IsaacAsimov Posted May 13, 2012 Author Share Posted May 13, 2012 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 tring for the same thing with pi? I've been trying to learn Latex all day on Sat. May 12, sometimes using a guide I got from the Net, and a lot of trial and error. I figured out most of it, but I haven't figured out what the \mathop command does, or how to separate the formulas on different lines on the screen. Can anybody help? Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted May 13, 2012 Share Posted May 13, 2012 If you want to separate formulas on different lines, do something like this: [math]3x^2[/math] [math]\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/math] Hit the Reply button on my post to see the code I used. I've never used mathop, so I don't know what it does. Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 13, 2012 Author Share Posted May 13, 2012 (edited) I have tried to use integrals to compute pi: [math]\mathop A(n) \rightarrow{\int_a^b} f(x) dx[/math] as [math] n\mathop\rightarrow \infty[/math] [math] = 2{\int\limits_{-1}^{1}}\sqrt{1-x^2}dx[/math] From p. 209 of Freshman Calculus: [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] a=1: [math]=\frac{1}{2}\arcsin{x}+\frac{x}{2}\sqrt{1-x^2}[/math] [math] = \left. 2 \bigg(\right|_{-1}^{1}\bigg) \frac{1}{2}\arcsin x + \frac{x}{2}\sqrt{1-x^2}[/math] [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] [math] = 2 [((\frac{1}{2})(\frac{\pi}{2})+\frac{1}{2}(0))-((\frac{1}{2})(\frac{-\pi}{2})-(\frac{1}{2})(0))] [/math] [math] = 2 \bigg(\frac{\pi}{4}+\frac{\pi}{4}\bigg)[/math] [math] = 2 \bigg(\frac{2\pi}{4}\bigg)[/math] [math] = \frac{4\pi}{4}[/math] [math] = \pi[/math] I have found that the integral equates to pi, but the formula doesn't tell me the value of pi. Edited May 13, 2012 by IsaacAsimov Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 13, 2012 Author Share Posted May 13, 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] What does the \mathop command do? Link to comment Share on other sites More sharing options...

studiot Posted May 13, 2012 Share Posted May 13, 2012 Sorry I have absolutely no idea. See my previous post it is a straight copy and paste. Perhaps some one who knows about these things will help? I think this forum uses something called MathML but I'm not sure - With the settings I now have in MathType to work in this website (SF) I ahve to edit the tags manually from [math] to [Tex] or [iTex]. Link to comment Share on other sites More sharing options...

studiot Posted May 13, 2012 Share Posted May 13, 2012 (edited) Please note that the function [math]y = \sqrt {(1 - {x^2})} [/math] Is the equation of a semicircle from x=-1 to x=+1. So you should have deduced that the area of a semicircle of radius 1 and diameter 2 is pi/2 and indeed pi is twice this integral as you say. Edited May 13, 2012 by studiot Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 16, 2012 Author Share Posted May 16, 2012 Given: [math]y=\sqrt{1-x^2}[/math] =SQR(1-X*X) a=0, b=1, n=10 [math] \Delta x = \frac{b-a}{n}=\frac{1-0}{10}=\frac{1}{10} = 0.1[/math] [math] \begin{matrix} I & X & Y & T \\ 0 & 0 & 1 & 1\\ 1 & 0.1 & 0.99499 & 1.99499\\ 2 & 0.2 & 0.97980 & 2.97478\\ 3 & 0.3 & 0.95394 & 3.92872\\ 4 & 0.4 & 0.91652 & 4.84524\\ 5 & 0.5 & 0.86602 & 4.84524\\ 6 & 0.6 & 0.8 & 6.51126\\ 7 & 0.7 & 0.71414 & 7.22541\\ 8 & 0.8 & 0.6 & 7.82541\\ 9 & 0.9 & 0.43589 & 8.26130 \end{matrix} [/math] [math] \pi=\Delta x(T)(4)=0.1(8.26130)(4)=3.30452[/math] Link to comment Share on other sites More sharing options...

mathematic Posted May 16, 2012 Share Posted May 16, 2012 What is T? Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 16, 2012 Author Share Posted May 16, 2012 What is T? T is the total of all the y values up to that point. Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 16, 2012 Author Share Posted May 16, 2012 Given: [math]y=\sqrt{1-x^2}[/math] =SQR(1-X*X) a=0, b=1, n=10 [math]\Delta x = \frac{b-a}{n}=\frac{1-0}{10}=\frac{1}{10} = 0.1[/math] [math] \begin{matrix} I & X & Y & T \\ 0 & 0 & 1 & 1 \\ 1 & 0.1 & 1.98997 & 2.98997 \\ 2 & 0.2 & 1.95959 & 4.94957 \\ 3 & 0.3 & 1.90788 & 6.85745 \\ 4 & 0.4 & 1.83303 & 8.69048 \\ 5 & 0.5 & 1.73205 & 10.42253 \\ 6 & 0.6 & 1.6 & 12.02253 \\ 7 & 0.7 & 1.42829 & 13.45081 \\ 8 & 0.8 & 1.2 & 14.65081 \\ 9 & 0.9 & 0.87178 & 15.52259 \\ 10 & 1 & 0 & 15.52259 \\ \end{matrix} [/math] [math]\pi=\frac{\Delta x}{2}(T)(4)=\frac{0.1}{2}(15.52259)(4) = 3.10452[/math] This is fairly close to the actual value of pi. Link to comment Share on other sites More sharing options...

