18 replies to this topic

[aruwin]

Can someone give me some hints on how to do this?

Calculate pi:
-Use the formula given to calculate the numerical integration
-use the trapezoidal rule to write the program
-change the divisional number, and present the way the margin of error of the calculus numerical value changes. (if divisional number is multiplied by 2, 4, 8 times....show how the margin of error changes)

This is the formula given to calculate the numerical intergration
 pi.JPG   11.89K   15 downloads

Someone give me the guideline to do this. :(

Attached Files

•  formula.JPG   30.17K   48 downloads

[WingedPanther]

The summation is going to be implemented as a for loop. The values of x_i will need to be calculated, based on the number of x's you intend to use (they will range from x_0 to x_n with a distance of 2/n between them).

[Gerg? Magyar]

Try it:

#include <iostream>

#include <iomanip>


int main() {

  unsigned int decP;

  unsigned int denom=3;

  float ourPi=4.0f;

  bool addFlop=true;

  cout << "How many loop iterations? ";

  cin >> decP;

  for (unsigned int i=1;i<=decP;i++) {

    if (addFlop) {

      ourPi-=(4.0/denom);

      addFlop=false;

      denom+=2;

    } else {

      ourPi+=(4.0/denom);

      addFlop=true;

      denom+=2;

    }

  }

  cout << setiosflags(ios::showpoint) << setiosflags(ios::fixed) << setprecision(9);

  cout << "Pi calculated with " << decP << " iterations is: ";

  cout << ourPi << endl;

}

[aruwin]

Gerg? Magyar said:

Try it:

#include <iostream>

#include <iomanip>


int main() {

  unsigned int decP;

  unsigned int denom=3;

  float ourPi=4.0f;

  bool addFlop=true;

  cout << "How many loop iterations? ";

  cin >> decP;

  for (unsigned int i=1;i<=decP;i++) {

    if (addFlop) {

      ourPi-=(4.0/denom);

      addFlop=false;

      denom+=2;

    } else {

      ourPi+=(4.0/denom);

      addFlop=true;

      denom+=2;

    }

  }

  cout << setiosflags(ios::showpoint) << setiosflags(ios::fixed) << setprecision(9);

  cout << "Pi calculated with " << decP << " iterations is: ";

  cout << ourPi << endl;

}

But I don't understand this format.I use <stdio.h>

[WingedPanther]

Are you using C or C++? cout is basically printf in C++.

[Gerg? Magyar]

I wrote to VS 2010.
I hope to this you thought.

#include "stdafx.h"

#include <iostream>

#include <conio.h>

#include <math.h>


using namespace std;


int main()

{

	double x=0;

	long double pi;

	double i;

	cout << " To how many terms? ";

	cin >> i;


	for (long int n = 2; n <= i; n++) { 

	  x += 1.0/(n*n);

	} 

	pi =sqrt(6*(1+x));

	cout << " Estimated PI value to " << i << " terms: "<< pi; 


	getch();

} 

Or this:
http://upload.wikime...45e9445becc.png

[WingedPanther]

No. You are summing up 1/n^2, which has nothing to do with the function of interest.

Given that you are integrating sqrt(4-x^2), I would expect to see a reference to that SOMEWHERE in your code.

[aruwin]

WingedPanther said:

No. You are summing up 1/n^2, which has nothing to do with the function of interest.

Given that you are integrating sqrt(4-x^2), I would expect to see a reference to that SOMEWHERE in your code.

I don't learn about cout, I just learn how to use printf.
Exactly, how do I write the code for the integration of sqrt(4-x^2)??

[WingedPanther]

I would create a function called f that will accept a float x and return sqrt(4-x^2).

Do you understand what the summation formula means?

[aruwin]

WingedPanther said:

I would create a function called f that will accept a float x and return sqrt(4-x^2).

Do you understand what the summation formula means?

Nope,I don't understand :(

[Gerg? Magyar]

Read it:
http://educ.jmu.edu/.../more on pi.pdf
Pseudo code:

for m from 1 to 12 do

 for n from 1 to 12 do

  x:=’x’: a:=’a’: b:=’b’: c:=’c’:

  sol:=int(x^m*(1-x)^n*(a+b*x+c*x^2)/(1+x^2),x=0..1):

  out:=op(sol): eq1:=0: eq2:=0: eq3:=0:

  for i from 1 to nops(sol) do

   if type(out[i],Or(name,rational&*name,name&*rational)) then

    eq1:=eq1+out[i]:

    elif type(out[i]/Pi,Or(name,rational&*name,name&*rational)) then

    eq2:=eq2+out[i]/Pi:

   else

    eq3:=eq3+out[i]/log(2):

   end if;

  end do:

  assign(solve({eq1=355/113,eq2=-1,eq3=0})):

  assume(x>=0,x<=1); print(m,n,a,b,c,is(a+b*x+c*x^2>=0));

 end do:

end do:

[aruwin]

Gerg? Magyar said:

Read it:
http://educ.jmu.edu/.../more on pi.pdf
Pseudo code:

for m from 1 to 12 do

 for n from 1 to 12 do

  x:=’x’: a:=’a’: b:=’b’: c:=’c’:

  sol:=int(x^m*(1-x)^n*(a+b*x+c*x^2)/(1+x^2),x=0..1):

  out:=op(sol): eq1:=0: eq2:=0: eq3:=0:

  for i from 1 to nops(sol) do

   if type(out[i],Or(name,rational&*name,name&*rational)) then

    eq1:=eq1+out[i]:

    elif type(out[i]/Pi,Or(name,rational&*name,name&*rational)) then

    eq2:=eq2+out[i]/Pi:

   else

    eq3:=eq3+out[i]/log(2):

   end if;

  end do:

  assign(solve({eq1=355/113,eq2=-1,eq3=0})):

  assume(x>=0,x<=1); print(m,n,a,b,c,is(a+b*x+c*x^2>=0));

 end do:

end do:

But where is the code for the integral function for pi given??