IsaacAsimov Posted April 30, 2012 Share Posted April 30, 2012 (edited) Area of Circumscribed (around a circle) n-gon = n/tan[(n-2)(180)/(2n)] square: Find: A_{s}, pi Given: r=1, n=4 A_{s}=n/tan[(n-2)(180)/(2n)]=4/tan[(4-2)(180)/(2x4)]=4/tan[2(180)/8]=4/tan45 = 4 hexagon: Find: A_{h}, pi Given: r=1, n=6 A_{h}=6/tan[(6-2)(180)/(2X6)]=6/tan[(4)(180)/12]=6/tan60 = 3.464 decagon: Find: A_{d}, pi Given: r=1, n=10 A_{d}=10/tan[(10-2)(180)/(2x10)]=10/tan[8(180)/20]=10/tan72 = 3.249 hectagon: Find: A_{h} Given: r=1, n=100 A_{h}=100/tan[(100-2)(180)/(2x100)]=100/tan[98(180)/200]=100/tan88.2 = 3.14263 chiliagon: Find: A_{c} Given: n=1000 A_{c}=1000/tan[(1000-2)(180)/(2x1000)]=1000/tan[998(180)/2000]=1000/tan89.82 = 3.1416 megagon: Find: A_{m} Given: n=1E6 A_{m}=1E6/tan[(1E6-2)(180)/(2x1E6)]=1E6/tan[(999998)(180)/2E6]=1E6/tan89.99982 = 3.141592654 = pi Edited April 30, 2012 by IsaacAsimov 1 Link to comment Share on other sites More sharing options...

John Cuthber Posted April 30, 2012 Share Posted April 30, 2012 Cool, but how do you calculate tan(x)? Link to comment Share on other sites More sharing options...

imatfaal Posted May 1, 2012 Share Posted May 1, 2012 MacLaurin Series? [math]\tan x = \sum_{n=0 }^{\infty} \frac{Bernouli_{2n}(-4)^n(1-4^n)}{(2n)!}x^{2n-1}[/math] Link to comment Share on other sites More sharing options...

Bignose Posted May 2, 2012 Share Posted May 2, 2012 3.141592654 = pi A small nit, but the above isn't right. 3.141592654 is not equal to pi, but is a close approximation 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 86280 34825 34211 70679 is the approximation to 100 digits, but it is not equal. Nit picking? Yes. Correct and important to realize that the above are both just approximations? I say also, yes. Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 2, 2012 Author Share Posted May 2, 2012 Cool, but how do you calculate tan(x)? Calculating tan(x) is easy - just use a calculator or a computer. Link to comment Share on other sites More sharing options...

imatfaal Posted May 2, 2012 Share Posted May 2, 2012 Calculating tan(x) is easy - just use a calculator or a computer. Why not cut out a step and use the calculator or computer to tell you pi? Surely the infinite series is the way to go in order to maintain the simple definition rather than use the blackbox of a calculator Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 3, 2012 Author Share Posted May 3, 2012 Why not cut out a step and use the calculator or computer to tell you pi? Surely the infinite series is the way to go in order to maintain the simple definition rather than use the blackbox of a calculator The whole purpose of the exercise was to calculate pi using circumscribed polygons. Using a calculator for the tan(x) function makes it a little easier. I'm tired of getting negative feedback from you people! Don't you think I deserve a little praise for my mathematical efforts? Link to comment Share on other sites More sharing options...

Bignose Posted May 4, 2012 Share Posted May 4, 2012 I'm tired of getting negative feedback from you people! Don't you think I deserve a little praise for my mathematical efforts? Easy there big fella. I don't think that there is a single thing in this thread that is negative. Just discussing things. Maybe there hasn't been a great deal of 'huzzahs' simply because this isn't exactly brand new. Consider the Dutchman Ludolph van Ceulen who spent most of his life computing pi based on a million million million sided polygon. He reported the value of pi to 35 digits based on that calculation in 1610. So, your calculation isn't exactly new. People have been doing it for literally hundreds of years now. Link to comment Share on other sites More sharing options...

studiot Posted May 4, 2012 Share Posted May 4, 2012 (edited) Some interesting historical values for pi The Rhind Papyrus (1650BC) [math]{\left( {\frac{{16}}{9}} \right)^2} = 3.1604[/math] The Bible 3 Archimedes (250BC) [math]3\frac{{10}}{{71}} < \pi < 3\frac{{10}}{{70}}[/math] Tsu Chung Chich (AD 450) [math]\frac{{355}}{{113}} = 3.1415929[/math] Edited May 4, 2012 by studiot Link to comment Share on other sites More sharing options...