mathematic Posted May 17, 2012 Share Posted May 17, 2012 Your method is distorting things. You are using 10 points for 10 rectangles. You need to adjust the end points (using 1/2 the value). The result will be closer. Link to comment Share on other sites More sharing options...

Daedalus Posted May 17, 2012 Share Posted May 17, 2012 (edited) I wouldn't call this a detailed analysis of 1/4 of a circle using rectangles when the underlying principle is based on the Riemann sum with only ten partitions. [math]f(x)=\sqrt{1-x^2}, \ \ \ \ \Delta x = \frac{b-a}{n}, \ \ \ \ x_k = a + k \ \Delta x[/math] The area of the region [math]R[/math] can be approximated using [math]n[/math] rectangles such that: [math]R_n = \sum_{k=0}^{n-1} f(x_k) \ \Delta x \ \ \ \ \text{or} \ \ \ \ R_n = \Delta x \ \sum_{k=0}^{n-1} f(x_k)[/math] Since integrating [math]f(x)=\sqrt{1-x^2}[/math] on the interval [math][0, 1][/math] is 1/4 the area of a circle with radius one, multiplying the above Riemann sum by 4 will approach the value [math]\pi[/math] as [math]n[/math] aproaches [math]\infty[/math]: [math] 4 \times \lim_{n \to \infty} R_n \ = \ 4 \times \lim_{n \to \infty} \ \sum_{k=0}^{n-1} f(x_k) \ \Delta x \ = \ 4 \times \int_0^1 f(x) \ dx \ = \ 4 \times \frac{\pi}{4} \ = \ \pi[/math] You can use the mid-point rule, the trapezoidal rule, or Simpson's rule to get better approximations when dealing with a finite number of rectangles. Edited May 17, 2012 by Daedalus 1 Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 18, 2012 Author Share Posted May 18, 2012 I wouldn't call this a detailed analysis of 1/4 of a circle using rectangles when the underlying principle is based on the Riemann sum with only ten partitions. [math]f(x)=\sqrt{1-x^2}, \ \ \ \ \Delta x = \frac{b-a}{n}, \ \ \ \ x_k = a + k \ \Delta x[/math] The area of the region [math]R[/math] can be approximated using [math]n[/math] rectangles such that: [math]R_n = \sum_{k=0}^{n-1} f(x_k) \ \Delta x \ \ \ \ \text{or} \ \ \ \ R_n = \Delta x \ \sum_{k=0}^{n-1} f(x_k)[/math] Since integrating [math]f(x)=\sqrt{1-x^2}[/math] on the interval [math][0, 1][/math] is 1/4 the area of a circle with radius one, multiplying the above Riemann sum by 4 will approach the value [math]\pi[/math] as [math]n[/math] aproaches [math]\infty[/math]: [math] 4 \times \lim_{n \to \infty} R_n \ = \ 4 \times \lim_{n \to \infty} \ \sum_{k=0}^{n-1} f(x_k) \ \Delta x \ = \ 4 \times \int_0^1 f(x) \ dx \ = \ 4 \times \frac{\pi}{4} \ = \ \pi[/math] You can use the mid-point rule, the trapezoidal rule, or Simpson's rule to get better approximations when dealing with a finite number of rectangles. I know that the integral formula equates to pi, however I was trying to find the decimal value of pi. Link to comment Share on other sites More sharing options...

