Maclaurin Series of Combinations Of Functions

# Maclaurin Series of Combinations Of Functions

Recall from the Frequently Used Maclaurin Series page the following common Maclaurin series:

• The Geometric Series: $\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + ...$, for $-1 < x < 1$
• The Derivative of the Geometric Series: $\frac{1}{(1 - x)^2} = \sum_{n=0}^{\infty} nx^{n-1} = x + 2x + 3x^2 + ...$, for $-1 < x < 1$.
• The Antiderivative of the Geometric Series: $-\ln (1 - x) = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = x + \frac{x^2}{2} + \frac{x^3}{3} + ...$, for $-1 ≤ x < 1$.
• Inverse Tangent Function: $\tan ^{-1} x = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - ...$, for $-1 ≤ x ≤ 1$.
• Euler Exponential Function: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + ...$, for $(-\infty, \infty)$.
• Natural Logarithm: $\ln (1 + x) = \sum_{n=0}^{\infty} (-1)^{n+1} \frac{x^n}{n}$.
• Sine Function: $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$, for $(-\infty, \infty)$.
• Cosine Function: $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2} + \frac{x^4}{4} - ...$, for $(-\infty, \infty)$.

Using some substitutions, we can easily obtain a bunch of other Maclaurin series for some somewhat more complicated functions as we demonstrate below.

## Example 1

Find a Maclaurin series representation of the function $e^{2x^2/5}$.

We have that the Maclaurin series for $e^x$ is given for all $x \in \mathbb{R}$ by:

(1)
\begin{align} \quad e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + .. \end{align}

Substituting in $\frac{2x^2}{5}$ for $x$ gives us a Maclaurin series for $e^{\frac{2x^2}{5}}$:

(2)
\begin{align} \quad e^{\frac{2x^2}{5}} = \sum_{n=0}^{\infty} \frac{\left ( \frac{2x^2}{5} \right )^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n x^{2n}}{5^n n!} \end{align}

## Example 2

Find a Maclaurin series representation for the function $x \tan^{-1} (7x)$.

We have that the Maclaurin series for the inverse tangent function is given for $-1 ≤ x < 1$ by:

(3)
\begin{align} \quad \tan ^{-1} x = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1} \end{align}

Substituting $7x$ for $x$ gives us that:

(4)
\begin{align} \quad \tan^{-1} (7x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} (7x)^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} 7^{n+1}x^{2n+1} \end{align}

Now multiplying the equation above by $x$ gives us:

(5)
\begin{align} \quad x \tan^{-1} (7x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} 7^{n+1}x^{2n+2} \end{align}

Note that this Maclaurin series converges for $\mid 7x \mid < 1$, that is $\mid x \mid < \frac{1}{7}$, which is a relatively small interval.