# 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

$$\begin{array}{l}f(x)\end{array}$$
up to order n may be found using Series
$$\begin{array}{l}[f, {x, 0, n}]\end{array}$$

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 $$\begin{array}{l}e^{x}\end{array}$$ $$\begin{array}{l}\sum_{k=0}^{\infty}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+…..\end{array}$$ $$\begin{array}{l}sin\;x\end{array}$$ $$\begin{array}{l}\sum_{k=0}^{\infty}(-1)^{2}=\frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\frac{x^{7}}{7!}+…..\end{array}$$ $$\begin{array}{l}cos\;x\end{array}$$ $$\begin{array}{l}\cos x=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\ldots\end{array}$$ $$\begin{array}{l}\frac{1}{1-x}\end{array}$$ $$\begin{array}{l}\sum_{k=0}^{\infty}x^{k}=1+x+x^{2}+x^{3}+….(if-1 ### Solved Examples Question 1: Expanding \(\begin{array}{l}e^{x}\end{array}$$
: Find the Maclaurin Series expansion of
$$\begin{array}{l}f(x)=e^{x}\end{array}$$

Solution:

Recalling that the derivative of the exponential function is

$$\begin{array}{l}{f}'(x)=e^{x}\end{array}$$
In fact, all the derivatives are
$$\begin{array}{l}e^{x}\end{array}$$
.

$$\begin{array}{l}{f}'(0)=e^{0}=1\end{array}$$

$$\begin{array}{l}{f}”(0)=e^{0}=1\end{array}$$

$$\begin{array}{l}{f}”'(0)=e^{0}=1\end{array}$$

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:

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