 5 ways to use currency data
Contents
## Introduction
Currencies are the lifeblood of global markets and economies, and currency data provides a unique window into the factors that drive inflation, interest rates, and growth. Here are five ways you can use currency data to gain insight into markets:
## Exchange rate analysis
Exchange rates are important to businesses because they can affect the cost of imports and exports. Businesses also use exchange rate data to predict future trends, compare the value of currencies, and make decisions regarding foreign investment.
There are several ways you can analyze currency data:
• Price patterns – You can look at historical exchange rates to see how they’ve changed over time. You could also use price patterns by comparing current prices with historic price patterns as well or when a market was experiencing extreme volatility in its exchange rate.
• Comparison – You can compare different currencies against each other using charts or tables that show all available currencies at once instead of just one currency versus another (e.g., US dollar vs. Euro).
## Currency correlation analysis
Currency correlation analysis is a method of studying how two or more currencies move in relation to each other. It can be done on historical data, with the aim of predicting future trends and so optimize trading strategies. You can use it to understand the relationship between the prices of two or more currencies and look for any patterns that may give you an advantage over other traders.
There are many ways to get currency correlation data, but one of the easiest methods is through Google Sheets — all you need is a free Google account and some basic knowledge about how spreadsheets work. To start, enter your desired currency pairs into cell A1 using this format: “USD” + “EUR”, where USD indicates that one unit will equal one US dollar and EUR indicates that one unit will equal one euro (for other currency pairs like GBP-USD use GBP instead). Then multiply this value by 100 so that there are no decimal places; next, copy these values into cells B2 through B5 so that they correspond with their corresponding cell (A1).
View More : Are We in a Bull Market Yet?
Now enter these formulas into columns C through F: =IFERROR(VLOOKUP(A3,”0C2EF2A7-D76F-47B3-9D23-F0958B8C9EDE!\$B\$1:\$B\$29″),IFERROR(VLOOKUP(A4,”0C2EF2A7-D76F-47B3-9D23-F0958B8C9EDE!\$C\$1:\$C\$29″),IFERROR(VLOOKUP(A5,”0C2EF2A7-D76F-47B3-9D23-F0958B8C9EDE!\$D\$1:\$D\$29″),IFERROR(“-“)))&””. Alternatively, you can pull data from vendors like fxapi.com.
## Multi-currency portfolios
• Multi-currency portfolios are a good way to diversify your investments.
• A multi-currency portfolio can help you hedge against risk.
• A multi-currency portfolio gives you exposure to different markets.
• A multi-currency portfolio will diversify your investments by investing in multiple currencies, which helps reduce risk and protect against global economic or political events that negatively impact one currency but not another (for example: The depreciation of the pound sterling after Brexit).
## Commodities pricing
Commodity prices are usually quoted in US dollars, so you can use the currency data to look at how each commodity is priced. For example, let’s say that a coffee farmer sold 500 pounds of raw coffee beans for \$1 per pound in December 2019. If that price is up from the previous month’s price of \$0.90 per pound and down from an average for November of \$1.10 per pound, then it would appear that demand for coffee was down during this particular time period due to a decrease in production costs and competition from other agricultural products like corn or soybeans which have been experiencing record high prices this year due to drought conditions across various regions around Earth where they’re grown.”
## Economic and policy analysis
Exchange rate data is used in economic and policy analysis. For example, exchange rates can be used to measure the relative strength of a country’s currency against another. If an economy’s central bank wants to boost exports, it might try to weaken its own currency by selling assets or cutting interest rates. The price of assets (such as stocks, bonds, and real estate) depends on what investors expect future exchange rates will be. This means that changes in expectations about future exchange rates can cause significant fluctuations in asset prices. For instance, when the European Central Bank said it would stop buying government bonds issued by some countries earlier this year (taking with one hand what it had given with another), bond yields rose sharply because investors became nervous about whether these countries could now borrow money at affordable rates from other investors like pension funds or insurance companies.
View More : Exploring the Formation and Success of Swiss Companies
## Currencies are key to economies and markets
Currencies are the most liquid asset in the world, and currency exchange rates are used to price goods and services, value assets, measure economic growth, and measure inflation.
## Conclusion
Currencies are one of the most complex yet important assets in the market, and currency data is a vital source of information for investors. In order to understand how to use this data, you must first understand what it is and how it works. At its core, currency data consists of exchange rates between currencies in relation to each other. These rates fluctuate constantly based on economic factors such as inflation rates or trade balances between countries as well as political events like elections or geopolitical turmoil over time periods ranging from weeks all the way up until decades!
YesNo
### Shankar
Shankar is a tech blogger who occasionally enjoys penning historical fiction. With over a thousand articles written on tech, business, finance, marketing, mobile, social media, cloud storage, software, and general topics, he has been creating material for the past eight years. | 1,254 | 6,035 | {"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} | 3.03125 | 3 | CC-MAIN-2024-33 | latest | en | 0.904537 |
https://www.mathnasium.com/chattanooga/news/multiplication-tips | 1,696,361,598,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00495.warc.gz | 969,889,032 | 11,189 | Get Started by Finding a Local Center
# Multiplication Tips!
Aug 28, 2015 | Chattanooga
Many of our young students have been earnestly filling in their times tables. Here are four of many important skills we teach to pave the way toward multiplication mastery:
1) Skip Counting: We have children count by a number starting at 0. This skill is easy to practice at home. "0, 4, 8, 12... 0, 6, 12, 18,..."
2) Adding 10 First: Skip counting is all fine and dandy when it comes to the small numbers, but what about the bigger ones? One trick we teach is to add 10 first. This works great for adding 8, 9, 11, and 12. Since 12 is 10+2, we could go up 10 and then up 2. Since 9 is 10-1, we could go up 10 and down 1. Kids enjoy getting fast at this. You could have them count by one of the numbers above starting at any number for practice at home: "6, 14, 22, 30... 4, 13, 22, 31..."
3) Multiplication is Repeated Addition: When kids are first learning multiplication, we rarely use the phrase what is five times three. Instead, we say "what is five three times?" Phrasing it this way gets kids into the mindset of multiplication as a process rather than a fact. They might not know 5 times 3, but they can often respond "five, ten, FIFTEEN!" when asked what five three times is.
4) Launching Points: Once children have developed a good understanding of multiplication as repeated addition and "memorized" some of their multiplication facts, we start to prompt them to reason as follows: "If five 5s is 25, then six 5s is 25 + 5. So six times five is 30." "If ten 8s is 80, then nine 8s is 80 - 8. So nine times 8 is 72." We can begin to use known multiplication facts as launching points to discover other multiplication facts.
##### Locations near
Could not find Center, try again | 489 | 1,784 | {"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} | 4.34375 | 4 | CC-MAIN-2023-40 | latest | en | 0.949133 |
https://www.teachoo.com/9080/2869/Example-8/category/Converting-fraction-to-percent/ | 1,725,706,167,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00184.warc.gz | 965,686,954 | 21,117 | Converting fraction to percent
Chapter 7 Class 7 Comparing Quantities
Concept wise
### Transcript
Example 2 Out of 25 children in a class, 15 are girls. What is the percentage of girls? Total number of children = 25 Number of girls = 15 Percentage of girls = (𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒈𝒊𝒓𝒍𝒔)/(𝑻𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒄𝒉𝒊𝒍𝒅𝒓𝒆𝒏) × 100% = 15/25 × 100% = 𝟑/𝟓 × 100% = 3 × 20% = 60% | 206 | 357 | {"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} | 4.09375 | 4 | CC-MAIN-2024-38 | latest | en | 0.834821 |
http://www.popflock.com/learn?s=Thermal_fluctuations | 1,611,393,322,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703536556.58/warc/CC-MAIN-20210123063713-20210123093713-00532.warc.gz | 168,077,799 | 22,468 | Thermal Fluctuations
Get Thermal Fluctuations essential facts below. View Videos or join the Thermal Fluctuations discussion. Add Thermal Fluctuations to your PopFlock.com topic list for future reference or share this resource on social media.
Thermal Fluctuations
Atomic diffusion on the surface of a crystal. The shaking of the atoms is an example of thermal fluctuations. Likewise, thermal fluctuations provide the energy necessary for the atoms to occasionally hop from one site to a neighboring one. For simplicity, the thermal fluctuations of the blue atoms are not shown.
In statistical mechanics, thermal fluctuations are random deviations of a system from its average state, that occur in a system at equilibrium.[1] All thermal fluctuations become larger and more frequent as the temperature increases, and likewise they decrease as temperature approaches absolute zero.
Thermal fluctuations are a basic manifestation of the temperature of systems: A system at nonzero temperature does not stay in its equilibrium microscopic state, but instead randomly samples all possible states, with probabilities given by the Boltzmann distribution.
Thermal fluctuations generally affect all the degrees of freedom of a system: There can be random vibrations (phonons), random rotations (rotons), random electronic excitations, and so forth.
Thermodynamic variables, such as pressure, temperature, or entropy, likewise undergo thermal fluctuations. For example, for a system that has an equilibrium pressure, the system pressure fluctuates to some extent about the equilibrium value.
Only the 'control variables' of statistical ensembles (such as the number of particules N, the volume V and the internal energy E in the microcanonical ensemble) do not fluctuate.
Thermal fluctuations are a source of noise in many systems. The random forces that give rise to thermal fluctuations are a source of both diffusion and dissipation (including damping and viscosity). The competing effects of random drift and resistance to drift are related by the fluctuation-dissipation theorem. Thermal fluctuations play a major role in phase transitions and chemical kinetics.
## Central limit theorem
The volume of phase space ${\displaystyle {\mathcal {V}}}$, occupied by a system of ${\displaystyle 2m}$ degrees of freedom is the product of the configuration volume ${\displaystyle V}$ and the momentum space volume. Since the energy is a quadratic form of the momenta for a non-relativistic system, the radius of momentum space will be ${\displaystyle {\sqrt {E}}}$ so that the volume of a hypersphere will vary as ${\displaystyle {\sqrt {E}}^{2m}}$ giving a phase volume of
${\displaystyle {\mathcal {V}}={\frac {(C\cdot E)^{m}}{\Gamma (m+1)}},}$
where ${\displaystyle C}$ is a constant depending upon the specific properties of the system and ${\displaystyle \Gamma }$ is the Gamma function. In the case that this hypersphere has a very high dimensionality, ${\displaystyle 2m}$, which is the usual case in thermodynamics, essentially all the volume will lie near to the surface
${\displaystyle \Omega (E)={\frac {\partial {\mathcal {V}}}{\partial E}}={\frac {C^{m}\cdot E^{m-1}}{\Gamma (m)}},}$
where we used the recursion formula ${\displaystyle m\Gamma (m)=\Gamma (m+1)}$.
The surface area ${\displaystyle \Omega (E)}$ has its legs in two worlds: (i) the macroscopic one in which it is considered a function of the energy, and the other extensive variables, like the volume, that have been held constant in the differentiation of the phase volume, and (ii) the microscopic world where it represents the number of complexions that is compatible with a given macroscopic state. It is this quantity that Planck referred to as a 'thermodynamic' probability. It differs from a classical probability inasmuch as it cannot be normalized; that is, its integral over all energies diverges--but it diverges as a power of the energy and not faster. Since its integral over all energies is infinite, we might try to consider its Laplace transform
${\displaystyle {\mathcal {Z}}(\beta )=\int _{0}^{\infty }e^{-\beta E}\Omega (E)\,dE,}$
which can be given a physical interpretation. The exponential decreasing factor, where ${\displaystyle \beta }$ is a positive parameter, will overpower the rapidly increasing surface area so that an enormously sharp peak will develop at a certain energy ${\displaystyle E^{\star }}$. Most of the contribution to the integral will come from an immediate neighborhood about this value of the energy. This enables the definition of a proper probability density according to
${\displaystyle f(E;\beta )={\frac {e^{-\beta E}}{{\mathcal {Z}}(\beta )}}\Omega (E),}$
whose integral over all energies is unity on the strength of the definition of ${\displaystyle {\mathcal {Z}}(\beta )}$, which is referred to as the partition function, or generating function. The latter name is due to the fact that the derivatives of its logarithm generates the central moments, namely,
${\displaystyle \langle E\rangle =-{\frac {\partial \ln {\mathcal {Z}}}{\partial \beta }},\qquad \ \langle (E-\langle E\rangle )^{2}\rangle =(\Delta E)^{2}={\frac {\partial ^{2}\ln {\mathcal {Z}}}{\partial \beta ^{2}}},}$
and so on, where the first term is the mean energy and the second one is the dispersion in energy.
The fact that ${\displaystyle \Omega (E)}$ increases no faster than a power of the energy ensures that these moments will be finite.[2] Therefore, we can expand the factor ${\displaystyle e^{-\beta E}\Omega (E)}$ about the mean value ${\displaystyle \langle E\rangle }$, which will coincide with ${\displaystyle E^{\star }}$ for Gaussian fluctuations (i.e. average and most probable values coincide), and retaining lowest order terms result in
${\displaystyle f(E;\beta )={\frac {e^{-\beta E}}{{\mathcal {Z}}(\beta )}}\Omega (E)\approx {\frac {\exp\{-(E-\langle E\rangle )^{2}/2\langle (\Delta E)^{2}\rangle \}}{\sqrt {2\pi (\Delta E)^{2}}}}.}$
This is the Gaussian, or normal, distribution, which is defined by its first two moments. In general, one would need all the moments to specify the probability density, ${\displaystyle f(E;\beta )}$, which is referred to as the canonical, or posterior, density in contrast to the prior density ${\displaystyle \Omega }$, which is referred to as the 'structure' function.[2] This is the central limit theorem as it applies to thermodynamic systems.[3]
If the phase volume increases as ${\displaystyle E^{m}}$, its Laplace transform, the partition function, will vary as ${\displaystyle \beta ^{-m}}$. Rearranging the normal distribution so that it becomes an expression for the structure function and evaluating it at ${\displaystyle E=\langle E\rangle }$ give
${\displaystyle \Omega (\langle E\rangle )={\frac {e^{\beta (\langle E\rangle )\langle E\rangle }{\mathcal {Z}}(\beta (\langle E\rangle ))}{\sqrt {2\pi (\Delta E)^{2}}}}.}$
It follows from the expression of the first moment that ${\displaystyle \beta (\langle E\rangle )=m/\langle E\rangle }$, while from the second central moment, ${\displaystyle \langle (\Delta E)^{2}\rangle =\langle E\rangle ^{2}/m}$. Introducing these two expressions into the expression of the structure function evaluated at the mean value of the energy leads to
${\displaystyle \Omega (\langle E\rangle )={\frac {\langle E\rangle ^{m-1}m}{{\sqrt {2\pi m}}m^{m}e^{-m}}}}$.
The denominator is exactly Stirling's approximation for ${\displaystyle m!=\Gamma (m+1)}$, and if the structure function retains the same functional dependency for all values of the energy, the canonical probability density,
${\displaystyle f(E;\beta )=\beta {\frac {(\beta E)^{m-1}}{\Gamma (m)}}e^{-\beta E}}$
will belong to the family of exponential distributions known as gamma densities. Consequently, the canonical probability density falls under the jurisdiction of the local law of large numbers which asserts that a sequence of independent and identically distributed random variables tends to the normal law as the sequence increases without limit.
The expressions given below are for systems that are close to equilibrium and have negligible quantum effects.[4]
### Single variable
Suppose ${\displaystyle x}$ is a thermodynamic variable. The probability distribution ${\displaystyle w(x)dx}$ for ${\displaystyle x}$ is determined by the entropy ${\displaystyle S}$:
${\displaystyle w(x)\propto \exp \left(S(x)\right).}$
If the entropy is Taylor expanded about its maximum (corresponding to the equilibrium state), the lowest order term is a Gaussian distribution:
${\displaystyle w(x)={\frac {1}{\sqrt {2\pi \langle x^{2}\rangle }}}\exp \left(-{\frac {x^{2}}{2\langle x^{2}\rangle }}\right).}$
The quantity ${\displaystyle \langle x^{2}\rangle }$ is the mean square fluctuation.[4]
### Multiple variables
The above expression has a straightforward generalization to the probability distribution ${\displaystyle w(x_{1},x_{2},\ldots ,x_{n})dx_{1}dx_{2}\ldots dx_{n}}$:
${\displaystyle w=\prod _{i,j=1\ldots n}{\frac {1}{\left(2\pi \right)^{n/2}{\sqrt {\langle x_{i}x_{j}\rangle }}}}\exp \left(-{\frac {x_{i}x_{j}}{2\langle x_{i}x_{j}\rangle }}\right),}$
where ${\displaystyle \langle x_{i}x_{j}\rangle }$ is the mean value of ${\displaystyle x_{i}x_{j}}$.[4]
### Fluctuations of the fundamental thermodynamic quantities
In the table below are given the mean square fluctuations of the thermodynamic variables ${\displaystyle T,V,P}$ and ${\displaystyle S}$ in any small part of a body. The small part must still be large enough, however, to have negligible quantum effects.
${\displaystyle \Delta T}$ ${\displaystyle \Delta V}$ ${\displaystyle \Delta S}$ ${\displaystyle \Delta P}$ ${\displaystyle \Delta T}$ ${\displaystyle {\frac {k_{\rm {B}}}{C_{V}}}T^{2}}$ ${\displaystyle 0}$ ${\displaystyle k_{\rm {B}}T}$ ${\displaystyle {\frac {k_{\rm {B}}}{C_{V}}}T^{2}\left({\frac {\partial P}{\partial T}}\right)_{V}}$ ${\displaystyle \Delta V}$ ${\displaystyle 0}$ ${\displaystyle -k_{\rm {B}}T\left({\frac {\partial V}{\partial P}}\right)_{T}}$ ${\displaystyle T\left({\frac {\partial V}{\partial T}}\right)_{P}}$ ${\displaystyle -k_{\rm {B}}T}$ ${\displaystyle \Delta S}$ ${\displaystyle k_{\rm {B}}T}$ ${\displaystyle T\left({\frac {\partial V}{\partial T}}\right)_{P}}$ ${\displaystyle C_{P}}$ ${\displaystyle 0}$ ${\displaystyle \Delta P}$ ${\displaystyle {\frac {k_{\rm {B}}}{C_{V}}}T^{2}\left({\frac {\partial P}{\partial T}}\right)_{V}}$ ${\displaystyle -k_{\rm {B}}T}$ ${\displaystyle 0}$ ${\displaystyle -k_{\rm {B}}T\left({\frac {\partial P}{\partial V}}\right)_{S}}$
## Notes
1. ^ In statistical mechanics they are often simply referred to as fluctuations.
2. ^ a b Khinchin 1949
3. ^ Lavenda 1991
4. ^ a b c d Landau 1985
## References
• Khinchin, A. I. (1949). Mathematical Foundations of Statistical Mechanics. Dover Publications. ISBN 0-486-60147-1.
• Lavenda, B. H. (1991). Statistical Physics: A Probabilistic Approach. Wiley-Interscience. ISBN 0-471-54607-0.
• Landau, L. D.; Lifshitz, E. M. (1985). Statistical Physics, Part 1 (3rd ed.). Pergamon Press. ISBN 0-08-023038-5. | 2,881 | 11,097 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 74, "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} | 2.765625 | 3 | CC-MAIN-2021-04 | latest | en | 0.867567 |
https://math.stackexchange.com/questions/4425546/question-on-an-improper-integral-of-the-form-int-0afx-adx | 1,696,114,258,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00649.warc.gz | 417,158,960 | 37,387 | # Question on an improper integral of the form $\int_0^af(x,a)dx$
The question I'm about to ask isn't covered in our lectures, so pardon my ignorance.
Suppose $$f:(0,a]\times [c,d]$$. How is uniform convergence of the improper integral $$\int_0^af(x,a)dx$$ defined?
Just to give an insight of what I've (hope so) learnt so far, I'm going to write down some definitions and results (and proofs) I think are relevant for functions $$f:[a,+\infty)\times [c,d]$$ that I found in the script by prof. Šime Ungar from 2004. It might be available here.
$$\underline{\boldsymbol{\text{ definition 1: }}}$$
An improper integral $$\int_a^\infty f(x)dx$$ is defined as $$\lim_{b\to\infty}\int_a^b f(x)dx$$ under the condition that $$f$$ is integrable on $$[a,b],\forall b>a$$ and that the limit exists. In this case we also say the improper integral $$\int_a^\infty f(x)dx$$ converges. That is equivalent to the following condition: $$f$$ is integrable on $$[a,b],\forall b>a$$ and $$(\forall\varepsilon>0)(\exists a_0>a), \left|\int_b^c f(x)dx\right|<\varepsilon,\forall c>b>a_0.$$
Now, suppose we have a function $$f:[a,+\infty)\times S\to\Bbb R,$$ where $$S\subseteq\Bbb R$$ is an arbitrary set. Here we can observe convergence of the integral $$\displaystyle\int_a^b f(x,y)dx,\forall y\in S.$$ In this situation, the following definition makes sense:
$$\underline{\boldsymbol{\text{definition 2:}}}$$
We say the improper integral $$\int_a^\infty f(x,y)dx$$ converges uniformly on $$S$$ if the integral converges $$\forall y\in S$$ and if $$\lim_{b\to\infty}\left(\sup_{y\in S}\left|\int_b^\infty f(x,y)dx\right|\right)=0.$$
$$\underline{\boldsymbol{\text{result 1:}}}$$
Suppose $$\int_a^b f(x,y)dx$$ exists $$\forall b>a$$ and $$\color{red}{\forall y\in S}.$$ Then, the improper integral $$\int_a^\infty f(x,y)dx$$ converges uniformly on $$S$$ if and only if $$(\forall\varepsilon>0)(\exists a_0>a), \color{red}{\sup_{y\in S}}\left|\int_b^c f(x,y)dx\right|<\varepsilon,\forall c>b\ge a_0.$$
$$\boldsymbol{\text{ proof: }}$$
$$\boxed{\Rightarrow}$$ Necessity of the conditions in the theorem is obvious.
$$\boxed{\Leftarrow}$$ Let's prove thr sufficiency. From the conditions of the theorem, it first follows that, $$\forall y\in S,$$ the integral $$\int_a^\infty f(x,y)dx$$ converges. Furthermore, for $$y\in S,$$ since $$\left|\int_b^c\right|<\varepsilon$$ whenever $$c>b>a_0,$$ letting $$c\to\infty,$$ we have $$\left|\int_b^\infty f(x,y)dx\right|\color{red}\le\varepsilon.$$ But, as this holds $$\color{red}{\forall y\in S},$$ it is also true that $$\color{red}{\sup_{y\in S}}\left|\int_b^\infty f(x,y)dx\right|\le\varepsilon$$ and the claim follows.
Something I believe is important (I'll write down my motivation in the end):
$$\underline{\boldsymbol{\text{Weierstrass M-test:}}}$$
Suppose $$\int_a^b f(x,y)dx$$ exists $$\forall b>a.$$ If there is a function $$M:[a,+\infty)\to\Bbb R$$ s. t. $$|f(x,y)|\le M(x),\forall x\in[a,+\infty)$$ and $$\forall y\in S$$ and if $$\int_a^\infty M(x)dx$$ converges, then $$\int_a^\infty f(x,y)dx$$ converges (absolutely and) uniformly on $$S$$.
I've gone through the proof of the discrete version. If needed, I'll analyze this more carefully.
Last result with the proof, I promise.
$$\underline{\boldsymbol{\text{result 2: }}}$$
Suppose $$f:[a,+\infty)\times [c,d]$$ is continuous and that the following statements are true:
1. $$\exists y\in [c,d]$$ s. t. the improper integral $$\int_a^\infty f(x,y)dx$$ converges
2. $$\partial_2f(x,y)$$ exists and is continuous on $$[a,+\infty)\times [c,d]$$
3. $$\int_a^\infty\partial_2f(x,y)dx$$ converges uniformly on $$[c,d].$$
Then, the improper integral $$\int_a^\infty f(x,y)dx$$ exists and converges uniformly on $$[c,d],$$ and the function $$F(y):=\int_a^\infty f(x,y)dx$$ is differentiable on $$[c,d]$$ and $$\color{purple}{F'(y)=\int_a^\infty\partial_2 f(x,y)dx}\forall y\in [c,d].$$
$$\boldsymbol{\text{ proof: }}$$
Suppose that $$\int_a^\infty f(x,y_0)dx$$ converges. Apart from that, we also know that $$\forall b>a$$ and $$\forall y\in [c,d]$$ the integral $$\int_a^b f(x,y)dx.$$
Because of $$2,\forall\varepsilon>0,\exists a_0$$ s. t. $$a_0 and at the same time $$\left|\int_b^g f(x,y_0)dx\right|<\varepsilon.$$
\begin{aligned}\left|\int_b^g(f(x,y)-f(x,y_0))dx\right|&=\left|\int_b^g\int_{y_0}^y\partial_2f(x,t)dtdx\right|\\&=\left|\int_{y_0}^y\int_b^g\partial_2f(x,t)dxdt\right|\\&\le\int_{y_0}^y\left|\int_b^g\partial_2f(x,t)dx\right|dt\\&\le (d-c)\varepsilon,\end{aligned} whenever $$a_0
\begin{aligned}\implies\left|\int_b^gf(x,y)dx\right|&\le\left|\int_b^g(f(x,y)-f(x,y_0))dx\right|+\left|\int_b^g f(x,y_0)dx\right|\\&\le(d-c)\varepsilon+\varepsilon\\&=(1+d-c)\varepsilon,\forall y\in [c,d]\end{aligned} it follows that $$\int_a^\infty f(x,y)dx$$ exists and converges uniformly on $$[c,d].$$
Now, let's define a function $$\color{purple}{G(y):=\int_a^\infty\partial_2f(x,y)dx}.$$ Then $$G$$ is continuous (a result proven in the script priorly). It remains to notice that $$\int_c^ydt\int_a^\infty\partial_2f(x,t)dx=\int_a^\infty dx\int_c^y\partial_2f(x,t)dt.$$
In all the results, we used the fact the set was unbounded. In old materials, I've come across the following task:
Prove that the function $$F(a)=\int_0^{1/a}\frac{e^{ax}-1}xdx$$ is constant.
I might be wrong, but I tried to use the substitution $$x=\frac1t$$ in order to get to an unbounded interval and then use the results above. This got a bit messy, so I was wondering if it makes sense to even consider the $$\boldsymbol{\text{ result 2}}$$ and write something as $$F'(a)=\int_0^{1/a}\partial_2 f(x,a)dx.$$ I'm primarily interested in justification for that.
I apologize for the length of my post and thank you for reading in advance!
• If I understand you correctly, you want to differentiate w.r.t $a$. Notice that $\int_0^{a+h}f(x,a+h)dx -\int_0^{a}f(x,a) = \left( \int_0^a f(x,a+h)dx - \int_0^{a}f(x,a) \right) + \int_a^{a+h}f(x,a+h)$. Next consider that if $f$ is continuous in a neighbourhood of $(a,a)$ then $(1/h)\int_a^{a+h}f(x,a+h)dx \approx f(a,a)$ using uniform continuity. So under mild conditions, you only really need to be worried about the term in the parentheses. Apr 12, 2022 at 0:36
• @MarkoKarbevski, thank you for the comment. It makes sense. (: Apr 12, 2022 at 5:12
$$\int_0^\frac{1}{a} \frac{e^{ax}-1}{x}\:dx \xrightarrow{u = ax} \int_0^1 \frac{e^u-1}{u}\:du$$
which has no $$a$$ dependence at all. | 2,228 | 6,417 | {"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": 83, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-40 | latest | en | 0.716614 |
http://aimsciences.org/article/doi/10.3934/dcdsb.2012.17.2431 | 1,563,941,765,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195530385.82/warc/CC-MAIN-20190724041048-20190724063048-00396.warc.gz | 5,590,769 | 12,555 | # American Institute of Mathematical Sciences
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October 2012, 17(7): 2431-2449. doi: 10.3934/dcdsb.2012.17.2431
## Dynamical bifurcation of the two dimensional Swift-Hohenberg equation with odd periodic condition
1 Department of Mathematics and Research Institute for Basic Sciences, Kyung Hee University, 1 Hoegi-Dong, Dongdaemun-Gu, Seoul, 130-701, South Korea 2 Department of Mathematics, National Taiwan University, Taipei, 10617
Received June 2011 Revised March 2012 Published July 2012
In this article, we study the stability and dynamic bifurcation for the two dimensional Swift-Hohenberg equation with an odd periodic condition. It is shown that an attractor bifurcates from the trivial solution as the control parameter crosses the critical value. The bifurcated attractor consists of finite number of singular points and their connecting orbits. Using the center manifold theory, we verify the nondegeneracy and the stability of the singular points.
Citation: Jongmin Han, Chun-Hsiung Hsia. Dynamical bifurcation of the two dimensional Swift-Hohenberg equation with odd periodic condition. Discrete & Continuous Dynamical Systems - B, 2012, 17 (7) : 2431-2449. doi: 10.3934/dcdsb.2012.17.2431
##### References:
show all references
##### References:
[1] Jongmin Han, Masoud Yari. Dynamic bifurcation of the complex Swift-Hohenberg equation. Discrete & Continuous Dynamical Systems - B, 2009, 11 (4) : 875-891. doi: 10.3934/dcdsb.2009.11.875 [2] Yuncherl Choi, Taeyoung Ha, Jongmin Han, Doo Seok Lee. Bifurcation and final patterns of a modified Swift-Hohenberg equation. Discrete & Continuous Dynamical Systems - B, 2017, 22 (7) : 2543-2567. doi: 10.3934/dcdsb.2017087 [3] Toshiyuki Ogawa, Takashi Okuda. Bifurcation analysis to Swift-Hohenberg equation with Steklov type boundary conditions. Discrete & Continuous Dynamical Systems - A, 2009, 25 (1) : 273-297. doi: 10.3934/dcds.2009.25.273 [4] Masoud Yari. Attractor bifurcation and final patterns of the n-dimensional and generalized Swift-Hohenberg equations. Discrete & Continuous Dynamical Systems - B, 2007, 7 (2) : 441-456. doi: 10.3934/dcdsb.2007.7.441 [5] J. Burke, Edgar Knobloch. Multipulse states in the Swift-Hohenberg equation. Conference Publications, 2009, 2009 (Special) : 109-117. doi: 10.3934/proc.2009.2009.109 [6] Peng Gao. Averaging principles for the Swift-Hohenberg equation. Communications on Pure & Applied Analysis, 2020, 19 (1) : 293-310. doi: 10.3934/cpaa.2020016 [7] Ling-Jun Wang. The dynamics of small amplitude solutions of the Swift-Hohenberg equation on a large interval. Communications on Pure & Applied Analysis, 2012, 11 (3) : 1129-1156. doi: 10.3934/cpaa.2012.11.1129 [8] Yanfeng Guo, Jinqiao Duan, Donglong Li. Approximation of random invariant manifolds for a stochastic Swift-Hohenberg equation. Discrete & Continuous Dynamical Systems - S, 2016, 9 (6) : 1701-1715. doi: 10.3934/dcdss.2016071 [9] Shengfu Deng. Periodic solutions and homoclinic solutions for a Swift-Hohenberg equation with dispersion. Discrete & Continuous Dynamical Systems - S, 2016, 9 (6) : 1647-1662. doi: 10.3934/dcdss.2016068 [10] John Burke, Edgar Knobloch. Normal form for spatial dynamics in the Swift-Hohenberg equation. Conference Publications, 2007, 2007 (Special) : 170-180. doi: 10.3934/proc.2007.2007.170 [11] Andrea Giorgini. On the Swift-Hohenberg equation with slow and fast dynamics: well-posedness and long-time behavior. Communications on Pure & Applied Analysis, 2016, 15 (1) : 219-241. doi: 10.3934/cpaa.2016.15.219 [12] Chaoqun Huang, Nung Kwan Yip. Singular perturbation and bifurcation of diffuse transition layers in inhomogeneous media, part II. Networks & Heterogeneous Media, 2015, 10 (4) : 897-948. doi: 10.3934/nhm.2015.10.897 [13] Chaoqun Huang, Nung Kwan Yip. Singular perturbation and bifurcation of diffuse transition layers in inhomogeneous media, part I. Networks & Heterogeneous Media, 2013, 8 (4) : 1009-1034. doi: 10.3934/nhm.2013.8.1009 [14] Robert Skiba, Nils Waterstraat. The index bundle and multiparameter bifurcation for discrete dynamical systems. Discrete & Continuous Dynamical Systems - A, 2017, 37 (11) : 5603-5629. doi: 10.3934/dcds.2017243 [15] Camillo De Lellis, Emanuele Spadaro. Center manifold: A case study. Discrete & Continuous Dynamical Systems - A, 2011, 31 (4) : 1249-1272. doi: 10.3934/dcds.2011.31.1249 [16] Thai Son Doan, Martin Rasmussen, Peter E. Kloeden. The mean-square dichotomy spectrum and a bifurcation to a mean-square attractor. Discrete & Continuous Dynamical Systems - B, 2015, 20 (3) : 875-887. doi: 10.3934/dcdsb.2015.20.875 [17] Tian Ma, Shouhong Wang. Attractor bifurcation theory and its applications to Rayleigh-Bénard convection. Communications on Pure & Applied Analysis, 2003, 2 (4) : 591-599. doi: 10.3934/cpaa.2003.2.591 [18] Claudia Valls. The Boussinesq system:dynamics on the center manifold. Communications on Pure & Applied Analysis, 2005, 4 (4) : 839-860. doi: 10.3934/cpaa.2005.4.839 [19] Jixun Chu, Pierre Magal. Hopf bifurcation for a size-structured model with resting phase. Discrete & Continuous Dynamical Systems - A, 2013, 33 (11&12) : 4891-4921. doi: 10.3934/dcds.2013.33.4891 [20] Jean-Philippe Lessard, Evelyn Sander, Thomas Wanner. Rigorous continuation of bifurcation points in the diblock copolymer equation. Journal of Computational Dynamics, 2017, 4 (1&2) : 71-118. doi: 10.3934/jcd.2017003
2018 Impact Factor: 1.008 | 1,776 | 5,673 | {"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} | 2.515625 | 3 | CC-MAIN-2019-30 | latest | en | 0.752291 |
https://api.kde.org/kstars/html/robuststatistics_8h_source.html | 1,725,839,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651035.2/warc/CC-MAIN-20240908213138-20240909003138-00058.warc.gz | 83,432,660 | 19,053 | # Kstars
robuststatistics.h
1/*
2 SPDX-FileCopyrightText: 2022 Sophie Taylor <sophie@spacekitteh.moe>
3
5*/
6
7// The purpose of Robust Statistics is to provide a more robust way of calculating typical statistical measures
8// such as averages. More information is available here: https://en.wikipedia.org/wiki/Robust_statistics
9// The basic idea is that traditional measures of "location" such as the mean are susceptible to outliers in the
10// distribution. More information on location is available here: https://en.wikipedia.org/wiki/Location_parameter
11// The same applies to measures of scale such as variance (more information available here:
12// https://en.wikipedia.org/wiki/Scale_parameter)
13//
14// Robust Statistics uses underlying GNU Science Library (GSL) routines and provides a wrapper around these routines.
15// See the Statistics section of the GSL documentation: https://www.gnu.org/software/gsl/doc/html/statistics.html
16//
17// Currently the following are provided:
18// Location: LOCATION_MEAN - calculate the mean of input data
19// LOCATION_MEDIAN - calculate the median
20// LOCATION_TRIMMEDMEAN - discard a specified fraction of high/low values before calculating the mean
21// LOCATION_GASTWIRTH - use the Gastwirth estimator based on combining different quantiles
22// see https://www.gnu.org/software/gsl/doc/html/statistics.html#gastwirth-estimator
23// LOCATION_SIGMACLIPPING - single step sigma clipping routine to sigma clip outliers fron the
24// input data and calculate the mean from the remaining data
25//
26// Scale: SCALE_VARIANCE - variance
27// SCALE_MAD - median absolute deviation is the median of absolute differences between each
28// datapoint and the median.
29// SCALE_BWMV - biweight midvariance is a more complex calculate detailed here
30// https://en.wikipedia.org/wiki/Robust_measures_of_scale#The_biweight_midvariance
31// implemenated internally - not provided by GSL
32// SCALE_SESTIMATOR - Qn and Sn estimators are pairwise alternative to MAD by Rousseeuw and Croux
33// SCALE_QESTIMATOR https://www.researchgate.net/publication/221996720_Alternatives_to_Median_Absolute_Deviation
34// SCALE_PESTIMATOR - Pairwise mean scale estimator (Pn). This is the interquartile range of the
35// pairwise means. Implemented internally, not provided by GSL.
36//
37// Where necessary data is sorted by the routines and functionality to use a user selected array sride is included.
38// C++ Templates are used to provide access to the GSL routines based on the datatype of the input data.
39
40#pragma once
41
42#include <limits>
43#include <vector>
44#include <cmath>
45#include <QObject>
46#include <QVector>
47
48#include "gslhelpers.h"
49
50namespace Mathematics::RobustStatistics
51{
52
53template<typename Base>
54double Pn_from_sorted_data(const Base sorted_data[],
55 const size_t stride,
56 const size_t n,
57 Base work[],
58 int work_int[]);
59
60enum ScaleCalculation { SCALE_VARIANCE, SCALE_BWMV, SCALE_SESTIMATOR, SCALE_QESTIMATOR, SCALE_MAD, SCALE_PESTIMATOR };
61
62
63enum LocationCalculation { LOCATION_MEAN, LOCATION_MEDIAN, LOCATION_TRIMMEDMEAN, LOCATION_GASTWIRTH,
64 LOCATION_SIGMACLIPPING
65 };
66
67struct SampleStatistics
68{
69 constexpr SampleStatistics() : location(0), scale(0), weight(std::numeric_limits<double>::infinity()) {}
70 constexpr SampleStatistics(const double location, const double scale, const double weight) :
71 location(location), scale(scale), weight(weight) {}
72 double location;
73 double scale;
74 double weight;
75};
76
77//template<typename Base=double>
78//struct Estimator
79//{
80// virtual double Estimate(std::vector<Base> data, const size_t stride = 1)
81// {
82// std::sort(Mathematics::GSLHelpers::make_strided_iter(data.begin(), stride),
83// Mathematics::GSLHelpers::make_strided_iter(data.end(), stride));
84// return EstimateFromSortedData(data, stride);
85// }
86// virtual double Estimate(const std::vector<Base> &data, const size_t stride = 1)
87// {
88// auto sorted = data;
89// std::sort(Mathematics::GSLHelpers::make_strided_iter(sorted.begin(), stride),
90// Mathematics::GSLHelpers::make_strided_iter(sorted.end(), stride));
91// return EstimateFromSortedData(sorted, stride);
92// }
93
94// virtual double EstimateFromSortedData(const std::vector<Base> &data, const size_t stride = 1) = 0;
95//};
96
97template<typename Base> struct LocationEstimator;
98
99template<typename Base = double>
100struct ScaleEstimator //: public virtual Estimator<Base>
101{
102 LocationEstimator<Base> locationEstimator;
103 virtual double EstimateScale(std::vector<Base> data, const size_t stride = 1)
104 {
105 std::sort(Mathematics::GSLHelpers::make_strided_iter(data.begin(), stride),
106 Mathematics::GSLHelpers::make_strided_iter(data.end(), stride));
107 return EstimateScaleFromSortedData(data, stride);
108 }
109 virtual double EstimateScale(const std::vector<Base> &data, const size_t stride = 1)
110 {
111 auto sorted = data;
112 std::sort(Mathematics::GSLHelpers::make_strided_iter(sorted.begin(), stride),
113 Mathematics::GSLHelpers::make_strided_iter(sorted.end(), stride));
114 return EstimateScaleFromSortedData(sorted, stride);
115 }
116
117 virtual double ConvertScaleToWeight(const double scale)
118 {
119 return 1 / (scale * scale);
120 }
121
122 virtual double EstimateScaleFromSortedData(const std::vector<Base> &data, const size_t stride = 1) = 0;
123};
124
125template <typename Base = double>
126struct MAD : public virtual ScaleEstimator<Base>
127{
128 virtual double EstimateScale(const std::vector<Base> &data, const size_t stride = 1) override
129 {
130 auto const size = data.size();
131 auto work = std::make_unique<double[]>(size / stride);
132 return Mathematics::GSLHelpers::gslMAD(data.data(), stride, size, work.get());
133 }
134 virtual double EstimateScale(std::vector<Base> data, const size_t stride = 1) override
135 {
136 return EstimateScale(data, stride);
137 }
138 virtual double EstimateScaleFromSortedData(const std::vector<Base> &data, const size_t stride = 1) override
139 {
140 return EstimateScale(data, stride);
141 }
142};
143template <typename Base = double>
144struct Variance : public virtual ScaleEstimator<Base>
145{
146 virtual double EstimateScaleFromSortedData(const std::vector<Base> &data, const size_t stride = 1) override
147 {
148 return Mathematics::GSLHelpers::gslVariance(data.data(), stride, data.size());
149 }
150 virtual double ConvertScaleToWeight(const double variance) override
151 {
152 return 1 / variance;
153 }
154};
155
156template <typename Base = double>
157struct SnEstimator : public virtual ScaleEstimator<Base>
158{
159 virtual double EstimateScaleFromSortedData(const std::vector<Base> &data, const size_t stride = 1) override
160 {
161 auto size = data.size();
162 auto work = std::make_unique<Base[]>(size);
163 return Mathematics::GSLHelpers::gslSnFromSortedData(data.data(), stride, size, work.get());
164 }
165};
166
167template <typename Base = double>
168struct QnEstimator : public virtual ScaleEstimator<Base>
169{
170 virtual double EstimateScaleFromSortedData(const std::vector<Base> &data, const size_t stride = 1) override
171 {
172 auto size = data.size();
173 auto work = std::make_unique<Base[]>(3 * size);
174 auto intWork = std::make_unique<Base[]>(5 * size);
175 return Mathematics::GSLHelpers::gslQnFromSortedData(data.data(), stride, size, work.get(), intWork.get());
176 }
177};
178
179template <typename Base = double>
180struct PnEstimator : public virtual ScaleEstimator<Base>
181{
182 virtual double EstimateScaleFromSortedData(const std::vector<Base> &data, const size_t stride = 1) override
183 {
184 auto size = data.size();
185 auto work = std::make_unique<Base[]>(3 * size);
186 auto intWork = std::make_unique<Base[]>(5 * size);
187 return Pn_from_sorted_data(data.data(), stride, size, work.get(), intWork.get());
188 }
189};
190
191
192template <typename Base = double>
193struct BiweightMidvariance : public virtual ScaleEstimator<Base>
194{
195 virtual double EstimateScaleFromSortedData(const std::vector<Base> &data, const size_t stride = 1) override
196 {
197 auto const size = data.size();
198 auto const theData = data.data();
199 auto work = std::make_unique<double[]>(size / stride);
200 auto const adjustedMad = 9.0 * Mathematics::GSLHelpers::gslMAD(theData, stride, size, work.get());
201 auto const median = this->locationEstimator.EstimateLocationFromSortedData(theData,
202 stride); //gslMedianFromSortedData(theData, stride, size);
203
204 auto const begin = Mathematics::GSLHelpers::make_strided_iter(data.cbegin(), stride);
205 auto const end = Mathematics::GSLHelpers::make_strided_iter(data.cend(), stride);
206
207 auto ys = std::vector<Base>(size / stride);
208 std::transform(begin, end, ys.begin(), [ = ](Base x)
209 {
210 return (x - median) / adjustedMad;
211 });
212 auto const indicator = [](Base y)
213 {
214 return std::fabs(y) < 1 ? 1 : 0;
215 };
216 auto const top = std::transform_reduce(begin, end, ys.begin(), 0, std::plus{},
217 [ = ](Base x, Base y)
218 {
219 return indicator(y) * pow(x - median, 2) * pow(1 - pow(y, 2), 4);
220 });
221 auto const bottomSum = std::transform_reduce(begin, end, 0, std::plus{},
222 [ = ](Base y)
223 {
224 return indicator(y) * (1.0 - pow(y, 2)) * (1.0 - 5.0 * pow(y, 2));
225 });
226 // The -1 is for Bessel's correction.
227 auto const bottom = bottomSum * (bottomSum - 1);
228
229 return (size / stride) * (top / bottom);
230 }
231
232 virtual double ConvertScaleToWeight(const double variance) override
233 {
234 return 1 / variance;
235 }
236};
237
238template<typename Base = double>
239struct LocationEstimator //: public virtual Estimator<Base>
240{
241 ScaleEstimator<Base> scaleEstimator;
242 virtual double EstimateLocation(std::vector<Base> data, const size_t stride = 1)
243 {
244 std::sort(Mathematics::GSLHelpers::make_strided_iter(data.begin(), stride),
245 Mathematics::GSLHelpers::make_strided_iter(data.end(), stride));
246 return EstimateLocationFromSortedData(data, stride);
247 }
248 virtual double EstimateLocation(const std::vector<Base> &data, const size_t stride = 1)
249 {
250 auto sorted = data;
251 std::sort(Mathematics::GSLHelpers::make_strided_iter(sorted.begin(), stride),
252 Mathematics::GSLHelpers::make_strided_iter(sorted.end(), stride));
253 return EstimateLocationFromSortedData(sorted, stride);
254 }
255 virtual double EstimateLocationFromSortedData(const std::vector<Base> &data, const size_t stride = 1) = 0;
256};
257
258template<typename Base = double>
259struct StatisticalCoEstimator : public virtual LocationEstimator<Base>, public virtual ScaleEstimator<Base>
260{
261 private:
262 LocationEstimator<Base> locationEstimator;
263 ScaleEstimator<Base> scaleEstimator;
264 public:
265 StatisticalCoEstimator(LocationEstimator<Base> locationEstimator, ScaleEstimator<Base> scaleEstimator)
266 : locationEstimator(locationEstimator), scaleEstimator(scaleEstimator) {}
267 virtual double EstimateLocationFromSortedData(const std::vector<Base> &data, const size_t stride = 1) override
268 {
269 auto const stats = EstimateStatisticsFromSortedData(data, stride);
270 return stats.location;
271 }
272 virtual double EstimateScaleFromSortedData(const std::vector<Base> &data, const size_t stride = 1) override
273 {
274 auto const stats = EstimateStatisticsFromSortedData(data, stride);
275 return stats.scale;
276 }
277 virtual SampleStatistics EstimateStatisticsFromSortedData(const std::vector<Base> &data, const size_t stride = 1)
278 {
279 auto const location = EstimateLocationFromSortedData(data, stride);
280 auto const scale = EstimateScaleFromSortedData(data, stride);
281 auto const weight = ConvertScaleToWeight(scale);
282
283 return SampleStatistics{location, scale, weight};
284 }
285};
286
287/**
288 * @short Computes a estimate of the statistical scale of the input sample.
289 *
290 * @param scaleMethod The estimator to use.
291 * @param data The sample to estimate the scale of.
292 * @param stride The stide of the data.
293 */
294template<typename Base = double>
295double ComputeScale(const ScaleCalculation scaleMethod, std::vector<Base> data,
296 const size_t stride = 1)
297{
298 if (scaleMethod != SCALE_VARIANCE)
299 std::sort(data.begin(), data.end());
300 return ComputeScaleFromSortedData(scaleMethod, data, stride);
301}
302//[[using gnu : pure]]
303template<typename Base = double>
304double ComputeScaleFromSortedData(const ScaleCalculation scaleMethod,
305 const std::vector<Base> &data,
306 const size_t stride = 1);
307/**
308 * @short Computes a estimate of the statistical location of the input sample.
309 *
310 * @param scaleMethod The estimator to use.
311 * @param data The sample to estimate the location of.
312 * @param sortedData The sorted data.
313 * @param trimAmount When using the trimmed mean estimate, the percentage quartile to remove either side.
314 * When using sigma clipping, the number of standard deviations a sample has to be from the mean
315 * to be considered an outlier.
316 * @param stride The stide of the data. Note that in cases other than the simple mean, the stride will only apply
317 * AFTER sorting.
318 */
319template<typename Base = double>
320double ComputeLocation(const LocationCalculation locationMethod, std::vector<Base> data,
321 const double trimAmount = 0.25, const size_t stride = 1)
322{
323 if (locationMethod != LOCATION_MEAN)
324 std::sort(data.begin(), data.end());
325 return ComputeLocationFromSortedData(locationMethod, data, trimAmount, stride);
326}
327
328//[[using gnu : pure]]
329template<typename Base = double>
330double ComputeLocationFromSortedData(const LocationCalculation locationMethod,
331 const std::vector<Base> &data, const double trimAmount = 0.25, const size_t stride = 1);
332
333/**
334 * @short Computes a weight for use in regression.
335 *
336 * @param scaleMethod The estimator to use.
337 * @param data The sample to estimate the scale of.
338 * @param sortedData The sorted data.
339 * @param stride The stide of the data. Note that in cases other than the simple variance, the stride will only
340 * apply AFTER sorting.
341 */
342template<typename Base = double>
343double ComputeWeight(const ScaleCalculation scaleMethod, std::vector<Base> data, const size_t stride = 1)
344{
345 if (scaleMethod != SCALE_VARIANCE)
346 std::sort(data.begin(), data.end());
347 return ComputeWeightFromSortedData(scaleMethod, data, stride);
348}
349//[[using gnu : pure]]
350template<typename Base = double>
351double ComputeWeightFromSortedData(const ScaleCalculation scaleMethod, const std::vector<Base> &data,
352 const size_t stride = 1)
353{
354 auto const scale = ComputeScaleFromSortedData(scaleMethod, data, stride);
355 return ConvertScaleToWeight(scaleMethod, scale);
356}
357
358SampleStatistics ComputeSampleStatistics(std::vector<double> data,
359 const LocationCalculation locationMethod = LOCATION_TRIMMEDMEAN,
360 const ScaleCalculation scaleMethod = SCALE_QESTIMATOR,
361 double trimAmount = 0.25,
362 const size_t stride = 1);
363
364[[using gnu : pure]]
365constexpr double ConvertScaleToWeight(const ScaleCalculation scaleMethod, double scale)
366{
367 // If the passed in scale is zero or near zero return a very small weight rather than infinity.
368 // This fixes the situation where, e.g. there is a single datapoint so scale is zero and
369 // weighting this datapoint with infinite weight is incorrect (and breaks the LM solver)
370 switch (scaleMethod)
371 {
372 // Variance, biweight midvariance are variances.
373 case SCALE_VARIANCE:
374 [[fallthrough]];
375 case SCALE_BWMV:
376 return (scale < 1e-10) ? 1e-10 : 1 / scale;
377 // MAD, Sn, Qn and Pn estimators are all calibrated to estimate the standard deviation of Gaussians.
379 [[fallthrough]];
380 case SCALE_SESTIMATOR:
381 [[fallthrough]];
382 case SCALE_QESTIMATOR:
383 [[fallthrough]];
384 case SCALE_PESTIMATOR:
385 return (scale < 1e-10) ? 1e-10 : 1 / (scale * scale);
386 }
387 return -1;
388}
389
390
391} // namespace Mathematics
QVariant location(const QVariant &res)
const QList< QKeySequence > & begin()
const QList< QKeySequence > & end()
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KDE's Doxygen guidelines are available online. | 4,034 | 16,332 | {"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} | 2.953125 | 3 | CC-MAIN-2024-38 | latest | en | 0.685759 |
https://math.stackexchange.com/questions/3194414/extending-knuth-up-arrow-hyperoperations-to-non-positive-values?noredirect=1 | 1,566,215,391,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00199.warc.gz | 548,693,127 | 27,708 | # Extending Knuth up-arrow/hyperoperations to non-positive values [duplicate]
So... I had the silly idea to extend Knuth's up-arrow notation so that it included zero and negative arrows. It is normally defined as \begin{align*} a \uparrow b & = a^b \\ a \uparrow^n b & = \underbrace{a \uparrow^{n - 1} (a \uparrow^{n - 1} (\dots(a \uparrow^{n - 1} a) \dots ))}_{b\text{ copies of } a} \end{align*} so, basically the hyperoperation sequence starting from exponentiation. For now, I will only consider $$a,b > 0$$.
If we try to go backwards from $$a \uparrow b$$, the "trivial" extension (letting down arrows represent negative up arrows, because why the heck not) is: \begin{align*} a \;b & = a \cdot b \\ a \downarrow b & = a + b \\ a \downarrow \downarrow b & = \text{see below} \end{align*} \\ \vdots But I had trouble coming up with an expression for $$a \downarrow \downarrow \downarrow b$$.
Maybe it doesn't exist. Alternatively, maybe there is a way of defining $$a \; b$$ (zero arrows) such that it does exist. So my question is: Is there an extension of Knuth's up-arrow notation such that $$a \downarrow^n b$$ exists for all $$n \geq 3$$?
Edit: Welp, I messed this question up. I initially thought $$a \downarrow \downarrow b = a + 1$$ was correct, but it is actually $$b + 1$$. So I thought I had an example of an extension when I did not. I have modified the question accordingly.
An extension would define $$a \uparrow^n b$$ for each $$n \leq 0$$ which satisfies the recursive definition of the notation.
Edit 2: Okay, turns out $$a \downarrow \downarrow b = b + 1$$ isn't correct either, as this would imply $$a \downarrow b = a + b - 1$$. For example, $$4 \downarrow 3 = 4 \downarrow \downarrow (4 \downarrow \downarrow 4) = 4 \downarrow \downarrow (4 + 1) = (4 + 1) + 1 = 6 = 4 + 3 - 1$$. But it is really close; perhaps we need an exception, such as \begin{align*} a \downarrow \downarrow b = \begin{cases}b + 1 & \text{if } a < b \\ b + 2 & \text{if } a = b \end{cases}\end{align*}. The case $$a > b$$ does not show up when evaluating $$a \downarrow b$$, but it will be need to be defined if we try to extend further to $$a \downarrow \downarrow \downarrow b$$. For instance, we could abuse the fact that the case $$a > b$$ is allowed to be anything, and let $$a \downarrow \downarrow b = b + 1 + \left\lfloor \frac{a}{b} \right\rfloor,$$ but finding $$a \downarrow \downarrow \downarrow b$$ may be intractable as a result. | 758 | 2,446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-35 | latest | en | 0.833075 |
https://physics.stackexchange.com/questions/481727/would-color-confinement-apply-in-higher-dimensions | 1,618,579,752,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066981.0/warc/CC-MAIN-20210416130611-20210416160611-00109.warc.gz | 547,439,404 | 39,651 | # Would Color Confinement apply in higher dimensions?
As I understand it color confinement comes from the fact that as the distance between two color charges increases the color potential energy increases, instead of decreasing, and the energy needed to pull two quarks apart is the same as the energy needed to create two new quarks. The way the color potential energy between two color charges is related to the fact that gluons themselves have color charge. For the electric potential energy between two electric charges the relationship between distance and potential energy depends on the number of dimensions, and for the spacetime curvature around a massive body the relationship between the distance from the massive body and the spacetime curvature depends on the number of dimensions, but I'm not sure if the relationship between distance and color potential energy depends on the number of dimensions?
Would color confinement apply in $$n+1$$ dimensions, in which $$n>3$$, or could color charges be free particles in higher dimensions?
Confinement is a non-perturbative phenomenon, not visible in a small-coupling expansion, so non-perturbative methods are needed in order to address this question. One relatively well-developed non-perturbative method uses numerical calculations in which continuous spacetime is replaced by a discrete lattice. Lattice calculations are easier when fermions (quarks) are not included, and they are also easier when the number of colors is two (gauge group SU(2)) instead of three. Probably for these reasons, published results are relatively abundant for QCD without quarks and with only two colors, including some results for five-dimensional spacetime.
This answer cites some theoretical evidence regarding the fate of confinement in higher dimensions, but it doesn't explain the underlying reason. That would be a tall order, because the reason for confinement even in the most important case of four dimensions is still not completely understood, as reviewed in Greensite (2011), An Introduction to the Confinement Problem.
## Lattice results for 5-d QCD without quarks
To extract predictions that are relevant to continuous spacetime from models formulated on a discrete lattice, the model's parameters are tuned to make the correlation length much larger than the lattice spacing — nominally infinitely larger. Such a divergence of the correlation length occurs near second-order phase transitions. According to the review [1], numerical studies of five-dimensional QCD with two colors and without quarks shows a first-order phase transition separating a confinement phase from a deconfined (Coulomb) phase. (See figure 2 in [1].) In other words, according to this numerical evidence, higher-dimensional QCD exhibits both confinement and non-confinement, at least without quarks, depending on the value of the coupling constant. However, the higher-dimensional theory doesn't necessarily have a continuum limit. According to page 11 in [2],
...the phase diagram of the $$d = 5$$ SU(2) Yang–Mills theory on the lattice does not contain a second order phase transition or a critical point where a five-dimensional continuum theory can be defined non-perturbatively...
In the context of a small-coupling expansion, higher-dimensional QCD is non-renormalizable (in the power-counting sense), suggesting that it might not have a continuum limit [2]. The small-coupling expansion may not be a reliable guide for that question, but this suggestion is at least consistent with the numerical evidence.
The paper [3], which claims to be the first lattice study of five-dimensional gauge theory with three colors (gauge group SU(3) but still without quarks), finds a similar structure: both a confined phase and a deconfined phase, separated from each other by a first-order transition (no continuum limit).
The question of the existence of a five-dimensional continuum limit is not yet settled, though. The paper [3] says,
The existence of the second order critical end point even for the SU(2) gauge theory is still under investigation..., and we need the large lattice data to show it.
## The effect of dynamical quarks
What happens to this picture when quarks are included? I don't know of any lattice studies of higher-dimensional QCD with dynamical quarks, but the small-coupling expansion in four-dimensional QCD indicates that asymptotic freedom disappears when the number of quark flavors is sufficiently large. If the loss of asymptotic freedom entails a loss of confinement (?), then this indicates that adding more quarks to the theory decreases the chances that the theory is confining. That's a pretty loose argument, but it suggests that the existence of a confining phase in QCD without quarks is at least a necessary condition for the existence of a confining phase with quarks. In this sense, the lattice evidence cited above isn't completely irrelevant to the question; but as far as I know, a definitive answer to the question is not yet available.
## References:
[1] "Extra-dimensional models on the lattice," https://arxiv.org/abs/1605.04341
[2] "Lattice Simulations of 10d Yang-Mills toroidally compactified to 1d, 2d and 4d," https://arxiv.org/abs/1612.06395
[3] "Phase structure of pure SU(3) lattice gauge theory in 5 dimensions," https://arxiv.org/abs/1403.6277 | 1,112 | 5,345 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-17 | longest | en | 0.937702 |
https://math.stackexchange.com/questions/3029449/the-general-proposition-of-fermat | 1,547,922,688,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583680452.20/warc/CC-MAIN-20190119180834-20190119202834-00136.warc.gz | 589,542,521 | 33,639 | # The general proposition of Fermat
In his letter to Frenicle, dated 18th October, 1640, Fermat states the following (Point 8, translated) :
If you subtract $$2$$ from a square, the remaining value cannot be divided by a prime which is greater than a square by $$2$$
For example, take for a square $$1,000,000$$, from which, subtracted by two, remains $$999,998$$. I say that the given remainder cannot be divided by $$11$$ or by $$83$$, by $$227$$, and so on.
You can prove the same rule for odd squares and, if I wanted, I would give you the lovely and general rule; but I'm content with having only indicated it to you.
In other words, numbers of the form $$x^2 -2$$ are not divisible by primes of the form $$a^2 + 2$$, where $$x$$ and $$a$$ are integers.
Questions :
$$1)$$ What is the general rule Fermat is talking about?
$$2)$$ Are there any modern references to this problem?
$$3)$$ How would Fermat have proved it?
• I believe he considered 3 cases : 1) $a^2+2$ is prime, 2) $a^2+2$ is odd, 3) $a^2+2$ is any integer (general case) – rsadhvika Dec 7 '18 at 4:19
• $x^2-2 = m*(a^2+2)$ has no solutions over positive integers, general case/rule – rsadhvika Dec 7 '18 at 4:21
• Well, $a^2+2$ primes are always $2$ or $4k+3$ primes (the converse is of course not true). Not sure how that helps though – YiFan Dec 7 '18 at 6:04
• Could the general rule be that numbers of the form $x^2-n$ are not divisible by primes of the form $a^2+n$? – tyobrien Dec 7 '18 at 6:20
Ad 3): If $$p = a^2 + 2$$ divides $$x^2 - 2$$, then $$p$$ divides $$x^2 - 2 + p = x^2 + a^2$$. But primes $$p = 4n+3$$ cannot divide sums of two squares without dividing the squares themselves. This observation is due to Weil.
Let $$p=a^2+2$$ be a prime. Suppose $$a>0$$ as for $$a=0$$, the statement clearly does not hold. Therefore, $$p\equiv 3 \pmod 8$$. Thus, the congruence $$x^2\equiv 2\pmod p$$ has no solutions$$(2$$ is a quadratic residue modulo $$p$$ if and only if $$p\equiv ±1\pmod 8)$$. | 646 | 1,981 | {"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": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-04 | longest | en | 0.893409 |
https://edutube.hccs.edu/tag?tagid=halves | 1,718,239,243,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00758.warc.gz | 204,411,066 | 32,043 | # Search for tag: "halves"
#### Dosages 9.9
9. On hand is an 8 oz. can Sustacal. How many ml water will be needed to make the Sustacal a 2/3 strength feeding?
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#### Dosages q2.14
1/2 tsp = ? mL
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#### Dosages q2.11
Gr. I ss = ? mg
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#### Dosages q1.11
.5 tsp = ? mL
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#### 1314 5.3.5
Cramer's Rule
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#### 1314 4.3.16
Example 5
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#### 1314 4.3.2
Example 1
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#### 1314 4.2.2
If 2^x = 2^4 then x = 4
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#### 1314 3.5.11
Graph f(x) = (2x + 1) / (x - 3)
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Graph f(x) = 2x^3 + 5x^2 - x - 6
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#### 1314 2.4.6
Parallel and Perpendicular Lines
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#### 1314 2.4.5
Vertical and Horizontal Lines
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#### 1314 2.1.14
2x^2 + 2y^2 - 6x + 10y = 1
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#### 1314 2.1.7
The midpoint formula
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#### 1314 1.7.7
Solve x^2 - x - 12 < 0
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#### 1314 1.6.7
Solve (x + 1)^(2/3) - (x + 1)^(1/3) - 2 = 0
From Marisol Montemayor 0 likes 0 | 611 | 1,365 | {"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} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.581531 |
https://admin.clutchprep.com/chemistry/practice-problems/36802/if-a-solution-has-a-temperature-of-355-k-what-is-its-temperature-in-degrees-cels-1 | 1,600,693,637,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201699.38/warc/CC-MAIN-20200921112601-20200921142601-00672.warc.gz | 263,060,623 | 23,886 | Problem: If a solution has a temperature of 355 K, what is its temperature in degrees celsius? a. 165°C b. 628°C c. 179°C d. 82°C
🤓 Based on our data, we think this question is relevant for Professor Crawford's class at GGC.
Problem Details
If a solution has a temperature of 355 K, what is its temperature in degrees celsius?
a. 165°C
b. 628°C
c. 179°C
d. 82°C
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Energy, Heat and Temperature concept. You can view video lessons to learn Energy, Heat and Temperature. Or if you need more Energy, Heat and Temperature practice, you can also practice Energy, Heat and Temperature practice problems.
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Based on our data, we think this problem is relevant for Professor Crawford's class at GGC. | 287 | 1,244 | {"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} | 2.59375 | 3 | CC-MAIN-2020-40 | latest | en | 0.942341 |
https://docplayer.net/20632098-Mathematical-induction-lecture-10-11.html | 1,642,975,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00397.warc.gz | 252,429,297 | 28,263 | # Mathematical Induction. Lecture 10-11
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1 Mathematical Induction Lecture 10-11
2 Menu Mathematical Induction Strong Induction Recursive Definitions Structural Induction
3 Climbing an Infinite Ladder Suppose we have an infinite ladder: 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we can reach the next rung. From (1), we can reach the first rung. Then by applying (2), we can reach the second rung. Applying (2) again, the third rung. And so on. We can apply (2) any number of times to reach any particular rung, no matter how high up. This example motivates proof by mathematical induction.
4 Principle of Mathematical Induction Principle of Mathematical Induction: To prove that P(n) is true for all positive integers n, we complete these steps: Basis Step: Show that P(1) is true. Inductive Step: Show that P(k) P(k + 1) is true for all positive integers k. To complete the inductive step, assuming the inductive hypothesis that P(k) holds for an arbitrary integer k, show that must P(k + 1) be true. Climbing an Infinite Ladder Example: Basis Step: By (1), we can reach rung 1. Inductive Step: Assume the inductive hypothesis that we can reach rung k. Then by (2), we can reach rung k + 1. Hence, P(k) P(k + 1) is true for all positive integers k. We can reach every rung on the ladder.
5 Logic and Mathematical Induction Mathematical induction can be expressed as the rule of inference (P(1) k (P(k) P(k + 1))) n P(n), where the domain is the set of positive integers. In a proof by mathematical induction, we don t assume that P(k) is true for all positive integers! We show that if we assume that P(k) is true, then P(k + 1) must also be true. Proofs by mathematical induction do not always start at the integer 1. In such a case, the basis step begins at a starting point b where b is an integer. We will see examples of this soon.
6 Why Mathematical Induction is Valid? Mathematical induction is valid because of the well ordering property. Proof: Suppose that P(1) holds and P(k) P(k + 1) is true for all positive integers k. Assume there is at least one positive integer n for which P(n) is false. Then the set S of positive integers for which P(n) is false is nonempty. By the well-ordering property, S has a least element, say m. We know that m can not be 1 since P(1) holds. Since m is positive and greater than 1, m 1 must be a positive integer. Since m 1 < m, it is not in S, so P(m 1) must be true. But then, since the conditional P(k) P(k + 1) for every positive integer k holds, P(m) must also be true. This contradicts P(m) being false. Hence, P(n) must be true for every positive integer n.
7 Proving a Summation Formula by Mathematical Induction Example: Show that: Solution: BASIS STEP: P(1) is true since 1(1 + 1)/2 = 1. INDUCTIVE STEP: Assume true for P(k). The inductive hypothesis is Under this assumption,
8 Exercise: Show that the sum of first n positive odd numbers is n 2. We will do it on the board!
9 Proving Inequalities Example: Use mathematical induction to prove that n < 2 n for all positive integers n. Solution: Let P(n) be the proposition that n < 2 n. Basis Step: (1) is true since 1 < 2 1 = 2. Inductive Step: Assume P(k) holds, i.e., k < 2 k, for an arbitrary positive integer k. Must show that P(k + 1) holds. Since by the inductive hypothesis, k < 2 k, it follows that: k + 1 < 2 k k + 2 k = 2 2 k = 2 k+1 Therefore n < 2 n holds for all positive integers n.
10 Proving Inequalities Example: Use mathematical induction to prove that 2 n < n!, for every integer n 4. Solution: Let P(n) be the proposition that 2 n < n!. Basis: P(4) is true since 2 4 = 16 < 4! = 24. Inductive Step: Assume P(k) holds, i.e., 2 k < k! for an arbitrary integer k 4. To show that P(k + 1) holds: 2 k+1 = 2 2 k < 2 k! (by the inductive hypothesis) < (k + 1)k! = (k + 1)! Therefore, 2 n < n! holds, for every integer n 4. Note: The basis step is P(4), since P(0), P(1), P(2), and P(3) are all false.
11 Example Example: Use mathematical induction to prove that n 3 n is divisible by 3, for every positive integer n. Solution: Let P(n) be the proposition that 3 (n 3 n). Basis: P(1) is true since = 0, which is divisible by 3. Induction: Assume P(k) holds, i.e., k 3 k is divisible by 3, for an arbitrary positive integer k. To show that P(k + 1) follows: (k + 1) 3 (k + 1) = (k 3 + 3k 2 + 3k + 1) (k + 1) = (k 3 k) + 3(k 2 + k) By the inductive hypothesis, the first term (k 3 k) is divisible by 3 and the second term is divisible by 3 since it is an integer multiplied by 3. So by part (i) of Theorem 1 in Section 4.1, (k + 1) 3 (k + 1) is divisible by 3. Therefore, n 3 n is divisible by 3, for every integer positive integer n.
12 Strong Induction Strong Induction: To prove that P(n) is true for all positive integers n, where P(n) is a propositional function, complete two steps: Basis Step: Verify that the proposition P(1) is true. Inductive Step: Show the conditional statement [P(1) P(2) P(k)] P(k + 1) holds for all positive integers k. Strong Induction is sometimes called the second principle of mathematical induction or complete induction.
13 Strong induction tells us that we can reach all rungs if: 1. We can reach the first rung of the ladder. 2. For every integer k, if we can reach the first k rungs, then we can reach the (k + 1)st rung. To conclude that we can reach every rung by strong induction: BASIS STEP: P(1) holds INDUCTIVE STEP: Assume P(1) P(2) P(k) holds for an arbitrary integer k, and show that P(k + 1) must also hold. We will have then shown by strong induction that for every positive integer n, P(n) holds, i.e., we can reach the nth rung of the ladder.
14 Proof using Strong Induction Example: Suppose we can reach the first and second rungs of an infinite ladder, and we know that if we can reach a rung, then we can reach two rungs higher. Prove that we can reach every rung. Solution: Prove the result using strong induction. BASIS STEP: We can reach the first step. INDUCTIVE STEP: The inductive hypothesis is that we can reach the first k rungs, for any k 2. We can reach the (k + 1)st rung since we can reach the (k 1)st rung by the inductive hypothesis. Hence, we can reach all rungs of the ladder.
15 Strong vs Mathematical Induction We can always use strong induction instead of mathematical induction. But there is no reason to use it if it is simpler to use mathematical induction. In fact, the principles of mathematical induction, strong induction, and the wellordering property are all equivalent. Sometimes it is clear how to proceed using one of the three methods, but not the other two.
16 Example Example: Show that if n is an integer greater than 1, then n can be written as the product of primes. Solution: Let P(n) be the proposition that n can be written as a product of primes. BASIS STEP: P(2) is true since 2 itself is prime. INDUCTIVE STEP: The inductive hypothesis is P(j) is true for all integers j with 2 j k. To show that P(k + 1) must be true under this assumption, two cases need to be considered: If k + 1 is prime, then P(k + 1) is true. Otherwise, k + 1 is composite and can be written as the product of two positive integers a and b with 2 a b < k + 1. By the inductive hypothesis a and b can be written as the product of primes and therefore k + 1 can also be written as the product of those primes. Hence, it has been shown that every integer greater than 1 can be written as the product of primes.
17 Proof using Strong Induction Example: Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps. Solution: Let P(n) be the proposition that postage of n cents can be formed using 4-cent and 5-cent stamps. BASIS STEP: P(12), P(13), P(14), and P(15) hold. P(12) uses three 4-cent stamps. P(13) uses two 4-cent stamps and one 5-cent stamp. P(14) uses one 4-cent stamp and two 5-cent stamps. P(15) uses three 5-cent stamps. INDUCTIVE STEP: The inductive hypothesis states that P(j) holds for 12 j k, where k 15. Assuming the inductive hypothesis, it can be shown that P(k + 1) holds. Using the inductive hypothesis, P(k 3) holds since k To form postage of k + 1 cents, add a 4-cent stamp to the postage for k 3 cents. Hence, P(n) holds for all n 12.
18 Recursive Definitions and Structural Induction
19 Recursively Defined Functions Definition: A recursive or inductive definition of a function consists of two steps. BASIS STEP: Specify the value of the function at zero. RECURSIVE STEP: Give a rule for finding its value at an integer from its values at smaller integers. A function f(n) is the same as a sequence a 0, a 1,, where a i, where f(i) = a i.
20 Example: Fibonacci Sequence Define the Fibonacci sequence, f 0,f 1,f 2,, by: Initial Conditions: f 0 = 0, f 1 = 1 Recurrence Relation: f n = f n-1 + f n-2 f -6 f -5 f -4 f -3 f -2 f -1 f 0 f 1 f 2 f 3 f 4 f 5 f
21 Recursively Defined Sets and Structures Recursive definitions of sets have two parts: The basis step specifies an initial collection of elements. The recursive step gives the rules for forming new elements in the set from those already known to be in the set. Sometimes the recursive definition has an exclusion rule, which specifies that the set contains nothing other than those elements specified in the basis step and generated by applications of the rules in the recursive step. We will always assume that the exclusion rule holds, even if it is not explicitly mentioned. We will later develop a form of induction, called structural induction, to prove results about recursively defined sets.
22 Examples Factorial of n n! = 1 if n = 0 n! = n (n-1)!, otherwise Sum of first n odd numbers S n S n = 1 if n = 1 S n = S n-1 + (2n 1), otherwise Length of a string s Σ * : len(s) len(s) = 0 if s = ε len(sa) = len(s) + 1 if s Σ* and a Σ. Sorting n numbers: SORT(<a1,, an>) <a1> = SORT(<a1,, an>) if n = 1 SORT(<a1,, an>) = <min(<a1,, an>), SORT <a1,, an> - < min(<a1,, an>)>)
23 String Concatenation Definition: Two strings can be combined via the operation of concatenation. Let Σ be a set of symbols and Σ* be the set of strings formed from the symbols in Σ. We can define the concatenation of two strings, denoted by, recursively as follows. BASIS STEP: If w Σ*, then w ε = w. RECURSIVE STEP: If w 1 Σ* and w 2 Σ* and x Σ, then w (w 2 x)= (w 1 w 2 )x. Often w 1 w 2 is written as w 1 w 2. If w 1 = abra and w 2 = cadabra, the concatenation w 1 w 2 = abracadabra.
24 Balanced Parentheses Example: Give a recursive definition of the set of balanced parentheses P. Solution: BASIS STEP: () P RECURSIVE STEP: If w P, then () w P, (w) P and w () P. Show that (() ()) is in P. Why is ))(() not in P?
25 Well-Formed Formulae in Propositional Logic Definition: The set of well-formed formulae in propositional logic involving T, F, propositional variables, and operators from the set {,,,, }. BASIS STEP: T,F, and s, where s is a propositional variable, are well-formed formulae. RECURSIVE STEP: If E and F are well formed formulae, then ( E), (E F), (E F), (E F), (E F), are well-formed formulae. Examples: ((p q) (q F)) is a well-formed formula. pq is not a well formed formula.
26 Structural Induction Definition: To prove a property of the elements of a recursively defined set, we use structural induction. BASIS STEP: Show that the result holds for all elements specified in the basis step of the recursive definition. RECURSIVE STEP: Show that if the statement is true for each of the elements used to construct new elements in the recursive step of the definition, the result holds for these new elements. The validity of structural induction can be shown to follow from the principle of mathematical induction.
27 Example: Full Binary Trees Definition: The set of full binary trees can be defined recursively by these steps. BASIS STEP: There is a full binary tree consisting of only a single vertex r. RECURSIVE STEP: If T 1 and T 2 are disjoint full binary trees, there is a full binary tree, denoted by T 1 T 2, consisting of a root r together with edges connecting the root to each of the roots of the left subtree T 1 and the right subtree T 2.
