Pi Approximation

Hey,

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This program uses n terms of the the infinite series:

pi*pi = 12*(1 - 1/(2*2) + 1/(3*3) - 1/(4*4) + 1/(5*5) - ...
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I know that I should use the loop (for) but how can I represent this series to use it with (for) ...The user can enter any number of n terms to approximate Pi.

HELP

Well the red numbers would be your loop counter.
The blue numbers just make the entire series regular for nice easy code.

pi*pi = 12*(1/(1*1) - 1/(2*2) + 1/(3*3) - 1/(4*4) + 1/(5*5) - ...

x = 1
Then in the loop
x *= -1;
will generate the numerator which toggles +1 -1

I'm not sure exactly what you want here, but something like this would get up to 100 numbers and store them in an array. If zero is entered then it would break the loop.

Code:

---

int n[100];
int i;

for(i=0; i<100; i++)
{
    scanf("&#37;i", &n[i]);
    if(! n[i]) break;
}

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[edit]Also if you are expecting your compiler to interpret you numbers in your equation as floats you might want to add a decimal point. IE:

pi*pi = 12.0*(1.0/(1.0*1.0) - 1.0/(2.0*2.0) + 1.0/(3.0*3.0) - 1.0/(4.0*4.0) + 1.0/(5.0*5.0)

Otherwise I think something like 1/(5*5) would come out as 0.

Quote:

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Originally Posted by Salem

Well the red numbers would be your loop counter.
The blue numbers just make the entire series regular for nice easy code.

pi*pi = 12*(1/(1*1) - 1/(2*2) + 1/(3*3) - 1/(4*4) + 1/(5*5) - ...

x = 1
Then in the loop
x *= -1;
will generate the numerator which toggles +1 -1

---

Salem I'm a beginner to C and just started couple of elementry programs....

I want from the user to enter the number of n (terms) to approximate pi
let's say :


Enter n
Equation.............
loop.............. I'M LOST HERE !! .......how can I combine n with the equation and the loop
...........
printf.........

Can you read a number from the user?

Can you write a for loop which prints from 1 to n ?

yes I can do it

OK, so enhance it to print the result of say n*n
Then perhaps the result of 1/(n*n)
Eventually, you'll get there.