Maclaurin Series Formula

A Maclaurin series, named after Scottish mathematician Colin Maclaurin, is a Taylor series expansion of a function about 0. All of the terms in a Maclaurin series are non-negative integer powers of the variable. Maclaurin theorem is a version of Taylor's theorem, sometimes known as Taylor's power series expansion, in which c = 0 and is a series expansion of a function around zero.

The Maclaurin theorem or standard maclaurin series is give as follows: 

\[f(x) - f(a) + f^{1}(a)(x-a) + \frac{f^{11}(a)}{2!}(x-a)^{2} + \frac{f^{111}(a)}{3!}(x-a)^{3} + \frac{f^{1V}(a)}{4!}(x-a)^{4} +......\]

The Maclaurin series formula is given as follows:

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

Where n! is the factorial of n.

a is the real or complex number.

\[f^{n}(a)\] is the nth derivative of f evaluated at point a.

Here we will study the formula for the Maclaurin series of common mathematical functions.

Exponential Function

The Maclaurin series expansion for the exponential function \[ e^{x}\] is given as follows:

\[ e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + .....\].

The Maclaurin series formula for an exponential function is given as follows:

\[ e^{x} = \sum_{n=0}^{\infty} \frac{(x)^{n}}{n!}\].

The exponential function converges for all the values of x.

Natural Logarithm Function

The Maclaurin series expansion for the natural logarithm function ln(1+x) and ln(1-x)  is given as follows:

\[ ln(1 - x) = -x * \frac{x^{2}}{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4} - .....\].

\[ ln(1 + x) = -x * \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + .....\].

The Maclaurin series formula for natural logarithm function ln(1+x) and ln(1-x) is given as follows:

\[ln(1 - x) = -\sum_{n = 0}^{\infty} \frac{x^{n}}{n}\]

\[ln(1 + x) = \sum_{n = 0}^{\infty} (-1)^{n + 1} \frac{x^{n}}{n}\]

The natural logarithm function ln(1+x) and ln(1-x) converges for |x| < 1.

Geometric Series

The Maclaurin series expansion for the geometric series and its derivatives is given as follows:

\[\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + x^{4} + .....\]

\[\frac{1}{(1 - x)^{2}} = 1 + x + 2x^{2} + 3x^{3} + 4x^{4} + .....\]

\[\frac{1}{(1 - x)^{3}} = 1 + 3x + 6x^{2} + 10x^{3} + 15x^{4} + .....\]

The Maclaurin series formula for the geometric series and its derivatives is given as follows:

\[\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n}\]

\[\frac{1}{(1 - x)^{2}} = \sum_{n = 1}^{\infty} nx^{n - 1}\]

\[\frac{1}{(1 - x)^{3}} = \sum_{n = 2}^{\infty} \frac{(n - 1)n}{2}x^{n - 2}\]

The geometric series and its derivatives converge for \[|x| < 1\].

Trigonometric Functions

The Maclaurin expansion for the trigonometric functions sin x, cos x and tan x  is given as follows:

\[sin x = x -  \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + .....\]

\[cos x = 1 -  \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + .....\]

\[tan x = x +  \frac{x^{3}}{3!} + \frac{2x^{5}}{15} - \frac{17x^{7}}{315} + .....\]

The Maclaurin series formula for the trigonometric functions sin x, cos x and tan x  is given as follows:

\[ sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n + 1)!}x^{2n + 1}\].

\[ cos x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}x^{2n}\].

\[ tan x = \sum_{n=1}^{\infty} \frac{(-4)^{n}(1 - 4^{n})}{(2n)!}x^{2n - 1}\].

Here the sin x and cos x converges for all values of x and tan x converges for \[\left | x \right | < \frac{\pi}{2}\].

Maclaurin Series Examples

1. Find the Maclaurin series expansion of the function \[e^{5x}\].

Ans: The Maclaurin series expansion for the exponential function ex is given as follows:

\[e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} - \frac{x^{4}}{4!} + .....\]

Here the given function is \[f(x) = e^{5x}\]. So expanding the function \[e^{5x}\] using Maclaurin series formula we get

\[e^{5x} = 1 + 5x + \frac{25x^{2}}{2!} + \frac{125x^{3}}{3!} - \frac{625x^{4}}{4!} + .....\]

\[e^{5x} = \sum_{n=0}^{\infty} \frac{(5x)^{n}}{(n)!}\]

Therefore, the Maclaurin series expansion of the function \[e^{5x}\] is \[\sum_{n=0}^{\infty} \frac{(5x)^{n}}{(n)!}\].

2. Find the Maclaurin series expansion of the function sin 2x.

Ans: The Maclaurin expansion for the trigonometric function sin x  is given as follows:

\[sin x = x -  \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + .....\]

Here the given function is f(x) = sin 2x. So expanding sin 2x using Maclaurin series formula we get

\[sin 2x = 2x -  \frac{8x^{3}}{3!} + \frac{32x^{5}}{5!} - \frac{128x^{7}}{7!} + .....\]

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

Therefore, the Maclaurin series expansion of the function sin 2x is \[\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n + 1)!}2x^{2n + 1}\]

3. Find the Maclaurin series expansion of the function cos 5x.

Ans: The Maclaurin expansion for the trigonometric function cos x  is given as follows:

\[cos 5x = 1 -  \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + .....\]

Here the given function is f(x) = cos 5x. So expanding cos 5x using Maclaurin series formula we get

\[cos 5x = 1 -  \frac{25x^{2}}{2!} + \frac{625x^{4}}{4!} + \frac{15625x^{6}}{6!} + .....\]

\[cos 5x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}5x^{2n}\]

Therefore, the Maclaurin series expansion of the function cos 5x is \[\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}5x^{2n}\].

Conclusion

A Maclaurin series is a power series that is used to approximate a function f(x) for input values near zero if the values of the function's successive derivatives at zero are known. If the value of the function and all of its derivatives are known at a single point, the Maclaurin series may be used to determine the value of the entire function at every point. Approximations of the function can be found in the partial sums of the series.