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Let f(x) = e^{x}

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) = e^{x}

Differentiating we get,

f^{‘}(x) = e^{x} = f^{”}(x) = f^{”’}(x) = …. ,\(\forall\) x

At x = 0,

f(0) = e^{0} = 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 e^{x}

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 e^{x} converges to its Maclaurin’s series

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