28 Example: Full Binary Trees Definition: The height h(t) of a full binary tree T is defined recursively as follows: BASIS STEP: The height of a full binary tree T consisting of only a root r is h(t) = 0. RECURSIVE STEP: If T 1 and T 2 are full binary trees, then the full binary tree T = T 1 T 2 has height h(t) = 1 + max(h(t 1 ),h(t 2 )). The number of vertices n(t) of a full binary tree T satisfies the following recursive formula: BASIS STEP: The number of vertices of a full binary tree T consisting of only a root r is n(t) = 1. RECURSIVE STEP: If T 1 and T 2 are full binary trees, then the full binary tree T = T 1 T 2 has the number of vertices n(t) = 1 + n(t 1 ) + n(t 2 ).
29 Example: Full Binary Trees 2 Theorem: If T is a full binary tree, then n(t) 2 h(t)+1 1. Proof: Use structural induction. BASIS STEP: The result holds for a full binary tree consisting only of a root, n(t) = 1 and h(t) = 0. Hence, n(t) = = 1. RECURSIVE STEP: Assume n(t 1 ) 2 h(t1)+1 1 and also n(t 2 ) 2 h(t2)+1 1 whenever T 1 and T 2 are full binary trees. n(t) = 1 + n(t 1 ) + n(t 2 ) (by recursive formula of n(t)) 1 + (2 h(t1)+1 1) + (2 h(t2)+1 1) (by inductive hypothesis) 2 max(2 h(t1)+1,2 h(t2)+1 ) 1 = 2 2 max(h(t1),h(t2))+1 1 (max(2 x, 2 y )= 2 max(x,y) ) = 2 2 h(t) 1 (by recursive definition of h(t)) = 2 h(t)+1 1
30 Generalized Induction Generalized induction is used to prove results about sets other than the integers that have the well-ordering property. For example, consider an ordering on N N, ordered pairs of nonnegative integers. Specify that (x 1,y 1 ) is less than or equal to (x 2,y 2 ) if either x 1 < x 2, or x 1 = x 2 and y 1 <y 2. This is called the lexicographic ordering. Strings are also commonly ordered by a lexicographic ordering. The next example uses generalized induction to prove a result about ordered pairs from N N.
31 2 Example: Suppose that a m,n a 0,0 = 0 and Generalized Induction is defined for (m,n) N N by Show that a m,n = m + n(n + 1)/2 is defined for all (m,n) N N. Solution: Use generalized induction. BASIS STEP: a 0,0 = 0 = 0 + (0 1)/2 INDUCTIVE STEP: Assume that a m,n = m + n (n + 1)/2 whenever(m,n ) is less than (m,n) in the lexicographic ordering of N N. If n = 0, by the inductive hypothesis we can conclude a m,n = a m 1,n + 1 = m 1+ n(n + 1)/2 + 1 = m + n(n + 1)/2. If n > 0, by the inductive hypothesis we can conclude am,n = am 1,n + 1 = m + n(n 1)/2 +n = m + n(n + 1)/2.
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http://www.longrangehunting.com/threads/true-muzzle-velocity.61081/ | 1,503,513,883,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886123312.44/warc/CC-MAIN-20170823171414-20170823191414-00137.warc.gz | 613,918,866 | 18,074 | # True muzzle velocity
Discussion in 'Rifles, Bullets, Barrels & Ballistics' started by Medicucho, Sep 14, 2010.
1. ### MedicuchoMember
Messages:
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Joined:
Sep 30, 2009
Hello everybody!does anybody know the equation which permits you to calculate the real muzzle velocity from a recorded velocity in a chronograph which is set at a certain distance from the muzzle? I know that at 10-15ft the difference isn´t significant but I will really thank to know the formula
Thank you!!!!
2. ### ss7mmWriters Guild
Messages:
3,707
Joined:
Jun 11, 2005
The info below is from a Google search.
=============================================
There is a formula that will closely estimate this loss at normal chronograph distances:
You will need a scientific calculator to do the calculation. If you don't have one. Window's calculator will work. Click Start ~ All Programs ~ Accessories ~ Calculator to bring it up. Then click the "View" menu and select "Scientific".
On a scientific calculator, the "e" function is the inversion of "ln" (log number) function and that is activated by pushing "shift" or "inv" + the "ln" key.
Now:
Let CD = The Distance, in feet, from the muzzle to the center of the chronograph screens.
Let BC = The Ballistic Coefficient of the bullet.
Let CV = The average of the shot chronograph values.
Let MV = The Muzzle Velocity calculated from the Chronograph Velocity.
Then
e( CD / BC / 8816) x CV = MV
Here's how to do it.
CD = 7 feet.
BC = 0.300
CV = 1500 fps
On your calculator, enter the following, being sure to enter the parentheses "( )" characters in the equation. The (*) may also be entered as (X) on a hand held calculator and the "=" may be entered as "EXE" or "Calc"
"shift" + "ln" ( 7 / 0.3 / 8816 ) * 1500 =
This should return an answer of 1503.975 fps
Let's do my 280 Remington shooting a Hornady 139 gr. bullet with average chronograph values of 2818 fps
Chronograph distance is 15 feet.
Ballistic Coefficient of the Hornady Bullet is 0.392
e( 15 / 0.392 / 8816 ) x 2818 =
This returns a Muzzle Velocity of 2830 fps
3. ### MedicuchoMember
Messages:
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Sep 30, 2009
Thanks! I´ve seen that formula in a webpage but if you use it and compare the results with the outputs of the JBM page you realize that it is only for G1 model because if you use a G7 model BC the outputs don´t match. Notwithstanding, thanks a lot!
I´ve found in this link exterior ballistics another probably solution but the equations are too small and I am not able to distinguish the variables very well, please help!
4. ### SBruceWell-Known Member
Messages:
1,786
Joined:
Oct 31, 2009
SS7MM
I was able to get the same answers you did in your example with my calculator.
However, when I use the Nightforce Program or the older Sierra Program; the answers derived from the calculator don't always match the printed velocities of either program. I tried a few different bullets and BC's, and the errors or differences between the calculator and the program varied from 0 or dead on to as much as 6 fps.
Not sure which is absolutely correct.....?? Assuming that the programs us a more complex equation and that they take into account more variables??
Messages:
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The easiest way to do this is to use a regular ballistics program and set the range step size to one yard. Iterate on the muzzle velocity input until the velocity at 3, 4, or 5 yards (9, 12, or 15 feet) is what you measured with your chronograph. This is much easier than trying to apply a great big long equation.
-Bryan
6. ### MedicuchoMember
Messages:
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Sep 30, 2009
Thanks! that sounds to be the easiest way to do it. Regardless, I would like to know the equation inspite of the fact that it´s too long. I am more or less good at maths and I would like to have a deeper knowledge about it. Thank you for all your contributions , they are very appreciated!
7. ### ss7mmWriters Guild
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3,707
Joined:
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Why don't you just contact Sierra and ask them if they'd give you the formula as you can't read what's on their site. Since they have posted the formula on the web I don't see why they wouldn't give you the formula in a format that you can read.
8. ### Hector RodriguezNew Member
Messages:
3
Joined:
Feb 2, 2015
I just use Aardvark Muzzle Velocity app on my smart phone to calculate the true muzzle velocity.
9. ### pyroducksx3Well-Known Member
Messages:
782
Joined:
Sep 27, 2009
Chronos lie! It doesn't really matter if it's 15 ft away all the chrono does is get you close enough to use your ballistic program. Once you start shooting out you will be able to adjust your fps to match the data you are collecting. So in my opinion trying to figure this out is a waste of time.
10. ### J E CustomWell-Known Member
Messages:
7,313
Joined:
Jul 29, 2004
If you have a chronograph that has a proof screen it will tell you the velocity loss between screens
and with the known distance from the muzzle you can calculate the muzzle velocity based on the loss in velocity in 2 feet increments. (It wont be exact, but it will be close).
One other method is to buy a Magneto Speed. (It measures velocity @ the muzzle).
When all else fails and/or you don't have a chronograph,you can shoot at least 3 rounds at 100,200,300,400 and 500 or 600 and measure drop at the center of all groups and compare it to a drop chart that has velocity and trajectory numbers. In any case this should be done to assure that your data(velocity, BCs Of bullet, and MOA are all correct before taking a very long shot.
Not very scientific, But they work.
J E CUSTOM
11. ### MikecrWell-Known Member
Messages:
4,269
Joined:
Aug 10, 2003
I don't think this is true.
To begin, timing between any screens is an average of velocities to cause that timing. That is velocity passing the 'start' screen is not actually the velocity passing the 'stop' screen.
And I believe the 'proof' screen is still taken to the same 'start' screen, and integrated at twice the distance.
That said, I'm pretty sure my Oehler velocity difference between primary & proof channels represents the inescapable averaging of passing velocities(causing error, and printing as different).
The proof channel is only to validate the primary channel, and not showing velocity at one distance versus another distance. | 1,571 | 6,314 | {"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} | 3.484375 | 3 | CC-MAIN-2017-34 | latest | en | 0.891032 |
https://cyberphysics.co.uk/Q&A/KS3/moments/A3.html | 1,726,193,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651506.7/warc/CC-MAIN-20240913002450-20240913032450-00601.warc.gz | 166,945,661 | 10,730 | # Questions on Moments
Q3. The drawing shows a very old set of scales. It can be used to check the weights of silver coins.
(a) Jack puts a silver coin in pan X. There is nothing in pan Y.
In which direction does pan X move?
Down
anticlockwise
(1 mark)
(b) The table shows the weights of five silver coins.
Silver Coin A B C D E weight in mN 112 106 120 112 98
Jack puts one coin in each pan of the scales.
Which two coins will make the scales balance?
(Give the letters).
A and D
(1 mark)
(c) Coin A is placed in pan X, and coin C is placed in pan Y.
In which direction does pan X now move?
up
accept clockwise
(1 mark)
(d) In another experiment, coin B is placed in pan X, and one of the other coins is placed in pan Y.
Pan X goes down. Which coin is in pan Y?
Give the letter.
E
(1 mark)
(e) Jack knows the weights of the five silver coins in the table. He also has a gold coin.
Explain how he could use the silver coins and the scales to find the approximate weight of the gold coin.
Put the gold coin in one pan and put the coins on in turn to find out which are lighter/heavier. If all are lighter he should then start to put on combinations of coins until he has a lowest value that is heavier and highest one that is lighter. He then knows the coin's weight is somewhere between the two values.
(2 marks)
Total 6 marks | 338 | 1,355 | {"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} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.935421 |
https://www.bartleby.com/questions-and-answers/given-that-lim-x0-cos-x-1-which-of-the-following-statements-is-true-a-if-orcos-x-1or-is-very-small-t/056ffac8-b8b2-4e17-9295-8d8597115f38 | 1,585,686,731,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370503664.38/warc/CC-MAIN-20200331181930-20200331211930-00394.warc.gz | 838,062,150 | 27,295 | # Given that limx→0cos x = 1, which of the following statements is true?(a) If |cos x − 1| is very small, then x is close to 0.(b) There is an E > 0 such that if if 0 < |cos x − 1| < E, then |x| < 10−5 .(c) There is a δ > 0 such that if 0 < |x| < δ, then |cos x − 1| < 10−5.(d) There is a δ > 0 such that if 0 < |x − 1| < δ, then |cos x| < 10−5.
Question
1 views
Given that lim
x→0
cos x = 1, which of the following statements is true?
(a) If |cos x − 1| is very small, then x is close to 0.
(b) There is an E > 0 such that if if 0 < |cos x − 1| < E, then |x| < 10−5 .
(c) There is a δ > 0 such that if 0 < |x| < δ, then |cos x − 1| < 10−5.
(d) There is a δ > 0 such that if 0 < |x − 1| < δ, then |cos x| < 10−5.
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Tagged in | 382 | 1,034 | {"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} | 3.15625 | 3 | CC-MAIN-2020-16 | latest | en | 0.893773 |
https://www.jiskha.com/display.cgi?id=1353901343 | 1,516,406,389,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888302.37/warc/CC-MAIN-20180119224212-20180120004212-00559.warc.gz | 935,609,629 | 4,708 | # physics
posted by .
When you stand on your tiptoes, your feet pivot about your ankle. As shown in the figure, the forces on your foot are an upward force on your toes from the floor, a downward force on your ankle from the lower leg bone, and an upward force on the heel of your foot from your Achilles tendon. Suppose a 61 woman stands on tiptoes with the sole of her foot making a 25 angle with the floor. Assume that each foot supports half her weight.
Sorry I can't add a picture to this b/c cramster doesn't have my book but....
From the toe to the ankle pivot is 15cm and from the toe to the Achillies tendon it is 20cm.
a)What upward force does the Achilles tendon exert on the heel of her foot N?
b)The tension in the Achilles tendon will cause it to stretch. If the Achilles tendon is 15 long and has a cross-section area of 110 , by how much will it stretch under this force?
• physics -
Many of your numbers require dimensions. For example, what do you mean by "a 61 woman"?
• physics -
883N
• physics -
927
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I'm trying to model a subset of the MALD dataset (language related) using Gaussian Distribution.
One of the variables I'm working with is 'Number of Phones' (NumPhones in MALD) in a given word or pseudoword (Item in MALD) and I'm trying to explore two things with my model: 1) relation between reaction time (numerical continuous data) and lexical status (categorical; IsWord in MALD), which says whether a word is an actual word or a pseudoword and 2) is this effect is impacted by the number of phones in the target word/pseudoword.
For 2), I'm not sure if number of phones (ranging from 1-17) in a word is a categorical ordinal or numerical discrete data? I learnt that counts are numerical but in order to answer 2), I'd need to apply some treatment coding to this variable (number of phones), for which I require it to be categorical, do I not? Is there such a thing as contrast coding (treatment) for numerical variables?
The reason why I think it is categorical despite being countable is that the numbers in this variable represent the number of phones in a word, which could be seen as a category (?). As in, for whichever words/pseudowords the 'number of phones' is '10', we can say that those words are in the "category of 10-phone words"?
Yet, it seems straightforward to just take it as a numerical discrete data - but how does one do treatment coding for numerical variables? Won't there simply be too many (17) levels? And what would be the "reference level"? Intuitively, it should be 0-phones but there are no words with "0-phones" (in real life or in my data).
• You write multiple times that this is a number. Why do you think it makes sense to code it as a category? If you code it as numerical, then "10-phone words" will already be treated the same. Are you concerned about possible nonlinearities? If so, take a look at splines. Commented Dec 5, 2023 at 9:32
• I'm hesitant to take it as numerical discrete data because I'd need to apply contrast coding on it, which as I understand is typically only done for categorical data: vasishth.github.io/bayescogsci/book/ch-contr.html Besides, even if I take NumPhones as categorical, that will leave me with 17 levels (a lot). And I can't take 0 as a reference level because there are no "0-phone' words. Or do you think it's appropriate to take '1-phone' words as the 0 reference level and compare all other cases to this? This would be done to explore question (2) in my post... Commented Dec 5, 2023 at 9:37
• You have it the wrong way around: if your data are categorical, then you may want to use contrast coding (or other way to represent it). If your data is numerical, then you just feed it in as-is, or potentially spline transform it. Why do you think you need to contrast code? Commented Dec 5, 2023 at 9:38
• I thought contrast coding was necessary to find the effect of one variable (typically a categorical one) on another effect (the effect of lexical status on reaction times, in my case). Is this not true? How else does one study the effect of NumPhones on reaction times, affected by lexical status? (that is, NumPhones is a variable that hypothetically has an influence on how RTs are affected by lexical status). Commented Dec 5, 2023 at 9:44
• That question is really orthogonal to the coding of predictors. What you describe sounds like an interaction between lexical status and number of phones in a model for reaction times. Take a look at the interaction plots at the Wikipedia page. Plot RTs (y axis) against number of phones (x axis) separately by lexical status. You can plot observed means, or model fits. For the latter, you can just use number of phones as is (leading to linear plots), or consider splines. Commented Dec 5, 2023 at 9:52
Some basics wrt psycholinguistics may help CV participants here (the OP undoubtedly knows this). First, the field posits that linguistic perception is categorical, "the grouping of like items along a continuum (1)."
So, wrt your first question, for decades applied psycholinguisticians have been studying the perceptual boundaries between phones in the context of reaction time, e.g., here Subcategorical phonetic mismatches slow phonetic judgments (2) and here Categorical perception of English /r/ and /l/ by Japanese bilinguals (3).
Your second question sounds like a much more recent area of study made possible by databases such as MALD but why would there be any ordinal ranking to a count of the number of phones?
There are many ways to code categorical information, most of which are limited to a non-masssive number of possible levels. One approach which is not is impact coding which has been used to code categorical information as massive as zip codes (4).
What you describe sounds like an interaction between lexical status and number of phones (do you mean phonemes?) in a model for reaction times.
Take a look at the interaction plots at the Wikipedia page. Plot RTs (y axis) against number of phones (x axis) separately by lexical status. You can plot observed means, or model fits. For the latter, you can just use number of phones as is (leading to linear plots), or consider splines. Splines are a good choice if you suspect nonlinearities.
Alternatively, you could take a look at moderator-mediator analysis. You can again use spline transforms on the number of phones.
Contrast coding is not relevant for numerical predictors. If your data are categorical, then you may want to use contrast coding (or other way to represent it). If your data is numerical, then you just feed it in as-is, or potentially spline transform it. | 1,297 | 5,657 | {"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} | 2.6875 | 3 | CC-MAIN-2024-38 | latest | en | 0.934692 |
https://ask.learncbse.in/t/to-construct-a-ray-diagram-we-use-two-rays-of-light-which-are-so-chosen-that-it-is-easy-to-determine-their-directions-after-reflection-from-the-mirror/24693 | 1,696,400,856,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511361.38/warc/CC-MAIN-20231004052258-20231004082258-00470.warc.gz | 132,406,237 | 4,021 | # To construct a ray diagram we use two rays of light which are so chosen that it is easy to determine their directions after reflection from the mirror
To construct a ray diagram we use two rays of light which are so chosen that it is easy to determine their directions after reflection from the mirror. Choose these two rays and state the path of these rays after reflection from a concave mirror. Use these two rays to find the nature and position of the image of an object placed at a distance of 15 cm from a concave mirror of focal length 10 cm.
Answer:
The candidate may choose the following rays:
1. A ray parallel to the principal axis, after reflection, will pass through the principal focus of a concave mirror.
2. A ray passing through the centre of curvature of a concave mirror after reflection is reflected back along the same path.
Object distance, u = - 15 cm, focal length,f= - 10 cm, image distance, v = ?
Apply mirror formula,
The image is formed at a distance of 30 cm in front of the mirror (negative sign means object and image are on the same side). It is real and inverted. | 239 | 1,102 | {"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} | 3.265625 | 3 | CC-MAIN-2023-40 | latest | en | 0.926108 |
https://mail.haskell.org/pipermail/haskell-cafe/2003-July/004691.html | 1,670,145,497,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00491.warc.gz | 429,817,088 | 3,135 | functional graph
Ron de Bruijn rondebruijn@yahoo.com
Tue, 15 Jul 2003 13:35:11 -0700 (PDT)
```--- Hal Daume <t-hald@microsoft.com> wrote:
> >> I think that the
> >> difficulties you are
> >> facing are from the fact that you are trying to
> >> express a purely
> >> functional "updatable" graph.
> >
> > Should I understand from this, that this is a
> > difficult problem and that there exist no easy way
> to
> > do this at the moment?
>
> Well, I think he was being a bit ironic. You can
> either have a
> functional graph or an updatable graph, but not
> both. It sounds like
> either would do and given that later in this message
> you discuss that
> you don't like imperativeness, then a functional
> graph (ala the
> Functional Graph Library -- FGL) would suit you
> perfectly.
>
> school
> > yet(first year student)... Although it is
> somewhere in
> > my textbooks...
>
> They're very simple. You have a collection of
> nodes, N, and a
> collection of edges between elements of N. Nodes
> and edges can have
> labels. So for instance you could have Teachers and
> Classes as the
> nodes and the edges could be between the teacher and
> the class(es) which
> go together.
>
> > To solve my problem... I need an mutable array,
> but in
> > Haskell my program will then be unbelievable
> > imperative.
>
> No, you just need a graph :).
>
> > I have been following the discussion about the
> graph
> > and so, but the arrow classes and so on, are like
> > something too sophisticated for me at this moment
> I
> > think, although I have some idea of how they work,
> I
> > really don't know when to apply them. I just found
> > out, how Monads work, so...
>
> THe discussion about arrows is a complete
> divergence. You don't need
>
> > But I certainly believe that my problem must be
> fixed
> > before it's easy to program in Haskell.
>
> Theorem: this problem is already fixed.
> Proof: it is easy to program in Haskell. QED.
>
> :)
>
> - Hal
I have read the chapter in my textbook :) and I have
read and played with the FGL module.
I must say the FGL module looks impressive, but I find
way of doing it for me was, to have different
datatypes for Teacher Subject and so on.
But it looks the Graph class works on types a and b
where a is the Node type where b is the Edge type, or
just the other way around :), I don't have the code
right now.
The above rules out my way of representing the data,
or doesn't it (i.e. is there some fancy, magical way
of doing this) ?
The second intuitive solution is to take a datatype
like
data GraphNode|Teacher etc
|Subject etc
|Student etc
Should this be the way to do it?
And then yet another question: I need some kind of way
In example in my C++ program I used a couple of
array's with in each array pointers to the
StudentObjects, TeacherObjects.
So I had the notion of the first teacher, the second
teacher, the third and so on(in 0(1)), although I only
need the notion of did I test this possibility and
give me the next of Student/Teacher/whatever.
I must be able to put it somewhere, but where?
In the Graph itself there isn't really space for it,
unless I create a complete timetable in the beginning
with all Studentnodes pointing to this first teacher.
But now I have to be able to change the Graph in such
a way that I can change the first node where an
error(schedulingerror) occurs, (for example a "room is
full" error) to let the Student point to the next
room(see the notion above).
But then I yet have another problem. How can I
guarantee, to do a complete path traversal, each time
in exactly the same way?
Greets Ron
__________________________________
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SBC Yahoo! DSL - Now only \$29.95 per month!
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``` | 978 | 3,723 | {"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} | 2.71875 | 3 | CC-MAIN-2022-49 | latest | en | 0.954857 |
https://m.scirp.org/papers/112552 | 1,643,034,373,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304570.90/warc/CC-MAIN-20220124124654-20220124154654-00549.warc.gz | 414,445,997 | 29,017 | Estimation of Aggregate Losses of Secondary Cancer Using PH-OPPL and PH-TPPL Distributions
Abstract: Kenyan insurance firms have introduced insurance policies of chronic illnesses like cancer; however, they have faced a huge challenge in the pricing of these policies as cancer can transit into different stages, which consequently leads to variation in the cost of treatment. This has made the estimation of aggregate losses of diseases which have multiple stages of transitions such as cancer, an area of interest of many insurance firms. Mixture phase type distributions can be used to solve this setback as they can in-cooperate the transition in the estimation of claim frequency while also in-cooperating the heterogeneity aspect of claim data. In this paper, we estimate the aggregate losses of secondary cancer cases in Kenya using mixture phase type Poisson Lindley distributions. Phase type (PH) distributions for one and two parameter Poisson Lindley are developed as well their compound distributions. The matrix parameters of the PH distributions are estimated using continuous Chapman Kolmogorov equations as the disease process of cancer is continuous while severity is modeled using Pareto, Generalized Pareto and Weibull distributions. This study shows that aggregate losses for Kenyan data are best estimated using PH-OPPL-Weibull model in the case of PH-OPPL distribution models and PH-TPPL-Generalized Pareto model in the case of PH-TPPL distribution models. Comparing the two best models, PH-OPPL-Weibull model provided the best fit for secondary cancer cases in Kenya. This model is also recommended for different diseases which are dynamic in nature like cancer.
1. Introduction
Aggregate losses are estimated by in-cooperating both claim frequency and claim severity distributions. Pavel (2010) [1] reviewed methods used to calculate distributions of aggregate losses. Robertson (1992) [2] applied Discrete Fourier Transform in estimation of aggregate losses from frequency and severity distributions. Ronoet al. (2020) [3] developed compound distribution to model extreme natural disasters in Kenya. Mohamed et al. (2010) [4] introduced use of simulation approach in estimation of aggregate losses which can be employed when frequency and severity distribution cannot be combined to derive a compound distribution. Aggregate loss distributions are based on collective risk model expressed as:
${S}_{N}=\underset{i=1}{\overset{N}{\sum }}\text{ }\text{ }{X}_{i}$ (1)
where:
${X}_{i}$ is the severity distribution and N is the claim count distribution. The distribution of N in this paper is considered to follow mixed PH Poisson distributions.
Phase type distributions are constructed, when mixture distributions are convoluted resulting to an interrelated Poisson process occurring in phases. Phase type distributions were introduced way back by Erlang (1909) [5] and it has been advanced by Marcel F. Neuts (1981) [6] and Assussen (2003) [7] among others. Mogens Bladt (2005) [8] introduced phase type distributions in risk theory while O’cinneide (2017) [9] highlighted on Phase type distributions as well as their invariant polytopes. Wu et al. (2010) [10] developed phase type distributions when frequency distributions followed Panjer class $\left(a,b,0\right)$ while Kok et al. (2010) [11] used phase type distributions of Panjer class $\left(a,b,1\right)$ to model claim frequency.
Markov chains were introduced by Andrei Markov (1856-1922). Nurul et al. (2019) [12] proposed a simple forecasting model of predicting the future air quality using Markov chains which in-cooperated the Markov chains as an operator of evaluating pollution distribution in the long run. Yajuan et al. (2018) [13] used Markov chains to model demand for stations in Bike sharing systems. In this study, the concept of Markov chains is used to determine the matrices of the phase type distributions used in modeling claim frequency.
Frequency data is used to model occurrences in different areas such as engineering, insurance, biology etc. Poisson distribution is often used to model count data; however, it is based on the assumption that variance to mean ratio is unity (equi-dispersion) which is not applicable to real data; hence, it is considered as an inflexible model. Most real life data either experience over dispersion where variance exceeds the mean or under dispersion where the mean exceeds the variance which can be modeled using Poisson mixtures [14]. Poisson Lindley distributions are perfect examples of Poisson mixtures where characteristics of Poisson distribution follow some characteristics of Lindley distribution. One parameter Poisson Lindley which can model over dispersed data was introduced by Sankaran (1970) [15] while Shanker and Mishra (2014) [16] developed two parameter Poisson Lindley which further research has justified that it can model over dispersed data.
In the insurance sector, when calculating aggregate losses for chronic diseases which have various stages like cancer the claim frequency distributions considered do not in-cooperate the different stages of such diseases. In-cooperating phase type distributions solve this short coming of ordinary distributions. Further considering mixed phase type distributions improves modeling of claim frequency data as it considers the heterogeneity aspect of claim data. In this paper, we develop PH one parameter Poisson Lindley distribution and PH two parameter Poisson Lindley distributions where the mixing distribution follows PH Lindley distribution. The resulting PH distributions are used to model claim numbers of secondary cancers in Kenya. Section 1 has a brief introduction to Poisson distributions and Poisson Lindley distributions.
The structure of this paper is as follows: Section 2 will discuss construction of phase type distribution using PH Lindley distributions which will later be applied in modeling of the aggregate losses. Compound distributions from the frequency and severity distributions are developed in Section 3. Aggregate losses for the data are estimated using Discrete Fourier Transforms and the results discussed in Section 4 and Section 5 outlines the conclusions.
2. Proposed Phase Type Poisson Lindley Distributions
In this section we develop phase type distributions for one parameter Poisson Lindley and two parameter Poisson Lindley. Phase type Poisson Lindley distributions are derived when the mixing distribution follow phase type Lindley distribution.
2.1. Phase Type One Parameter Poisson Lindley Distribution
Definition 1. A random variable X is said to be a phase type one parameter Poisson Lindley distribution if it follows:
$X|\lambda ~Po\left( \lambda \right)$
$\lambda |\Lambda ~PH-OPL\left( \Lambda \right)$
for $\lambda >0$ and $\Lambda$ is $m\ast m$ matrix.
Theorem 1. If $X~PH-OPPL$ distribution then the probability distribution function of X is:
$f\left(x;\Lambda \right)=\stackrel{\to }{\gamma }\frac{{\Lambda }^{2}}{{\left(I+\Lambda \right)}^{x+3}}\left\{\left(x+2\right)I+\Lambda \right\}{\stackrel{\to }{1}}^{\text{T}}$ (2)
where $\Lambda$ is $M\ast M$ and I is an identity matrix.
Proof:
If $X|\lambda ~Po\left(\lambda \right)$ and $\lambda |\Lambda ~PH-OPL\left(\Lambda \right)$, then the pdf of variable X is expressed as;
$P\left(x\right)={\int }_{0}^{\infty }\text{ }\text{ }Pr\left(x|\lambda \right)f\left(\lambda ;\Lambda \right)\text{d}\lambda$
where $f\left(\lambda ;\Lambda \right)$ is $PH-OPL\left(\Lambda \right)$.
$\begin{array}{c}P\left(x\right)={\int }_{0}^{\infty }\frac{{\text{e}}^{-\lambda }{\lambda }^{x}}{x!}\frac{{\Lambda }^{2}}{I+\Lambda }\left(1+\lambda \right){\text{e}}^{-\Lambda \lambda }\text{ }\lambda >0,\Lambda =M\ast M\\ =\frac{{\Lambda }^{2}}{I+\Lambda }{\int }_{0}^{\infty }\left[\frac{{\lambda }^{x}}{x!}{\text{e}}^{-\lambda \left(I+\Lambda \right)}+\frac{{\lambda }^{x+1}}{x!}{\text{e}}^{-\lambda \left(I+\Lambda \right)}\right]\\ =\stackrel{\to }{\gamma }\frac{{\Lambda }^{2}}{{\left(I+\Lambda \right)}^{x+3}}\left\{\left(x+2\right)I+\Lambda \right\}{\stackrel{\to }{1}}^{\text{T}}\end{array}$ (3)
Properties of Phase Type One Parameter Poisson Lindley Distribution
The rth moments of PH-OPPL distribution is given by:
$\begin{array}{c}E\left({X}^{r}\right)={\int }_{0}^{\infty }\text{ }\text{ }{x}^{r}f\left(x,\Lambda \right)\text{d}x\\ =\frac{{\Lambda }^{2}}{I+\Lambda }{\int }_{0}^{\infty }\text{ }\text{ }{x}^{r}{\text{e}}^{-\Lambda x}\left(1+x\right)\text{d}x\\ =\frac{{\Lambda }^{2}}{I+\Lambda }\left[\frac{\Gamma \left(x+1\right)}{{\Lambda }^{x+1}}+\frac{\Gamma \left(x+2\right)}{{\Lambda }^{x+2}}\right]\\ =\stackrel{\to }{\gamma }\frac{x!\left[\left(x+1\right)I+\Lambda \right]}{\Lambda +I}{\stackrel{\to }{1}}^{T}\end{array}$ (4)
The expectation and variance of PH-OPPL distribution can be easily obtained from Equation (4) as:
1) Expectation
$E\left(x\right)=\frac{1!\left[\left(1+1\right)I+\Lambda \right]}{\Lambda \left(\Lambda +I\right)}=\stackrel{\to }{\gamma }\frac{2I+\Lambda }{\Lambda \left(\Lambda +I\right)}{\stackrel{\to }{1}}^{\text{T}}$ (5)
2) Variance
$\begin{array}{c}Var\left(x\right)=\frac{2!\left[\left(2+1\right)I+\Lambda \right]}{{\Lambda }^{2}\left(\Lambda +I\right)}-{\left\{\frac{2I+\Lambda }{\Lambda \left(\Lambda +I\right)}\right\}}^{2}\\ =\stackrel{\to }{\gamma }\frac{2I+4\Lambda +{\Lambda }^{2}}{{\left(\Lambda +I\right)}^{2}}+\frac{2I+\Lambda }{\Lambda \left(\Lambda +I\right)}{\stackrel{\to }{1}}^{\text{T}}\end{array}$ (6)
The probability generating function of PH-OPPL distribution is given by:
$\begin{array}{c}G\left(s\right)={\int }_{0}^{\infty }\text{ }\text{ }{\text{e}}^{\lambda \left(1-s\right)}\frac{{\Lambda }^{2}}{I+\Lambda }\left(1+\lambda \right){\text{e}}^{-\Lambda \lambda }\text{d}\lambda \\ =\frac{{\Lambda }^{2}}{I+\Lambda }\left[{\int }_{0}^{\infty }\text{ }\text{ }\lambda {\text{e}}^{-\lambda \left(I+\Lambda -sI\right)}\text{d}\lambda +{\int }_{0}^{\infty }\text{ }\text{ }{\text{e}}^{-\lambda \left(\Lambda +I-sI\right)}\text{d}\lambda \right]\\ =\stackrel{\to }{\gamma }\frac{{\Lambda }^{2}}{I+\Lambda }\left[\frac{\Lambda +\left(2-s\right)I}{{\left[\Lambda +\left(1-s\right)I\right]}^{2}}\right]{\stackrel{\to }{1}}^{\text{T}}\end{array}$ (7)
The parameter $\Lambda$ of PH-OPPL distribution is estimated using continuous Chapman-Kolmogorov equation.
2.2. Phase Type Two Parameter Poisson Lindley Distribution
Definition 2. A random variable X is said to be a phase type two parameter Poisson Lindley distribution if it follows:
$X|\lambda ~Po\left( \lambda \right)$
$\lambda |\Lambda ,\alpha ~PH-TPL\left(\Lambda ,\alpha \right)$
for $\alpha >0,\lambda >0$ and $\Lambda$ is $M\ast M$ matrix.
Theorem 2. If $X~PH-TPPL$ distribution then the probability density function of X is expressed as:
$f\left(x;\Lambda ,\alpha \right)=\stackrel{\to }{\gamma }\frac{{\Lambda }^{2}}{{\left(I+\Lambda \right)}^{x+2}}\left[I+\frac{\left(\alpha +x\right)I}{\alpha \Lambda +I}\right]{\stackrel{\to }{1}}^{\text{T}}$ (8)
where $\alpha >0$, $\Lambda$ is $M\ast M$ and I is an identity matrix.
Proof:
If $X|\lambda ~Po\left(\lambda \right)$ and $\lambda |\Lambda ,\alpha ~PH-TPL\left(\Lambda ,\alpha \right)$, then the pdf of variable X is given by;
$P\left(x\right)={\int }_{0}^{\infty }\text{ }\text{ }Pr\left(X=x|\lambda \right)f\left(\lambda ;\Lambda ,\alpha \right)\text{d}\lambda$
where $f\left(\lambda ;\Lambda ,\alpha \right)$ is $PH-TPL\left(\Lambda ,\alpha \right)$.
$\begin{array}{c}P\left(x\right)={\int }_{0}^{\infty }\frac{{\text{e}}^{-\lambda }{\lambda }^{x}}{x!}\frac{{\Lambda }^{2}}{I+\Lambda \alpha }\left(\alpha +\lambda \right){\text{e}}^{-\Lambda \lambda }\text{d}\lambda \text{ }\lambda >0,\text{\hspace{0.17em}}\Lambda =M\ast M\\ =\frac{{\Lambda }^{2}}{I+\Lambda \alpha }{\int }_{0}^{\infty }\frac{\alpha {\lambda }^{x}}{x!}{\text{e}}^{-\lambda \left(I+\Lambda \right)}\text{d}\lambda +{\int }_{0}^{\infty }\frac{{\lambda }^{x+1}}{x!}{\text{e}}^{-\lambda \left(I+\Lambda \right)}\text{d}\lambda \\ =\frac{{\Lambda }^{2}}{\alpha \Lambda +I}\left[\frac{\alpha \text{ }\Gamma \left(x+1\right)I}{x!{\left(I+\Lambda \right)}^{x+1}}+\frac{\Gamma \left(x+2\right)}{x!{\left(I+\Lambda \right)}^{x+2}}\right]\\ =\stackrel{\to }{\gamma }\frac{{\Lambda }^{2}}{{\left(I+\Lambda \right)}^{x+2}}\left[I+\frac{\left(\alpha +x\right)I}{\alpha \Lambda +I}\right]{\stackrel{\to }{1}}^{\text{T}}\end{array}$ (9)
Properties of Phase Type Two Parameter Poisson Lindley Distribution
The rth moments of PH-TPPL distribution is given by:
$\begin{array}{c}E\left({X}^{r}\right)={\int }_{0}^{\infty }\text{ }\text{ }{x}^{r}f\left(x,\Lambda ,\alpha \right)\text{d}x\\ ={\int }_{0}^{\infty }\left[\underset{x=0}{\overset{\infty }{\sum }}\text{ }\text{ }{x}^{r}\frac{{\text{e}}^{-\lambda }{\lambda }^{x}}{x!}\right]\frac{{\Lambda }^{2}}{I+\alpha \Lambda }\left(\alpha +\lambda \right){\text{e}}^{-\Lambda \lambda }\text{d}\lambda \\ =\frac{{\Lambda }^{2}}{I+\Lambda \alpha }{\int }_{0}^{\infty }\text{ }\text{ }{\lambda }^{r}\Lambda {\text{e}}^{-\Lambda \lambda }\text{d}\lambda +{\int }_{0}^{\infty }\text{ }\text{ }{\lambda }^{r+1}{\text{e}}^{-\Lambda \lambda }\text{d}\lambda \\ =\frac{{\Lambda }^{2}}{I+\Lambda \alpha }\left[\alpha \frac{\Gamma \left(r+1\right)I}{{\Lambda }^{r+1}}+\frac{\Gamma \left(r+2\right)}{{\Lambda }^{r+2}}\right]\\ =\stackrel{\to }{\gamma }\frac{\Gamma \left(r+1\right)I}{{\Lambda }^{r}}\frac{\alpha \Lambda +\left(r+1\right)I}{\alpha \Lambda +I}{\stackrel{\to }{1}}^{\text{T}}\end{array}$ (10)
The expectation and variance of PH-TPPL distribution can be easily obtained from Equation (10) as:
1) Expectation
$\begin{array}{c}E\left(x\right)=\frac{{\Lambda }^{2}}{\alpha \alpha +I}{\int }_{0}^{\infty }\text{ }\text{ }\lambda \left(\alpha +\lambda \right){\text{e}}^{-\Lambda \lambda }\text{d}\lambda \\ =\stackrel{\to }{\gamma }\frac{2I+\Lambda \alpha }{\Lambda \left(\Lambda \alpha +I\right)}{\stackrel{\to }{1}}^{\text{T}}\end{array}$ (11)
2) Variance
$Var\left(x\right)=E\left({x}^{2}\right)-{\left[E\left(x\right)\right]}^{2}$
$\begin{array}{c}E\left({x}^{2}\right)=\frac{\alpha \Lambda +2I}{\Lambda \left(\alpha \Lambda +I\right)}+\frac{2\left(\alpha \Lambda +3I\right)}{{\Lambda }^{2}\left(\alpha \Lambda +I\right)}\\ =\stackrel{\to }{\gamma }\frac{\alpha \Lambda +2I}{\Lambda \left(\alpha \Lambda +I\right)}+\frac{2\left(\alpha \Lambda +3I\right)}{{\Lambda }^{2}\left(\alpha \Lambda +I\right)}-{\left[\frac{2I+\Lambda \alpha }{\Lambda \left(\Lambda \alpha +I\right)}\right]}^{2}{\stackrel{\to }{1}}^{\text{T}}\end{array}$ (12)
The probability generating function of PH-TPPL distribution is given by:
$\begin{array}{c}G\left(s\right)=\frac{{\Lambda }^{2}}{{\left(\Lambda +I\right)}^{2}}\underset{x=0}{\overset{\infty }{\sum }}{\left[\frac{sI}{\Lambda +I}\right]}^{x}\\ \text{\hspace{0.17em}}+\frac{{\Lambda }^{2}}{{\left(\Lambda +I\right)}^{2}\left(\alpha \Lambda +I\right)}\underset{0}{\overset{\infty }{\sum }}\left(\alpha +x\right){\left[\frac{s}{\Lambda +I}\right]}^{x}\\ =\stackrel{\to }{\gamma }\frac{\alpha \Lambda \left[\Lambda +\left(1-s\right)I\right]+{\Lambda }^{2}}{\left(\alpha \Lambda +I\right){\left[\Lambda +\left(1-s\right)I\right]}^{2}}{\stackrel{\to }{1}}^{\text{T}}\end{array}$ (13)
The value of $\Lambda$ is known hence the value of $\alpha$ can be obtained from Equation (11) if the value of $E\left(x\right)$ is known.
2.3. Shape of Probability Function of PH-OPPL and PH-TPPL Distributions
Matrix $\Lambda$ was determined using continuous Chapman-Kolmogorov equation for cancer data in Kenya and the values of $\gamma$ is the stationary probabilities obtained using the formula ${\pi }_{k}={\pi }_{0}{\Lambda }^{k}$. The values of $\Lambda$ for three state Markov model represents cancer patients who transit from Healthy-Leukemia-Dead states, four state Markov model represents patients who transit from Healthy-Liver-Colon-Dead states, five state Markov model represents Healthy-Stomach-Pharynx-Colon-Dead states and six state Markov model represents patients transiting from Healthy-Oesophagus-Stomach-Lung-Kidney-Dead states. The values of $\Lambda$ for different states are:
$\left[\begin{array}{ccc}0.8783& 0.1217& 0\\ 0& 0.3938& 0.6062\\ 0& 0& 1.0000\end{array}\right]\left[\begin{array}{cccc}0.7900& 0.2100& 0& 0\\ 0& 0.2898& 0.7102& 0\\ 0& 0& 0.8985& 0.1015\\ 0& 0& 0& 1.0000\end{array}\right]$
$\left[\begin{array}{ccccc}0.8364& 0.1636& 0& 0& 0\\ 0& 0.3892& 0.6108& 0& 0\\ 0& 0& 0.6688& 0.3312& 0\\ 0& 0& 0& 0.5524& 0.4476\\ 0& 0& 0& 0& 1.0000\end{array}\right]\left[\begin{array}{cccccc}0.4851& 0.5149& 0& 0& 0& 0\\ 0& 0.1223& 0.8777& 0& 0& 0\\ 0& 0& 0.1533& 0.8467& 0& 0\\ 0& 0& 0& 0.4410& 0.5590& 0\\ 0& 0& 0& 0& 0.8668& 0.1332\\ 0& 0& 0& 0& 0& 1.0000\end{array}\right]$
The shape of probability function of phase type one parameter Poisson Lindley is expressed as:
Figure 1 shows that phase type one parameter Poisson Lindley is a long tailed distribution.
The shape of probability function of phase type two parameter Poisson Lindley is expressed as:
Figure 2 shows that phase type two parameter Poisson Lindley is a long tailed distribution.
(a) (b) (c) (d)
Figure 1. Pdf plots of PH-OPPL for different values of Λ.
(a) (b) (c) (d)
Figure 2. Pdf plots of PH-TPPL for different values of Λ.
3. Compound Phase Type Distribution
Compound distribution in the actuarial field is the total loses in the group of insurance policies. In this section we develop compound phase type distributions (CPHD) which can be used to model secondary cancer cases.
Definition 3. Let N be a r.v with probability generating function $F\left(S\right)$ and ${X}_{1},\cdots ,{X}_{N}$ be a set of iid random variable with a common probability generating function $G\left(S\right)$ and is independent of N, then the probability generating function of the compound distribution is expressed as:
$H\left(S\right)=F\left[G\left(S\right)\right]$ (14)
Unlike ordinary compound distributions which do not consider transition phases of diseases, (CPHD) in-cooperates the transition states. Probability generating functions of compound distributions can be derived by convolution of probability generating function of two distributions as shown in Equation (14).
Theorem 3 (Compound one parameter Poisson Lindley distribution). If the pgf of $N~PH-OPPL\left(\Lambda \right)$ the compound pgf of N is:
$H\left(S\right)=\stackrel{\to }{\gamma }\frac{{\Lambda }^{2}}{I+\Lambda }\left[\frac{\Lambda +\left(2-{L}_{x}\left[G\left(S\right)\right]\right)I}{{\left[\Lambda +\left(1-{L}_{x}\left[G\left(S\right)\right]\right)I\right]}^{2}}\right]{\stackrel{\to }{1}}^{\text{T}}$ (15)
where ${L}_{x}\left[G\left(S\right)\right]$ is the Laplace transform of the severity distribution as most continuous distributions their pgf is not available.
Proof:
$\begin{array}{c}H\left(S\right)=F\left[G\left(S\right)\right]=F\left[{L}_{x}\left[G\left(S\right)\right]\right]\\ =\stackrel{\to }{\gamma }\frac{{\Lambda }^{2}}{I+\Lambda }\left[\frac{\Lambda +\left(2-{L}_{x}\left[G\left(S\right)\right]\right)I}{{\left[\Lambda +\left(1-{L}_{x}\left[G\left(S\right)\right]\right)I\right]}^{2}}\right]{\stackrel{\to }{1}}^{\text{T}}\end{array}$ (16)
Theorem 4 (Compound two parameter Poisson Lindley distribution). If the pgf of $N~PH-TPPL\left(\Lambda \right)$ the compound pgf of N is:
$H\left(S\right)=\stackrel{\to }{\gamma }\frac{\alpha \Lambda \left[\Lambda +\left(1-{L}_{x}\left[G\left(S\right)\right]\right)I\right]+{\Lambda }^{2}}{\left(\alpha \Lambda +I\right){\left[\Lambda +\left(1-{L}_{x}\left[G\left(S\right)\right]\right)I\right]}^{2}}{\stackrel{\to }{1}}^{\text{T}}$ (17)
where ${L}_{x}\left[G\left(S\right)\right]$ is as defined in theorem (3).
Proof:
$\begin{array}{l}H\left(S\right)=F\left[G\left(S\right)\right]=F\left[{L}_{x}\left[G\left(S\right)\right]\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\stackrel{\to }{\gamma }\frac{\alpha \Lambda \left[\Lambda +\left(1-{L}_{x}\left[G\left(S\right)\right]\right)I\right]+{\Lambda }^{2}}{\left(\alpha \Lambda +I\right){\left[\Lambda +\left(1-{L}_{x}\left[G\left(S\right)\right]\right)I\right]}^{2}}{\stackrel{\to }{1}}^{\text{T}}\end{array}$ (18)
The continuous distributions considered in this research are; Weibull, Pareto and Generalized Pareto distributions hence their Laplace transforms will be derived and replaced in Equations (16) and (18) to get the pgf of their compound distribution using PH-OPPL and PH-TPPL distributions respectively. The Laplace transform of Weibull, Pareto and Generalized Pareto are derived as:
1) Weibull distribution
${L}_{x}\left(S\right)=E\left[{\text{e}}^{-sx}\right]$
$\begin{array}{c}{L}_{x}G\left(S\right)={\int }_{0}^{\infty }\text{ }\text{ }{\text{e}}^{-sx}\frac{\beta }{\alpha }{\left(\frac{x}{\alpha }\right)}^{\beta -1}{\text{e}}^{-{\left(\frac{x}{\alpha }\right)}^{\beta }}\text{d}x\\ =\frac{\beta }{\alpha }{\int }_{0}^{\infty }{\left(\frac{x}{\alpha }\right)}^{\beta -1}{\text{e}}^{-\frac{x}{\alpha }\left[s\alpha +{\left(\frac{x}{\alpha }\right)}^{\beta -1}\right]}\\ =\frac{\beta }{\alpha }\frac{\Gamma \beta }{{\left[s\alpha +{\left(\frac{x}{\alpha }\right)}^{\beta -1}\right]}^{\beta }}\end{array}$ (19)
2) Pareto distribution
${L}_{x}\left(S\right)=E\left[{\text{e}}^{-sx}\right]$
$\begin{array}{c}{L}_{x}G\left(S\right)=\alpha {\beta }^{\alpha }{\int }_{0}^{\infty }\frac{{\text{e}}^{-sx}}{{\left(x+\beta \right)}^{\alpha +1}}\text{d}x=\frac{\alpha }{\beta }{\int }_{0}^{\infty }\text{ }\text{ }{\text{e}}^{-\beta x}\underset{k=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}-\left(\alpha +1\right)\\ k\end{array}\right){\left(\frac{x}{\beta }\right)}^{k}\text{d}x\\ =\frac{\alpha }{\beta }\underset{k=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{k}\frac{\Gamma \left(\alpha +k\right)}{k!\Gamma \alpha }\frac{k!}{{\beta }^{2k+1}}=\underset{k=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{k}\frac{\alpha }{\Gamma \alpha }\frac{\Gamma \left(\alpha +k\right)}{{\beta }^{2k+2}}\end{array}$ (20)
3) Generalized Pareto distribution
${L}_{X}\left(S\right)=E\left[{\text{e}}^{-sx}\right]$
$\begin{array}{c}{L}_{X}G\left(S\right)={\int }_{0}^{\infty }\text{ }\text{ }{\text{e}}^{-sx}\frac{{x}^{\alpha -1}}{\beta \left(\alpha ,\gamma \right){\left(x+\lambda \right)}^{\alpha +\gamma }}\text{d}x\\ =\frac{1}{{\lambda }^{\gamma }\beta \left(\alpha ,\gamma \right)}{\int }_{0}^{\infty }\text{ }\text{ }{x}^{\alpha }{\text{e}}^{-sx}\underset{k=0}{\overset{\infty }{\sum }}\left(\begin{array}{c}\alpha +\gamma \\ k\end{array}\right){\frac{x}{\lambda }}^{k}\text{d}x\\ =\frac{1}{{\lambda }^{\gamma }\beta \left(\alpha ,\gamma \right)}\underset{k=0}{\overset{\infty }{\sum }}\frac{-\left(\alpha +\gamma \right)}{{\lambda }^{k}}{\int }_{0}^{\infty }\text{ }\text{ }{x}^{\gamma +k+1-1}{\text{e}}^{-sx}\text{d}x\\ =\frac{1}{{\lambda }^{\gamma }\beta \left(\alpha ,\gamma \right)}\underset{k=0}{\overset{\infty }{\sum }}\frac{-\left(\alpha +\gamma \right)}{{\lambda }^{k}}\frac{\Gamma \alpha +k}{{s}^{\alpha +k}}\end{array}$ (21)
Replacing Equations (19), (20) and (21) in Equation (16) the pgf of the compound distributions of PH-one parameter Poisson Lindley with Weibull, Pareto and Generalized Pareto respectively are:
1) Compound PH-OPPL-Weibull distribution
$H\left(S\right)=\stackrel{\to }{\gamma }\frac{{\Lambda }^{2}}{I+\Lambda }\left[\frac{\Lambda +\left(2-\frac{\beta }{\alpha }\frac{\Gamma \beta }{{\left[s\alpha +{\left(\frac{x}{\alpha }\right)}^{\beta -1}\right]}^{\beta }}\right)I}{{\left[\Lambda +\left(1-\frac{\beta }{\alpha }\frac{\Gamma \beta }{{\left[s\alpha +{\left(\frac{x}{\alpha }\right)}^{\beta -1}\right]}^{\beta }}\right)I\right]}^{2}}\right]{\stackrel{\to }{1}}^{\text{T}}$ (22)
2) Compound PH-OPPL-Pareto distribution
$H\left(S\right)=\stackrel{\to }{\gamma }\frac{{\Lambda }^{2}}{I+\Lambda }\left[\frac{\Lambda +\left(2-\underset{k=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{k}\frac{\alpha }{\Gamma \alpha }\frac{\Gamma \left(\alpha +k\right)}{{\beta }^{2k+2}}\right)I}{{\left[\Lambda +\left(1-\underset{k=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{k}\frac{\alpha }{\Gamma \alpha }\frac{\Gamma \left(\alpha +k\right)}{{\beta }^{2k+2}}\right)I\right]}^{2}}\right]{\stackrel{\to }{1}}^{\text{T}}$ (23)
3) Compound PH-OPPL-Generalized Pareto distribution
$H\left(S\right)=\stackrel{\to }{\gamma }\frac{{\Lambda }^{2}}{I+\Lambda }\left[\frac{\Lambda +\left(2-\frac{1}{{\lambda }^{\gamma }\beta \left(\alpha ,\gamma \right)}\underset{k=0}{\overset{\infty }{\sum }}\frac{-\left(\alpha +\gamma \right)}{{\lambda }^{k}}\frac{\Gamma \alpha +k}{{s}^{\alpha +k}}\right)I}{{\left[\Lambda +\left(1-\frac{1}{{\lambda }^{\gamma }\beta \left(\alpha ,\gamma \right)}\underset{k=0}{\overset{\infty }{\sum }}\frac{-\left(\alpha +\gamma \right)}{{\lambda }^{k}}\frac{\Gamma \alpha +k}{{s}^{\alpha +k}}\right)I\right]}^{2}}\right]{\stackrel{\to }{1}}^{\text{T}}$ (24)
Replacing Equations (19), (20) and (21) in Equation (18) the pgf of the compound distributions of PH-two parameter Poisson Lindley with Weibull, Pareto and Generalized Pareto respectively are:
1) Compound PH-TPPL-Weibull distribution
$H\left(S\right)=\stackrel{\to }{\gamma }\frac{\alpha \Lambda \left[\Lambda +\left(1-\frac{\beta }{\alpha }\frac{\Gamma \beta }{{\left[s\alpha +{\left(\frac{x}{\alpha }\right)}^{\beta -1}\right]}^{\beta }}\right)I\right]+{\Lambda }^{2}}{\left(\alpha \Lambda +I\right){\left[\Lambda +\left(1-\frac{\beta }{\alpha }\frac{\Gamma \beta }{{\left[s\alpha +{\left(\frac{x}{\alpha }\right)}^{\beta -1}\right]}^{\beta }}\right)I\right]}^{2}}{\stackrel{\to }{1}}^{\text{T}}$ (25)
2) Compound PH-TPPL-Pareto distribution
$H\left(S\right)=\stackrel{\to }{\gamma }\frac{\alpha \Lambda \left[\Lambda +\left(1-\underset{k=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{k}\frac{\alpha }{\Gamma \alpha }\frac{\Gamma \left(\alpha +k\right)}{{\beta }^{2k+2}}\right)I\right]+{\Lambda }^{2}}{\left(\alpha \Lambda +I\right){\left[\Lambda +\left(1-\underset{k=0}{\overset{\infty }{\sum }}{\left(-1\right)}^{k}\frac{\alpha }{\Gamma \alpha }\frac{\Gamma \left(\alpha +k\right)}{{\beta }^{2k+2}}\right)I\right]}^{2}}{\stackrel{\to }{1}}^{\text{T}}$ (26)
3) Compound PH-TPPL-Generalized Pareto distribution
$H\left(S\right)=\stackrel{\to }{\gamma }\frac{\alpha \Lambda \left[\Lambda +\left(1-\frac{1}{{\lambda }^{\gamma }\beta \left(\alpha ,\gamma \right)}\underset{k=0}{\overset{\infty }{\sum }}\frac{-\left(\alpha +\gamma \right)}{{\lambda }^{k}}\frac{\Gamma \alpha +k}{{s}^{\alpha +k}}\right)I\right]+{\Lambda }^{2}}{\left(\alpha \Lambda +I\right){\left[\Lambda +\left(1-\frac{1}{{\lambda }^{\gamma }\beta \left(\alpha ,\gamma \right)}\underset{k=0}{\overset{\infty }{\sum }}\frac{-\left(\alpha +\gamma \right)}{{\lambda }^{k}}\frac{\Gamma \alpha +k}{{s}^{\alpha +k}}\right)I\right]}^{2}}{\stackrel{\to }{1}}^{\text{T}}$ (27)
4. Data Analysis, Results and Discussions
4.1. Severity and Frequency Probabilities
The cancer data considered in this research is obtained from a medical facility in Kenya. The cancer transitions states considered are Healthy-Leukemia-Dead states for 3 state model, Healthy-Liver-Colon-Dead states for four state model, Healthy-Stomach-Pharynx-Colon-Dead states for five state model and Healthy-Oesophagus-Stomach-Lung-Kidney-Dead states for six state models. The values of $\Lambda$ for the data are obtained using continuous Chapman-Kolmogorov equations expressed as:
${p}_{ij}\left({\varphi }_{A},{\gamma }_{t}+{\Psi }_{d}\right)=\underset{k=1}{\overset{n}{\sum }}\text{ }\text{ }{p}_{ik}\left({\varphi }_{A},{\gamma }_{t}\right){p}_{kj}\left({\gamma }_{t},{\gamma }_{t}+{\Psi }_{d}\right)$
$\begin{array}{l}\underset{{\Psi }_{d}\to 0}{lim}\frac{{p}_{ij}\left({\varphi }_{A},{\gamma }_{t}+{\Psi }_{d}\right)-{p}_{ij}\left({\varphi }_{A},{\gamma }_{t}\right)}{{\Psi }_{d}}\\ =\underset{\Psi \to 0}{lim}\frac{{p}_{ik}\left({\varphi }_{A},{\gamma }_{t}\right){p}_{kj}\left({\gamma }_{t},{\gamma }_{t}+{\Psi }_{d}\right)-{p}_{ij}\left({\varphi }_{A},{\gamma }_{t}\right)\left[1-{p}_{jj}\left({\gamma }_{t},{\gamma }_{t}+{\Psi }_{d}\right)\right]}{{\Psi }_{d}}\end{array}$
$\frac{\partial }{\partial {\gamma }_{t}}{p}_{ij}\left({\varphi }_{A},{\gamma }_{t}\right)=\underset{k=1}{\overset{n}{\sum }}\text{ }\text{ }{p}_{ik}\left({\varphi }_{A},{\gamma }_{t}\right){\Im }_{kj}-{p}_{ij}\left({\varphi }_{A},{\gamma }_{t}\right){\Im }_{j}$
${p}_{ij}\left({\varphi }_{A},{\gamma }_{t}+{\Psi }_{d}\right)=\underset{k=1}{\overset{n}{\sum }}\text{ }\text{ }{p}_{ik}\left({\varphi }_{A},{\gamma }_{t}\right){p}_{kj}\left({\gamma }_{t},{\gamma }_{t}+{\Psi }_{d}\right)$
${p}_{ij}\left({\varphi }_{A},{\gamma }_{t}\right)=1-{\text{e}}^{-{\Im }_{ij}\left({\varphi }_{A}\right)t}$ (28)
where:
$\underset{{\Psi }_{d}\to 0}{\mathrm{lim}}\frac{{p}_{kj}\left({\gamma }_{t},{\gamma }_{t}+{\Psi }_{d}\right)}{{\Psi }_{d}}={\Im }_{kj}$
$\underset{\Psi \to 0}{\mathrm{lim}}\frac{1-{p}_{jj}\left({\gamma }_{t},{\gamma }_{t}+{\Psi }_{d}\right)}{{\Psi }_{d}}={\Im }_{j}$
The values of $\Lambda$ for three, four, five and six state using the data obtained were as shown in Section 2.3.
The severity distributions considered in this research are Weibull, Pareto and Generalized Pareto distributions. DFT requires severity probabilities to be discrete hence they will be discretized using method of mass rounding which is expressed as:
$\begin{array}{l}{f}_{0}={F}_{J}\left(\frac{h}{2}\right)\\ {f}_{x}={F}_{J}\left(xh+\frac{h}{2}\right)-{F}_{J}\left(xh+\frac{h}{2}\right)\text{ }x=1,2,3,\cdots \\ {f}_{m}=1-{F}_{J}\left(mh-\frac{h}{2}\right)\end{array}$
The pdf of Wei-bull, Pareto and Generalized Pareto distributions respectively are expressed as;
$\begin{array}{l}f\left(x\right)=\frac{\beta }{\alpha }{\left(\frac{x}{\alpha }\right)}^{\beta -1}{\text{e}}^{-{\left(\frac{x}{\alpha }\right)}^{\beta }};\text{\hspace{0.17em}}\text{\hspace{0.17em}}x>0;\text{\hspace{0.17em}}a,b>0\\ f\left(x\right)=\frac{\alpha {\beta }^{\alpha }}{x+{\beta }^{\alpha +1}}\\ f\left(x\right)=\frac{{x}^{\alpha -1}{\lambda }^{\gamma }\Gamma \left(\alpha +\gamma \right)}{\Gamma \gamma \Gamma \alpha {\left(x+\lambda \right)}^{\alpha +\gamma }}\end{array}$
The frequency and severity probabilities for secondary cancer cases are: (Table 1).
Table 1. Claim frequency and severity probabilities.
4.2. Discrete Fourier Transform
There are different numerical methods used in estimation of aggregate losses such as; Monte Carlo, Panjer recursive model, Fourier transforms and Direct Numerical Integration. Panjer recursive model is applicable when the claim frequency distributions follow either Panjer class $\left(a,b,0\right)$ or class $\left(a,b,1\right)$. In this study we will consider Discrete Fourier Transform (DFT) in estimation of the aggregate losses. Robertson (1992) applied Fourier transforms in computation of aggregate losses [2]. Pavel (2010) [1] reviewed these numerical methods and concluded that each method had it strength and weaknesses hence they should be chosen according to the study. DFT mostly preferred as it is arguably said to be the most elegant and powerful technique in evaluation of aggregate loss probabilities when claim amount ${X}_{i}$ is both discrete and continuous [17].
The algorithm of DFT of aggregate losses requires computation of DFT of frequency and DFT of severity separately.
Definition 4 (Discrete Fourier Transform). Let ${X}_{n}$ be the severity or frequency distribution of the claim data. For any discrete function ${X}_{k}$ the Discrete Fourier transform is the mapping;
${X}_{k}=\underset{n=0}{\overset{N-1}{\sum }}\text{ }\text{ }X\left(n\right){\text{e}}^{\frac{-i2\Pi kn}{N}}\text{ }k=0,1,2,\cdots ,N-1$ (29)
Expression (29) is very complex to work with hence to reduce its complexity we apply Euler’s formula and it becomes:
$X\left(k\right)=\underset{n=0}{\overset{N-1}{\sum }}\text{ }\text{ }X\left(n\right)\mathrm{cos}\left(\frac{2\Pi kn}{N}\right)+i\mathrm{sin}\left(\frac{2\Pi kn}{N}\right)$
$X\left(k\right)=\underset{n=0}{\overset{N-1}{\sum }}\text{ }\text{ }X\left(n\right){W}_{N}^{kn}$ (30)
which is the DFT of the severity or frequency probabilities. The severity and frequency probabilities are of length 8 and hence the matrix W must be a primitive 8th root of unity therefore Equation (30) can be rewritten as:
$X\left(k\right)=\underset{N=0}{\overset{7}{\sum }}\text{ }\text{ }X\left(n\right){W}_{N}^{kn}$ (31)
The frequency or severity probabilities will be padded with equal number of zero’s as its elements in order to perform no wrap convolution. The DFT algorithm is as follows:
1) Multiply the matrix ${W}_{N}^{kn}$ with the frequency or severity probabilities to get the DFT of frequency or severity probabilities.
2) Compute DFT of DFT of frequency and severity by multiplying DFT of frequency probabilities with the DFT of the severity probabilities and consequently multiplying the resulting vector with the matrix ${W}_{N}^{kn}$.
3) Select the values without the complex i and divide each value by the number of elements in the vector of frequency or severity distribution and arrange the resulting probabilities in reverse except for the first probability.
4) Values corresponding to original frequency and severity values are the aggregate loss probabilities.
The values of aggregate loss probabilities using DFT are:
Table 2. Aggregate loss probabilities.
The values of Table 2 can be represented graphically as:
(a) (b) (c)
Figure 3. Aggregate loss probabilities.
Figure 3(a) shows aggregate loss probabilities using PH-OPPL distribution with severity distributions and it indicates that PH-OPPL with Weibull and Pareto were similar to the actual aggregate loss probabilities while PH-OPPL with generalized Pareto distribution overestimate the aggregate losses for six state model. Figure 3(b) shows aggregate loss probabilities using PH-TPPL distribution with Pareto and generalized Pareto provided a better fit for secondary cancer data while PH-TPPL with Weibull overestimated the aggregate losses. However, PH-OPPL with Weibull and PH-TPPL with generalized Pareto provided a better fit compared to PH-OPPL-Pareto model and PH-TPPL Pareto respectively hence they are compared in Figure 3(c) indicating that PH-OPPL with Weibull provided the best fit for aggregate loss data of secondary cancers in Kenya. PH-OPPL-Weibull model can be used to provide better estimates of aggregate losses for secondary cancer data in Kenya.
5. Conclusion
Mixed phase type distributions are developed to model secondary cancer cases in Kenya. Unlike ordinary distributions which do not in-cooperate the transition of different states, the distributions proposed here take into consideration transition states while modeling claim frequency data. The distributions are based on Poisson and Lindley distributions, where PH-OPPL-Weibull provided the best for PH-OPPL models while PH-TPPL-Generalized Pareto provided the best fit for PH-TPPL models. This model improves estimation of aggregate loses as it in-cooperates transition probabilities of different states of cancer as well as heterogeneous aspect of claim data. This greatly improves estimation of insurance policies for diseases which transit to different state such as cancer hence improving the financial positions of the insurance firms as it will improve estimation of its reserves. This model, however, is only applicable in risk theory for diseases which have multiple transitions states. Further research can be done on this study factoring in patients who were censored in this study and also the same study can be carried out for disease such as HIV-AID which has transition states.
Data Availability
The data used to support the findings of this study can be availed upon request.
Cite this paper: Mwende, C. , Weke, P. , Bundi, D. and Ottieno, J. (2021) Estimation of Aggregate Losses of Secondary Cancer Using PH-OPPL and PH-TPPL Distributions. Open Journal of Statistics, 11, 838-853. doi: 10.4236/ojs.2021.115049.
References
[1] Shevchenko, P.V. (2010) Calculation of Aggregate Loss Distribution. Journal of Operational Risk, 5, 3-40.
[2] Robertson, J. (1992) The Computation of Aggregate Loss Distribution.
[3] Rono, A., Ogutu, C. and Weke, P. (2020) On Compound Distributions for Natural Disaster Modelling in Kenya. International Journal of Mathematics and Mathematical Sciences, 2020, Article ID: 9398309.
https://doi.org/10.1155/2020/9398309
[4] Mohamed, M.A., Razali, A.M. and Ismail, N. (2010) Approximation of Aggregate Losses Using Simulation. Journal of Mathematics and Statistics, 6, 233-239.
https://doi.org/10.3844/jmssp.2010.233.239
[5] Erlang, A. (1909) Sandsynlighedsregning og telefonsamtaler. Nyttidsskriftfor Matematik, 20, 33-39.
[6] Neuts, M.F. (1981) Matrix-Geometric Solutions in Stochastic Models. Series in the Mathematical Sciences. Johns Hopkins University Press, Baltimore.
[7] Assussen, S. (2003) Applied Probability and Queues. Springer-Verag, New York.
[8] Mogens, B. (2005) A Review of Phase-Type Distribution and Their Used in Risk Theory. Austin Bulletin, 35, 145-161.
https://doi.org/10.2143/AST.35.1.583170
[9] O’Cinneide, C. (2017) Phase-Type Distributions and Invariant Polytopes.
[10] Wu, X.Y. and Li, S.M. (2010) Matrix-Form Recursions for a Family of Compound Distributions. Austin Bulletin, 40, 351-368.
https://doi.org/10.2143/AST.40.1.2049233
[11] Kok, S. and Wu, X. (2010) Matrix-Form Recursive Evaluation of the Aggregate Claims Distribution Revisited. Centre for Actuarial Studies, Department of Economics, The University of Melbourne, Melbourne.
[12] Nurul, N.Z., Mahmod, O., Rajalingam, S., Hanita, D., Lazim, A. and Evizal, A.K. (1981) Markov Chain Model Development for Forecasting Air Pollution Index of Miri, Sarawak. Journal of Sustainability, 11, 5190.
https://doi.org/10.3390/su11195190
[13] Zhou, Y.J., Wang, L.L., Zhong, R. and Tan, Y.L. (2018) A Markov Chain Based Demand Prediction Model for Stations in Bike Sharing Systems. Journal of Mathematical Problems in Engineering, 2018, Article ID: 8028714.
https://doi.org/10.1155/2018/8028714
[14] Das, K.K., Ahmed, I. and Bhattacharjee, S. (2018) A New Three-Parameter Poisson-Lindley Distribution for Modelling Over-Dispersed Count Data. International Journal of Applied Engineering Research, 13, 16468-16477.
[15] Sankaran, M. (1970) The Discrete Poisson-Lindley Distribution. Biometrics, 26, 145-149.
https://doi.org/10.2307/2529053
[16] Shanker, R. and Mishra, A. (2014) A Two-Parameter Poisson-Lindley Distribution. International Journal of Statistics and Systems, 9, 79-85.
[17] Kemeny, J.G. and Snell, J.L. (2016) Finite Markov Chains. Springer-Verlag, Princeton.
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08-25-2019, 07:26 PM #21
Anthony
Join Date: Feb 2005
Location: Berkeley, CA
Re: Conditional Injury with Knowing Your Own Strength
Quote:
Originally Posted by Raekai [*]How do I make it work with ×10 is +20?
You'll need to remake the tables based on 20xlog10 instead of 30xlog10 (easy with a spreadsheet, otherwise requires you to remember values). Also need to redo the addition and subtraction table and wound thresholds.
__________________
My GURPS site and Blog.
08-25-2019, 08:45 PM #22 RyanW ☣ Join Date: Sep 2004 Location: Southeast NC Re: Conditional Injury with Knowing Your Own Strength Regarding armor, my thoughts were along these lines. Given that +1 Wound Potential is about +50% damage, that means that armor stops all of WP X allows through about 1/3 of X+1, 1/2 of X+2, 2/3 of X+3, and 4/5 of X+4. On this scale: 1/3 = -3 1/2 = -2 1/3 = -1 4/5 is between -1 and -0. As long as you're consistent, you could round it up or down. Regarding +3 ST = +1 WP, I assumed damage scaled with the square root of BL (roughly the same as RAW, but I did some figuring involving force, mass, and distance over which acceleration occurs to make sure the number wasn't unreasonable). On a logarithmic scale, taking a root becomes dividing, so the square root of damage means half the WP. +3 ST = ×2 BL, therefore +3 ST = ×1.5 damage = +1 WP. Edit: I don't have access to the original KYOS right now. It occurs to me that I may have made modifications to it. __________________ RyanW Gondor has no king. Gondor needs no king. Gondor doesn't have to show you any stinking king. Last edited by RyanW; 08-26-2019 at 10:01 AM.
08-26-2019, 01:42 AM #23 dataweaver Join Date: Aug 2004 Location: the frozen wastelands of Southern California Re: Conditional Injury with Knowing Your Own Strength Here's an alternative to the Robustness Threshold and Wound Potential Tables that's based on +20=×10 instead of the current +6=×10: HP/dmg: RT/WP | 1: 0 2: 6 3: 9 4: 12 5: 14 6: 16 7: 17 8: 18 9: 19 10: 20 11: 21 –13: 22 –15: 23 –17: 24 –19: 25 –21: 26 –24: 27 –27: 28 –30: 29 –34: 30 –38: 31 –42: 32 –47: 33 –53: 34 –59: 35 –65: 36 –75: 37 –85: 38 –95: 39 –105: 40 –115: 41 –135: 42 –150: 43 –170: 44 –190: 45 –210: 46 –240: 47 –270: 48 –300: 49 –340: 50 –380: 51 –420: 52 –470: 53 –530: 54 –590: 55 –650: 56 –750: 57 –850: 58 –950: 59 Entries starting with a – are ranges that start one higher than the previous entry: so “–13” is actually “12–13”, and “–950” is actually “851–950”. You can extend this table further by multiplying HP/dmg by 10 and adding 20 to RT/WP. The rule for applying DR (actually, log-subtraction) is: WP vs. LogDR: WP modifier 0 or less: no damage +1: -21 +2: -13 +3: -10 +4: -8 +5: -7 +6: -6 +7: -5 +8: -4 +9–+10: -3 +11–+13: -2 +14–+19: -1 +20 or more: -0 That is, find out how much the WP exceeds the “LogDR”, and reduce the effective WP by an amount based on the difference. LogDR comes from DR the same way that RT comes from HP. Damage ranges (e.g., 1d, 2d, 2d×10, etc.) have a 16-point spread: 1d goes from WP0 to WP16; 2d goes from WP6 to WP22; 3d6 goes from WP9 to WP25; 4d goes from WP12 to WP27; 5d goes from WP14 to WP29; 6d goes from WP16 to WP31; 7d goes from WP17 to WP32; 8d goes from WP18 to WP34; and so on. That is, the maximum WP you can achieve is 15 higher than the minimum you can achieve. Meanwhile, the average damage you can roll always translates to a WP that's 11 higher than the minimum, and 4 less than the maximum. You can get a reasonable approximation of this by rolling 5d, adding the top three dice to the average WP of the attack, and subtracting 14 from the result. That's your damage roll. __________________ Point balance is a myth.[1][2][3][4] Last edited by dataweaver; 08-26-2019 at 02:04 AM.
08-26-2019, 01:38 PM #24
Raekai
World's Worst Detective
Join Date: May 2011
Location: Columbus, Ohio
Re: Conditional Injury with Knowing Your Own Strength
Quote:
Originally Posted by Anthony You'll need to remake the tables based on 20xlog10 instead of 30xlog10 (easy with a spreadsheet, otherwise requires you to remember values). Also need to redo the addition and subtraction table and wound thresholds.
That makes sense. Thanks!
Quote:
Originally Posted by RyanW Regarding armor, my thoughts were along these lines. Given that +1 Wound Potential is about +50% damage, that means that armor stops all of WP X allows through about 1/3 of X+1, 1/2 of X+2, 2/3 of X+3, and 4/5 of X+4. On this scale: 1/3 = -3 1/2 = -2 1/3 = -1 4/5 is between -1 and -0. As long as you're consistent, you could round it up or down. Regarding +3 ST = +1 WP, I assumed damage scaled with the square root of BL (roughly the same as RAW, but I did some figuring involving force, mass, and distance over which acceleration occurs to make sure the number wasn't unreasonable). On a logarithmic scale, taking a root becomes dividing, so the square root of damage means half the WP. +3 ST = ×2 BL, therefore +3 ST = ×1.5 damage = +1 WP. Edit: I don't have access to the original KYOS right now. It occurs to me that I may have made modifications to it.
This looks interesting. It's a bit easier for me to understand, which is always a plus! And this all lines up with your thread, right? You did mention using a "variant on the KYOS system". How does the variant differ? And how much would have to be done for it to work with KYOS as-is? (Your "+3 ST = ×2 BL" checks out. Though, KYOS says, "Extending the “thrust = swing-2” found at ST 10 across the board implies that swinging multiplies force by about 1.5", so I would assume that +2 ST = ×1.5 damage, but I could be very wrong. Of course, it could be the other way around, and we could then say that, with your math, thrust = swing-3.)
Quote:
Originally Posted by dataweaver Here's an alternative to the Robustness Threshold and Wound Potential Tables that's based on +20=×10 instead of the current +6=×10: [...] Damage ranges (e.g., 1d, 2d, 2d×10, etc.) have a 16-point spread: 1d goes from WP0 to WP16; 2d goes from WP6 to WP22; 3d6 goes from WP9 to WP25; 4d goes from WP12 to WP27; 5d goes from WP14 to WP29; 6d goes from WP16 to WP31; 7d goes from WP17 to WP32; 8d goes from WP18 to WP34; and so on. That is, the maximum WP you can achieve is 15 higher than the minimum you can achieve. Meanwhile, the average damage you can roll always translates to a WP that's 11 higher than the minimum, and 4 less than the maximum. You can get a reasonable approximation of this by rolling 5d, adding the top three dice to the average WP of the attack, and subtracting 14 from the result. That's your damage roll.
