Finding Pi in C++Date: 10/10/2001 at 19:24:38 From: Chris Subject: Finding PI in C++ Hi, I have to write a program in C++ to find the first 100 digits in Pi. I'm supposed to follow the sequence: 4+(4/3)-(4/5)+(4/7)-(4/9)... etc. I have no idea where to begin except to use a WHILE or FOR loop. Any ideas? Thanks. Chris Date: 10/11/2001 at 09:51:24 From: Doctor Paul Subject: Re: Finding PI in C++ First, I think you have the addition and subtraction backward: Pi = 4 - (4/3) + (4/5) - (4/7) + ... I've never seen this formula for Pi before, but it is nonetheless true. I was convinced that your formula wasn't true because as I summed the series, I didn't get anything close to Pi. I'm not familiar with C++, but I'm assuming it has the value of Pi to 100 decimal places built in (or you can find it). If not, I'm not sure how you would be expected to do the assignment. All you need to do is compare the value of the sum of the series at any given term to the known value of Pi. For example: If the difference between the two is less than .01, then you are guaranteed *one* (not two) digit of accuracy. So if you want 2 digits of accuracy, continue summing terms of the series until the difference is less than .001. And so forth. Something kind of interesting occurs: I told Maple to sum the terms of the series as long as the difference between the values was > .01 Maple did this and it stopped after the 100th term. So after the 100th term in the series, the sum was < .01 away from the actual value of Pi. Then I told Maple to sum the terms of the series as long as the difference between the values was > .001 Maple did this and it stopped after the 1000th term. So after the 1000th term in the series, the sum was < .001 away from the actual value of Pi. So after the 1000th term I was guaranteed 2 digits of accuracy. Two digits of accuracy actually came much earlier in the series, but there was no guarantee until the difference between the two number was < .001. It seems that if I want to guarantee that the two numbers will be less than 10(-n) apart, I need to have the program sum 10^n terms in the series. If this is the case, then this series is converging to Pi very slowly. So I wonder if you would have to compute 10^101 terms in the series to guarantee 100 digits of accuracy. Maybe I'm not doing something correctly here because 10^101 is a VERY large number. I doubt your program could compute that many iterations in your lifetime. Even if your computer could compute 1 million terms a second (mine does about 10 per second), your program would take 3.16 x 10^87 years to run. Finally, here's the program I wrote. Yours will have to be different because the syntax won't be the same. But this should give you an idea. > restart; > i:=2; i := 2 > t:=1; t := 1 > pie:=0; pie := 0 > count:=0; count := 0 > while evalf(abs(pie-Pi)) > .01 do > pie := pie + (-1)^i * 4/t: > t:= t + 2: > i:=i + 1: > print(evalf(pie)): > count:=count + 1: > od: the variable pie keeps track of the sum of the series. The (-1)^i term makes it add and subtract with alternating terms. And the variable count isn't necessary - it just keeps track of how many terms the series has to sum before it reaches the desired level of accuracy. If you need more, there are *many* more formulas for Pi at Eric Weisstein's World of Mathematics: http://mathworld.wolfram.com/PiFormulas.html I'm sure some of them can be implemented into a programming language. I hope this is helpful. Please write back and let me know how it turns out. I'm curious now. I'd especially like to know if you can confirm or deny my conjecture that it takes 10^(n+1) iterations to guarantee n digits of accuracy. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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