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

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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

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