Awesome! Just to clarify, are the alternative to the Robustness Threshold Table and Wound Potential Table based off of standard HP or logarithmic HP (as in, HP = ST from KYOS)? It's hard for me to keep track of who's doing what where. I ask because I was surprised that the bands would balloon so much if based on logarithmic HP. Of course, that's just my intuition—again, math is not my strong suit—because RyanW uses +3 ST = +1 WP, so I'd imagine that +3 HP = +1 RT (again, assuming that HP has been converted like ST in KYOS).
Actually, it seems like my guess is correct. HP 34 is RT 30, and [20 × log(34)] is ~30. Just in case I found a one-off coincidence, I checked with other values. HP 190 is RT 45, and [20 × log(190)] is ~45. I don't understand what I did, but I did it. So, it almost seems like, if you converted HP with the KYOS ST formula, then ~RT would just be HP + 10. Then, at that point, couldn't you just knock the +10 off of each side? Unless I'm just making stuff up, which is very possible.
Also, would it be beneficial to also switch the Conditional Effects Table over to ×10 = +20? Or does it really matter?
08-26-2019, 02:31 PM #25
Anthony
Join Date: Feb 2005
Location: Berkeley, CA
Re: Conditional Injury with Knowing Your Own Strength
Quote:
Originally Posted by Raekai Actually, it seems like my guess is correct. HP 34 is RT 30, and [20 × log(34)] is ~30. Just in case I found a one-off coincidence, I checked with other values. HP 190 is RT 45, and [20 × log(190)] is ~45. I don't understand what I did, but I did it. So, it almost seems like, if you converted HP with the KYOS ST formula, then ~RT would just be HP + 10. Then, at that point, couldn't you just knock the +10 off of each side? Unless I'm just making stuff up, which is very possible.
It's kinda useful to have the zero point be at 1 hp. Otherwise there is no particular reason not to do it.
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08-26-2019, 02:32 PM #26 dataweaver Join Date: Aug 2004 Location: the frozen wastelands of Southern California Re: Conditional Injury with Knowing Your Own Strength I'm operating on the premise that HP increase roughly at the same rate as damage, on the notion that maintaining a parity between damage and HP will avoid obvious problems of disparities between what you can dish out and what you can take. Thing is, I'm not assuming that damage increases linearly with ST; that's kind of the whole point of Conditional Injury, to put damage and HP on a geometric progression with respect to ST. Remember that weight is also being recalibrated (by KYOS) to be on an exponential scale; so the relationship between weight and HP is still something to the effect of HP=weight^n; I'm just not sure off the top of my head what n is. I think it's ˝. On the other hand, I'm trying to keep RT, WP, and the equivalent counterpart to DR directly proportional to ST; and as much as possible, I'm trying to get all of the mechanics to work with them instead of dmg, HP, and DR: if I can do that, then I can render moot the HP-to-RT and dmg-to-WP tables, except for the purpose of converting existing material to RT and WP. Which reminds me: another table that needs to be changed is the Conditional Effect Table. The Severity column needs to be rescaled so that what's currently -6 becomes -20, and what's currently +6 becomes +20. That's a simple multiple, though: just divide by 3 and multiply by 10. So ±6 becomes ±20, ±5 becomes ±16, ±4 becomes ±13, ±3 becomes ±10, ±2 becomes ±6, and ±1 becomes ±3. Likewise, all of the Injury and Severity modifiers need to be inflated to for the new scale: e.g., the Target Composition modifiers for Unloving go from -2, -3, and -4 to -6, -10, and -13. Heck, all modifiers need to be recalibrated to the 20-step scale. __________________ Point balance is a myth.[1][2][3][4]
08-26-2019, 03:19 PM #27 Anthony Join Date: Feb 2005 Location: Berkeley, CA Re: Conditional Injury with Knowing Your Own Strength The advantages of +30/x10:You can just look up the mass of an object on the KYoS table and convert directly to a toughness rating. A 1 lb unliving object (KYoS -3) has 4 hp (LogD 18). A 1,000 lb unliving object (KYoS 27) has 40 hp (LogD 48). It interacts nicely with the range/speed chart. For example, if you want to divide damage by distance, that's just a modifier of -10 + 5x range modifier. If you want to multiply by speed, add 10 - 5x speed modifier (e.g. collision damage for an unliving object in KYoD works out to 3d + Mass - 5xRSM - 23). You can do the same sort thing with the +20/x10, but you wind up multiplying by 10/3, which is annoying. In the typical range of PC HP (8-18) you can just add 20 to the linear value and be correct within error margins. The disadvantage, however, is that GURPS 4e did a lot of things based on quadratic scaling, and in vanilla GURPS lifting ability actually varies with 2/3 power of mass, so a 1,000x heavier creature (with 10x the ST) actually only has 100x the BL (+20). The collision case is particularly inconsistent because you can get 10x damage by either multiplying speed by 10 (100x the energy) or by multiplying mass by 1,000 (1,000x the energy). The thing this is leaving out is mostly wound width: the 1,000x mass collision might make hole of the same depth as the 10x speed collision, but the hole is much wider. The only attacks in G4e that have a concept of wound with are piercing attacks, but it actually applies to everything. In the case of ST-based attacks, a 1,000x more massive creature might not apply 1,000 the force -- but it probably applies 100x the force over 10x the distance, for a net of x1,000 energy. __________________ My GURPS site and Blog.
08-26-2019, 04:02 PM #28 dataweaver Join Date: Aug 2004 Location: the frozen wastelands of Southern California Re: Conditional Injury with Knowing Your Own Strength Personally, Is be more inclined to rework KYOS to use +12=×10, then rework CI to +24=×10. Using a multiple of six gives you easy squares and cubes. __________________ Point balance is a myth.[1][2][3][4]
08-26-2019, 05:29 PM #29
Anthony
Join Date: Feb 2005
Location: Berkeley, CA
Re: Conditional Injury with Knowing Your Own Strength
Quote:
Originally Posted by dataweaver Personally, Is be more inclined to rework KYOS to use +12=×10
Problem with that is nasty numbers. On a +10=x10 scale, there are several values that are very close to integer multipliers -- using two significant figures, the values are 1, 1.3, 1.6, 2.0, 2.5, 3.2, 4.0, 5.0, 6.3, 7.9, which gives you x2, x4, and x5 as integer. By comparison, on the +12=x10 scale, the values are 1, 1.2, 1.5, 1.8, 2.2, 2.6, 3.2, 3.8, 4.6, 5.6, 6.8, 8.3, none of which are particularly nice numbers.
If I were making bigger changes, I'd probably switch the range/speed chart to decibel (+10=x10) or 2dB (+5 = x10) rather than changing KYoS away from dB, as converting to dB is generally easier than the RSM -- for example, 1 AU = 1.6e+11 yards, which can be converted to dB by looking up 1.6 and adding 11x10 (112 dB relative to 1 yard). On the RSM, you instead look up 1.6 and add 11x6 (+67 relative to one yard, +65 relative to 2 yards).
__________________
My GURPS site and Blog.
08-26-2019, 05:55 PM #30
dataweaver
Join Date: Aug 2004
Location: the frozen wastelands of Southern California
Re: Conditional Injury with Knowing Your Own Strength
Quote:
Originally Posted by Anthony Problem with that is nasty numbers. On a +10=x10 scale, there are several values that are very close to integer multipliers -- using two significant figures, the values are 1, 1.3, 1.6, 2.0, 2.5, 3.2, 4.0, 5.0, 6.3, 7.9, which gives you x2, x4, and x5 as integer. By comparison, on the +12=x10 scale, the values are 1, 1.2, 1.5, 1.8, 2.2, 2.6, 3.2, 3.8, 4.6, 5.6, 6.8, 8.3, none of which are particularly nice numbers.
That's less of a concern than you might think, since we're actually more interested in ranges of values than exact values. You're not looking at, say, the number 1.21 as you're looking at the range (1.10, 1.33].
And when you're working in an environment where squares and cubes are likely to come up frequently, the fact that squaring a twelve-step pattern gives you a six-step pattern, and coming it gives you a four-step pattern, is much nicer than how the decibel scale behaves in similar conditions.
__________________
Point balance is a myth.[1][2][3][4]
Tags conditional injury, hit points, knowing your own strength, kyos, logarithm
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Frequency: the number of complete waves
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# Formula To Move Contents From One Cell To The Next
I'm working on a speadsheet that has thousands of line items with lots of great information, but not in the fields I need them to be in. Is there a formula to move for example cell contents A26 to B25 and D26 to F27? Also I have contents in one cell that I need to break apart is there a formula for that too? I searced the help and the net an came up with nothing. Maybe I'm not wording it correctly, but I'd appreciate the help.
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## Similar Topics
Afternoon all,
I'm hoping someone can help me.
I need to be able to clear the contents of cells in column C, D, E, H if the value of column Q equals HELLO for arguments sake.
Clearing the contents of an entire row is not a problem, however when it comes to resizing rows I seem to come unstuck.
Any help would be appreciated.
Hello,
I'm trying to set up a basic formula to clear out unwanted cells. Basically, if the cell is not equal a number, I'd like it to be cleared of any information. I would rather not use a space, because I have text that is overlapping between cells and would like it to be legible.
Here's the basic formula:
=if(A1>0, A1, ???)
Any help would be great. Thanks!
I would like the contents of a selected active cell to be displayed in a certain other cell, say R4. When another cell is selected and active, that cell's contents should be displayed.
I have an array of 9 columns and 20 rows with equity symbols and conditionally formatted to show the severity of price movement in either direction.
Instead of typing in the value of whatever cell a trader is looking at I would like for them to just be able to click on the cell and have that symbol copied into R4 so some detailed information can be displayed for them.
Hope I'm explaining this at least somewhat clearly.
Thanks.
Hi there,
I'm very new to Excel, and I'm having trouble figuring a few things out. Hopefully this will be very easy for you guys!
In Sheet 1, I have a column of cells that I would like to also appear in Sheet 2. If I add a new row to the column in Sheet 1, I would also like it to be updated automatically in Sheet 2. Currently, I can get it to show the contents of individual cells from Sheet 1 in Sheet 2 by using this formula in the formula bar for each cell in Sheet 2:
=Sheet1!A3 (or whichever cell it is)
That's fine, but I'd like to just have a formula that will reproduce the entire column (ie. without a fixed range, as new rows are going to be added to the column).
If anyone could point me in the right direction, I'd be very very grateful. Thanks.
Hey guys,
I'm trying to write a macro which prints to PDF and saves the file name as the contents of a cell. I've been looking through all the posts currently on this forum to get something working. I'm using the following code -
Sub PrintPDF()
Filename = "C:\Documents and Settings\samb\My Documents\" & ActiveSheet.Range("Z1").Value
SendKeys Filename & "{ENTER}", False
ActiveWindow.SelectedSheets.PrintOut Copies:=1, ActivePrinter:= _
End Sub
The macro prints to PDF, but then it stops at the Save As stage, where I have to manually enter the name of the file and click Save. I want the macro to automatically name the file with the contents of cell Z1. I then want it to then automatically press enter. Any ideas where I'm going wrong? Any help would be much appreciated!
Great tip! But one thing: On my machine (Excel 2008 Mac), the values in the formula cell are not correctly calculated simply by dragging. Instead, the values are equal to the cell above (the first entry formula value). To get the correct value, I need to click in the formula bar and then hit enter. (I only discovered this after an hour of tinkering, figuring I had botched the formula!)
Are there any ways around this so that it updates upon dragging the formula?
I have a problem sometimes. I will click on a cell to add information.
The cell turns blue and then wherever I move, it highlights those to.
No matter where I go on the page. If I Alt-Tab and work in another
program on my computer, that excel page keeps highlighting wherever
I move even in those other programs (I know this sounds confusing).
When I return to excel thousands and thousands of cells are blue.
The biggest problem is that the highlighting won't turn off, no matter
what. I can't select anything from the tool bars, do any work on the sheet or close the program.
I can close it only with the task manager but when I open it again,
the cursor is still stuck in the highlighting mode and won't perform any
other functions.
Do you think this is a problem with my computer, the excel program? I have changed my mouse and this didn't help.
Is there some shortcut to turn off this highlight feature other than restarting
my computer. Which is the only current way I can get rid of it.
Hi there,
I am currently using 'Activecell.Offset(1,0).Select' to move down one cell at a time when I click on a button.
The problem I have now is that if someone was to filter by something then the 'next cell down' could be hidden behind the filter (by that I mean it didn't meet the filter criteria).
Is there anyway to move down to the next row, even if that row does not follow on Sequentially .
Any help would be greatly appreciated.
Regards,
James
Hi,
I am entering lots of family history data into a spreadsheet. At the simplest I have columns (in cells A1 B1 C1) the headings, Surname, Forename, Year. Right now, I enter in cells A2 B2 and C2 say: Smith <Tab> John <Tab> 1555 <Enter, move mouse to the A column in the next row down). What I want to happen is when I have entered the last data in a row and pressed <Enter> I move automatically to the A-column in the next row down.
Is this possible?
Regards and a Merry Christmas to all
Wibs
I would like to create a formula that removes a specific character if it appears in a cell. In this case, if the text in the name cell starts with * or #, remove it. Otherwise, keep the contents intact. Examples:
Code:
```Text in Cell Desired Results
*Bobby Abreu Bobby Abreu
#Erick Aybar Erick Aybar
Jason Bartlett Jason Bartlett
```
Is there a formula that will get me where I want to be?
Thanks!
Chuck
I often use if statements to return empty cells, for example:
=IF(a1=0,"ERROR","")
The trailing "" returns an empty cell. The problem is, it is not TRULY empty. If I fill that formula down, I cannot, for example, jump from one "ERROR" cell to the next by hitting Ctrl arrow-down because it seems to think that these empty cells have contents.
Is there a way to designate, in such a formula as above, to return a TRULY empty cell?
Thanks
hi,
I would like a formula that would list all the items in row B that match the criteria in row A. the first cell with formula would list the first item, the second cell with the formula would list the next item, and so forth. Also, column B might have a duplicates that should be listed. Is this possible? I cannot manipulate the order of the original items (ie, filters) because this data is being used to derive other formulas.
I have a name convention in one cell that has the last name first serperated by a comma with the first name last. How can I reverse this to show the first name first and the last name last in one cell?
I have a graph with various information in it. One is a line graph that tracks hours used in that department per month. The problem is that when it gets to the last month, and there is not yet data for future months, the line goes down to Zero (leavin this big diagonal line that makes the data look funny). I need this line to stop on the last month there is data for. (I do not manually enter the data, it is a formula that I do not wish to delete.) I know I can manually move the data that the line is pulling to make it stop on the last month, but I have a graph for 36 different departments and that's a pain to have to manually adjust 36 graphs each month. Is there a way to tell the line to stop if there is no data? Thanks!!
Today I ran into an odd problem. I typed in values for column A rows 1 through 10 then values for column B rows 1 through 10. then in column C, I made the formula C1=A1/B1. The math was correct it showed 542 in the C1 cell. So I dragged that formula down and it showed 542 in all column C cells which is not correct. And when I went to check to see if the formula was correctly dragged it was. For instance, the formula in C2 is =A2/B2 however the value of that cell showed 542 which was not the correct math/value.
But it gets even more weird. When I click on the Column C cells and then it shows the formula up top in the formula bar and if I put my cursor anywhere in the formula bar and hit Return the formula does not change however the correct value then appears in the Column C cell. It is like the act of putting the cursor in the formula activates it to work properly but until it is activated that C cell shows the value of the cell which it was dragged down from.
This is quite bizarre. Has anyone ever seen this before? I have no idea what is going on. I ran a scan for viruses and none were found. I tried it on several new/different spreadsheets but it keeps happening.
Thanks for any tips on this.
Posted this on the Ozgrid forums, but haven't gotten any help yet, so I thought I'd try here too!
I've been having a strange problem lately. I have a fairly lengthy macro that works perfectly most of the time. Occasionally it will run as expected but as soon as the macro ends, excel becomes unresponsive to mouse-clicks. When I click anywhere (trying to select a cell, or an excel menu item...clicking anywhere in excel) I'll get the a 'ding' system sound and nothing will happen. BUT, if I use the keyboard arrow keys, I can see that the active cell selection moves accordingly. Then it gets really strange - when I have a cell highlighted and press any key to input text, it gets duplicated. So if I press "s' it will input "ss" into the cell, and then when I press enter it will auto-move to the next cell down, but nothing ends up getting saved into the previous cell.
I don't understand what's going on at all. I can't think of anything in my macro that would have these kinds of effects. I've made sure that screenupdating is turned back on at the end of every procedure. If I go into the VBE, I can manually run procedures and they all work fine. The only way I've been able to get back to normal is by force closing excel altogether and re-opening. Any ideas?
______________________
Still having trouble, and the same thing is happening with this workbook on two different computers, so I don't think it's a hardware, or OS specific issue. When it gets locked up like this, I can still do anything in the VBE (edits cells, run macros, etc.) with no problems. If I'm in the excel window, I can click alt on the keyboard and the shortcut keys for the menu come up, but I can't go deeper than that by clicking the letter shortcuts, they do nothing. If I use the delete key to delete the contents of a cell, then it gets deleted. But if I type anything else (numbers, letters, or symbols) then it types 2 instances of the key every time, yet when I hit enter, nothing changes in the cell. I also can't really bring focus to the excel window if something else (ie. the VBE) is on top of it, clicking into the excel window just gives the little system 'ding' sound and nothing happens.
If I hit the save button in the VBE (since I can't click anything in the excel window), then it seems to snap out of it and go back to normal. I tried searching for anything simmilar to this and can't find anything...
Any help would be much appreciated. This is driving me nuts!
I recently moved into a new office and I am using a brand new computer on our network. When trying to edit an existing file, I'm not able to select a single cell and type. After clicking on a cell, if I move the mouse at all (even without pressing the left button) it just continues highlighting cells no matter where I move the mouse. I am also not able to click on the tool bars at all. I have to actually Ctrl-Alt_del to get out of the program. I'm sure this is something simple, but I don't know that much about this program. Any help would be greatly appreciated.
Column 1 has roughly 20 rows of information. Cell C1 has the formula =A1.
Is there a formula so that when I drag C1 horizontally into D1, E1, F1, ..., the values placed in each cell will be =A2, =A3, =A4, ...
I do not want to transpose the values from column 1 into C1, D1,.... I want these cells to have a formula that links them up to column 1's values
Thanks
I have a spreadsheet with simple (addition/subtraction) formulas. The file is quite large and the formulas are too. All of a sudden, the formulas stopped working except when I double click in the cell containing the actual formula. For example, if I enter "2" in each cell, A2 and B2, cell B3 should reflect "4" because there is a formula in cell B3 which totals cells A2 and B2. Only by double clicking on cell B3 will the program actually calculate. I'm totally perplexed and so is my IT contact. Anyone?
First time in this forum. Hi all. I have a challenging question, is there a way either by VBA or manually (preferably both, if possible) to actually unite the X amount of numbers that are in a cell given the contents is alphanumeric? I'll give you the following examples to see if you can understand what I' referring to?
DATA output should be
asd67,h876 --------> 67876
2,3,ujdj5&34 -------> 23534
909k86m34 --------> 9098634
Hope this makes sense? | 3,317 | 14,202 | {"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} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.929829 |
https://www.teachstarter.com/us/teaching-resource/integers-worksheet-integer-number-line-cut-and-paste/ | 1,713,530,333,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817398.21/warc/CC-MAIN-20240419110125-20240419140125-00133.warc.gz | 930,821,234 | 70,734 | teaching resource
# Integers Worksheet - Integer Number Line Cut and Paste
• Updated: 27 Apr 2023
Practice identifying positive and negative numbers on a number line with a cut-and-paste integers worksheet
• Editable: Google Slides
• Non-Editable: PDF
• Pages: 1 Page
Tag #TeachStarter on Instagram for a chance to be featured!
teaching resource
# Integers Worksheet - Integer Number Line Cut and Paste
• Updated: 27 Apr 2023
Practice identifying positive and negative numbers on a number line with a cut-and-paste integers worksheet
• Editable: Google Slides
• Non-Editable: PDF
• Pages: 1 Page
Practice identifying positive and negative numbers on a number line with a cut-and-paste integers worksheet
## Let’s Locate Positive and Negative Numbers on a Number Line!
Are your students working their way through the opening lessons in your unit on positive and negative numbers? Are they struggling a bit? We can help!
Studies show that repeated practice in different manners is the best way for students to master a skill. We’ve created this activity to help your students practice their positive and negative number skills on a number line.
This worksheet presents the student with a series of statements telling them what number/number line they must look for on page 2 of the document. They will then cut out the matching number line and paste it into its correct location on the grid. | 308 | 1,416 | {"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} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.824109 |
http://physics.stackexchange.com/questions/tagged/gravity+astrophysics | 1,411,424,762,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657137698.52/warc/CC-MAIN-20140914011217-00056-ip-10-234-18-248.ec2.internal.warc.gz | 218,114,916 | 24,855 | # Tagged Questions
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# Math Revolution Approach (PS)
Author Message
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6815
GMAT 1: 760 Q51 V42
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Re: Math Revolution Approach (PS) [#permalink]
### Show Tags
15 Feb 2018, 03:40
[GMAT math practice question]
How many arrangements of the letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’ in a straight line are possible?
A. 60
B. 120
C. 180
D. 240
E. 420
=>
The seven letters ‘s’, ‘u’, ‘c’, ‘c’, ‘e’, ‘s’, ‘s’ include three s’s and two c’s.
The number of permutations is 7! / (3!*2!) = 420.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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Re: Math Revolution Approach (PS) [#permalink]
### Show Tags
16 Feb 2018, 04:08
[GMAT math practice question]
A committee has 10 members. How many different selections of a chairman, an editor, and a secretary can be made from the members of the committee?
A. 180
B. 360
C. 540
D. 720
E. 810
=>
There are 10 ways of choosing the chairman. Once the chairman has been chosen, there are 9 remaining possibilities for the editor. Once both the chairman and the editor have been chosen, 8 choices remain for the secretary. So, the total number of choices for these three positions is:
10*9*8 = 720.
This may also be written as
10P3 = 10 * 9 * 8 = 720.
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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6815
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Re: Math Revolution Approach (PS) [#permalink]
### Show Tags
18 Feb 2018, 18:23
[GMAT math practice question]
The length and width of a certain rectangle are in the ratio 4 to 5. The area of the rectangle is 2,000 square units. If the length and width of the rectangle are both increased by 10 units, what is the increase in the area of the rectangle?
A. 800
B. 900
C. 1,000
D. 1,100
E. 1,250
=>
Suppose the length is L = 4k and the width is W = 5k for some k. Then
L*W = 4k*5k = 20k^2 = 2,000.
k^2 = 100 and k = 10.
Thus, L = 4k = 40 and W = 5k = 50.
The area of the rectangle after increasing both the length and the width is (L+10)(W+10) = (40 + 10)(50 + 10) = 50*60 = 3,000 square units.
Thus, the increase in the area is 3,000 – 2,000 = 1,000 square units.
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Re: Math Revolution Approach (PS) [#permalink]
### Show Tags
18 Feb 2018, 18:25
[GMAT math practice question]
Attachment:
2.16.png [ 4.66 KiB | Viewed 459 times ]
What is the value of x?
A. 1.5
B. 3.0
C. 4.5
D. 5.0
E. 6.0
=>
The two triangles are similar since they have the same angles. Consequently, each pair of corresponding sides is in the same proportion: 39:3 = (18+x):x.
Rearranging this proportion yields
39x = 3(18+x)
39x = 3*18 + 3x
36x = 54
x = 1.5
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Math Revolution GMAT Instructor
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Re: Math Revolution Approach (PS) [#permalink]
### Show Tags
21 Feb 2018, 00:06
[GMAT math practice question]
The points A, B, C and D lie on the number line in that order. If AB=BC/3, AC=(4/7)CD, A=-1, and D=10, what is the value of B?
A. -2
B. -1
C. 0
D. 1
E. 2
=>
Attachment:
2.21.png [ 1015 Bytes | Viewed 445 times ]
Let BC = 3d.
Then AB = BC/3 = d and AC = 4d.
Also, AC = 4d = (4/7)CD and CD = 7d.
Therefore,
AD = AC + CD = 4d + 7d = 11d = 10 – (-1) = 11.
So,
d = 1, and
B = -1 + d = -1 + 1 = 0.
_________________
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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6815
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GPA: 3.82
Re: Math Revolution Approach (PS) [#permalink]
### Show Tags
22 Feb 2018, 01:11
[GMAT math practice question]
Let A=2^{50}, B=3^{30}, and C=5^{20}. Which of the following is true?
A. A<B<C
B. A<C<B
C. C<A<B
D. B<C<A
E. C<B<A
=>
If we wish to compare these numbers, we need to either make their bases the same or make their exponents the same. In this case, it is easiest to make all exponents the same as follows:
A=2^{50} = (2^5)^{10} = 32^{10}
B=3^{30} = (3^3)^{10} = 27^{10}
C=5^{20} = (5^2)^{10} = 25^{10}
Since 32 > 27 > 25, we must have A > B > C.
_________________
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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6815
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GPA: 3.82
Re: Math Revolution Approach (PS) [#permalink]
### Show Tags
25 Feb 2018, 17:40
[GMAT math practice question]
Which of the following is closest to 11^5*9^5–7*10^7?
A. 10^2
B. 10^7
C. 10^8
D. 10^9
E. 10^10
=>
Since 11 is close to 10 and 9 is close to 10, 11^5*9^5 is close to 10^5*10^5 = 10^10.
Since 7*10^7 is a relatively small number compared with 10^10, 11^5*9^5 – 7*10^7 is approximately equal to 10^10.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6815
GMAT 1: 760 Q51 V42
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Re: Math Revolution Approach (PS) [#permalink]
### Show Tags
25 Feb 2018, 17:42
[GMAT math practice question]
What is the units digit of 2^105+3^105+4^105+5^105?
A. 3
B. 4
C. 5
D. 6
E. 7
=>
The units digit of any integer raised to the exponent 5 is the same as the units digit of the integer:
0^1 and 0^5 have remainder 0
1^1 and 1^5 have remainder 1
2^1 and 2^5 have remainder 2
9^1 and 9^5 have remainder 9
when divided by 10.
Therefore, the units digit of 2^105+3^105+4^105+5^105 is same as the units digit of 2 + 3 + 4 + 5 = 14.
It is 4
_________________
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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6815
GMAT 1: 760 Q51 V42
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Re: Math Revolution Approach (PS) [#permalink]
### Show Tags
26 Feb 2018, 01:35
[GMAT math practice question]
When both positive integers a and b are divided by 9, their remainders are 7. What is the reminder when a2b is divided by 9?
A. 4
B. 3
C. 2
D. 1
E. 0
=>
The best way to approach remainder questions is to plug in a number. We shall plug in 7 for both a and b because its remainder, when divided by 9, is 7, as required by the question. Then a^2b=7^27=343, and the remainder of 343 when it is divided by 9 is 1.
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Re: Math Revolution Approach (PS) [#permalink]
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28 Feb 2018, 00:29
[GMAT math practice question]
The 20% acid liquid solution is produced by adding a gallons of 10% acid liquid solution to b gallons of 50% acid liquid solution. To get 10 gallons of 20% acid liquid solution, how many gallons of 50% acid liquid solution are needed?
A. 1
B. 1.5
C. 2
D. 2.5
E. 3
=>
We have a + b = 10.
( 0.1a + 0.5b ) / 10 = 0.2
⇔ 0.1a + 0.5b = 2
⇔ 0.1(10-b) + 0.5b = 2
⇔ 1 – 0.1b + 0.5b = 2
⇔ 0.4b = 1
⇔ b = 2.5
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Re: Math Revolution Approach (PS) [#permalink]
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01 Mar 2018, 02:49
[GMAT math practice question]
Which of the following is 6/7 times as far from 7/3 as is 17/6 from 2/3?
A. 88/21
B. 21/88
C. 77/20
D. 20/77
E. 78/21
=>
We have |x – 7/3| = (6/7)|17/6 – 2/3|.
|x - 7/3| = (6/7)(13/6)
x – 7/3 = ±(6/7)(13/6) = ±13/7
x = ±13/7 + 7/3
x = (39 + 49)/21 = 88/21 or x = (-39 + 49)/21=10/21
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Re: Math Revolution Approach (PS) [#permalink]
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02 Mar 2018, 00:21
[GMAT math practice question]
When x/y = 2.6, (x-y)/(x+y)=?
A. 2/7
B. 3/8
C. 4/9
D. 5/9
E. 7/10
=>
( x – y ) / ( x + y ) = ( x/y – 1 ) / ( x/y + 1 ) = ( 2.6 – 1 ) / ( 2.6 + 1 ) = 1.6 / 3.6 = 16 / 36 = 4 / 9
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Re: Math Revolution Approach (PS) [#permalink]
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04 Mar 2018, 18:25
[GMAT math practice question]
There are 4 machines with the same work rate. If it took k hours for 3 machines to work and did k-2 hours for 4 machines when working together, what is the value of k?
A. 4
B. 5
C. 6
D. 7
E. 8
=>
Work Amount = Number of machines * Work Rate * Time
If r is the machine’s work rate, we have 3rk = 4r(k-2) or 3k = 4k -8.
Thus k = 8.
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Re: Math Revolution Approach (PS) [#permalink]
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04 Mar 2018, 18:28
[GMAT math practice question]
John traveled the entire 50 km trip. If he traveled the first 20 km of the trip at a constant rate 40 km per hour and the remaining trip at a constant rate 20 km per hour, what is his average speed, in km per hour?
A. 20 km/h
B. 23 km/h
C. 25 km/h
D. 26 km/h
E. 27 km/h
=>
Ave Speed = ( Total Distance ) / ( Total Time )
The total distance = 50 km.
Total time is 20 / 40 + 30 / 20 = 0.5 + 1.5 = 2
The average speed is 50 / 2 = 25.
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Re: Math Revolution Approach (PS) [#permalink]
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07 Mar 2018, 02:40
[GMAT math practice question]
If n!/(n-2)!<100, what is the greatest possible value of n?
A. 8
B. 9
C. 10
D. 11
E. 12
=>
We must have
n! / (n-2)! = n(n-1) < 100.
If n = 10, n(n-1) = 10*9 = 90 < 100.
If n = 11, n(n-1) = 11*10 = 110 > 100.
10 is the greatest value of n for which n!/(n-2)! < 100.
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Re: Math Revolution Approach (PS) [#permalink]
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08 Mar 2018, 01:26
[GMAT math practice question]
10 members of a society, including members A and B, attended a meeting. 2 of these members were selected to form a committee. What is the probability that the 2 members selected were members A and B?
A. 1/10
B. 1/15
C. 1/20
D. 1/30
E. 1/45
=>
The number of committees of 2 people that include members A and B is equal to the number of ways that 2 people can be selected from 2 people, which is 2C2 = 1.
The number different possible committees of 2 people, is equal to the number of ways that 2 people can be selected from 10 people, which is 10C2 = (10*9) / (1*2) = 45.
The probability that the committee contains members A and B is 2C2 / 10C2 = 1/45.
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Re: Math Revolution Approach (PS) [#permalink]
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08 Mar 2018, 23:22
[GMAT math practice question]
The terms of the sequence {An}, where n is a positive integer, satisfy A1=81, A2=82, A3=83, and An+3=An+4. Which of the following cannot be a value of An?
A. 801
B. 802
C. 803
D. 804
E. 805
=>
The terms of the sequence can be divided into three groups:
A1 = 81, A4 = 85, A7 = 89, … : These have a remainder of 1 when they are divided by 4.
A2 = 82, A5 = 86, A8 = 90, … : These have a remainder of 2 when they are divided by 4.
A3 = 83, A6 = 87, A9 = 91, … : These have a remainder of 3 when they are divided by 4.
No term of the sequence is a multiple of 4.
Since 804 is a multiple of 4, it cannot be a term of the sequence.
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Re: Math Revolution Approach (PS) [#permalink]
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11 Mar 2018, 17:33
[GMAT math practice question]
If neither x nor y is divisible by 3, which of the following could be the value of x^2+y^2?
A. 333
B. 334
C. 335
D. 336
E. 337
=>
Consider the squares of the integers that are not divisible by 3:
1^2 = 1, 2^2 = 4, 4^2 = 16, 5^2 = 25, 7^2 = 49, 8^2=64, ….
They all have a remainder of 1 when they are divided by 3.
Thus, the sum of the squares of two integers which are not divisible by 3 must have a remainder of 2 when it is divided by 3.
The only answer choice having this property is 335.
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Re: Math Revolution Approach (PS) [#permalink]
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11 Mar 2018, 17:35
[GMAT math practice question]
Reading from the right, how many consecutive zeros, starting with the units digit, appear in 29!?
A. 5
B. 6
C. 7
D. 8
E. 9
=>
To find the number of 0’s ending 29!, we need to count the numbers of 2’s and 5’s in the prime factorization of 29!.
Since the number 2’s is greater than the number of 5’s, we only need to count the number of 5’s in the prime factorization of 29!.
The factors of 5 are contributed by 5, 10, 15, 20 and 25. Each of 5, 10, 15 and 20 contributes one 5, while 25 contributes two 5s to the prime factorization.
Thus, there are 6 copies of 5 in the prime factorization of 29!, giving rise to 6 consecutive 0’s at the end of 29!.
Note: The actual value of 29! is 8841761993739701954543616000000.
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Re: Math Revolution Approach (PS) [#permalink]
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13 Mar 2018, 23:24
[GMAT math practice question]
If (n+2)!= n!(an^2+bn+c), then abc=?
A. 2
B. 3
C. 4
D. 6
E. 8
=>
(n+2)! = (n+2)(n+1)n! = (n^2 + 3n + 2)n!
So,
a = 1, b = 3, and c = 2.
Thus, abc = 6.
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Re: Math Revolution Approach (PS) &nbs [#permalink] 13 Mar 2018, 23:24
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# Math Revolution Approach (PS)
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Florida
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Manager
Joined: 16 May 2007
Posts: 145
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24 Sep 2007, 20:43
In the United States, of the people who moved from one state to another when they retired, the percentage who retired to Florida has decreased by three percentage points over the past ten years. Since many local businesses in Florida cater to retirees, this decline is likely to have a noticeably negative economic effect on these businesses.
Which of the following, if true, most seriously weakens the argument?
A. Florida attracts more people who move from one state to another when they retire than does any other state.
B. The number of people who move out of Florida to accept employment in other states has increased over the past ten years.
C. There are far more local businesses in Florida that cater to tourists than there are local businesses that cater to retirees.
D. The total number of people who retired and moved to another state for their retirement has increased significantly over the past ten years.
E. The number of people who left Florida when they retired to live in another state was greater last year than it was ten years ago.
CEO
Joined: 29 Mar 2007
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24 Sep 2007, 21:57
cruiser wrote:
In the United States, of the people who moved from one state to another when they retired, the percentage who retired to Florida has decreased by three percentage points over the past ten years. Since many local businesses in Florida cater to retirees, this decline is likely to have a noticeably negative economic effect on these businesses.
Which of the following, if true, most seriously weakens the argument?
A. Florida attracts more people who move from one state to another when they retire than does any other state.
B. The number of people who move out of Florida to accept employment in other states has increased over the past ten years.
C. There are far more local businesses in Florida that cater to tourists than there are local businesses that cater to retirees.
D. The total number of people who retired and moved to another state for their retirement has increased significantly over the past ten years.
E. The number of people who left Florida when they retired to live in another state was greater last year than it was ten years ago.
This one is D.
I still wrestle with this choice b/c I don't see how this increase gurantees that these retirees will go to florida.
However, a 3% decrease may be countered by a 20% increase. So overall the businesses are still ok.
This question is posted on here somewhere with a better explanation that I have.
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24 Sep 2007, 23:09
D looks good. Without knowing the absolute figure of the number of people moving from one state to another when they retire, we do not know who representative 3% is.
Manager
Joined: 16 May 2007
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25 Sep 2007, 05:48
I had picked C thinking that if there are more buisnesses who cater to general tourists than to retirees than the economy will not be affected.
Anyways, you guys are correct.
OA is D.
Thanks.
[#permalink] 25 Sep 2007, 05:48
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For discussion of specific patterns or specific families of patterns, both newly-discovered and well-known.
### Re: Soup search results
Here's 6 15-bit still lifes
15.647:
x = 129, y = 22, rule = B3/S23$95bobo$96b2o$96bo3bo$100bobo$100b2o$2bobo52bo$3b2o53bo$3bo10bo41b3o62b2o$12b2o107bo$5b3o5b2o12b2o18b2o6b3o9b2o18b2o12bo15b2o3bo$7bo19bo2bo16bo2bo4bo11bo2bob2o13bo2bob2o6b2o15bo2bobo$6bo22b2o18b2o5bo12b2obobo14b2obobo5bobo16b2obobo$73bo19bo29b2o$10bo42bo$9b2o42b2o$9bobo40bobo!
15.652:
x = 155, y = 17, rule = B3/S2385bo$86bo$84b3o$o45bobo19b2o18b2o$b2o3bobo5bo18b2o12b2o4b2o12bo2bo2b2o12bo2bo2b2o14b2o2b2o14b2o2b2o14b2o2b2o$2o5b2o4bo17bo2bo12bo3bo2bo13b2obo2bo13b2obo2bo14bobo2bo14bobo2bo14bobo2bo$7bo5b3o15b2o18b2o18b2o18b2o18b2o18b2o18b2o$28b2obo13bo2b2obo19bo19bo19bo19bo19bo$14bo13b2o2bo10bobo2b2o2bo19bo19bo19bo19bo19bo$4b2o7b2o16b2o11b2o5b2o18b2o18b2o18b2o18b2o3bo16bo$5b2o6bobo31bo87bo18bo$4bo42b2o86b3o15b2o$46bobo82b2o$7b2o41b2o78bobo$7bobo40bobo79bo3b2o$7bo42bo86b2o$136bo!
15/658:
x = 112, y = 31, rule = B3/S23$83bo$84bo$82b3o2$80bo17bo$81bo16bobo$79b3o16b2o5$97bo$54bo40b2o$15bo36b2o42b2o$13b2o38b2o$14b2o83bo$56bo12bo19bo8b2o8b2o$55b2o12bo19bo8bobo7bo$55bobo11bo19bo19bo$23b2o18b2o18b2o18b2o18b2o3b2o$22bo2bobo14bo2bobo14bo2bobo14bo2bobo14bo2bobo$3b2o2b3o12b2o2b2o14b2o2b2o14b2o2b2o14b2o2b2o14b2o2b2o$2bobo2bo$4bo3bo3$9b2o$8b2o$10bo!
15.668:
x = 119, y = 34, rule = B3/S235$87bo$54bo33bo$54bobo21bo7b3o9bo$54b2o2b3o16bobo17bobo$58bo18bo2bo16bo2bo$59bo18b2o18b2o3bo$93bo9bobo$72b3o18bo9b2o$93bo$95b3o$95bo$96bo4bo$101bobo$11bo89b2o10b2o$10bo8bo93bobo$10b3o5bo9b2o18b2o18b2o18b2o18b2o6bo$18b3o7bobo17bobo17bobo17bobo17bobo3b2o$4bobo24bob2obo14bob2obo14bob2obo14bob2obo14bobo$5b2o8bo16b2ob2o15b2ob2o15b2ob2o15b2ob2o15b2o$5bo8b2o6bo$14bobo4b2o$3b3o15bobo77bo$5bo5b3o86b2o$4bo8bo86bobo$12bo! 15.708: x = 96, y = 45, rule = B3/S2312$57bo$58b2obo$50bobo4b2o2bobo$6b2o43b2o8b2o4bo$5bobo2b2o39bo14bo$7bo2bobo46bo6b3o$10bo38b2o6b2o3b3o$50b2o6b2o2bo13bo$49bo13bo11bobo$34b2o18b2o18bobo$34bo19bo19bo$35bo19bo19bo$32b4o16b4o16b4o$32bo19bo19bo$33bo19bo19bo$32b2o18b2o18b2o$9b2o$9bobo$5b2o2bo$4bobo$6bo$12b3o$12bo$13bo4$27b2o$27bobo$27bo!
15.712:
x = 117, y = 36, rule = B3/S236$8bo$7bo$7b3o51bo$62bo$60b3o2$63bobo$14bo3bobo36bo5b2o$6b2o7b2ob2o35bobo6bo$7b2o5b2o3bo10b2o18b2o4b2o17b2o18b2o$6bo24bo19bo24bo19bo$31bobo17bobo7bobo12bobo3b2o12bobo3b2o$15b2o15b2o18b2o7b2o14b2o3bo14b2o3bo$16b2o16b2o18b2o6bo16b2obo16b2obo$15bo18bobo17bobo22bobo17bobo$35bo19bo24bo$22bo$21b2o55b2ob2o$21bobo53bobobobo$79bobo! Bob Shemyakin BobShemyakin Posts: 205 Joined: June 15th, 2014, 6:24 am ### Re: Soup search results BobShemyakin wrote:Here's 6 15-bit still lifes ... These syntheses yield a few others as well. Row 1: 15.649 from 8. Row 2: 15.652: The base 13.212 can be made from 9 reducing this to 12, which also gives 15.431 from 13 Row 3: 15.652 and 15.925 can also be made from related 13.211 from 12. Row 4: 15.658 from 10. EDIT: Row 5: From the converter used in 15.668, we also get 15.1035 from 12: x = 147, y = 176, rule = B3/S23obo$boo$bo$$95bo96bo4bobo94b3o4boo102bo$$103b3o$103bo$104bo$34bo19bo19bo19bo18boo$33bobo17bobo17bobo17bobo17bo$34boo18boo18boo18boo18b3obo$18bobo96boo$18boo14boo18boo18boo18boo18boo$19bo14bo19bo19bo19bo19bo$35bo19bo19bo19bo19bo$34boo18boo18boo18boo18boo3$12bo$11bo$11b3o3$10b3o20boo18boo$10bo22boo18boo$11bo37b3o$51bo$50bo12$23bo$21boo$22boo$32bo$30boo$31boo3$o$boo$oo3$58boo$57bobo3bo37bobo$25bo33boboo34bo3boo$26bo35boo34boobbo$24b3o21boo18boo12boo4boo7boo9boo18boo$10boo37bo10bo8bo12bobo4bo19bo19bo$9bobo32boobo12boobboobo17bobo16boobo16boobo$11bo11b3o18bob3o10bobobbob3o17b3o15bob3o14bobb3o$25bo23bo19bo19bo19bo13boo4bo$24bo23boo18boo18boo12bo5boo18boo$97bo3boo$95bobo3bobo$96boo3$3boo19b3o$bbobo19bo$4bo20bo5$28boo$27boo$29bo11$bbobo$3boo$3bo8bo$10bobo88bobo$11boo89boo$102bo$19bo41bobo$6bo13boo35bo3boo37boo$7bo11boo4bo32boobbo38boo20boo$5b3o17bobo20boo7boo9boo12boo4boo10bo7boo13bo4boo$b3o21boo17boo3bo14boo3bo12bobo4bo14boo3bo14bo4bo$3bo40bobbo16bobbo17bobo16bobbo17bobo$bbo43b3o17b3o17b3o17b3o17b3o$31bo17bo19bo19bo19bo19bo$30boo16boo18boo18boo18boo18boo$30bobo4$27b3o$27bo$28bo$3boo8b3o$bbobo10bo$4bo9bo15$93bo$92bo$92b3o$44bo45bo$43bo44bobo$obo40b3o43boo$boo38bo$bo37bobo42bo$40boo40bobo$83boo$13bobo48bo29bo$8bo4boo49bo29bo20boo$8bobo3bo49bo29bo18bobbo$8boo17boo18boo18boo28boo14boobboo$28bo19bo19bo29bo19bo$27bo19bo19bo29bo19bo$8b3o15bo19bo19bo29bo19bo$8bo17bobo17bobo17bobo27bobo17bobo$9bo17boo18boo18boo28boo18boo4$94bo$93boo$93bobo5$132boo$128bobboo$46bo79bobo4bo$o3bobo39bobo78boo$booboo14boobo16boobobboo12booboboo13booboboo13booboo15booboo15booboobo$oo3bo13bobboo15bobboo5boo8bobbooboo12bobbooboo6boo4bobboobo13bobboobo13bobbooboo$19boo18boo8bobo7boo18boo8boobbobo3boo5bo12boo5bo12boo$49bo40boobo13bo19bo$3o86bo16boo18boo$bbo$bo4bo118bo$5boo117boo$5bobo116bobo$$78bo78boo77bobo! mniemiec Posts: 878 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results Great Job! Which still-lifes are still left? I've got stuck on 15.310. I couldn't synthesise it, or some smaller still-lifes that can be used to make it (in a 16 glider sythesis) in few enough gliders. x = 14, y = 8, rule = B3/S23210b2o2b2o3bobobo3bo3b2o3bob2o4bobo! Goldtiger997 Posts: 351 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Soup search results Goldtiger997 wrote:I've got stuck on 15.310. ... By doing it in a different order, it can be done with 15: x = 155, y = 18, rule = B3/S2382bo4bo76bobbobo32bo43b3o3boo33boo37boo10bobo26bobo49boo28boo5bo50bo28bo6bobo22bo19bo5bo37boo8bo12bobo17bobo3boo5bo18boo18boo18boo18boo18boobo5boo11bobo17bobo4bobo3boo15bobobo5bobo7bobobo10boo3bobobo10boo3bobobo10boo3bobobobboo3bobo10boo18boo11bobo14boo9boo7boo14bo3boo14bo3boo14bo3booboo15boo18boo28boo11bo6boo16boboo16boboo16boboo17bobo17bobo27bobo17bobo17bobo17bobo17bobo18bo19bo29bo19bo19bo19booo3bobooboo120boobooo3bobo118bobobobo127bobo! mniemiec Posts: 878 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results mniemiec wrote:Row 2: 15.652: The base 13.212 can be made from 9 reducing this to 12, which also gives 15.431 from 13 Actually, the base (as well as a variant with mango) can be made from only seven gliders: x = 48, y = 44, rule = B3/S2320bo3bo21b2obobo17b2o20b2o2b2o19bo2bo37b2oboobo37bob3ob2o42bo24b2o3bo14b2o24b2o2bo28b3o231b3o25b2o4bo9b3o12bo2bo4bo9bo15b2o10bo28b2o9b2o8bo1120bo3bo12bo8b2obobo17b2o11bo8b2o2b2o19bo11b3o26b2obo9bo30bob3o10bo34bo8b3o13b2o3bo14bobo24b2o2bo15bo2bo28b3o14bobo46bo31b3o31bo25bo6bo25b2o24bobo! I Like My Heisenburps! (and others) Extrementhusiast Posts: 1669 Joined: June 16th, 2009, 11:24 pm Location: USA ### Re: Soup search results I wrote:Row 2: 15.652: The base 13.212 can be made from 9 reducing this to 12, which also gives 15.431 from 13 Extrementhusiast wrote:Actually, the base (as well as a variant with mango) can be made from only seven gliders: ... This then also reduces 15.926 to 11: x = 31, y = 14, rule = B3/S23bbo3bob3o$$6bo$5bo$bo3b3o$bboo20boo$boo6boo13bo4boo$10bo14bo4bo$5boobo17bobo$bbobbob3o17b3o$obo7bo19bo$boo6boo18boo! mniemiec Posts: 878 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results mniemiec wrote: Goldtiger997 wrote:I've got stuck on 15.310. ... By doing it in a different order, it can be done with 15: x = 155, y = 18, rule = B3/S2382bo$4bo76bo$bbobo32bo43b3o$3boo33boo$37boo10bobo26bobo$49boo28boo5bo$50bo28bo6bobo$22bo19bo5bo37boo$8bo12bobo17bobo3boo5bo18boo18boo18boo18boo18boo$bo5boo11bobo17bobo4bobo3boo15bobobo5bobo7bobobo10boo3bobobo10boo3bobobo10boo3bobobo$bboo3bobo10boo18boo11bobo14boo9boo7boo14bo3boo14bo3boo14bo3boo$boo15boo18boo28boo11bo6boo16boboo16boboo16boboo$17bobo17bobo27bobo17bobo17bobo17bobo17bobo$18bo19bo29bo19bo19bo19bo$oo3bo$booboo120booboo$o3bobo118bobobobo$127bobo!
Due to that alternate order, here is a 13 glider synthesis
x = 61, y = 57, rule = B3/S23bo$2bo$3o4$54bo$53bo$53b3o3$58bo$45bo12bobo$43b2o13b2o$27bo16b2o$28b2o$27b2o4$27bo$25bobo$26b2o3bo15bo$29b2o15b2o$30b2o14bobo$41b2o$41bobo$41bo6$38b2o$37b2o$39bo15$3o$2bo$bo48b2o$49b2o$51bo3b2o$54b2o$56bo! Goldtiger997 Posts: 351 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Soup search results Goldtiger997 wrote: mniemiec wrote: Goldtiger997 wrote:I've got stuck on 15.310. ... By doing it in a different order, it can be done with 15: ... Due to that alternate order, here is a 13 glider synthesis ... here is a 11 glider synthesis: x = 115, y = 30, rule = B3/S233$15bo$15bobo$15b2o3$39bo$38bo$38b3o2$35bobo$4bo4bo26b2o5bo$5b2o2bobo24bo6bobo$4b2o3b2o32b2o$30b2o18b2o18b2o18b2o18b2o$27bobobo5bobo7bobobo10b2o3bobobo10b2o3bobobo10b2o3bobobo$27b2o9b2o7b2o14bo3b2o14bo3b2o14bo3b2o$7b2o6b3o7b2o11bo6b2o16bob2o16bob2o16bob2o$8b2o5bo8bobo17bobo17bobo17bobo17bobo$7bo8bo8bo19bo19bo19bo2$83b2ob2o$82bobobobo$84bobo!
Bob Shemyakin
BobShemyakin
Posts: 205
Joined: June 15th, 2014, 6:24 am
### Re: Soup search results
Here's another 15-bit still-life
x = 101, y = 94, rule = B3/S235bobo$6b2o$6bo$bo90bo$2bo88bo$3o88b3o18$91bo$90bo$90b3o12$55bobo$51b2o2b2o$51bobo2bo$51bo4$52b3o$52bo$53bo$61b2o$44bo16bobo$42bobo16bo$43b2o13b2o$57bobo$59bo4$53b2o$52b2o$54bo$46b2o$47b2o$46bo16$98b2o$98bobo$98bo12$3b2o$4b2o$3bo! @ mniemiec & BobShemyakin: How many are left? I haven't found a suitable glider synthesis for 15.337 yet, so is there an alternative way of using smaller still-lifes to construct it, other than the one below? x = 148, y = 17, rule = B3/S2339bo42bobo$40bo41b2o$38b3o42bo$76bo$3bo38bo34bo$bobo5bo30b2o33b3o$2b2o3b2o32b2o$8b2o50b2o10b3o5b2o17bobo14bo2bobo19bo$11b2o13b2o18b2o12bobo3b2o6bo5bobo3b2o11b2obo3b2o6bobo2b2obo3b2o12bobo3b2o$11bobo9bobobo10bobo2bobobo14bo2bobo5bo8bo2bobo14bo2bobo3b2o2b2o5bo2bobo9bobo2bo2bobo$b2o8bo11b2o14b2o2b2o17bobo17bobo17bobo7b2o8bobo12b2o3bobo$obo36bo23bo19bo19bo7bo7bo3bo19bo$2bo115b2o$40b3o75bobo$40bo74b2o$41bo72bobo$116bo! Goldtiger997 Posts: 351 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Soup search results Goldtiger997 wrote:Here's another 15-bit still-life x = 101, y = 94, rule = B3/S235bobo$6b2o$6bo$bo90bo$2bo88bo$3o88b3o18$91bo$90bo$90b3o12$55bobo$51b2o2b2o$51bobo2bo$51bo4$52b3o$52bo$53bo$61b2o$44bo16bobo$42bobo16bo$43b2o13b2o$57bobo$59bo4$53b2o$52b2o$54bo$46b2o$47b2o$46bo16$98b2o$98bobo$98bo12$3b2o$4b2o$3bo!
@ mniemiec & BobShemyakin: How many are left?
I haven't found a suitable glider synthesis for 15.337 yet, so is there an alternative way of using smaller still-lifes to construct it, other than the one below?
x = 148, y = 17, rule = B3/S2339bo42bobo$40bo41b2o$38b3o42bo$76bo$3bo38bo34bo$bobo5bo30b2o33b3o$2b2o3b2o32b2o$8b2o50b2o10b3o5b2o17bobo14bo2bobo19bo$11b2o13b2o18b2o12bobo3b2o6bo5bobo3b2o11b2obo3b2o6bobo2b2obo3b2o12bobo3b2o$11bobo9bobobo10bobo2bobobo14bo2bobo5bo8bo2bobo14bo2bobo3b2o2b2o5bo2bobo9bobo2bo2bobo$b2o8bo11b2o14b2o2b2o17bobo17bobo17bobo7b2o8bobo12b2o3bobo$obo36bo23bo19bo19bo7bo7bo3bo19bo$2bo115b2o$40b3o75bobo$40bo74b2o$41bo72bobo$116bo!
If you remove 4 glider get 13.214 (11G-> 10G):
x = 77, y = 67, rule = B3/S237$52bo$51bo$51b3o12$16bobo$12b2o2b2o$12bobo2bo$12bo4$13b3o$13bo$14bo$22b2o$5bo16bobo$3bobo16bo$4b2o13b2o50b2o$18bobo50bo$20bo51bo$69b3o$68bo$69bo$14b2o54bobo$13b2o56b2o$15bo$7b2o$8b2o$7bo16$59b2o$59bobo$59bo!
15.337 replace last converter:
x = 148, y = 16, rule = B3/S2339bo42bobo$40bo41b2o$38b3o42bo$76bo$3bo38bo34bo$bobo5bo30b2o33b3o$2b2o3b2o32b2o72bo$8b2o50b2o10b3o5b2o17bobo14bo2bobo19bo$11b2o13b2o18b2o12bobo3b2o6bo5bobo3b2o11b2obo3b2o6b3o2b2obo3b2o12bobo3b2o$11bobo9bobobo10bobo2bobobo14bo2bobo5bo8bo2bobo14bo2bobo14bo2bobo9bobo2bo2bobo$b2o8bo11b2o14b2o2b2o17bobo17bobo17bobo17bobo12b2o3bobo$obo36bo23bo19bo19bo13b2o4bo19bo$2bo109bobo2bobo$40b3o70b2o2bo$40bo72bo$41bo! The same applies to 15,320 from 13G: x = 130, y = 62, rule = B3/S23$31bo$30bo$30b3o73bo$102bo2b2o$19bobo78bobo2bobo$20b2o54bo19bo4b2o13bo$20bo54bobo17bobo17bobo3b2o$74bo2bo16bo2bo16bo2bo2bobo$73bo3bob2o12bo3bob2o2b3o7bo3bobo$40b3o30b2o3bobo12b2o3bobo2bo9b2o3bo$23bo11b2o5bo61bo$23b2o9bobo4bo$22bobo11bo$42b2o$41b2o$29bo13bo$30bo$28b3o2$31b3o$33bo$32bo27$55b2o$55bobo$55bo6$b3o$3bo$2bo!
And 15.345 from 14G:
x = 166, y = 20, rule = B3/S2398bobo$98b2o$99bo$obo88bo57bo$b2o86bobo53bo2b2o$bo88b2o10bo40bobo2bobo$102bobo2bo36b2o$102b2o2b2o56b2o$25b2o18b2o8bo9b2o18b2o19bobo6b2o18b2o18b2o6bobo$26bo19bo8bobo8bo19bo29bo5b2o12bo5b2o2b3o7bo5bo$7bo18bob2o16bob2o5b2o9bob2ob2o13bob2ob2o23bob2obobo12bob2obobo2bo9bob2obo$6bo20bobo17bobo2b2o13bobob2o14bobob2o5b2o17bob2o16bob2o6bo9bob2o$6b3o43bobo42b2o$52bo46bo$8bo$7b2o$7bobo$b2o$2b2o$bo! More 15.503 from 14G on base 14.558 by Goldtiger997: x = 106, y = 45, rule = B3/S232$26bobo$26b2o27b2o18b2o18b2o$27bo28bo2bo16bo2bo16bo2bo$40b2o2b3o9bobobo15bobobo15bobobo$24b2o15b2obo10b2o2bo15b2o2bo15b2o2b2o$25b2o13bo4bo10bo19bo5b2o12bo$24bo30bo19bo5bobo3bo7bo$16bobo36b2o18b2o6bo2b2o7b2o$17b2o67bobo$17bo3$28b2o$27bobo2b3o$14bo14bo2bo$14b2o17bo$13bobo2b3o2b2o$20bo2bobo$19bo3bo10$18b2o$19b2o$18bo8$4b2o$3bobo$5bo!
According to my data passed over a quarter (the easiest) way. Left 76 still lifes cost > 1glider/bit and 39 cost = 1glider/bit. The biggest cost 15.441 = 31/15glider/bit.
Bob Shemyakin
BobShemyakin
Posts: 205
Joined: June 15th, 2014, 6:24 am
### Re: Soup search results
GoldTiger997 wrote:Here's another 15-bit still-life ...
BobShemyakin wrote:If you remove 4 glider get 13.214 (11G-> 10G): ...
One cleanup glider can be shaved off both syntheses: 13.234 (not 13.214) and 15.320:
x = 135, y = 45, rule = B3/S2351bo65boo$51bobo59bo3bobo$51boo61boobo$43bo44bo19bo4boo13bo$44boo41bobo17bobo17bobo3boo$43boo41bobbo16bobbo6bo9bobbobbobo$30boo28boo23bo3boboo12bo3boboobboo8bo3bobo$12b3o15boo14boo12boo23boo3bobo12boo3bobobbobo7boo3bo$7boo5bo32boo$6bobo4bo32bo$8bo$14boo$13boo$bo13bo$bbo$3o22bo7bo21bo7bo$24bobo6bo20bobo6bo$3b3o18bobo6bo20bobo6bo$5bo19bo29bo$4bo23$74bo$73boo$73bobo!
BobShemyakin wrote:More 15.503 from 14G on base 14.558 by Goldtiger997: ...
I had added this improvement to my database on 2016-10-04, in the process of adding 14.558; also, 15.502 (with boat flipped the other way)
GoldTiger997 wrote:@ mniemiec & BobShemyakin: How many are left?
BobShemyakin wrote:According to my data passed over a quarter (the easiest) way.
Left 76 still lifes cost > 1glider/bit and 39 cost = 1glider/bit.
I haven't yet included Bob Shemyakin's completed list of 15-bit still-lifes (except for a few that improve older ones), but from the others, I have 12 =15 gliders and 72 >15 gliders:
Equals: 15.369 15.386 15.405 15.406 15.440 15.447 15.485 15.487 15.598 15.653 15.873 15.1060
Greater: 15.297 15.359 15.370 15.371 15.373 15.376 15.377 15.379 15.381 15.382
15.383 15.385 15.388 15.389 15.390 15.391 15.395 15.402 15.403 15.407
15.410 15.413 15.417 15.418 15.435 15.438 15.441 15.445 15.446 15.451
15.455 15.458 15.470 15.473 15.475 15.476 15.477 15.478 15.480 15.483
15.486 15.489 15.494 15.497 15.501 15.506 15.512 15.544 15.619 15.624
15.731 15.759 15.834 15.835 15.839 15.840 15.845 15.860 15.866 15.868
15.869 15.887 15.918 15.919 15.928 15.929 15.952 15.986 15.1001 15.1013
15.1032 15.1064
mniemiec
Posts: 878
Joined: June 1st, 2013, 12:00 am
### Re: Soup search results
We interrupt your regular scheduled glider syntheses for an unrelated question... does anybody recognize this xp6, which just showed up for the second time in D8_1? (EricABQ saw the first occurrence on Catagolue back in April 2015.)
Sample soups:
Eric's:
x = 31, y = 31, rule = B3/S23obbobboobobbooobooobboboobbobbo$bbbobooooooooobbbooooooooobobbb$bbobooobbbobobooobobobbbooobobb$oobobbbbobbooboboboobbobbbboboo$bbobooobbbbooobbbooobbbbooobobb$booboooobbooooooooooobbooooboob$oooboobbbbbbooboboobbbbbboobooo$oobbbobobboobobbboboobbobobbboo$bobobbbbobobobooobobobobbbbobob$oobbbbbbbbbboobbboobbbbbbbbbboo$boobboboobbbbooboobbbboobobboob$bobooobobbbbbbbobbbbbbbobooobob$oooooooboobbooooooobboobooooooo$oobbooooboobooobooobooboooobboo$oboobobbobobooboboobobobboboobo$bbobboobobbooboooboobboboobbobb$oboobobbobobooboboobobobboboobo$oobbooooboobooobooobooboooobboo$oooooooboobbooooooobboobooooooo$bobooobobbbbbbbobbbbbbbobooobob$boobboboobbbbooboobbbboobobboob$oobbbbbbbbbboobbboobbbbbbbbbboo$bobobbbbobobobooobobobobbbbobob$oobbbobobboobobbboboobbobobbboo$oooboobbbbbbooboboobbbbbboobooo$booboooobbooooooooooobbooooboob$bbobooobbbbooobbbooobbbbooobobb$oobobbbbobbooboboboobbobbbboboo$bbobooobbbobobooobobobbbooobobb$bbbobooooooooobbbooooooooobobbb$obbobboobobbooobooobboboobbobbo!
Mine:
x = 31, y = 31, rule = B3/S23obbobbboobbbobobobobbboobbbobbo$bbobbbbbbobobbobobbobobbbbbbobb$boobbobobbooobbobbooobbobobboob$obboboobobbboobbboobbboboobobbo$bbbbbbbbbobbbbobobbbbobbbbbbbbb$bboobbbbbbobooooooobobbbbbboobb$bbbobbbbobbooobbbooobbobbbbobbb$obobbbbobobbbooooobbbobobbbbobo$obbobbobboboobbbbboobobbobbobbo$bobbobboobbbbobbbobbbboobbobbob$bbobbobbbbobbbobobbbobbbbobbobb$boobbbobobbbbbobobbbbbobobbboob$obooboobobbbbbbbbbbbbbobooboobo$bbbobooobobbbbooobbbbobooobobbb$oobboobobbooboboboboobboboobboo$bbobbobobbbbbooooobbbbbobobbobb$oobboobobbooboboboboobboboobboo$bbbobooobobbbbooobbbbobooobobbb$obooboobobbbbbbbbbbbbbobooboobo$boobbbobobbbbbobobbbbbobobbboob$bbobbobbbbobbbobobbbobbbbobbobb$bobbobboobbbbobbbobbbboobbobbob$obbobbobboboobbbbboobobbobbobbo$obobbbbobobbbooooobbbobobbbbobo$bbbobbbbobbooobbbooobbobbbbobbb$bboobbbbbbobooooooobobbbbbboobb$bbbbbbbbbobbbbobobbbbobbbbbbbbb$obboboobobbboobbboobbboboobobbo$boobbobobbooobbobbooobbobobboob$bbobbbbbbobobbobobbobobbbbbbobb$obbobbboobbbobobobobbboobbbobbo!
Living proof that a little knowledge is a dangerous thing.
Catagolue: Apple Bottom • Life Wiki: Apple Bottom • Twitter: @_AppleBottom_
Proud member of the Pattern Raiders!
Apple Bottom
Posts: 842
Joined: July 27th, 2015, 2:06 pm
### Re: Soup search results
On BobShemyakin's database, 15.236 was listed as having as having a 15 glider synthesis, so I made a 7 glider one.
x = 37, y = 46, rule = B3/S235bo$3bobo$4b2o16$34b3o$28bo5bo$27b2o6bo$12bobo12bobo$13b2o$13bo2$14b3o$16bo$15bo3$6b2o$7b2o$6bo11$3o$2bo$bo! However, mniemiec wrote:I haven't yet included Bob Shemyakin's completed list of 15-bit still-lifes (except for a few that improve older ones), but from the others, I have 12 =15 gliders and 72 >15 gliders: Equals: 15.369 15.386 15.405 15.406 15.440 15.447 15.485 15.487 15.598 15.653 15.873 15.1060 Greater: 15.297 15.359 15.370 15.371 15.373 15.376 15.377 15.379 15.381 15.382 15.383 15.385 15.388 15.389 15.390 15.391 15.395 15.402 15.403 15.407 15.410 15.413 15.417 15.418 15.435 15.438 15.441 15.445 15.446 15.451 15.455 15.458 15.470 15.473 15.475 15.476 15.477 15.478 15.480 15.483 15.486 15.489 15.494 15.497 15.501 15.506 15.512 15.544 15.619 15.624 15.731 15.759 15.834 15.835 15.839 15.840 15.845 15.860 15.866 15.868 15.869 15.887 15.918 15.919 15.928 15.929 15.952 15.986 15.1001 15.1013 15.1032 15.1064 15.236 isn't listed there. Does that mean my synthesis was unnecessary, is it just a bookkeeping error between the two databases? EDIT: I noticed as well that 15.297 is listed above, but BobShemyakin's database has this synthesis: x = 146, y = 21, rule = B3/S23129bo$bo47bo68bo8b2o$2bo40bo5bobo25bo41b2o7b2o$3o41b2o3b2o27bo39b2o$43b2o31b3o48bo$35bo24b2o18b2o44b2o$2o34bo2b2o18bo2bo16bo2bo18b2o18b2o3bobo15b2o$b2o8bo22b3obobo19b2obo16b2obo17bobo17bobo19bobo$o8b2o29bo22bo19bo19bo19bo18bo$10b2o11b2o18b2o18b2o18b2o18b2o12b2o4b2o18b2o$24bo10b2o7bo19bo19bo19bo13b2o4bo19bo$23bo12b2o5bo19bo19bo19bo13bo5bo19bo$22bo12bo6bo19bo19bo19bo19bo19bo$21bo19bo19bo19bo19bo19bo19bo$21bobo17bobo17bobo17bobo17bobo17bobo17bobo$22b2o18b2o18b2o18b2o18b2o18b2o18b2o3$7b2o$6b2o$8bo! EDIT2: Here's 15.359 in 11 gliders: x = 33, y = 33, rule = B3/S23obo$b2o$bo3$14bobo$14b2o$15bo$9bobo$10b2o$10bo11bo$17bo3bo$18b2ob3o$8bo8b2o$b3o3bo$3bo3b3o$2bo$5b2o$4bobo8b3o$6bo8bo$11bo4bo$11b2o$10bobo8$30b3o$30bo$31bo!
Goldtiger997
Posts: 351
Joined: June 21st, 2016, 8:00 am
Location: 11.329903°N 142.199305°E
### Re: Soup search results
Goldtiger997 wrote:In BobShemyakin's database, 15.236 was listed as having as having a 15 glider synthesis, so I made a 7 glider one. ...
The shuttle can be made from 3 gliders, reducing this to 6:
x = 30, y = 25, rule = B3/S23o$boo$oo$$12bo12bobobbobo7boobboo3bo21boo25bobo4b3o19bobo4bo23bo5bo21bo26bo6bo19bobo6boo19bobo5bobo20boo67b3o9bo8bo! Goldtiger997 wrote:15.236 isn't listed there. Does that mean my synthesis was unnecessary, is it just a bookkeeping error between the two databases? As I had mentioned earlier, I am attempting to create a complete database, and finally finished all the bookkeeping for the 14-bit still-lifes a few years ago, but only had about half the 15-bit ones synthesized. Bob Shemyakin synthesized all the rest, plus he found many other, cheaper syntheses of some of the smaller ones. I am in the process of assimilating all his data, but there are over 2000 syntheses I need to examine in order to do so, so it will take a while before my list is up to date. Thus, my own list is missing about half of the 15-bit ones, including many of the expensive ones. (I also have the 700 or so difficult 18-bit still-lifes to assimilate, and those are in a more difficult format). Goldtiger997 wrote:I noticed as well that 15.297 is listed above, but BobShemyakin's database has this synthesis: ... This is because the first step, 10.21, had improved, but I had not yet updated all the derived syntheses. Goldtiger997 wrote:Here's 15.359 in 11 gliders: ... Very nice! mniemiec Posts: 878 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results Goldtiger997 wrote:Here's 15.359 in 11 gliders: x = 33, y = 33, rule = B3/S23obob2obo314bobo14b2o15bo9bobo10b2o10bo11bo17bo3bo18b2ob3o8bo8b2ob3o3bo3bo3b3o2bo5b2o4bobo8b3o6bo8bo11bo4bo11b2o10bobo830b3o30bo31bo! There is an impossible glider crossing in that pattern but it can be fixed by making the pond sufficiently early. chris_c Posts: 775 Joined: June 28th, 2014, 7:15 am ### Re: Soup search results chris_c wrote: Goldtiger997 wrote:Here's 15.359 in 11 gliders: x = 33, y = 33, rule = B3/S23obob2obo314bobo14b2o15bo9bobo10b2o10bo11bo17bo3bo18b2ob3o8bo8b2ob3o3bo3bo3b3o2bo5b2o4bobo8b3o6bo8bo11bo4bo11b2o10bobo830b3o30bo31bo! There is an impossible glider crossing in that pattern but it can be fixed by making the pond sufficiently early. 10 gliders: x = 102, y = 41, rule = B3/S2346bobo47b2o47bo9bo2bo97b2o3o94b2o2bo31b2o28b2o33bo2b2o31b2o4b2o13bo8b2o4b2o28bobo7b2o27bo2bo13b2o11bo2bo26b2o2bo2b3ob2o28bo2bo12b2o12bo2bo29b2o4bo3bo28b2o20b2o6b2o3bo51bo3bobo56b2obo55b2o569b3o69bo57bo12bo57b2o56bobo884b3o84bo85bo! Iteration of sigma(n)+tau(n)-n [sigma(n)+tau(n)-n : OEIS A163163] (e.g. 16,20,28,34,24,44,46,30,50,49,11,3,3, ...) : 965808 is period 336 (max = 207085118608). AbhpzTa Posts: 373 Joined: April 13th, 2016, 9:40 am Location: Ishikawa Prefecture, Japan ### Re: Soup search results Known reaction? x = 10, y = 11, rule = B3/S23b2oo3boo2bob3o47b2o6bo2bo6bobo7bo! LifeWiki: Like Wikipedia but with more spaceships. [citation needed] BlinkerSpawn Posts: 1592 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's ### Re: Soup search results mniemiec wrote:As I had mentioned earlier, I am attempting to create a complete database, and finally finished all the bookkeeping for the 14-bit still-lifes a few years ago, but only had about half the 15-bit ones synthesized. Bob Shemyakin synthesized all the rest, plus he found many other, cheaper syntheses of some of the smaller ones. I am in the process of assimilating all his data, but there are over 2000 syntheses I need to examine in order to do so, so it will take a while before my list is up to date. Thus, my own list is missing about half of the 15-bit ones, including many of the expensive ones. (I also have the 700 or so difficult 18-bit still-lifes to assimilate, and those are in a more difficult format). Ah, I see. Well according to BobShemyakin's database, 15.297 takes 15 gliders like this: x = 79, y = 20, rule = B3/S2317bobo40bo17b2o41bobo6bo11bo41b2o7bo5b3o51bo57bobo3bo58b2o2bo62b3o258bobo3bo11bo17b2o18b2o7b2o3b2o4b2o9bobo18bo19bo8bo3bobo4bo6bo10b2o8bo9bob2o16bob2o16bob2o2bobo20bobo8b2obobo14b2obobo14b2obo2bo20b2o13b2o18b2o18b2o17b2o5b2o4b2o3b2o44b2o3o3b2o2bobo5bo39bo2b2o2bo2bo6bo45b2o3bobo55bobo! A different order yields a better synthesis from 11 gliders: x = 50, y = 50, rule = B3/S2347bobo47b2o6bo41bo7bo5b3o1430bobo30b2o31bo333bo31bobo32b2o233b3o33bo34bo916b2o17b2o2b2o16bo3bobo22bo24b2o36b3o8bobo36bo10bo37bo25b3o3o4bo2bo3bobo! I believe that now all 15-bit still-lifes from 15.1 to 15.300 have syntheses in less than 15 gliders. Goldtiger997 Posts: 351 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Soup search results Goldtiger997 wrote:... I believe that now all 15-bit still-lifes from 15.1 to 15.300 have syntheses in less than 15 gliders. With the exception of 15.182 (22G): x = 161, y = 29, rule = 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! and 15.252 (16G): x = 118, y = 21, rule = B3/S2353bo37bo7bo43bobo36bo8bo10bo32b2o6bo29b3o6b3o10bobo37bo19b2o38b3o26b3o90bo89bo10bobo22bo19bo11b2o21bobo17bobo17bo19bo14b2o3bo11bo21bobo17bobo3bo13bobo10b2o5bobo13bo3bobo33bo19bo5bobo11bo2bo8bobo5bo2bo13bo2bo2bo31bobo17bobo5b2o10bobo3bo9bo3bobo3bo13bobo3bo11b3o3bobo10bobo17bobo17bobo3b2o12bobo3b2o14bo3b2o13bo3b2o12bo19bo19bo19bo12bo5bo38bo56bo56b3o26b3obo85bob2o51b3o29bo6b2oobo53bo36bobo55bo37bo! Bob Shemyakin BobShemyakin Posts: 205 Joined: June 15th, 2014, 6:24 am ### Re: Soup search results Goldtiger997 wrote:Well according to BobShemyakin's database, 15.297 takes 15 gliders like this: ... A different order yields a better synthesis from 11 gliders: ... This is actually 15.287. In my collection, I try, for multi-part objects, I try, whenever possible, to provide the smallest synthesis, as well as all syntheses beginning with any of the parts. This facilitates future, larger incremental syntheses in which those parts may be already attached to something else. Such techniques would yield both your synthesis and his, with yours being the cheaper path (and, at least at present, there is no "natural" atomic synthesis that is better.) I believe that now all 15-bit still-lifes from 15.1 to 15.300 have syntheses in less than 15 gliders.[/quote] By examining Bob Shemyakin's most recent still15.rar archive, I found the following additional still-lifes that aren't on my lists: Equals: 15.365 15.402 15.410 15.431 15.446 15.487 15.506 15.516 15.751 15.786 15.839 15.855 15.890 15.919 15.931 15.934 15.939 15.942 15.949 15.1024 15.1033 15.1034 15.1036 15.1088 15.1103 15.1207 Greater: 15.182 15.252 15.304 15.416 15.457 15.507 15.508 15.562 15.578 15.620 15.698 15.723 15.791 15.797 15.840 15.848 15.849 15.889 15.923 15.947 15.960 15.983 15.1012 15.1035 15.1065 (15.182 and 15.252 remain unoptimized, from 22 and 16 gliders respectively (EDIT: as Bob just pointed out), and 15.304 from 16 for the next batch.) mniemiec Posts: 878 Joined: June 1st, 2013, 12:00 am ### Re: Soup search results A weird pulsar variant in D4_+1: x = 31, y = 31, rule = 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! I think a synthesis is possible based on this for the [cis x2]-two-quadrant and double-mirrored four-quadrant versions. x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)$$x_1=\eta xV^*_\eta=c^2\sqrt{\Lambda\eta}K=\frac{\Lambda u^2}2P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$http://conwaylife.com/wiki/A_for_all Aidan F. Pierce A for awesome Posts: 1555 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 ### Re: Soup search results From here, for 15.297: x = 8, y = 13, rule = B3/S233b2o3bobo5b2o4b2o34b3o4bo2bo5b3o2b2oo2bob2o! For 15.889: x = 12, y = 18, rule = B3/S2310bo8b2o9b2o6boobo2obo5bobo3b2o4b2o3bo6bo2bo26b2o5bobo7bo! EDIT: 15.182 in 7G: x = 38, y = 22, rule = B3/S2337bo35b2o36b2o10bobo11b2o11bo2obob2obo215bo13b2o14b2o17b3o17bo6bo11bo6b2o5bobo16bo15b2o15bobo! 15.624: x = 27, y = 25, rule = B3/S2325bo24bo11b2o11b3o10b2o11bo514b3o215bo13bobo13b2o7b2o6bob2o6bo3bo7b3o8bo42ob2oo! Everything but the R in 9G: x = 33, y = 31, rule = B3/S2331bo30bo30b3o714bo15bo12bo13b3o10b2o27b2o47bobo8b2o8bo6bo15b2o14bobo25bo4bo9b2o13b2o3b2o8bobo13bobo2bobo10bo42ob2oo! LifeWiki: Like Wikipedia but with more spaceships. [citation needed] BlinkerSpawn Posts: 1592 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's ### Re: Soup search results mniemiec wrote:(15.182 and 15.252 remain unoptimized, from 22 and 16 gliders respectively (EDIT: as Bob just pointed out), and 15.304 from 16 for the next batch.) Oh sorry, I must have missed those. BlinkerSpawn wrote:15.182 in 7G: x = 38, y = 22, rule = B3/S2337bo35b2o36b2o10bobo11b2o11bo2obob2obo215bo13b2o14b2o17b3o17bo6bo11bo6b2o5bobo16bo15b2o15bobo! Nice! Here is 15.252 and 15.304 both in 12 gliders. x = 111, y = 121, rule = B3/S23106bo105bo105b3o4073bo73bobo44bo28b2o45b2o44b2o1156bo54bobo55b2o257bo57bobo57b2o1455b3o2b2o57bo2bobo56bo3bo1613b2o14b2o13bo73b3o5bo4bo3b2oobo2bo53o2bobo106b2o108bobo108bo! x = 51, y = 32, rule = B3/S2337bobo37b2o38bo15bo13bobo14b2o319bo20bo18b3o35bo33bobo2bobo34b2o2b2o39bo213b3o15bo8bo6b3o14bo9b2o5bo23bobo6bo317b2o4b2o24b2o16bobo3bobo23b2o18bo5bo25bo53o2bobo! Now all 15-bit still-lifes from 15.1 to 15.300 have syntheses in less than 15 gliders! Goldtiger997 Posts: 351 Joined: June 21st, 2016, 8:00 am Location: 11.329903°N 142.199305°E ### Re: Soup search results BlinkerSpawn wrote:15.182 in 7G: ... Row 1: The above gives 15.624 from 13. Row 2: 15.465 from 11, giving 15.889 from 17. x = 129, y = 89, rule = B3/S2361bo61bobo61boo$$39boo18boo$39boo18boo5$104bo$98bo5bobo$99boo3boo$98boo$$7bo5bobo119booo5boo24booboboo13booboboo13booboboo13booboboo6bobo14boobobboboo28boboboobo12boboboobo12boboboobo12boboboobo6boo14boboboooo6bobo20bobbo16bobbo16bobbo16bobbo11bo14bobbo8boo22bobo17bobo17bobo17bobo27bobo9bo23bo19bo19bo19bo8bo20bo17boo82boo6bo16boo83bobo4boo18bo89bobo98b3o100bo3boo94bo4boo3bo313boo12boo14bo1625bo25bobo25boo$$bbo109bo$obo17bobo89bobo$boo17boo90boo$21bo88bo$104bo3bobo$105bo3boo$69bobo31b3o$69boo40bo$12bo57bo38boo4bo$11bo98boo3bobo$11b3o57boo42boo$9bo60boo54boo$7bobo36bo19bo5bo13boo18boo19bo$8boo32boobobo14boobobo14boobobo14boobobo14boobo$18bo22boboboo14boboboo14boboboo14boboboo5boo7boboboo$16boo23bobbo16bobbo16bobbo16bobbo7bobo6bobbo$17boo23bobo17bobo17bobo17bobo7bo9bobo$43bo19bo19bo19bo19bo10$16bo$15boo$15bobo3$10bo16boo$10boo14boo$9bobo16bo!
mniemiec
Posts: 878
Joined: June 1st, 2013, 12:00 am
### Re: Soup search results
mniemiec wrote:
Row 1: The above gives 15.624 from 13.
