for

[poorman]

if limit is 4

how to use for calculat this program?


#include<stdio.h>

int main(void)
{
double tpt =0;
int i, limit;

for(i=0; i<=limit; i++)
{
1 1 1 1 1
tot += - + - + - + - + -
1*1 1*2 1*2*3 1*2*3*4 (limit-1)
(4-1)
1*2*3
}

[Prelude]

>1 1 1 1 1 tot += - + - + - + - + - 1*1 1*2 1*2*3 1*2*3*4 (limit-1) (4-1) 1*2*3
What is this? This isn't even *close* to a valid expression.

-Prelude

[Shiro]

Hmmm, saw this post too late. There was another posting by you with this question. Can you tell us in words what you try to do?

[Brian]

Perhaps you should set limit to a value before comparing it.

[poorman]

Euler's equation

[Govtcheez]

> Euler's equation

Um... okay.... Still doesn't explain your problem... Almost nothing in your code is valid C.

[Brian]

"Bad programming = Compiler errors" -- Euler's equation.

[poorman]

What's wrong in this program?
How to fix it?

use loop to terminates when the difference between
two succesive values of e differ by less than
0.0000001

#include<stdio.h>

void getLimit(int *);
void calculation(int);

int main(void)
{
int limit;

getLimit(&limit);
calculation(limit);

return 0;
}

/**************************GET LIMIT********************/

void getLimit(int *limit)
{
printf("Enter limit: ");
scanf("%d", limit);

return;
}

/**************************CALCULATION************* *******/

void calculation(int limit)
{
int i, j, p, flag=0;
double sum, diff, psum=1;

for(i=1; i<limit; i++)
{
p=1;

for(j=1; j<=i; j++)
p=p*j;

sum=psum + (1/p);
diff=psum-sum;

if(diff < 0.0000001)
{
flag=1;
break;
}

if(flag==1)
printf("Prev-next:%.2lf %.2lf %.2lf\n", psum, sum, diff);
else
printf("Num of iteration: %d, e=%.5lf\n", limit, sum);
}
return;
}

[Shiro]

Your program looks a bit unstructured. Take some more effort in designing your algorithm.

Second, there are many formula's of Euler. But I think your pointing at the formula to calculate e.

e = sigma (x=0->inf) (1 / x!)

Which you need to implement as

e = sigma (x=0->limit) (1 / x!)

Which is

e = 1 / 0! + 1 / 1! + 1 / 2! + ... + 1 / limit!

As you can see there are two main calculations to be made

1. Sum
2. Faculty

A sum can be calculated by using a for-loop and faculty can be a function:

e = 0
for (x = 0; x < limit; x++)
{
    e += 1 / fac (x);
}

The faculty function is a recursive one

x! = x * (x-1) * (x-2) * ... * 3 * 2 * 1

So you see the end-step is reached if x is 1 and the recursion step is x * (x - 1). The function is now easily created:

int fac (int i)
{
    if (i == 1 || i == 0)
    {
        return 1;
    }
    else
    {
        return (i * fac (i - 1));
    }
}