# C Exercises: Calculate e raise to the power x using sum of first n terms of Taylor Series

## C Programming Mathematics: Exercise-24 with Solution

Write a C program to calculate e raised to the power of x using the sum of the first n terms of the Taylor Series.

From Wikipedia,

In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point.

Example:

The Taylor series for any polynomial is the polynomial itself.

The above expansion holds because the derivative of e^{x} with respect to x is also e^{x}, and e^{0} equals 1.

This leaves the terms (x − 0)^{n} in the numerator and n! in the denominator for each term in the infinite sum.

Example:

Input: n = 25

float x = 5.0

Output: e^x = 148.413162

**Sample Solution:**

**C Code:**

```
#include <stdio.h>
#include <stdlib.h>
float Taylor_exponential(int n, float x)
{
float exp_sum = 1;
for (int i = n - 1; i > 0; --i )
exp_sum = 1 + x * exp_sum / i;
return exp_sum;
}
int main(void)
{
int n = 25;
float x = 5.0;
if (n>0 && x>0)
{
printf("value of n = %d and x = %d ", n, x );
printf("\ne^x = %f",Taylor_exponential(n,x));
}
}
```

Sample Output:

value of n = 25 and x = 1968710504 e^x = 148.413162

**Flowchart: **

**C Programming Code Editor:**

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## C Programming: Tips of the Day

** What does (x ^ 0x1) != 0 mean?**

The XOR operation (x ^ 0x1) inverts bit 0. So the expression effectively means: if bit 0 of x is 0, or any other bit of x is 1, then the expression is true.

Conversely the expression is false if x == 1.

So the test is the same as:

if (x != 1)

and is therefore (arguably) unnecessarily obfuscated.

Ref :https://bit.ly/2NIisQM

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