calculate pi

[btfrenchy]

Hey guys,
my professor wants me to write a program that calculates pi to about 15 digits, and i am just stumped. He wants us to use the formula pi=4arctan(1/5)-arctan(1/239), but i cant just call the function, i have to write my own. Here is what he gave us in class;

#define _USE_MATH_DEFINES
#include <iostream>
#include <cmath>
using namespace std;

int main ()
{
double oldval, newval, fact, xtothen, x;
int n, i;

cout << "Enter x: ";
cin >> x;

newval = 0;
oldval = 1;
n = 0;

while(oldval != newval)
{
oldval = newval;

//compute n**x
xtothen = 1;
for (i = 1; i<=n; i++)
xtothen *= x;

//compute n!
fact = 1;
for (i = 1; i<=n; i++)
fact +=i;

newval = oldval + xtothen / fact;

n++;

}

cout << "The answer is " << newval << endl;

return 0;


}

he said we should be able to use this to write it. Would anyone be kind enough to help me out? I'm so stumped... Thank you!

[JLBorges]

> he said we should be able to use this to write it

What he gave you in class computes

1 + x + x²/2! + x³/3! + x⁴/4! + ...

This is the Maclaurin series for e^x; when x==1, this becomes

e = 1 + x + 12/2! + 13/3! + 14/4! + ...

and can be used to compute the value of e.


What you are required to do is to use Machin's formula

PI/4 = 4 * arctan(1/5) - arctan(1/239)

to compute the value of PI;

the Maclaurin series expansion for arctan(x) is:

arctan(x) = x - x3/3 + x5/5 - x7/7 + x9/9 - ...

https://ccrma.stanford.edu/~jos/pasp/Arctangent_Series_Expansion.html

So write a function to compute arctan(x) to the required accuracy using this series, and use this function to calculate the value of PI.

[gelatine]

heres what i have but when i return k it makes an int of it and that int becomes 0 because the rounding. so the ouput is 0. if someone knows how to fix this tell me please.

#include "stdafx.h"
#include <cstdlib>
#include <cmath>
#include <stdio.h>

int arctan(float x) 
{
	float i,k;
	for(i=1;i<=31;i=i+2)
	{
      if(int((x+1))%4==0)
	  {
		  k=k-(pow(x,i)/i);
	  }
	  else
	  {
		  k=k+(pow(x,i)/i);
	  }
	}
	return(k);
}

int main() 
{
long double pi,arc1,arc2;
arc1=arctan(1/5);
arc2=arctan(1/239);
pi=(16*arc1)-(4*arc2);
printf("pi= %Lf \n",pi);
system("pause");
return 0;
}

[histrungalot]

// If you don't put the point zero 1/5 integer is zero
// and 1/239 is zero
arc1=arctan(1.0/5.0);
arc2=arctan(1.0/239.0);

Also, init k to zero float i,k(0); and make the return type a float or double but not int.

Edit: I don't think the if statement is working like you think. Its always false.
You just want to ping-pong between 1 and -1, so try

float val(1.0);
for (...) {
// Just put the val in the right spot
val *= -1.0;
}

I get

pi= 3.141592706 

[gelatine]

@btfrenchy
here is the full code:

#include <cstdlib>
#include <cmath>
#include <stdio.h>

float arctan(float x) 
{
	float i,k,val(1.0);
	for(i=1;i<=99;i=i+2)
	{
      
		  k=k+(val*(pow(x,i)/i));
		  val*=-1;
	
	}
	return(k);
}

int main() 
{
long double pi,arc1,arc2;
arc1=arctan(1.0/5.0);
arc2=arctan(1.0/239.0);
pi=((4*arc1)-(arc2))*4;
printf("pi= %.7Lf \n",pi);//if the length is higher then 7 the numbers are wrong
system("pause");
return 0;
}

[JLBorges]

In the series

arctan(x) = x - x3/3 + x5/5 - x7/7 + x9/9 - ...

Let:

termi = numreratori / denominatori

The following recurrence relation holds:

termi+1 = numreratori * ( x * -x ) / ( denominatori + 2 )

I would think that your professor expects you to exploit this recurrence relation (the class room code does use the recurrence relation in the expansion of e).

[btfrenchy]

thanks guys, this is kind of helpful. I'm still a little confused, you helped me a ton. Thank you.