Row 2: 15.465 from 11, giving 15.889 from 17.
x = 129, y = 89, rule = B3/S2361bo$61bobo$61boo$$39boo18boo39boo18boo5104bo98bo5bobo99boo3boo98boo$$7bo$5bobo119boo$o5boo24booboboo13booboboo13booboboo13booboboo6bobo14boobobbo$boo28boboboobo12boboboobo12boboboobo12boboboobo6boo14boboboo$oo6bobo20bobbo16bobbo16bobbo16bobbo11bo14bobbo$8boo22bobo17bobo17bobo17bobo27bobo$9bo23bo19bo19bo19bo8bo20bo$17boo82boo6bo$16boo83bobo4boo$18bo89bobo$98b3o$100bo$3boo94bo$4boo$3bo3$13boo$12boo$14bo16$25bo$25bobo$25boobbo109bo$obo17bobo89bobo$boo17boo90boo$21bo88bo$104bo3bobo$105bo3boo$69bobo31b3o$69boo40bo$12bo57bo38boo4bo$11bo98boo3bobo$11b3o57boo42boo$9bo60boo54boo$7bobo36bo19bo5bo13boo18boo19bo$8boo32boobobo14boobobo14boobobo14boobobo14boobo$18bo22boboboo14boboboo14boboboo14boboboo5boo7boboboo$16boo23bobbo16bobbo16bobbo16bobbo7bobo6bobbo$17boo23bobo17bobo17bobo17bobo7bo9bobo$43bo19bo19bo19bo19bo10$16bo$15boo$15bobo3$10bo16boo$10boo14boo$9bobo16bo! 15.889 in 7G: x = 26, y = 29, rule = B3/S2323bobo$23b2o$24bo3$o$b2o$2o5$14bo$13b2o$13bobo6$4bo15b3o$4bobo13bo$4b2o2b3o10bo$8bo$9bo2$2b2o$b2o\$3bo!
Iteration of sigma(n)+tau(n)-n [sigma(n)+tau(n)-n : OEIS A163163] (e.g. 16,20,28,34,24,44,46,30,50,49,11,3,3, ...) :
965808 is period 336 (max = 207085118608).
AbhpzTa
Posts: 373
Joined: April 13th, 2016, 9:40 am
Location: Ishikawa Prefecture, Japan
PreviousNext | 15,842 | 33,231 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-09 | longest | en | 0.426565 |
https://simplywall.st/stocks/us/semiconductors/nasdaq-powi/power-integrations/news/heres-what-power-integrations-inc-s-nasdaqpowi-p-e-is-telling-us/ | 1,555,950,601,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578558125.45/warc/CC-MAIN-20190422155337-20190422181337-00128.warc.gz | 558,595,738 | 15,437 | # Here’s What Power Integrations, Inc.’s (NASDAQ:POWI) P/E Is Telling Us
The goal of this article is to teach you how to use price to earnings ratios (P/E ratios). We’ll show how you can use Power Integrations, Inc.’s (NASDAQ:POWI) P/E ratio to inform your assessment of the investment opportunity. Power Integrations has a P/E ratio of 29.79, based on the last twelve months. That means that at current prices, buyers pay \$29.79 for every \$1 in trailing yearly profits.
### How Do I Calculate A Price To Earnings Ratio?
The formula for P/E is:
Price to Earnings Ratio = Price per Share ÷ Earnings per Share (EPS)
Or for Power Integrations:
P/E of 29.79 = \$70.78 ÷ \$2.38 (Based on the year to December 2018.)
### Is A High P/E Ratio Good?
A higher P/E ratio means that buyers have to pay a higher price for each \$1 the company has earned over the last year. All else being equal, it’s better to pay a low price — but as Warren Buffett said, ‘It’s far better to buy a wonderful company at a fair price than a fair company at a wonderful price.’
### How Growth Rates Impact P/E Ratios
When earnings fall, the ‘E’ decreases, over time. That means unless the share price falls, the P/E will increase in a few years. A higher P/E should indicate the stock is expensive relative to others — and that may encourage shareholders to sell.
Notably, Power Integrations grew EPS by a whopping 155% in the last year. Unfortunately, earnings per share are down 7.1% a year, over 5 years.
### How Does Power Integrations’s P/E Ratio Compare To Its Peers?
The P/E ratio indicates whether the market has higher or lower expectations of a company. The image below shows that Power Integrations has a higher P/E than the average (17.3) P/E for companies in the semiconductor industry.
Its relatively high P/E ratio indicates that Power Integrations shareholders think it will perform better than other companies in its industry classification. The market is optimistic about the future, but that doesn’t guarantee future growth. So investors should delve deeper. I like to check if company insiders have been buying or selling.
### Remember: P/E Ratios Don’t Consider The Balance Sheet
Don’t forget that the P/E ratio considers market capitalization. That means it doesn’t take debt or cash into account. Theoretically, a business can improve its earnings (and produce a lower P/E in the future), by taking on debt (or spending its remaining cash).
Such spending might be good or bad, overall, but the key point here is that you need to look at debt to understand the P/E ratio in context.
### Power Integrations’s Balance Sheet
Power Integrations has net cash of US\$229m. That should lead to a higher P/E than if it did have debt, because its strong balance sheets gives it more options.
### The Bottom Line On Power Integrations’s P/E Ratio
Power Integrations has a P/E of 29.8. That’s higher than the average in the US market, which is 17.5. With cash in the bank the company has plenty of growth options — and it is already on the right track. So it is not surprising the market is probably extrapolating recent growth well into the future, reflected in the relatively high P/E ratio.
Investors should be looking to buy stocks that the market is wrong about. People often underestimate remarkable growth — so investors can make money when fast growth is not fully appreciated. So this free visualization of the analyst consensus on future earnings could help you make the right decision about whether to buy, sell, or hold.
Of course, you might find a fantastic investment by looking at a few good candidates. So take a peek at this free list of companies with modest (or no) debt, trading on a P/E below 20.
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading. | 954 | 4,305 | {"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} | 2.78125 | 3 | CC-MAIN-2019-18 | latest | en | 0.940397 |
https://www.cpalms.org/PreviewCourse/Preview/13046?isShowCurrent=false%27 | 1,709,576,370,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476464.74/warc/CC-MAIN-20240304165127-20240304195127-00012.warc.gz | 703,748,352 | 55,456 | # Algebra 1 for Credit Recovery (#1200315)
Version for Academic Year:
## General Course Information and Notes
### Version Description
Credit Recovery courses are credit bearing courses with specific content requirements defined by Next Generation Sunshine State Standards and/or Florida Standards. Students enrolled in a Credit Recovery course must have previously attempted the corresponding course (and/or End-of-Course assessment) since the course requirements for the Credit Recovery course is exactly the same as the previously attempted corresponding course. For example, Geometry (1206310) and Geometry for Credit Recovery (1206315) have identical content requirements. It is important to note that Credit Recovery courses are not bound by Section 1003.436(1) (a), Florida Statutes, requiring a minimum of 135 hours of bona fide instruction (120 hours in a school/district implementing block scheduling) in a designed course of study that contains student performance standards, since the students have previously attempted successful completion of the corresponding course. Additionally, Credit Recovery courses should ONLY be used for credit recovery, grade forgiveness, or remediation for students needing to prepare for an End-of-Course assessment retake.
### General Notes
The fundamental purpose of this course is to formalize and extend the mathematics that students learned in the middle grades. The critical areas, called units, deepen and extend understanding of linear and exponential relationships by contrasting them with each other and by applying linear models to data that exhibit a linear trend, and students engage in methods for analyzing, solving, and using quadratic functions. The Standards for Mathematical Practice apply throughout each course and, together with the content standards, prescribe that students experience mathematics as a coherent, useful, and logical subject that makes use of their ability to make sense of problem situations.
Unit 1- Relationships Between Quantities and Reasoning with Equations: By the end of eighth grade, students have learned to solve linear equations in one variable and have applied graphical and algebraic methods to analyze and solve systems of linear equations in two variables. Now, students analyze and explain the process of solving an equation. Students develop fluency writing, interpreting, and translating between various forms of linear equations and inequalities, and using them to solve problems. They master the solution of linear equations and apply related solution techniques and the laws of exponents to the creation and solution of simple exponential equations.
Unit 2- Linear and Exponential Relationships: In earlier grades, students define, evaluate, and compare functions, and use them to model relationships between quantities. In this unit, students will learn function notation and develop the concepts of domain and range. They explore many examples of functions, including sequences; they interpret functions given graphically, numerically, symbolically, and verbally, translate between representations, and understand the limitations of various representations. Students build on and informally extend their understanding of integer exponents to consider exponential functions. They compare and contrast linear and exponential functions, distinguishing between additive and multiplicative change. Students explore systems of equations and inequalities, and they find and interpret their solutions. They interpret arithmetic sequences as linear functions and geometric sequences as exponential functions.
Unit 3- Descriptive Statistics: This unit builds upon students prior experiences with data, providing students with more formal means of assessing how a model fits data. Students use regression techniques to describe and approximate linear relationships between quantities. They use graphical representations and knowledge of the context to make judgments about the appropriateness of linear models. With linear models, they look at residuals to analyze the goodness of fit.
Unit 4- Expressions and Equations: In this unit, students build on their knowledge from unit 2, where they extended the laws of exponents to rational exponents. Students apply this new understanding of number and strengthen their ability to see structure in and create quadratic and exponential expressions. They create and solve equations, inequalities, and systems of equations involving quadratic expressions.
Unit 5- Quadratic Functions and Modeling: In this unit, students consider quadratic functions, comparing the key characteristics of quadratic functions to those of linear and exponential functions. They select from among these functions to model phenomena. Students learn to anticipate the graph of a quadratic function by interpreting various forms of quadratic expressions. In particular, they identify the real solutions of a quadratic equation as the zeros of a related quadratic function. Students expand their experience with functions to include more specialized functions absolute value, step, and those that are piece wise-defined.
English Language Development ELD Standards Special Notes Section:
Teachers are required to provide listening, speaking, reading and writing instruction that allows English language learners (ELL) to communicate information, ideas and concepts for academic success in the content area of Mathematics. For the given level of English language proficiency and with visual, graphic, or interactive support, students will interact with grade level words, expressions, sentences and discourse to process or produce language necessary for academic success. The ELD standard should specify a relevant content area concept or topic of study chosen by curriculum developers and teachers which maximizes an ELL’s need for communication and social skills. To access an ELL supporting document which delineates performance definitions and descriptors, please click on the following link:
https://cpalmsmediaprod.blob.core.windows.net/uploads/docs/standards/eld/ma.pdf
### General Information
Course Number: 1200315
Course Path:
Abbreviated Title: ALG 1 CR
Number of Credits: One (1) credit
Course Length: Credit Recovery (R)
Course Type: Elective Course
Course Level: 2
Course Status: Course Approved
Grade Level(s): 9,10,11,12
## Educator Certifications
One of these educator certification options is required to teach this course.
## Student Resources
Vetted resources students can use to learn the concepts and skills in this course.
## Original Student Tutorials
Multistep Factoring: Quadratics:
Learn how to use multistep factoring to factor quadratics in this interactive tutorial.
This is part 5 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Solving Systems of Linear Equations Part 7: Word Problems:
Learn to solve word problems represented by systems of linear equations, algebraically and graphically, in this interactive tutorial.
This part 7 in a 7-part series. Click below to explore the other tutorials in the series.
Type: Original Student Tutorial
Factoring Polynomials when "a" Does Not Equal 1, Snowflake Method:
Learn to factor quadratic trinomials when the coefficient a does not equal 1 by using the Snowflake Method in this interactive tutorial.
This is part 4 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Solving Systems of Linear Equations Part 6: Writing Systems from Context:
Learn how to create systems of linear equations to represent contextual situations in this interactive tutorial.
This part 6 in a 7-part series. Click below to explore the other tutorials in the series.
Type: Original Student Tutorial
Solving Systems of Linear Equations Part 5: Connecting Algebraic Methods to Graphing:
Learn to solve systems of linear equations by connecting algebraic and graphing methods in this interactive tutorial.
This part 5 in a 7-part series. Click below to explore the other tutorials in the series.
Type: Original Student Tutorial
Factoring Quadratics When the Coefficient a Does Not Equal 1: The Box Method:
Learn how to factor quadratic polynomials when the leading coefficient (a) is not 1 by using the box method in this interactive tutorial.
This is part 3 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
The Diamond Game: Factoring Quadratics when a = 1:
Learn how to factor quadratics when the coefficient a = 1 using the diamond method in this game show-themed, interactive tutorial.
This is part 1 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Solving Systems of Linear Equations Part 4: Advanced Elimination:
Learn to solve systems of linear equations using advanced elimination in this interactive tutorial.
This part 4 in a 7-part series. Click below to explore the other tutorials in the series.
Type: Original Student Tutorial
Solving Systems of Linear Equations Part 3: Basic Elimination:
Learn to solve systems of linear equations using basic elimination in this interactive tutorial.
This part 3 in a 7-part series. Click below to explore the other tutorials in the series.
Part 1: Solving Systems of Linear Equations Part 1: Using Graphs
Part 2: Solving Systems of Linear Equations Part 2: Substitution
Part 4: Solving Systems of Linear Equations Part 4: Advanced Elimination (Coming soon)
Part 5: Solving Systems of Linear Equations Part 5: Connecting Algebraic Methods to Graphing (Coming soon)
Part 6: Solving Systems of Linear Equations Part 6: Writing Systems from Context (Coming soon)
Part 7: Solving Systems of Linear Equations Part 7: Word Problems (Coming soon)
Type: Original Student Tutorial
Highs and Lows Part 2: Completing the Square:
Learn the process of completing the square of a quadratic function to find the maximum or minimum to discover how high a dolphin jumped in this interactive tutorial.
This is part 2 of a 2 part series. Click HERE to open part 1.
Type: Original Student Tutorial
Highs and Lows Part 1: Completing the Square:
Learn the process of completing the square of a quadratic function to find the maximum or minimum to discover how high a dolphin jumped in this interactive tutorial.
This is part 1 of a 2 part series. Click HERE to open Part 2.
Type: Original Student Tutorial
Identifying Parts of Linear Expressions:
Learn to identify and interpret parts of linear expressions in terms of mathematical or real-world contexts in this original tutorial.
Type: Original Student Tutorial
Movies Part 2: What’s the Spread?:
Follow Jake along as he relates box plots with other plots and identifies possible outliers in real-world data from surveys of moviegoers' ages in part 2 in this interactive tutorial.
This is part 2 of 2-part series, click HERE to view part 1.
Type: Original Student Tutorial
Solving Systems of Linear Equations Part 2: Substitution:
Learn to solve systems of linear equations using substitution in this interactive tutorial.
This part 2 in a 7-part series. Click below to explore the other tutorials in the series.
Part 1: Solving Systems of Linear Equations Part 1: Using Graphs
Part 3: Solving Systems of Linear Equations Part 3: Basic Elimination (Coming soon)
Part 4: Solving Systems of Linear Equations Part 4: Advanced Elimination (Coming soon)
Part 5: Solving Systems of Linear Equations Part 5: Connecting Algebraic Methods to Graphing (Coming soon)
Part 6: Solving Systems of Linear Equations Part 6: Writing Systems from Context (Coming soon)
Part 7: Solving Systems of Linear Equations Part 7: Word Problems (Coming soon)
Type: Original Student Tutorial
Movies Part 1: What's the Spread?:
Follow Jake as he displays real-world data by creating box plots showing the 5 number summary and compares the spread of the data from surveys of the ages of moviegoers in part 1 of this interactive tutorial.
This is part 1 of 2-part series, click HERE to view part 2.
Type: Original Student Tutorial
Exponential Functions Part 3: Decay:
Learn about exponential decay as you calculate the value of used cars by examining equations, graphs, and tables in this interactive tutorial.
Type: Original Student Tutorial
Solving Systems of Linear Equations Part 1: Using Graphs:
Learn how to solve systems of linear equations graphically in this interactive tutorial.
Type: Original Student Tutorial
Linear Functions: Jobs:
Learn how to interpret key features of linear functions and translate between representations of linear functions through exploring jobs for teenagers in this interactive tutorial.
Type: Original Student Tutorial
Exponential Functions Part 2: Growth:
Learn about exponential growth in the context of interest earned as money is put in a savings account by examining equations, graphs, and tables in this interactive tutorial.
Type: Original Student Tutorial
Exponential Functions Part 1:
Learn about exponential functions and how they are different from linear functions by examining real world situations, their graphs and their tables in this interactive tutorial.
Type: Original Student Tutorial
Functions, Functions, Everywhere: Part 2:
Continue exploring how to determine if a relation is a function using graphs and story situations in this interactive tutorial.
This is the second tutorial in a 2-part series. Click HERE to open Part 1.
Type: Original Student Tutorial
Travel with Functions:
Learn how to evaluate and interpret function notation by following Melissa and Jose on their travels in this interactive tutorial.
Type: Original Student Tutorial
Functions, Functions Everywhere: Part 1:
What is a function? Where do we see functions in real life? Explore these questions and more using different contexts in this interactive tutorial.
This is part 1 in a two-part series on functions. Click HERE to open Part 2.
Type: Original Student Tutorial
Solving Rational Equations: Using Common Denominators:
Learn how to solve rational functions by getting common denominators in this interactive tutorial.
Type: Original Student Tutorial
Dilations...The Effect of k on a Graph:
Visualize the effect of using a value of k in both kf(x) or f(kx) when k is greater than zero in this interactive tutorial.
Type: Original Student Tutorial
Solving Rational Equations: Cross Multiplying:
Learn how to solve rational linear and quadratic equations using cross multiplication in this interactive tutorial.
Type: Original Student Tutorial
Solving Inequalities and Graphing Solutions Part 2:
Learn how to solve and graph compound inequalities and determine if solutions are viable in part 2 of this interactive tutorial series.
Click HERE to open Part 1.
Type: Original Student Tutorial
Writing Equations in Two Variables:
Learn how to write equations in two variables in this interactive tutorial.
Type: Original Student Tutorial
Solving Inequalities and Graphing Solutions: Part 1:
Learn how to solve and graph one variable inequalities, including compound inequalities, in part 1 of this interactive tutorial series.
Click HERE to open Part 2.
Type: Original Student Tutorial
Reflections...The Effect of k on a Graph:
Learn how reflections of a function are created and tied to the value of k in the mapping of f(x) to -1f(x) in this interactive tutorial.
Type: Original Student Tutorial
Translations...The Effect of k on the Graph:
Explore translations of functions on a graph that are caused by k in this interactive tutorial. GeoGebra and interactive practice items are used to investigate linear, quadratic, and exponential functions and their graphs, and the effect of a translation on a table of values.
Type: Original Student Tutorial
The Year-Round School Debate: Identifying Faulty Reasoning – Part Two:
This is Part Two of a two-part series. Learn to identify faulty reasoning in this interactive tutorial series. You'll learn what some experts say about year-round schools, what research has been conducted about their effectiveness, and how arguments can be made for and against year-round education. Then, you'll read a speech in favor of year-round schools and identify faulty reasoning within the argument, specifically the use of hasty generalizations.
Make sure to complete Part One before Part Two! Click HERE to launch Part One.
Type: Original Student Tutorial
The Year-Round School Debate: Identifying Faulty Reasoning – Part One:
Learn to identify faulty reasoning in this two-part interactive English Language Arts tutorial. You'll learn what some experts say about year-round schools, what research has been conducted about their effectiveness, and how arguments can be made for and against year-round education. Then, you'll read a speech in favor of year-round schools and identify faulty reasoning within the argument, specifically the use of hasty generalizations.
Make sure to complete both parts of this series! Click HERE to open Part Two.
Type: Original Student Tutorial
Evaluating an Argument – Part Four: JFK’s Inaugural Address:
Examine President John F. Kennedy's inaugural address in this interactive tutorial. You will examine Kennedy's argument, main claim, smaller claims, reasons, and evidence.
In Part Four, you'll use what you've learned throughout this series to evaluate Kennedy's overall argument.
Make sure to complete the previous parts of this series before beginning Part 4.
• Click HERE to launch Part One.
• Click HERE to launch Part Two.
• Click HERE to launch Part Three.
Type: Original Student Tutorial
Evaluating an Argument – Part Three: JFK’s Inaugural Address:
Examine President John F. Kennedy's inaugural address in this interactive tutorial. You will examine Kennedy's argument, main claim, smaller claims, reasons, and evidence. By the end of this four-part series, you should be able to evaluate his overall argument.
In Part Three, you will read more of Kennedy's speech and identify a smaller claim in this section of his speech. You will also evaluate this smaller claim's relevancy to the main claim and evaluate Kennedy's reasons and evidence.
Make sure to complete all four parts of this series!
• Click HERE to launch Part One.
• Click HERE to launch Part Two.
• Click HERE to launch Part Four.
Type: Original Student Tutorial
It's a Slippery Slope!:
Learn what slope is in mathematics and how to calculate it on a graph and with the slope formula in this interactive tutorial.
Type: Original Student Tutorial
Finding the Zeros of Quadratic Functions:
Learn to find the zeros of a quadratic function and interpret their meaning in real-world contexts with this interactive tutorial.
Type: Original Student Tutorial
Introduction to Polynomials, Part 2 - Adding and Subtracting:
Learn how to add and subtract polynomials in this online tutorial. You will learn how to combine like terms and then use the distribute property to subtract polynomials.
This is part 2 of a two-part lesson. Click below to open part 1.
Type: Original Student Tutorial
Introduction to Polynomials: Part 1:
Learn how to identify monomials and polynomials and determine their degree in this interactive tutorial.
This is part 1 in a two-part series. Click here to open Part 2.
Type: Original Student Tutorial
Ready for Takeoff! -- Part Two:
This is Part Two of a two-part tutorial series. In this interactive tutorial, you'll practice identifying a speaker's purpose using a speech by aviation pioneer Amelia Earhart. You will examine her use of rhetorical appeals, including ethos, logos, pathos, and kairos. Finally, you'll evaluate the effectiveness of Earhart's use of rhetorical appeals.
Be sure to complete Part One first. Click here to launch PART ONE.
Type: Original Student Tutorial
Ready for Takeoff! -- Part One:
This is Part One of a two-part tutorial series. In this interactive tutorial, you'll practice identifying a speaker's purpose using a speech by aviation pioneer Amelia Earhart. You will examine her use of rhetorical appeals, including ethos, logos, pathos, and kairos. Finally, you'll evaluate the effectiveness of Earhart's use of rhetorical appeals.
Click here to launch PART TWO.
Type: Original Student Tutorial
Expository Writing: Eyes in the Sky (Part 4 of 4):
Practice writing different aspects of an expository essay about scientists using drones to research glaciers in Peru. This interactive tutorial is part four of a four-part series. In this final tutorial, you will learn about the elements of a body paragraph. You will also create a body paragraph with supporting evidence. Finally, you will learn about the elements of a conclusion and practice creating a “gift.”
This tutorial is part four of a four-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Expository Writing: Eyes in the Sky (Part 3 of 4):
Learn how to write an introduction for an expository essay in this interactive tutorial. This tutorial is the third part of a four-part series. In previous tutorials in this series, students analyzed an informational text and video about scientists using drones to explore glaciers in Peru. Students also determined the central idea and important details of the text and wrote an effective summary. In part three, you'll learn how to write an introduction for an expository essay about the scientists' research.
This tutorial is part three of a four-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Data and Frequencies:
Learn to define, calculate, and interpret marginal frequencies, joint frequencies, and conditional frequencies in the context of the data with this interactive tutorial.
Type: Original Student Tutorial
Finding the Maximum or Minimum of a Quadratic Function:
Learn to complete the square of a quadratic expression and identify the maximum or minimum value of the quadratic function it defines. In this interactive tutorial, you'll also interpret the meaning of the maximum and minimum of a quadratic function in a real world context.
Type: Original Student Tutorial
Graphing Linear Inequalities:
Learn to graph linear inequalities in two variables to display their solutions as you complete this interactive tutorial.
Type: Original Student Tutorial
The Radical Puzzle:
Learn to rewrite products involving radicals and rational exponents using properties of exponents in this interactive tutorial.
Type: Original Student Tutorial
Comparing Mitosis and Meiosis:
Compare and contrast mitosis and meiosis in this interactive tutorial. You'll also relate them to the processes of sexual and asexual reproduction and their consequences for genetic variation.
Type: Original Student Tutorial
Creating Exponential Functions:
Follow as we construct an exponential function from a graph, from a table of values, and from a description of a relationship in the real world in this interactive tutorial.
Type: Original Student Tutorial
Changing Rates:
Learn how to calculate and interpret an average rate of change over a specific interval on a graph in this interactive tutorial.
Type: Original Student Tutorial
Justifiable Steps:
Learn how to explain the steps used to solve multi-step linear equations and provide reasons to support those steps with this interactive tutorial.
Type: Original Student Tutorial
Finding Solutions on a Graph:
Learn to determine the number of possible solutions for a linear equation with this interactive tutorial.
Type: Original Student Tutorial
Solving an Equation Using a Graph:
Follow as we learn why the x-coordinate of the point of intersection of two functions is the solution of the equation f(x) = g(x) in this interactive tutorial.
Type: Original Student Tutorial
Cancer: Mutated Cells Gone Wild!:
Explore the relationship between mutations, the cell cycle, and uncontrolled cell growth which may result in cancer with this interactive tutorial.
Type: Original Student Tutorial
Writing Inequalities with Money, Money, Money:
Write linear inequalities for different money situations in this interactive tutorial.
Type: Original Student Tutorial
Factoring Polynomials Using Special Cases:
Learn how to factor quadratic polynomials that follow special cases, difference of squares and perfect square trinomials, in this interactive tutorial.
This is part 2 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
## Educational Games
Solving Inequalities: Inequalities and Graphs of Inequalities:
In this challenge game, you will be solving inequalities and working with graphs of inequalities. Use the "Teach Me" button to review content before the challenge. During the challenge you get one free solve and two hints! After the challenge, review the problems as needed. Try again to get all challenge questions right! Question sets vary with each game, so feel free to play the game multiple times as needed! Good luck!
Type: Educational Game
Timed Algebra Quiz:
In this timed activity, students solve linear equations (one- and two-step) or quadratic equations of varying difficulty depending on the initial conditions they select. This activity allows students to practice solving equations while the activity records their score, so they can track their progress. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Educational Game
Algebra Four:
In this activity, two students play a simulated game of Connect Four, but in order to place a piece on the board, they must correctly solve an algebraic equation. This activity allows students to practice solving equations of varying difficulty: one-step, two-step, or quadratic equations and using the distributive property if desired. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the Java applet.
Type: Educational Game
## Educational Software / Tool
Two Way Frequency Excel Spreadsheet:
This Excel spreadsheet allows the educator to input data into a two way frequency table and have the resulting relative frequency charts calculated automatically on the second sheet. This resource will assist the educator in checking student calculations on student-generated data quickly and easily.
Steps to add data: All data is input on the first spreadsheet; all tables are calculated on the second spreadsheet
1. Modify column and row headings to match your data.
2. Input joint frequency data.
3. Click the second tab at the bottom of the window to see the automatic calculations.
Type: Educational Software / Tool
## Lesson Plan
Do Credit Cards Make You Gain Weight? What is Correlation, and How to Distinguish It from Causation:
This lesson introduces the students to the concepts of correlation and causation, and the difference between the two. The main learning objective is to encourage students to think critically about various possible explanations for a correlation, and to evaluate their plausibility, rather than passively taking presented information on faith. To give students the right tools for such analysis, the lesson covers most common reasons behind a correlation, and different possible types of causation.
Type: Lesson Plan
## Perspectives Video: Experts
Jumping Robots and Quadratics:
Jump to it and learn more about how quadratic equations are used in robot navigation problem solving!
Type: Perspectives Video: Expert
Mathematically Exploring the Wakulla Caves:
The tide is high! How can we statistically prove there is a relationship between the tides on the Gulf Coast and in a fresh water spring 20 miles from each other?
Download the CPALMS Perspectives video student note taking guide.
Type: Perspectives Video: Expert
MicroGravity Sensors & Statistics:
Statistical analysis played an essential role in using microgravity sensors to determine location of caves in Wakulla County.
Download the CPALMS Perspectives video student note taking guide.
Type: Perspectives Video: Expert
Problem Solving with Project Constraints:
It's important to stay inside the lines of your project constraints to finish in time and under budget. This NASA systems engineer explains how constraints can actually promote creativity and help him solve problems!
Type: Perspectives Video: Expert
## Perspectives Video: Professional/Enthusiasts
Base 16 Notation in Computing:
Listen in as a computing enthusiast describes how hexadecimal notation is used to express big numbers in just a little space.
Download the CPALMS Perspectives video student note taking guide.
Type: Perspectives Video: Professional/Enthusiast
Unit Conversions:
Get fired up as you learn more about ceramic glaze recipes and mathematical units.
Type: Perspectives Video: Professional/Enthusiast
Making Candy: Illuminating Exponential Growth:
No need to sugar coat it: making candy involves math and muscles. Learn how light refraction and exponential growth help make candy colors just right!
Type: Perspectives Video: Professional/Enthusiast
Correlation and Causation in a Scientific Study:
Watching this video will cause your critical thinking skills to improve. You might also have a great day, but that's just correlation.
Download the CPALMS Perspectives video student note taking guide.
Type: Perspectives Video: Professional/Enthusiast
## Problem-Solving Tasks
Quadrupling Leads to Halving:
Students explore the structure of the operation s/(vn). This question provides students with an opportunity to see expressions as constructed out of a sequence of operations: first taking the square root of n, then dividing the result of that operation into s.
Type: Problem-Solving Task
Speed Trap:
The purpose of this task is to allow students to demonstrate an ability to construct boxplots and to use boxplots as the basis for comparing distributions.
Type: Problem-Solving Task
Musical Preferences:
This problem solving task asks students to make deductions about what kind of music students like by examining a table with data.
Type: Problem-Solving Task
Haircut Costs:
This problem could be used as an introductory lesson to introduce group comparisons and to engage students in a question they may find amusing and interesting.
Type: Problem-Solving Task
Golf and Divorce:
This is a simple task addressing the distinction between correlation and causation. Students are given information indicating a correlation between two variables, and are asked to reason out whether or not a causation can be inferred.
Type: Problem-Solving Task
Texting and Grades II:
The purpose of this task is to assess ability to interpret the slope and intercept of the least squares regression line in context.
Type: Problem-Solving Task
Coffee and Crime:
This problem solving task asks students to examine the relationship between shops and crimes by using a correlation coefficient.
Type: Problem-Solving Task
Random Walk III:
The task provides a context to calculate discrete probabilities and represent them on a bar graph.
Type: Problem-Solving Task
Algae Blooms:
In this example, students are asked to write a function describing the population growth of algae. It is implied that this is exponential growth.
Type: Problem-Solving Task
As the Wheel Turns:
In this task, students use trigonometric functions to model the movement of a point around a wheel and, through space. Students also interpret features of graphs in terms of the given real-world context.
Type: Problem-Solving Task
What functions do two graph points determine?:
This problem solving task challenges students to find the linear, exponential and quadratic functions based on two points.
Type: Problem-Solving Task
US Population 1982-1988:
This problem solving task asks students to predict and model US population based on a chart of US population data from 1982 to 1988.
Type: Problem-Solving Task
US Population 1790-1860:
This problem solving task asks students to solve five exponential and linear function problems based on a US population chart for the years 1790-1860.
Type: Problem-Solving Task
Two Points Determine an Exponential Function II:
This problem solving tasks asks students to find the values of points on a graph.
Type: Problem-Solving Task
Two Points Determine an Exponential Function I:
This problem solving task asks students to graph a function and find the values of points on a graph.
Type: Problem-Solving Task
Taxi!:
This simple conceptual problem does not require algebraic manipulation, but requires students to articulate the reasoning behind each statement.
Type: Problem-Solving Task
Sandia Aerial Tram:
The task provides an opportunity for students to engage in detailed analysis of the rate of change of the elevation.
Type: Problem-Solving Task
Rumors:
This problem is an exponential function example that uses the real-world problem of how fast rumors spread.
Type: Problem-Solving Task
Rising Gas Prices - Compounding and Inflation:
The purpose of this task is to give students an opportunity to explore various aspects of exponential models (e.g., distinguishing between constant absolute growth and constant relative growth, solving equations using logarithms, applying compound interest formulas) in the context of a real world problem with ties to developing financial literacy skills. In particular, students are introduced to the idea of inflation of prices of a single commodity, and are given a very brief introduction to the notion of the Consumer Price Index for measuring inflation of a body of goods.
Type: Problem-Solving Task
Newton's Law of Cooling:
The coffee cooling experiment is a popular example of an exponential model with immediate appeal. The model is realistic and provides a good context for students to practice work with exponential equations.
Type: Problem-Solving Task
Linear or exponential?:
This task gives a variation of real-life contexts which could be modeled by a linear or exponential function. The key distinguishing feature between the two is whether the change by equal factors over equal intervals (exponential functions), or by a constant increase per unit interval (linear functions). The task could either be used as an assessment problem on this distinction, or used as an introduction to the differences between these very important classes of functions.
Type: Problem-Solving Task
Linear Functions:
This task requires students to use the fact that on the graph of the linear function h(x) = ax + b, the y-coordinate increases by a when x increases by one. Specific values for a and b were left out intentionally to encourage students to use the above fact as opposed to computing the point of intersection, (p,q), and then computing respective function values to answer the question.
Type: Problem-Solving Task
Solution Sets:
The typical system of equations or inequalities problem gives the system and asks for the graph of the solution. This task turns the problem around. It gives a solution set and asks for the system that corresponds to it. The purpose of this task is to give students a chance to go beyond the typical problem and make the connections between points in the coordinate plane and solutions to inequalities and equations. Students have to focus on what the graph is showing. When you are describing a region, why does the inequality have to go one way or another? When you pick a point that clearly lies in a region, what has to be true about its coordinates so that it satisfies the associated system of inequalities?
Type: Problem-Solving Task
Quinoa Pasta 3:
This task is an example of a mathematical modeling problem (SMP 4) and it also illustrates SMP 1 (Making sense of a problem). Students are only told that there are two ingredients in the pasta and they have a picture of the box. It might even be better to just show the picture of the box, or to bring in the box and ask the students to pose the question themselves. The brand of pasta is quite commonly available at supermarkets or health food stores such as Whole Foods and even at Amazon.com. The box has the nutritional label and a reference to the website where the students can find other information about the ingredients
Type: Problem-Solving Task
Quinoa Pasta 2:
This task has some aspects of a mathematical modeling problem (SMP 4) and it also illustrates SMP 1 (Making sense of a problem). Students are given all the relevant information on the nutritional labels, but they have to figure out how to use this information. They have to come up with the idea that they can set up two equations in two unknowns to solve the problem.
Type: Problem-Solving Task
Pairs of Whole Numbers:
This task addresses A-REI.3.6, solving systems of linear equations exactly, and provides a simple example of a system with three equations and three unknown. Two (of many) methods for solving the system are presented. The first takes the given information to make three equations in three unknowns which can then be solved via algebraic manipulation to find the three numbers. The second solution is more clever, creating a single equation in three unknowns from the given information. This equation is then combined with the given information about the sums of pairs of numbers to deduce what the third number is. In reality, this solution is not simpler than the first: rather it sets up a slightly different set of equations which can be readily solved (the key being to take the sum of the three equations in the first solution). It provides a good opportunity for the instructor to show different methods for solving the same system of linear equations.
Type: Problem-Solving Task
How does the solution change?:
The purpose of this task is to continue a crucial strand of algebraic reasoning begun at the middle school level (e.g, 6.EE.5). By asking students to reason about solutions without explicitly solving them, we get at the heart of understanding what an equation is and what it means for a number to be a solution to an equation. The equations are intentionally very simple; the point of the task is not to test technique in solving equations, but to encourage students to reason about them.
Type: Problem-Solving Task
Do two points always determine an exponential function?:
This problem complements the problem "Do two points always determine a linear function?'' There are two constraints on a pair of points R1 and R2 if there is an exponential function f(x) = ae^bx whose graph contains R1 and R2.
Type: Problem-Solving Task
Do two points always determine a linear function?:
This problem complements the problem "Do two points always determine a linear function?'' There are two constraints on a pair of points R1 and R2 if there is an exponential function f(x) = ae^bx whose graph contains R1 and R2. First, the y-coordinates of R1 and R2 cannot have different signs, that is it cannot be that one is positive while the other is negative. This is because the function g(x) = ex takes only positive values. Consequently, f(x) = ae^bx cannot take both positive and negative values. Furthermore, the only way aebx can be zero is if a = 0 and then the function is linear rather than exponential. As long as the y-coordinates of R1 and R2 are non-zero and have the same sign, there is a unique exponential function f(x) = ae^bx whose graph contains R1 and R2.
Type: Problem-Solving Task
Comparing Exponentials:
This task gives students an opportunity to work with exponential functions in a real world context involving continuously compounded interest. They will study how the base of the exponential function impacts its growth rate and use logarithms to solve exponential equations.
Type: Problem-Solving Task
Carbon 14 Dating, Variation 2:
This exploratory task requires the student to use properties of exponential functions in order to estimate how much Carbon 14 remains in a preserved plant after different amounts of time.
Type: Problem-Solving Task
Carbon 14 Dating in Practice II:
This problem introduces the method used by scientists to date certain organic material. It is based not on the amount of the Carbon 14 isotope remaining in the sample but rather on the ratio of Carbon 14 to Carbon 12. This ratio decreases, hypothetically, at a constant exponential rate as soon as the organic material has ceased to absorb Carbon 14, that is, as soon as it dies.
Type: Problem-Solving Task
Carbon 14 Dating in Practice I:
In the task "Carbon 14 Dating" the amount of Carbon 14 in a preserved plant is studied as time passes after the plant has died. In practice, however, scientists wish to determine when the plant died, and as this task shows, that is not possible with a simple measurement of the amount of Carbon 14 remaining in the preserved plant.
Type: Problem-Solving Task
Basketball Rebounds:
This task involves a fairly straightforward decaying exponential. Filling out the table and developing the general formula is complicated only by the need to work with a fraction that requires decisions about rounding and precision.
Type: Problem-Solving Task
A Saturating Exponential:
This task provides an interesting context to ask students to estimate values in an exponential function using a graph.
Type: Problem-Solving Task
Population and Food Supply:
In this task students use verbal descriptions to construct and compare linear and exponential functions and to find where the two functions intersect (F-LE.2, F-LE.3, A-REI.11).
Type: Problem-Solving Task
Fishing Adventures 3:
This task is the last in a series of three tasks that use inequalities in the same context at increasing complexity in 6th grade, 7th grade and in HS algebra. Students write and solve inequalities, and represent the solutions graphically.
Type: Problem-Solving Task
Braking Distance:
This task provides an exploration of a quadratic equation by descriptive, numerical, graphical, and algebraic techniques. Based on its real-world applicability, teachers could use the task as a way to introduce and motivate algebraic techniques like completing the square, en route to a derivation of the quadratic formula.
Type: Problem-Solving Task
Cash Box:
The given solutions for this task involve the creation and solving of a system of two equations and two unknowns, with the caveat that the context of the problem implies that we are interested only in non-negative integer solutions. Indeed, in the first solution, we must also restrict our attention to the case that one of the variables is further even. This aspect of the task is illustrative of mathematical practice standard MP4 (Model with mathematics), and crucial as the system has an integer solution for both situations, that is, whether or not we include the dollar on the floor in the cash box or not.
Type: Problem-Solving Task
Accurately weighing pennies II:
This task is a somewhat more complicated version of "Accurately weighing pennies I'' as a third equation is needed in order to solve part (a) explicitly. Instead, students have to combine the algebraic techniques with some additional problem-solving (numerical reasoning, informed guess-and-check, etc.) Part (b) is new to this task, as with only two types of pennies the weight of the collection determines how many pennies of each type are in the collection. This is no longer the case with three different weights but in this particular case, a collection of 50 is too small to show any ambiguity. This is part of the reason for part (c) of the question where the weight alone no longer determines which type of pennies are in the roll. This shows how important levels of accuracy in measurement are as the answer to part (b) could be different if we were to measure on a scale which is only accurate to the nearest tenth of a gram instead of to the nearest hundredth of a gram.
Type: Problem-Solving Task
A Cubic Identity:
Solving this problem with algebra requires factoring a particular cubic equation (the difference of two cubes) as well as a quadratic equation. An alternative solution using prime numbers and arithmetic is presented.
Type: Problem-Solving Task
Two Squares are Equal:
This classroom task is meant to elicit a variety of different methods of solving a quadratic equation (A-REI.4). Some are straightforward (for example, expanding the square on the right and rearranging the equation so that we can use the quadratic formula); some are simple but clever (reasoning from the fact that x and (2x - 9) have the same square); some use tools (using a graphing calculator to graph the functions f(x) = x^2 and g(x) = (2x-90)^2 and looking for values of x at which the two functions intersect). Some solution methods will work on an arbitrary quadratic equation, while others (such as the last three) may have difficulty or fail if the quadratic equation is not given in a particular form, or if the solutions are not rational numbers.
Type: Problem-Solving Task
Accurately weighing pennies I:
This problem involves solving a system of algebraic equations from a context: depending how the problem is interpreted, there may be one equation or two. The main work in parts (a) and (b) is in setting up the equation(s) appropriately. Question (c) is more subtle and it requires thinking carefully about the accuracy available in a particular measurement (weight). The first two parts of this task could be used for instructional or assessment purposes while the third part should strictly be implemented for instructional purposes.
Type: Problem-Solving Task
Same Solutions?:
The purpose of this task is to provide an opportunity for students to reason about equivalence of equations. The instruction to give reasons that do not depend on solving the equation is intended to focus attention on the transformation of equations as a deductive step.
Type: Problem-Solving Task
Exponential growth versus linear growth I:
The purpose of this task it to have students discover how (and how quickly) an exponentially increasing quantity eventually surpasses a linearly increasing quantity. Students' intuitions will probably have them favoring Option A for much longer than is actually the case, especially if they are new to the phenomenon of exponential growth. Teachers might use this surprise as leverage to segue into a more involved task comparing linear and exponential growth.
Type: Problem-Solving Task
Exponential Functions:
This task requires students to use the fact that the value of an exponential function f(x) = a · b^x increases by a multiplicative factor of b when x increases by one. It intentionally omits specific values for c and d in order to encourage students to use this fact instead of computing the point of intersection, (p,q), and then computing function values to answer the question.
Type: Problem-Solving Task
Equal Factors over Equal Intervals:
This problem assumes that students are familiar with the notation x0 and ?x. However, the language "successive quotient" may be new.
Type: Problem-Solving Task
Equal Differences over Equal Intervals 2:
This task assumes that students are familiar with the ?x and ?y notations. Students most likely developed this familiarity in their work with slope.
Type: Problem-Solving Task
Equal Differences over Equal Intervals 1:
An important property of linear functions is that they grow by equal differences over equal intervals. In F.LE Equal Differences over Equal Intervals 1, students prove this for equal intervals of length one unit, and note that in this case the equal differences have the same value as the slope.
Type: Problem-Solving Task
In the Billions and Linear Modeling:
This problem-solving task asks students to examine if linear modeling would be appropriate to describe and predict population growth from select years.
Type: Problem-Solving Task
In the Billions and Exponential Modeling:
This problem-solving task provides students an opportunity to experiment with modeling real data by using population growth rates from the past two centuries.
Type: Problem-Solving Task
Interesting Interest Rates:
This problem-solving task challenges students to write expressions and create a table to calculate how much money can be gained after investing at different banks with different interest rates.
Type: Problem-Solving Task
Illegal Fish:
This problem-solving task asks students to describe exponential growth through a real-world problem involving the illegal introduction of fish into a lake.
Type: Problem-Solving Task
Doubling your money:
This task asks students to write equations to predict how much money will be in a savings account at the end of each year, based on different factors like interest rates.
Type: Problem-Solving Task
Identifying Functions:
This problem-solving emphasizes the expectation that students know linear functions grow by constant differences over equal intervals and exponential functions grow by constant factors over equal intervals.
Type: Problem-Solving Task
Exponential growth versus polynomial growth:
This problem solving task shows that an exponential function takes larger values than a cubic polynomial function provided the input is sufficiently large. This resource also includes standards alignment commentary and annotated solutions.
Type: Problem-Solving Task
Exponential growth versus linear growth II:
This task asks students to calculate exponential functions with a base larger than one.
Type: Problem-Solving Task
Your Father:
This is a simple task touching on two key points of functions. First, there is the idea that not all functions have real numbers as domain and range values. Second, the task addresses the issue of when a function admits an inverse, and the process of "restricting the domain" in order to achieve an invertible function.
Type: Problem-Solving Task
Yam in the Oven:
The purpose of this task is to give students practice interpreting statements using function notation. It can be used as a diagnostic if students seem to be having trouble with function notation, for example mistakenly interpreting f(x) as the product of f and x.
Type: Problem-Solving Task
Which Function?:
The task addresses knowledge related to interpreting forms of functions derived by factoring or completing the square. It requires students to pay special attention to the information provided by the way the equation is represented as well as the sign of the leading coefficient, which is not written out explicitly, and then to connect this information to the important features of the graph.
Type: Problem-Solving Task
Warming and Cooling:
This task is meant to be a straight-forward assessment task of graph reading and interpreting skills. This task helps reinforce the idea that when a variable represents time, t = 0 is chosen as an arbitrary point in time and positive times are interpreted as times that happen after that.
Type: Problem-Solving Task
Using Function Notation I:
This task addresses a common misconception about function notation.
Type: Problem-Solving Task
Throwing Baseballs:
This task could be used for assessment or for practice. It allows students to compare characteristics of two quadratic functions that are each represented differently, one as the graph of a quadratic function and one written out algebraically. Specifically, students are asked to determine which function has the greatest maximum and the greatest non-negative root.
Type: Problem-Solving Task
The Random Walk:
This task requires interpreting a function in a non-standard context. While the domain and range of this function are both numbers, the way in which the function is determined is not via a formula but by a (pre-determined) sequence of coin flips. In addition, the task provides an opportunity to compute some probabilities in a discrete situation. The task could be used to segue the discussion from functions to probability, in particular the early standards in the S-CP domain.
Type: Problem-Solving Task
The Parking Lot:
The purpose of this task is to investigate the meaning of the definition of function in a real-world context where the question of whether there is more than one output for a given input arises naturally. In more advanced courses this task could be used to investigate the question of whether a function has an inverse.
Type: Problem-Solving Task
Domains:
The purpose of this task to help students think about an expression for a function as built up out of simple operations on the variable and understand the domain in terms of values for which each operation is invalid (e.g., dividing by zero or taking the square root of a negative number).
Type: Problem-Solving Task
Cell Phones:
This simple task assesses whether students can interpret function notation. The four parts of the task provide a logical progression of exercises for advancing understanding of function notation and how to interpret it in terms of a given context.
Type: Problem-Solving Task
Average Cost:
This task asks students to find the average, write an equation, find the domain, and create a graph of the cost of producing DVDs.
Type: Problem-Solving Task
Springboard Dive:
The problem presents a context where a quadratic function arises. Careful analysis, including graphing of the function, is closely related to the context. The student will gain valuable experience applying the quadratic formula and the exercise also gives a possible implementation of completing the square.
Type: Problem-Solving Task
Weed Killer:
The principal purpose of the task is to explore a real-world application problem with algebra, working with units and maintaining reasonable levels of accuracy throughout. Students are asked to determine which product will be the most economical to meet the requirements given in the problem.
Type: Problem-Solving Task
The High School Gym:
This task asks students to consider functions in regard to temperatures in a high school gym.
Type: Problem-Solving Task
The Customers:
The purpose of this task is to introduce or reinforce the concept of a function, especially in a context where the function is not given by an explicit algebraic representation. Further, the last part of the task emphasizes the significance of one variable being a function of another variable in an immediately relevant real-life context.
Type: Problem-Solving Task
Telling a Story with Graphs:
In this task students are given graphs of quantities related to weather. The purpose of the task is to show that graphs are more than a collection of coordinate points; they can tell a story about the variables that are involved, and together they can paint a very complete picture of a situation, in this case the weather. Features in one graph, like maximum and minimum points, correspond to features in another graph. For example, on a rainy day, the solar radiation is very low, and the cumulative rainfall graph is increasing with a large slope.
Type: Problem-Solving Task
Random Walk II:
These problems form a bridge between work on functions and work on probability. The task is better suited for instruction than for assessment as it provides students with a non-standard setting in which to interpret the meaning of functions. Students should carry out the process of flipping a coin and modeling this Random Walk in order to develop a sense of the process before analyzing it mathematically.
Type: Problem-Solving Task
Points on a graph:
This task is designed to get at a common student confusion between the independent and dependent variables. This confusion often arises in situations like (b), where students are asked to solve an equation involving a function, and confuse that operation with evaluating the function.
Type: Problem-Solving Task
Pizza Place Promotion:
This tasks asks students to use functions to predict the price of a pizza on a specific day and find which day the pizza would be cheapest according to a promotion.
Type: Problem-Solving Task
Parabolas and Inverse Functions:
This problem is a simple de-contextualized version of F-IF Your Father and F-IF Parking Lot. It also provides a natural context where the absolute value function arises as, in part (b), solving for x in terms of y means taking the square root of x^2 which is |x|.This task assumes students have an understanding of the relationship between functions and equations.
Type: Problem-Solving Task
Oakland Coliseum:
This deceptively simple task asks students to find the domain and range of a function from a given context. The function is linear and if simply looked at from a formulaic point of view, students might find the formula for the line and say that the domain and range are all real numbers.
Type: Problem-Solving Task
Logistic Growth Model, Explicit Version:
This problem introduces a logistic growth model in the concrete settings of estimating the population of the U.S. The model gives a surprisingly accurate estimate and this should be contrasted with linear and exponential models.
Type: Problem-Solving Task
Logistic Growth Model, Abstract Version:
This task is for instructional purposes only and students should already be familiar with some specific examples of logistic growth functions. The goal of this task is to have students appreciate how different constants influence the shape of a graph.
Type: Problem-Solving Task
How Is the Weather?:
This task can be used as a quick assessment to see if students can make sense of a graph in the context of a real world situation. Students also have to pay attention to the scale on the vertical axis to find the correct match. The first and third graphs look very similar at first glance, but the function values are very different since the scales on the vertical axes are very different. The task could also be used to generate a group discussion on interpreting functions given by graphs.
Type: Problem-Solving Task
Equations and Formulas:
In this task, students will use inverse operations to solve the equations for the unknown variable or for the designated variable if there is more than one.
Type: Problem-Solving Task
Writing Constraints:
The purpose of this task is to give students practice writing a constraint equation for a given context. Instruction accompanying this task should introduce the notion of a constraint equation as an equation governing the possible values of the variables in question (i.e., "constraining" said values). In particular, it is worth differentiating the role of constraint equations from more functional equations, e.g., formulas to convert from degrees Celsius to degree Fahrenheit. The task has students interpret the context and choose variables to represent the quantities, which are governed by the constraint equation and the fact that they are non-negative (allowing us to restrict the graphs to points in the first quadrant only).
The four parts are independent and can be used as separate tasks.
Type: Problem-Solving Task
Interpreting the Graph:
The purpose of this task is to help students learn to read information about a function from its graph, by asking them to show the part of the graph that exhibits a certain property of the function. The task could be used to further instruction on understanding functions or as an assessment tool, with the caveat that it requires some amount of creativity to decide how to best illustrate some of the statements.
Type: Problem-Solving Task
Bernardo and Sylvia Play a Game:
This task presents a simple but mathematically interesting game whose solution is a challenging exercise in creating and reasoning with algebraic inequalities. The core of the task involves converting a verbal statement into a mathematical inequality in a context in which the inequality is not obviously presented, and then repeatedly using the inequality to deduce information about the structure of the game.
Type: Problem-Solving Task
Dimes and Quarters:
Students are given a word problem that can be solved by using a pair of linear equations. This task does not actually require that the student solve the system but that they recognize the pairs of linear equations in two variables that would be used to solve the system. This is an important step in the process of solving systems.
Type: Problem-Solving Task
Regular Tessellations of the Plane:
This task examines the ways in which the plane can be covered by regular polygons in a very strict arrangement called a regular tessellation. These tessellations are studied here using algebra, which enters the picture via the formula for the measure of the interior angles of a regular polygon (which should therefore be introduced or reviewed before beginning the task). The goal of the task is to use algebra in order to understand which tessellations of the plane with regular polygons are possible.
Type: Problem-Solving Task
Compounding with a 100% Interest Rate:
This task provides an approximation, and definition, of e, in the context of more and more frequent compounding of interest in a bank account. The approach is computational.
Type: Problem-Solving Task
Building a quadratic function from f(x) = x^2:
This task aims for students to understand the quadratic formula in a geometric way in terms of the graph of a quadratic function.
Type: Problem-Solving Task
Building an explicit quadratic function by composition:
This task is intended for instruction and to motivate "Building a general quadratic function." This task assumes that the students are familiar with the process of completing the square.
Type: Problem-Solving Task
Checking a Calculation of a Decimal Exponent:
In this example, students use properties of rational exponents and other algebraic concepts to compare and verify the relative size of two real numbers that involve decimal exponents.
Type: Problem-Solving Task
A Sum of Functions:
In this example, students are given the graph of two functions and are asked to sketch the graph of the function that is their sum. The intent is that students develop a conceptual understanding of function addition.
Type: Problem-Solving Task
Dinosaur Bones:
The purpose of this task is to illustrate through an absurd example the fact that in real life quantities are reported to a certain level of accuracy, and it does not make sense to treat them as having greater accuracy.
Type: Problem-Solving Task
Bus and Car:
This task operates at two levels. In part it is a simple exploration of the relationship between speed, distance, and time. Part (c) requires understanding of the idea of average speed, and gives an opportunity to address the common confusion between average speed and the average of the speeds for the two segments of the trip.
At a higher level, the task addresses MAFS.912.N-Q.1.3, since realistically neither the car nor the bus is going to travel at exactly the same speed from beginning to end of each segment; there is time traveling through traffic in cities, and even on the autobahn the speed is not constant. Thus students must make judgments about the level of accuracy with which to report the result.
Type: Problem-Solving Task
Accuracy of Carbon 14 Dating I:
This task examines, from a mathematical and statistical point of view, how scientists measure the age of organic materials by measuring the ratio of Carbon 14 to Carbon 12. The focus here is on the statistical nature of such dating.
Type: Problem-Solving Task
Accuracy of Carbon 14 Dating II:
This task examines, from a mathematical and statistical point of view, how scientists measure the age of organic materials by measuring the ratio of Carbon 14 to Carbon 12. The focus here is on the statistical nature of such dating.
Type: Problem-Solving Task
Fuel Efficiency:
The problem requires students to not only convert miles to kilometers and gallons to liters but they also have to deal with the added complication of finding the reciprocal at some point.
Type: Problem-Solving Task
How Much Is a Penny Worth?:
This task asks students to calculate the cost of materials to make a penny, utilizing rates of grams of copper.
Type: Problem-Solving Task
Forms of Exponential Expressions:
There are many different ways to write exponential expressions that describe the same quantity, in this task the amount of a radioactive substance after t years. Depending on what aspect of the context we need to investigate, one expression of the quantity may be more useful than another. This task contrasts the usefulness of four equivalent expressions. Students first have to confirm that the given expressions for the radioactive substance are equivalent. Then they have to explain the significance of each expression in the context of the situation.
Type: Problem-Solving Task
Runner's World:
Students are asked to use units to determine if the given statement is valid.
Type: Problem-Solving Task
Harvesting the Fields:
This is a challenging task, suitable for extended work, and reaching into a deep understanding of units. Students are given a scenario and asked to determine the number of people required to complete the amount of work in the time described. The task requires students to exhibit , Make sense of problems and persevere in solving them. An algebraic solution is possible but complicated; a numerical solution is both simpler and more sophisticated, requiring skilled use of units and quantitative reasoning. Thus the task aligns with either MAFS.912.A-CED.1.1 or MAFS.912.N-Q.1.1, depending on the approach.
Type: Problem-Solving Task
Calculating the Square Root of 2:
This task is intended for instructional purposes so that students can become familiar and confident with using a calculator and understanding what it can and cannot do. This task gives an opportunity to work on the notion of place value (in parts [b] and [c]) and also to understand part of an argument for why the square root of 2 is not a rational number.
Type: Problem-Solving Task
Throwing a Ball:
Students manipulate a given equation to find specified information.
Type: Problem-Solving Task
Paying the Rent:
Students solve problems tracking the balance of a checking account used only to pay rent. This simple conceptual task focuses on what it means for a number to be a solution to an equation, rather than on the process of solving equations.
Type: Problem-Solving Task
Buying a Car:
Students extrapolate the list price of a car given a total amount paid in states with different tax rates. The emphasis in this task is not on complex solution procedures. Rather, the progression of equations, from two that involve different values of the sales tax, to one that involves the sales tax as a parameter, is designed to foster the habit of looking for regularity in solution procedures, so that students don't approach every equation as a new problem but learn to notice familiar types.
Type: Problem-Solving Task
Graphs of Compositions:
This task addresses an important issue about inverse functions. In this case the function f is the inverse of the function g but g is not the inverse of f unless the domain of f is restricted.
Type: Problem-Solving Task
Planes and Wheat:
In this resource, students refer to given information which defines 5 variables in the context of real world government expenses. They are then asked to write equations based upon specific known values for some of the variables. The emphasis is on setting up, rather than solving, the equations.
Type: Problem-Solving Task
Crude Oil and Gas Mileage:
This task asks students to write expressions for various problems involving distance per units of volume.
Type: Problem-Solving Task
Flu on Campus:
The context of this example is the spread of a flu virus on campus and the related sale of tissue boxes sold. Students interpret the composite function and determine values simply by using the tables of values.
Type: Problem-Solving Task
Building a General Quadratic Function:
In this resource, a method of deriving the quadratic formula from a theoretical standpoint is demonstrated. This task is for instructional purposes only and builds on "Building an explicit quadratic function."
Type: Problem-Solving Task
Compounding with a 5% Interest Rate:
This task develops reasoning behind the general formula for balances under continuously compounded interest. While this task itself specifically address the standard (F-BF), building functions from a context, an auxiliary purpose is to introduce and motivate the number e, which plays a significant role in the (F-LE) domain of tasks.
Type: Problem-Solving Task
Profit of a Company:
This task compares the usefulness of different forms of a quadratic expression. Students have to choose which form most easily provides information about the maximum value, the zeros and the vertical intercept of a quadratic expression in the context of a real world situation. Rather than just manipulating one form into the other, students can make sense out of the structure of the expressions.
(From Algebra: Form and Function, McCallum et al., Wiley 2010)
Type: Problem-Solving Task
Increasing or Decreasing? Variation 2:
The purpose of this task is to help students see manipulation of expressions as an activity undertaken for a purpose.
Variation 1 of this task presents a related more complex expression already in the correct form to answer the question.
The expression arises in physics as the reciprocal of the combined resistance of two resistors in parallel. However, the context is not explicitly considered here.
Type: Problem-Solving Task
Downhill:
This task would be especially well-suited for instructional purposes. Students will benefit from a class discussion about the slope, y-intercept, x-intercept, and implications of the restricted domain for interpreting more precisely what the equation is modeling.
Type: Problem-Solving Task
Ice Cream:
This task illustrates the process of rearranging the terms of an expression to reveal different aspects about the quantity it represents, precisely the language being used in standard MAFS.912.A-SSE.2.3. Students are provided with an expression giving the temperature of a container at a time t, and have to use simple inequalities (e.g., that 2t>0 for all t) to reduce the complexity of an expression to a form where bounds on the temperature of a container of ice cream are made apparent.
Type: Problem-Solving Task
Sum of Even and Odd:
Students explore and manipulate expressions based on the following statement:
A function f defined for -a < x="">< a="" is="" even="" if="" f(-x)="f(x)" and="" is="" odd="" if="" f(-x)="-f(x)" when="" -a="">< x="">< a.="" in="" this="" task="" we="" assume="" f="" is="" defined="" on="" such="" an="" interval,="" which="" might="" be="" the="" full="" real="" line="" (i.e.,="" a="">
Type: Problem-Solving Task
Graphs of Quadratic Functions:
Students compare graphs of different quadratic functions, then produce equations of their own to satisfy given conditions.
This exploration can be done in class near the beginning of a unit on graphing parabolas. Students need to be familiar with intercepts, and need to know what the vertex is. It is effective after students have graphed parabolas in vertex form (y=a(x–h)2+k), but have not yet explored graphing other forms.
Type: Problem-Solving Task
Equivalent Expressions:
This is a standard problem phrased in a non-standard way. Rather than asking students to perform an operation, expanding, it expects them to choose the operation for themselves in response to a question about structure. Students must understand the need to transform the factored form of the quadratic expression (a product of sums) into a sum of products in order to easily see a, the coefficient of the x2 term; k, the leading coefficient of the x term; and n, the constant term.
Type: Problem-Solving Task
Radius of a Cylinder:
Students are asked to interpret the effect on the value of an expression given a change in value of one of the variables.
Type: Problem-Solving Task
Mixing Fertilizer:
Students examine and answer questions related to a scenario similar to a "mixture" problem involving two different mixtures of fertilizer. In this example, students determine and then compare expressions that correspond to concentrations of various mixtures. Ultimately, students generalize the problem and verify conclusions using algebraic rather than numerical expressions.
Type: Problem-Solving Task
Mixing Candies:
Students are asked to interpret expressions and equations within the context of the amounts of caramels and truffles in a box of candy.
Type: Problem-Solving Task
Kitchen Floor Tiles:
This problem asks students to consider algebraic expressions calculating the number of floor tiles in given patterns. The purpose of this task is to give students practice in reading, analyzing, and constructing algebraic expressions, attending to the relationship between the form of an expression and the context from which it arises. The context here is intentionally thin; the point is not to provide a practical application to kitchen floors, but to give a framework that imbues the expressions with an external meaning.
Type: Problem-Solving Task
Extending the Definitions of Exponents, Variation 2:
The goal of this task is to develop an understanding of rational exponents (MAFS.912.N-RN.1.1); however, it also raises important issues about distinguishing between linear and exponential behavior (MAFS.912.F-LE.1.1c) and it requires students to create an equation to model a context (MAFS.912.A-CED.1.2).
Type: Problem-Solving Task
Delivery Trucks:
This resource describes a simple scenario which can be represented by the use of variables. Students are asked to examine several variable expressions, interpret their meaning, and describe what quantities they each represent in the given context.
Type: Problem-Solving Task
Traffic Jam:
This resource poses the question, "how many vehicles might be involved in a traffic jam 12 miles long?"
This task, while involving relatively simple arithmetic, promps students to practice modeling (MP4), work with units and conversion (N-Q.1), and develop a new unit (N-Q.2). Students will also consider the appropriate level of accuracy to use in their conclusions (N-Q.3).
Type: Problem-Solving Task
Animal Populations:
In this task students interpret the relative size of variable expressions involving two variables in the context of a real world situation. All given expressions can be interpreted as quantities that one might study when looking at two animal populations.
Type: Problem-Solving Task
Computations with Complex Numbers:
This resource involves simplifying algebraic expressions that involve complex numbers and various algebraic operations.
Type: Problem-Solving Task
Operations with Rational and Irrational Numbers:
This task has students experiment with the operations of addition and multiplication, as they relate to the notions of rationality and irrationality.
Type: Problem-Solving Task
Seeing Dots:
The purpose of this task is to identify the structure in the two algebraic expressions by interpreting them in terms of a geometric context. Students will have likely seen this type of process before, so the principal source of challenge in this task is to encourage a multitude and variety of approaches, both in terms of the geometric argument and in terms of the algebraic manipulation.
Type: Problem-Solving Task
Selling Fuel Oil at a Loss:
The task is a modeling problem which ties in to financial decisions faced routinely by businesses, namely the balance between maintaining inventory and raising short-term capital for investment or re-investment in developing the business.
Type: Problem-Solving Task
Felicia's Drive:
This task provides students the opportunity to make use of units to find the gas needed (). It also requires them to make some sensible approximations (e.g., 2.92 gallons is not a good answer to part (a)) and to recognize that Felicia's situation requires her to round up. Various answers to (a) are possible, depending on how much students think is a safe amount for Felicia to have left in the tank when she arrives at the gas station. The key point is for them to explain their choices. This task provides an opportunity for students to practice MAFS.K12.MP.2.1: Reason abstractly and quantitatively, and MAFS.K12.MP.3.1: Construct viable arguments and critique the reasoning of others.
Type: Problem-Solving Task
Transforming the Graph of a Function:
This problem solving task examines, in a graphical setting, the impact of adding a scalar, multiplying by a scalar, and making a linear substitution of variables on the graph of the function f. This resource also includes standards alignment commentary and annotated solutions.
Type: Problem-Solving Task
Temperature Conversions:
Unit conversion problems provide a rich source of examples both for composition of functions (when several successive conversions are required) and inverses (units can always be converted in either of two directions).
Type: Problem-Solving Task
Susita's Account:
This task asks students to determine a recursive process from a context. Students who study computer programming will make regular use of recursive processes.
Type: Problem-Solving Task
Summer Intern:
This task asks students to use proportions of mass and volume to create ideal brine for saltwater fish tanks. It also asks students to compare graphs.
Type: Problem-Solving Task
Skeleton Tower:
This problem is a quadratic function example. The other tasks in this set illustrate MAFS.912.F.BF.1.1.a in the context of linear, exponential, and rational functions.
Type: Problem-Solving Task
Medieval Archer:
The task addresses the first part of standard MAFS.912.F-BF.2.3: "Identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative)."
Type: Problem-Solving Task
Growing Coffee:
This task is designed to make students think about the meaning of the quantities presented in the context and choose which ones are appropriate for the two different constraints presented. In particular, note that the purpose of the task is to have students generate the constraint equations for each part (though the problem statements avoid using this particular terminology), and not to have students solve said equations. If desired, instructors could also use this task to touch on such solutions by finding and interpreting solutions to the system of equations created in parts (a) and (b).
Type: Problem-Solving Task
Lake Algae:
The purpose of this task is to introduce students to exponential growth. While the context presents a classic example of exponential growth, it approaches it from a non-standard point of view.
Type: Problem-Solving Task
Kimi and Jordan:
In the middle grades, students have lots of experience analyzing and comparing linear functions using graphs, table, symbolic expressions, and verbal descriptions. In this task, students may choose a representation that suits them and then reason from within that representation.
Type: Problem-Solving Task
The Canoe Trip, Variation 2:
The primary purpose of this task is to lead students to a numerical and graphical understanding of the behavior of a rational function near a vertical asymptote, in terms of the expression defining the function.
Type: Problem-Solving Task
The Canoe Trip, Variation 1:
The purpose of this task is to give students practice constructing functions that represent a quantity of interest in a context, and then interpreting features of the function in the light of the context. It can be used as either an assessment or a teaching task.
Type: Problem-Solving Task
Identifying Even and Odd Functions:
This task asks students to determine whether a the set of given functions is odd, even, or neither.
Type: Problem-Solving Task
Calories in a Sports Drink:
This problem involves the meaning of numbers found on labels. When the level of accuracy is not given we need to make assumptions based on how the information is reported. An unexpected surprise awaits in this case, however, as no reasonable interpretation of the level of accuracy makes sense of the information reported on the bottles in parts (b) and (c). Either a miscalculation has been made or the numbers have been rounded in a very odd way.
Type: Problem-Solving Task
## Student Center Activity
Method to Multiplying Polynomials:
This video will demonstrate how to multiply polynomials.
Type: Student Center Activity
## Tutorials
Solving Quadratic Equations Using the Quadratic Formula:
You will learn in this video how to solve Quadratic Equations using the Quadratic Formula.
Type: Tutorial
Learning How to Complete the Square:
You will learn int his video how to solve the Quadratic Equation by Completing the Square.
Type: Tutorial
Example 3: Solving Systems by Elimination:
This video is an example of solving a system of linear equations by elimination where the system has infinite solutions.
Type: Tutorial
Solving Systems of Linear Equations with Elimination Example 1:
This video shows how to solve a system of equations through simple elimination.
Type: Tutorial
Inconsistent Systems of Equations:
This video explains how to identify systems of equations without a solution.
Type: Tutorial
Addition and Subtraction of Polynomials:
This video tutorial shows students: the standard form of a polynomial, how to identify polynomials, how to determine the degree of a polynomial, how to add and subtract polynomials, and how to represent the area of a shape as an addition or subtraction of polynomials.
Type: Tutorial
Example 2: Solving Systems by Elimination:
This video shows how to solve systems of equations by elimination.
Type: Tutorial
Solving Quadratic Equations by Square Roots:
In this video tutorial students will learn how to solve quadratic equations by square roots.
Type: Tutorial
Addition Elimination Example 1:
This video is an introduction to the elimination method of solving a system of equations.
Type: Tutorial
Example 3: Solving Systems by Substitution:
This example demonstrates solving a system of equations algebraically and graphically.
Type: Tutorial
Substitution Method Example 2:
This video demonstrates a system of equations with no solution.
Type: Tutorial
The Substitution Method:
This video shows how to solve a system of equations using the substitution method.
Type: Tutorial
Systems of Equations Word Problems Example 1:
This video demonstrates solving a word problem by creating a system of linear equations that represents the situation and solving them using elimination.
Type: Tutorial
Graphing systems of equations:
In this tutorial, students will learn how to solve and graph a system of equations.
Type: Tutorial
Solving system of equations by graphing:
This tutorial shows students how to solve a system of linear equations by graphing the two equations on the same coordinate plane and identifying the intersection point.
Type: Tutorial
Solving a system of equations by graphing:
This tutorial shows how to solve a system of equations by graphing. Students will see what a no solution system of equations looks like in a graph.
Type: Tutorial
Solving a system of equations using substitution:
This tutorial shows how to solve a system of equations using substitution.
Type: Tutorial
Graph the solution to a system of inequalities.:
This video will demonstrate how to graph the solution to a system of inequalities.
Type: Tutorial
Solving a literal equation:
Students will learn to solve a literal equation.
Type: Tutorial
Solving Percentage Problems with Linear Equations:
Many real world problems involve involve percentages. This lecture shows how algebra is used in solving problems of percent change and profit-and-loss.