Daedalus Posted May 19, 2012 Share Posted May 19, 2012 (edited) You can always use the Riemann sum to approximate [math]\pi[/math]. Although, it may not be the most efficient method: [math]R_n=\sum_{k=0}^{n-1} f(x_k) \, \Delta x \approx \pi[/math] [math]4 \times R_1 \ = 4.000000000[/math] [math]4 \times R_2 \ = 3.732050808[/math] [math]4 \times R_3 \ = 3.584220045[/math] [math]4 \times R_4 \ = 3.495709068[/math] [math]4 \times R_5 \ = 3.437048829[/math] [math]4 \times R_6 \ = 3.395316356[/math] [math]4 \times R_7 \ = 3.364086195[/math] [math]4 \times R_8 \ = 3.339818144[/math] [math]4 \times R_9 \ = 3.320407605[/math] [math]4 \times R_{10} = 3.304518326[/math] [math]...[/math] [math]4 \times R_{1000} = 3.143555467[/math] You might be interested in these methods: http://en.wikipedia....tions_of_%CF%80 In 1910, the Indian mathematician Srinivasa Ramanujan found several rapidly converging infinite series of pi, including which computes a further eight decimal places of pi with each term in the series. His series are now the basis for the fastest algorithms currently used to calculate pi. ... In 1989, the Chudnovsky brothers correctly computed pi to over a 1 billion decimal places on the supercomputer IBM 3090 using the following variation of Ramanujan's infinite series of pi: Edited May 19, 2012 by Daedalus Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 19, 2012 Author Share Posted May 19, 2012 [math]T_n=\frac{\Delta x}{2}[f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)][/math] [math]\text{Rectangle and Trapezoid Methods:}[/math] [math] \begin{matrix} \text{number of rectangles} & \text{value of pi} & \text{number of trapezoids} & \text{value of pi} \\ 10 & 3.30452 & 10 & 3.10452 \\ 100 & 3.16042 & 100 & 3.14042 \\ 1000 & 3.14356 & 1000 & 3.14156 \end{matrix} [/math] Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 20, 2012 Author Share Posted May 20, 2012 Thank you for showing me those two approximation of pi formulas. I would use them, but my C64 computer doesn't have a factorial function. I am very fond of pi, and I like calculating pi, as long as it's not too complicated. Marlon S. Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 20, 2012 Author Share Posted May 20, 2012 Here's a program I wrote in Structured Basic on my C64 computer: 100 REM COMPUTING PI USING RECTANGLES 110 REM DOTS REPRESENT INDENTATIONS 120 CALL INIT 140 CALL MAIN 160 CALL OUTPUT 180 END 200 : 220 PROC INIT 240 .....TI$="000000" 260 .....A=0:B=1 280 .....N=10 300 ENDPROC 320 : 340 PROC MAIN 360 .....DX=(B-A)/N 380 .....PRINT:PRINT"DX =";DX 400 .....X=-DX 420 .....I=-1 440 .....PRINT:PRINT" I ";" X ";" Y ";" T ":PRINT 460 .....LOOP 480 ........I=I+1 500 ........X=X+DX 520 ........Y=SQR(1-X*X) 540 ........T=T+Y 560 ........PRINT I;X;Y;T 580 .....UNTIL I=N-1 600 .....PI=DX*T*4 620 ENDPROC 640 : 660 PROC OUTPUT 680 .....PRINT:PRINT"PI =";PI 700 .....PRINT:PRINT"TIME TAKEN = ";TI$ 720 ENDPROC Link to comment Share on other sites More sharing options...

insane_alien Posted May 20, 2012 Share Posted May 20, 2012 Thank you for showing me those two approximation of pi formulas. I would use them, but my C64 computer doesn't have a factorial function. surely you could just make a factorial function for your C64, right? thats the beauty of programming, if there isn't a function for it, just program one! computers are turing machines so they can perform any process (at more or less efficiency than others) if you really really wanted to you could run global weather simulations on a C64. they'd be so slow as to be useless and the hardware would probably fail before the second iteration completed but you could get the exact same results as the super computers they run it on. Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 20, 2012 Author Share Posted May 20, 2012 Another fun formula (due to Wallis) is When I tried to use this formula, I got: m=10: [math]a=\frac{384(2(10)-2)}{945(2(10)-1)}\sqrt{2(10)}=\frac{384(18)}{945(19)}\sqrt{20}=1.72160[/math] [math]a=\sqrt{\frac{\pi}{2}}[/math] [math]a^2=\frac{\pi}{2}[/math] [math]\pi=2a^2=2(1.72160)^2=5.92784[/math] which is clearly wrong m=100: [math]a=\frac{384(2(100)-2)}{945(2(100)-1)}\sqrt{2(100)}=\frac{384(198)}{945(199)}\sqrt{200}=5.71777[/math] [math]\pi=2a^2=2(5.71777)^2=65.38574[/math] which is also clearly wrong What am I doing wrong? Link to comment Share on other sites More sharing options...

Bignose Posted May 20, 2012 Share Posted May 20, 2012 (edited) I can't seem to find the formula posted above as "Wallis' formula" http://en.wikipedia.org/wiki/Wallis_product has a somewhat different form. Your formula looks like it may be missing some terms... Furthermore, I think that 2*4*6*8*...*(2m-2) means that you multiple every even number together until you get to 2m-2 i.e. if m = 20, 2m-2 = 38, so the numerator is 2*4*6*8*10*12*14*16*18*20*22*24*26*28*30*32*34*36*38. The denominator looks the same, just with odd numbers. Edited May 20, 2012 by Bignose Link to comment Share on other sites More sharing options...

