Subject: Calculus

# Maclaurin Series

Colin Maclaurin (above) is a famous mathematician known for his work in Maclaurin series.

Taylor series provides an approximate for a certain function f(x) at some value x = c. A particular case where c = 0 is called the Maclaurin series, named after a Scottish mathematician, Colin Maclaurin.
The Taylor series in the form

f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}\left (x-c \right )^2 + \frac{f'''(c)}{3!}\left (x-c \right )^3+...

Becomes;

f(x)=f(0)+f'(0)(x)+\frac{f''(0)}{2!}\left (x \right )^2 + \frac{f'''(0)}{3!}\left (x \right )^3+...=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n

where f’(0) is the first derivative evaluated at x=0, f’’(0) is the second derivative, and so on. This is now the Maclaurin series.

### Illustration #1

Find the Maclaurin Series expansion for the following function

f(x) = sin x

Solution:

We first list the derivatives of f(x) evaluated at x=0.

f(x)=\sin{x}
,

therefore f(0)=0.

f'(x)=\cos{x}
,

so f'(0)=1.

f''{x}=-\sin{x}

so f''(0)=0.

f'''(x)=-\cos{x}
,

so f'''(0)=-1.

f''''(x)=\sin{x}=f{x}
,

so f''''(0)=0.

We can observe that the pattern will continue forever. We then substitute these values to the Maclaurin series:

f(x)=f(0)+f'(0)(x)+\frac{f''(0)}{2!}\left (x \right )^2 + \frac{f'''(0)}{3!}\left (x \right )^3+...

Which then becomes;

f(x)=\sin{x}= 0 + (1)(x)+(0)(x)^2-\frac{1}{6}x^3+(0)x^4+\frac{1}{120}x^5+...

Thus, the Mclaurin expansion for the function \sin{x} is

f(x)=\sin{x}= x-\frac{1}{6}x^3+\frac{1}{120}x^5+...

### Illustration #2

Find the Maclaurin Series expansion for the function

f(x) = cos x
.

Solution: We first list the derivatives of f(x) evaluated at x=0.

f(x)=\cos{x}
,

therefore f(0)=1.

f'(x)=-\sin{x}
,

so f'(0)=0.

f''{x}=-\cos{x}

so f''(0)=-1.

f'''(x)=\sin{x}
,

so f'''(0)=-0.

f''''(x)=\cos{x}=f{x}
,

so f''''(0)=1.

We can observe that the pattern is just repeated. We then substitute these values to the Maclaurin series:

f(x)=f(0)+f'(0)(x)+\frac{f''(0)}{2!}\left (x \right )^2 + \frac{f'''(0)}{3!}\left (x \right )^3+...

Which then becomes;

f(x)=\cos{x}= 1 + (0)(x)-\frac{1}{2}(x)^2+(0)x^3+\frac{1}{24}x^4+(0)x^5+...

Thus, the Mclaurin expansion for the function \cos{x} is

f(x)=\cos{x}= 1-\frac{1}{2}x^2+\frac{1}{24}x^4+...

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