# Find the Maclaurin series for ex and prove  that it represents ex for all x.

Solution:

Let f(x) = ex

By Maclaurin expansion we have,

f(x) = f(0) + x f'(o) + $$\frac{x^2}{2!}$$f(0) + $$\frac{x^3}{3!}$$f”’(0) + …

Here,

f(x) = ex

Differentiating we get,

f(x) = ex = f(x) = f”’(x) = ….          ,$$\forall$$ x

At x = 0,

f(0) = e0 = 1 = f(0) = f(0) = f”’(0) = ….      ,$$\forall$$ x

Then (i) becomes

$$f(x) = 1 + x.1 + \frac{x^2}{2!}.1 + \frac{x^3}{3!}+ ….$$

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

This is the Maclaurin’s series of ex

Also,

$$f(x) = 1 + x.1 + \frac{x^2}{2!}.1 + \frac{x^3}{3!}+ …. + \frac{x^n}{n!} + R_n(x)$$

Where $$R_n(x)$$ is the remainder of the series. Since,

$$\left | R_n(x)) \right | \leq \frac{x^{n+1}}{(n + 1)!)}\to 0 as nn \to ∞$$

This shows that ex converges to its Maclaurin’s series