In mathematics, series is the sum of finite or infinite number of terms. Finite series defined its last terms whereas infinite series continue till infinity (indefinite). In short we can say that, Series is the result of summing up all the terms of any sequence. In mathematics we deal with various series like arithmetic series, geometric series, tayors series etc. Maclaurin series usually are a form of series enlargement through which all terms are positive integer powers of the variable.

## Definition

Maclaurin series is a special case of the Taylor series. This Taylor series is useful for approximating operates as soon as a single can easily distinguish a compact parameter.

Consider a function g has a power series expansion at b, then the Tayor series of the functon g at b is as follow:

g(y) = $\sum_{n=0}^{\infty}$ $\frac{g^n(b) (y - b)^n}{n!}$

= g(b) + $\frac{g'(b)(y - b)}{1!}$ + $\frac{g''(b)(y - b)^2}{2!}$ + $\frac{g'''(b)(y - b)^3}{3!}$ + $\frac{g''''(b)(y - b)^4}{4!}$ + .....

Special case: Put b = 0 in the above formula, Tayor series becomes:

g(y) = $\sum_{n=0}^{\infty}$ $\frac{g^n(0) (y - 0)^n}{n!}$

= g(0) + $\frac{g'(0)(y - 0)}{1!}$ + $\frac{g''(0)(y - 0)^2}{2!}$ + $\frac{g'''(0)(y - 0)^3}{3!}$ + $\frac{g''''(0)(y - 0)^4}{4!}$ + .....

This is the special case of Tayor's series and named as Maclaurin series.

## Formula

Maclaurin series is generally referred to as following Scottish mathematician Colin Maclaurin. It is a expansion of Taylor series of a function about zero. For example, e$^x$ is equal to the sum of its Taylor series at zero.
Maclaurin series of the function g at zero is as follow:

g(y) = $\sum_{n=0}^{\infty}$ $\frac{g^n(0) (y - 0)^n}{n!}$

= g(0) + $\frac{g'(0)(y - 0)}{1!}$ + $\frac{g''(0)(y - 0)^2}{2!}$ + $\frac{g'''(0)(y - 0)^3}{3!}$ + $\frac{g''''(0)(y - 0)^4}{4!}$ + .....

## Maclaurin series for sinx

Maclaurin series for trigonometric function sin y is represents sin y for all y.

Let g(y) = sin y

 g(y) = sin y g(0) = sin 0 = 0 g'(y) = cos y g'(0) = cos 0 = 1 g''(y) = -sin y g''(0) = -sin 0 = 0 g'''(y) = -cos y g'''(0) = -cos 0 = - 1 g''''(y) = sin y g''''(0) = sin 0 = 0 g'''''(y) = cos y g'''''(0) = cos 0 = 1

We can see that derivatives repeats in a cycle of four.
This implies, Maclaurin series for sine function is:

g(0) + $\frac{g'(0)y}{1!}$ + $\frac{g''(0)y^2}{2!}$ + $\frac{g'''(0)y^3}{3!}$ + $\frac{g''''(0)y^4}{4!}$ + ................

= 0 + $\frac{y}{1!}$ + $\frac{0 \times y^2}{2!}$ + $\frac{-1 \times y^3}{3!}$ + $\frac{0 \times y^4}{4!}$ + .............
= y - $\frac{y^3}{3!}$ + .....

Sin y is equal to the sum of its Maclaurin series. Thus

Sin y = y - $\frac{y^3}{3!}$ + $\frac{y^5}{5!}$ - $\frac{y^7}{7!}$ + ..... =

$\sum_{n=0}^{\infty}$ (-1)$^n$ $\frac{y^{2n+1}}{(2n+1)!}$ for all y.

Maclaurin series for sin x: (Replace y by x)

Sin x = x - $\frac{x^3}{3!}$ + $\frac{x^5}{5!}$ - $\frac{x^7}{7!}$ + ..... =

$\sum_{n=0}^{\infty}$ (-1)$^n$ $\frac{x^{2n+1}}{(2n+1)!}$

## Maclaurin series for cosx

Prepare the calculation in two columns:

Let g(y) = cos y

 g(y) = cos y g(0) = cos 0 = 1 g'(y) = -sin y g'(0) = -sin 0 = 0 g''(y) = -cos y g''(0) = -cos 0 = -1 g'''(y) = sin y g'''(0) = sin 0 = 0 g''''(y) = cos y g''''(0) =  cos 0 = 1 g'''''(y) = -sin y g'''''(0) = -sin 0 = 0

Derivatives of function cosine are repeat in a cycle of four.