Type: Tutorial
Introduction to the Coordinate Plane:
In this video, you will learn about Rene Descartes, and how he bridged the gap between algebra and geometry.
Type: Tutorial
Subtracting Polynomials with Multiple Variables:
This video explains how to subtract polynomials with multiple variables and reinforces how to distribute a negative number.
Type: Tutorial
Squaring a Binomial:
This video covers squaring a binomial with two variables. Students will be given the area of a square.
Type: Tutorial
Dependent and independent variables exercise: graphing the equation:
It's helpful to represent an equation on a graph where we plot at least 2 points to show the relationship between the dependent and independent variables. Watch and we'll show you.
Type: Tutorial
Trolls, tolls, and systems of equations:
This video tutorial discusses how to create a system of equations.
Type: Tutorial
Solving Basic Systems Using the Elimination Method:
This 8 minute video will show step-by-step directions for using the elimination method to solve a system of linear equations.
Type: Tutorial
Constructing an Equations with Two Variables - Yoga Plan:
This video provides a real-world scenario and step-by-step instructions to constructing equations using two variables. Possible follow-up videos include Plotting System of Equations - Yoga Plan, Solving System of Equations with Substitution - Yoga Plan, and Solving System of Equations with Elimination - Yoga Plan.
Type: Tutorial
Example: Evaluating expressions with 2 variables:
Evaluating Expressions with Two Variables
Type: Tutorial
How to evaluate an expression with variables:
Learn how to evaluate an expression with variables using a technique called substitution (or "plugging in").
Type: Tutorial
Multiplying And Dividing With Inequalities:
This video discusses multiplication and division of inequalities with negative numbers to solve the inequality.
Type: Tutorial
What is a variable?:
Our focus here is understanding that a variable is just a letter or symbol (usually a lower case letter) that can represent different values in an expression. We got this. Just watch.
Type: Tutorial
Power of a Power Property:
This tutorial demonstrates how to use the power of a power property with both numerals and variables.
Type: Tutorial
Calculating Mixtures of Solutions:
This lecture shows how algebra is used to solve problems involving mixtures of solutions of different concentrations.
Type: Tutorial
Solving Inconsistent or Dependent Systems:
When solving a system of linear equations in x and y with a single solution, we get a unique pair of values for x and y. But what happens when try to solve a system with no solutions or an infinite number of solutions?
Type: Tutorial
Inconsistent, Dependent, and Independent Systems:
Systems of two linear equations in two variables can have a single solution, no solutions, or an infinite number of solutions. This video gives a great description of inconsistent, dependent, and independent systems. A consistent independent system of equations will have one solution. A consistent dependent system of equations will have infinite number of solutions, and an inconsistent system of equations will have no solution. This tutorial also provides information on how to distinguish a given system of linear equations as inconsistent, independent, or dependent system by looking at the slope and intercept.
Type: Tutorial
Vertical Line Test:
A graph in Cartesian coordinates may represent a function or may only represent a binary relation. The "vertical line test" is a visual way to determine whether or not a graph represents a function.
Type: Tutorial
Solving Systems of Equations by Elimination:
Systems of two equations in x and y can be solved by adding the equations to create a new equation with one variable eliminated. This new equation can then be solved to find the value of the remaining variable. That value is then substituted into either equation to find the value of other variable.
Type: Tutorial
Solving Systems of Equations by Substitution:
A system of two equations in x and y can be solved by rearranging one equation to represent x in terms of y, and "substituting" this expression for x in the other equation. This creates an equation with only y which can then be solved to find y's value. This value can then be substituted into either equation to find the value of x.
Type: Tutorial
Special Products of Binomials:
The video tutorial discusses about two typical polynomial multiplications. First, squaring a binomial and second, product of a sum and difference.
Type: Tutorial
Multiplying Bionomials:
Binomials are the polynomials with two terms. This tutorial will help the students learn about the multiplication of binomials. In multiplication, we need to make sure that each term in the first set of parenthesis multiplies each term in the second set.
Type: Tutorial
Linear Equations in One Variable:
This lesson introduces students to linear equations in one variable, shows how to solve them using addition, subtraction, multiplication, and division properties of equalities, and allows students to determine if a value is a solution, if there are infinitely many solutions, or no solution at all. The site contains an explanation of equations and linear equations, how to solve equations in general, and a strategy for solving linear equations. The lesson also explains contradiction (an equation with no solution) and identity (an equation with infinite solutions). There are five practice problems at the end for students to test their knowledge with links to answers and explanations of how those answers were found. Additional resources are also referenced.
Type: Tutorial
Linear Inequalities:
Upon completing this lesson, the student should be able to use the addition, subtraction, multiplication, and division properties of equality to solve linear inequalities, write the answer to an inequality using interval notation and draw a graph to give a visual answer to an inequality problem.
The lesson begins with explanations of inequality signs and interval notation and then moves on to demonstrate addition/subtraction and multiplication/division properties of equality. The site demonstrates a strategy for solving linear inequalities and presents three problems for students to practice what they have learned.
There is also a link to a previous tutorial which covers solving linear equations of one variable for students who need the review.
Type: Tutorial
## Video/Audio/Animations
What is a Function?:
This video will demonstrate how to determine what is and is not a function.
Type: Video/Audio/Animation
Solving Quadratic Equations using Square Roots:
This video will demonstrate how to solve a quadratic equation using square roots.
Type: Video/Audio/Animation
Relations and Functions:
This video demonstrates how to determine if a relation is a function and how to identify the domain.
Type: Video/Audio/Animation
Real-Valued Functions of a Real Variable:
Although the domain and codomain of functions can consist of any type of objects, the most common functions encountered in Algebra are real-valued functions of a real variable, whose domain and codomain are the set of real numbers, R.
Type: Video/Audio/Animation
Roots and Unit Fraction Exponents:
Exponents are not only integers. They can also be fractions. Using the rules of exponents, we can see why a number raised to the power " one over n" is equivalent to the nth root of that number.
Type: Video/Audio/Animation
Rational Exponents:
Exponents are not only integers and unit fractions. An exponent can be any rational number expressed as the quotient of two integers.
Type: Video/Audio/Animation
Simplifying Radical Expressions:
Radical expressions can often be simplified by moving factors which are perfect roots out from under the radical sign.
Type: Video/Audio/Animation
Solving Mixture Problems with Linear Equations:
Mixture problems can involve mixtures of things other than liquids. This video shows how Algebra can be used to solve problems involving mixtures of different types of items.
Type: Video/Audio/Animation
Using Systems of Equations Versus One Equation:
When should a system of equations with multiple variables be used to solve an Algebra problem, instead of using a single equation with a single variable?
Type: Video/Audio/Animation
Systems of Linear Equations in Two Variables:
The points of intersection of two graphs represent common solutions to both equations. Finding these intersection points is an important tool in analyzing physical and mathematical systems.
Type: Video/Audio/Animation
Why the Elimination Method Works:
This chapter presents a new look at the logic behind adding equations- the essential technique used when solving systems of equations by elimination.
Type: Video/Audio/Animation
Domain and Range of Binary Relations:
Two sets which are often of primary interest when studying binary relations are the domain and range of the relation.
Type: Video/Audio/Animation
Point-Slope Form:
The point-slope form of the equation for a line can describe any non-vertical line in the Cartesian plane, given the slope and the coordinates of a single point which lies on the line.
Type: Video/Audio/Animation
Two Point Form:
The two point form of the equation for a line can describe any non-vertical line in the Cartesian plane, given the coordinates of two points which lie on the line.
Type: Video/Audio/Animation
Linear Equations in the Real World:
Linear equations can be used to solve many types of real-word problems. In this episode, the water depth of a pool is shown to be a linear function of time and an equation is developed to model its behavior. Unfortunately, ace Algebra student A. V. Geekman ends up in hot water anyway.
Type: Video/Audio/Animation
Solving Literal Equations:
Literal equations are formulas for calculating the value of one unknown quantity from one or more known quantities. Variables in the formula are replaced by the actual or 'literal' values corresponding to a specific instance of the relationship.
Type: Video/Audio/Animation
Example of Solving for a Variable - Khan Academy:
This video takes a look at rearranging a formula to highlight a quantity of interest.
Type: Video/Audio/Animation
Basic Linear Function:
This video demonstrates writing a function that represents a real-life scenario.
Type: Video/Audio/Animation
Quadratic Functions 2:
This video gives a more in-depth look at graphing quadratic functions than previously offered in Quadratic Functions 1.
Type: Video/Audio/Animation
MIT BLOSSOMS - Fabulous Fractals and Difference Equations :
This learning video introduces students to the world of Fractal Geometry through the use of difference equations. As a prerequisite to this lesson, students would need two years of high school algebra (comfort with single variable equations) and motivation to learn basic complex arithmetic. Ms. Zager has included a complete introductory tutorial on complex arithmetic with homework assignments downloadable here. Also downloadable are some supplemental challenge problems. Time required to complete the core lesson is approximately one hour, and materials needed include a blackboard/whiteboard as well as space for students to work in small groups. During the in-class portions of this interactive lesson, students will brainstorm on the outcome of the chaos game and practice calculating trajectories of difference equations.
Type: Video/Audio/Animation
Graphing Lines 1:
Khan Academy video tutorial on graphing linear equations: "Algebra: Graphing Lines 1"
Type: Video/Audio/Animation
Fitting a Line to Data:
Khan Academy tutorial video that demonstrates with real-world data the use of Excel spreadsheet to fit a line to data and make predictions using that line.
Type: Video/Audio/Animation
Averages:
This Khan Academy video tutorial introduces averages and algebra problems involving averages.
Type: Video/Audio/Animation
## Virtual Manipulatives
Adding and Subtracting Polynomials:
This resource will assess students' understanding of addition and subtraction of polynomials.
Type: Virtual Manipulative
Solving Quadratics By Taking The Square Root:
This resource can be used to assess students' understanding of solving quadratic equation by taking the square root. A great resource to view prior to this is "Solving quadratic equations by square root' by Khan Academy.
Type: Virtual Manipulative
Slope Slider:
In this activity, students adjust slider bars which adjust the coefficients and constants of a linear function and examine how their changes affect the graph. The equation of the line can be in slope-intercept form or standard form. This activity allows students to explore linear equations, slopes, and y-intercepts and their visual representation on a graph. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Linear Function Machine:
In this activity, students plug values into the independent variable to see what the output is for that function. Then based on that information, they have to determine the coefficient (slope) and constant(y-intercept) for the linear function. This activity allows students to explore linear functions and what input values are useful in determining the linear function rule. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the Java applet.
Type: Virtual Manipulative
Graphing Lines:
Allows students access to a Cartesian Coordinate System where linear equations can be graphed and details of the line and the slope can be observed.
Type: Virtual Manipulative
Box Plot:
In this activity, students use preset data or enter in their own data to be represented in a box plot. This activity allows students to explore single as well as side-by-side box plots of different data. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the Java applet.
Type: Virtual Manipulative
Data Flyer:
Using this virtual manipulative, students are able to graph a function and a set of ordered pairs on the same coordinate plane. The constants, coefficients, and exponents can be adjusted using slider bars, so the student can explore the affect on the graph as the function parameters are changed. Students can also examine the deviation of the data from the function. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Function Matching:
This is a graphing tool/activity for students to deepen their understanding of polynomial functions and their corresponding graphs. This tool is to be used in conjunction with a full lesson on graphing polynomial functions; it can be used either before an in depth lesson to prompt students to make inferences and connections between the coefficients in polynomial functions and their corresponding graphs, or as a practice tool after a lesson in graphing the polynomial functions.
Type: Virtual Manipulative
Normal Distribution Interactive Activity:
With this online tool, students adjust the standard deviation and sample size of a normal distribution to see how it will affect a histogram of that distribution. This activity allows students to explore the effect of changing the sample size in an experiment and the effect of changing the standard deviation of a normal distribution. Tabs at the top of the page provide access to supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Function Flyer:
In this online tool, students input a function to create a graph where the constants, coefficients, and exponents can be adjusted by slider bars. This tool allows students to explore graphs of functions and how adjusting the numbers in the function affect the graph. Using tabs at the top of the page you can also access supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Advanced Data Grapher:
This is an online graphing utility that can be used to create box plots, bubble graphs, scatterplots, histograms, and stem-and-leaf plots.
Type: Virtual Manipulative
Number Cruncher:
In this activity, students enter inputs into a function machine. Then, by examining the outputs, they must determine what function the machine is performing. This activity allows students to explore functions and what inputs are most useful for determining the function rule. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Curve Fitting:
With a mouse, students will drag data points (with their error bars) and watch the best-fit polynomial curve form instantly. Students can choose the type of fit: linear, quadratic, cubic, or quartic. Best fit or adjustable fit can be displayed.
Type: Virtual Manipulative
Equation Grapher:
This interactive simulation investigates graphing linear and quadratic equations. Users are given the ability to define and change the coefficients and constants in order to observe resulting changes in the graph(s).
Type: Virtual Manipulative
Line of Best Fit:
This manipulative allows the user to enter multiple coordinates on a grid, estimate a line of best fit, and then determine the equation for a line of best fit.
Type: Virtual Manipulative
Histogram Tool:
This virtual manipulative histogram tool can aid in analyzing the distribution of a dataset. It has 6 preset datasets and a function to add your own data for analysis.
Type: Virtual Manipulative
Multi Bar Graph:
This activity allows the user to graph data sets in multiple bar graphs. The color, thickness, and scale of the graph are adjustable which may produce graphs that are misleading. Users may input their own data, or use or alter pre-made data sets. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
Histogram:
In this activity, students can create and view a histogram using existing data sets or original data entered. Students can adjust the interval size using a slider bar, and they can also adjust the other scales on the graph. This activity allows students to explore histograms as a way to represent data as well as the concepts of mean, standard deviation, and scale. This activity includes supplemental materials, including background information about the topics covered, a description of how to use the application, and exploration questions for use with the java applet.
Type: Virtual Manipulative
## Parent Resources
Vetted resources caregivers can use to help students learn the concepts and skills in this course. | 20,462 | 104,189 | {"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} | 2.8125 | 3 | CC-MAIN-2024-10 | latest | en | 0.920746 |
https://blog.purplepixie.org/tag/c/ | 1,642,964,600,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.5/warc/CC-MAIN-20220123172206-20220123202206-00423.warc.gz | 195,314,462 | 18,047 | # Working With Big Numbers
Recently a friend asked me a some questions:
“97 raised to the power of 242 has equalled infinity on every calculator I’ve used, but it’s not infinite just very big. Why do they say infinity? And what is 97^242?”
The answer to the first part is easy; precision (and hence maximum values) are limited. Since this isn’t a basic intro to binary and computing we won’t go into it but just say in the good old days this would have resulted in the number simply wrapping. Modern devices and systems detect this overflow and show a special case result, usually Infinity/Inf or sometimes Not-a-Number/NaN.
97^242 in R
97^242 in Matlab
Above are two common tools (R and Matlab) both running on 64-bit Linux and overflowing for 97^242. The difference in overflow can be seen in that the OSX Calculator can handle 97^71 but overflows at 96^72, whereas both Matlab and R will handle 97^166 but not 97^156.
Ok well Google has a calculator function so maybe we can just ask it for 97^242?
Alas, no. But maybe if we trick it with 97+242 to get it’s calculator up and then use that?
Nope. So how do we go about trying to calculate 97^242?
There are approaches we can use to estimate it (one of the most promising being looking for differences between powers of 100 and 97 then extrapolating, maybe something for another blog post) but we want an exact answer. As shown by Matlab/R and common logic built-in types are just too small and will overflow.
The solution comes, as with so much in life, from GNU in the GNU Multiple Precision Arithmetic Library aka GNU MP Bignum.
Once installed (yum install gmp-devel on Fedora) it’s just a case of hacking together some C to calculate the result (note the below isn’t supposed to be efficient, more transparent):
Build this with: gcc bignum.c -lgmp -o bignum
And voila:
Which for those who can’t be bothered to count digits is rounded as 6.29E+480.
Though of course after all this loading of libraries and writing of C it turns out that although Google couldn’t answer it, naturally Wolfram|Alpha could:
Wolfram|Alpha Calculates 97^242 with Ease!
# Variable Length Arguments in C++, Java, and PHP
Normally in software development we define methods with a given number of parameters (and their type in some languages). Quite often however we want to be able to deal with different numbers of arguments and there are two widely used approaches; different methods and default parameters.
Different methods relies on the concept that the call to function is matched not just on the name of the method but also the count (and type if applicable) of the parameters. So if we wanted a method that could accept one or two integers in C++ we could define two methods:
So if we called SomeFunction(1) the first would be used, SomeFunction(1,2) would use the second.
Default parameters allows us to define some of the parameters as optional and their default values if not passed so the definition:
Would accept one or two integer parameters. SomeMethod(1,2) would have a=1 and b=2 whereas SomeMethod(1) would use the default value and so b=0.
This is all very good and highly useful in a variety of situations but suppose you wanted to handle any number of parameters, from a very small set (or zero) to a large number. Using either of these techniques would require a lot of additional coding, creating a method for each length or the longest set of default parameters imaginable.
This is where the concept of variable length arguments for a method comes in; we want to be able to define a method and accept an arbitrary number of arguments which it can process (please note in most if not all cases the best option for this would be to pass something like a Vector in C++ or an array in PHP, but best practice is not the point of the exercise).
Let’s consider a problem.
We want to have a LineShape function. This function takes a series of Points (a simple class just containing an X and Y coordinate). In a proper system it would then start with the first point and draw a line to each consecutive one but for our example we just want it to print a list of the points it will draw to/from in order.
This could be two points (a single line) or a complex shape of an undetermined total number of points (again note the caveat above that a Vector/List/Array would be the best and safest way to do this in TRW).
So for our implementation we need:
• A simple Point class
• A method (LineShape) that takes an arbitrary number of Points and prints out the coordinates
• Code to create a set of Points and pass them to LineShape
How to do this varies from language to language and, as you might assume, it’s hardest and most dangerous in C++ (because of it’s lack of type safety), slightly easier in Java and PHP (Java because of it’s high type safety and PHP because of it’s lack of any enforced typing).
Variable Length Arguments in C++
To implement in C++ we make use of the va_ functionality provided in stdarg.h. The function is defined as taking the number of parameters passed (int) and then the parameters themselves represented by “…”.
We read the parameter count and then iterate through reading each in turn with va_arg and specifying the type to be used. Note in C++ you must specify the number of parameters being passed when calling the method.
There you have it in C++ (well actually using C libraries); but don’t do it (see above).
Variable Length Arguments in Java
In Java it’s a lot easier as the functionality is built-in to the language. Additionally you don’t need to pass the number of parameters and also the type is determined for the entire set of parameters (in our case Point).
Note that in order for Point to be instantiated as non-static it must be in a seperate file (Point.java).
So our Point class is:
And the main Var.java contains:
So in Java we just need to declare a method with Type… name and then iterate through the array in a for loop.
Variable Length Arguments in PHP
PHP isn’t quite as built-in as Java (an actual language construct) but PHP natively provides functions to support variable length parameters to methods such as func_num_args (number of arguments) and func_get_args (arguments as an array). | 1,383 | 6,231 | {"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} | 3.59375 | 4 | CC-MAIN-2022-05 | latest | en | 0.923514 |
https://stacks.math.columbia.edu/tag/0GT5 | 1,696,453,463,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511406.34/warc/CC-MAIN-20231004184208-20231004214208-00711.warc.gz | 581,594,718 | 6,127 | Lemma 20.55.3. In Situation 20.55.2 let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. The following are equivalent
1. the subsheaf $\mathcal{F}[\mathcal{I}] \subset \mathcal{F}$ of sections annihilated by $\mathcal{I}$ is zero,
2. the subsheaf $\mathcal{F}[\mathcal{I}^ n]$ is zero for all $n \geq 1$,
3. the multiplication map $\mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{F}$ is injective,
4. for every open $U \subset X$ such that $\mathcal{I}|_ U = \mathcal{O}_ U \cdot f$ for some $f \in \mathcal{I}(U)$ the map $f : \mathcal{F}|_ U \to \mathcal{F}|_ U$ is injective,
5. for every $x \in X$ and generator $f$ of the ideal $\mathcal{I}_ x \subset \mathcal{O}_{X, x}$ the element $f$ is a nonzerodivisor on the stalk $\mathcal{F}_ x$.
Proof. Omitted. $\square$
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Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```atomic mass unit = 1.66044E-27 kilogram (mass) ``` Related Measurements: Try converting from "AMU" to as (Roman as), assay ton, Babylonian shekel, catty (Chinese catty), dinar (Arabian dinar), dram troy (troy dram), English carat, funt (Russian funt), gram, Greek obolos, Israeli shekel mass, kwan (Japanese kwan), pennyweight troy (troy pennyweight), pfund (German pfund), quintal, scruple (apothecary scruple), short hundredweight (avoirdupois short hundredweight), Spanish quintal, UK quintal (British quintal), uncia (Roman uncia), or any combination of units which equate to "mass" and represent mass. Sample Conversions: AMU = 1.39E-28 arroba (Mexican arroba), 7.63E-27 bes (Roman bes), 8.30E-24 carat troy (troy carat), 2.75E-27 catty (Chinese catty), 4.88E-30 cotton bale Egypt, 1.83E-23 crith, 4.27E-25 dram ap (apothecary dram), 4.07E-27 funt (Russian funt), 1.69E-25 hyl, 2.77E-27 kin (Japanese kin), 3.87E-27 mina (Greek mina), .99135481 neutron mass (neutron rest mass), 5.86E-26 oz (ounce), 2.92E-24 Roman obolus, 4.07E-29 Roman talent, 1.46E-24 scrupulum (Roman scrupulum), 1.83E-30 short ton, 2.61E-28 stone, 3.66E-29 UK quintal (British quintal), 6.10E-26 uncia (Roman uncia).
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1. ## Indefinite Integration: Sub
The resale value of a certain industrial machine decreases at a rate that depends on its age. When the machine is t years old, the rate at which its value is changing is $-880e^{- \frac{t}{5}}$ dollars per year. If the machine was originally worth $5,200, how much will it be worth when it is 10 years old? Here's what I did: Let $u = - \frac{t}{5} $ Let $dx = 5 du $ $= - \int 880e^u \times 5 du$ $= - 4400 e^{- \frac{t}{5}} + c$ $5200 = -4400e^{0} + c$ $5200 + 4400 = c$ $c = 9600$ $f(10) = -4400e^{-2} + 9600$ $f(10) = 9004.52$ ...and I'm sure this is wrong since the value has to be less than the initial value and my answer is greater than that value... Please show and explain to me how to get the right answer! 2. value of the machine in 10 years ... $5200 - \int_0^{10} 800e^{-\frac{t}{5}} \, dt = 1741.34$ 3. ... the answer is$1,395.48 and I don't know how to get this
4. Originally Posted by skeeter
value of the machine in 10 years ...
$5200 - \int_0^{10} 800e^{-\frac{t}{5}} \, dt = 1741.34$
my mistake ... rate coefficient should be 880 instead of 800.
$5200 - \int_0^{10} 880e^{-\frac{t}{5}} \, dt = 1395.48$ | 423 | 1,199 | {"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": 13, "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} | 3.8125 | 4 | CC-MAIN-2016-50 | longest | en | 0.835798 |
https://www.quizover.com/course/section/signal-to-noise-ratio-logarithms-by-openstax | 1,529,955,983,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868876.81/warc/CC-MAIN-20180625185510-20180625205510-00397.warc.gz | 872,253,275 | 24,817 | # 0.9 Logarithms
Page 1 / 4
This module is part of a collection of modules intended for use by preengineering students enrolled in MATH 108 at the University of Texas at El Paso. This module addresses some applications of logarithms in several fields of engineering. Examples are presented.
## Introduction
This module is intended to present some areas of engineering in which logarithms are used. By reading the material and solving the associated problems, you will learn about some important applications of logarithms in engineering.
## Decibels
The decibel ( dB ) is a logarithmic unit that indicates the ratio of a physical quantity relative to a specified or implied reference level. The decibel is used for a wide variety of measurements in science and engineering, most prominently in acoustics, electronics, communications, radar, sonar and control systems.
Decibels are frequently used as a means to express the power ratio for physical systems. It is computed by multiplying the factor 10 by the base 10 logarithm of the ratio of the quantities under consideration. Equation (1) shows the computation that is used to express the ratio of two powers using decibels
${L}_{\text{DB}}=\text{10}{\text{log}}_{\text{10}}\left(\frac{{P}_{2}}{{P}_{1}}\right)$
Gain of an Amplifier: We will begin our discussion of decibels with an application in the field of electronics. An amplifier is an electronic device that is capable of boosting the power present in an input signal to produce an output signal with more power. It can be thought of as a black box as shown in Figure 1.
In practical cases, the ratio of the power in the output signal to the power in the input signal is a positive quantity whose value is greater than unity. The decibel measurement of this ratio of power is often called the gain of the amplifier and is given as
$\text{Gain}=\text{10}{\text{log}}_{\text{10}}\left(\frac{{P}_{\text{output}}}{{P}_{\text{input}}}\right)\text{dB}$
Question: An electronic signal is passed through an amplifier. Suppose that the power present in the signal at the input to the amplifier is 10 W. The power present in the signal at the output of the amplifier is 20 W. Express the gain of the amplifier in decibels.
We can use equation (2) to easily express the gain of the amplifier in terms of decibels
$\text{Gain}=\text{10}{\text{log}}_{\text{10}}\left(\frac{\text{20}W}{\text{10}W}\right)=\text{10}{\text{log}}_{\text{10}}\left(2\right)=3\text{.}\text{01}\text{dB}\approx 3\text{dB}$
## Signal to noise ratio
Electrical signals are often corrupted by a random phenomenon known as noise when they are transmitted from one point to another . Because it is impossible to know the exact value of the noise at any point in time, it is often becomes difficult to extract the orignal signal at the receiver without the application of some form of signal processing algorithm such as a filter . The situation is depicted in Figure 2.
A common figure of merit of communication systems is the signal-to-noise ratio . Communication systems that are characterized by high signal-to-noise ratios are in general superior to those that are characterized by low signal-to-noise ratios.
By definition the signal-to-noise ratio or SNR is given as the ratio of the power in a signal divided by the power in the noise that is responsible for corrupting the signal. The signal-to-noise ratio can be expressed in decibels as follows
Black market what is the meaning
where there is illegal transactions of goods or services. where there illegal goods or services are sold.
Azhar
Scarcity is the major problem of any economy which is the limited resources an economy has in order to produce its country's unlimited wants/needs. Therefore, scarcity is the key term beyond the study of Economics.
what is supply
Like the demand, the supply demonstrates the quantities that will be sold at a certain price. But unlike the demand, the supply relationship shows an upward slope. This means that the higher the price, the higher the quantity supplied. Producers supply more at a higher
Nazneen
or we can say that By supply we mean various quantatirs of a commodity which a producer will offer for sale at different possible prices over a given period of time
Nazneen
What is scarcity?
having unlimited needs and wants in a world of limited resources
William
what are the indicator of over population
Inefficiency of natural resources
Low PCI .
abhinav
resources are limited and wants are unlimited
Nazneen
limited resources
Nazneen
what product market?
explain demand
what is economics
Murali
Economics is the social science that studies the production, distribution and consumption of goods and services.Economics focuses on the behaviour and interactions of economic agents and how economies work.
Preeti
ok
SAGAR
Pls what is national income?
Ishmael
demand is the number of buyers. as the price of goods is increase, the buyers will not consider to buy the goods, so the demand will decreases
Pung
National Income is the aggregate monetary value of all final goods and services produced in an economy in an year .
abhinav
Why pecularities of land has no economic significance, What the economic bad, effictive demand?
James
Demand is a quantity of goods that a consumer is willing to buy at different prices over a given period of time
Nazneen
Define law of demand
Nazneen
hi
Bless
deadweight loss is the allocative inefficiency.... when the equilibrium for good or services is not achived.
Azhar
hi
Azhar
hi
azaloo
hii too
Jackson
Hello
Micah
how are you all?
Azhar
I am here been sick but here
Amanda
may you get well soon Amanda
Azhar
What are the implications of classical Economics?
Micah
What are the implications of classical Economics school of thought?
Micah
Dead weight loss is an inefficient allocation of resources, especially through taxation or restrictions.
Ezeigwe
Classical Economics had an influenced by ancient economic principle and theory.
Ezeigwe
yec
Yasinta
nice
Amanda
hello please what is localization of industry
price is tantalisingly the only factor determining demand which can be analyze the view
Price is tantalizingly the only factor determining demand which can be analyze the view
Sinit
what is scarcity
the state of being scarce or in short supply; shortage.
Abdirahin
what is meant by an abnormal demand curve?
what is microeconomics
micronomics can be define as that part of Economics that deals with small scale business. e.g. House hold stuff
Amodu
Micro economics is the study of individuals, households and firms' behavior un decision making and allocation of resources
Christian
what is underemployment
It is the situation where the available resources are not used to it optimum
Bless
state the law of diminishing returns
Bless
Bless, the law of diminishing returns state that one point, adding a single worker will result in a decrease of production.
David
it is a situation where by people are employ but work under their potential
Akua
David please go into details kk
Bless
Akua pls when you say people alone? what about facilities?
Bless
Definition of underemployment. 1 : the condition in which people in a labor force are employed at less than full-time or regular jobs or at jobs inadequate with respect to their training or economic needs
Boahen
OK that is good
Set
I am new here
Taimoor
what does law of demand says?
Samuel
with a diagram explain fairy elastic demand
Samuel
comprehensive answer for public finance is the money that a government has available to spend from taxes and borrowing.
what then comprises of public opinion
Raphael
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
Got questions? Join the online conversation and get instant answers! | 2,297 | 10,013 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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} | 4.4375 | 4 | CC-MAIN-2018-26 | longest | en | 0.90543 |
http://www.questionotd.com/2006/05/going-to-orchard.html?showComment=1148577900000 | 1,590,561,409,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392141.7/warc/CC-MAIN-20200527044512-20200527074512-00129.warc.gz | 204,031,581 | 22,103 | ## Thursday, May 25, 2006
### Going to the Orchard
The local orchard has a strange pricing structure. Each bag holds seven apples for which they charge five cents. If you want an eighth apple, you get charged 15 cents. So, if you want to buy 23 apples, you would have to pay 15 cents for three bags and 30 cents for the remaining two for a total of 45 cents.
My question is which costs more: 10 apples, 30 apples or 50 apples?
There's no real puzzle or hard math here, but the answer may surprise you.
1. they all cost the same
neat!
2. Ooh that was weird. They all cost the same: 50 cents.
3. And 70 also cost the same. But there is a big jump for 90.
Likewise, 6, 26, 46, 66, 86, 106, and 126 all cost the same.
When you add 20 apples, you are adding 3 bags, but taking away one single (so +3*5¢ - 15¢), sunless you have a multiple of 7 already, in which case you add 2 bags and 6 singles.
Leave your answer or, if you want to post a question of your own, send me an e-mail. Look in the about section to find my e-mail address. If it's new, I'll post it soon.
Please don't leave spam or 'Awesome blog, come visit mine' messages. I'll delete them soon after.
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https://www.numbersaplenty.com/22666 | 1,642,608,087,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00306.warc.gz | 962,691,835 | 3,320 | Search a number
22666 = 271619
BaseRepresentation
bin101100010001010
31011002111
411202022
51211131
6252534
7123040
oct54212
934074
1022666
1116036
121114a
13a417
148390
156ab1
hex588a
22666 has 8 divisors (see below), whose sum is σ = 38880. Its totient is φ = 9708.
The previous prime is 22651. The next prime is 22669. The reversal of 22666 is 66622.
It is a sphenic number, since it is the product of 3 distinct primes.
22666 is a modest number, since divided by 666 gives 22 as remainder.
It is a plaindrome in base 10, base 12 and base 16.
It is a zygodrome in base 10.
It is not an unprimeable number, because it can be changed into a prime (22669) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 796 + ... + 823.
It is an arithmetic number, because the mean of its divisors is an integer number (4860).
222666 is an apocalyptic number.
22666 is a deficient number, since it is larger than the sum of its proper divisors (16214).
22666 is a wasteful number, since it uses less digits than its factorization.
22666 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1628.
The product of its digits is 864, while the sum is 22.
The square root of 22666 is about 150.5523164883. The cubic root of 22666 is about 28.3003383695.
Subtracting 22666 from its reverse (66622), we obtain a triangular number (43956 = T296).
It can be divided in two parts, 226 and 66, that added together give a palindrome (292).
The spelling of 22666 in words is "twenty-two thousand, six hundred sixty-six".
Divisors: 1 2 7 14 1619 3238 11333 22666 | 500 | 1,673 | {"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} | 3.484375 | 3 | CC-MAIN-2022-05 | latest | en | 0.900283 |
https://denisegaskins.com/2018/08/15/mathematics-is-worthy/?replytocom=281651 | 1,686,357,601,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656869.87/warc/CC-MAIN-20230609233952-20230610023952-00145.warc.gz | 236,184,956 | 32,881 | # Mathematics Is Worthy
“When I began my college education, I still had many doubts about whether I was good enough for mathematics. Then a colleague said the decisive words to me: it is not that I am worthy to occupy myself with mathematics, but rather that mathematics is worthy for one to occupy oneself with.”
Rózsa Péter
Mathematics is beautiful
essay in The Mathematical Intelligencer
### Rózsa Péter and the Curious Students
I would like to win over those who consider mathematics useful, but colourless and dry — a necessary evil…
No other field can offer, to such an extent as mathematics, the joy of discovery, which is perhaps the greatest human joy.
The schoolchildren that I have taught in the past were always attuned to this, and so I have also learned much from them.
It never would have occurred to me, for instance, to talk about the Euclidean Algorithm in a class with twelve-year-old girls, but my students led me to do it.
I would like to recount this lesson.
What we were busy with was that I would name two numbers, and the students would figure out their greatest common divisor. For small numbers this went quickly. Gradually, I named larger and larger numbers so that the students would experience difficulty and would want to have a procedure.
I thought that the procedure would be factorization into primes.
They had still easily figured out the greatest common divisor of 60 and 48: “Twelve!”
But a girl remarked: “Well, that’s just the same as the difference of 60 and 48.”
“That’s a coincidence,” I said and wanted to go on.
But they would not let me go on: “Please name us numbers where it isn’t like that.”
“Fine. 60 and 36 also have 12 as their greatest common divisor, and their difference is 24.”
Another interruption: “Here the difference is twice as big as the greatest common divisor.”
“All right, if this will satisfy all of you, it is in fact no coincidence: the difference of two numbers is always divisible by all their common divisors. And so is their sum.”
Certainly that needed to be stated in full, but having done so, I really did want to move on.
However, I still could not do that.
A girl asked: “Couldn’t they discover a procedure to find the greatest common divisor just from that?”
They certainly could! But that is precisely the basic idea behind the Euclidean Algorithm!
So I abandoned my plan and went the way that my students led me.
— Rózsa Péter
quoted at the MacTutor History of Mathematics Archive
### For Further Exploration
Note: When the video narrator says “Greatest Common Denominator,” he really means “Greatest Common Divisor.”
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if x, y, and n are pos. integers, is (x/y)^n greater than
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if x, y, and n are pos. integers, is (x/y)^n greater than [#permalink]
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27 Jan 2008, 17:13
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if x, y, and n are pos. integers, is (x/y)^n greater than 1000?
1. x=y^3 and n>y
2. x>5y and n>x
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27 Jan 2008, 17:29
I would say B.
1. x=y^3 and n>y
Lets assume n = y.
(y^3 / y ) ^y = y^2y.
Not sufficient.
2. x>5y and n>x
Lets just assume that X = 5y and n = x.
(x/y)^n = (5y / y ) ^ 5y
If y is 1, 5 ^ 5 > 1000.
When x > 5y and n> x, (x/y)^n will certainly be greater than 1000.
#2 is sufficient.
Whats OA ?
_________________
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When you come to the end of your rope, tie a knot and hang on.
Re: DS: x&y [#permalink] 27 Jan 2008, 17:29
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Proportion
0
44
1
Suppose that a^2 varies inversely with b^3. If a = 7 when b = 4, find the value of a^2 when b = 6.
Feb 17, 2022
#1
+23183
+2
Since a2 varies inversely with b3: a2 = k / b3.
a = 7, b = 4: 72 = k / 43 ---> 49 = k / 64 ---> k = 49 / 64
So, the equation is: a2 = 49 / (64 · b3)
When b = 6: a2 = 49 / (64 · 63) ---> a2 = ....
Feb 18, 2022 | 205 | 397 | {"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} | 3.703125 | 4 | CC-MAIN-2022-21 | longest | en | 0.172714 |
https://vaughnclimenhaga.wordpress.com/ | 1,685,490,459,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00181.warc.gz | 674,810,347 | 98,944 | ## Non-uniform specification for piecewise monotonic interval maps
1. Coding spaces for a class of interval maps
The starting point for this post is Franz Hofbauer’s paper “On intrinsic ergodicity of piecewise monotonic transformations with positive entropy”, which is actually two papers; both appeared in Israel J. Math, one in 1979 and one in 1981. He proves existence and uniqueness of the measure of maximal entropy (MME) for a map ${f\colon [0,1]\to [0,1]}$ provided the following are true:
1. there are disjoint intervals ${J_1,\dots, J_d}$ such that ${\bigcup_i J_i = [0,1]}$ and ${f|_{J_i}}$ is continuous and strictly monotonic;
2. the partition into these intervals is generating — that is, ${\bigcup_{m=0}^\infty f^{-m} (\partial J)}$ is dense in ${[0,1]}$, where ${\partial J}$ denotes the set of all endpoints of the intervals ${J_i}$;
3. ${h_{\mathrm{top}}(f)>0}$;
4. ${f}$ is topologically transitive.
Hofbauer’s proof proceeds by modeling ${f}$ using a countable-state Markov shift, and actually gives quite a bit of information even without topological transitivity, including a decomposition of the non-wandering set into transitive pieces, each of which supports at most one MME.
I want to describe an approach to these maps using non-uniform specification. These maps admit a natural symbolic coding on the finite alphabet ${A := \{1,\dots, d\}}$ obtained by tracking which interval ${J_i}$ an orbit lands in at each iterate, and this is the lens through which we will study them.
To make this more precise, suppose that ${f}$ satisfies Condition 1 (piecewise monotonicity), and let ${{\mathcal L}}$ denote the set of all words ${w\in A^* = \bigcup_{n\geq 0} A^n}$ such that the following interval is non-empty:
\displaystyle \begin{aligned} I(w) &:= J_{w_1} \cap f^{-1}(J_{w_2}) \cap \cdots \cap f^{-(|w|-1)}(J_{w_{|w|}}) \\ &= \{ x\in [0,1] : f^{k-1}(x) \in J_{w_k} \text{ for all } 1\leq k\leq |w| \}. \end{aligned} \ \ \ \ \ (1)
The letter “I” can be understood as standing for “interval” or for “initial”, in the sense that an orbit segment ${(x, f(x),\dots, f^{n-1}(x))}$ is coded by the word ${w\in A^n}$ if and only if its initial point lies in ${I(w)}$. In this case we have
$\displaystyle f^n(w) \in F(w) := f^n(I(w)), \ \ \ \ \ (2)$
where the “F” in this notation can be understood as standing for “following” (or “final” if you prefer). Observe that the intervals ${\{ I(w) : w\in {\mathcal L}_n\}}$ are precisely the intervals on which ${f^n}$ is continuous and monotonic, and thus each ${F(w)}$ is an interval as well.
Exercise 1 Suppose that ${f}$ is uniformly expanding on each ${J_i}$ in the sense that there is ${\lambda > 1}$ such that ${|f(x) - f(y)| \geq \lambda |x-y|}$ whenever ${x,y\in J_i}$, and prove that in this case Condition 2 (generating) is satisfied.
Exercise 2 Suppose that ${f}$ satisfies Conditions 1 and 2. Show that ${{\mathcal L}}$ is the language of a shift space ${X \subset A^{\mathbb N}}$, and that there is a semi-conjugacy ${\pi\colon X\to [0,1]}$ defined by
$\displaystyle \pi(z) = \bigcup_{n\geq 1} \overline{I(z_1 \cdots z_n)}.$
(The fact that this intersection is a single point relies on Condition 2.) Show that there is a countable set ${C\subset X}$ such that ${\pi|_{X\setminus C}}$ is 1-1.
Exercise 3 Show that if ${f}$ is uniformly expanding on each ${J_i}$, then Condition 3 (positive entropy) is satisfied, and in particular ${(X,\sigma)}$ has positive topological entropy ${h>0}$.
Under Conditions 1–3, there is a 1-1 correspondence between positive entropy invariant measures for ${f}$ and positive entropy invariant measures for ${(X,\sigma)}$, because the countable set ${C}$ can support only zero entropy measures. Thus to study the MMEs for ${f}$, it suffices to study the MMEs for the shift space ${X}$.
In general one cannot expect this shift space to be Markov. However, it turns out that in many cases it does have the non-uniform specification property described in my last post, which will lead to an alternate proof of intrinsic ergodicity (that is, existence of a unique MME). One of my motivations for writing out the details of this here is the hope that these techniques might eventually be applicable to more general hyperbolic systems with singularities, in particular to billiard systems; piecewise expanding interval maps provide a natural sandbox in which to experiment with the approach before grappling with the difficulties introduced by the presence of a stable direction.
To this end, the goal for this post is to exhibit collections of words ${{\mathcal G},{\mathcal C} \subset {\mathcal L}}$ with the following properties.
• Decomposition: For every ${w\in {\mathcal L}}$, there are ${u\in {\mathcal G}}$ and ${v\in {\mathcal C}}$ such that ${w=uv}$.
• Specification: There is ${\tau\in {\mathbb N}}$ such that for every ${v,w\in {\mathcal G}}$, there is ${u\in {\mathcal L}}$ with ${|u|\leq \tau}$ such that ${vuw\in {\mathcal G}}$.
• Entropy gap: ${\limsup_{n\to\infty} \frac 1n \log \# {\mathcal C}_n < h_{\mathrm{top}}(f)}$.
Once these are proved, the main theorem from the previous post guarantees existence of a unique MME. (The specification property there requires ${|u|=\tau}$, but this was merely to simplify the exposition; with a little more work one can establish the result under the weaker condition here.)
2. Exploring ${\beta}$-transformations
Rather than stating a definitive result immediately, let us go step-by-step — in particular, we will eventually add another condition on ${f}$ (beyond Conditions 1–3) to replace topological transitivity, and the initial definition of ${{\mathcal G}}$ and ${{\mathcal C}}$ will only cover certain examples, with a modification required to handle the general case.
Start by considering the ${\beta}$-transformation ${f(x) = \beta x \pmod 1}$ when ${\beta = \frac 12 (1+\sqrt{5})}$, whose first three iterates are shown in Figure 2. Labeling the branches with the alphabet ${A = \{0,1\}}$, we see that ${I(0) = [0,\frac 1\beta]}$ and ${I(1) = [\frac 1\beta,1]}$; in the figure, these intervals are marked on both the horizontal and vertical axes. Clearly ${F(0) = [0,1] = I(0) \cup I(1)}$, and for this specific choice of ${\beta}$ we have ${f(1) = \beta - 1 = \frac 1\beta}$, so ${F(1) = [0,\frac 1\beta] = I(0)}$.
Looking at the second picture in the figure, which shows the graph of ${f^2}$, we can see that
$\displaystyle F(00) = I(0) \cup I(1), \quad F(01) = I(0), \quad F(10) = I(0) \cup I(1).$
The third picture shows a similar phenomenon for ${f^3}$, with ${F(w) = I(0) \cup I(1)}$ when ${w \in \{000,010,100\}}$, and ${F(w) = I(0)}$ when ${w\in \{001,101\}}$.
Exercise 4 For this particular map ${f}$, show by induction that ${F(w) = I(0)}$ if ${w\in {\mathcal L}}$ ends in a ${1}$, and ${F(w) = I(0) \cup I(1)}$ if ${w\in {\mathcal L}}$ ends in a ${0}$. In particular, we have ${F(w) = F(w_{|w|})}$ for all ${w\in {\mathcal L}}$. Use this to deduce that ${{\mathcal L}}$ consists of precisely those words ${w\in \{0,1\}^*}$ that do not contain ${11}$ as a subword. (Thus ${X}$ is a transitive SFT and has a unique MME.)
The key step in this exercise is the observation that given ${a\in A}$ and ${w\in {\mathcal L}}$ such that ${aw\in {\mathcal L}}$, the definition of the intervals ${I}$ and ${F}$ in (1) and (2) gives
\displaystyle \begin{aligned} I(aw) &= J_a \cap f^{-1} (J_{w_1}) \cap f^{-2}(J_{w_2}) \cap \cdots \cap f^{-|w|}(J_{w_{|w|}}) \\ &= J_a \cap f^{-1}\big(J_{w_1} \cap f^{-1}(J_{w_2} \cap \cdots \cap f^{-(|w|-1)}(J_{w_{|w|}})\big) \\ &= I(a) \cap f^{-1}(I(w)), \end{aligned}
and thus
$\displaystyle F(aw) = f^{|aw|}(I(aw)) = f^{|w|}\big( f(I(a) \cap f^{-1}(I(w)) \big) = f^{|w|}(F(a) \cap I(w)).$
In particular, if ${F(a) \supset I(w)}$, then we have ${F(a) \cap I(w) = I(w)}$, and thus
$\displaystyle F(aw) = f^{|w|}(I(w)) = F(w).$
A similar argument replacing ${a}$ with a word ${v\in {\mathcal L}}$ (and the first iterate ${f}$ with ${f^{|v|}}$) gives the following lemma, whose detailed proof is left as an exercise.
Lemma 1 Given any map ${f}$ satisfying Conditions 1 and 2, and any ${v,w\in {\mathcal L}}$, we have
$\displaystyle I(vw) = I(v) \cap f^{-|v|}(I(w)) \quad\text{and}\quad F(vw) = f^{|w|}(F(v) \cap I(w)). \ \ \ \ \ (3)$
In particular, if ${F(v) \supset I(w)}$, then ${vw\in {\mathcal L}}$ and ${F(vw) = F(w)}$.
Now consider the map ${f(x) = \beta x \pmod 1}$ for ${\frac 12(1+\sqrt{5}) < \beta < 2}$; Figure 2 shows the first three iterates. Observe that we once again have ${F(0) = I(0) \cup I(1)}$ and ${F(1) \supset I(0)}$, so that the same argument as in Exercise 4 shows that if ${w\in \{0,1\}^*}$ is any word that does not contain ${11}$ as a subword, then ${w\in {\mathcal L}}$. However, this is no longer a complete description of the language, since as the second picture in Figure 2 illustrates, we have ${I(11) \neq \emptyset}$ (equivalently, ${F(1) \cap I(1) \neq \emptyset}$, as seen in the first picture) — consequently ${11\in {\mathcal L}}$.
In Exercise 4, every word ${w\in {\mathcal L}}$ had the property that ${F(w) = F(w_{|w|})}$; writing ${w = va}$ for ${v\in {\mathcal L}}$ and ${a\in A}$, we can rewrite this property as ${F(va) = F(a)}$, which by Lemma 1 occurs if and only if ${F(v) \supset I(a)}$. These two properties can be seen on Figure 2 as follows.
• ${F(va) = F(a)}$: the vertical interval spanned by the graph of ${f^{|va|}}$ on ${I(va)}$ is the same as the vertical interval spanned by the graph of ${f}$ on ${I(a)}$.
• ${F(v) \supset I(a)}$:
on the horizontal interval ${I(v)}$, the graph of ${f^{|v|}}$ crosses the vertical interval ${I(a)}$ completely.
Looking at Figure 2, we see that now these properties are satisfied for some words, but not for all. Let us write
\displaystyle \begin{aligned} {\mathcal G} &= \{ va \in {\mathcal L} : v\in {\mathcal L}, a\in A, F(va) = F(a) \} \\ &= \{va \in {\mathcal L} : v\in {\mathcal L}, a\in A, F(v) \supset I(a) \}. \end{aligned} \ \ \ \ \ (4)
Exercise 5 Referring to Figure 2, prove that ${{\mathcal G}_2 = \{00,01,10\}}$ and that ${{\mathcal G}_3 = \{000,001,010,100,101\}}$. In particular, ${11}$, ${011}$, and ${110}$ lie in ${{\mathcal L}}$ but not in ${{\mathcal G}}$. Show that on the other hand, ${1100 \in {\mathcal G}}$ (at least for the value of ${\beta}$ illustrated), so that words in ${{\mathcal G}}$ can have ${11}$ as a subword.
The notation ${{\mathcal G}}$ suggests that we are going to try to prove the specification property for this collection of words. The first step in this direction is to prove the following Markov-type property.
Lemma 2 Let ${f}$ satisfy Conditions 1 and 2, and let ${{\mathcal G}}$ be given by (4). If ${v,w\in {\mathcal G}}$ have the property that ${v_{|v|} = w_1}$, then ${v.w\in {\mathcal G}}$, where ${v.w := v_1 \cdots v_{|v|} w_2 w_3 \cdots w_{|w|}}$ (concatenation with the repeated symbol only appearing once).
Proof: Write ${v = ua}$ where ${u\in {\mathcal L}}$ and ${a\in A}$, so that ${v.w = uw}$. By Lemma 1, we have
$\displaystyle F(v.w) = F(uw) = f^{|w|}(F(u) \cap I(w)).$
Since ${v = ua \in {\mathcal G}}$ and ${w_1 = a}$, we have ${F(u) \supset I(a) \supset I(w)}$, so ${F(u) \cap I(w) = I(w)}$ and we conclude that
$\displaystyle F(v.w) = f^{|w|}(I(w)) = F(w) = F(w_{|w|}),$
where the last equality uses the fact that ${w\in {\mathcal G}}$. This shows that ${v.w\in {\mathcal G}}$. $\Box$
In the specific case of a ${\beta}$-transformation with ${\frac 12(1+\sqrt{5}) < \beta < 2}$, we can refer to Figure 2 and Exercise 5 to see that ${{\mathcal G}}$ contains the words ${00}$, ${01}$, ${10}$, and ${101}$. In particular, for any ${v,w\in {\mathcal G}}$, we can choose ${u}$ to be the one of these four words that begins with the last symbol in ${v}$, and ends with the first symbol in ${w}$. Then ${u\in {\mathcal G}}$ and ${|u|\leq 3}$; applying Lemma 2 twice gives ${v.u.w\in {\mathcal G}}$, which proves the specification property.
3. Obtaining a specification property
In the general case of an interval map satisfying Conditions 1 and 2, we would like to play the same game, proving specification for ${{\mathcal G}}$ by mimicking the proof of specification for a topologically transitive SFT: given ${a,b\in A}$, we would like to let ${u=u(a,b)\in {\mathcal G}}$ be such that ${u_1 = a}$ and ${u_{|u|} = b}$; then given any ${v,w\in {\mathcal G}}$, we can put ${u=u(v_{|v|},w_1)}$ and apply Lemma 2 twice to get ${v.u.w\in {\mathcal G}}$. Since ${A}$ is finite, this would prove that ${{\mathcal G}}$ has specification with ${\tau = \max\{ |u(a,b)| : a,b\in A \}}$.
But how can we guarantee that such a ${u(a,b)}$ exists for every ${a,b\in A}$? This does not follow from Conditions 1–3, and indeed one can easily produce examples satisfying these conditions for which there are multiple MMEs and the collection ${{\mathcal G}}$ defined in (4) does not have specification, because there are some pairs ${a,b}$ for which no such ${u}$ exists.
At this point it is tempting to simply add existence of the required words ${u(a,b)}$ as an extra condition, replacing topological transitivity. But it is worth working a little harder. Consider the map ${f(x) = \alpha + \beta x \pmod 1}$, three iterates of which are illustrated in Figure 3 for ${\alpha = 0.1}$ and ${\beta = 1.4}$.
For this ${f}$, we see from the first picture that neither of ${F(0)}$ or ${F(1)}$ contains ${I(0)}$, and thus there can be no word ${v\in {\mathcal L}}$ such that ${F(v) \supset I(0)}$, since we always have ${F(v) \subset F(v_{|v|})}$. It follows that ${{\mathcal G}}$ does not contain any words ending in ${0}$, which rules out a proof of specification along the lines above.
The story is different if instead of single letters, we look at words of length two. Observe from the second picture that
$\displaystyle F(00) \supset I(01) \cup I(10), \quad F(01)\supset I(00), \quad F(10) \supset F(01). \ \ \ \ \ (5)$
This suggests that we should modify the definition of ${{\mathcal G}}$ in (4) to require that the follower set depends only on the last two symbols (instead of the last one). More generally, we can fix ${k\in {\mathbb N}}$ and define
\displaystyle \begin{aligned} {\mathcal G}^{(k)} &= \{ vq \in {\mathcal L} : v,q\in {\mathcal L}, |q|=k, F(vq) = F(q) \} \\ &= \{vq \in {\mathcal L} : v,q\in {\mathcal L}, |q|=k, F(v) \supset I(q) \}. \end{aligned} \ \ \ \ \ (6)
We have the following generalization of Lemma 2.
Lemma 3 If ${vq,wr\in {\mathcal G}^{(k)}}$ (with ${|q| = |r| = k}$) have the property that ${w_1 \cdots w_k = q}$, then ${vwr\in {\mathcal G}^{(k)}}$.
Proof: Using Lemma 1, we have
$\displaystyle F(vwr) = f^{|wr|}(F(v) \cap I(wr)),$
and since ${vq\in {\mathcal G}^{(k)}}$, we have ${F(v) \supset I(q) \supset I(wr)}$, so ${F(v) \cap I(wr) = I(wr)}$, giving
$\displaystyle F(vwr) = f^{|wr|}(I(wr)) = F(wr) = F(r)$
since ${w\in {\mathcal G}}$. This shows that ${vwr\in {\mathcal G}}$. $\Box$
Now we can prove that ${{\mathcal G}^{(2)}}$ has specification for the map ${f}$ in Figure 3; as described above, it suffices to prove that if ${q,r}$ are any two words from the set ${\{00,01,10\}}$, then there is some ${qur\in {\mathcal G}^{(2)}}$. But this follows quickly from the fact that the transitions ${00\to 10 \to 01 \to 00}$ are all allowed by (5), and that any word obtained by concatenating allowable transitions must lie in ${{\mathcal G}^{(k)}}$ by iterating Lemma 3.
This discussion motivates the following.
Definition 4 Given ${k\in {\mathbb N}}$, define a directed graph whose vertices are the words in ${{\mathcal L}_k}$ and whose edges are given by the condition that ${q\to r}$ exactly when ${F(q) \supset I(r)}$. Say that ${f}$ has Property ${(T_k)}$ if this graph is strongly connected, meaning that given any two ${q,r\in {\mathcal L}_k}$, there is a path in the graph from ${q}$ to ${r}$.
Property ${(T_k)}$ is playing the role of topological transitivity, but the exact relationship is not quite clear yet. I’ll have more to say about this in my next post, and the formulation here is almost certainly not the “optimal” one, but since this is a blog and not a paper, let’s stick with this property for the time being so that we can formulate an actual result and not get lost in the weeds.
Proposition 5 Let ${f}$ satisfy Conditions 1 and 2, and Property ${(T_k)}$ for some ${k\in {\mathbb N}}$. Then ${{\mathcal G}^{(k)}}$ has specification.
Proof: By Property ${(T_k)}$, for every ${q,r\in {\mathcal L}_k}$, there are ${p^1,\dots, p^\ell \in {\mathcal L}_k}$ such that ${q \to p^1 \to p^2 \to \cdots \to p^\ell \to r}$. Thus ${qp^1, p^1p^2, \dots, p^{\ell-1} p^\ell, p^\ell r\in {\mathcal G}^{(k)}}$. Applying Lemma 3 repeatedly shows that ${qp^1\cdots p^\ell r \in {\mathcal G}^{(k)}}$. Write ${u = u(q,r) = p^1\cdots p^\ell}$, so ${qur \in {\mathcal G}^{(k)}}$.
Now let ${\tau = \max \{ |u(q,r)| : q,r\in {\mathcal L}_k \}}$. Given any ${vq,rw\in {\mathcal G}}$ with ${|q|=|r|=k}$, let ${u = u(q,r)}$, so that ${qur \in {\mathcal G}}$. One application of Lemma 3 gives ${vqur \in {\mathcal G}}$, and a second application gives ${vqurw\in {\mathcal G}}$. Since ${|u| \leq \tau}$, this completes the proof. $\Box$
4. Bounding the entropy of suffixes
Let ${f}$ be an interval map satisfying Conditions 1–3 and Property ${(T_k)}$. Proposition 5 shows that ${{\mathcal G}^{(k)}}$ has specification, so it remains to find ${{\mathcal C} \subset {\mathcal L}}$ such that the decomposition and entropy gap conditions are satisfied.
For simplicity, let us first discuss the case ${k=1}$. Given ${w\in {\mathcal L}}$, we want to write ${w = uv}$ where ${u\in {\mathcal G}}$ and ${v}$ lies in some yet-to-be-determined collection ${{\mathcal C} \subset {\mathcal L}}$. Since this collection should be “small”, it makes sense to ask for ${u}$ to be as long as possible, and thus it is reasonable to take ${u = w_1 \cdots w_m}$, where ${m}$ is the largest index in ${\{1,\dots, |w|\}}$ with the property that
$latex \displaystyle w_1\cdots w_m \in {\mathcal G}. \ \ \ \ \ (7)&fg=000000$
In particular, for every ${\ell \in \{m+1,\dots, |w|\}}$, we have
$latex \displaystyle w_1 \cdots w_\ell \notin {\mathcal G}. \ \ \ \ \ (8)&fg=000000$
By Lemma 2, this implies that ${w_m \cdots w_\ell \notin {\mathcal G}}$: indeed, if we had ${w_m \cdots w_\ell \in {\mathcal G}}$, then using (7) and Lemma 2 would give ${w_1 \cdots w_\ell \in {\mathcal G}}$, contradicting (8).
This shows that if we put ${a = w_m}$, then ${w_{m+1} \cdots w_{|w|}}$ lies in the collection
$latex \displaystyle {\mathcal C}(a) := \{ v\in {\mathcal L} : av\in {\mathcal L}, av_1\cdots v_\ell \notin {\mathcal G} \ \forall 1\leq \ell \leq |v| \}. \ \ \ \ \ (9)&fg=000000$
Taking ${{\mathcal C} = \bigcup_{a\in A} {\mathcal C}(a)}$ gives the desired decomposition. But what is the entropy of ${{\mathcal C}}$? First note that if
${vb \in {\mathcal C}_{n+1}(a)}$ for some ${v\in {\mathcal L}}$ and ${a,b \in A}$, then ${v\in {\mathcal C}_n(a)}$, so we can bound ${\#{\mathcal C}_{n+1}(a)}$ in terms of ${\#{\mathcal C}_n(a)}$ by controlling how many choices of ${b}$ there are for each ${v \in {\mathcal C}_n(a)}$.
The key is in the observation that ${F(av)}$ and ${I(b)}$ are intervals that intersect in more than one point (since ${avb \in {\mathcal L}}$), but that ${F(av) \not\supset I(b)}$ (since ${avb \notin {\mathcal G}}$). The only way for this to happen is
if the interior of ${I(b)}$ contains one of the endpoints of ${F(av)}$. Since the interiors of the intervals ${I(b)}$ are disjoint, we see that for any ${v\in {\mathcal C}_n(a)}$ there are at most two choices of ${b}$ for which ${vb \in {\mathcal C}_{n+1}(a)}$. It follows that
$\displaystyle \#{\mathcal C}_{n+1}(a) \leq 2 \#{\mathcal C}_n(a),$
and iterating this gives
$\displaystyle \#{\mathcal C}_n(a) \leq 2^n \quad\Rightarrow\quad \#{\mathcal C}_n \leq 2^n (\# A)$
We conclude that
$\displaystyle \limsup_{n\to\infty} \frac 1n \log \#{\mathcal C}_n \leq \log 2.$
If ${h_{\mathrm{top}}(f) > \log 2}$, then this provides the needed entropy gap, and we are done.
But what if ${h_{\mathrm{top}}(f)}$ lies in ${(0,\log 2]}$? This is where we need to use ${{\mathcal G}^{(k)}}$ instead of ${{\mathcal G}}$. Given any ${k\in {\mathbb N}}$ and ${q\in {\mathcal L}_k}$, let
$latex \displaystyle {\mathcal C}^{(k)}(q) := \{v \in {\mathcal L} : qv\in {\mathcal L}, q v_1 \cdots v_\ell \notin {\mathcal G}^{(k)} \ \forall 1\leq \ell \leq |v| \}. \ \ \ \ \ (10)&fg=000000$
Following the same idea as in the case ${k=1}$, we see that every word in ${{\mathcal C}^{(k)}_{n+k}(q)}$ is of the form ${vp}$, where ${v\in {\mathcal C}^{(k)}_n(q)}$ and ${|p| = k}$ are such that the intervals ${F(qv)}$ and ${I(p)}$ intersect in more than one point (since ${qvp \in {\mathcal L}}$) but ${F(qv) \not\supset I(p)}$ (since ${qvp \notin {\mathcal G}^{(k)}}$). The only way for this to happen is if the interior of ${I(p)}$ contains one of the endpoints of ${F(qv)}$. Since the interiors of the intervals ${I(p)}$ are disjoint (as ${p}$ ranges over ${{\mathcal L}_k}$), we see that for any ${v\in {\mathcal C}^{(k)}_n(q)}$ there are at most two choices of ${p\in {\mathcal L}_k}$ for which ${vp \in {\mathcal C}^{(k)}_{n+1}(q)}$. It follows that
$\displaystyle \#{\mathcal C}^{(k)}_{n+k}(q) \leq 2 \#{\mathcal C}^{(k)}_n(q),$
and iterating this gives
$\displaystyle \#{\mathcal C}^{(k)}_n(a) \leq 2^{\lfloor n/k\rfloor} (\#{\mathcal L}_k) \quad\Rightarrow\quad \#{\mathcal C}^{(k)}_n \leq 2^{n/k} (\#{\mathcal L}_k)^2$
We conclude that
$\displaystyle \limsup_{n\to\infty} \frac 1n \log \#{\mathcal C}^{(k)}_n \leq \frac 1k\log 2 < h_{\mathrm{top}}(f) \ \ \ \ \ (11)$
as long as we choose ${k > \log(2) / h_{\mathrm{top}}(f)}$.
Together with Proposition 5, this proves the following.
Theorem 6 Let ${f}$ be an interval map satisfying Conditions 1–3 and Property ${(T_k)}$ for some ${k > \log(2) / h_{\mathrm{top}}(f)}$. Then the coding space ${X}$ for ${f}$ has a language with a decomposition satisfying the specification and entropy gap conditions; in particular, ${f}$ has a unique MME.
Some natural questions present themselves at this point.
• How does Property ${(T_k)}$ relate to the topological transitivity condition from Hofbauer’s paper?
• Does having Property ${(T_k)}$ for some value of ${k}$ tell us anything about having it for other values? (If so, one might hope to be able to remove the condition on ${k}$ in Theorem 6.)
• Is there a more general version of Property ${(T_k)}$ that we can use in Proposition 5 and Theorem 6?
I’ll address these in the next post. For now I will conclude this post by emphasizing the idea in the proof of (11): “if we go ${k}$ iterates in each step, then since each step only has two ways to be bad, we can bound the entropy of always-bad things by ${\frac 1k \log 2}$”. This same idea plays a key role in the proof of the growth lemma for dispersing billiards (although in that setting one needs to replace “two ways” with “${k}$ ways” and the bound becomes ${\frac 1k \log k}$); see Chapter 5 of the book “Chaotic Billiards” by Chernov and Markarian, or Section 5 of “On the measure of maximal entropy for finite horizon Sinai Billard maps” by Baladi and Demers (JAMS, 2020) (in particular, compare equation (5.4) there with (11) above). The fact that taking ${k}$ large makes this estimate smaller than the entropy is an example of the phenomenon “growth beats fragmentation”, which is important in studying systems with singularities.
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## An improved non-uniform specification result
My last post (nearly two years ago!) described Bowen’s proof that a shift space with the specification property has a unique measure of maximal entropy. A little over ten years ago, Dan Thompson and I gave a version of Bowen’s proof that turned out to be well-suited for applications to non-uniformly hyperbolic systems; an overview of the approach and of various extensions and applications since then, especially to equilibrium states for geodesic flows, can be found in our survey article in “Thermodynamic Formalism: CIRM Jean-Morlet Chair, Fall 2019” (Springer LNM, volume 2290). Today I want to return to the original setting of shift spaces, as described in our original paper (Israel J. Math, 2012), which I will refer to as [CT] from now on.
Our main result in [CT] contained an assumption that always bothered me (Condition (III) in Theorem C, see below for a description). I really had the feeling that the result should remain true even without it, but we couldn’t figure out how to remove it. Recently, Maria Jose Pacifico, Fan Yang, and Jiagang Yang have managed to remove this hypothesis in a new paper, which I will refer to as [PYY]. They consider flows rather than shift spaces, and it turns out that for flows with fixed points, including the Lorenz attractors that they are interested in, verifying the “extra assumption” (a version of which does indeed appear in my 2016 paper with Dan Thompson on non-uniform specification for flows) actually presents real problems, so their ability to prove a uniqueness result without requiring this condition has not just aesthetic but also practical value.
The arguments for flows are substantially more technical than those for shift spaces, and the proofs involve a steady stream of Bowen balls, adapted partitions, separated sets, two-scale partition sums, bounded distortion estimates, etc., which somewhat obscure the core innovation in [PYY] that brings the improvement over [CT]. The purpose of this post is to translate the argument from [PYY] back to the symbolic setting in order to clarify exactly where this improvement lies; the results I describe below can all be formally deduced from [PYY], but I will give complete proofs of everything that does not appear already in [CT].
Basic notation for shift spaces, etc., will be as in my previous post. To formulate the (improved) uniqueness result, we need the notion from [CT] of a decomposition of the language ${{\mathcal L}}$ of a shift space ${X}$: this is a choice of three collections of words ${{\mathcal C}^p, {\mathcal G}, {\mathcal C}^s \subset {\mathcal L}}$ such that every ${w\in {\mathcal L}}$ can be decomposed as a prefix’ from ${{\mathcal C}^p}$, followed by a good core’ from ${{\mathcal G}}$, followed by a suffix’ from ${{\mathcal C}^s}$. Formally, for every ${w\in {\mathcal L}}$, there are ${u^p \in {\mathcal C}^p}$, ${v\in {\mathcal G}}$, and ${u^s\in {\mathcal C}^s}$ such that ${w = u^p v u^s}$. See [CT] for examples of shift spaces with decompositions satisfying the conditions in the following theorem. (Spoiler: ${\beta}$-shifts and ${S}$-gap shifts.)
Theorem 1 Suppose ${X}$ is a shift space on a finite alphabet with topological entropy ${h>0}$ whose language ${{\mathcal L}}$ has a decomposition ${{\mathcal C}^p,{\mathcal G},{\mathcal C}^s}$ satisfying the following conditions.
(I) Specification for ${{\mathcal G}}$. There is ${\tau\in{\mathbb N}}$ such that for all ${w^1,w^2,\dots,w^n\in {\mathcal G}}$, there are ${u^1,\dots,u^{n-1}\in {\mathcal L}_\tau}$ such that ${w^1 u^1 w^2 u^2 \cdots u^{n-1} w^n\in{\mathcal L}}$.
(II) Entropy gap for ${{\mathcal C}^{p,s}}$. ${\limsup_{n\rightarrow\infty} \frac 1n \log \#({\mathcal C}^p \cup {\mathcal C}^s)_n < h}$.
Then ${X}$ has a unique MME.
In the proofs (both in [CT] and in [PYY]) one must work not just with ${{\mathcal G}}$ but also with the collections
$\displaystyle {\mathcal G}(M) = \{u^pvu^s : u^{p,s}\in {\mathcal C}^{p,s}, v\in {\mathcal G}, |u^{p,s}| \leq M \}.$
Observe that ${{\mathcal L} = \bigcup_M {\mathcal G}(M)}$; if you are familiar with Pesin theory for non-uniformly hyperbolic diffeomorphisms, it is reasonable to think of this as an analogue of the decomposition of a (non-uniformly) hyperbolic set into regular level sets. (If you are not familiar with Pesin theory, you should probably just ignore that last sentence.) In [CT], we required the following extra condition.
(III) For every ${M\in{\mathbb N}}$, there exists ${t}$ such that given ${w\in {\mathcal G}(M)}$, there exist words ${q,r\in {\mathcal L}}$ with length ${\leq t}$ such that ${qwr \in {\mathcal G}}$.
This condition guarantees that each of the collections ${{\mathcal G}(M)}$ has (a slightly weaker version of) specification as well, with gap size allowed to depend on ${M}$, and it appears in this form in all of our later papers; this is the condition that turns out to present problems for Lorenz attractors, and indeed for general flows with fixed points, and this is what motivated the authors of [PYY].
Let me mention in passing that there is also a symbolic setting where (III) becomes problematic; a few years ago I wrote a paper with Ronnie Pavlov (ETDS, 2019) where we studied shift spaces with “one-sided almost specification with bounded mistake function” and proved uniqueness of the MME. We were able to verify conditions (I) and (II) for these examples, but not condition (III) (see the end of §2.3 in that paper), and ultimately needed to use a slightly different set of conditions I had formulated in a paper constructing countable-state Markov models for shifts with non-uniform specification (Comm. Math. Phys., 2018). These conditions are a little more complicated, and the resulting proofs are a lot more complicated, so it is nice to know now that the uniqueness results in my paper with Ronnie Pavlov do in fact follow from the simpler and shorter proof of Theorem 1.
To see where [PYY] makes the improvement leading to a proof of Theorem 1 without requiring (III), start by recalling the overall structure of Bowen’s proof as I outlined it in my last post: first one uses specification to construct a Gibbs measure ${\mu}$, then one proves that ${\mu}$ is ergodic and that there can be no mutually singular MME ${\nu\perp \mu}$. The proof in [CT] follows similar lines, and roughly speaking, Condition (III) was only required for the second half. More precisely, using only Conditions (I) and (II), [CT] still contains a proof of the following.
Proposition 2 There exists an MME ${\mu}$ that has the following Gibbs-type property: there is ${c>0}$ such that
$\displaystyle \mu[w] \geq ce^{-|w| h} \text{ for all } w\in {\mathcal G}. \ \ \ \ \ (1)$
Moreover, there is ${m\in {\mathbb N}}$ such that
$\displaystyle \mu([u] \cap \sigma^{-(|u| + m)}[v]) \geq c e^{-(|u| + |v|) h} \text{ for all } u,v\in {\mathcal G}. \ \ \ \ \ (2)$
Proof: See Lemmas 5.1–5.10 in [CT] for the proofs of the uniform counting bounds and the construction of a measure satisfying (1); note the sentence after the proof of Lemma 5.10 emphasizing that Condition (III) was not used. The proof that this measure also satisfies the “two-step Gibbs bound” (2) is contained in the proof of [CT, Lemma 5.15], even though the statement of that lemma is a little different. (In fact one can take ${m}$ to be arbitrarily large, going to infinity along a sequence with bounded gaps, but we only need (2) for a single value of ${m}$.) $\Box$
The remainder of the proof in [CT] uses (2) to establish ergodicity and (1) to rule out the existence of a mutually singular MME. More precisely, this is done by using the fact under Condition (III), Proposition 2 still holds with ${{\mathcal G}}$ replaced by ${{\mathcal G}(M)}$ (although the constant ${c}$ is allowed to depend on ${M}$). One also uses the fact that various sets that appear in the arguments can be well-approximated by unions of cylinders corresponding to words in ${{\mathcal G}(M)}$.
Without Condition (III), there is no guarantee that Proposition 2 holds for ${{\mathcal G}(M)}$, and so instead [PYY] must work with the “good cores” of words, and must approximate the relevant sets with unions of cylinders from ${{\mathcal G}}$, not ${{\mathcal G}(M)}$. The two lemmas that make this precise both require a certain truncation operation’ to state properly: given ${w\in {\mathcal L}}$ and ${i,j\in{\mathbb N}}$ with ${i+j \leq |w|}$, let
$\displaystyle \varphi_{i,j}(w) = w_{(i,|w|-j)} = w_{i+1} w_{i+2} \cdots w_{|w| - j - 1}$
denote the word obtained by deleting the first ${i}$ and last ${j}$ symbols from ${w}$. Given a collection of words ${{\mathcal D}_n \subset {\mathcal L}_n}$, we will write ${[{\mathcal D}_n] = \bigcup_{w\in {\mathcal D}_n} [w] \subset X}$ for the union of the corresponding cylinders.
Lemma 3 There exist ${\alpha>0}$ and ${M\in {\mathbb N}}$ such that if ${\nu}$ is any MME and ${{\mathcal D}_n \subset {\mathcal L}_n}$ satisfies ${\nu([{\mathcal D}_n]) \geq \frac 12}$, then there are ${i,j \in \{0,1,\dots M\}}$ (depending on ${n}$ and ${{\mathcal D}}$) such that ${\#(\varphi_{i,j}({\mathcal D}_n) \cap {\mathcal G}) \geq \alpha e^{nh}}$.
(In fact one can replace ${\frac 12}$ with any ${\gamma \in (0,1)}$, at the cost of possibly needing to use a different ${\alpha}$ and ${M}$.)
Proof: This very nearly follows from two lemmas in [CT]:
• Lemma 5.7: For all ${\gamma \in (0,1)}$ there exists ${c_1 > 0}$ such that if ${\nu}$ is any MME and ${\nu({\mathcal D}_n) \geq \gamma}$, then ${\#{\mathcal D}_n \geq c_1 e^{nh}}$.
• Lemma 5.6: For every ${0, there exists ${M\in{\mathbb N}}$ such that if ${\#{\mathcal D}_n \geq c_1 e^{nh}}$, then ${\#({\mathcal D}_n \cap {\mathcal G}(M)) \geq c_2 e^{nh}}$. (The formulation in [CT] is slightly different, but the proof there gives this statement.)
Combining these gives ${M}$ such that
$\displaystyle \#({\mathcal D}_n \cap {\mathcal G}(M)) \geq c_2 e^{nh} \text{ whenever } \nu({\mathcal D}_n) \geq \tfrac12. \ \ \ \ \ (3)$
The remaining step is not done in [CT] but is done in [PYY, Lemma 5.2]. Observe that for every ${w\in {\mathcal D}_n \cap {\mathcal G}(M)_n}$ there are ${i,j\in \{0,1,\dots, M\}}$ such that ${\varphi_{i,j}(w) \in {\mathcal G}}$; moreover, for any ${v\in {\mathcal G}_{n-(i+j)}}$, there are at most ${(\#{\mathcal L}_i)(\#{\mathcal L}_j) \leq (\#{\mathcal L}_M)^2}$ words ${w\in {\mathcal D}_n \cap {\mathcal G}(M)}$ with ${\varphi_{i,j}(w) = v}$. It follows that
\displaystyle \begin{aligned} \#({\mathcal D}_n \cap {\mathcal G}(M)) &\leq \sum_{i,j=0}^M (\#{\mathcal L}_M)^2\#(\varphi_{i,j}({\mathcal D}_n) \cap {\mathcal G}) \\ &\leq (\#{\mathcal L}_M)^2(M+1)^2 \max_{i,j} \#(\varphi_{i,j}({\mathcal D}_n) \cap {\mathcal G}). \end{aligned}
Combining this with (3) gives
$\displaystyle (\#{\mathcal L}_M)^2(M+1)^2 \max_{i,j} \#(\varphi_{i,j}({\mathcal D}_n) \cap {\mathcal G}) \geq c_2 e^{nh},$
which proves Lemma 3 with ${\alpha = (\#{\mathcal L}_M)^{-2}(M+1)^{-2} c_2}$. $\Box$
From now on, we fix ${\alpha}$ and ${M}$ as given by Lemma 3. The key innovation for approximating sets using ${{\mathcal G}}$ (instead of ${{\mathcal G}(M)}$) is in [PYY, Lemma 7.3], which implies the following in our simple setting.
