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Maclaurin Series Formula

A Maclaurin series is a function that has expansion series that gives the sum of derivatives of that function. The Maclaurin series of a function $f(x)$ up to order n may be found using Series $[f,  {x, 0, n}]$. 

It is a special case of Taylor series when x = 0. The Maclaurin series is given by

\[\large f(x)=f(x_{0})+{f}'(x_{0})(x-x_{0})+\frac{{f}”(x_{0})}{2!}(x-x_{0})^{2}+\frac{{f}”'(x_{0})}{3!}(x-x_{0})^{3}+…..\]

The Maclaurin series formula is

\[\large f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(x_{0})}{n!}(x-x_{0})\]

Where,
f(xo), f’(xo), f’‘(xo)……. are the successive differentials when xo = 0.

Function Maclaurin Series
 $e^{x}$  $\sum_{k=0}^{\infty}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+…..$
 $sin\;x$  $\sum_{k=0}^{\infty}(-1)^{2}=\frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\frac{x^{7}}{7!}+…..$
 $cos\;x$  $\sum_{k=0}^{\infty}(-1)^{2}=\frac{x^{2k}}{(2k)!}=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\frac{x^{6}}{6!}+…..$
 $\frac{1}{1-x}$  $\sum_{k=0}^{\infty}x^{k}=1+x+x^{2}+x^{3}+….(if-1<x<1)$
  $ln(1+x)$  $\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^{k}}{k}=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+….(-if-1<x\leq 1)$

Solved Examples

Question 1: Expanding $e^{x}$ : Find the Maclaurin Series expansion of $f(x)=e^{x}$

Solution:

Recalling that the derivative of the exponential function is ${f}'(x)=e^{x}$ In fact, all the derivatives are $e^{x}$ .

${f}'(0)=e^{0}=1$
${f}”(0)=e^{0}=1$
${f}”'(0)=e^{0}=1$

We see that all the derivatives, when evaluated at x = 0, give us the value 1.

Also, f(0)=1, so we can conclude the Maclaurin Series expansion will be simply:

$e^{x}\approx 1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\frac{1}{24}x^{4}+\frac{1}{120}x^{5}+….$

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