Help with apppproximate value of pi = sqrt(6* (1/(1)2 + 1/(2)2 + 1/(3)2+ 1/(4)2 + 1/(

[pyramid]

Hello
I am working on one of my lab for this week, which calculates the approximate value of pi. Listed below is the actual problem, which I am posting here, so that you can see the different requirements for the program.

The program is simple. I used the ‘for loop’ to define the number of increments, then set up a formula to calculate the pi. My codes compile with no problem, but the result is incorrect. One of the test data I used was 100 (terms), which should result in 3.13592904 (or something close to this). My codes return a result of 19.7060531358098.

I would appreciate any help from you guys. Thanks.



pi = sqrt(6* (1/(1)2 + 1/(2)2 + 1/(3)2+ 1/(4)2 + 1/(5)2 +...... + 1/(n)2)) where n is the number of terms.

Your problem for this lab is to write a program (using a for loop) that approximates pi using the above series. Allow the user to specify the number, n, of terms to be used. Try your program with n = 100, then 1,000, then 10,000.

Note how the approximation becomes better and better as you increase the number of terms. For comparison purposes, the correct value of pi to 15 decimal places is 3.141592653589793. You won’t get that kind of accuracy with your program, but to get the best accuracy possible make your integer datatypes long and your floating point datatypes long double.

Important: In order to run correctly in the Automatic Grading System, you must set your output precision to 15 and use the correct datatypes for all your variables.

#include <iostream>
#include <iomanip>
 
using namespace std;
 
int main()
{
    long double terms;     
    double pi;
    long int count;
 
    // Prompt the user to enter the number of terms
       cout << "Specify the number of terms to use:   " << endl;
       cin >> terms; 
 
    // Set up increments
 
       for (count =1; count <= terms; count++)
 
    // formula that will approximate pi based on the user input
 
       {
           pi = pi + (1.0 / (2.0 * count - 1));
       }   
            pi = pi * 6;
 
     cout << setprecision (15) << "The approximate value of pi is:   " << pi << endl;
 
system ("pause");
 
return 0;
} // end main

[Ganon11]

pi = pi + (1.0 / (2.0 * count - 1));

I believe this line is giving you trouble. Specifically, then denominator of the fraction. You write 2.0 * count - 1, which will perform (2.0 * count) - 1, though you might have meant 2.0 * (count - 1). However, from your problem specification, I don't think you have to have the -1 - that is, I think it should just be 2.0 * count.

[pyramid]

pi = pi + (1.0 / (2.0 * count - 1));

I believe this line is giving you trouble. Specifically, then denominator of the fraction. You write 2.0 * count - 1, which will perform (2.0 * count) - 1, though you might have meant 2.0 * (count - 1). However, from your problem specification, I don't think you have to have the -1 - that is, I think it should just be 2.0 * count.

Thanks.

     pi = pi + (1.0 / (2.0 * count));

.

I changed the code and removed the "-1". The result is 15.5621325529189 (using 100 as terms). It is still incorrect, but it's closer to the correct answer. I worked on this all day yesterday and I just can't figure out why it's not working.

Also, I apologize that I didn't put the codes in block. I just learned that today.

Thanks.

[Ganon11]

Don't worry too much about the code tags - just go ahead and read our Posting Guidelines when you get the chance.

Anyway, I think I found your problem. In the assignment specification, it says to take the square root of your final answer, which I don't see you doing.

However, if your answer is ~15, then your final result will probably be closer to 4 than 3...It will be yet closer to your answer, though.

[pyramid]

Don't worry too much about the code tags - just go ahead and read our Posting Guidelines when you get the chance.

Anyway, I think I found your problem. In the assignment specification, it says to take the square root of your final answer, which I don't see you doing.

However, if your answer is ~15, then your final result will probably be closer to 4 than 3...It will be yet closer to your answer, though.

I added the sqrt * pi, and the result is as you said, around 4...... I still cannot submit because it's not running the program completely (with correct results). I don't understand why it's not working.

I appreciate your input. Please let me know if you can think of anything else that may be causing the problem. I will continue to work on the codes. Thanks.

[Ganon11]

I got it!

I think you are reading your problem incorrectly, and when you copied it to the forums, it was also copied incorrectly. We have been reading the denominator as 2(1), then 2(2), 2(3),...,2(n). But I think it's actually (1)^2, (2)^2, (3)^2,...,n^2. Making this change and using terms = 100, I got 3.13207653180911 as desired from my own sample program.

As an aside, make sure you make pi a long double as requested in your problem specification.