Lemma 4 Given any mutually disjoint invariant Borel probability measures ${\nu_1,\nu_2}$ on ${X}$, and any ${\beta>0}$, there is a collection of words ${{\mathcal U} \subset {\mathcal L}}$ and ${n_0\in {\mathbb N}}$ such that
1. ${\nu_1([{\mathcal U}_n]) \geq 1-\beta}$ for all ${n}$, and
2. ${\nu_2([\varphi_{i,j} {\mathcal U}_n]) \leq \beta}$ for all ${i,j,n}$ satisfying ${\lfloor n/2 \rfloor \geq i}$ and ${\lceil n/2 \rceil - j \geq n_0}$, in particular, for all ${n \geq 2(n_0 + \max(i,j))}$.
Proof: Since ${\nu_1 \perp \nu_2}$ and both are invariant, there are disjoint invariant sets ${P_1, P_2 \subset X}$ such that ${\nu_\ell(P_\ell) = 1}$ for ${\ell=1,2}$. By inner regularity there are compact sets ${K_\ell \subset P_\ell}$ such that ${\nu_\ell(K_\ell) \geq 1-\beta}$.
(If one only wanted the results to hold for ${i=j=0}$, then one could at this point put ${{\mathcal U} = \{w\in {\mathcal L} : [w] \cap K_1 \neq \emptyset\}}$ and be done; this is a fairly standard argument and is what we used in [CT], where in fact we did not even bother to spell it out so explicitly. The extension to general ${i,j}$ — which is needed in order to leverage Lemma 3, as we will see later — requires a little more care in the choice of ${{\mathcal U}}$, and this is where [PYY] is quite clever in strengthening the usual argument.)
Now given ${n\in{\mathbb N}}$, let ${{\mathcal U}_n = \{ w \in {\mathcal L}_n : [w] \cap \sigma^{-\lfloor n/2 \rfloor} K_1 \neq \emptyset \}}$. It follows immediately that ${[{\mathcal U}_n] \supset \sigma^{-\lfloor n/2 \rfloor}K_1}$, and since ${\nu_1}$ is invariant we get
$\displaystyle \nu_1[{\mathcal U}_n] \geq \nu_1(\sigma^{-\lfloor n/2 \rfloor} K_1) = \nu_1(K_1) \geq 1-\beta,$
which proves the first claim. For the second claim, we take ${n_0}$ large enough that no ${n_0}$-cylinder intersects both ${K_1}$ and ${K_2}$ (this is possible since these are compact disjoint sets). Then for every ${i,j,n}$ as in the statement, and every ${w\in {\mathcal U}_n}$, the word ${v := \varphi_{\lfloor n/2 \rfloor, j}(w)}$ has ${[v] \supset \sigma^{\lfloor n/2 \rfloor} [w]}$, which intersects ${K_1}$, so ${[v] \cap K_1 \neq \emptyset}$, and moreover ${|v| = (n-j) - \lfloor n/2 \rfloor \geq n_0}$, so ${[v] \cap K_2 = \emptyset}$; that is, ${[v] \subset K_2^c}$. We conclude that
$\displaystyle \sigma^{\lfloor n/2 \rfloor - i}[\varphi_{i,j}(w)] \subset [v] \subset K_2^c.$
Taking a union over all ${w\in {\mathcal U}_n}$ gives ${\sigma^{\lfloor n/2 \rfloor - i}[\varphi_{i,j}({\mathcal U}_n)] \subset X \setminus K_2}$, and now invariance of ${\nu_2}$ gives
$\displaystyle \nu_2([\varphi_{i,j}{\mathcal U}_n]) \leq \nu_2(\sigma^{-(\lfloor n/2 \rfloor - i)}(X\setminus K_2)) = \nu_2(X\setminus K_2) \leq \beta,$
which completes the proof of Lemma 4. $\Box$
With Lemmas 3 and 4 in hand, we can now follow [PYY] in completing the second part of the uniqueness argument by showing that there is no MME ${\nu\perp \mu}$, and that ${\mu}$ is ergodic. The first of these follows [PYY, Proposition 7.2]. Recall that ${c}$ is the constant in the lower Gibbs bounds (1) and (2), and ${\alpha,M}$ are given by Lemma 3.
Proposition 5 There is no MME ${\nu\perp \mu}$.
Proof: Suppose for a contradiction that there is another MME ${\nu\perp \mu}$. Fix ${\beta < \min(\frac 12, c\alpha)}$, and apply Lemma 4 with ${\nu_1 = \nu}$ and ${\nu_2 = \mu}$ to get ${{\mathcal U}\subset {\mathcal L}}$ and ${n_0\in{\mathbb N}}$ such that ${\nu([{\mathcal U}_n]) \geq 1-\beta}$ for all ${n}$, and
$\displaystyle \mu([\varphi_{i,j} {\mathcal U}_n]) \leq \beta \ \ \ \ \ (4)$
for all ${i,j,n}$ satisfying ${n \geq 2(n_0 + \max(i,j))}$.
Since ${\beta<\frac 12}$, we have ${\nu([{\mathcal U}_n]) > \frac 12}$ for all ${n}$, and since ${\nu}$ is an MME, Lemma 3 gives ${i,j\in \{0,1,\dots, M\}}$ (depending on ${n}$) with
$\displaystyle \#(\varphi_{i,j}({\mathcal U}_n) \cap {\mathcal G}) \geq \alpha e^{nh};$
thus the Gibbs property on ${{\mathcal G}}$ gives
$\displaystyle \mu([\varphi_{i,j} ({\mathcal U}_n) \cap {\mathcal G}]) \geq c e^{-(n-(i+j)) h} \#(\varphi_{i,j}({\mathcal U}_n) \cap {\mathcal G}) \geq c e^{(i+j) h} \alpha \geq c\alpha.$
For all ${n\geq 2(n_0 + M)}$ and this choice of ${i,j \in \{0,1,\dots, M\}}$, we also have (4), which leads to
$\displaystyle \beta \geq \mu([\varphi_{i,j} {\mathcal U}_n]) \geq \mu([\varphi_{i,j} ({\mathcal U}_n) \cap {\mathcal G}]) \geq c\alpha.$
This contradicts our choice of ${\beta}$ and completes the proof of Proposition 5. $\Box$
The proof of ergodicity follows [PYY, Proposition 7.4].
Proposition 6 ${\mu}$ is ergodic.
Proof: Suppose for a contradiction that ${P\subset X}$ is invariant and ${0 < \mu(P) < 1}$. Let ${\nu}$ and ${\nu'}$ be the normalizations of ${\mu}$ restricted to ${P}$ and ${P^c}$, respectively. Fix ${\beta < \min (\frac 12, \frac 12 c\alpha^2)}$, and apply Lemma 4 twice — once with ${\nu_1=\nu}$ and ${\nu_2=\nu'}$, and once with the roles reversed — to obtain ${{\mathcal U},{\mathcal U}' \subset {\mathcal L}}$ and ${n_0 \in {\mathbb N}}$ such that ${\nu([{\mathcal U}_n]) \geq 1-\beta}$ and ${\nu'([{\mathcal U}'_n]) \geq 1-\beta}$ for every ${n}$, and moreover
$\displaystyle \nu([\varphi_{i,j}{\mathcal U}_n']) \leq \beta \text{ and } \nu'([\varphi_{i,j}{\mathcal U}_n]) \leq \beta \ \ \ \ \ (5)$
for all ${i,j,n}$ satisfying ${n \geq 2(n_0 + \max(i,j))}$.
Observe that each of ${\nu}$ and ${\nu'}$ is an MME (since entropy depends affinely on the measure, and the MME ${\mu}$ is a convex combination of ${\nu}$ and ${\nu'}$). Thus we can apply Lemma 3 to both ${{\mathcal U}}$ and ${{\mathcal U}'}$, obtaining ${i,j,i',j' \in \{0,1,\dots, M\}}$ (depending on ${n}$) such that the collections
$\displaystyle {\mathcal V}(n) := \varphi_{i,j}({\mathcal U}_n) \cap {\mathcal G} \subset {\mathcal L}_{n-(i+j)} \text{ and } {\mathcal V}'(n) := \varphi_{i,j}({\mathcal U}_n')\cap {\mathcal G} \subset {\mathcal L}_{n-(i'+j')}$
both have cardinality at least ${\alpha e^{nh}}$.
Let ${k = n-(i+j) + m}$, so that ${|v| + m = k}$ for every ${v\in {\mathcal V}(n)}$. Given any such ${v}$, and any ${v' \in {\mathcal V}'(n)}$, the “two-step” Gibbs property in (2) gives
$\displaystyle \mu([v] \cap \sigma^{-k}[v']) \geq c e^{-(|v| + |v'|)h} \geq c e^{-2nh}.$
Summing over all such ${v}$ and ${v'}$ gives
$\displaystyle \mu([{\mathcal V}(n)] \cap \sigma^{-k} [{\mathcal V}'(n)]) \geq c e^{-2nh} (\#{\mathcal V}(n)) (\#{\mathcal V}'(n)) \geq c \alpha^2. \ \ \ \ \ (6)$
We will get an upper bound on this measure and derive a contradiction. Given any ${x\in [{\mathcal V}(n)] \cap \sigma^{-k}[{\mathcal V}'(n)]}$, we either have ${x\in P^c}$ or ${x\in P = \sigma^{-k} P}$ (since ${P}$ is invariant), and thus
$\displaystyle [{\mathcal V}(n)] \cap \sigma^{-k}[{\mathcal V}'(n)] \subset ([{\mathcal V}(n)] \cap P^c) \cup \sigma^{-k} ([{\mathcal V}'(n)] \cap P).$
Observe that whenever ${n\geq 2(n_0 + M)}$, (5) gives
$\displaystyle \mu([{\mathcal V}(n)] \cap P^c) = \nu'([{\mathcal V}(n)]) \mu(P^c) \leq \nu'([\varphi_{i,j}({\mathcal U}_n)]) \leq \beta$
and similarly
$\displaystyle \mu(\sigma^{-k}([{\mathcal V}'(n)] \cap P)) = \mu([{\mathcal V}'(n)] \cap P) \leq \beta,$
so that
$\displaystyle \mu([{\mathcal V}(n)] \cap \sigma^{-k}[{\mathcal V}'(n)]) \leq 2\beta.$
Comparing this with (6) gives ${2\beta \geq c\alpha^2}$, contradicting our choice of ${\beta}$ and completing the proof of Proposition 6. $\Box$
Posted in ergodic theory, theorems, Uncategorized | 1 Comment
## Specification and the measure of maximal entropy
These are notes for a talk I am giving in Jon Chaika’s online working seminar in ergodic theory. The purpose of the talk is to outline Bowen’s proof of uniqueness of the measure of maximal entropy for shift spaces with the specification property. Bowen’s approach extends much more broadly than this: to non-symbolic systems (assuming expansivity); to equilibrium states for non-zero potential functions (assuming a bounded distortion property); and to non-uniformly hyperbolic systems using the notion of “obstructions to specification and expansivity” developed by Dan Thompson and myself (see some notes here and here, and videos here). In these notes, though, I want to give the bare bones of the argument in the simplest possible setting, to make the essential structure as clear as possible. I gave an alternate argument in a previous post; here I am giving Bowen’s original argument, although I do not necessarily follow his presentation. I should also point out that this argument differs from the construction of the MME, or more generally the equilibrium state, in Bowen’s monograph, which uses the Ruelle operator.
1. Setting and result
Let ${A}$ be a finite set; the alphabet. Then the full shift ${A^{\mathbb N}}$ is the set of infinite sequences of symbols from ${A}$; this is a compact metric space with ${d(x,y) = e^{-n(x,y)}}$, where ${n(x,y) = \min \{ n : x_n \neq y_n\}}$. The shift map ${\sigma\colon A^{\mathbb N}\rightarrow A^{\mathbb N}}$ is defined by ${\sigma(x)_n = x_{n+1}}$. A shift space is a closed set ${X\subset A^{\mathbb N}}$ with ${\sigma(X) = X}$.
Example 1 The best example to keep in mind through this talk is a topological Markov shift: fix ${d\in{\mathbb N}}$ and put ${A = \{1,\dots, d\}}$; then fix a ${d\times d}$ transition matrix ${T}$ with entries in ${\{0,1\}}$, and write
$\displaystyle i\rightarrow j \text{ if } T_{ij} = 1, \quad i\not\rightarrow j \text{ if } T_{ij} = 0. \ \ \ \ \ (1)$
Define ${X = \{ x\in A^{\mathbb N} : x_n \rightarrow x_{n+1}}$ for all ${n\}}$. This is a topological Markov shift (TMS). It can be viewed in terms of a directed graph with vertex set ${A}$ and edges given by (1): ${X}$ consists of all sequences that label infinite paths on the graph.
The TMS is mixing or primitive if there is ${N\in{\mathbb N}}$ such that ${(T^N)_{ij} > 0}$ for all ${i,j}$. Equivalently, the graph is strongly connected and the set of loop lengths on the graph has gcd ${1}$.
Given a shift space ${X}$, consider
\displaystyle \begin{aligned} \mathcal{M} &= \{ \text{Borel probability measures on }X\}, \\ \mathcal{M}_\sigma &= \{ \mu \in \mathcal{M} : \sigma_* \mu := \mu\circ \sigma^{-1} = \mu \}, \\ \mathcal{M}_\sigma^e &= \{ \mu \in \mathcal{M}_\sigma : \mu \text{ is ergodic} \}. \end{aligned}
The set ${\mathcal{M}_\sigma}$ of invariant measures is extremely large for a mixing TMS (and more generally for systems with some sort of hyperbolic behavior), and it is important to identify “distinguished” invariant measures. One way of doing this is via the variational principle
$\displaystyle h_{\mathrm{top}}(X) = \sup \{ h(\mu) : \mu \in \mathcal{M}_\sigma \}.$
The next section recalls the definitions of the topological and measure-theoretic entropies in this setting. A measure achieving the supremum is a measure of maximal entropy (MME).
We will see that every mixing TMS has a unique MME, via a more general result. Given ${n\in{\mathbb N}_0}$ and ${w\in A^n}$, let ${[w] = wX \cap X}$ be the set of sequences in ${X}$ that start with the word ${w}$ (juxtaposition denotes concatenation); call this the cylinder of ${w}$. Define the language of ${X}$ by
$\displaystyle \mathcal{L}_n = \{ w\in A^n : [w] \neq \emptyset \}, \qquad \mathcal{L} = \bigcup_{n\in {\mathbb N}_0} \mathcal{L}_n.$
Definition 1 ${X}$ has specification if there is ${\tau\in \mathbb{N}_0}$ such that for all ${v,w\in \mathcal{L}}$ there is ${u\in \mathcal{L}_\tau}$ such that ${vuw\in \mathcal{L}}$.
Exercise 1 Prove that every mixing TMS has specification.
Theorem 2 (Bowen) If ${X}$ has specification, then it has a unique MME.
2. Entropies
2.1. Topological entropy
Every word in ${\mathcal{L}_{m+n}}$ is of the form ${vw}$ for some ${v\in \mathcal{L}_m}$ and ${w\in \mathcal{L}_n}$; thus
$\displaystyle \mathcal{L}_{m+n} \subset \mathcal{L}_m \mathcal{L}_n \quad\Rightarrow\quad \#\mathcal{L}_{m+n} \leq \#\mathcal{L}_m \#\mathcal{L}_n.$
This means that the sequence ${c_n := \log \#\mathcal{L}_n}$ has the sub-additivity property
$\displaystyle c_{m+n} \leq c_m + c_n. \ \ \ \ \ (2)$
Exercise 2 Prove Fekete’s lemma: for any sequence satisfying (2), ${\lim_{n\rightarrow\infty} \frac 1n c_n}$ exists and is equal to ${\inf_n \frac 1n c_n}$ (a priori it could be ${-\infty}$).
We conclude that the topological entropy ${h_{\mathrm{top}}(X) := \lim_{n\rightarrow\infty} \frac 1n \log \#\mathcal{L}_n}$ exists for every shift space. This quantifies the growth rate of the total complexity of the system.
Exercise 3 Prove that ${h_{\mathrm{top}}(X)}$ is the box dimension of ${X}$ (easy). Then prove that it is also the Hausdorff dimension of ${X}$ (harder). (Both of these facts rely very strongly on the fact that ${d(\sigma x,\sigma y)/d(x,y)}$ is globally constant whenever ${x,y}$ are close; for general systems where the amount of expansion may vary, the definition of topological entropy is more involved and the relationship to dimension is more subtle, although it is worth noting that a 1973 paper of Bowen in TAMS gives a definition of topological entropy that is analogous to Hausdorff dimension.)
2.2. Measure-theoretic entropy
Recall the motivation from information theory for the definition of entropy: given ${p\in (0,1]}$, let ${I(p) = -\log p}$. This can be interpreted as the information associated to an event with probability ${p}$; note that it is monotonic (the less likely an event is, the more information we gain by learning that it happened) and that ${I(pq) = I(p) + I(q)}$.
Now define ${\phi}$ on ${[0,1]}$ by ${\phi(p) = pI(p) = -p\log p}$ (and ${\phi(0) = 0}$); this can be interpreted as the expected amount of information associated to an event with probability ${p}$.
Exercise 4 Show that ${\phi}$ is concave.
Given ${N\in{\mathbb N}}$, let ${\Delta = \Delta_N = \{ \bar{p} = (p_1,\dots, p_N) : p_i \geq 0 }$ and ${\sum p_i \leq 1\}}$ be the set of sub-probability vectors with ${N}$ components. Define
$\displaystyle H(\bar{p}) = \sum_{i=1}^N \phi(p_i); \ \ \ \ \ (3)$
this can be interpreted as the expected information associated to a collection of mutually exclusive events with probabilities ${p_1,\dots, p_N}$.
Exercise 5 Show that ${H(\bar{p}) \leq \log N}$ for all ${\bar{p} \in \Delta_N}$, with equality if and only if ${p_i = \frac 1N}$ for all ${i}$.
Given ${\mu\in \mathcal{M}_\sigma}$, we have for each ${n}$ a probability vector with ${\#\mathcal{L}_n}$ components; writing ${\mu(w) = \mu([w])}$ for convenience, the entropy (expected information) associated to this vector is
$\displaystyle H_n(\mu) := \sum_{w\in \mathcal{L}_n} \phi(\mu(w)). \ \ \ \ \ (4)$
Lemma 3 ${H_{m+n}(\mu) \leq H_m(\mu) + H_n(\mu)}$
Proof:
\displaystyle \begin{aligned} H_{m+n}(\mu) &\leq \sum_{v\in \mathcal{L}_m} \sum_{w\in \mathcal{L}_n} \mu(vw) I \Big( \frac{\mu(vw)}{\mu(v)} \cdot \mu(v) \Big) \\ &= \sum_{w\in \mathcal{L}_n} \sum_{v\in \mathcal{L}_m} \mu(v) \phi \Big( \frac{\mu(vw)}{\mu(v)} \Big) + \sum_{v\in \mathcal{L}_m} \sum_{w\in \mathcal{L}_n} \mu(vw) I(\mu(v)) \\ &\leq \sum_{w\in \mathcal{L}_n} \phi \Big( \sum_{v\in \mathcal{L}_m} \mu(vw) \Big) + \sum_{v\in \mathcal{L}_m} \phi(\mu(v))\\ &= H_n(\mu) + H_m(\mu), \end{aligned}
where the first line is by definition, the second is since ${I(pq) = I(p) + I(q)}$, the third uses concavity of ${\phi}$, and the fourth uses invariance of ${\mu}$ to get ${\sum_{v\in \mathcal{L}_m} \mu(vw) = \mu(w)}$. $\Box$
This lemma has the following intuitive interpretation: the expected information from the first ${m+n}$ symbols is at most the expected information from the first ${m}$ symbols plus the expected information from the next ${n}$ symbols.
Now Fekete’s lemma implies that the following measure-theoretic entropy exists:
$\displaystyle h(\mu) := \lim_{n\rightarrow\infty} \frac 1n H_n(\mu). \ \ \ \ \ (5)$
2.3. Variational principle
Using Exercise 5 we see that ${H_n(\mu) \leq c_n = \log \#\mathop{\mathcal L}_n}$, with equality if and only if ${\mu(w) = 1/\#\mathop{\mathcal L}_n}$ for all ${w\in \mathop{\mathcal L}_n}$. This immediately proves that
$\displaystyle h(\mu) \leq h_{\mathrm{top}}(X) \text{ for all } \mu \in \mathcal{M}_\sigma, \ \ \ \ \ (6)$
and suggests that in order to have equality we should look for a measure with
$\displaystyle \mu(w) \approx \frac 1{\#\mathop{\mathcal L}_n} \approx e^{-n h_{\mathrm{top}}(X)} \text{ for all } w\in \mathcal{L}_n. \ \ \ \ \ (7)$
The following makes this more precise.
Definition 4 ${\mu\in \mathcal{M}_\sigma}$ is a Gibbs measure if there are ${h,c,C>0}$ such that for all ${n\in {\mathbb N}}$ and ${w\in \mathcal{L}_n}$, we have ${ce^{-nh} \leq \mu(w) \leq C e^{-nh}}$.
Now Theorem 2 is a consequence of the following two results, which we prove below.
Theorem 5 If ${\mu}$ is an ergodic Gibbs measure for ${X}$, then ${h=h_{\mathrm{top}}(X) = h(\mu)}$ and ${\mu}$ is the unique MME.
Theorem 6 If ${X}$ has specification, then it has an ergodic Gibbs measure.
In fact the construction of ${\mu}$ below always gives equality in (6), without relying on the specification property (or obtaining uniqueness), but we will not prove this.
2.4. Convex combinations
Before embarking on the proof of Theorems 5 and 6, we establish a general property of entropy that will be important.
Lemma 7 Given ${\bar{p},\bar{q} \in \Delta_N}$ and ${s,t \in [0,1]}$ with ${s+t = 1}$, we have
$\displaystyle sH(\bar{p}) + tH(\bar{q}) \leq H(s\bar{p} + t\bar{q}) \leq sH(\bar{p}) + tH(\bar{q}) + \log 2. \ \ \ \ \ (8)$
Proof: The first inequality follows immediately from concavity of ${\phi}$. For the second inequality we first observe that
$\displaystyle H(s\bar{p}) = \sum_{i=1}^N sp_i I(sp_i) = \sum_{i=1}^N \big(sp_i I(p_i) + sp_i I(s)\big) = sH(\bar{p}) + \phi(s) \sum_{i=1}^N p_i. \ \ \ \ \ (9)$
Then we use monotonicity of ${I}$ to get
\displaystyle \begin{aligned} H(s\bar{p} + t\bar{q}) &= \sum s p_i I(sp_i + tq_i) + tq_i I(sp_i + tq_i) \\ &\leq \sum s p_i I(sp_i) + \sum tq_i I(tq_i) \\ & = sH(\bar{p}) + tH(\bar{q}) + \phi(s) \sum p_i + \phi(t) \sum q_i, \end{aligned}
where the last equality uses (9) twice; this proves (8). $\Box$
Applying Lemma 7 to the probability vectors associated to two measures ${\nu,\mu \in \mathcal{M}_\sigma}$, we see that
$\displaystyle s H_n(\nu) + t H_n(\mu) \leq H_n(s\nu + t\mu) \leq s H_n(\nu) + t H_n(\mu) + \log 2$
for all ${n}$: dividing by ${n}$ and sending ${n\rightarrow\infty}$ gives
$\displaystyle h(s\nu + t\mu) = s h(\nu) + t h(\mu). \ \ \ \ \ (10)$
Remark 1 It is worth mentioning the deeper fact that a version of (10) holds for infinite convex combinations (even uncountable ones given by an integral); this is due to Konrad Jacobs, see Section 9.6 of “Foundations of Ergodic Theory” by Viana and Oliveira.
3. Gibbs implies uniqueness
To prove Theorem 5, start by observing that the lower Gibbs bound gives
$\displaystyle 1 = \mu(X) = \sum_{w\in \mathcal{L}_n} \mu(w) \geq ce^{-nh} \#\mathcal{L}_n \quad\Rightarrow\quad \#\mathcal{L}_n \leq c^{-1} e^{nh}, \ \ \ \ \ (11)$
and thus ${h_{\mathrm{top}}(X) \leq h}$. Meanwhile, the upper Gibbs bound gives
$\displaystyle H_n(\mu) = \sum_{w\in \mathcal{L}_n} \mu(w) I(\mu(w)) \geq \sum_{w\in \mathcal{L}_n} \mu(w) I(C e^{-nh}) = I(C e^{-nh}) = nh - \log C,$
and thus ${h(\mu) \geq h}$, so we conclude that ${h_{\mathrm{top}}(X) = h = h(\mu)}$. It remains to show that every ${\nu \in \mathcal{M}_\sigma}$ with ${\nu \neq \mu}$ has ${h(\nu) < h}$.
First observe that by (10), we can restrict our attention to the case when ${\nu\perp \mu}$. Indeed, given any ${\nu \in \mathcal{M}_\sigma}$, the Lebesgue decomposition theorem gives ${\nu = s\nu_1 + t\nu_2}$ for some ${\nu_1 \perp \mu}$ and ${\nu_2 \ll \mu}$, and (10) gives ${h(\nu) = sh(\nu_1) + th(\nu_2)}$. By ergodicity we must have ${\nu_2 = \mu}$ and thus ${s>0}$, so if ${h(\nu_1) < h}$ then the same is true of ${h(\nu)}$.
Now consider ${\nu \perp \mu}$. Then there is a Borel set ${D\subset X}$ with ${\nu(D) = 0}$ and ${\mu(D) = 1}$, and this in turn gives ${\mathcal{D}_n \subset \mathcal{L}_n}$ such that
$\displaystyle \nu(\mathcal{D}_n) \rightarrow 0 \text{ and } \mu(\mathcal{D}_n) \rightarrow 1 \text{ as } n\rightarrow\infty. \ \ \ \ \ (12)$
Let ${\nu|_{\mathcal{D}_n}}$ denote the normalization of ${\nu}$ after restricting to words in ${\mathcal{D}_n}$, and similarly for ${\nu|_{\mathcal{D}_n^c}}$. Recall from Fekete’s lemma and subadditivity of ${H_n(\nu)}$ that ${\frac 1n H_n(\nu) \geq h(\nu)}$ for all ${n}$. Then we get
\displaystyle \begin{aligned} nh(\nu) &\leq H_n(\nu) = H_n \big(\nu(\mathcal{D}_n) \nu|_{\mathcal{D}_n} + \nu(\mathcal{D}_n^c) \nu|_{\mathcal{D}_n^c} \big) \\ &\leq \nu(\mathcal{D}_n) H_n(\nu|_{\mathcal{D}_n}) + \nu(\mathcal{D}_n^c) H_n(\nu|_{\mathcal{D}_n^c}) + \log 2 \\ &\leq \nu(\mathcal{D}_n) \log \#\mathcal{D}_n + \nu(\mathcal{D}_n^c) \log \#\mathcal{D}_n^c + \log 2 \\ &\leq \nu(\mathcal{D}_n) \log(c^{-1} e^{nh} \mu(\mathcal{D}_n)) + \nu(\mathcal{D}_n^c) \log(c^{-1} e^{nh} \mathcal{D}_n^c) + \log 2 \\ &= nh - \log c + \log 2 + \nu(\mathcal{D}_n) \log \mu(\mathcal{D}_n) + \nu(\mathcal{D}_n^c) \log \mu(\mathcal{D}_n^c), \end{aligned}
where the second line uses Lemma 7, the third line uses Exercise 5, and the fourth line uses the lower Gibbs bound as in (11). We conclude that
$\displaystyle n(h(\nu) - h) \leq \log (2/c) + \nu(\mathcal{D}_n) \log \mu(\mathcal{D}_n) + \nu(\mathcal{D}_n^c) \log \mu(\mathcal{D}_n^c),$
and as ${n\rightarrow\infty}$ the right-hand side goes to ${-\infty}$ by (12), which implies that ${h(\nu) < h}$, completing the proof.
4. Specification implies Gibbs
Now we outline the proof of Theorem 6. This comes in three steps: (1) uniform counting bounds; (2) construction of a Gibbs measure; (3) proof of ergodicity.
4.1. Uniform counting bounds
From now on we write ${h = h_{\mathrm{top}}(X)}$ for convenience. Fekete’s lemma gives ${\frac 1n \log \#\mathcal{L}_n \geq h}$ for all ${n}$, or equivalently ${\#\mathcal{L}_n \geq e^{nh}}$. This can also be deduced by writing
$\displaystyle \#\mathcal{L}_{kn} \leq (\#\mathcal{L}_n)^k \quad\Rightarrow\quad \frac 1{kn} \log \#\mathcal{L}_{kn} \leq \frac 1n \log\# \mathcal{L}_n$
and sending ${k\rightarrow\infty}$ so that the left-hand side goes to ${h}$ (this is basically part of the proof of Fekete’s lemma).
To get an upper bound on ${\#\mathcal{L}_n}$ we need the specification property, which gives a 1-1 map ${\mathcal{L}_m\times \mathcal{L}_n \rightarrow \mathcal{L}_{m+n+\tau}}$, so that ${\#\mathcal{L}_{m+n+\tau} \geq \#\mathcal{L}_m \#\mathcal{L}_n}$. Then one can either apply Fekete’s lemma to ${b_n := -\log \# \mathcal{L}_{n+\tau}}$, or observe that
$\displaystyle \#\mathcal{L}_{k(n+\tau)} \geq (\#\mathcal{L}_n)^k \quad\Rightarrow\quad \frac 1{k(n+\tau)} \log \#\mathcal{L}_{k(n+\tau)} \geq \frac 1{n+\tau} \log\# \mathcal{L}_n.$
Sending ${k\rightarrow\infty}$ the left-hand side goes to ${h}$, and so combined with the lower bound above we get the uniform counting bounds
$\displaystyle e^{nh} \leq \#\mathcal{L}_n \leq e^{\tau h} e^{nh}. \ \ \ \ \ (13)$
4.2. A Gibbs measure
There is a standard procedure for constructing an MME: let ${\nu_n \in \mathcal{M}}$ be any sequence of (not necessarily invariant) Borel probability measures such that ${\nu_n(w) = 1/\#\mathcal{L}_n}$ for all ${w\in \mathcal{L}_n}$, and then put
$\displaystyle \mu_n = \frac 1n \sum_{k=0}^{n-1} \sigma_*^k \nu_n = \frac 1n \sum_{k=0}^{n-1} \nu_n \circ \sigma^{-k}.$
Since ${\mathcal{M}}$ is weak* compact, there is a weak* convergent subsequence ${\mu_{n_j}}$.
Exercise 6 Show that ${\mu := \lim_{j\rightarrow\infty} \mu_{n_j}}$ is ${\sigma}$-invariant (${\mu\circ \sigma^{-1} = \mu}$).
The preceding exercise is basically the proof of the Krylov-Bogolyubov theorem, and does not require any properties of the measures ${\nu_n}$ beyond the fact that they are Borel probability measures.
One can prove that ${\mu}$ is an MME whether or not ${X}$ has specification, but we will use specification to directly prove the stronger Gibbs property.
Given any ${w\in \mathcal{L}_m}$ and any ${\tau \leq k \leq n-\tau}$, we bound ${\nu_n(\sigma^{-k}[w])}$ by estimating how many words in ${\#\mathcal{L}_n}$ are of the form ${uwv}$ for some ${u \in \mathcal{L}_k}$ and ${v\in \mathcal{L}_{n-m-k}}$. Arguments similar to those in the uniform counting bounds show that
$\displaystyle \#\mathcal{L}_{k-\tau} \#\mathcal{L}_{n-k-m-\tau} \leq (\text{\# of such words}) \leq \#\mathcal{L}_k \#\mathcal{L}_{n-k-m},$
where the first inequality requires specification. Dividing by ${\#\mathcal{L}_n}$ and using the uniform counting bounds (13) gives
$\displaystyle \nu_n(\sigma^{-k}[w]) \leq \frac{\#\mathcal{L}_k \#\mathcal{L}_{n-k-m}}{\#\mathcal{L}_n} \leq \frac{e^{\tau h} e^{k h} e^{\tau h} e^{(n-k-m)h}}{e^{nh}} = e^{2\tau h} e^{-mh};$
using a similar estimate from below we get
$\displaystyle e^{-3\tau h} e^{-mh} \leq \nu_n(\sigma^{-k}[w]) \leq e^{2\tau h} e^{-mh}. \ \ \ \ \ (14)$
Averaging over all ${k}$ and sending ${n\rightarrow\infty}$ along the subsequence ${n_j}$ gives
$\displaystyle e^{-3\tau h} e^{-mh} \leq \mu(w) \leq e^{2\tau h} e^{-mh}, \ \ \ \ \ (15)$
so ${\mu}$ is a Gibbs measure.
4.3. Ergodicity
To prove that ${\mu}$ is ergodic, start by fixing ${v,w\in \mathcal{L}}$, with lengths ${|v|}$ and ${|w|}$, respectively. Given ${j\gg |v|}$ and ${k\ll n}$, follow the same procedure as above to estimate the number of words in ${\mathcal{L}_n}$ with the form ${xvywz}$, where ${x\in \mathcal{L}_k}$, ${y\in \mathcal{L}_{j - |v|}}$, and ${z\in \mathcal{L}_{n - k - j - |w|}}$, and obtain the bounds
$\displaystyle \frac{\#\mathcal{L}_{k-\tau} \#\mathcal{L}_{j-|v|-\tau} \#\mathcal{L}_{n - k - j - |w| - \tau}}{\#\mathcal{L}_n} \leq \sigma_*^k \nu_n([v] \cap \sigma^{-j} [w]) \leq \frac{\#\mathcal{L}_k \#\mathcal{L}_{j-|v|} \#\mathcal{L}_{n - k - j - |w|}}{ \#\mathcal{L}_n}.$
Averaging over ${k}$, sending ${n\rightarrow\infty}$, and using the uniform counting bounds (13) gives
$\displaystyle e^{-4\tau h} e^{-|v|h} e^{-|w|h} \leq \mu([v] \cap \sigma^{-j}[w]) \leq e^{3\tau h} e^{-|v|h} e^{-|w|h}.$
Using the Gibbs bounds (15) gives
$\displaystyle e^{-8\tau h} \mu(v) \mu(w) \leq \mu([v] \cap \sigma^{-j}[w]) \leq e^{9\tau h} \mu(v) \mu(w). \ \ \ \ \ (16)$
Exercise 7 Given Borel sets ${V,W \subset X}$, approximate ${V}$ and ${W}$ with cylinders and use (16) to get
$\displaystyle e^{-8\tau h} \mu(V) \mu(W) \leq \varliminf_{j\rightarrow\infty} \mu(V \cap \sigma^{-j}W) \leq \varlimsup_{j\rightarrow\infty} \mu(V \cap \sigma^{-j}W) \leq e^{9\tau h} \mu(V) \mu(W).$
Now if ${E \subset X}$ is invariant, then taking ${V=E}$ and ${W = E^c}$ in Exercise 7, the lower bound gives ${e^{-8\tau h} \mu(E)(1-\mu(E)) =0}$, so ${\mu(E)}$ is either ${0}$ or ${1}$. This proves ergodicity and completes the proof of Theorem 6.
Remark 2 In fact, the upper bound in Exercise 7 can be used to show that ${\mu}$ is mixing; see Proposition 20.3.6 in Katok and Hasselblatt.
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## Cohomologous functions and the Livsic theorem
Given a dynamical system ${f\colon X\rightarrow X}$ (typically ${X}$ is a compact metric space and ${f}$ is continuous), we commonly encounter real-valued functions ${\varphi\colon X\rightarrow {\mathbb R}}$ in one of the following two roles.
1. An observable function represents a measurement of the system, so that the sequence of functions ${\{\varphi\circ f^n\}}$ represents measurements made at different times, about which we want to make predictions. In the cases we are interested in, these predictions will be probabilistic and will be in terms of the Birkhoff averages ${\frac 1n S_n\varphi(x)}$, where ${S_n\varphi(x) = \sum_{k=0}^{n-1} \varphi(f^k x)}$ is the ${n}$th Birkhoff sum.
2. A potential function is used to assign weights to different trajectories of the system for purposes of selecting an invariant measure with specific dynamical or geometric properties. Generally the orbit segment ${x, f(x), \dots, f^{n-1}(x)}$ is assigned a weight given by ${e^{S_n\varphi(x)}}$. A potential function also assigns weights to invariant measures by integration; in particular, we can study equilibrium measures for ${(X,f,\varphi)}$, which are invariant probability measures maximizing ${h_\mu(f) + \int\varphi\,d\mu}$. (Here ${h_\mu}$ is Kolmogorov–Sinai entropy.)
Before moving on we remark that ${\varphi\colon X\rightarrow[0,\infty)}$ can also play the role of a density function, especially if we are looking for an invariant measure that is absolutely continuous with respect to some reference measure, but for today we will focus on the two roles described above.
Let ${\mathcal{M}_f(X)}$ denote the set of ${f}$-invariant Borel probability measures on ${X}$. Then each ${\varphi \in C(X)}$ induces a map ${\hat\varphi\colon \mathcal{M}_f(X) \rightarrow {\mathbb R}}$ by ${\hat\varphi(\mu) = \int\varphi\,d\mu}$ as in the second item above. It is natural to ask when two functions ${\varphi,\psi \in C(X)}$ have ${\hat\varphi=\hat\psi}$. One immediate observation to make is that the defintion of ${f}$-invariance implies that ${\int\varphi \,d\mu = \int\varphi\circ f \,d\mu}$ for all ${\varphi\in C(X)}$ and ${\mu\in \mathcal{M}_f(X)}$, so that in particular we have ${\hat\varphi = \widehat{\varphi\circ f}}$. Thus the function ${\psi = \varphi - \varphi\circ f}$ has ${\hat\psi = \hat \varphi - \widehat{\varphi\circ f} = 0}$. We call such a function ${\psi}$ a coboundary; we have just shown that
Proposition 1 if ${\psi}$ is a coboundary, then it integrates to ${0}$ with respect to any invariant measure.
Moreover, since integration is linear in ${\varphi}$, we have
Proposition 2 if two functions ${\varphi}$ and ${\psi}$ differ by a coboundary, then ${\hat\varphi=\hat\psi}$, ie., ${\int\varphi\,d\mu = \int\psi\,d\mu}$ for every ${\mu\in \mathcal{M}_f(X)}$.
If ${\varphi}$ and ${\psi}$ differ by a coboundary, we say that they are cohomologous; in this case we can write ${\varphi = \psi + h - h\circ f}$ for some ${h\in C(X)}$, which we call the transfer function. Note that ${\varphi}$ is a coboundary if and only if it is cohomologous to the zero function.
Remark 1 For a discussion of how this is connected to the notion of cohomology in algebraic topology, see Terry Tao’s blog post from December 2008.
It is natural to ask whether cohomology is the only mechanism by which two functions ${\varphi,\psi\in C(X)}$ can have ${\hat\varphi=\hat\psi}$; in other words, do the converses of Propositions 1 and 2 hold?
In general the answer is no, as one can quickly see by letting ${f}$ be an irrational circle rotation, so that the only invariant measure is Lebesgue, and there are many continuous functions on the circle that have Lebesgue integral ${0}$ but are not coboundaries. When ${f}$ has hyperbolic behavior, however, the story is quite different, and this is what we will discuss in this post.
If ${f}$ is a diffeomorphism and ${X}$ is a topologically transitive locally maximal hyperbolic set, then ${f\colon X\rightarrow X}$ satisfies the following result.
Theorem 3 (Closing Lemma) For every ${\epsilon>0}$ there exists ${\delta>0}$ such that if ${x\in X}$ and ${n\geq 0}$ are such that ${d(f^n(x),x) < \delta}$, then there exists ${y\in X}$ such that ${f^n(y) = y}$ and ${d(f^k(y), f^k(x)) < \epsilon}$ for all ${0\leq k < n}$.
Moreover, in this case every Hölder continuous ${\varphi\colon X\rightarrow {\mathbb R}}$ has the following property, introduced by Walters in 1978.
Theorem 4 (Walters Property) For every ${\zeta>0}$ there exist ${\epsilon>0}$ such that if ${x,y\in X}$ and ${n\geq 0}$ are such that ${d(f^k(x),f^k(y)) < \epsilon}$ for all ${0\leq k < n}$, then ${|S_n\varphi(x) - S_n\varphi(y)| < \zeta}$.
Both the Closing Lemma and the Walters Property also hold when ${X}$ is a subshift of finite type and ${\varphi}$ is Hölder continuous.
Using these properties we can formulate and prove an important result.
Theorem 5 (Livsic Theorem) Let ${X}$ be a compact metric space, ${f\colon X\rightarrow X}$ a continuous map satisfying the Closing Lemma and possessing a point whose orbit is dense, and ${\varphi\colon X\rightarrow{\mathbb R}}$ a continuous function satisfying the Walters Property. Then ${\varphi}$ is a coboundary if and only if for every periodic point ${x = f^p(x) \in X}$, we have ${S_p \varphi(x) = 0}$.
The forward implication is immediate: if ${\varphi}$ is a coboundary, then ${\int\varphi\,d\mu = 0}$ for every invariant ${\mu}$, in particular for ${\mu = \frac 1p \sum_{k=0}^{p-1} \delta_{f^k x}}$. Equivalently one can observe that Birkhoff sums of coboundaries have the following behavior: if ${\varphi(x) = h(x) - h(f x)}$, then
$\displaystyle S_n\varphi(x) = \sum_{k=0}^{n-1} \big(h(f^k x) - h(f^{k+1} x)\big) = h(x) - h(f^n x), \ \ \ \ \ (1)$
and consequently ${S_p \varphi(x) = h(x) - h(f^p x) = 0}$ whenever ${x=f^p x}$. We must prove the reverse implication, that if ${S_p\varphi(x)=0}$ whenever ${x=f^p x}$, then ${\varphi}$ is a coboundary. To this end note that if ${h}$ is a continuous (hence uniformly continuous) transfer function for ${\varphi}$, then (1) immediately determines ${h}$ along the entire forward orbit of a point ${y}$ by
$\displaystyle h(f^n y) = h(y) + S_n\varphi(y). \ \ \ \ \ (2)$
(A similar procedure defines ${h}$ along the backward orbit if ${f}$ is invertible.) If ${y}$ is a point whose orbit is dense in ${X}$, then this determines ${h}$ on all of ${X}$ by continuity. Thus it suffices to prove that (2) defines a uniformly continuous function on the orbit of ${y}$. To this end, given ${\zeta>0}$, let ${\epsilon = \epsilon(\zeta)>0}$ be given by the Walters Property, and then let ${\delta = \delta(\epsilon)>0}$ be given by the Closing Lemma. Suppose that ${w,z}$ are two points on the orbit of ${y}$ such that ${d(w,z) < \delta}$. Since ${w,z}$ lie on the same orbit, there is ${n\in {\mathbb N}}$ such that ${f^n(z)=w}$ or ${f^n(w)=z}$. Without loss of generality we assume the first case, so that ${d(z,f^n z) = d(z,w) < \delta}$. By the Closing Lemma there is a periodic point ${y=f^n y}$ such that ${d(f^k y, f^k z) < \epsilon}$ for all ${0\leq k < n}$. By (2) and the Walters Property, this implies that
$\displaystyle |h(w) - h(z)| = |h(f^n z) - h(z)| = |S_n\varphi(z)| \leq |S_n\varphi(x)| + \zeta = \zeta,$
where the last equality uses the fact that Birkhoff sums of ${\varphi}$ vanish around periodic orbits. This proves uniform continuity of ${h}$ on the orbit of ${y}$, so ${h}$ extends uniformly continuously to ${X}$, and it is an easy exercise using continuity of both sides of the equation ${\varphi = h - h\circ f}$ to verify that this relationship continues to hold on all of ${X}$, so that ${\varphi}$ is indeed a coboundary. This completes the proof of the Livsic Theorem.
Let ${X,f}$ be as in the Livsic Theorem; then for any ${\varphi,\psi}$ satisfying the Walters Property, the following are equivalent:
1. ${\varphi = \psi + h - h\circ f}$ for some ${h\in C(X)}$;
2. ${\int\varphi\,d\mu = \int\psi\,d\mu}$ for all ${\mu\in\mathcal{M}_f(X)}$;
3. ${\int\varphi\,d\mu = \int\psi\,d\mu}$ for all periodic orbit measures.
Here is one final remark on a specific example of cohomologous potentials: one immediately sees that ${\int \frac 1nS_n\varphi \,d\mu = \int \varphi\,d\mu}$ for every ${\mu\in\mathcal{M}_f(X)}$, and so under the conditions of the Livsic Theorem one can conclude that each Birkhoff average ${\frac 1nS_n\varphi}$ is cohomologous to ${\varphi}$. In fact this can be shown directly, without using the Livsic Theorem, by putting
\displaystyle \begin{aligned} h(x) &= \frac 1n\big((n-1) \varphi(x) + (n-2)\varphi(f x) + (n-3)\varphi(f^2 x) + \cdots + \varphi(f^{n-2} x)\big) \\ &= \frac 1n\sum_{k=0}^{n-1} (n-1-k) \varphi(f^{k-1} x). \end{aligned}
Then we have
\displaystyle \begin{aligned} h(x) - h(fx) &= \frac 1n\big((n-1) \varphi(x) + (n-2)\varphi(f x) + \cdots + \varphi(f^{n-2} x)\big) \\ &\quad\quad - \frac 1n\big((n-1) \varphi(f x) + (n-2)\varphi(f^2 x) + \cdots + \varphi(f^{n-1} x)\big) \\ &= \frac{n-1}n \varphi(x) - \frac 1n\big( \varphi(fx) + \varphi(f^2x) + \cdots + \varphi(f^{n-1} x) \big) \\ &= \varphi(x) - \frac 1n S_n\varphi(x), \end{aligned}
which demonstrates that ${\varphi}$ and ${\frac 1nS_n\varphi}$ are cohomologous.
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## Gibbs measures have local product structure
Let ${M}$ be a compact smooth manifold and ${f\colon M\rightarrow M}$ a transitive Anosov ${C^{1+\alpha}}$ diffeomorphism. If ${\mu}$ is an ${f}$-invariant Borel probability measure on ${M}$ that is absolutely continuous with respect to volume, then the Hopf argument can be used to show that ${\mu}$ is ergodic. In fact, recently it has been shown that even stronger ergodic properties such as multiple mixing can be deduced; for example, see Coudène, Hasselblatt, and Troubetzkoy [Stoch. Dyn. 16 (2016), no. 2].
A more general class of measures with strong ergodic properties is given by the theory of thermodynamic formalism developed in the 1970s: given any Hölder continuous potential ${\varphi\colon M\rightarrow{\mathbb R}}$, the quantity ${h_\mu(f) + \int\varphi\,d\mu}$ is maximized by a unique invariant Borel probability measure ${\mu = \mu_\varphi}$, which is called the equilibrium state of ${\varphi}$; the maximum value of the given quantity is the topological pressure ${P = P(\varphi)}$. The unique equilibrium state has the Gibbs property: for every ${\varepsilon>0}$ there is a constant ${K>0}$ such that
$\displaystyle K^{-1} \leq \frac{\mu(B_n(x,\varepsilon))}{e^{S_n\varphi(x) - nP}} \leq K, \ \ \ \ \ (1)$
where ${B_n(x,\varepsilon) = \{y\in M : d(f^kx,f^ky) < \varepsilon\ \forall 0\leq k < n\}}$ is the Bowen ball around ${x}$ of order ${n}$ and radius ${\varepsilon}$, and we write ${S_n\varphi(x) = \sum_{k=0}^{n-1} \varphi(f^kx)}$ for the ${n}$th ergodic sum along the orbit of ${x}$.
Historically, strong ergodic properties (mixing, K, Bernoulli) for equilibrium states have been established using methods such as Markov partitions rather than via the Hopf argument. However, in the more general non-uniformly hyperbolic setting, it can be difficult to extend these symbolic arguments, and so it is interesting to ask whether the Hopf argument can be applied instead, even if it only recovers some of the strong ergodic properties. The key property of absolutely continuous measures that is needed for the Hopf argument is the fact that they have a local product structure, which we define below. It was shown by Haydn [Random Comput. Dynam. 2 (1994), no. 1, 79–96] and by Leplaideur [Trans. Amer. Math. Soc. 352 (2000), no. 4, 1889–1912] that in the uniformly hyperbolic setting, the equilibrium states ${\mu_\varphi}$ have local product structure when ${\varphi}$ is Hölder continuous; thus one could apply the Hopf argument to them.
This post contains a direct proof that any measure ${\mu}$ with the Gibbs property (1) has local product structure; see Theorem 3 below. (This will be a bit of a longer post, since we need to recall several different concepts and then do some non-trivial technical work.) Since Bowen’s proof of uniqueness of equilibrium states using specification [Math. Systems Theory 8 (1974/1975), no. 3, 193–202] establishes the Gibbs property, this means that the equilibrium states produced this way could be addressed with the Hopf argument (I haven’t carried out the details yet, so I claim no formal results here). I should point out, though, that even without the use of Markov partitions, Ledrappier showed that these measures have the K property, which in particular implies multiple mixing. Since multiple mixing is the strongest thing we might hope to get from the Hopf argument, my primary motivation for the present approach is that Dan Thompson and I recently generalized Bowen’s result to systems satisfying a certain non-uniform specification property [Adv. Math. 303 (2016), 745–799], and the unique equilibrium states we obtain satisfy a non-uniform version of the Gibbs property (1), so it is reasonable to hope that they also have local product structure and can be studied using the Hopf argument; but this is beyond the scope of this post and will be addressed in a later paper.
1. Local product structure
Before defining local product structure for ${\mu}$, we recall some definitions. Since ${f}$ is Anosov, every ${x\in M}$ has local stable and unstable manifolds ${W^{s,u}_x}$, which have the following properties.
1. There are ${C \geq 1}$ and ${\lambda < 1 }$ such that for all ${x\in M}$, ${y\in W^s_x}$, and ${n\geq 0}$, we have ${d(f^n x, f^n y) \leq C\lambda^n d(x,y)}$; a similar contraction bound holds going backwards in time when ${y\in W^u_x}$.
2. There is ${\delta>0}$ such that if ${d(f^n x, f^ny) \leq \delta}$ for all ${n\geq 0}$, then ${y\in W^s_x}$; similarly for ${W^u_x}$ with ${n\leq 0}$.
3. There is ${\varepsilon>0}$ such that if ${d(x,y) \leq \varepsilon}$, then ${W_x^s \cap W_y^u}$ is a single point, which we denote ${[x,y]}$. Moreover, there is a constant ${Q}$ such that ${d([x,y],x) \leq Q d(x,y)}$, and similarly for ${d([x,y],y)}$.
A set ${R\subset M}$ is a rectangle if it has diameter ${\leq\varepsilon}$ and is closed under the bracket operation: in other words, for every ${x,y\in R}$, the intersection point ${[x,y] = W_x^s \cap W_y^u}$ exists and is contained in ${R}$.
Lemma 1 For every ${x\in M}$, there is a rectangle ${R}$ containing ${x}$.
Proof: Write ${V_x^s = \{y\in W_x^s : d(x,y) \leq \frac{\varepsilon}{2(Q+1)}\}}$ and similarly for ${V_x^u}$. Consider the set ${R = \{[y,z] : y\in V_x^u, z\in V_x^s\}}$ and observe that for every ${[y,z]\in R}$ we have
$\displaystyle d([y,z],x) \leq d([y,z],y) + d(y,x) \leq (Q+1)d(y,x) \leq \tfrac\varepsilon2.$
Thus ${R}$ has diameter ${\leq \varepsilon}$, and for every ${[y,z], [y',z']\in R}$ we have
$\displaystyle [[y,z],[y',z']] = W_{[y,z]}^s \cap W_{[y',z']}^u = W^s_y \cap W^u_{z'} = [y,z'] \in R,$
so ${R}$ is indeed a rectangle. $\Box$
Given ${x\in M}$ and a rectangle ${R\ni x}$, let ${V^{s,u}_x = W^{s,u}_x \cap R}$. Then ${R}$ can be recovered from ${V^{s,u}_x}$ via the method in the proof above, as the image of the map ${[\cdot,\cdot] \colon V_x^u \times V_x^s \rightarrow M}$. Given measures ${\nu^{s,u}}$ on ${V^{s,u}_x}$, let ${\nu^u\otimes\nu^s}$ be the pushforward of ${\nu^s\times \nu^u}$ under this map; that is, for every pair of Borel sets ${A\subset V_x^u}$ and ${B\subset V_x^s}$, we put ${(\nu^u\otimes \nu^s)(\{[y,z] : y\in A, z\in B\}) = \nu^u(A)\nu^s(B)}$.
Definition 2 A measure ${\mu}$ has local product structure with respect to ${W^{s,u}}$ if for every ${x\in M}$ and rectangle ${R\ni x}$ there are measures ${\nu^{s,u}}$ on ${V^{s,u}_x}$ such that ${\mu|_R \ll \nu^u \otimes \nu^s}$.
Theorem 3 Let ${f\colon M\rightarrow M}$ be a ${C^1}$ Anosov diffeomorphism, ${\varphi\colon M\rightarrow {\mathbb R}}$ a Hölder continuous function, and ${\mu}$ an ${f}$-invariant Borel probability measure on ${M}$ satisfying the Gibbs property (1) for some ${P\in {\mathbb R}}$. Then ${\mu}$ has local product structure in the sense of Definition 2. Moreover, there is ${\bar K \geq 1}$ such that for all ${R}$, the measures ${\nu^{s,u}}$ can be chosen so that the Radon–Nikodym derivative ${\psi = \frac{d\mu}{d(\nu^u\otimes \nu^s)}}$ satisfies ${\bar K^{-1} \leq \psi \leq \bar K}$ at ${\mu}$-a.e. point.
A quick side remark: here the diffeomorphism is only required to be ${C^1}$. The reason for the ${C^{1+\alpha}}$ hypothesis at the beginning of this post was so that the geometric potential ${\varphi(x) = -\log |\det Df|_{E^u_x}|}$ is Hölder continuous and its unique equilibrium state (the absolutely continuous invariant measure if it exists, or more generally the SRB measure) has the Gibbs property; this may not be true if ${f}$ is only ${C^1}$.
2. Conditional measures
In order to prove Theorem 3, we must start by recalling the notion of conditional measures; see Coudène’s book (especially Chapters 14 and 15) or Viana’s notes for more details than what is provided here.
Let ${(X,\mathop{\mathcal A},\mu)}$ be a Lebesgue space. A partition of ${X}$ is a map ${\xi \colon X\rightarrow \mathop{\mathcal A}}$ such that for ${\mu}$-a.e. ${x,y\in X}$, the sets ${\xi(x)}$ and ${\xi(y)}$ either coincide or are disjoint. Write ${\Xi = \{\xi(x) : x\in X\}}$ for the set of partition elements, and say that the partition ${\xi}$ is finite if ${\Xi}$ is finite.
Given a finite partition ${\xi}$, it is easy to define conditional measures ${\mu_{\xi(x)}}$ on the set ${\xi(x)}$ for ${\mu}$-a.e. ${x}$ by writing
$\displaystyle \mu_{\xi(x)}(A) = \frac{\mu(A \cap \xi(x))}{\mu(\xi(x))} \ \ \ \ \ (2)$
when ${\mu(\xi(x))>0}$, and ignoring those partition elements with zero measure. One can recover the measure ${\mu}$ from its conditional measures by the formula
$\displaystyle \mu(A) = \sum_{C\in \Xi} \mu_C(A) \mu(C). \ \ \ \ \ (3)$
If we write ${\hat\mu}$ for the measure on ${\Xi}$ defined by putting ${\hat\mu(\{C\}) = \mu(C)}$ for all ${C\in \Xi}$, then (3) can be written as
$\displaystyle \mu(A) = \int_\Xi \mu_C(A) \,d\hat\mu(C). \ \ \ \ \ (4)$
Even when the partition ${\xi}$ is infinite, one may still hope to obtain a formula along the lines of (4).
Example 1 Let ${X}$ be the unit square, ${\mu}$ be two-dimensional Lebesgue measure, and ${\xi(x)}$ be the horizontal line through ${x}$. Then Fubini’s theorem gives (4) by taking ${\mu_C}$ to be Lebesgue measure on horizontal lines, and defining ${\hat\mu}$ on ${\Xi}$, the set of horizontal lines in ${[0,1]^2}$, in one of the two following (equivalent) ways:
1. given ${E \subset \Xi}$, let ${\hat\mu(E) = \mu(\bigcup_{C\in E} C)}$;
2. identify ${\Xi}$ with the interval ${\{0\}\times [0,1]}$ on the ${y}$-axis, and define ${\hat\mu}$ as the image of one-dimensional Lebesgue measure on this interval.
Note that ${\hat\mu}$ must satisfy the first of these no matter what the partition ${\xi}$ is, while the second is a convenient description of ${\hat\mu}$ in this particular example.
A similar-looking example (and the one which is most relevant for our purposes) comes by letting ${R\subset M}$ be a rectangle and letting ${\xi(y) = V_y^u = W_y^u\cap R}$ be the partition into local unstable leaves. To produce conditional measures ${\mu_y^u = \mu_{V_y^u}}$, we need to use the fact that the partition ${\xi}$ is measurable. This means that there is a sequence of finite partitions ${\xi_n}$ that refines to ${\xi}$ in the sense that for ${\mu}$-a.e. ${y}$, we have
$\displaystyle \xi_{n+1}(y) \subset \xi_n(y) \text{ for all } n, \text{ and } \xi(y) = \textstyle\bigcap_{n=1}^\infty \xi_n(y).$
Lemma 4 Given any rectangle ${R}$, the partition of ${R}$ into local unstable leaves is measurable.
Proof: Fix ${x\in R}$ and let ${\{\eta_n(y)\}}$ be a refining sequence of finite partitions of ${V_x^s}$ with the property that ${\bigcap_n \eta_n(y) = \{y\}}$ for all ${y\in V_x^s}$; then let ${\xi_n(y) = \bigcup_{z\in \eta_n(y)} V_z^u}$, and we are done. $\Box$
Whenever ${\xi}$ is a measurable partition of a compact metric space ${X}$, we can define the conditional measures ${\mu_{\xi(y)}}$ as the limits of the conditional measures ${\mu_{\xi_n(y)}}$. Indeed, one can show (we omit the proofs) that for ${\mu}$-a.e. ${y}$, the limit ${I_y(\varphi) := \lim_{n\rightarrow\infty} \int \varphi \,d\mu_{\xi_n(y)}}$ exists for every continuous ${\varphi\colon X\rightarrow{\mathbb R}}$ and defines a continuous linear functional ${I_y\in C(X)^*}$; the corresponding measures ${\mu_{\xi(y)}}$ satisfy
$\displaystyle \int\varphi\,d\mu = \int_\Xi \int_C \varphi\,d\mu_C \,d\hat\mu(C) \ \ \ \ \ (5)$
for every ${\varphi\in C(X)}$, where once again we put ${\hat\mu(E) = \mu(\bigcup_{C\in E} C)}$.
The key result that we will need to describe properties of ${\mu_y^u}$ when ${\xi}$ is the partition of ${R}$ into local unstable leaves is that for ${\mu}$-a.e. ${y\in R}$ and every ${\varphi\in C(X)}$, we have
$\displaystyle \int \varphi\,d\mu_y^u = \lim_{n\rightarrow\infty} \frac 1{\mu(\xi_n(y))}\int_{\xi_n(y)} \varphi\,d\mu, \ \ \ \ \ (6)$
where ${\xi_n}$ is any sequence of finite partitions that refines to ${\xi}$.
In order to establish the local product structure for ${\mu}$ that is claimed by Theorem 3, we will show that the measures ${\mu_y^u}$ vary in an absolutely continuous manner as ${y}$ varies within ${R}$. That is, we consider for every ${x,y\in R}$ the holonomy map ${\pi_{x,y} \colon V_x^u \rightarrow V_{y}^u}$ defined by moving along local stable manifolds, so
$\displaystyle \pi_{x,y}(z) = [z,y] = V_z^s \cap V_{y}^u \text{ for all } z\in V_x^u.$
Our goal is to use the Gibbs property (1) for ${\mu}$ to prove that for every rectangle ${R}$ and ${\mu}$-a.e. ${x,y\in R}$, the conditional measures ${\mu_x^u}$ and ${\mu_{y}^u}$ satisfy
$\displaystyle (\pi_{x,y})_* \mu_x^u \ll \mu_{y}^u \text{ with } \bar{K}^{-1} \leq \frac{d(\pi_{x,y})_* \mu_x^u}{d\mu_{y}^u} \leq \bar{K}. \ \ \ \ \ (7)$
Once this is established, we can proceed as follows. Consider a rectangle ${R}$ with the decomposition ${\xi}$ into local unstable manifolds, let ${x\in R}$ be such that (7) holds for ${\mu}$-a.e. ${y\in R}$, and then identify ${\Xi}$ with ${V_x^s}$, as in the second characterization of ${\hat\mu}$ in Example 1. Let ${\nu^s}$ be the measure on ${V_x^s}$ corresponding to ${\hat\mu}$ under this identification, and let ${\nu^u = \mu_x^u}$. Given ${y\in V_x^s}$, let ${\psi_y = \frac{d(\pi_{x,y})_* \mu_x^u}{d\mu_{y}^u} \colon V_y^u \rightarrow [\bar{K}^{-1},\bar{K}]}$, so that in particular we have
$\displaystyle \int_{V_y^u} \varphi\,d\mu_y^u = \int_{V_y^u} \frac{\varphi}{\psi_y}\, d(\pi_{x,y})_*\mu_x^u = \int_{V_x^u} \frac{\varphi([z,y])}{\psi_y([z,y])} \,d\mu_x^u$
for every continuous ${\varphi\colon R\rightarrow {\mathbb R}}$. Then by (5), we have
\displaystyle \begin{aligned} \int\varphi\,d\mu &= \int_{\Xi} \int_C \varphi \,d\mu_C \,d\hat\mu(C) = \int_{V_x^s} \int_{V_y^u} \varphi \,d\mu_y^u \,d\nu^s(y) \\ &= \int_{V_x^s} \int_{V_x^u} \frac{\varphi([z,y])}{\psi_y([z,y])} \,d\nu^u(z) \,d\nu^s(y). \end{aligned}
By Definition 2, this shows that ${\mu}$ has local product structure with respect to ${W^{s,u}}$. Thus in order to prove Theorem 3, it suffices to shows that Gibbs measures satisfy the absolute continuity property (7).
It is worth noting quickly that our use of the term “absolute continuity” here has a rather different meaning from another common concept, which is that of a measure with “absolutely continuous conditional measures on unstable manifolds”. This latter notion is essential for the definition of SRB measures (indeed, in the Anosov setting it is the definition), and involves comparing ${\mu_x^u}$ to volume measure on ${V_x^u}$, instead of to the pushforwards of other conditional measures under holonomy.
3. Adapted partitions
In order to prove the absolute continuity property (7), we need to obtain estimates on ${\mu_y^u}$. We start by getting estimates on ${\mu}$ from the Gibbs property (1), and then using these to get estimates on ${\mu_y^u}$ using (6).
We will need a family of partitions of ${R}$ that refines to the partition into points. Fix a reference point ${q\in R}$, and suppose we have chosen partitions ${\eta_n^s}$ of ${V_q^s}$ and ${\eta_m^u}$ of ${V_q^u}$ for ${m,n\geq 0}$. Then we can define a partition ${\xi_{m,n}}$ of ${R}$ by taking the direct product of these two partitions, using the foliations of ${R}$ by local stable and unstable leaves: that is, we put
$\displaystyle \xi_{m,n}(y) = \{z\in R : \eta_m^u([z,q]) = \eta_m^u([y,q]) \text{ and } \eta_n^s([q,z]) = \eta_n^s([q,y]) \} \ \ \ \ \ (8)$
In order to obtain information on ${\mu(\xi_{m,n}(y))}$ using the Gibbs property (1), we need to put an extra condition on the partitions we use; we need to them to be adapted, meaning that each partition element both contains a Bowen ball and is contained within a larger Bowen ball. Most of the ideas here are fairly standard in thermodynamic formalism, but it is important for us to work separately on the stable and unstable manifolds, then combine the two, so we describe things explicitly. Fix ${\varepsilon>0}$. Given ${y\in V_q^u}$ and ${m\geq 0}$, let
\displaystyle \begin{aligned} B_m^u(y,\varepsilon) &= \{y'\in V_q^u : d(f^ky' , f^k y) \leq \varepsilon\ \forall 0\leq k \leq m \} \\ &= \{y'\in R : d(f^k y', f^ky) \leq \varepsilon\ \forall -\infty < k \leq m \}. \end{aligned}
Similarly, given ${z\in V_q^s}$ and ${n\geq 0}$, let
\displaystyle \begin{aligned} B_n^s(z,\varepsilon) &= \{z'\in V_q^s : d(f^{-k} z', f^{-k}z) \leq \varepsilon\ \forall 0\leq k \leq n\} \\ &= \{z'\in R : d(f^k z', f^kz) \leq \varepsilon\ \forall -n\leq k < \infty\}. \end{aligned}
Given ${\varepsilon > 0}$, we say that the partitions ${\eta_m^u}$ are ${\varepsilon}$-adapted if for every partition element ${\eta_m^u(y')}$ there is ${y\in V_q^u}$ such that
$\displaystyle B_m^u(y,\varepsilon/2) \subset \eta_m^u(y') \subset B_m^u(y,\varepsilon).$
We make a similar definition for ${\eta_n^s}$ using ${B_n^s}$. Note that we can produce an ${\varepsilon}$-adapted sequence of partitions ${\eta_n^s}$ as follows:
1. say that ${E\subset V_q^s}$ is ${(n,\varepsilon)}$-separated if for every ${x\neq y \in E}$ there is ${0\leq k\leq n}$ such that ${d(f^{-k}x,f^{-k}y) > \varepsilon}$;
2. let ${E \subset V_q^s}$ be a maximal ${(n,\varepsilon)}$-separated set and observe that ${\bigcup_{x\in E} B_n^s(x,\varepsilon) = V_q^s}$, while the sets ${\{B_n^s(x,\varepsilon/2) : x\in E \}}$ are disjoint;
3. enumerate ${E}$ as ${x_1,\dots, x_\ell}$, and build an ${\varepsilon}$-adapted partition ${\eta_n^s}$ by considering the sets
$\displaystyle C_j = B_n^s(x_j,\varepsilon/2) \cup \Big( V_q^s \setminus \bigcup_{i>j} B_n^s(x_i,\varepsilon)\Big). \ \ \ \ \ (9)$
Lemma 5 Let ${\mu}$ be a measure satisfying the Gibbs property (1). Then there are ${\varepsilon > 0}$ and ${K_0>0}$ such that if ${\eta_m^u}$ and ${\eta_n^s}$ are ${\varepsilon}$-adapted partitions of ${V_q^u}$ and ${V_q^s}$, then the product partition ${\xi_{m,n}}$ defined in (8) satisfies
$\displaystyle K_0^{-1} \leq \frac{\mu(\xi_{m,n}(x))}{e^{-(n+m)P + \sum_{k=-n}^{m-1} \varphi(f^k x)}} \leq K_0 \ \ \ \ \ (10)$
for every ${x\in R}$.
Proof: First we show that the upper bound in (10) holds whenever ${\varepsilon>0}$ is sufficiently small. Fix ${\varepsilon'>0}$ such that the Gibbs bound (1) holds for Bowen balls of radius ${2\varepsilon'}$, and such that ${\varepsilon'}$ is significantly smaller than the size of any local stable or unstable leaf. It suffices to show that for every ${x\in R}$ we have ${\xi_{m,n}(x) \subset B_{-n,m}(y,2\varepsilon') := f^{-n}(B_{n+m}(f^n y, 2\varepsilon'))}$ for some ${y}$, since then (1) gives the upper bound, possibly with a different constant; note that replacing ${x}$ with ${y}$ in the denominator changes the quantity by at most a constant factor, using the fact that ${\varphi}$ is Hölder continuous together with some basic properties of Anosov maps. (See, for example, Section 2 of this previous post for a proof of this Bowen property.)
Given ${x,y\in M}$ such that ${x}$ and ${y}$ lie on the same local stable leaf with ${d(x,y) \geq \varepsilon'}$, let
$\displaystyle r^u(x,y) = \inf\{d(x',y') : x'\in W_x^u, y'\in W_y^u \cap W_{x'}^s\}$
be the closest that any holonomy along local unstable leaves can bring ${x}$ and ${y}$. Note that ${r^u}$ is positive and continuous in ${x}$ and ${y}$; by compactness there is ${\bar{r}^u > 0}$ such that ${r^u(x,y) \geq \bar{r}^u}$ for all ${x,y}$ as above. In particular, this means that if ${x,y}$ are the images under an (unstable) holonomy of some ${x',y'}$ with ${d(x',y') < \bar{r}^u}$, then we must have ${d(x,y) <\varepsilon}$.
Choose ${\bar{r}^s}$ similarly for stable holonomies, and fix ${\varepsilon <\min(\bar{r}^s,\bar{r}^u)}$. Fix ${y\in V_q^u}$, ${z\in V_q^s}$, ${y'\in B_m^u(y,\varepsilon)}$, and ${z'\in B_n^s(z,\varepsilon)}$. Then for every ${-n\leq k\leq m}$ we have
$\displaystyle d(f^k[y',z'],f^k[y,z]) \leq d(f^k[y',z'],f^k[y',z]) + d(f^k[y',z],f^k[y,z]). \ \ \ \ \ (11)$
These distances can be estimated by observing that ${f^k[y',z']}$ and ${f^k[y',z]}$ are the images of ${f^k z'}$ and ${f^kz}$ under a holonomy map along local unstables, and similarly ${f^k[y',z]}$ and ${f^k[y,z]}$ are images of ${y}$ and ${y'}$ under a holonomy map along local stables. Then our choice of ${\varepsilon}$ shows that both quantities in the right-hand side of (11) are ${\leq \varepsilon'}$, which gives the inclusion we needed. This proves the upper bound in (10); the proof of the lower bound is similar. $\Box$
4. A refining sequence of adapted partitions
Armed with the formula from Lemma 5, we may look at the characterization of conditional measures in (6) and try to prove that the absolute continuity bound (7) holds whenever ${x,y\in R}$ both satisfy (6). There is one problem before we do this, though; the formula in (6) requires that the sequence of partitions ${\eta_n^s}$ be refining, and there is no a priori reason to expect that the adapted partitions produced by the simple argument before Lemma 5 refine each other. To get this additional property, we must do some more work. To the best of my knowledge, the arguments here are new.
4.1. Strategy
Start by letting ${E_n \subset V_q^s}$ be a maximal ${(n, 10\varepsilon)}$-separated set. We want to build a refining sequence of adapted partitions using the sets ${E_n}$, where instead of using Bowen balls with radius ${\varepsilon/2}$ and ${\varepsilon}$ we will use Bowen balls with radius ${\varepsilon}$ and ${20\varepsilon}$; this does not change anything essential about the previous section. We cannot immediately proceed as in the argument before Lemma 5, because if we are not careful when choosing the elements of the partition ${\eta_n^s}$, their boundaries might cut the Bowen balls ${B_{n'}^s(x,\varepsilon)}$ for ${x\in E_{n'}}$ and ${n' > n}$, spoiling our attempt to build an adapted partition ${\eta_{n'}^s}$ that refines ${\eta_n^s}$.
A first step will be to only build ${\eta_n^s}$ for some values of ${n}$. More precisely, we fix ${\ell\in {\mathbb N}}$ such that ${C\lambda^\ell < \frac 13}$, so that for every ${x\in M}$ and ${y\in W_x^s}$, we have ${d(f^\ell x, f^\ell y) < \frac 13 d(x,y)}$; then writing
$\displaystyle d_n^s(x,y) := \max_{0\leq k \leq n} d(f^{-k}x, f^{-k}y), \ \ \ \ \ (12)$
we have the following for every ${n,t\in {\mathbb N}}$:
$\displaystyle d_n^s(x,y) \leq 3^{-t} d_{n+t\ell}^s(x,y) \text{ whenever } d_{n+t\ell}^s(x,y) < \varepsilon. \ \ \ \ \ (13)$
We will only build adapted partitions for those ${n}$ that are multiples of ${\ell}$. We need to understand when the Bowen balls associated to points in the sets ${E_n}$ can overlap. We write ${\mathcal{E} = \{(x,n) \in V_q^s \times \ell{\mathbb N} : x\in E_n \}}$.
Definition 6 An ${\varepsilon}$-path between ${(x,n)\in \mathcal{E}}$ and ${(x',n')\in \mathcal{E}}$ is a sequence ${\{(z_i, k_i) \in \mathcal{E} : 0\leq i\leq m \}}$ with ${(z_0,k_0) = (x,n)}$ and ${(z_m,k_m) = (x',n')}$ such that ${B_{k_i}^s(z_i,\varepsilon) \cap B_{k_{i+1}}^s(z_{i+1},\varepsilon) \neq \emptyset}$ for all ${0\leq i < m}$. A subset ${\mathcal{C} \subset \mathcal{E}}$ is ${\varepsilon}$-connected if there is an ${\varepsilon}$-path between any two elements of ${\mathcal{C}}$.
Given ${n\in {\mathbb N}}$, let ${\mathcal{E}_{\geq n} = \{(x,k)\in \mathcal{E} : k\geq n\}}$. In the next section we will prove the following.
Proposition 7 If ${\mathcal{C} \subset \mathcal{E}_{\geq n}}$ is ${\varepsilon}$-connected and ${(x,n)\in \mathcal{C}}$, then
$\displaystyle U(\mathcal{C}) := \bigcup_{(y,k) \in \mathcal{C}} B_k^s(y,\varepsilon) \subset B_n^s(x,5\varepsilon). \ \ \ \ \ (14)$
For now we show how to build a refining sequence of adapted partitions assuming (14) by modifying the construction in (9). The key is that we build our partitions ${\eta_n^s}$ so that for every ${\varepsilon}$-connected ${\mathcal{C} \subset \mathcal{E}_{\geq n}}$, the set ${U(\mathcal{C})}$ is completely contained in a single element of ${\eta_n^s}$.
