approximate PI help

hello all

my current project is a program to approximate π (<sorry: its ugly but is pi) up to 1/99
w/ the usually taylor series representation, eg (π=4x(1-1/3+1/5...etc)

so far my code is

Code:

#include<stdio.h>
#include<stdafx.h>
#include"conio.h"
#include<math.h>

int main()
{
double n;
int i=0;

  while(i<100)
  {
  n=(((-1)^i)/(2*i+1));
  i++;
  }

printf(" PI is &#37;lf (debug %d i)", n, i);
getch();
return (0);
}

which runs but my n seems to stay 0 always... i was wondering if im approaching the problem too simply (quite simple myself) or if i am screwing up somthing (possibly ^ , i tried pow(x,x) config but that ran me into syntax.... anyways thank you for help in advance.

In C, ^ is bitwise XOR. You indeed need pow().

Furthermore, you aren't faithfully representing the summation; there are two errors. Try using a debugger to step through each line while watching the variables.

thanks zx

i am still a bit unsure of this, ive seen this stuff only on forums and never been taught in class.

is the pow() done in the code in the loop or is it done in the printf (eg... printf("%f is power or 4 to the 5th", pow(4,5));...) as ive seen examples of both but not usually explained well.

and would it need to be used as a function (im thinking not as functions arnt usually neccessary just easier to keep track of code and reuse.

i seem to become lost easily on the simplest things and im wondering if its my fault or my teachers :P

You need it anywhere you plan to raise something to a power. And of course it's a function.

Originally Posted by musifuger

im wondering if its my fault or my teachers :P

Given the fallout when some of those teachers post on this very forum, it's fair to wonder.

This one i fail syntax because pow, i still cant find an example that clears it up for me.

Code:

#include<stdio.h>
#include<stdafx.h>
#include"conio.h"
#include<math.h>


int main(void)
{
double n=0.0;
int i=0;

	while(i<100)
	{
	n= pow((-1), i)/((2*i)+1);
	++i;
	}

printf(" PI is %lf (debug %d i)", n, i);
getch();
return (0);
}

and i have tried to play w/ it this way also but i get "Error 1 error C2296: '/' : illegal, left operand has type 'void' f:\microsoft visual studio 8\projects\7 pi\7 pi\7 pi.cpp 21"

and im sure its because i dont understand the pow also...

Code:

#include<stdio.h>
#include<stdafx.h>
#include"conio.h"
#include<math.h>

void
pow1(double x, int i)
{
	pow(x, i);
}

int main(void)
{
double n=0.0 ,x=-1;
int i=0;

	while(i<100)
	{
	n= pow1(x, i)/(2*i+1);
	++i;
	}

printf(" PI is %lf (debug %d i)", n, i);
getch();
return (0);
}

i have seen a few posts talking about making the pow thru a loop, and not sure if this is easier i have tried to get it going but sadly have lost my code for this as of now: /

sorry to be thick of skull but any help would help... also for the debug im still getting 0 for n even w/ a print for 'n' inside the while loop.

You're better off not having the pow1 function (especially since it's declared wrong for the purpose intended).

Code:

	n= pow1(x, i)/(2*i+1);

With that snippet above, each time through the loop, you are setting "n" to the latest and greatest calculation. I presume you don't want to do this, but you want to actually adjust "n" by the amount you calculate.

Also, don't you need to flip x each loop too?

Todd

Originally Posted by Todd Burch

You're better off not having the pow1 function (especially since it's declared wrong for the purpose intended).

Code:

	n= pow1(x, i)/(2*i+1);

With that snippet above, each time through the loop, you are setting "n" to the latest and greatest calculation. I presume you don't want to do this, but you want to actually adjust "n" by the amount you calculate.

Also, don't you need to flip x each loop too?

Todd

I'm not familiar with pow1 either, although there is a powl. Todd's right about the n, but not the x, since we do actually want to calculate (-1)^i. Of course, any sane person wouldn't use pow, but would flip instead....

ahh i believe it should be n+=(pow(-1, i)/(2*i+1));
and yes -1^n should alternate the sign as is done in a taylor series (looks for old calc book)

i guess its not only pow im wondering about now as in my class has been guessing over most of our structure for c this semester.

am i correct in thinking i can use pow the way i have it in the line above?
and as to the pow1 i thought they were just names given to different prototypes? i hope they arnt keywords or im much more lost than already thought.

That's how pow works, yes.

powl returns a long double, instead of just a plain old double, in case you have big numbers.

thanks tabstop
i shouldnt need powl for this application.

i am getting an error of "Error 1 error C2668: 'pow' : ambiguous call to overloaded function f:\microsoft visual studio 8\projects\7 pi\7 pi\7 pi.cpp 16"

going to sleep since i doubt i can fix this tired, again thanks for th help.... and ill post my code again just so you can see what else it could be doing.

Code:

int main()
{
double n=0.0;
int i=0;

	while(i<100)
	{
	n+=(pow(-1, i)/(2*i+1)); 
	++i;
	}

printf(" PI is &#37;lf (debug %d i)", n, i);
getch();
return (0);
}

>> i am getting an error of "Error 1 error C2668: 'pow' : ambiguous call to overloaded function [...]"

pow is declared in several ways. The compiler can choose more than one of them and be correct, which is ambiguous, and thus, you get the error.

That's because you're using a C++ compiler: instead of having different names (like pow and powl), it uses one overloaded name (there's several different pow functions). But: none of them take integers. So, change -1 to -1.0. (Alternatively: use a C compiler.)

do not use pow function at all

on step i you have value of (-1)^i

just multiply it on (-1) to get the value of (-1)^(i+1) - you do not need power here...
even simple var which will be fliiped on each iteration from x=-x; will be enough

sweet the -1 to -1.0 did it, thanks so much, i figured it would a simple fix that i wouldnt have guessed in many years.

now i just gotta play w/ the formula to get half a bit closer to actual pi.

thanks alot