John Cuthber Posted May 5, 2012 Share Posted May 5, 2012 Imatfaaal, could I trouble you to define the bit in your equation that says "Bernoulli" please? Link to comment Share on other sites More sharing options...

studiot Posted May 5, 2012 Share Posted May 5, 2012 (edited) Imatfaal is using Bernoulli numbers and their connection to pi through series expansions. http://mathworld.wol...ulliNumber.html Other useful numbers are Euler's numbers and Stirlings formula go well Edit In this case we have The bernoulli numbers B_{n} are the coeffiecients of the series [math]\frac{z}{{{e^z} - 1}} + \frac{z}{2} = \sum\limits_0^\infty {{B_{2m}}} \frac{{{z^{2m}}}}{{\left( {2m} \right)!}}[/math] for even indices, with B_{1} = -1/2 and with all other odd terms zero. This evaluates even values of the zeta function [math]\zeta (2m) = {\left( { - 1} \right)^{m + 1}}{B_{2m}}\frac{{{{\left( {2\pi } \right)}^{2m}}}}{{2.\left( {2m} \right)!}}[/math] leading to [math]\zeta (6) = \frac{{{\pi ^6}}}{{945}}[/math] Edited May 5, 2012 by studiot 1 Link to comment Share on other sites More sharing options...

imatfaal Posted May 5, 2012 Share Posted May 5, 2012 JC - I wasn't quite sure what the accepted short form for the Bernoulli Numbers (or more accurately the even-index - ie non-zero - Bernoullis) was - so I put it in full as Studiot correctly explained. The accepted form is B_{2m} for the even-index I was just reading up the Wikipage on them and read the following trivia that is pretty cool Ada Lovelace's note G on the analytical engine from 1842 describes an algorithm for generating Bernoulli numbers with Babbage's machine.^{[3]} As a result, the Bernoulli numbers have the distinction of being the subject of the first computer program. The wikipage has some good stuff - both the various definitions and some interesting commentary http://en.wikipedia.org/wiki/Bernoulli_number Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 5, 2012 Author Share Posted May 5, 2012 This is something I got out of Freshman Calculus, p. 255 A_{n}=delta x(f(x_{0})+f(x_{1})+f(x_{n-1})) Let delta x=(b-a)/n, x_{0}=a, x_{1}=a+delta x, x_{k}=a+k delta x for k=1,2,...,n I will be using the formula for a circle, x^{2}+y^{2}=1, or y=sqrt(1-x^{2}), evaluated between 0 and 1, so it will be a quarter of a circle. Let a=0, b=1, n=10 delta x = (b-a)/n=(1-0)/10=1/10=0.1 x_{0}=a=0, x_{1}=a+delta x=0+0.1=0.1, x_{2}=a+k delta x=0+2(0.1)=0.2,... Using a calculator: A_{n}=0.1(f(0)+f(1)+f(2)+...+f(9)) =0.1(sqrt(1-0^{2})+sqrt(1-0.1^{2})+sqrt(1-0.2^{2})+sqrt(1-0.3^{2})+sqrt(1-0.4^{2})+sqrt(1-0.5^{2})+sqrt(1-0.6^{2})+sqrt(1-0.7^{2})+sqrt(1-0.8^{2})+sqrt(1-0.9^{2})) =0.1(1+0.99499+0.9798+0.95394+0.9165+0.8660+0.8+0.71414+0.6+0.43589) =0.1(8.26126) =0.826126 x4 = 3.30 which is close to pi, but not very close Link to comment Share on other sites More sharing options...

John Cuthber Posted May 5, 2012 Share Posted May 5, 2012 So, rather than summing an infinite series to get pi , you sum an infinite series to get e then plug that into the infinite series to get the Bernoulli number, then you use that to calculate another infinite series to get tan(x). It seems a lot of trouble to go to. Link to comment Share on other sites More sharing options...

imatfaal Posted May 5, 2012 Share Posted May 5, 2012 The whole OP struck me as deliberately difficult - so I saw no reason not to pile on the complexity. Link to comment Share on other sites More sharing options...

studiot Posted May 5, 2012 Share Posted May 5, 2012 Another fun formula (due to Wallis) is [math]\mathop {\lim }\limits_{m \to \infty } \frac{{2.4.6.8.\left( {2m - 2} \right)}}{{3.5.7.9.\left( {2m - 1} \right)}}\sqrt {2m} = \sqrt {\frac{\pi }{2}} [/math] Link to comment Share on other sites More sharing options...

