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*(*x*_{o}), *f’*(*x*_{o}), *f’*‘(*x*_{o})……. are the successive differentials when *x*_{o} = 0.

Function | Maclaurin Series |

$e^{x}$ | $\sum_{k=0}^{\infty}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+…..$ |

$sin\;x$ | $\sum_{k=0}^{\infty}(-1)^{2}=\frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\frac{x^{7}}{7!}+…..$ |

$cos\;x$ | \(\cos x=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\ldots\) |

$\frac{1}{1-x}$ | $\sum_{k=0}^{\infty}x^{k}=1+x+x^{2}+x^{3}+….(if-1<x<1)$ |

$ln(1+x)$ | \(\ln (1+x)=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{n}}{n}=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots\) |

### 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:

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