This implies, we can write the Maclaurin series as follows:

g(0) + $\frac{g'(0)y}{1!}$ + $\frac{g''(0)y^2}{2!}$ + $\frac{g'''(0)y^3}{3!}$ + $\frac{g''''(0)y^4}{4!}$ + ................

= 1 + $\frac{0 \times y}{1!}$ + $\frac{-1 \times y^2}{2!}$ + $\frac{0 \times y^3}{3!}$ + $\frac{1 \times y^4}{4!}$ + .............

= 1 - $\frac{y^2}{2!}$ + $\frac{y^4}{4!}$.....

Maclaurin series for cos y = 1 - $\frac{y^2}{2!}$ + $\frac{y^4}{4!}$..... = $\sum_{n=0}^{\infty}$ (-1)$^n$ $\frac{y^{2n}}{(2n)!}$

## Maclaurin series for tanx

Deriving the actual Maclaurin series for tanx is usually a simple method. It is a lot more of the exercise within distinguishing while using the chain rule to find the derivatives.

Let g(x) = tan x

g(0) = tan 0 = 0

g'(x) = 1 + tan$^2$x

g'(0) = 1 + tan$^2$(0) = 1 + 0 = 1

g''(x) = 2tan x + 2tan$^3$ x

g''(0) = 2tan (0) + 2tan$^3$ (0) = 0

g''' (x) = 2 + 8tan$^2$ x + 6tan$^4$ x

g''' (0) = 2 + 8tan$^2$ (0) + 6tan$^4$ (0) = 2
.......

Now g(x) = g(0) +  $\frac{g'(0)x}{1!}$ + $\frac{g''(0)x^2}{2!}$ + $\frac{g'''(0)x^3}{3!}$ + $\frac{g''''(0)x^4}{4!}$ + ................

Substituting all the value, we have

Tan x =  x + $\frac{2x^3}{3!}$ + .......

Maclaurin series for tan x is
Tan x =  x + $\frac{2x^3}{3!}$ + $\frac{16x^5}{7!}$.......

## How to Find Maclaurin Series

Follow below steps to find the maclourin series for any function:

1) Find the derivatives of function of every order. Evaluate every derivate at zero.

2)
Insert all the values in the maclourin series formula.

3)
Simplify the expression and get your result.

## Examples

Examples on Maclaurin series are illustrated below:

Example 1: Find the Maclaurin series of the function g(y) = $e^y$.

Solution: Given function is g(y) = e$^y$

Find the nth derivative of the function:

g$^1$(y) = $e^y$

g$^2$(y) = $e^y$

g$^3$(y) = $e^y$
.......
.......
then g$^n$(y) = $e^y$

Value of nth derivative at y = 0 is

g$^n$(0) = $e^0$ = 1 $\forall$ n

Thus the Taylor series for g at 0 (which is Maclaurin series) is

$\sum_{n=0}^{\infty}$ $\frac{g^n(0) y^n}{n!}$ = $\sum_{n=0}^{\infty}$ $\frac{y^n}{n!}$ = 1 + $\frac{y}{1!}$ +$\frac{y^2}{2!}$ + $\frac{y^3}{3!}$ + $\frac{y^4}{4!}$ + ..........

Which is required series.

Example 2: Find the Maclaurin series for g(y) = (y + 1)$^m$, where m is a real number.

Solution:
Find the nth derivative of g(y) = (y + 1)$^m$ and check the result for y = 0.

g(y) = (y + 1)$^m$

g'(y) = m(y + 1)$^{m-1}$

at y = 0

g'(0.) = m(0 + 1)$^{m-1}$ = m

g''(y) = m(m-1)(y + 1)$^{m-2}$

at y = 0

g''(0.) = m(m-1)(0 + 1)$^{m-2}$ = m(m-1)
......
......
g$^{(n)}$(y) = m(m-1).....(m-n+1)(y + 1)$^{m-n}$

at y = 0

g$^{(n)}$(y) = m(m-1).....(m-n+1)(0 + 1)$^{m-n}$ = m(m-1).....(m-n+1)

This implies Maclaurin series for g(y) = (y + 1)$^m$ is:

$\sum_{n=0}^{\infty}$ $\frac{g^{(n)}(0)y^n}{n!}$ = 1 + $\sum_{n=1}^{\infty}$ $\frac{m(m-1).....(m-n+1)y^n}{n!}$