Bignose Posted May 6, 2012 Share Posted May 6, 2012 (edited) which is close to pi, but not very close Sure. The more rectangles you use, the better. Also, you can use trapezoids, and polynomials, and .... lots of choices. Numerical integration is still an active area of research to maximize accuracy with the minimum number of function evaluations. If you can save even a few % over the current methods, you will get significant attention, considering how many function calls those large supercomputers do every second. Edited May 6, 2012 by Bignose Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 7, 2012 Author Share Posted May 7, 2012 Another fun formula (due to Wallis) is [math]\mathop {\lim }\limits_{m \to \infty } \frac{{2.4.6.8.\left( {2m - 2} \right)}}{{3.5.7.9.\left( {2m - 1} \right)}}\sqrt {2m} = \sqrt {\frac{\pi }{2}} [/math] That is an interesting formula. I'm trying to figure out how to insert special scientific symbols such as square root, right arrow, infinity symbol, etc. I can see the notation you used in your post, but it looks sort of like a programming language, and I'm not familiar with that language. Is there a list of commands that I could get so I could type up a post using that notation? Marlon S. Link to comment Share on other sites More sharing options...

imatfaal Posted May 7, 2012 Share Posted May 7, 2012 Hi Marlon It's called Latex - there is a tutorial here. http://www.scienceforums.net/topic/3751-quick-latex-tutorial/ Best way to learn is to click on someone else post and you will see the code you need to use. you need to surround the formula with the tags [math] formula here [/math] There are also loads of sites on the web that will help Link to comment Share on other sites More sharing options...

studiot Posted May 7, 2012 Share Posted May 7, 2012 Hello Marlon, Yes it is written in LaTex, I think. From my point of view I use a (paid for) program called MathType which allows direct typing of mathematical formulae. I then copy and paste into the reply window and the program does the rest. When I started here a couple of weeks ago it didn't work and I had to change the tags from [tex] to [math] manually but something has been tweaked and it now happens automatically. This procedure works with many forums. Unfortunately similar actions in MathCad or Microsoft Equation editor do not have the desired effect. Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 9, 2012 Author Share Posted May 9, 2012 Here is a sneaky method of estimating pi: Draw a circle, with a radius of 1, let a be a small angle, and let x be the opposite side inscribed in the circle. sin a = x, and pi is approximately (360 sin a)/(2a), or pi=(180 sin a)/a, as a gets small. Angle a is measured in degrees. Try the formula on a few values of a, using a calculator: a............................pi .01.........................3.14154029392740 .0001......................3.14159264835381 .000001...................3.14159265358927 Question: Why is the formula so accurate? Answer: My calculator converts the angle to radians, then uses several terms of an infinite series to estimate the sin. Here is the equation in radians: pi=(pi sin a)/a. We are using pi to estimate pi. Pi=3.14159265358979 is hard coded into my calculator. We are using pi with 15 digits of accuracy to estimate pi to 13 digits of accuracy. At least it works! Link to comment Share on other sites More sharing options...

mathematic Posted May 9, 2012 Share Posted May 9, 2012 sin x = x - (x^3)/6 + (x^5)/120 - .... You see that for the values of a you were using the error term for sina = a gets very small. Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 12, 2012 Author Share Posted May 12, 2012 (edited) I have tried to use integrals to compute pi:<br><br>[math]\mathop A(n) \rightarrow{\int_a^b} f(x) dx[/math] as [math] n\mathop\rightarrow \infty[/math]<br>[math] = 2{\int\limits_{-1}^{1}}\sqrt{1-x^2}dx[/math]<br>From p. 209 of Freshman Calculus:<br><br>[math]\mathop{\int}\sqrt{a^2-x^2}dx = \frac{a^2}{2}\arcsin{\frac{x}{a}}+\frac{x}{2}\sqrt{a^2-x^2}+C [/math]<br>a=1: [math]=\frac{1}{2}\arcsin{x}+\frac{x}{2}\sqrt{1-x^2}[/math]<br>[math] = \left. 2 \bigg(\right|_{-1}^{1}\bigg) \frac{1}{2}\arcsin x + \frac{x}{2}\sqrt{1-x^2}[/math]<br>[math] = 2 \bigg[\bigg(\frac{1}{2}\arcsin1+\frac{1}{2}\sqrt{1-1}\bigg)-\bigg(\frac{1}{2}\arcsin-1-\frac{1}{2}\sqrt{1-1}\bigg)\bigg] [/math]<br>[math] = 2 [((\frac{1}{2})(\frac{\pi}{2})+\frac{1}{2}(0))-((\frac{1}{2})(\frac{-\pi}{2})-(\frac{1}{2})(0))] [/math]<br><br> Edited May 13, 2012 by IsaacAsimov Link to comment Share on other sites More sharing options...

mathematic Posted May 12, 2012 Share Posted May 12, 2012 You need to supply more details. A(n), f(x), integral limits (a,b)? Link to comment Share on other sites More sharing options...

studiot Posted May 12, 2012 Share Posted May 12, 2012 (edited) I have tried to use integrals to compute pi: So what happened? I remember reading a short story by your best friend Arthur C Clark. The story was about a monastery in Tibet where the monks had been working for thousands of years to enumerate all the names of God. Then they bought a computer and it was able to complete the task in hours. Then...... But that would be telling! Are you trying for the same thing with pi? Edited May 13, 2012 by studiot Link to comment Share on other sites More sharing options...

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