Suppose we have built a partition ${\eta_{n-\ell}^s}$ with this property; we need to construct a partition ${\eta_n^s}$ that refines ${\eta_{n-\ell}^s}$, still has this property, and also has the property that every partition element ${A}$ has some ${(x,n)\in \mathcal{E}}$ such that ${B_n^s(x,\varepsilon) \subset A \subset B_n^s(x,20\varepsilon)}$. To this end, let ${Z}$ be an element of the partition ${\eta_{n-\ell}^s}$, and enumerate the ${\varepsilon}$-connected components of ${\mathcal{E}_{\geq n}}$ as ${\mathcal{C}_1,\dots, \mathcal{C}_m, \mathcal{D}_1,\mathcal{D}_2,\dots}$, where each ${\mathcal{C}_j}$ contains some ${(x_j,n)}$, while each ${\mathcal{D}_i}$ is in fact a subset of ${\mathcal{E}_{\geq n+\ell}}$.
It follows from (14) that ${U(\mathcal{C}_j) \subset B_n^s(x_j,5\varepsilon)}$ for all ${1\leq j\leq m}$. Given ${i\in {\mathbb N}}$, we can observe that ${U(\mathcal{D}_i) \subset B_{n_i}^s(y_i,5\varepsilon) \subset B_n^s(y_i,5\varepsilon)}$ for some ${(y_i,n_i)\in \mathcal{E}}$ with ${n_i > n}$. Moreover, the sets ${\{B_n^s(x_j,10\varepsilon) : 1\leq j\leq m\}}$ cover ${Z}$, so every ${U(\mathcal{D}_i)}$ must intersect some set ${B_n^s(x_j,10\varepsilon)}$, and hence be contained in some ${B_n^s(x_j,20\varepsilon)}$. Given ${i\in {\mathbb N}}$, let ${J(i) = \min \{j : U(\mathcal{D}_i) \subset B_n(x_i,20\varepsilon)\}}$. Then for each ${1\leq j\leq m}$, let ${I_j = \{i\in {\mathbb N} : J(i) = j\}}$. Define sets ${A_1,\dots, A_m}$ by
$\displaystyle A_j = U(\mathcal{C}_j) \cup \Big( \bigcup_{i\in I_j} U(\mathcal{D}_i) \Big) \cup \Big( Z\setminus \bigcup_{j' > j} B_n(x_{j'},20\varepsilon) \Big).$
It is not hard to verify that the sets ${A_j}$ form a partition of ${Z}$ such that ${B_n^s(x_j,\varepsilon) \subset A_j \subset B_n^s(x_j,20\varepsilon)}$, and such that every ${\varepsilon}$-connected component ${\mathcal{C}}$ of ${\mathcal{E}_{\geq n}}$ that has ${U(\mathcal{C}) \cap Z \neq\emptyset}$ is completely contained in some ${A_j}$. Repeating this procedure for the other elements of ${\eta_{n-\ell}^s}$ produces the desired ${\eta_n^s}$.
4.2. Proof of the proposition
Now we must prove Proposition 7. Let ${\mathcal{E} \subset V_q^s\times \ell{\mathbb N}}$ be as in the previous section, so that in particular, every ${(x,n)\in \mathcal{E}}$ has that ${n}$ is a multiple of ${\ell}$, and given any ${(x,n)\neq (y,n)\in \mathcal{E}}$ we have ${d_n^s(x,y) \geq 10\varepsilon}$.
The following concept is essential for our proof.
Definition 8 Given an ${\varepsilon}$-path ${\{(z_i,k_i)\}_{i=0}^m \subset \mathcal{E}}$, a ford is a pair ${0\leq i k_i}$ for all ${i. Think of “fording a river'' to get from ${B_{k_i}^s(z_i,\varepsilon)}$ to ${B_{k_j}^s(z_j,\varepsilon)}$ by going through deeper levels of the ${\varepsilon}$-path; the alternative is that ${k_a \leq k_i}$ for some ${i, in which case ${B_{k_a}^s(z_a,\varepsilon)}$ is a sort of “bridge'' between ${z_i}$ and ${z_j}$.
Lemma 9 Suppose that ${\{(z_i,k_i)\}_{i=0}^m \subset \mathcal{E}}$ is an ${\varepsilon}$-path without any fords, and that ${k_i \geq k_0}$ for all ${1\leq i \leq m}$. Then ${d_{k_0}^s(z_0,z_m) \leq 4\varepsilon}$.
Proof: By the definition of ${\varepsilon}$-path, for every ${0\leq a < m}$ there is a point ${y_a \in B_{k_a}^s(z_a,\varepsilon) \cap B_{k_{a+1}}^s(z_{a+1},\varepsilon)}$. Thus
\displaystyle \begin{aligned} d_{k_0}^s(z_0,z_m) &\leq d_{k_0}^s(z_0,y_0) + \Big( \sum_{a=1}^{m-1} d_{k_0}^s(y_{a-1}, z_a) + d_{k_0}^s(z_a,y_a) \Big) + d_{k_0}^s(y_{m-1},z_m) \\ &\leq 2\varepsilon + \sum_{a=1}^{m-1} (\tfrac 13)^{k_a - k} \big( d_{k_a}^s(y_{a-1}, z_a) + d_{k_a}^s(z_a,y_a) \big) \\ &\leq 2\varepsilon \Big( 1 + \sum_{a=i+1}^{j-1} (\tfrac13)^{k_a - k} \Big), \end{aligned} \ \ \ \ \ (15)
where the second inequality uses (13). Now we need to estimate how often different values of ${k_a}$ can appear. Let ${A = \{1, \dots, m-1\}}$; we claim that for every ${t\in {\mathbb N}}$, we have
$\displaystyle \#\{a\in A : k_a = k_0 + t\ell \} \leq 2^{t-1}. \ \ \ \ \ (16)$
First note that ${k_a > k_0}$ for all ${a\in A}$, because otherwise ${(0,a)}$ would be a ford. Since the ${\varepsilon}$-path has no fords, every ${i with ${k_i = k_j}$ must be separated by some ${a}$ with ${k_a < k_i}$, and we conclude that for every ${t\in {\mathbb N}}$, we have
\displaystyle \begin{aligned} \#\{a \in A : &\, k_a = k_0 + t\ell\} \leq 1 + \#\{a\in A : k_0 < k_a < k_0 + t\ell\} \\ &= 1 + \sum_{q=1}^{t-1} \#\{a\in A : k_a = k_0 + q\ell\}. \end{aligned}
For ${t=1}$, this establishes (16) immediately. If (16) holds up to ${t-1}$, then this gives
$\displaystyle \#\{a\in A : k_a = k_0 + t\ell\} \leq 1 + \sum_{q=1}^{t-1} 2^{q-1} = 2^{t-1},$
which proves (16) for all ${t\in {\mathbb N}}$ by induction. Combining (15) and (16), we have
\displaystyle \begin{aligned} d_k^s(z_i,z_j) &\leq 2\varepsilon \Big( 1 + \sum_{t=1}^\infty \Big(\frac13 \Big)^t \#\{a\in A : k_a = k_i + t\ell\}\Big) \\ &\leq 2\varepsilon + \varepsilon \sum_{t=1}^\infty \Big(\frac23 \Big)^t =4\varepsilon, \end{aligned}
which proves Lemma 9. $\Box$
Now we show that the ${10\varepsilon}$-separation condition rules out the existence of fords entirely.
Lemma 10 If ${\{(z_i,k_i)\}_{i=0}^m \subset \mathcal{E}}$ is an ${\varepsilon}$-path, then it has no fords.
Proof: Fix an ${\varepsilon}$-path ${\{(z_i,k_i)\}_{i=0}^m}$. Denote the set of fords by
$\displaystyle I = \{(i,j)\in \{0,1,\dots,m\}^2 : i < j \text{ and } k_i = k_j \leq k_a \text{ for all } i
our goal is to prove that ${I}$ is empty. Put a partial ordering on ${I}$ by writing
$\displaystyle (i',j') \preceq (i,j)\ \Leftrightarrow\ [i',j'] \subset [i,j] \ \Leftrightarrow\ i \leq i' < j' \leq j.$
If ${I}$ is non-empty, then since it is finite it must contain some element ${(i,j)}$ that is minimal with respect to this partial ordering. In particular, the ${\varepsilon}$-path ${(z_i,k_i),\dots, (z_j,k_j)}$ contains no fords, and so by Lemma 9 we have ${d_{k_i}^s(z_i,z_j) \leq 4\varepsilon}$, contradicting the assumption that ${d_{k_i}^s(z_i,z_j) \geq 10\varepsilon}$ (since ${E_k}$ is ${(k,10\varepsilon)}$-separated), and we conclude that ${I}$ must be empty, proving the lemma. $\Box$
Now we prove Proposition 7. Let ${\mathcal{C} \subset \mathcal{E}_{\geq n}}$ be ${\varepsilon}$-connected and fix ${(x,n) \in \mathcal{C}}$. Given any ${(y,k) \in \mathcal{C}}$, there is an ${\varepsilon}$-path ${\{(z_i,k_i)\}_{i=0}^m}$ such that ${(z_0,k_0) = (x,n)}$ and ${(z_m, k_m) = (y,k)}$. By Lemma 10, this path has no fords, and so Lemma 9 gives ${d_n^s(x,y) \leq 4\varepsilon}$. We conclude that ${B_n^s(y,\varepsilon) \subset B_n^s(x,5\varepsilon)}$, which proves Proposition 7.
5. Completion of the proof
Thanks to the previous section, we can let ${\eta_m^u}$ and ${\eta_n^s}$ be ${\varepsilon}$-adapted partitions of ${V_q^u}$ and ${V_q^s}$ such that ${\eta_{n'}^s}$ refines ${\eta_n^s}$ whenever ${n'>n}$ and ${n,n'}$ are both multiples of ${\ell}$. By Lemma 5, we have good lower and upper estimates on ${\mu(\xi_{m,n}(x))}$ for all ${x\in R}$, where ${\xi_{m,n}}$ is the product partition defined in (8). By (6), there is ${R'\subset R}$ with ${\mu(R\setminus R')=0}$ such that for every ${x\in R'}$, we have
$\displaystyle \int \psi\,d\mu_x^u = \lim_{n\rightarrow\infty} \frac 1{\mu(\xi_{0,n\ell}(x))}\int_{\xi_{0,n\ell}(x)} \psi\,d\mu \ \ \ \ \ (17)$
for every continuous ${\psi\colon M\rightarrow {\mathbb R}}$. The next step towards proving Theorem 3 is the following result.
Proposition 11 There is a constant ${K_1}$ such that for any ${x,y\in R'}$ and every continuous ${\psi\colon V_y^u \rightarrow (0,\infty)}$, we have ${\int \psi\,d(\pi_{x,y})_*\mu_x^u \leq K_1 \int\psi\,d\mu_y^u}$, where ${\pi_{x,y} \colon V_x^u \rightarrow V_y^u}$ is the holonomy map along local unstables.
Proof: Start by fixing for each ${m\in {\mathbb N}}$ a set ${D_m^q \subset V_q^u}$ such that every element of ${\eta_m^s}$ contains exactly one point in ${D_m^q}$. Then let ${D_m^x = \pi_{q,x}D_m^q}$ for every ${x\in R}$.
Given a positive continuous function ${\psi}$, there is ${\delta>0}$ such that if ${d(z,z')<\delta}$, then ${\psi(z) \leq 2\psi(z')}$. Thus for all sufficiently large ${m,n}$, we have ${\psi(z) \leq 2\psi(z')}$ whenever ${\xi_{m,n\ell}(z) = \xi_{m,n\ell}(z')}$. For any such ${m,n}$ and any ${x,y\in R'}$, we conclude that
\displaystyle \begin{aligned} \int_{\xi_{0,n\ell}(x)} \psi\circ \pi_{x,y}\,d\mu &= \sum_{z\in D_m^x} \int_{\xi_{m,n\ell}(z)} \psi\circ \pi_{x,y}\,d\mu \leq \sum_{z\in D_{m}^x} 2\psi(\pi_{x,y}z) \mu(\xi_{m,n\ell}(z)) \\ &\leq \sum_{z\in D_{m}^x} 2\psi(\pi_{x,y}z) K_0 e^{-(n+m)P + S_{n,m}\varphi(z)}, \end{aligned}
where we write ${S_{n,m}\varphi(z) = \sum_{k=-n\ell}^{m-1} \varphi(f^kz)}$ and where the last inequality uses Lemma 5. Similarly,
$\displaystyle \mu(\xi_{0,n\ell}(x)) = \sum_{z\in D_m^x} \mu(\xi_{m,n\ell}(z)) \geq K_0^{-1} e^{-(n+m) P + S_{n,m}\varphi(f^kz)},$
and we conclude from (6) that
$\displaystyle \int \psi\circ \pi_{x,y} \,d\mu_x^u \leq \varlimsup_{n\rightarrow\infty} \frac{\sum_{z\in D_m^x} 2\psi(\pi_{x,y}z) K_0 e^{S_{n,m}\varphi(z)}}{\sum_{z\in D_m^x} K_0^{-1} e^{S_{n,m}\varphi(z)}}. \ \ \ \ \ (18)$
Note that since ${\varphi}$ is Hölder continuous, there are ${\beta>0}$ and ${|\varphi|_\beta > 0}$ such that ${|\varphi(z)-\varphi(z')| \leq |\varphi|_\beta d(z,z')^\beta}$ for all ${z,z'\in M}$. Thus given any ${z\in V_x^u}$ and any ${n\geq 0}$, we have
\displaystyle \begin{aligned} \Big|\sum_{k=-n\ell}^{-1} \varphi(f^kx) &- \sum_{k=-n\ell}^{-1} \varphi(f^kz)\Big| \leq \sum_{k=1}^{n\ell} |\varphi(f^kx) - \varphi(f^kz)| \\ &\leq \sum_{k=1}^{n\ell} |\varphi|_\beta (C\lambda^k \varepsilon)^\beta < |\varphi|_\beta (C\varepsilon)^\beta \sum_{k=1}^\infty (\lambda^\beta)^k =: L < \infty. \end{aligned} \ \ \ \ \ (19)
Thus (18) gives
\displaystyle \begin{aligned} \int \psi \,d(\pi_{x,y})_*\mu_x^u &\leq 2K_0^2 \varlimsup_{n\rightarrow\infty} \frac{\sum_{z\in D_m^x} \psi(\pi_{x,y}z) e^{L \sum_{k=-n\ell}^{-1} \varphi(f^k x)} e^{S_m\varphi(z)}} {\sum_{z\in D_m^x} e^{L^{-1} \sum_{k=-n\ell}^{-1} \varphi(f^k x)} e^{S_m\varphi(z)}} \\ &\leq 2K_0^2 e^{2L} \frac{\sum_{z\in D_m^x} \psi(\pi_{x,y}z) e^{S_m\varphi(z)}}{\sum_{z\in D_m^x} e^{S_m\varphi(z)}}. \end{aligned} \ \ \ \ \ (20)
A similar set of computations for ${\mu_y^u}$ shows that
$\displaystyle \int\psi\,d\mu_y^u \geq \frac 1{2K_0^2 e^{2L}} \frac{\sum_{z'\in D_m^y} \psi(z') e^{S_m\varphi(z')}}{\sum_{z'\in D_m^y} e^{S_m\varphi(z')}}. \ \ \ \ \ (21)$
Since ${D_m^y = \pi_{x,y} D_m^x}$, we can rewrite the sums over ${D_m^y}$ as sums over ${D_m^x}$; for example,
$\displaystyle \sum_{z' \in D_m^y} e^{S_{m}\varphi(z')} = \sum_{z\in D_m^x} e^{S_{m}\varphi(\pi_{x,y} z)} \leq e^L \sum_{z\in D_m^x} e^{S_m\varphi(z)},$
where the inequality uses the fact that the estimate (19) also holds for forward ergodic averages of two points on the same local stable manifold. Using a similar estimate for the numerator in (21) gives
$\displaystyle \int \psi\,d\mu_y^u \geq \frac 1{2K_0^2 e^{4L}} \frac{\sum_{z\in D_m^x} \psi(\pi_{x,y}z) e^{S_m\varphi(z)}}{\sum_{z\in D_m^x} e^{S_m\varphi(z)}}.$
Together with (20), this gives
$\displaystyle \int \psi \,d(\pi_{x,y})_*\mu_x^u \leq 4K_0^4 e^{6L} \int\psi\,d\mu_y^u,$
which completes the proof of Proposition 11. $\Box$
To complete the proof of Theorem 3, we first observe that for every open set ${U \subset V_y^u}$, there is a sequence of continuous functions ${\psi_n\colon V_y^u \rightarrow (0,1]}$ that converge pointwise to the indicator function ${\mathbf{1}_U}$; applying Proposition 11 to these functions and using the dominated convergence theorem gives
\displaystyle \begin{aligned} d(\pi_{x,y})_*\mu_x^u(U) &= \int \mathbf{1}_U \,d(\pi_{x,y})_*\mu_x^u = \lim_{n\rightarrow\infty} \int \psi_n \,d(\pi_{x,y})_*\mu_x^u \\ &\leq \lim_{n\rightarrow\infty} K_1 \int\psi_n \,d\mu_y^u = K_1 \int \mathbf{1}_U \,d\mu_y^u = K_1\mu_y^u(U). \end{aligned}
Then for every measurable ${E \subset V_y^u}$, we have
\displaystyle \begin{aligned} d(\pi_{x,y})_*\mu_x^u(E) &= \inf \{ d(\pi_{x,y})_*\mu_x^u(U) : U \supset E \text{ is open} \} \\ &\leq K_1 \inf \{ \mu_y^u(U) : U \supset E \text{ is open} \} = K_1 \mu_y^u(E). \end{aligned}
This proves that ${d(\pi_{x,y})_*\mu_x^u \ll \mu_y^u}$ and that the Radon–Nikodym derivative is ${\leq K_1}$ ${\mu_y^u}$-a.e. The lower bound ${K_1^{-1}}$ follows since the argument is symmetric in ${x}$ and ${y}$. This proves (7) and thus completes the proof of Theorem 3.
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## Alpha-beta shifts
Given ${\beta>1}$, the interval map ${T\colon [0,1]\rightarrow [0,1]}$ given by ${T(x) = \beta x \pmod 1}$ is naturally semi-conjugate to the ${\beta}$-shift ${\Sigma_\beta \subset \{0,1,\dots,\lceil\beta\rceil-1\}^{\mathbb N}}$. The shift space ${\Sigma_\beta}$ admits a natural description in terms of the lexicographic order, and another in terms of a countable-state directed graph. These have been used to obtain a fairly complete description of the dynamics of the ${\beta}$-transformation ${T}$, including existence, uniqueness, and strong statistical properties of equilibrium states for Hölder potentials.
In this post I want to explain how one can give similar descriptions (both in terms of lexicographic order and in terms of a countable-state Markov structure) for the coding spaces associated to the interval map ${T(x) = \alpha + \beta x \pmod 1}$. Just as with the ${\beta}$-shifts, these “${\alpha}$${\beta}$ shifts” arise from piecewise expanding interval maps, and thus can be studied using the general machinery developed by Hofbauer, but I believe that they are worth studying a little more carefully as an intermediate case between the ${\beta}$-shifts, where the fact that ${0}$ is a `safe symbol’ can be used to prove various results that are still open for ${\alpha}$${\beta}$ shifts, and the more general setting where the description of the shift space requires more bookkeeping and it can be harder to see what is going on.
1. Coding spaces for interval maps
We start by recalling the general notion of a coding space for a map of the interval. Say that a map ${T\colon [0,1]\rightarrow [0,1]}$ is a piecewise expanding interval map if there is ${\lambda>1}$ and a partition of ${[0,1]}$ into finitely many intervals ${I_0,\dots, I_{d-1}}$ such that ${T}$ is continuous on each ${I_j}$ and ${C^1}$ on the interior of each ${I_j}$, with ${|T'| \geq \lambda}$. Note that we do not care whether the intervals are closed, open, or half of each.
Let ${A = \{0,1,\dots, d-1\}}$, and define a map ${h\colon [0,1]\rightarrow A^{\mathbb N}}$ by ${h(x)_n = a \in A}$ whenever ${T^n(x) \in I_a}$. Let ${\Sigma = \Sigma_T = \overline{h([0,1])}}$; we say that ${\Sigma}$ is the coding space for the interval map ${T}$ relative to the partition ${\{I_j\}_j}$. We can define a map ${\pi\colon \Sigma\rightarrow [0,1]}$ by the condition that ${\pi(h(x)) = x}$ for all ${x\in [0,1]}$ and ${\pi}$ is continuous. Then we have ${\pi\circ \sigma = T\circ \pi}$, so ${\pi}$ is a semiconjugacy between ${(\Sigma,\sigma)}$ and ${([0,1],T)}$.
Another way of interpreting this is the following: for each ${a\in A}$ there is a unique map ${S_a \colon \overline{T(I_a)} \rightarrow \overline{I_a}}$ such that ${S_a(T x) = x}$ for all ${x}$ in the interior of ${I_a}$. Note that ${S_a}$ is ${C^1}$ with ${|S_a'| \leq \lambda^{-1} < 1}$. Given any set ${E\subset [0,1]}$, write ${S_a(E) = S_a(E \cap \overline{T(I_a)})}$ for convenience; note that ${S_a(E)=\emptyset}$ if ${E}$ does not intersect ${\overline{T(I_a)}}$. Given a finite word ${w\in A^* = \bigcup_{n\geq 0} A^n}$, let ${J_w = S_w([0,1])}$, where ${S_w = S_{w_1} \circ \cdots S_{w_{|w|}}}$ and ${|w|}$ is the length of ${w}$. Roughly speaking, ${J_w}$ represents the set of points ${x\in [0,1]}$ for which ${x\in I_{w_1}}$, ${T(x) \in I_{w_2}}$, and so on. (This is not quite completely true because of problems at the endpoints of the intervals.)
The language of the shift ${\Sigma}$ is the set of all ${w\in A^*}$ such that ${[w] := \{x\in \Sigma : x_1 \cdots x_{|w|} = w\} \neq\emptyset}$. Write ${\mathcal{L}}$ for this collection of words; then ${\mathcal{L} = \{ w \in A^* : J_w \neq \emptyset\}}$, and we have ${\pi([w]) = J_w}$ for all ${w\in \mathcal{L}}$. Let ${\mathop{\mathcal L}_n = \{w\in \mathop{\mathcal L} : |w| = n\}}$; given ${w\in \mathop{\mathcal L}_n}$, the interval ${J_w}$ has length ${\leq \lambda^{-n}}$, and since for every ${x\in \Sigma}$ we have ${\{x\} = \bigcap_{n\geq 0} [x_1\cdots x_n]}$, we also have ${\{\pi(x)\} = \bigcap_{n\geq 0} J_{x_1 \cdots x_n}}$.
So far these considerations have been completely general, valid for any piecewise expanding interval map. Now we consider the specific transformations ${T_\beta(x) = \beta x \pmod 1}$ and ${T_{\alpha,\beta} = \alpha + \beta x \pmod 1}$, where ${z\pmod 1 := z - \lfloor z \rfloor}$ and we assume without loss of generality that ${\alpha \in [0,1)}$. These are piecewise expanding interval maps, where the natural partition to consider is the partition into maximal intervals of continuity. Thus for ${T_\beta}$, we have ${I_0 = [0,\frac 1\beta)}$, ${I_1 = [\frac 1\beta, \frac 2\beta)}$, and so on, with ${I_{d-1} = [\frac {d-1}\beta,1]}$, where ${d = \lceil\beta\rceil-1}$. For ${T_{\alpha,\beta}}$, we have ${I_0 = [0,\frac{1-\alpha}\beta)}$, ${I_1 = [\frac{1-\alpha}{\beta},\frac{2-\alpha}\beta)}$, and so on, with ${d = \lceil\alpha+\beta\rceil - 1}$ and ${I_{d-1} = [\frac{d-1+\alpha}{\beta},1]}$. We write ${\Sigma_\beta}$ and ${\Sigma_{\alpha,\beta}}$ for the coding spaces of these transformations relative to their natural partitions.
2. Description via lexicographic order
The full shift ${A^{\mathbb N}}$ carries a natural lexicographic order inherited from the usual order on ${A = \{0,1,\dots, d-1\}}$: given ${y\neq z\in A^{\mathbb N}}$, let ${n=n(y,z)}$ be minimal such that ${y\neq z}$, and write ${y\prec z}$ if ${y_n < z_n}$. Write ${y\preceq z}$ if ${y\prec z}$ or ${y=z}$. This is a total order on ${A^{\mathbb N}}$. The key observation for our purposes is that because ${T_\beta}$ and ${T_{\alpha,\beta}}$ are order-preserving on each of their intervals of continuity, their coding maps are also order-preserving in the sense that ${\pi(y) \leq \pi(z)}$ if and only if ${y\preceq z}$. (Note that we must use non-strict inequalities because the coding maps are not 1-1.)
2.1. The ${\beta}$-shifts
Fix ${\beta>1}$ and let ${T=T_\beta}$. Let ${{\mathbf b} = \sup \Sigma_\beta}$, where supremum is w.r.t. lexicographic order; the supremum exists because ${A^{\mathbb N}}$ has the least upper bound property, and is in ${\Sigma_\beta}$ because the ${\beta}$-shift is closed. Thus ${{\mathbf b} = b_1 b_2 b_3 \cdots}$ is the lexicographically maximal element of ${\Sigma_\beta}$; it would code the trajectory of the right endpoint of the unit interval if we replaced ${T_\beta}$ with the map ${T(x) = 1 + \beta x - \lceil \beta x \rceil}$, which agrees with ${T_\beta}$ everywhere except at the endpoints of the intervals of monotonicity. The points ${T^n(1) \in [0,1]}$ will play an important role in the following result.
Proposition 1 A sequence ${y\in A^{\mathbb N}}$ is in ${\Sigma_\beta}$ if and only if
$\displaystyle \sigma^n(y) \preceq {\mathbf b} \text{ for all } n=0,1,2,\dots. \ \ \ \ \ (1)$
Similarly, a word ${w\in A^*}$ is in ${\mathop{\mathcal L}(\Sigma_\beta)}$ if and only if
$\displaystyle w_{|w|-k+1}\cdots w_{|w|} \preceq b_1 \cdots b_k \text{ for all } 1\leq k\leq |w|. \ \ \ \ \ (2)$
Proof: Because ${\Sigma_\beta}$ is closed and ${\sigma}$-invariant, every ${y\in \Sigma_\beta}$ satisfies (1), and it follows immediately that every ${w\in \mathop{\mathcal L}(\Sigma_\beta)}$ satisfies (2).
To prove the converse direction, it suffices to show that every ${w\in A^*}$ which satisfies (2) is contained in ${\mathop{\mathcal L}}$; in other words, that ${J_w \neq \emptyset}$ for every such ${w}$. Equivalently, we can work with the following object: consider for each ${w \in A^*}$ the follower interval ${F_w = \overline{T^{|w|} (J_w)}}$; this can also be defined as the minimal interval with the property that ${S_w([0,1]\setminus F_w)=\emptyset}$. Note that ${F_w = \emptyset}$ whenever ${w\notin \mathop{\mathcal L}}$. Given ${w\in \mathop{\mathcal L}}$ and ${a\in A}$, observe that
$\displaystyle F_{wa} = \overline{T^{1+|w|}(J_{wa})} = \overline{T(T^{|w|}S_w S_a[0,1])} = \overline{T(T^{|w|}S_w I_a)} = \overline{T(F_w \cap I_a)}. \ \ \ \ \ (3)$
Now we can give a non-recursive description of ${F_w}$ that completes the proof of Proposition 1. If ${w\in A^*}$ satisfies (2), let
$\displaystyle k(w) = \max \{ k : w_{|w|-k+1} \cdots w_{|w|} = b_1 \cdots b_k \} \ \ \ \ \ (4)$
be the length of the longest prefix of ${{\mathbf b}}$ that appears as a suffix of ${w}$.
Lemma 2 If ${w\in A^*}$ satisfies (2), then ${F_w = [0,T^{k(w)}(1)]}$.
Proof: We go by induction in the length of ${w}$, using ${|w|=0}$ (so ${w}$ is the empty word and ${F_w = [0,1]}$) as the base case. Suppose we have some ${w\in A^n}$ satisfying (2), and that the lemma has been proved for all words with length ${. Write ${w = va}$ where ${|v| = n-1}$ and ${a\in A}$. Let ${j = k(v)}$, so that ${v=u b_1 \cdots b_j}$ for some word ${u}$. By the inductive hypothesis, we have ${F_v = [0,T^{j}(1)]}$ and hence ${v\in \mathop{\mathcal L}}$. Then by (3), we have
$\displaystyle F_w = F_{va} = \overline{T(F_v \cap I_a)} = \overline{T([0,T^{k(v)}(1)] \cap I_a)}.$
There are two cases to consider.
Case one: ${T^{k(v)}(1) \in I_a}$. In this case we have ${a=b_{j+1}}$ and ${k(w) = j+1}$, and ${F_w = \overline{T([\frac a\beta,T^{k(v)}(1)])} = [0,T^{k(v)+1}] = [0,T^{k(w)}(1)]}$.
Case two: ${T^{k(v)}(1) \notin I_a}$. In this case we must have ${a < b_{j+1}}$, otherwise we would have ${w=ub_1\cdots b_j a}$ and hence ${w_{|w|-j} \cdots w_{|w|} \succ b_1 \cdots b_{j+1}}$, violating (2). Thus ${I_a \subset F_v}$ and hence by (3), we have ${F_w = \overline{T(I_a)} = [0,1]}$. On the other hand, since ${a < b_{j+1}}$, we see that for every ${1\leq i\leq j}$ we have
$\displaystyle w_{|w|-j+i} \cdots w_{|w|}= b_i b_{i+1} \cdots b_j a \prec b_i \cdots b_j b_{j+1} \preceq b_1 \cdots b_{j-i+2},$
where the last inequality uses the fact that ${\sigma^{i-1}{\mathbf b} \preceq {\mathbf b}}$. We conclude that ${k(w) = 0}$, which completes the proof of Lemma 2. $\Box$
With Lemma 2 complete, Proposition 1 follows immediately since ${F_w\neq \emptyset}$ implies that ${w\in \mathop{\mathcal L}}$. $\Box$
A historical note: the transformation ${T_\beta}$ was introduced by Rényi in 1957 to study representations of real numbers in non-integer bases. The lexicographic characterization of ${\beta}$-shifts in Proposition 1 was given by Parry in 1960 (see Theorem 3 of that paper) using different methods than the proof here. This proof is essentially the same as the one given by Hofbauer in 1979 (see Theorem 2 there); Hofbauer’s approach has the advantage of dealing with all piecewise monotonic interval maps, and the argument given here for the ${\beta}$-shift is actually just a special case of his general result.
2.2. The ${\alpha}$${\beta}$ shifts
Instead of describing the full generality of Hofbauer’s result here, we describe how Proposition 1 adapts to the ${\alpha}$${\beta}$ shifts. This family of examples already reveals some of the challenges that arise when we go from ${\beta}$-shifts to more general piecewise expanding transformations.
Fix ${\alpha\in [0,1)}$ and ${\beta>1}$, then let ${T = T_{\alpha,\beta} \colon x \mapsto \alpha + \beta x \pmod 1}$, and let ${\Sigma = \Sigma_{\alpha,\beta}}$ be the coding space for ${T}$ with respect to its natural partition. Let ${{\mathbf a} = \inf \Sigma}$ and ${{\mathbf b} = \sup \Sigma}$ be the lexicographically minimal and maximal elements of ${\Sigma}$. Then every ${y\in \Sigma}$ has the property that
$\displaystyle {\mathbf a} \preceq \sigma^n y \preceq {\mathbf b} \text{ for all } n=0,1,2,\dots, \ \ \ \ \ (5)$
and every ${w\in \mathop{\mathcal L} = \mathop{\mathcal L}(\Sigma)}$ has the property that
$\displaystyle a_1 \cdots a_k \preceq w_{|w|-k+1} \cdots w_{|w|} \preceq b_1 \cdots b_k \text{ for all } 1\leq k \leq |w|. \ \ \ \ \ (6)$
As with the ${\beta}$-shifts, these conditions are both necessary and sufficient for membership in ${\Sigma}$ and ${\mathop{\mathcal L}}$. The key to proving this is the following analogue of Lemma 2, whose proof we omit since it is similar to the proof there (with just a little more bookkeeping).
Lemma 3 Given ${w\in A^*}$ satisfying (6), let
\displaystyle \begin{aligned} k_1(w) &= \max \{k : a_1 \cdots a_k = w_{|w|-k+1} \cdots w_{|w|} \}, \\ k_2(w) &= \max \{k : b_1 \cdots b_k = w_{|w|-k+1} \cdots w_{|w|} \}. \end{aligned} \ \ \ \ \ (7)
Then ${F_w = [\underline{T}^{k_1(w)}(0),\overline{T}^{k_2(w)}(1)]}$, where ${\underline{T}}$ and ${\overline{T}}$ agree with ${T_{\alpha,\beta}}$ on the interiors of the intervals ${I_a}$, and are defined at the endpoints by the condition that ${\underline{T}}$ is continuous from the right and ${\overline{T}}$ is continuous from the left.
3. A directed graph on ${{\mathbb N}}$
The lexicographic description of ${\Sigma_\beta}$ in Proposition 1 is appealing, but for many purposes it is more useful to have a description of the ${\beta}$-shift in terms of a countable-state directed graph, which goes back to work of Takahashi in 1973 and Hofbauer in 1978.
Start by considering the notion of follower set in a shift space, which is analogous to the follower intervals considered in the previous part. Given a shift space ${\Sigma}$ with language ${\mathop{\mathcal L}}$, consider for each ${w\in \mathop{\mathcal L}}$ the follower set
$\displaystyle F^w = \{y\in \Sigma : wy\in \Sigma \} = \sigma^{|w|}([w] \cap \Sigma). \ \ \ \ \ (8)$
One could also consider the set of words ${v\in \mathop{\mathcal L}}$ such that ${wv\in \mathop{\mathcal L}}$, but the definition in (8) will be more convenient for our purposes. It was shown by Weiss in 1973 that a shift can be represented via a labeled directed graph on finitely many vertices if and only if it has finitely many follower sets; that is, if ${\{F^w : w\in \mathop{\mathcal L}\}}$ is finite. Such shifts are called sofic. Similarly, ${\Sigma}$ can be represented via a labeled irreducible directed graph on countably many vertices if and only if it has countably many follower sets, and such shifts are called coded; see Fiebig and Fiebig 1992 and 2002.
Every ${\beta}$-shift is characterized by a sequence ${{\mathbf b}\in A^{\mathbb N}}$ with the property that ${\sigma^n{\mathbf b} \preceq{\mathbf b}}$ for all ${n\geq 0}$: as we saw in Proposition 1, we have
\displaystyle \begin{aligned} \Sigma_\beta &= \{y\in A^{\mathbb N} : \sigma^n y \preceq {\mathbf b} \text{ for all } n\geq 0 \}, \\ \mathop{\mathcal L}(\Sigma_\beta) &= \{w\in A^* : w_{|w|-k+1}\cdots w_{|w|} \preceq b_1 \cdots b_k \text{ for all } 1\leq k \leq |w| \}. \end{aligned}
Moreover, it follows from Lemma 2 that for every ${w\in \mathop{\mathcal L}(\Sigma_\beta)}$, the corresponding follower set in ${\Sigma_\beta}$ is
$\displaystyle F^w = \{y\in \Sigma : y\preceq \sigma^{k(w)}{\mathbf b}\} =: [\mathbf{0},\sigma^{k(w)}{\mathbf b}],$
where ${k(w)}$ is defined in (4) to be the length of the longest prefix of ${{\mathbf b}}$ that appears as a suffix or ${w}$. Thus ${F^w}$ is completely determined by ${k(w)}$, which can take countably many values, and since ${\beta}$-shifts are transitive this implies that they are coded. Thus they can be presented by a countable-state graph which supports a countable-state topological Markov chain (the Fischer cover).
Given any coded shift, the vertex set of the Fischer cover is the set of follower sets: ${V = \{F \subset \Sigma : F=F^w}$ for some ${w\in \mathop{\mathcal L}\}}$. A labeled edge is a triple ${(F,F',a)}$, where ${F,F'\in V}$ and ${a\in A}$; the labeled edge set of the Fischer cover is ${E = \{(F^w,F^{wa},a) : wa\in \mathop{\mathcal L}, a\in A\}}$; that is, there is an edge from ${F^w}$ to ${F^{wa}}$ whenever ${wa\in \mathop{\mathcal L}}$ and ${a\in A}$, and this edge is labeled with the symbol ${a}$. Note that there may be multiple edges between two vertices, each with a different label.
Say that a sequence ${y\in A^{\mathbb N}}$ labels a walk on the graph if there is a sequence of edges ${e_1,e_2,\dots}$ such that for every ${n}$,
1. ${e_n}$ is labeled with the symbol ${y_n}$, and
2. the target vertex of ${e_n}$ is the source vertex of ${e_{n+1}}$.
Then if ${y\in A^{\mathbb N}}$ labels a walk on this graph, there is a word ${w\in \mathop{\mathcal L}}$ such that ${w y_1 \cdots y_n \in \mathop{\mathcal L}}$ for all ${n}$, and hence ${y\in \Sigma}$. Conversely, every ${y\in \Sigma}$ labels a walk on the graph, so we can describe both ${\Sigma}$ and ${\mathop{\mathcal L}}$ in terms of walks on the graph.
For the ${\beta}$-shift, since the follower set ${F^w}$ is completely determined by ${k(w)}$, we can identify the vertex set of the Fischer cover with ${\{0,1,2,3,\dots\}}$, and observe that the two cases in Lemma 2 give the following two types of edges.
1. There is an edge from ${k}$ to ${k+1}$ whenever there are ${v\in \mathop{\mathcal L}}$ and ${a\in A}$ such that ${va\in \mathop{\mathcal L}}$, ${k(v)=k}$, and ${k(va) = k+1}$. This happens for every ${k}$ since we can take ${v=b_1 \cdots b_k}$ and ${a=b_{k+1}}$, so the edge from ${k}$ to ${k+1}$ is labeled with the symbol ${b_{k+1}}$.
2. There is an edge from ${k}$ to ${0}$ whenever there are ${v\in \mathop{\mathcal L}}$ and ${a\in A}$ such that ${va\in \mathop{\mathcal L}}$, ${k(v)=k}$, and ${k(va)=0}$. This happens whenever ${a < b_{k+1}}$, and so whenever ${b_{k+1} > 0}$ there are ${b_{k+1}}$ edges from ${k}$ to ${0}$ labeled with the symbols ${0,1,\dots, b_{k+1}-1}$.
For example, when ${{\mathbf b} = 21201\dots}$, the first part of the graph looks like this:
The vertex labeled 0 can be thought of as the base vertex; it corresponds to the follower set ${\Sigma}$, so every sequence ${y\in \Sigma}$ labels a walk starting at this vertex. The vertex labeled 1 corresponds to the follower set ${F^2 = \{y\in \Sigma : y\preceq \sigma{\mathbf b}\}}$; a sequence ${y\in \Sigma}$ labels a walk starting at this vertex if and only if | 42,405 | 125,280 | {"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": 1825, "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} | 2.875 | 3 | CC-MAIN-2023-23 | latest | en | 0.852404 |
https://dailywealth.com/articles/this-bull-market-is-near-the-end-of-the-deck/ | 1,725,752,357,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00882.warc.gz | 180,612,238 | 11,331 | This Bull Market Is Near the 'End of the Deck'...
Can you beat the odds in Vegas? According to former MIT math professor Edward O. Thorp, yes – and he actually figured out how...
He literally wrote the book on it – Beat the Dealer – in the early 1960s. It became a New York Times bestseller. The book explains how to reliably win the game of blackjack.
"I certainly wasn't motivated by the hope of making big money," Thorp said in his excellent autobiography, A Man for All Markets.
"What drew me was the chance of doing something people thought wasn't possible, to be a bit prankish – the fun of just pulling it off."
So how do you beat the house in blackjack?
At the beginning of a game, you can't. You have no advantage.
However... once you get close to the end of the deck, it's a completely different story...
As you approach the end of a deck of cards, your chances of winning can increase. You can change your strategy toward the end of some games to improve your odds of winning.
Here's what's going on...
The game of blackjack is also called "twenty-one." That's because your goal is to get as close as possible to 21 points – without going over that amount. You do this by collecting cards with different point values.
Ed discovered that if a large number of "face" cards haven't yet been played in the deck, then your odds of winning change.
The face cards in blackjack are Jacks, Queens, and Kings. They are all worth 10 points. (Therefore, 10s also count as face cards.)
Normally, if you have a total of 16 points in your hand, you would ask the dealer for another card. However, if a large number of face cards have not been played, then toward the end of the game, you are better off NOT asking for another card if you have 16.
Meanwhile, the dealer will ALWAYS take another card when he has 16 (and your buddies probably will, too).
Ed figured out that the odds of winning slightly tilt in your favor toward the end of the game. And that's when you need to increase your bets the most.
I tell you this because we have a similar situation in the financial markets today...
In normal times, you can't beat the market. You have no advantage.
But something extraordinary often happens late in the game. As you approach the end of the deck, your chances of winning can increase.
Likewise, as you approach the end of a great bull market in any asset, a cycle of higher and higher prices can get underway.
Whether it's the housing market a dozen years ago, the dot-com boom nearly 20 years ago, or, to a lesser extent, bitcoin in late 2017... the situation changes late in the game. Something changes in investors, and they push prices higher in a way that doesn't normally happen. A self-fulfilling boom kicks in.
Since 2009, stock prices have been going up. It is one of the greatest stock market booms in history. But we still haven't reached THAT moment – the moment where the game changes.
However, when it comes to this bull market, we are approaching the end of the deck. We have counted cards along the way. And now it's time. The game is changing.
I call this final phase the "Melt Up." And it means big gains in a relatively short period of time are possible.
So where will the biggest gains happen?
They typically happen in the highest-volatility stocks. That usually means tech stocks, biotech stocks, and more volatile foreign markets.
These are what's been beaten up in the recent market decline. But it's also where we'll see the biggest gains when the Melt Up comes roaring back.
Today, we're near the end of the game for this bull market. That gives us an advantage. We have a chance to earn big gains in a short period of time as a result. Don't miss it.
Good investing,
Steve
P.S. When I say "don't miss out," I mean it... Based on history, stocks could underperform for years once this boom is finally over. I've shared some of my favorite Melt Up recommendations with my True Wealth subscribers... along with how to get out before the crash. To learn more about how to access the stocks I believe will soar most in this final phase, click here.
Catch up on Steve's latest essays about this long bull market and the situation in stocks today:
INSIDE TODAY'S
#### NEW HIGHS OF NOTE LAST WEEK
Burlington Stores (BURL)... discount "one-stop shop"
Columbia Sportswear (COLM)... outdoor apparel and products
Lululemon Athletica (LULU)... innovative apparel
Tailored Brands (TLRD)... men's apparel
Guess (GES)... clothing and accessories
Estée Lauder (EL)... cosmetics
Tapestry (TPR)... luxury purses
Ruth's Hospitality (RUTH)... steak dinners
Bloomin' Brands (BLMN)... steak dinners
Boston Beer (SAM)... beer
Callaway Golf (ELY)... golf equipment
Southern Copper (SCCO)... copper
Northrop Grumman (NOC)... "offense" contractor
Ecolab (ECL)... sanitation technology
Medifast (MED)... weight-loss products
NEW LOWS OF NOTE LAST WEEK
GoPro (GPRO)... "cocktail party" stock
Kraft Heinz (KHC)... ketchup, cheese, and more
Walgreens Boots Alliance (WBA)... drugstores
Merck (MRK)... Big Pharma
Weatherford (WFT)... oilfield services | 1,179 | 5,085 | {"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} | 2.890625 | 3 | CC-MAIN-2024-38 | latest | en | 0.978995 |
https://sciencedocbox.com/Physics/123605558-Symplectic-varieties-and-poisson-deformations.html | 1,632,313,257,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057347.80/warc/CC-MAIN-20210922102402-20210922132402-00208.warc.gz | 538,126,787 | 27,762 | # Symplectic varieties and Poisson deformations
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1 Symplectic varieties and Poisson deformations Yoshinori Namikawa A symplectic variety X is a normal algebraic variety (defined over C) which admits an everywhere non-degenerate d-closed 2-form ω on the regular locus X reg of X such that, for any resolution f : X X with f 1 (X reg ) = X reg, the 2-form ω extends to a regular closed 2-form on X. There is a natural Poisson structure {, } on X determined by ω. Then we can introduce the notion of a Poisson deformation of (X, {, }). A Poisson deformation is a deformation of the pair of X itself and the Poisson structure on it. When X is not a compact variety, the usual deformation theory does not work in general because the tangent object T 1 X may possibly have infinite dimension, and moreover, infinitesimal or formal deformations do not capture actual deformations of non-compact varieties. On the other hand, Poisson deformations work very well in many important cases where X is not a complete variety. Denote by PD X the Poisson deformation functor of a symplectic variety. In this lecture, we shall study the Poisson deformation of an affine symplectic variety. The main result is: Theorem 1. Let X be an affine symplectic variety. Then the Poisson deformation functor PD X is unobstructed. A Poisson deformation of X is controlled by the Poisson cohomology HP 2 (X). When X has only terminal singularities, we have HP 2 (X) = H 2 ((X reg ) an, C), where (X reg ) an is the associated complex space with X reg. This description enables us to prove that PD X is unobstructed. But, in general, there is not such a direct, topological description of HP 2 (X). Let us explain our strategy to describe HP 2 (X). As remarked, HP 2 (X) is identified with PD X (C[ϵ]) where C[ϵ] is the ring of dual numbers over C. First, note that there is an open locus U of X where X is smooth, or is locally a trivial deformation of a (surface) rational double point at each p U. Let Σ be the singular locus of U. Note that X \ U has codimension 4 in X. Moreover, we have PD X (C[ϵ]) = PD U (C[ϵ]). Put TU 1 := an Ext1 (Ω 1 U an, O U an). As is well- 1
2 2 known, a (local) section of TU 1 corresponds to a 1-st order deformation of an U an. Let H be a locally constant C-modules on Σ defined as the subsheaf of TU 1 which consists of the sections coming from Poisson deformations of an U an. Now we have an exact sequence: 0 H 2 (U an, C) PD U (C[ϵ]) H 0 (Σ, H). Here the first term H 2 (U an, C) is the space of locally trivial 1 Poisson deformations of U. By the definition of U, there exists a minimal resolution π : Ũ U. Let m be the number of irreducible components of the exceptional divisor of π. A key result is: Proposition 2. The following equality holds: dim H 0 (Σ, H) = m. In order to prove Proposition 2, we need to know the monodromy action of π 1 (Σ) on H. The idea is to compare two sheaves R 2 π an C and H. Note that, for each point p Σ, the germ (U, p) is isomorphic to the product of an ADE surface singularity S and (C 2n 2, 0). Let S be the minimal resolution of S. Then, (R 2 π an C) p is isomorphic to H 2 ( S, C). A monodromy of R 2 π an C comes from a graph automorphism of the Dynkin diagram determined by the exceptional (-2)-curves on S. As is well known, S is described in terms of a simple Lie algebra g, and H 2 ( S, C) is identified with the Cartan subalgebra h of g; therefore, one may regard R 2 π an C as a local system of the C-module h (on Σ), whose monodromy action coincides with the natural action of a graph automorphism on h. On the other hand, H is a local system of h/w, where h/w is the linear space obtained as the quotient of h by the Weyl group W of g. The action of a graph automorphism on h descends to an action on h/w, which gives a monodromy action for H. This description of the monodromy enables us to compute dim H 0 (Σ, H). Proposition 2 together with the exact sequence above gives an upperbound of dim PD U (C[ϵ]) in terms of some topological data of X (or U). We shall prove Theorem 1 by using this upper-bound. The rough idea is the following. There is a natural map of functors PDŨ PD U induced by the resolution map Ũ U. The tangent space PD Ũ (C[ϵ]) to PD Ũ is identified with H 2 (Ũ an, C). We have an exact sequence 0 H 2 (U an, C) H 2 (Ũ an, C) H 0 (U an, R 2 π an C) 0, 1 More exactly, this means that the Poisson deformations are locally trivial as usual flat deformations of U an
3 3 and dim H 0 (U an, R 2 π an C) = m. In particular, we have dim H 2 (Ũ an, C) = dim H 2 (U an, C)+m. But, this implies that dim PDŨ(C[ϵ]) dim PD U (C[ϵ]). On the other hand, the map PDŨ PD U has a finite closed fiber; or more exactly, the corresponding map SpecRŨ SpecR U of pro-representable hulls, has a finite closed fiber. Since PDŨ is unobstructed, this implies that PD U is unobstructed and dim PDŨ(C[ϵ]) = dim PD U (C[ϵ]). Finally, we obtain the unobstructedness of PD X from that of PD U. Theorem 1 is only concerned with the formal deformations of X; but, if we impose the following condition (*), then the formal universal Poisson deformation of X has an algebraization. (*): X has a C -action with positive weights with a unique fixed point 0 X. Moreover, ω is positively weighted for the action. We shall briefly explain how this condition (*) is used in the algebraization. Let R X := lim R X /(m X ) n+1 be the pro-representable hull of PD X. Then the formal universal deformation {X n } of X defines an m X -adic ring A := lim Γ(X n, O Xn ) and let  be the completion of A along the maximal ideal of A. The rings R X and  both have the natural C -actions induced from the C -action on X, and there is a C -equivariant map R X Â. By taking the C -subalgebras of R X and  generated by eigen-vectors, we get a map C[x 1,..., x d ] S from a polynomial ring to a C-algebra of finite type. We also have a Poisson structure on S over C[x 1,..., x d ] by the second condition of (*). As a consequence, there is an affine space A d whose completion at the origin coincides with Spec(R X ) in such a way that the formal universal Poisson deformation over Spec(R X ) is algebraized to a C -equivariant map X A d. According to a result of Birkar-Cascini-Hacon-McKernan, we can take a crepant partial resolution π : Y X in such a way that Y has only Q- factorial terminal singularities. This Y is called a Q-factorial terminalization of X. In our case, Y is a symplectic variety and the C -action on X uniquely extends to that on Y. Since Y has only terminal singularities, it is relatively easy to show that the Poisson deformation functor PD Y is unobstructed. Moreover, the formal universal Poisson deformation of Y has an
4 4 algebraization over an affine space A d : Y A d. There is a C -equivariant commutative diagram Y X A d ψ A d We have the following. Theorem 3 (a) ψ is a finite Galois covering. (b) Y A d is a locally trivial deformation of Y. (c) The induced map Y t X ψ(t) is an isomorphism for a general point t A d. The Galois group of ψ is described as follows. Let Σ be the singular locus of X. There is a closed subset Σ 0 Σ such that X is locally isomorphic to (S, 0) (C 2n 2, 0) at every point p Σ Σ 0 where S is an ADE surface singularity. We have Codim X Σ 0 4. Let B be the set of connected components of Σ Σ 0. Let B B. Pick a point b B and take a transversal slice S B Y of B passing through b. In other words, X is locally isomorphic to S B (B, b) around b. S B is a surface with an ADE singularity. Put S B := π 1 (S B ). Then S B is a minimal resolution of S B. Put T B := S B (B, b) and T B := π 1 (T B ). Note that T B = S B (B, b). Let C i (1 i r) be the ( 2)-curves contained in S B and let [C i ] H 2 ( S B, R) be their classes in the 2-nd cohomology group. Then Φ := {Σa i [C i ]; a i Z, (Σa i [C i ]) 2 = 2} is a root system of the same type as that of the ADE-singularity S B. Let W be the Weyl group of Φ. Let {E i (B)} 1 i r be the set of irreducible exceptional divisors of π lying over B, and let e i (B) H 2 (X, Z) be their classes. Clearly, r r. If r = r, then we define W B := W. If r < r, the Dynkin diagram of Φ has a non-trivial graph automorphism. When Φ is of type A r with r > 1, r = [r + 1/2] and the Dynkin diagram has a graph automorphism τ of order 2. When Φ is of type D r with r 5, r = r 1 and the Dynkin diagram has a graph automorphism τ of order 2. When Φ is of type D 4, the Dynkin diagram has two different graph automorphisms of order 2 and 3. There are (1)
5 5 two possibilities of r; r = 2 or r = 3. In the first case, let τ be the graph automorphism of order 3. In the latter case, let τ be the graph automorphism of order 2. Finally, when Φ is of type E 6, r = 4 and the Dynkin diagram has a graph automorphism τ of order 2. In all these cases, we define W B := {w W ; τwτ 1 = w}. The Galois group of ψ coincides with W B. As an application of Theorem 3, we have Corollary 4: Let (X, ω) be an affine symplectic variety with the property (*). Then the following are equivalent. (1) X has a crepant projective resolution. (2) X has a smoothing by a Poisson deformation. Example 5 (i) Let O g be a nilpotent orbit of a complex simple Lie algebra. Let Õ be the normalization of the closure Ō of O in g. Then Õ is an affine symplectic variety with the Kostant-Kirillov 2-form ω on O. Let G be a complex algebraic group with Lie(G) = g. By [Fu], Õ has a crepant projective resolution if and only if O is a Richardson orbit (cf. [C- M]) and there is a parabolic subgroup P of G such that its Springer map T (G/P ) Õ is birational. In this case, every crepant resolution of Õ is actually obtained as a Springer map for some P. If Õ has a crepant resolution, Õ has a smoothing by a Poisson deformation. The smoothing of Õ is isomorphic to the affine variety G/L, where L is the Levi subgroup of P. Conversely, if Õ has a smoothing by a Poisson deformation, then the smoothing always has this form. (ii) In general, Õ has no crepant resolutions. But, by [Na 4], at least when g is a classical simple Lie algebra, every Q-factorial terminalization of Õ is given by a generalized Springer map. More explicitly, there is a parabolic subalgebra p with Levi decomposition p = n l and a nilpotent orbit O in l so that the generalized Springer map G P (n+ō ) Õ is a crepant, birational map, and the normalization of G P (n + Ō ) is a Q-factorial terminalization of Õ. By a Poisson deformation, Õ deforms to the normalization of G L Ō. Here G L Ō is a fiber bundle over G/L with a typical fiber Ō, and its normalization can be written as G L Õ with the normalization Õ of Ō. We can apply Theorem 3 to the Poisson deformations of an affine symplectic variety related to a nilpotent orbit in a complex simple Lie algebra.
6 6 Let g be a complex simple Lie algebra and let G be the adjoint group. For a parabolic subgroup P of G, denote by T (G/P ) the cotangent bundle of G/P. The image of the Springer map s : T (G/P ) g is the closure Ō of a nilpotent (adjoint) orbit O in g. Then the normalization Õ of Ō is an affine symplectic variety with the Kostant-Kirillov 2-form. If s is birational onto its image, then the Stein factorization T (G/P ) Õ Ō of s gives a crepant resolution of Õ. In this situation, we have the following commutative diagram G P r(p) G r(p) (2) k(p) k(p)/w where r(p) is the solvable radical of p, G r(p) is the normalization of the adjoint G-orbit of r(p) and k(p) is the centralizer of the Levi part l of p. Moreover, W := N W (L)/W (L), where L is the Levi subgroup of P and W (L) is the Weyl group of L. Theorem 6. The diagram above coincides with the C -equivariant commutative diagram of the universal Poisson deformations of T (G/P ) and Õ. Note that W has been extensively studied by Howlett and others. Another important example is a transversal slice of g. In the commutative diagram above, put p = b the Borel subalgebra. Then we have: G B b h π B ϕ g h/w. (3) Let x g be a nilpotent element of g and let O be the adjoint orbit containing x. Let V g be a transversal slice for O passing through x. Put V B := π 1 B (V). Denote by V (resp. Ṽ B ) the central fiber of V h/w (resp. G B b h). Note that ṼB is somorphic to the cotangent bundle T (G/B) of G/B, and ṼB V is a crepant resolution.
7 7 Theorem 7 The commutative diagram Ṽ B V ϕ V h h/w (4) is the C -equivariant commutative diagram of the universal Poisson deformations of ṼB and V if g is simply laced. When g is not simply-laced, Theorem 7 is no more true. In fact, Slodowy pointed out that the transversal slice V for a subregular nilpotent orbit of non-simply-laced g does not give the universal deformation. However, we have a criterion of the universality. Let ρ : A(O) GL(H 2 (π 1 B,0 (x), Q)) be the monodromy representation of the component group A(O) of O. Theorem 8. Let g be a comple simple Lie algebra which is not necessarilly simply-laced. Then the above commutative diagram is universal if and only if ρ is trivial. Department of Mathematics, Kyoto University,
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### Real and p-adic Picard-Vessiot fields
Spring Central Sectional Meeting Texas Tech University, Lubbock, Texas Special Session on Differential Algebra and Galois Theory April 11th 2014 Real and p-adic Picard-Vessiot fields Teresa Crespo, Universitat | 10,876 | 39,289 | {"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} | 2.90625 | 3 | CC-MAIN-2021-39 | latest | en | 0.890106 |
http://www.physicsforums.com/showpost.php?p=3578117&postcount=4 | 1,409,181,363,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500829916.85/warc/CC-MAIN-20140820021349-00254-ip-10-180-136-8.ec2.internal.warc.gz | 548,097,505 | 3,447 | View Single Post
Math
Emeritus
Why did you delete this? I wouldn't say you should assume associativity but certainly you can just note that multiplication of real numbers is associative and this is just a subset of the real numbers. To show closure write the product of two such numbers as $(a+ b\sqrt{2})(c+ d\sqrt{2})$ and actually do the multiplication. What do you get? Show that it can be written as $u+ v\sqrt{2}$ by showing what u and v must be. The identity is $1= 1+ 0\sqrt{2}$, of course.
And the multiplicative inverse of $a+ b\sqrt{2}$ is $1/(a+ b\sqrt{2})$. Rationalize the denominator to show how that can be written in the form $u+ v\sqrt{2}$. | 186 | 658 | {"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} | 3.21875 | 3 | CC-MAIN-2014-35 | latest | en | 0.959418 |
https://alex.state.al.us/cr_view.php?res_id=2295&res_id=2295&res_type=CR | 1,660,108,239,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571147.84/warc/CC-MAIN-20220810040253-20220810070253-00794.warc.gz | 118,131,832 | 7,431 | # ALEX Classroom Resource
## Finite State Automata
Classroom Resource Information
Title:
Finite State Automata
URL:
https://classic.csunplugged.org/finite-state-automata/
Content Source:
Other
CS Unplugged
Type: Learning Activity
Overview:
Computer programs often need to process a sequence of symbols such as letters or words in a document, or even the text of another computer program. Computer scientists often use a finite-state automaton to do this. A finite-state automaton (FSA) follows a set of instructions to see if the computer will recognize the word or string of symbols. We will be working with something equivalent to a FSA—treasure maps!
The goal of the students is to find Treasure Island. Friendly pirate ships sail along a fixed set of routes between the islands in this part of the world, offering rides to travelers. Each island has two departing ships, A and B, which you can choose to travel on. You need to find the best route to Treasure Island. At each island you arrive at you may ask for either ship A or B (not both). The person at the island will tell you where your ship will take you to next, but the pirates don’t have a map of all the islands available. Use your map to keep track of where you are going and which ship you have traveled on.
Content Standard(s):
Digital Literacy and Computer Science DLIT (2018) Grade: 5 2) Create an algorithm to solve a problem while detecting and debugging logical errors within the algorithm. Examples: Program the movement of a character, robot, or person through a maze. Define a variable that can be changed or updated. Unpacked Content Evidence Of Student Attainment:Students will: create an algorithm to solve a problem. detect and debug logical errors within an algorithm.Teacher Vocabulary:algorithm debug detect logical errorsKnowledge:Students know: an algorithm is a logical set of steps to solve a problem. detecting and debugging logical errors within an algorithm will ensure the algorithm serves to solve a problem successfully.Skills:Students are able to: create an algorithm to solve a problem while detecting and debugging logical errors within the algorithm.Understanding:Students understand that: debugging an algorithm is searching for logical errors within the algorithm. an algorithm is a set of steps to solve a problem. how to create an algorithm to solve a problem while detecting and debugging logical errors within the algorithm. Digital Literacy and Computer Science DLIT (2018) Grade: 5 3) Create an algorithm that is defined by simple pseudocode. Unpacked Content Evidence Of Student Attainment:Students will: create set of steps that is written in simple pseudocode.Teacher Vocabulary:algorithm pseudocodeKnowledge:Students know: simple pseudocode resembles language used to communicate with computers.Skills:Students are able to: create an algorithm that is written in simple pseudocode.Understanding:Students understand that: an algorithm that is written in simple pseudocode is similar to an algorithm written using a programming language. Digital Literacy and Computer Science DLIT (2018) Grade: 5 4) Create a simple pseudocode. Unpacked Content Evidence Of Student Attainment:Students will: create simple pseudocode.Teacher Vocabulary:pseudocodeKnowledge:Students know: how to write an algorithm in pseudocode.Skills:Students are able to: create simple pseudocode.Understanding:Students understand that: pseudocode is simple phrases of instruction that mimics the logic of a programming language. Digital Literacy and Computer Science DLIT (2018) Grade: 5 5) Develop and recommend solutions to a given problem and explain the process to an audience. Unpacked Content Evidence Of Student Attainment:Students will: develop and recommend solutions to a given problem. explain the development process to an audience.Teacher Vocabulary:processKnowledge:Students know: steps of the problem-solving process. many solutions exist to solve a problem.Skills:Students are able to: develop and recommend solutions to a given problem. share their process with others. Understanding:Students understand that: problems can have multiple solutions. Digital Literacy and Computer Science DLIT (2018) Grade: 6 5) Identify algorithms that make use of sequencing, selection or iteration. Examples: Sequencing is doing steps in order (put on socks, put on shoes, tie laces); selection uses a Boolean condition to determine which of two parts of an algorithm are used (hair is dirty? True, wash hair; false, do not); iteration is the repetition of part of an algorithm until a condition is met (if you're happy and you know it clap your hands, when you're no longer happy you stop clapping). Unpacked Content Evidence Of Student Attainment:Students will: find algorithms that demonstrate the three basic programming structures.Teacher Vocabulary:algorithm sequence selection iterationKnowledge:Students know: differences between the three basic programming structures.Skills:Students are able to: explain the differences in sequencing, selection, and iteration.Understanding:Students understand that: differences exist in sequencing, selection, and iteration. Digital Literacy and Computer Science DLIT (2018) Grade: 6 6) Identify steps in developing solutions to complex problems using computational thinking. Unpacked Content Evidence Of Student Attainment:Students will: use the problem solving or design thinking process to think logically through a previously solved complex problem.Teacher Vocabulary:computational thinkingKnowledge:Students know: how to define the problem. how to plan solutions. how to implement a plan. how to reflect on the results and process. how to iterate through the process again.Skills:Students are able to: identify the steps involved with formulating problems and solutions in a way that can be represented or carried with or without a computer.Understanding:Students understand that: computational thinking is formulating problems and solutions in a way that can be represented or carried out with or without a computer. Digital Literacy and Computer Science DLIT (2018) Grade: 6 7) Describe how automation works to increase efficiency. Example: Compare the amount of time/work to hand wash a car vs. using an automated car wash. Unpacked Content Evidence Of Student Attainment:Students will: explain how an automated activity or system increases productivity.Teacher Vocabulary:automationKnowledge:Students know: how automation works to increase efficiency.Skills:Students are able to: describe how automation increases efficiency.Understanding:Students understand that: automation works to increase efficiency. Digital Literacy and Computer Science DLIT (2018) Grade: 7 3) Create algorithms that demonstrate sequencing, selection or iteration. Examples: Debit card transactions are approved until the account balance is insufficient to fund the transaction = iteration, do until. Unpacked Content Evidence Of Student Attainment:Students will: create an algorithm using one of the three basic programming structures: sequencing, selections, or iterations.Teacher Vocabulary:algorithm sequence selection iterationKnowledge:Students know: how to use the programming structures to create algorithms and how many algorithms make use of all three programming structures.Skills:Students are able to: create and recognize various programming structures found in algorithms.Understanding:Students understand that: each structure sequencing, selections, and iterations have a purpose. Digital Literacy and Computer Science DLIT (2018) Grade: 7 6) Create and organize algorithms in order to automate a process efficiently. Example: Set of recipes (algorithms) for preparing a complete meal. Unpacked Content Evidence Of Student Attainment:Students will: use algorithms to automate a process such as sorting numbers in a random list or playing cards in a deck.Teacher Vocabulary:algorithmKnowledge:Students know: algorithms can be used to automate a process efficiently.Skills:Students are able to: use search and sort algorithms to automate organizing a set.Understanding:Students understand that: algorithms organized and applied to the appropriate task can significant increase proficiency. Digital Literacy and Computer Science DLIT (2018) Grade: 8 5) Discuss the efficiency of an algorithm or technology used to solve complex problems. Unpacked Content Evidence Of Student Attainment:Students will: examine a given artifact used to aid in problem solving. discuss the efficiency of that artifact in problem solving.Knowledge:Students know: that many solutions exist to solve a problem.Skills:Students are able to: communicate their opinion on the efficiency of problem solving methods.Understanding:Students understand that: while many solutions exist for a problem, some are better suited to meet specific needs, such as efficiency. Digital Literacy and Computer Science DLIT (2018) Grade: 8 6) Describe how algorithmic processes and automation increase efficiency. Unpacked Content Evidence Of Student Attainment:Students will: explain how algorithms and automation have and can increase efficiency.Teacher Vocabulary:algorithmic process automationKnowledge:Students know: how algorithmic processes and automation have increased efficiency.Skills:Students are able to: explain how algorithmic processes and automation increase efficiency.Understanding:Students understand that: automation is a useful tool for increasing efficiency. while many things can and have been automated, not everything can be automated using algorithmic processes.
Tags: algorithm, finite state automation, selection, sequence
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Comments
CS Unplugged is a free resource.
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Author: Aimee Bates | 1,944 | 10,023 | {"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} | 3.609375 | 4 | CC-MAIN-2022-33 | latest | en | 0.899691 |
https://plainmath.net/91158/the-volume-control-on-a-surround-sound-a | 1,670,021,241,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00600.warc.gz | 495,032,375 | 11,621 | # The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB. What is the ratio of the final sound intensity to the original sound intensity?
The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB.
What is the ratio of the final sound intensity to the original sound intensity?
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Given
${\beta }_{1}=23db$
${\beta }_{2}=61db$
The sound intensity level is given by
$\beta =10\mathrm{log}\left(\frac{I}{{I}_{0}}\right)$ where ${I}_{0}$ is intensity at reference level
Therefore the change in sound intensity level can be determined as
${\beta }_{2}-{\beta }_{1}=10\mathrm{log}\left(\frac{{I}_{2}}{{I}_{0}}\right)-10\mathrm{log}\left(\frac{{I}_{1}}{{I}_{0}}\right)$
$61-23=10\mathrm{log}\left(\frac{\frac{{I}_{2}}{{I}_{0}}}{\frac{{I}_{1}}{{I}_{0}}}\right)since\mathrm{log}a-\mathrm{log}b=\mathrm{log}\frac{a}{b}$
$38=10\mathrm{log}\left(\frac{{I}_{2}}{{I}_{1}}\right)$
$3.8=\mathrm{log}\left(\frac{{I}_{2}}{{I}_{1}}\right)$
$\frac{{I}_{2}}{{I}_{1}}={10}^{3.8}$
$\approx 6310$
Hence
The ratio of final sound intensity to original is approximately 6310
Result:
The ratio of final sound intensity to original is approximately 6310 | 453 | 1,519 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "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} | 4.09375 | 4 | CC-MAIN-2022-49 | latest | en | 0.79529 |
http://learnerstv.in/mixed-cost-definition-example-how-to-calculate/ | 1,670,393,704,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711150.61/warc/CC-MAIN-20221207053157-20221207083157-00526.warc.gz | 28,113,130 | 17,819 | Thus, the cost structure of an entire department can be said to be a https://www.bookstime.com/. This is also a key concern when developing budgets, since some mixed costs will vary only partially with expected activity levels, and so must be properly accounted for in the budget. A mixed cost is a cost that contains both fixed costs and variable costs. To calculate a mixed cost, one must determine the fixed and variable costs, and then add them together to get the total cost. For example, if a company’s monthly office space rent is \$10,000 and their monthly utilities bill is \$500, then their total monthly cost would be \$10,500. In this case, the rent would be the fixed cost and the utilities would be the variable cost. Unlike fixed costs, variable costs are directly related to the cost of production of goods or services.
• The fabric cost is \$10 per unit at every level of production.
• Make sure to note the period of time your fixed cost is for .
• A fixed cost is a cost that does not vary in the short term, irrespective of changes in production or sales levels or other measures of activity.
• It is because the price fluctuates in a stair-step sequence, often horizontally, then vertically, then horizontally, and so on.
You started a small coffee shop that specializes in gourmet roasted coffee beans. Your fixed costs are around \$1,800 per month, which includes your building lease, utility bills, and coffee roaster loan payment.
## Mixed Costs and Step Costs
Your average fixed cost can be used to see the level of fixed costs you’re required to pay for each unit you produce. Also known as “indirect costs” or “overhead costs,” fixed costs are the critical expenses that keep your business afloat. These expenses can’t be changed in the short-term, so if you’re looking for ways to make your business more profitable quickly, you should look elsewhere. Graphically, we can see that fixed costs are not related to the volume of automobiles produced by the company. No matter how high or low sales are, fixed costs remain the same. Some times the high and low levels of activity don’t coincide with the high and low amounts of cost.
They tend to be recurring, such as interest or rents being paid per month. This is in contrast to variable costs, which are volume-related and unknown at the beginning of the accounting year. Fixed costs have an effect on the nature of certain variable costs. This graph shows that the company can’t completely eliminate fixed costs. Even if the company does sell or produce a single product, there will still be fixed costs. Since mixed costs have characteristics of both fixed and variable costs, they are usually separated into segments in order to be graphed.
## Example 1 – Fixed vs. Variable Costs
The analysis of mixed cost primarily means identifying and bifurcating the fixed and variable components. The annual expense of operating an automobile is a mixed cost. Some of the expenses are fixed because they do not change in total as the number of annual miles change. These include insurance, parking fees, and some depreciation. Some of the expenses are variable since the total amount will increase when more miles are driven and will decrease when fewer miles are driven. The variable expenses include gas, oil, tires, and some depreciation.
In recent years, fixed costs gradually exceed variable costs for many companies. Firstly, automatic production increases the cost of investment equipment, including the depreciation and maintenance of old equipment. Secondly, labor costs are often considered as long-term costs. It is difficult to adjust human resources according to the actual work needs in short term. As a result, direct labor costs are now regarded as fixed costs. The a and b components of the mixed cost formula represent the fixed and variable costs, respectively. The x component of the mixed cost formula represents the number of units produced or activity level. | 787 | 3,961 | {"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} | 2.9375 | 3 | CC-MAIN-2022-49 | latest | en | 0.94951 |
https://de.scribd.com/doc/316103203/AA101-Proficiency-Test-Answer | 1,563,641,123,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526536.46/warc/CC-MAIN-20190720153215-20190720175215-00275.warc.gz | 360,326,670 | 55,399 | You are on page 1of 2
# 1.
Identify the ecliptic plane, what objects are located in the ecliptic plane. Explain or
diagram why these objects appear to move or not move, relative to the horizon, during the year.
Ecliptic Plane is the surface in which the Earth orbits around the sun.
In the ecliptic plane, there are:
## Planets in the solar system
Constellations in the Zodiac (Cc chm sao hong o)
They move relatively to the horizon during the year because:
Other planets in the solar systems move around the sun in different orbits in different
speeds thus they move relatively to Earth.
The Earth moves around the Sun thus during the year, at different positions of the Earth
in its orbit, we can observe different constellations.
2.
Site Polaris or during daylight be able to point to roughly where it would be located.
Would Polaris be located in the same place in the sky if we were at the equator? Explain why or
why not.
Polaris is 40o up from the Northern horizon if observed from Fort Collins.
At the equator, the Polaris appears at the Northern horizon, because on the celestial sphere,
Polaris is approximately directly above the Earths North Pole.
3.
Explain or diagram how the telescope used in the lab on the roof gathers light.
It gathers light by having a large, reflective lens at its bottom. The lens reflects light to a
triangular piece of glass which then redirects light to eyepiece and finally to the observers eyes.
4.
Explain or diagram why objects appear different in the telescope than with the naked eye.
This should include how different eyepieces change the magnification or field of view.
Object observed by telescope is different from by naked eyes in two ways: brighter and larger.
The lens in telescope is much larger than human eyes thus it gathers more light and lighten the
image. The lens-eyepiece system provides magnification and enlarge the image.
The magnification capability is measured by
Magnification = Lens Focal Length / Eyepiece Focal Length
Thus the smaller the focal length of eyepiece is, the more a telescope can magnify an object and
the smaller the field of view it has.
5.
Explain and diagram the current phase of the Moon. The explanation or diagram must
include the Sun, Earth, and Moon.
Nu cc bn i thi vo Th Ba tun sau 29/04/2014 th Trng ng New Moon Phase. Nhng
ngy khc c th tham kho website: http://stardate.org/nightsky/moon
6.
Be able to identify the closest star to Earth and diagram its motion in the sky.
