Taylor Series of Combinations of Functions Examples 1

# Taylor Series of Combinations of Functions Examples 1

On the Taylor Series of Combinations of Functions page, we use the following Maclaurin series to obtain Taylor series of various functions:

• 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 $(-\infty, \infty)$.

We will now look at some more examples of obtaining Taylor series from the Maclaurin series above.

## Example 1

Find a Taylor series representation of the function $f(x) = \cos^2 x$ about $\frac{\pi}{8}$.

Let $t = x - \frac{\pi}{8}$. Then $x = t + \frac{\pi}{8}$ and so:

(1)
\begin{align} \quad \cos^2 x = \cos^2 \left ( t + \frac{\pi}{8} \right ) \end{align}

We use the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$ to get that:

(2)
\begin{align} \quad \cos^2 \left ( t + \frac{\pi}{8} \right ) = \frac{1}{2} \left [ 1 + \cos \left ( 2t + \frac{\pi}{4} \right ) \right ] \end{align}

We will now use the sum identity $\cos (x + y) = \cos x \cos y - \sin x \sin y$ to get that:

(3)
\begin{align} \quad \frac{1}{2} \left [ 1 + \cos \left ( 2t + \frac{\pi}{4} \right ) \right ] = \frac{1}{2} \left [ 1 + \cos (2t) \cos \left ( \frac{\pi}{4} \right ) - \sin (2t) \sin \left ( \frac{\pi}{4} \right ) \right ] = \frac{1}{2} \left [ 1 + \frac{1}{\sqrt{2}}\cos (2t) - \frac{1}{\sqrt{2}}\sin (2t) \right ] \end{align}

The Maclaurin series for $\cos t$ and $\sin t$ are respectively:

(4)
\begin{align} \quad \cos t = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} t^{2n} \end{align}
(5)
\begin{align} \quad \sin t =\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}t^{2n+1} \end{align}

Therefore, the Maclaurin series for $\cos 2t$ and $\sin 2t$ are respectively:

(6)
\begin{align} \quad \cos 2t = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} 2^{2n} t^{2n} \end{align}
(7)
\begin{align} \quad \sin 2t =\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}2^{2n+1} t^{2n+1} \end{align}

Substituting these sums into what we obtained earlier and we have that:

(8)
\begin{align} \quad \frac{1}{2} \left [ 1 + \frac{1}{\sqrt{2}}\cos (2t) - \frac{1}{\sqrt{2}}\sin (2t) \right ] = \frac{1}{2} + \frac{1}{2 \sqrt{2}} \cos (2t) - \frac{1}{2 \sqrt{2}} \sin (2t) \\ \quad = \frac{1}{2} + \frac{1}{2 \sqrt{2}} \left [ \sum_{n=0}^{\infty} \left ( \frac{(-1)^n}{(2n)!} 2^{2n} t^{2n} - \frac{(-1)^n}{(2n+1)!}2^{2n+1} t^{2n+1} \right ) \right ] \\ \quad = \frac{1}{2} + \frac{1}{2 \sqrt{2}} \left [ \sum_{n=0}^{\infty} \left ( \frac{(-1)^n}{(2n)!} 2^{2n} \left ( x - \frac{\pi}{8} \right )^{2n} - \frac{(-1)^n}{(2n+1)!}2^{2n+1} \left ( x - \frac{\pi}{8} \right )^{2n+1} \right ) \right ] \end{align}