IsaacAsimov Posted May 21, 2012 Author Share Posted May 21, 2012 Thank you for clearing that up. I guess I'll have to use a computer instead of a calculator. I can't seem to find the formula posted above as "Wallis' formula" http://en.wikipedia..../Wallis_product has a somewhat different form. Your formula looks like it may be missing some terms... Furthermore, I think that 2*4*6*8*...*(2m-2) means that you multiple every even number together until you get to 2m-2 i.e. if m = 20, 2m-2 = 38, so the numerator is 2*4*6*8*10*12*14*16*18*20*22*24*26*28*30*32*34*36*38. The denominator looks the same, just with odd numbers. My calculator can handle numbers up to 69!, or 9.999...E+99, but using a calculator for the above even-odd formula is too time-consuming. I wrote a program for the even-odd formula on my C64 computer, and I got it right the first time! The only problem is that my computer can only handle numbers up to 1.7014E+38, or 33!, so I'm limited to m=10 for the even-odd formula, k=8 for the Ramanujan formula, and k=5 for the Chudnovsky formula. Link to comment Share on other sites More sharing options...

Xittenn Posted May 21, 2012 Share Posted May 21, 2012 (edited) By Arclength Approximation Riemann Form [math] f(x) = \sqrt { 1 - x^2 }\, ; \; f'(x) = - \frac{x}{\sqrt{1-x^2}}\, ; \; f''(x) = - \frac{1}{\sqrt{(1-x^2)^3}}\, ; \; f'''(x) = - \frac{3x}{\sqrt{(1-x^2)^5}}[/math] [math] S = \frac {\pi}{2} = \int_{0}^{1} \sqrt { 1 + [f'(x)]^2 } \, dx = \int_{0}^{1} \sqrt { 1 + [- \frac{x}{\sqrt{1-x^2}}]^2 } \, dx = \int_{0}^{1} \sqrt { 1 + \frac{x^2}{1-x^2} } \, dx [/math] [math] \Delta x = \frac{b - a}{n} = \frac{1}{n} [/math] [math] x_i^* = a + i \cdot \Delta x = \frac{i}{n} [/math] [math] x_{i-1}^* = a + (i - 1) \cdot \Delta x = \frac{i - 1}{n} [/math] [math] \bar{x} = \frac{x_i^* + x_{i-1}^*}{2} = \frac{(\frac{i}{n}) + (\frac{i - 1}{n})}{2} = \frac{2i - 1}{2n} [/math] [math] M_n \approx \sum_{i=1}^n f( \bar{x} ) \cdot \Delta x = \sum_{i=1}^n \sqrt { 1 + \frac{(\bar{x})^2}{1-(\bar{x})^2} } \cdot \frac{1}{n} = \frac{1}{n} \sum_{i=1}^n \sqrt { 1 + \frac{(\frac{2i - 1}{2n})^2}{1-(\frac{2i - 1}{2n})^2} } = \frac{1}{n} \sum_{i=1}^n \sqrt{\frac{4n^2}{4n^2 - 4i^2 + 4i - 1 }}[/math] [math] error = |S - M_n| \, ; \; K_2 = max | f''(x) | \, ; \; 0 \leq n = \sqrt{\frac{K_2(b - a)^3}{24(|S - M_n|)}} [/math] Edited May 21, 2012 by Xittenn Link to comment Share on other sites More sharing options...

mathematic Posted May 21, 2012 Share Posted May 21, 2012 Thank you for clearing that up. I guess I'll have to use a computer instead of a calculator. My calculator can handle numbers up to 69!, or 9.999...E+99, but using a calculator for the above even-odd formula is too time-consuming. I wrote a program for the even-odd formula on my C64 computer, and I got it right the first time! The only problem is that my computer can only handle numbers up to 1.7014E+38, or 33!, so I'm limited to m=10 for the even-odd formula, k=8 for the Ramanujan formula, and k=5 for the Chudnovsky formula. How about (2/3)(4/5)(6/7)..... instead of doing the numerator and denominator separately? Link to comment Share on other sites More sharing options...

doG Posted May 21, 2012 Share Posted May 21, 2012 Another fun formula (due to Wallis) is.... Did you mean this one? Link to comment Share on other sites More sharing options...

baxtrom Posted May 22, 2012 Share Posted May 22, 2012 You can use the mid-point rule, the trapezoidal rule, or Simpson's rule to get better approximations when dealing with a finite number of rectangles. Also Richardson extrapolation is worth mentioning here. If [math]I_n[/math] is the result using the trapezoidal method for n steps, then [math]I_{2n} + \frac{I_{2n}-I_n}{3}[/math] ..will give super-duper accuracy compared to [math]I_{2n}[/math] alone. Good way to improve results without too much added coding. Link to comment Share on other sites More sharing options...

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