The closest star to Earth is the Sun. Observed from Fort Collins, the Sun moves in the ecliptic
plane in the sky from East to West and draws a line across the sky that makes with the horizontal
line an angle of 40o. Images below. | 630 | 2,715 | {"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} | 2.671875 | 3 | CC-MAIN-2019-30 | latest | en | 0.93645 |
https://agda.github.io/agda-stdlib/v1.1/Function.Inverse.html | 1,675,630,557,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00794.warc.gz | 100,603,598 | 8,697 | ```------------------------------------------------------------------------
-- The Agda standard library
--
-- Inverses
------------------------------------------------------------------------
{-# OPTIONS --without-K --safe #-}
module Function.Inverse where
open import Level
open import Function using (flip)
open import Function.Bijection hiding (id; _∘_; bijection)
open import Function.Equality as F
using (_⟶_) renaming (_∘_ to _⟪∘⟫_)
open import Function.LeftInverse as Left hiding (id; _∘_)
open import Relation.Binary
open import Relation.Binary.PropositionalEquality as P using (_≗_; _≡_)
open import Relation.Unary using (Pred)
------------------------------------------------------------------------
-- Inverses
record _InverseOf_ {f₁ f₂ t₁ t₂}
{From : Setoid f₁ f₂} {To : Setoid t₁ t₂}
(from : To ⟶ From) (to : From ⟶ To) :
Set (f₁ ⊔ f₂ ⊔ t₁ ⊔ t₂) where
field
left-inverse-of : from LeftInverseOf to
right-inverse-of : from RightInverseOf to
------------------------------------------------------------------------
-- The set of all inverses between two setoids
record Inverse {f₁ f₂ t₁ t₂}
(From : Setoid f₁ f₂) (To : Setoid t₁ t₂) :
Set (f₁ ⊔ f₂ ⊔ t₁ ⊔ t₂) where
field
to : From ⟶ To
from : To ⟶ From
inverse-of : from InverseOf to
open _InverseOf_ inverse-of public
left-inverse : LeftInverse From To
left-inverse = record
{ to = to
; from = from
; left-inverse-of = left-inverse-of
}
open LeftInverse left-inverse public
using (injective; injection)
bijection : Bijection From To
bijection = record
{ to = to
; bijective = record
{ injective = injective
; surjective = record
{ from = from
; right-inverse-of = right-inverse-of
}
}
}
open Bijection bijection public
using (equivalence; surjective; surjection; right-inverse;
to-from; from-to)
------------------------------------------------------------------------
-- The set of all inverses between two sets (i.e. inverses with
-- propositional equality)
infix 3 _↔_ _↔̇_
_↔_ : ∀ {f t} → Set f → Set t → Set _
From ↔ To = Inverse (P.setoid From) (P.setoid To)
_↔̇_ : ∀ {i f t} {I : Set i} → Pred I f → Pred I t → Set _
From ↔̇ To = ∀ {i} → From i ↔ To i
inverse : ∀ {f t} {From : Set f} {To : Set t} →
(to : From → To) (from : To → From) →
(∀ x → from (to x) ≡ x) →
(∀ x → to (from x) ≡ x) →
From ↔ To
inverse to from from∘to to∘from = record
{ to = P.→-to-⟶ to
; from = P.→-to-⟶ from
; inverse-of = record
{ left-inverse-of = from∘to
; right-inverse-of = to∘from
}
}
------------------------------------------------------------------------
-- If two setoids are in bijective correspondence, then there is an
-- inverse between them
fromBijection :
∀ {f₁ f₂ t₁ t₂} {From : Setoid f₁ f₂} {To : Setoid t₁ t₂} →
Bijection From To → Inverse From To
fromBijection b = record
{ to = Bijection.to b
; from = Bijection.from b
; inverse-of = record
{ left-inverse-of = Bijection.left-inverse-of b
; right-inverse-of = Bijection.right-inverse-of b
}
}
------------------------------------------------------------------------
-- Inverse is an equivalence relation
-- Reflexivity
id : ∀ {s₁ s₂} → Reflexive (Inverse {s₁} {s₂})
id {x = S} = record
{ to = F.id
; from = F.id
; inverse-of = record
{ left-inverse-of = LeftInverse.left-inverse-of id′
; right-inverse-of = LeftInverse.left-inverse-of id′
}
} where id′ = Left.id {S = S}
-- Transitivity
infixr 9 _∘_
_∘_ : ∀ {f₁ f₂ m₁ m₂ t₁ t₂} →
TransFlip (Inverse {f₁} {f₂} {m₁} {m₂})
(Inverse {m₁} {m₂} {t₁} {t₂})
(Inverse {f₁} {f₂} {t₁} {t₂})
f ∘ g = record
{ to = to f ⟪∘⟫ to g
; from = from g ⟪∘⟫ from f
; inverse-of = record
{ left-inverse-of = LeftInverse.left-inverse-of (Left._∘_ (left-inverse f) (left-inverse g))
; right-inverse-of = LeftInverse.left-inverse-of (Left._∘_ (right-inverse g) (right-inverse f))
}
} where open Inverse
-- Symmetry.
sym : ∀ {f₁ f₂ t₁ t₂} →
Sym (Inverse {f₁} {f₂} {t₁} {t₂}) (Inverse {t₁} {t₂} {f₁} {f₂})
sym inv = record
{ from = to
; to = from
; inverse-of = record
{ left-inverse-of = right-inverse-of
; right-inverse-of = left-inverse-of
}
} where open Inverse inv
------------------------------------------------------------------------
-- Transformations
map : ∀ {f₁ f₂ t₁ t₂} {From : Setoid f₁ f₂} {To : Setoid t₁ t₂}
{f₁′ f₂′ t₁′ t₂′}
{From′ : Setoid f₁′ f₂′} {To′ : Setoid t₁′ t₂′} →
(t : (From ⟶ To) → (From′ ⟶ To′)) →
(f : (To ⟶ From) → (To′ ⟶ From′)) →
(∀ {to from} → from InverseOf to → f from InverseOf t to) →
Inverse From To → Inverse From′ To′
map t f pres eq = record
{ to = t to
; from = f from
; inverse-of = pres inverse-of
} where open Inverse eq
zip : ∀ {f₁₁ f₂₁ t₁₁ t₂₁}
{From₁ : Setoid f₁₁ f₂₁} {To₁ : Setoid t₁₁ t₂₁}
{f₁₂ f₂₂ t₁₂ t₂₂}
{From₂ : Setoid f₁₂ f₂₂} {To₂ : Setoid t₁₂ t₂₂}
{f₁ f₂ t₁ t₂} {From : Setoid f₁ f₂} {To : Setoid t₁ t₂} →
(t : (From₁ ⟶ To₁) → (From₂ ⟶ To₂) → (From ⟶ To)) →
(f : (To₁ ⟶ From₁) → (To₂ ⟶ From₂) → (To ⟶ From)) →
(∀ {to₁ from₁ to₂ from₂} →
from₁ InverseOf to₁ → from₂ InverseOf to₂ →
f from₁ from₂ InverseOf t to₁ to₂) →
Inverse From₁ To₁ → Inverse From₂ To₂ → Inverse From To
zip t f pres eq₁ eq₂ = record
{ to = t (to eq₁) (to eq₂)
; from = f (from eq₁) (from eq₂)
; inverse-of = pres (inverse-of eq₁) (inverse-of eq₂)
} where open Inverse
``` | 1,773 | 5,334 | {"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} | 2.65625 | 3 | CC-MAIN-2023-06 | longest | en | 0.565157 |
https://wiki.alquds.edu/?query=Levenshtein_coding | 1,679,756,851,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945333.53/warc/CC-MAIN-20230325130029-20230325160029-00049.warc.gz | 670,737,119 | 18,714 | # Levenshtein coding
Levenstein coding, or Levenshtein coding, is a universal code encoding the non-negative integers developed by Vladimir Levenshtein.[1][2]
## Encoding
The code of zero is "0"; to code a positive number:
1. Initialize the step count variable C to 1.
2. Write the binary representation of the number without the leading "1" to the beginning of the code.
3. Let M be the number of bits written in step 2.
4. If M is not 0, increment C, repeat from step 2 with M as the new number.
5. Write C "1" bits and a "0" to the beginning of the code.
The code begins:
Number Encoding Implied probability
0 `0` 1/2
1 `10` 1/4
2 `110 0` 1/16
3 `110 1` 1/16
4 `1110 0 00` 1/128
5 `1110 0 01` 1/128
6 `1110 0 10` 1/128
7 `1110 0 11` 1/128
8 `1110 1 000` 1/256
9 `1110 1 001` 1/256
10 `1110 1 010` 1/256
11 `1110 1 011` 1/256
12 `1110 1 100` 1/256
13 `1110 1 101` 1/256
14 `1110 1 110` 1/256
15 `1110 1 111` 1/256
16 `11110 0 00 0000` 1/4096
17 `11110 0 00 0001` 1/4096
To decode a Levenstein-coded integer:
1. Count the number of "1" bits until a "0" is encountered.
2. If the count is zero, the value is zero, otherwise
3. Start with a variable N, set it to a value of 1 and repeat count minus 1 times:
4. Read N bits, prepend "1", assign the resulting value to N
The Levenstein code of a positive integer is always one bit longer than the Elias omega code of that integer. However, there is a Levenstein code for zero, whereas Elias omega coding would require the numbers to be shifted so that a zero is represented by the code for one instead.
## Example code
### Encoding
```void levenshteinEncode(char* source, char* dest)
{
BitWriter bitwriter(dest);
{
if (num == 0)
bitwriter.outputBit(0);
else
{
int c = 0;
BitStack bits;
do {
int m = 0;
for (int temp = num; temp > 1; temp>>=1) // calculate floor(log2(num))
++m;
for (int i=0; i < m; ++i)
bits.pushBit((num >> i) & 1);
num = m;
++c;
} while (num > 0);
for (int i=0; i < c; ++i)
bitwriter.outputBit(1);
bitwriter.outputBit(0);
while (bits.length() > 0)
bitwriter.outputBit(bits.popBit());
}
}
}
```
### Decoding
```void levenshteinDecode(char* source, char* dest)
{
IntWriter intwriter(dest);
{
int n = 0;
while (bitreader.inputBit()) // potentially dangerous with malformed files.
++n;
int num;
if (n == 0)
num = 0;
else
{
num = 1;
for (int i = 0; i < n-1; ++i)
{
int val = 1;
for (int j = 0; j < num; ++j)
val = (val << 1) | bitreader.inputBit();
num = val;
}
}
intwriter.putInt(num); // write out the value
} | 877 | 2,501 | {"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} | 3.578125 | 4 | CC-MAIN-2023-14 | longest | en | 0.641812 |
http://math.stackexchange.com/questions/159950/sequence-of-square-integrable-functions?answertab=active | 1,461,982,067,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111581.11/warc/CC-MAIN-20160428161511-00190-ip-10-239-7-51.ec2.internal.warc.gz | 186,285,576 | 16,486 | # Sequence of square integrable functions
Let $\{f_{n}\}$ be a sequence of nonzero continuous functions on $\mathbb{R}$, which is uniformly bounded, uniformly Lipschitz on $\mathbb{R}$, and the derivative sequence $\{f_{n}'\}$ is also uniformly Lipschitz on $\mathbb{R}$, and $f_{n}\in L^{2}(\mathbb{R})$ for all $n$. If $\{f_{n}\}$ converges uniformly on any closed interval $I\subset \mathbb{R}$ to a continuous function $f$, does this imply that $f\in L^{2}(\mathbb{R})$. If not, what condition(s) the sequence $\{f_{n}\}$ must have to get such result?
-
Just curios: are you another incarnation of berry who posted this question: math.stackexchange.com/questions/159507/… ? Or are you just attending the same class? – user20266 Jun 18 '12 at 16:39
## 1 Answer
You need a bound on the $L^2$ norms of the $f_n$. Since you know that your sequence converges pointwise (even better) the same is true for $|f_n|^2$. You can then use Fatou's lemma applied to $|f_n|^2, |f|^2$ to conclude that the limit is in $L^2$.
-
if the sequence of $L^2$ norms is unbounded you are probably out of luck. I do admit that I do not have a counterexample right now. If you know that the sequence of $L^2$ -norms is unbounded you seem to have additional information about the $f_n$. Why don't you add this to your question, this may make it easier to provide an answer. – user20266 Jun 18 '12 at 17:13 | 408 | 1,386 | {"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} | 3.28125 | 3 | CC-MAIN-2016-18 | latest | en | 0.867362 |
https://se.mathworks.com/matlabcentral/cody/problems/44309-pi-digit-probability/solutions/1313786 | 1,586,282,915,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371803248.90/warc/CC-MAIN-20200407152449-20200407182949-00129.warc.gz | 686,436,755 | 15,938 | Cody
# Problem 44309. Pi Digit Probability
Solution 1313786
Submitted on 25 Oct 2017 by Eric G.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
N = 101; n = 3; y_correct = 0.1200; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match')))) % modified from the comment of Alfonso on https://www.mathworks.com/matlabcentral/cody/problems/44343
ans = 12 y = 0.1200
2 Pass
N = 201; n = 6; y_correct = 0.0750; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
ans = 15 y = 0.0750
3 Pass
N = 202; n = 6; y_correct = 0.0796; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
ans = 16 y = 0.0796
4 Pass
N = 203; n = 6; y_correct = 0.0792; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
ans = 16 y = 0.0792
5 Pass
N = 1001; n = 9; y_correct = 0.1050; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
ans = 105 y = 0.1050 | 607 | 1,590 | {"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} | 3.796875 | 4 | CC-MAIN-2020-16 | latest | en | 0.357004 |
https://www.storyboardthat.com/storyboards/dab24b39/jahseh | 1,556,058,495,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578613888.70/warc/CC-MAIN-20190423214818-20190424000818-00329.warc.gz | 824,416,893 | 17,049 | # jahseh
More Options: Make a Folding Card
#### Storyboard Text
• By using the midpoint formula I am exactly in the middle from my house and the farm.
• However, the midpoint between both locations is dangerous, and I'm crippled so I'm finna die.
• The midpoint formula takes both x and y coordinates, adds them, divides them by two and then gets a new coordinate(the midpoint)
• Tekashi 69 and Collin in the fresh whip.
• Yo tekashi 69, I used a coordinate plane of the city and the distance formula to calculate that from your house, we are 30 miles away from prison.
• In federal prison, with Bill Cosby, Big Chungus, and Shaggy
• I am measuring the area and perimeter of this cell for a future escape, you will be free to harass small children again!
• I used the perimeter formula (4s) and the area formula (s2) to help tekashi escape
• Meanwhile, outside of the prison cell. A cripple with a flamethrower
• I will burn a circle in the wall that has a circumference of 69.69
• But first I had to find the radius of the circle and then multiply it by 2pi.
• Yung Collin and treyway after robbing the bank
• YOO, we got bands!
• Now look Tekashi, If the cops catch us, then we won't get to buy those sexy lamborghinis.
• Tekashi you double-crossed me, you know I cant swim.
• The betrayal
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Video Marketing | 376 | 1,478 | {"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} | 2.984375 | 3 | CC-MAIN-2019-18 | latest | en | 0.864514 |
http://www.ck12.org/physics/Generators-and-Motors/lesson/Generators-and-Motors-PPC/ | 1,490,755,466,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190134.67/warc/CC-MAIN-20170322212950-00226-ip-10-233-31-227.ec2.internal.warc.gz | 455,162,570 | 34,322 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Generators and Motors
## Electrical or kinetic energy can be generated using principles of magnetism.
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Generators and Motors
Changing magnetic fields passing through a loop of wire generate currents in that wire; this is how electric power generators work. Likewise, a changing current in a wire will create a changing magnetic field; this is how speakers and electric motors work.
Since only a changing flux can produce an induced potential difference, one or more of the variables in the equation must be changing if any emf is to be produced. Specifically, the following can all induce a current in the loops of wire:
• Changing the direction or magnitude of the magnetic field.
• Changing the loops' orientation or area.
• Moving the loops out of the region with the magnetic field.
Key Equations
emf=ΔΦΔtFaraday's Law of Induction\begin{align*} emf = -\frac{\Delta \Phi}{\Delta t}&& \text{Faraday's Law of Induction} \end{align*} Φ=NBA Electromagnetic Flux\begin{align*} \Phi = N \vec{B} \cdot \vec{A} &&\text{ Electromagnetic Flux} \end{align*}
The direction of the induced current is determined as follows: the current will flow so as to generate a magnetic field that opposes the change in flux. This is called Lenz’s Law. Note that the electromotive force described above is not actually a force, since it is measured in Volts and acts like an induced potential difference. It was originally called that since it caused charged particles to move --- hence electromotive --- and the name stuck (it's somewhat analogous to calling an increase in a particle's gravitational potential energy difference a gravitomotive force).
#### Example
To create a simple generator, you are holding a single coil of wire in place on a table. You hold a bar magnet directly above the center of the wire and move it toward the wire and away from the wire repeatedly to create a current in loop. (a) As you bring the bar magnet toward the loop, which direction does the current flow? (b) As you move the magnet away from the loop, which direction does the current flow?
(a): As you bring the magnet closer to the loop, the magnetic field will increase in strength thus increasing the magnetic flux going into the table through the loop. In order to oppose this change in flux, the induced current will need to create a magnetic field that will point out of the table through the loop. We can use the first right hand rule, starting with the direction of them magnetic field instead of the direction of the current, to determine that the current will need to move in the CCW direction.
(b): When you move the magnet away from the loop, the magnetic field will be decreasing thus decreasing the magnetic flux going into the table through the loop. The induced current will need to create a magnetic field that is also pointed into the table through the loop to oppose this change in flux. Again, we use the second right hand rule to find that the current must go in the CW direction.
### Review
1. How is electrical energy produced in a dam using a hydroelectric generator? Explain in a short essay involving as many different ideas from physics as you need.
2. When building a generator why not make it with almost infinite number of wire loops, since the emf is proportional to N the number of loops? What is the trade off, so to speak.
3. Explain how a motor works.
4. Explain in your own words why the generator in example 1 spins the same direction upon discharge.
1. A simple method would be to use the water to spin a turbine, which would in turn spin a loop of wire through a constant stationary magnetic field.
2. The field lines of your magnetic field only reach as far as they curl around.
3. If we bend a current-carrying wire into a loop in such a way that the two sides of the loop are at right angles to a magnetic field supplied by magnets, then the sides of the loop will experience forces in opposite directions which will create a torque to rotate the loop.
4. When we look at the direction that the current is flowing and the direction of the magnetic field, we find that the resulting force from these two physical vector quantities result in a torque spinning the generator in the same direction during which it was charged.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes | 1,010 | 4,644 | {"found_math": true, "script_math_tex": 2, "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} | 4.09375 | 4 | CC-MAIN-2017-13 | latest | en | 0.83834 |
https://www.odoo.com/fr_FR/groups/community-59/community-21121492?mode=date&date_begin=2016-10-01&date_end=2016-11-01 | 1,532,042,094,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591332.73/warc/CC-MAIN-20180719222958-20180720002958-00026.warc.gz | 951,773,738 | 10,986 | # Community archives des diffusions par email
#### Re: Price on sale order
par
Alberto
- 01/11/2016 09:31:55
No
It is a total amount, some sales have products with taxes, and products without taxes
On Tue, Nov 1, 2016 at 5:12 AM, Dominique Chabord wrote:
```I guess you get it if you use a 0% tax rate. don't you ?
2016-11-01 2:12 GMT+01:00 Kristian Koci <kristian.koci@gmail.com>:
> Hi
>
> I'm going to explain what I'm trying to accomplish, so hopefuly somebody can
> help me.
>
> On sales lines, You got the total amount_untaxed and amount_tax
>
> These are self explanatory.
>
> However, I've added to fields to this form, which should take, total price
> sum of all products on sales line which are tax affected, and total price
> sum of all products on sales line which aren't tax affected.
>
> For example, I have 50 products on a sale.order, from these 50 products
>
> 40 are tax affected
> 10 aren't tax affected
>
> So these fields, should sum and show me,
>
> 1) The total sum of 40 tax affected products (with taxes included)
> 2) The total sum of 10 tax unaffected products
>
> So far I've tried with this method:
>
> @api.one
> @api.depends('invoice_line.price_subtotal', 'tax_line.amount')
> def extras(self):
> self.impo = sum(line.price_subtotal for line in self.invoice_line)
> if self.amount_tax and self.amount_tax > 0 else 0
> self.exe = sum(line.price_subtotal for line in self.invoice_line) if
> not self.amount_tax or self.amount_tax == 0 else 0
>
> rec=0
>
> 'impo' should sum all tax affected, while 'exe' should sum all tax
> unaffected products.
>
> But so far this isn't showing me the expected results.
>
> Actually only 'exe' is working, but it takes all lines with taxes and sum
> the up, like the amount_total field.
>
> How can I accomplish this?
>
> I hope I've explained myself
>
> Thank You
>
> --
> Kristian Koci
> Linux User #582221
> Public repository: https://github.com/kkoci
>
>
>
>
> _______________________________________________
> Mailing-List: https://www.odoo.com/groups/community-59
> Post to: mailto:community@mail.odoo.com
> Unsubscribe: https://www.odoo.com/groups?unsubscribe
--
Dominique Chabord - SISalp
Logiciel libre pour l'entreprise Tryton et open-source Odoo, OpenERP
18 avenue Beauregard 74960 Cran Gevrier
145A rue Alexandre Borrely 83000 Toulon
tel(repondeur) +33(0)950274960 fax +33(0)955274960 mob +33(0)622616438
http://sisalp.fr
http://openerp-online.fr
https://twitter.com/SISalp l'actualité de vos services en temps réel.```
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--
Kristian Koci
Linux User #582221
• #### Price on sale order
par
Alberto
- 31/10/2016 21:09:44 - 0 | 817 | 2,832 | {"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} | 2.53125 | 3 | CC-MAIN-2018-30 | longest | en | 0.877645 |
http://www.aqua-calc.com/one-to-all/torque/preset/kilogram-force-meter/38 | 1,409,701,304,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535923940.4/warc/CC-MAIN-20140909032200-00148-ip-10-180-136-8.ec2.internal.warc.gz | 640,109,350 | 8,044 | # One to all
## Torque units conversions
Unit
name value symbol reference
dyne centimeter 3 726 527 011.4 dyn⋅cm
dyne decimeter 372 652 701.14 dyn⋅dm
dyne meter 37 265 270.11 dyn⋅m
dyne millimeter 37 265 270 114 dyn⋅mm
gram force centimeter 3 800 000 gf-cm
gram force decimeter 380 000 gf-dm
gram force meter 38 000 gf-m
gram force millimeter 38 000 000 gf-mm
kilogram force centimeter 3 800 kgf-cm
kilogram force decimeter 380 kgf-dm
kilogram force meter 38 kgf-m
kilogram force millimeter 38 000 kgf-mm
newton centimeter 37 265.27 N⋅cm
newton decimeter 3 726.53 N⋅dm
newton meter 372.65 N⋅m
newton millimeter 372 652.7 N⋅mm
pound force foot 274.85 lbf-ft
pound force inch 3 298.25 lbf-in
#### Foods, Nutrients and Calories
Cookies, chocolate sandwich, with creme filling, special dietary contain(s) 461 calories per 100 grams or ≈3.527 ounces [ calories ]
#### Gravels and Substrates
CaribSea, Freshwater, Super Naturals, VooDoo River density is equal to 865 kg/m³ or 54 lb/ft³ with specific gravity of 0.865 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylinderquarter cylinder or in a rectangular shaped aquarium or pond
#### Materials and Substances
Nitromethane weigh(s) 1.137 gram per (cubic centimeter) or 0.657 ounce per (cubic inch) [ weight to volume | volume to weight | density ]
#### What is pound per US gallon?
The pound per US gallon density measurement unit is used to measure volume in US gallons in order to estimate weight or mass in pounds
#### What is radiation absorbed dose measurement?
Radiation exposure to effects of alpha (α), beta (β) and heavy particles, neutrons of unknown energy and high-energy protons, fission fragments, gamma (γ) and X- rays is a matter of public health. These radiation sources can be found in nature, and some are created by human activities, e.g. x rays. | 562 | 1,884 | {"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} | 2.84375 | 3 | CC-MAIN-2014-35 | latest | en | 0.632908 |
https://www.begellhouse.com/es/ebook_platform/4f54d35e2dc24988.html | 1,596,626,741,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735939.26/warc/CC-MAIN-20200805094821-20200805124821-00581.warc.gz | 610,326,637 | 7,691 | ISBN:
978-1-56700-217-1 (Imprimir)
978-1-56700-275-1 (En Línea)
# Principios Matemáticos de la Transferencia de Calor 传热的数学原理
K. N. Shukla
Karunya Institute of Technology and Sciences Coimbatore-641114, India
## Descripción
This book presents an investigative account of Mathematical Principles of Heat Transfer. It is concerned with three aspects of heat transfer analysis: theoretical development of conservation equations, analytical and numerical techniques of the solution, and the physical processes involved in the three basic modes of heat transfer, namely conduction, convection, and radiation. A concept of mathematical modeling is developed through the use of differential equations. In doing so, the well-posed boundary value problems are constructed and the solutions are attempted.
310 pages, © 2005
## Tabla de Contenidos:
Preface
Chapter 1 Basic Concepts of Heat Transfer
1.1 Basic Modes of Heat Transfer
1.1.1 Temperature field
1.2 Conduction
1.3 Thermal Conductivity
1.3.1 Thermal conductivities of gases
1.3.2 Thermal conductivities of liquids
1.3.3 Thermal conductivities of solids
1.4 Convection
1.6 Heat Transfer with Change of Phase
1.7 Units and Dimensions
References
Chapter 2 Conservation Equations
2.1 General Conservation Equation
2.2 Equation of Continuity
2.3 Equation of Motion
2.4 Equation of Energy
2.5 Equation of Entropy
References
Chapter 3 Similarity Theory and the Generalized Variables
3.1 Boundary Value Problem in Generalized Variables: A Mathematical Presentation
3.2 Method of Dimensionality
References
Chapter 4 Mathematical Methods for Boundary Value Problems
4.1 Separation of Variables
4.2 Integral Transform Method
4.3 Laplace Transform
4.3.1 Laplace transform-fundamental properties
4.3.2 Inversion theorem for Laplace transform
4.4 Fourier Kernals
4.5 The Hankel Transform
4.6 Finite Integral Fourier and Hankel Transforms
4.7 Limiting Cases of Laplace Transform
4.8 Green's Function for the Solution of Heat Conduction
4.9 Approximate Methods in the Solution of Heat Transfer Problems
4.9.1 Integral method
4.9.2 Variational method
4.9.3 Ritz method
4.9.4 Galerkin method
4.9.5 Least-squares method
References
Chapter 5 Numerical Methods in Heat Transfer
5.1 Finite Difference Method
5.2 Gauss Elimination Method
5.3 Gauss-Seidel Method of Iteration
5.4 Successive Overrelaxation Method
5.5 Derivative Type of Boundary Conditions
5.6 Stability: Analytical Treatment
5.7 Convergence: Analytical Treatment
5.8 Compatibility
5.9 Two-Dimensional Problem of Heat Conduction
5.9.1 Locally one-dimensional method
References
Chapter 6 Steady-State Heat Conduction
6.1 Heat Transfer in a Slab
6.1.1 Dirichlet boundary conditions
6.1.2 Temperature-dependent thermal conductivity
6.1.3 Composite slab
6.1.4 Newtonian boundary conditions
6.1.5 Mixed boundary conditions
6.2 Heat Transfer through a Cylindrical Wall
6.2.1 Dirichlet boundary conditions
6.2.2 Temperature-dependent thermal conductivity
6.2.3 Composite cylindrical wall
6.2.4 Newtonian boundary conditions
6.2.5 Mixed boundary conditions
6.3 Heat Transfer through a Spherical Wall
6.3.1 Dirichlet boundary conditions
6.3.2 Temperature-dependent thermal conductivity
6.4 Critical Thickness of Insulation
6.5 Heat Conduction through a Thin Rod
6.6 Extended Surface
6.6.1 Longitudinal fin
6.6.2 Rectangular profile
6.6.3 Optimum dimension
6.6.4 Rectangular fin of minimum weight
6.6.5 Efficiency of the fin
6.6.6 Longitudinal fin of triangular profile
6.6.7 Optimum dimension
6.7 Steady Heat Flow in a Rectangle
6.7.1 Conjugate functions
References
Chapter 7 Transient Heat Conduction
7.1 Infinite Plate
7.2 Infinite Cylinder
7.3 The Sphere
7.4 Composite Solids
7.4.1 For Γ= 0
7.4.2 For Γ= 1
7.4.3 For Γ= 2
7.5 Periodic Variation of Ambient Temperature
References
Chapter 8 Heat Conduction with Phase Change
8.1 Statement of Problem and Existence of Solution
8.2 State-of-the-Art Solution Technique
8.3 Solidification of a Semi-Infinite Liquid
8.4 Axisymmetric Melting
8.5 Spherical Melting
8.6 Dynamics of Melt Growth and Axisymmetric Melting
8.6.1 Energy equation
References
Chapter 9 Convection
9.1 Hydrodynamics and Thermal Boundary Layers
9.2 Similarity Solution for Boundary Layers
9.2.1 Skin friction and heat transfer
9.2.2 Laminar boundary layers by the integral method
9.3 Similarity Solution for Boundary Layers, u = cxm
9.4 Finite Difference Solution
9.4.1 Transient solution for a pipe flow
9.5 Natural Convection
9.5.1 Natural convection from an isothermal vertical plate
9.6 Turbulence
References
Chapter 10 Diffusive Processes
10.1 Conservation Equations
10.2 Isothermal Ternary System
10.3 Nonisothermal Binary System
10.4 Thermohaline Convection
10.4.1 Free boundaries with specified solute concentration and temperature
10.4.2 Diffusive convection
10.5 Chaos
10.6 Diffusion Processes in Material Processing and Microgravity Environments
References
Chapter 11 Radiation Heat Transfer
11.5 Radiant Interchange between Surfaces Separated by a Nonparticipating Medium
11.6 View Factor
11.6.1 Evaluation of integral
11.6.2 Contour integral representation
11.7 Electrical Network Analog for an Enclosure
11.8 Enclosures with Diffuse Gray Surfaces
11.9 Enclosure with Specularly Reflecting Surfaces
11.9.1 Solution for radiative transfer
11.10 Radiative Transfer in a Plane Layer | 1,422 | 5,297 | {"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} | 2.6875 | 3 | CC-MAIN-2020-34 | latest | en | 0.727121 |
https://dsp.stackexchange.com/questions/tagged/linear-systems?sort=newest&page=7 | 1,566,531,818,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317817.76/warc/CC-MAIN-20190823020039-20190823042039-00474.warc.gz | 437,256,174 | 40,436 | # Questions tagged [linear-systems]
A linear system operates on inputs with only linear operators so the response to a complex input can be analysed as the sum of the response to a set of simpler inputs. This mathematical property makes the analysis of linear systems much simpler than non-linear systems where this summation or superposition does not hold. Linear systems are generally further classified as time invariant, meaning that there characteristics do not change over time.
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215 views | 1,168 | 4,897 | {"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} | 3 | 3 | CC-MAIN-2019-35 | latest | en | 0.907713 |
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The field of Consciousness research is rapidly evolving. Abundant new techniques and strategies for human and non-human animal research have been developed. Consequently, more data is becoming readily available, and this calls for a periodic reevaluation of previously held preconceptions in this field. Studies of non-human animals have shown that homologous brain circuits correlated with conscious experience and perception can be selectively facilitated and disrupted to assess whether they are in fact necessary for those experiences
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On an exponential scale in meters, humans are located at the midpoint between the nanometer scale (1×10−9 m) (a strand of DNA is 3 nanometers thick) and the scale of stars ( the sun is 1.4 ×109 m in diameter.) Reaching “down”, what we try to do in nanotechnology, is just as difficult as reaching “up,” exploring the solar system with our probes. But, the journey is only beginning.
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Read More... | 474 | 2,336 | {"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} | 2.546875 | 3 | CC-MAIN-2017-51 | longest | en | 0.926939 |
https://byjus.com/question-answer/molar-heat-capacity-at-constant-pressure-for-a-non-linear-triatomic-gas-is-dfrac-5r/ | 1,721,406,259,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00479.warc.gz | 127,084,371 | 25,380 | 1
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Question
# Molar heat capacity at constant pressure for a non-linear triatomic gas is
A
3R2
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B
5R2
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C
4R
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D
2R
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Solution
## The correct option is C 4RFor a non-linear triatomic gas, it can have degree of freedom as as translational (3), rotational (2), vibrational (1). f=3A−B For non-linear triatomic, A=3 and B=3 ⇒Total degree of freedom will be, f=6 Since, Cp=(f2+1)R ∴Cp=(62+1)R=4R
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http://aolanswers.com/questions/how_many_acres_is_universal_studios_orlando_p495283136461135 | 1,419,200,913,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802772743.56/warc/CC-MAIN-20141217075252-00152-ip-10-231-17-201.ec2.internal.warc.gz | 9,028,697 | 19,837 | # how many acres is universal studios orlando?
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There are many companies that offer car rental services. You can hire a car for a more comfortable and enjoyable journey. There are many car rental and transportation services that you can book online or over the phone. You can avail their services and make traveling much easier.
### Answered:What is the Federal Government Per Diem rates for Orlando, FL, 2009
Go to www.DFAS.gov . That is the finance center for DOD. You can find pay rates and the Per Diem rates for Orlando as well as other locations throughout the world.
### Answered:What percentage of an acre?
YOU TAKE 120 X 300 = 36,000 DIVIDE THAT BY 43,560 =.826 OR 82.6% WWW.LBDINC.INFO
## Featured Questions
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## Be The First To Answer
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### Minor's Studio Portraits
Oh, heck!!! Just sue 'em for a couple of million dollars...get in line with the other people wanting a fast buck. Seriously...I would be flattered if a Portrait Studio wanted to share pictures of my child for the purpose of drumming up their business...unless the pictures were nudes. But then ...
### Where can i find 30% off link to universal studios orlando?
this link does not work any more do you have another one please
### Universe
Univere is anything and everything outside of one's self
### Are all online universities the same? Are they ...
there are many online colleges and universities which are being established now after the evolution of online education but one needs to ensure that the online college or university he or she enrolls in is accredited and is authenticated by the verifying bodies so that he or she does not jeopardizes ... | 655 | 2,900 | {"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} | 2.796875 | 3 | CC-MAIN-2014-52 | longest | en | 0.932744 |
https://www.emathhelp.net/calculators/differential-equations/inverse-laplace-transform-calculator/ | 1,675,008,571,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00769.warc.gz | 777,268,571 | 6,381 | # Inverse Laplace Transform Calculator
## Find the inverse Laplace transform
The calculator will try to find the Inverse Laplace transform of the given function.
Recall that $\mathcal{L}^{-1}(F(s))$ is such a function $f(t)$ that $\mathcal{L}(f(t))=F(s)$.
Usually, to find the Inverse Laplace transform of a function, we use the property of linearity of the Laplace transform. Just perform partial fraction decomposition (if needed), and then consult the table of Laplace transforms.
Enter the function F(s):
For the Laplace Transform, see Laplace Transform calculator.
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.
Your input: find $\mathcal{L}^{-1}\left(\frac{5}{s^{2} + 2 s + 10}\right)$
$\mathcal{L}^{-1}\left(\frac{5}{s^{2} + 2 s + 10}\right)=\frac{5 e^{- t} \sin{\left(3 t \right)}}{3}$ | 251 | 906 | {"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} | 2.984375 | 3 | CC-MAIN-2023-06 | latest | en | 0.6763 |
https://www.physicsforums.com/threads/centripetal-force-of-an-amusement-park-ride-problem.181208/ | 1,511,091,507,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805541.30/warc/CC-MAIN-20171119095916-20171119115916-00332.warc.gz | 845,250,429 | 14,523 | # Centripetal Force of an amusement park ride problem
1. Aug 21, 2007
### jcmtnez
1. The problem statement, all variables and given/known data
An amusement park ride consits of a rotating vertical cylinder with rough canvas walls. After the rider has entered and the cylinder is rotating sufficiently fast, the floor is dropped down, yet the rider does not slide down, The rider has a mass of 50 kilograms, the radius R of the cylinder is 5 meters, the angular velocity of the cylinder when rotating is 2 radians per second, and the coefficient of static friction between the rider and the wall of the cylinder is 0.6.
(b) Calculate the centripetal force on the rider when the cylinder and state what provides that force.
(c) Calculate the upward force that keeps the rider from falling when the floor is dropped down and state what provides that force.
(d) At the same rotational speed would a rider of twice the mass slide down the wall? Explain you answer.
2. Relevant equations
F=m*a
a=R*w^2
3. The attempt at a solution
I could derive the centripetal force by applying newton's second law and the formula for centripetal aceleration. I got 1000 N but i am not sure of my answer. And I dont understand very well what force is involved in pulling up the person when the floor is dropped down and how is friction involved.
Last edited: Aug 21, 2007
2. Aug 21, 2007
### rootX
It is not pulling up the person, but instead is acting against W, and is equals to W.
-It's frictional force.
-the person tends to go in a straight line, but the wall prevents him from doing so, by constantly applying a centripetal force on him
-gravity tends to pull him towards the earth, but since he is sticking to the wall, and there is friction between him and the wall, so he doesn't slide. | 420 | 1,788 | {"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} | 3.84375 | 4 | CC-MAIN-2017-47 | longest | en | 0.944572 |
https://pyproveit.github.io/Prove-It/packages/proveit/linear_algebra/inner_products/__pv_it/theorems/093b5159c37cc9f6ae2f5222aed15db4508ad04e0/theorem_expr.html | 1,660,205,623,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00553.warc.gz | 430,177,481 | 7,031 | # from the theory of proveit.linear_algebra.inner_products¶
see dependencies
In [1]:
import proveit
# Automation is not needed when only building an expression:
proveit.defaults.automation = False # This will speed things up.
proveit.defaults.inline_pngs = False # Makes files smaller.
# import the special expression
from proveit.linear_algebra.inner_products import inner_prod_linearity
In [2]:
# check that the built expression is the same as the stored expression
assert inner_prod_linearity.expr == stored_expr
assert inner_prod_linearity.expr._style_id == stored_expr._style_id
print("Passed sanity check: inner_prod_linearity matches stored_expr")
Passed sanity check: inner_prod_linearity matches stored_expr
In [3]:
# Show the LaTeX representation of the expression for convenience if you need it.
print(stored_expr.latex())
\forall_{K}~\left[\forall_{H \underset{{\scriptscriptstyle c}}{\in} \textrm{InnerProdSpaces}\left(K\right)}~\left[\forall_{a, b \in K}~\left[\forall_{x, y, z \in H}~\left(\left \langle \left(a \cdot x\right) + \left(b \cdot y\right), z\right \rangle = \left(\left(a \cdot \left \langle x, z\right \rangle\right) + \left(b \cdot \left \langle y, z\right \rangle\right)\right)\right)\right]\right]\right]
In [4]:
stored_expr.style_options()
Out[4]:
namedescriptiondefaultcurrent valuerelated methods
with_wrappingIf 'True', wrap the Expression after the parametersNoneNone/False('with_wrapping',)
condition_wrappingWrap 'before' or 'after' the condition (or None).NoneNone/False('with_wrap_after_condition', 'with_wrap_before_condition')
wrap_paramsIf 'True', wraps every two parameters AND wraps the Expression after the parametersNoneNone/False('with_params',)
justificationjustify to the 'left', 'center', or 'right' in the array cellscentercenter('with_justification',)
In [5]:
# display the expression information
stored_expr.expr_info()
Out[5]:
core typesub-expressionsexpression
0Operationoperator: 21
operand: 2
1ExprTuple2
2Lambdaparameter: 33
body: 3
3Operationoperator: 21
operand: 5
4ExprTuple5
5Lambdaparameter: 52
body: 7
6ExprTuple52
7Conditionalvalue: 8
condition: 9
8Operationoperator: 21
operand: 13
9Operationoperator: 11
operands: 12
10ExprTuple13
11Literal
12ExprTuple52, 14
13Lambdaparameters: 15
body: 16
14Operationoperator: 17
operand: 33
15ExprTuple67, 68
16Conditionalvalue: 19
condition: 20
17Literal
18ExprTuple33
19Operationoperator: 21
operand: 24
20Operationoperator: 36
operands: 23
21Literal
22ExprTuple24
23ExprTuple25, 26
24Lambdaparameters: 27
body: 28
25Operationoperator: 47
operands: 29
26Operationoperator: 47
operands: 30
27ExprTuple69, 70, 71
28Conditionalvalue: 31
condition: 32
29ExprTuple67, 33
30ExprTuple68, 33
31Operationoperator: 34
operands: 35
32Operationoperator: 36
operands: 37
33Variable
34Literal
35ExprTuple38, 39
36Literal
37ExprTuple40, 41, 42
38Operationoperator: 65
operands: 43
39Operationoperator: 53
operands: 44
40Operationoperator: 47
operands: 45
41Operationoperator: 47
operands: 46
42Operationoperator: 47
operands: 48
43ExprTuple49, 71
44ExprTuple50, 51
45ExprTuple69, 52
46ExprTuple70, 52
47Literal
48ExprTuple71, 52
49Operationoperator: 53
operands: 54
50Operationoperator: 62
operands: 55
51Operationoperator: 62
operands: 56
52Variable
53Literal
54ExprTuple57, 58
55ExprTuple67, 59
56ExprTuple68, 60
57Operationoperator: 62
operands: 61
58Operationoperator: 62
operands: 63
59Operationoperator: 65
operands: 64
60Operationoperator: 65
operands: 66
61ExprTuple67, 69
62Literal
63ExprTuple68, 70
64ExprTuple69, 71
65Literal
66ExprTuple70, 71
67Variable
68Variable
69Variable
70Variable
71Variable | 1,106 | 3,603 | {"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} | 2.734375 | 3 | CC-MAIN-2022-33 | longest | en | 0.367146 |
https://uk.answers.yahoo.com/question/index?qid=20201001042244AAX5gw8 | 1,603,855,659,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107896048.53/warc/CC-MAIN-20201028014458-20201028044458-00529.warc.gz | 569,354,410 | 26,207 | # What will happen to this formula if i=0?
I was thinking what would be present worth if there was no interest. but, it turned out t to be infinite, which i find funny.
am i right?
Relevance
• 4 weeks ago
First, this question should have been placed in the Business & Finance category,
not Astronomy & Space.
A = principal amount = \$10,000
n = number of payment periods = 48
i = interest rate per period = 0% = 0.0
P = present worth = to be determined
P = A{[(1 + i)^n - 1] / [i(1 + i)^n]}
P = \$10,000{[(1 + 0)^48 - 1] / [0(1 + 0)^48]}
In the formula that you have quoted, P does go to infinity if i = 0.
This is because the variable I is in the denomonator and as I goes
to zero, then P becomes bigger and bigger until it approaches
infinity. Are you sure that this formula is correct?
• Acetek
Lv 4
4 weeks ago
i is the square root of -1 it can never be 0
• Jim
Lv 7
4 weeks ago
no, but with no interest, it becomes simple principle payments | 286 | 964 | {"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} | 3.25 | 3 | CC-MAIN-2020-45 | latest | en | 0.958128 |
https://brainmass.com/math/ring-theory/ideals-factor-rings-prime-ideal-17287 | 1,642,380,769,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00650.warc.gz | 216,525,527 | 74,999 | Explore BrainMass
# Ideals and Factor Rings : Prime Ideal
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Problem:
Let R be a commutative ring.
Show that every maximal ideal of R is prime and if R is finite, show that every prime ideal is maximal.
ALSO is every prime ideal of Z(integers) maximal? Why?
https://brainmass.com/math/ring-theory/ideals-factor-rings-prime-ideal-17287
#### Solution Preview
Proof:
R is a commutative ring. Then M is an maximal ideal of R if and only if R/M is a field, P is a prime ideal if and only if R/P is an integral domain.
(1) If M is an maximal ideal, then R/M is a field. So if xy is in M, we consider x+M and y+M in R/M. Since R/M is a field, we have ...
#### Solution Summary
A proof involving a prime ideal is provided in the solution.
\$2.49 | 244 | 936 | {"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} | 3.625 | 4 | CC-MAIN-2022-05 | longest | en | 0.910688 |
https://www.jdsupra.com/legalnews/term-sheet-math-when-is-your-66-percen-76318/ | 1,513,075,198,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948515313.13/warc/CC-MAIN-20171212095356-20171212115356-00118.warc.gz | 730,428,336 | 31,468 | # Term Sheet Math — When Is Your 66 Percent Really 52 Percent?
by Foley & Lardner LLP
Contact
When negotiating valuation for a financing, an investor may conduct detailed due diligence and present you with a term sheet that reflects multiples, discounts, comparables, and so forth. In the end, you are negotiating for percentage — how much of the company will the investor get, and how much will you keep? Your investor is focused on maximizing return on investment. You are focused on keeping meaningful upside for your innovation and hard work.
For example, if you raise \$5 million on a \$10 million pre-money valuation, you will be giving the investor 33 percent of your company. You keep 66 percent. But that 66 percent may not be really be 66 percent even before you take into account any later dilution by subsequent rounds of investors. There are at least three reasons why.
First, investors often invest on a “fully diluted basis,” which includes an assumption that an unused stock option pool is actually issued and outstanding. The theory is that you will need to issue options to incentivize your employees to achieve the growth reflected in your business plan, and that you received a valuation based on your entire business plan. The result: Only you are diluted by the option pool.
Based on the example above, if the investor requires that a 10 percent stock option pool be included in the calculation of shares on a fully diluted basis, then the investor gets 33 percent, the option pool is 10 percent, and you keep 56 percent. Of course, the result is not as bad if the full option pool is not used prior to an exit event, but many option pools are completely used, and often increased, prior to an exit event.
Second, investors often receive a liquidation preference upon a sale of the company — in other words, a return of capital before anyone else gets paid. If the investor also requires a “participation” feature, then the investor receives a return of capital, and then participates with respect to the balance of the sale on an as-converted basis. The result: The investor receives a percentage of deal proceeds that is higher than the investor’s percentage of shares.
Based on the example above, assume a sale with \$100 million in proceeds. If the investor has a liquidation preference with a participation feature, then the investor receives \$5 million as a return of capital, and then another \$31.35 million (33 percent of \$95 million). The option pool participants receive \$9.5 million (10 percent of \$95 million). You receive \$53.2 million (56 percent of \$95 million). The result: You receive 53.2 percent of the proceeds. This shift in percentage gets worse for you if there are less deal proceeds, and better if there are more.
Third, investors often receive payment of accrued dividends on an exit event, and the dividends can add up. Dividends on \$5 million accruing at eight percent, compounded annually, amounts to another \$2.3 million after five years. The result, re-running the math based on the example above: You receive 51.91 percent of the deal proceeds.
When negotiating percentages, we caution against being too focused on incremental percentage points. With the right investment, you can end up with a slightly smaller slice of a much bigger pie. However, it’s important to understand all aspects of dilution so that you know, in the beginning, what you are really bargaining for.
DISCLAIMER: Because of the generality of this update, the information provided herein may not be applicable in all situations and should not be acted upon without specific legal advice based on particular situations.
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In the event there is a change in the corporate structure of JD Supra such as, but not limited to, merger, consolidation, sale, liquidation or transfer of substantial assets, JD Supra may, in its sole discretion, transfer, sell or assign information collected on and through the Service to one or more affiliated or unaffiliated third parties. | 1,554 | 7,781 | {"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} | 3.21875 | 3 | CC-MAIN-2017-51 | latest | en | 0.952041 |
https://devrycourses.com/product/devry-fin-515-week-7-project-capital-budgeting-analysis/ | 1,721,114,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514737.55/warc/CC-MAIN-20240716050314-20240716080314-00298.warc.gz | 192,351,520 | 41,876 | # Devry FIN 515 Week 7 Project Capital Budgeting Analysis
\$24
## Devry FIN 515 Week 7 Project Capital Budgeting Analysis
### FIN 515: Week 7 Project – Capital Budgeting Analysis
Once again, your team is the key financial management team for your company. The company’s CEO is now looking to expand its operations by investing in new property, plant, and equipment. Your team recently calculated the WACC for your company, which will now be useful in evaluating the project’s effectiveness. You are now asked to do some capital budgeting analysis that will determine whether the company should invest in these new plant assets.
The parameters for this project are:
Your team will be using the same company for this project that you used in the Week 6 project. The company is now looking to expand its operations and wants you to do some analysis using key capital budgeting tools to do this. The parameters for this project are as follows.
The firm is looking to expand its operations by 10% of the firm’s net property, plant, and equipment. (Calculate this amount by taking 10% of the property, plant, and equipment figure that appears on the firm’s balance sheet.)
The estimated life of this new property, plant, and equipment will be 12 years. The salvage value of the equipment will be 5% of the property, plant and equipment’s cost.
The annual EBIT for this new project will be 18% of the project’s cost.
The company will use the straight-line method to depreciate this equipment. Also assume that there will be no increases in net working capital each year. Use the same marginal tax rate that you used in the Week 6 project.
The hurdle rate for this project will be the WACC that you calculated in Week 6.
Deliverable for this Project
Prepare a narrated PowerPoint presentation using VoiceThread or Webex that will highlight the following items.
• Your calculations for the amount of property, plant, and equipment and the annual depreciation for the project
• Your calculations that convert the project’s EBIT to free cash flow for the 12 years of the project.
• The following capital budgeting results for the project
• Net present value
• Internal rate of return
• Discounted payback period.
• Your discussion of the results that you calculated above, including a recommendation for acceptance or rejection of the project | 490 | 2,345 | {"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} | 2.546875 | 3 | CC-MAIN-2024-30 | latest | en | 0.91845 |
https://physics-network.org/what-are-the-elements-of-projectile-motion/ | 1,670,310,697,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00146.warc.gz | 471,781,788 | 17,163 | # What are the elements of projectile motion?
There are the two components of the projectile’s motion – horizontal and vertical motion. And since perpendicular components of motion are independent of each other, these two components of motion can (and must) be discussed separately.
## What are the 4 types of projectile motion?
• Oblique projectile motion.
• Horizontal projectile motion.
• Projectile motion on an inclined plane.
## What are the 3 components of projectile motion?
The key components that we need to remember in order to solve projectile motion problems are: Initial launch angle, θ Initial velocity, u. Time of flight, T.
## What is projectile motion in physics PDF?
Projectile motion is a special case of two-dimensional motion. A particle moving in a vertical plane with an initial velocity and experiencing a free-fall (downward) acceleration, displays projectile motion.
## What force is acting on the projectile?
A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.
## What is the formula of projectile motion?
The equation for the distance traveled by a projectile being affected by gravity is sin(2θ)v2/g, where θ is the angle, v is the initial velocity and g is acceleration due to gravity.
## Why is it important to study projectile motion?
Understanding projectile motion is important to many engineering designs. Any engineered design that includes a projectile, an object in motion close to the Earth’s surface subject to gravitational acceleration, requires an understanding of the physics involved in projectile motion.
## What factors influence the projectile motion?
FACTORS AFFECTING PROJECTILE MOTION There are three main factors that affect the trajectory of an object or body in flight: the projection angle, magnitude of projection velocity and height of projection.
## How many types of projectile motion are there?
The following are the three types of projectile motion: Horizontal projectile motion. Oblique projectile motion. Projectile motion on an inclined plane.
## Which factor appears most important in projectile motion?
Take off Velocity. (This determines height and horizontal length of trajectory (range)). This is the single MOST IMPORTANT factor affecting the range of projectile.
## What is the path of a projectile called?
The path followed by a projectile is called its trajectory.
## What is projectile motion explain it in detail?
Projectile motion is the motion of an object thrown (projected) into the air. After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory.
## What is the formula of time of flight?
T tof = 2 ( v 0 sin θ 0 ) g . This is the time of flight for a projectile both launched and impacting on a flat horizontal surface.
## Which component of projectile is constant?
The horizontal velocity of a projectile is constant (a never changing in value), There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down, The vertical velocity of a projectile changes by 9.8 m/s each second, The horizontal motion of a projectile is independent of its vertical motion.
## What happens to the projectile if no force is applied?
Under these conditions the first law says that if an object is not pushed or pulled upon, its velocity will naturally remain constant. This means that if an object is moving along, untouched by a force of any kind, it will continue to move along in a perfectly straight line at a constant speed.
## Why is an airplane not a projectile?
Airplanes, guided missiles, and rocket-propelled spacecraft are sometimes also said to follow a trajectory. Since these devices are acted upon by the lift of wings and the thrust of engines in addition to the force of gravity, they are not really projectiles.
## What is the maximum height of projectile?
Thus, the maximum height of the projectile formula is, H = u 2 sin 2 θ 2 g .
## Which is not a projectile?
Answer:An aircraft taking off. Explanation: When aircraft take off it doesn’t follow a parabolic path. So, it is not a projectile motion.
## What do call the object moving in a projectile motion?
Projectile refers to an object that is in flight after being thrown or projected. In a projectile motion, the only acceleration acting is in the vertical direction which is acceleration due to gravity (g).
## What is velocity of projection?
Projection Velocity. The minimum velocity to throw an object vertically upwards to a definite height is called the projection velocity.”
## What is angle of projection?
The angle between the direction of projection and the horizontal drawn at that point is called the angle of projection.
## What two assumptions are needed for projectile motion?
Assumptions made in the study of projectile motion: (i) Acceleration due to gravity ‘g’ is constant both in magnitude and direction. (ii) There is no resistance due to air.
## How does projectile motion help in real life situations?
In real life, the projectile motion finds applications in sports. Playing basketball, football is examples of projectile motion in real life. While throwing a basketball into the basket, the player shoots the ball in such a way that the flight taken by the ball is in the form of a parabola. | 1,093 | 5,574 | {"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} | 3.84375 | 4 | CC-MAIN-2022-49 | latest | en | 0.905093 |
https://www.physicsforums.com/threads/finding-the-domain-and-range-without-a-calculator.637030/ | 1,591,388,913,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348502204.93/warc/CC-MAIN-20200605174158-20200605204158-00456.warc.gz | 854,449,512 | 15,358 | # Finding the domain and range without a calculator
## Homework Statement
Find the domain and range of the following function without the use of a calculator:
f(x) = sec (pi x/4)
## Homework Equations
As far as I know, this problem doesn't specifically require "equations". Therefore I am leaving this section blank. Not because I am a malicious poster that is trying to break the rules and upset the administration here, but simply because I don't know any equations for this problem.
## The Attempt at a Solution
I tried flipping this and finding the cosine of 4pi and graphing it. But I failed to come up with a valid graph and was puzzled. I could take a picture of my "graph", but I don't think anyone could read it. Help would be greatly appreciated here.
Related Precalculus Mathematics Homework Help News on Phys.org
For what values of x, sec(x) is not defined?
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## Homework Statement
Find the domain and range of the following function without the use of a calculator:
f(x) = sec (pi x/4)
## Homework Equations
As far as I know, this problem doesn't specifically require "equations". Therefore I am leaving this section blank. Not because I am a malicious poster that is trying to break the rules and upset the administration here, but simply because I don't know any equations for this problem.
## The Attempt at a Solution
I tried flipping this and finding the cosine of 4pi and graphing it. But I failed to come up with a valid graph and was puzzled. I could take a picture of my "graph", but I don't think anyone could read it. Help would be greatly appreciated here.
"I tried flipping this and finding the cosine of 4pi ..."
That won't work.
sec(θ) = 1/(cos(θ)). It's not equal to cos(1/θ), nor 1/(cos(1/θ)).
Consider the graph of $\displaystyle y=\cos\left(\frac{\pi}{4}x\right)\ .$
Where does this graph have zeros?
What are the maximum & minimum values for y on this graph? | 464 | 1,960 | {"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} | 3.75 | 4 | CC-MAIN-2020-24 | longest | en | 0.959164 |
http://doc.roboticslibrary.org/0.7.0/dd/d15/classrl_1_1math_1_1_circular_3_01_vector2_01_4.html | 1,576,414,957,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541308149.76/warc/CC-MAIN-20191215122056-20191215150056-00422.warc.gz | 39,761,823 | 6,084 | Robotics Library 0.7.0
rl::math::Circular< Vector2 > Class Template Reference
Circular segment function that maps from a time x to a point in 2D on a circular trajectory. More...
#include <CircularVector2.h>
Inheritance diagram for rl::math::Circular< Vector2 >:
[legend]
## Public Member Functions
Circular ()
virtual ~Circular ()
Circular< Vector2 > * clone () const
Real getAngle () const
Vector2 getAxisX () const
Vector2 getAxisY () const
Vector2 getCenter () const
Vector2 operator() (const Real &x, const ::std::size_t &derivative=0) const
Evaluates the circular segment function for a given x. More...
Public Member Functions inherited from rl::math::Function< Vector2 >
Function ()
virtual ~Function ()
Real duration () const
Reallower ()
const Reallower () const
Realupper ()
const Realupper () const
## Static Public Member Functions
static Circular< Vector2ThreePoints (const Vector2 &y0, const Vector2 &yi, const Vector2 &y1, const Real &x1=1)
Generates a circular segment function in 2D through three given points. More...
static Circular< Vector2ThreePointsAngle (const Vector2 &y0, const Vector2 &yi, const Vector2 &y1, const Real &angle, const Real &x1=1)
Generates a circular segment through three given points in 2D with a given segment angle. More...
## Protected Attributes
Real angle
Angle of circular motion. More...
Vector2 axisX
First axis of the circular motion. More...
Vector2 axisY
Second axis of the circular motion. More...
Vector2 center
Center of the circle. More...
Protected Attributes inherited from rl::math::Function< Vector2 >
Real x0
Real x1
## Detailed Description
### template<> class rl::math::Circular< Vector2 >
Circular segment function that maps from a time x to a point in 2D on a circular trajectory.
## ◆ Circular()
rl::math::Circular< Vector2 >::Circular ( )
inline
## ◆ ~Circular()
virtual rl::math::Circular< Vector2 >::~Circular ( )
inlinevirtual
## ◆ clone()
Circular* rl::math::Circular< Vector2 >::clone ( ) const
inlinevirtual
## ◆ getAngle()
Real rl::math::Circular< Vector2 >::getAngle ( ) const
inline
## ◆ getAxisX()
Vector2 rl::math::Circular< Vector2 >::getAxisX ( ) const
inline
## ◆ getAxisY()
Vector2 rl::math::Circular< Vector2 >::getAxisY ( ) const
inline
## ◆ getCenter()
Vector2 rl::math::Circular< Vector2 >::getCenter ( ) const
inline
## ◆ operator()()
Vector2 rl::math::Circular< Vector2 >::operator() ( const Real & x, const ::std::size_t & derivative = 0 ) const
inlinevirtual
Evaluates the circular segment function for a given x.
Note that only the first two derivatives are implemented, all higher orders will return NaN.
Implements rl::math::Function< Vector2 >.
## ◆ ThreePoints()
static Circular rl::math::Circular< Vector2 >::ThreePoints ( const Vector2 & y0, const Vector2 & yi, const Vector2 & y1, const Real & x1 = 1 )
inlinestatic
Generates a circular segment function in 2D through three given points.
The given points most not be (numerically close to) colinear.
## ◆ ThreePointsAngle()
static Circular rl::math::Circular< Vector2 >::ThreePointsAngle ( const Vector2 & y0, const Vector2 & yi, const Vector2 & y1, const Real & angle, const Real & x1 = 1 )
inlinestatic
Generates a circular segment through three given points in 2D with a given segment angle.
The given points must not be (numerically close to) colinear. Contrary to ThreePoints, where the circular segment ends in y1, the circular segments ends after a given angle. With this, a full circle can be constructed, given an angle of 2*pi. The given segment angle may be any real number, which allows multiple rotations.
## ◆ angle
Real rl::math::Circular< Vector2 >::angle
protected
Angle of circular motion.
Any real value is allowed.
## ◆ axisX
Vector2 rl::math::Circular< Vector2 >::axisX
protected
First axis of the circular motion.
Note that at time 0, the function value is (center + axisX).
## ◆ axisY
Vector2 rl::math::Circular< Vector2 >::axisY
protected
Second axis of the circular motion.
After a motion of 90 degrees, the function value is (center + axisY).
## ◆ center
Vector2 rl::math::Circular< Vector2 >::center
protected
Center of the circle.
The documentation for this class was generated from the following file: | 1,102 | 4,272 | {"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} | 2.515625 | 3 | CC-MAIN-2019-51 | longest | en | 0.474374 |
https://docs.3di.live/h_simulation_settings.html | 1,725,937,791,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00771.warc.gz | 183,517,788 | 9,699 | # Simulation settings
## Physical settings
Switch the use of advection in the 1D domain on (1) or off (0)
Switch the use of advection in the 2D domain on (1) or off (0)
## Numerical settings
There are various numerical settings that can either improve the solution under certain conditions, and some can speed up the computations and others will improve the stability. The various options are explained here.
### Settings for matrix solvers
There are several methods available to solve the matrix consisting of the unknown water levels. Depending on the application, these settings can speed up the simulation or make the solutions more accurate. As 3Di uses the so-called subgrid method, the system of equations becomes a weakly non-linear system. Therefore, we need the use of a Newton iteration method in combination with matrix solvers in order to actually solve this system of equations. The so-called nested-Newton method is needed when the application consists of closed profiles.
#### Newton iteration settings
[max_nonlin_iterations] (default = 20), [convergence_eps] (default = 1.0^-5), [use_of_nested_newton] (no default)
Maximum number of non linear iterations (max_nonlin_iterations) is the number the computational core will try to reach the convergence value of the Newton iteration. In cases where is extensive flooding and drying of areas, this number can be raised as it might need more iterations to find the correct solution. The Newton iteration needs a value that defines convergence. Initially, 3Di requires a much lower value, but in case the system has difficulties with finding a solution, it will loosen this requirement with a maximum of the by the user set convergence_eps.
Nested newton iterations are needed in case profiles in 1D are narrowing with height. Mathematically, in case d^2V/dzeta^2<0. This occurs, for example, a lot in sewerage systems. For these cases, the Newton iteration method does not guarantee a solution, so the system is split in two systems that do guarantee a solution. In case 3Di cannot find a solution it will always try, whether it can find a solution using the nested Newton method. However, in case one has an application that consists of many of these profiles it is faster to tel the system that it should always used the nested Newton method (use_of_nested_newton).
#### Maximum Degree
[max_degree](no default)
One of the methods to solve a matrix is by Gauss-Jordan elimination, substitution. Depending on the type of network, either 1D, 2D or 1D with many bifurcations or combination of those, this method is very efficient or not. It is also possible to solve parts of the system using this method and others with the other method. The efficiency of the solver depends on the network. For 1D simulations this is a very efficient solver, for 2D simulations it is less so.
[use_of_cg] (default =20) [convergence_cg] (default = 1.0^-9) [precon_cg] (default =1) [convergence_cg] (default=1.0^-9)
This is an iterative method to solve matrices. Therefore, also a convergence definition (convergence_cg) is required. It is possible to prepare this method to make it more efficient during the simulation. The system will then be preconditioned (precon_cg), this will take time in the initializing phase, but will safe time during the simulation itself. To limit the possible amount of iterations in order to guarantee swiftness of the solver, there can be put a maximum of iterations before the convergence threshold is loosened.
#### CFL condition
[cfl_strictness_factor_1d] (default=1.0) [cfl_strictness_factor_2d] (default=1.0)
There is a limit to the time step, called the CFL condition. This condition is due to the chosen discretization of the equations. It defined as cdt/dx<1. C is the velocity, defined as
(31)$C = |U| + \sqrt(gH)$
Often it is not necessary to be so strict, so sometimes the user can set this parameter which loosens the strictness of it. Consequently, stability can decrease.
#### Pump implicit ratio
[pump_implicit_ratio] (default=1, between 0 and 1)
Pumps will be switched on or off depending on the characteristics of the pump and the local water level. For water levels between the start and stop levels of the pump, the pump will drain at maximum capacity. For an optimal pump operation, the supply of water is in balance with or larger than the pump capacity. However, in real-life applications, the pump capacity is larger than the supply. This results in a pump that switches repetitively on and off during an event. Even though, this is a real-life issue and is known from observations, one does not always want to mimic this behaviour in a simulation. This behaviour can make the analysis of your results on water levels and discharges more difficult and as this triggers wave-like phenomenon in the water levels and flow, it can cause time step reductions.
The computational core of 3Di can make estimates of the available water to the pump and adjust the capacity based in these estimates. This will avoid the switching on/off of the pump unnecessarily. The pump capacity is not affected in cases where the supply is higher or equal to the capacity and in cases where the supply is that low that the water level should drop below the stop level. How strong this implicit behaviour is used in the simulation, can be set by the pump implicit ratio.
A pump_implicit_ratio of 0 means the computational core does not take the supply information into account. By setting it higher than zero, this information is taken into account more strongly according to the value. So, the pump capacity is adjusted based on the (expected) available water.
#### Thresholds
For numerical computation several tresholds are needed in the code, to avoid deficiencies due to a limited numerical accuracy. Generally this is to keep the behaviour consistent:
In order to determine the upstream method the direction of the flow is considered. To avoid the exact 0.0 m/s point we use a threshold given by flow_direction_threshold (default=1.0^-5).
We also use for various things a general threshold, this one is defined as general_numerical_threshold, the default is 1.0d-8.
### Limiters
A limiter is a general term used for certain aspects in numerical schemes that limit the effect of high gradients in flow or forcing. They are used to avoid strong oscillations, instabilities in the solution and to increase the accuracy. 3Di has various limiters implemented, which can be switched on or off.
#### Limiter for water level gradient
The limiter on the water level gradient allows the model to deal with unrealistically steep gradients. These can occur when there are, for example, jumps in the bottom. In such case the water is not forced by the difference in water level, as this gradient is limited to the actual depth. Therefore, a limiter function is part of the discretisation scheme. This setting exists for flow in the 1D domain and 2D domains.
Function where the ratio between water depth and water level gradient prescribes the behaviour.
(32)$\phi_(m+1) = min[ 1 , H / ( \sigma_(m+1) - \sigma_m ) ]$
#### Limiter for cross-sectional area
limiter_slope_crossectional_area_2d = 0 (default)
The Subgrid method assumes that the variation in water levels is much more gradual than variations in bottom elevation or bathymetry. Within a computational cell, the water level is assumed uniform, while the bottom elevation is allowed to vary. This assumption is not valid in sloping areas where water flowss down the slope as sheet flow. In such situations, the spatial variation of the water level has the same length scales as the bottom elevation. The uniform water level assumption can lead to overestimating the wet cross-sectional area at a computational cell edge and an underestimation of the friction. This would lead to an overestimation of the discharge. Therefore, 3Di uses limiters to correct the computed cross-sectional areas and the friction. These limiters are based on the sheet flow concept; in these sloping areas, it is assumed that the water depth is uniform within a flow domain instead assuming the water level to be uniform. The way this uniform water depth is calculated, depends on the limiter type that is chosen:
limiter_slope_crossectional_area_2d = 1
The limiter type 1 represents an accurate redefintion of the water depth, since the water is spread over two adjacent cells. This limiter is activated in case the downstream water depth is zero. Then two options are possible. In case of a large difference in water levels, the sum of upstream and downstream volume is divided by the total maximum surface area of the two cells. When the difference is smaller, the average water level of upstream and downstream is used. This makes the scheme mathematically second order.
limiter_slope_crossectional_area_2d = 2
The limiter type 2 is a very stable upstream method to redefine the water depth at the cell edge. It is assumed that the flow behaves as a thin sheet flow. Therefore, the depth is defined as the upstream volume divided by the maximum surface area of the upstream cell.
limiter_slope_crossectional_area_2d = 3, in combination with thin_layer_definition = xx [m]
The limiter type 3 provides a smooth transition from the default water depth to the altered one. This transition depends on the local depth and a user-defined thin water layer. In case the depth at the edge, based on the downstream water level, is larger than the thin water layer definition, the cross-sectional area is based on the uniform water level assumption. In case the downstream water level is below the thin water layer definition, then limiter 2 determines the cross-sectional area. Finally, if the downstream water level is within the thin water layer depth, these two types of cross-sections are weighted to define the new value (i.e., limiter type 3).
#### Limiter for friction depth
[limiter_slope_friction_2d] default = 0
In order to take high resolution depth and roughness variations into account to determine the friction, an estimate is made of the effective frictional depth. To determine this, the actual depth is needed. Similar to the limiter for the cross-sectional area, the actual depth in sloping areas is overestimated. In such case not only the depth to determine the cross-sectional area can be adjusted, but also the depth to determine the effective frictional depth. The friction can therefore be underestimated in sloping areas. Therefore, the same limiter can be used to determine the effective frictional depth by switching this limiter on. This limiter is obligatory in combination with the limiter_slope_crossectional_area_2d.
### Numerical settings for friction
There are several settings that affect the friction.
#### Friction shallow water correction
[friction_shallow_water_correction] (default =0) (possible values 0,1,2,3)
In case the friction assumptions based on the dominant friction balance structurally underestimates the friction, one can switch this setting on. This situation can occur in case the flow is mainly distributed based on continuity instead. In Figure 1, the difference between the two type of flows is shown. Such a situation occurs, for example, in a sloping area where filled canals are cutting through in cross-slope direction. When the correction is switched on, the friction is determined both in the classical way and based on averaged values of depth, velocity and roughness coefficients. The maximum friction computed by the two is used.
It is important to define a depth for which the friction is computed. When the friction shallow water correction is set to 2 or 3, it will define the depth similar to the cross-sectional area limiter. For the value 1 it will use the maximum depth at the edge of the cell.
#### Friction averaging
[frict_avg] (default = 0)
The roughness coefficient will be averaged within one cell.
#### Minimum friction velocity
minimum_friction_velocity [float], (default = 0.01 m/s)
In case a cell is flooded, there is a moment that initially there is no water, therefore no friction as the velocity is zero. Followed by a moment that there is a velocity. To assure a smooth transition and to avoid extreme accelerations of the flow, we define a sort of minimum amount of friction based on this velocity. Generally this is important only when a cell is flooded.
### Other numerical settings
#### Preissman slot
[preissmann_slot ] (default= 0.0 m^2)
A preissmann slot is often used to model flows in pipes. When the pipes are not completely filled, such flows can be modelled as free surface flows. However, when the discharges increase, the pipes are filled and the flow can become pressurised. Not all hyrdodynamic models are suited for these kind of flows. Therefore, to mimic the effects of pressurised flows, the water level can be allowed to rise higher than the upper limit of the cross section. In order to allow this, a narrow tube is added on top of the pipe (Figure 2). These tubes are generally quite narrow to allow the water level to rise, at a minimum cost of extra added volume. In 3Di this is not necessary, however it can be added to circular tubes. This can increase the stability at larger time steps. The way flow is computed in pipes is described here.
#### Integration method
[integration_method] (default=0)
There are various ways to discretise the equation. At the moment only first order semi implicit is supported and tested.
## Timestep settings
### Time step
The simulation time step that 3Di will use if timestep reduction is not required.
### Minimum time step
The smallest value that 3Di will reduce the time step to when applying timestep reduction. Setting this too high is not recommended.
### Maximum time step
When using time step stretch, the 3Di will use larger time steps when a stationary condition has been reached. The time step will not become larger than maximum time step.
### Use time step stretch
When switched on (1), once flow conditions are stationary, the time step will become larger (but no larger than maximum time step).
### Output time step
The time step for writing results to the Results 3Di. | 3,114 | 14,231 | {"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} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.903928 |
https://guillaumeboivin.com/what-is-the-neutral-wire-for.html | 1,656,135,600,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034170.1/warc/CC-MAIN-20220625034751-20220625064751-00374.warc.gz | 338,164,046 | 13,076 | # What is the neutral wire for?
Neutral is a circuit conductor that normally carries current back to the source. In the electrical trade, the conductor of a 2-wire circuit connected to the supply neutral point and earth ground is referred to as the “neutral”.
So, is the common wire the same as the neutral wire?
All neutral wires of the same earthed (grounded) electrical system should have the same electrical potential, because they are all connected through the system ground. Neutral conductors are usually insulated for the same voltage as the line conductors, with interesting exceptions.
What is the Cu wire?
Copper is the electrical conductor in many categories of electrical wiring. Copper wire is used in power generation, power transmission, power distribution, telecommunications, electronics circuitry, and countless types of electrical equipment.
## Can you get a shock from the neutral wire?
So even the current returns through neutral (only from a connected load that completes the current flow circuit) you touching the neutral with a 0V cant get you a shock. But its not safe to touch neutral wire! It is possible that the path to ground on neutral is not very good.
## Is a neutral wire hot?
Ground: The bare wire is called the ground wire. Like the neutral wire, the ground wire is also connected to an earth ground. However, the neutral and ground wires serve two distinct purposes. The neutral wire forms a part of the live circuit along with the hot wire.
## What happens if you touch live and neutral?
If a fault occurs where the live wire connects to the case, the earth wire allows a large current to flow through the live and earth wires. This overheats the fuse which melts and breaks the circuit. If a faulty live wire touches the inside of the plastic case there’s little risk as the case is an insulator.
## What is the purpose of the neutral?
The purpose of the neutral wire is to complete the 120volt AC circuit by providing the path back to the electrical panel where the neutral wire is connected and bonded to the earth ground. The neutral is an insulated wire because it is part of the circuit which flows electrical current.
## Do you need a neutral?
With 120V wiring in the US, you have a center tapped transformer with two hots that total 240V. The neutral is that center tap, which combined with only one of the hots gives you 120V. Use both of the hots and you have 240V. The only need for the neutral is to get a 120V circuit.
## What color is live wire?
Colours of inner wires within a cablecolourwireblueneutralbrownlivegreen and yellow stripesearth
## What will happens if we connect neutral to ground?
A second problem with connecting the ground to the neutral happens if your neutral wire breaks between the outlet and your service entrance. Given a ground to neutral connection, this will cause the chassis of your device to be at the “hot” voltage, which is very dangerous.
## What is the voltage on the neutral wire?
If the neutral-to-ground is 120V and the hot-to-ground is a few volts or less, then the hot and neutral wires are reversed (Fig. 1). Neutral-to-ground connection. Some neutral-to-ground voltage should be present under load conditions, typically 2V or less.
## What is the point of the earth wire?
The earth wire creates a safe route for the current to flow through if the live wire touches the casing. You would get an electric shock if the live wire inside an appliance, such as a cooker, came loose and touched the metal casing.
## What is the purpose of the ground wire?
The ground wire is an additional path for electrical current to return safely to the ground without danger to anyone in the event of a short circuit. In that instant, the short would cause the current to flow through the ground wire, causing a fuse to blow or a circuit breaker to trip.
## Why do you need to separate grounds and neutrals?
If we bond the ground wire to the neutral in the sub-panel, current will flow on both the neutral AND on the ground wire. Which means that if you do not keep the ground wires separate from the neutral wires, you will be allowing return currents to flow on the ground wires back to the main panel.
## Is the common hot or neutral?
The “common” is the “neutral” or “ground” wire, depending on the type of circuit. In normal US residential wiring, you’ll have a black “hot” wire, a white “neutral” or “common” wire, and a green or bare “ground” wire.
## Can you connect a ground wire to the neutral?
This gives a small voltage between the grounded metal parts of devices connected to it and true ground if phases are unbalanced, which is clearly suboptimal, but if you have a faulty device where a hot wire touches the case, that will at least blow the fuse. It is however wrong to connect ground to neutral in the wall.
## What should be the neutral to earth voltage?
The voltage drop in the neutral wire carrying 15a will be 6 volts. 25 volts that’s 150 feet of #14 gauge wire for a 15a circuit. Voltage between live and neutral is 240v what will be the. min max voltage earth phase if it allinterview perfect earthing.
## What is the neutral wire on a light switch?
The most common requirement of any hardwired automated light switch is a neutral wire. This is a diagram of a switch with a neutral. The black “hot” connection is broken to turn the light on/off, the white “neutral” connection completes the circuit. The bare (hopefully) solid copper wire is the ground.
## What is the red electrical wire for?
Red electrical wire indicates the secondary live wires in a 220-volt circuit, used in some types of switch legs and in the interconnection between smoke detectors that are hard-wired into the power system. You can connect a red wire to another red wire or to a black wire.
## What is AC supply?
A common source of DC power is a battery cell in a flashlight. The abbreviations AC and DC are often used to mean simply alternating and direct, as when they modify current or voltage. The usual waveform of alternating current in most electric power circuits is a sine wave.
## What is the neutral wire in 3 phase?
In a symmetrical three-phase four-wire, wye system, the three phase conductors have the same voltage to the system neutral. The voltage between line conductors is √3 times the phase conductor to neutral voltage: The currents returning from the customers’ premises to the supply transformer all share the neutral wire.
## What is the difference between earthing and grounding and neutral?
The term “Earthing means that the circuit is physically connected to the ground and it is Zero Volt Potential to the Ground (Earth) but in case of “Grounding” the circuit is not physically connected to ground, but its potential is zero(where the currents are algebraically zero) with respect to other point, which is
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