# Program to find ex in an efficient way in Python

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Suppose we have a number n. We have to find $e^{x}$ efficiently, without using library functions. The formula for $e^{x}$ is like

$$e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$

So, if the input is like x = 5, then the output will be 148.4131 because e^x = 1 + 5 + (5^2/2!) + (5^3/3!) + ... = 148.4131...

To solve this, we will follow these steps −

• fact := 1
• res := 1
• n := 20 it can be large for precise results
• nume := x
• for i in range 1 to n, do
• res := res + nume/fact
• nume := nume * x
• fact := fact *(i+1)
• return res

## Example

Let us see the following implementation to get better understanding −

def solve(x):
fact = 1
res = 1
n = 20
nume = x

for i in range(1,n):
res += nume/fact
nume = nume * x
fact = fact * (i+1)
return res

x = 5
print(solve(x))

5

## Output

143
Updated on 12-Oct-2021 14:38:19