[Banfa]

I think you are reading your problem incorrectly, and when you copied it to the forums, it was also copied incorrectly. We have been reading the denominator as 2(1), then 2(2), 2(3),...,2(n). But I think it's actually (1)^2, (2)^2, (3)^2,...,n^2. Making this change and using terms = 100, I got 3.13207653180911 as desired from my own sample program.

I think you are right the original series SUM 1..N ( 1 / 2N ) does not converge, it diverges as can be proved with the Integral test for convergence.

Since it diverges sqrt( 6 * SUM) also diverges and therefore reaches no limiting value and N -> infinity.

[pyramid]

Thank you guys. I worked on this more last night, and finally gave up around 10 p.m. -

I'm really excited about the results you got when you made the change. I will change the codes using the guidelines you provided. I'll post here again once I get my change completed.

Thanks again.

[pyramid]

The description of the problem is actually the same as what the instructor gave us ( i copied and pasted here). I am so frustrated because the program should be simple, but yet, I can't seem to generate the correct codes. Granted that I'm new to this class, I should at least understand the instructions.

I changed my codes from the original codes using (1)^2, (2)^2, (3)^2,...,n^2.,. I tested using 100 as 'n' of terms, and the result = 3.69166666666667 which is incorrect (but very close...... )

       for (count =1; count <= terms; count++)
       {
           pi = 6 * 1.0/2 + 1.0/4 + 1.0/6 + 1.0/8 + 1.0/10 + 1.0/ (terms * 2.0);
       }
        pi = pi;

long terms;
long double pi;
long int count;


Thanks for all your help.

[Ganon11]

What are you doing inside that for...loop? It looks like you're trying to write the entire series in one line, which would ruin the value of using a for...loop in the first place. You should only be adding one term to pi each execution of the loop.

You were very close before with your program. As Banfa and I have determined, the series is not (1 / 2N), it is (1 / N^2).

I was able to get the program working correctly by taking your sample code and changing one line.

[pyramid]

I think I have been working on this problem too long. I still cannot get the codes to work. I changed the codes to:

for (count =1; count <= terms; count++)
 
        {  
         pi = (1.0 /(terms * terms)*2);
         }
        pi = sqrt * pi; 
        pi= pi * 6;

Using terms = 100, the result I got was 2.01232066038189e-310 which is incorrect.

I appreciate all the help.

Thank you

[Ganon11]

You shouldn't be multiplying the denominator by 2 at all. Ignore what your assignment said, it did not mean to multiply by two, it meant to square the term. Get that *2 out of there and see if it works.

[pyramid]

This is the revised codes. I am still unable to get the correct results.

for (count =1; count <= terms; count++);
       {          
       pi = (1.0 /(terms * terms));                                   
       }
       pi = sqrt * pi;
       pi = pi * 6;
 
 

I set all the datatypes as outlined on my lab:
long int terms
long int count;
long double pi;
double sqrt;

using terms=100, the result I'm getting is 4.64600985669798e+265.

Thanks.

[RRick]

You got a couple of problems.

First it is not pi that's being summed up in the loop, but an intermediate value based only on the inverse of the squared counter. You need to maintain a separate value for that value and have the loop modify it.

Second, got your sqrt and times 6 reversed.

[pyramid]

Got it.

A big thanks to all of you.

I modified the codes based on your input and it worked!. Thanks again.

[vpascuzzi]

I have been fooling around with this little app and cannot get it to work. I've tried debugging in everyway, and my algorithm seems to be working, but I am getting the wrong results. I am talking, way off. This should be so simple for me, but I am a little rusty on my C++.

#include <iostream>
#include <iomanip>
#include <conio.h>
#include <math.h>
using namespace std;
 
int main(){
    long double terms;
    float denom = 0, numer = 0, pi = 0;
    long int count;
 
    // Get user input for n terms
    cout << "Please enter number of terms: ";
    cin >> terms;
 
    // Setup for...loop to count
    for ( count = 1; count <= terms; count ++ )
        denom += ( pow( count, 2 ) );
 
    cout << denom << endl;
 
    numer = 1 / denom;
 
    cout << numer << endl;
 
    pi = ( numer * 6 );
 
    cout << pi << endl;
 
    pi = sqrt( pi );
 
    // Output value of pi
    cout << "The value of pi is " << pi;
 
    getch();
 
    return 0;
} // end main

Any ideas, please help me! This has been driving me crazy all day at work.

/me sounds lame...

[Ganon11]

You are incorrectly assuming that 1/a + 1/b + 1/c + ...+ 1/n is 1/(a + b + c + ... + n), so you can't calculate the sum in separate steps like this. Instead of having denom and numer. calculate the fraction at each step of the for loop and add it to a running sum.

[JosAH]

Don't they teach basic calculus and algebra anymore nowadays?

kind regards,

Jos