Taylor Series of Combinations of Functions

# Taylor Series of Combinations of Functions

Recall from the Frequently Used Maclaurin Series page the following 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 $(-\infty, \infty)$.

We will now see that we can obtain Taylor series of functions using these Maclaurin series just by changing variables.

## Example 1

Find the Taylor series of the function $f(x) = \ln x$ in powers of $x - 4$.

We have that:

(1)
\begin{align} \quad \ln x = \ln (x + 4 - 4) = \ln (4 + x - 4) = \ln \left (4(1 +\frac{x - 4}{4} \right ) \end{align}

Let $t = \frac{x - 4}{4}$. Then form above, we have that:

(2)
\begin{align} \quad \ln x = \ln (4(1 + t)) = \ln (4) + \ln (1 + t) \end{align}

We have that the Maclaurin series for the natural logarithm is given by:

(3)
\begin{align} \quad \ln (4) + \ln (1 + t) = \ln (4) + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{t^n}{n} \end{align}

Substituting back $t = \frac{x - 4}{4}$ into the series above gives us that:

(4)
\begin{align} \quad \ln(4) + \ln (1 + t) = \ln (4) + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x - 4)^n}{4^n n} \end{align}

## Example 2

Find the Taylor series of the function $f(x) = \cos x$ in powers of $(x - \pi)$.

We have that:

(5)
\begin{align} \quad \cos (x) = - \cos(x - \pi) \end{align}

We have that the Maclaurin series for $\cos x$ is:

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

Substituting $x - \pi$ for $x$ and negating the result gives us:

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

Find a Taylor series representation for the function $e^{2x + 3}$ in powers of $x + 1$.

Let $t = x + 1$. Then $x = t - 1$ and so:

(8)
\begin{align} \quad e^{2x + 3} = e^{2(t - 1) + 3)} = e^{2t - 2 + 3} = e^{2t + 1} = e \cdot e^{2t} \end{align}

The Maclaurin series for $e^{t}$ is:

(9)
\begin{align} \quad e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!} \end{align}

Therefore the Maclaurin series for $e^{2t}$ is:

(10)
\begin{align} \quad e^{2t} = \sum_{n=0}^{\infty} \frac{(2t)^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n t^n}{n!} \end{align}

Multiplying both sides by $e$ gives us:

(11)
\begin{align} \quad e^{2t + 1} = \sum_{n=0}^{\infty} e \frac{2^n t^n}{n!} \end{align}

Substituting $t = x + 1$ back in and we have that:

(12)
\begin{align} \quad e^{2x + 3} = \sum_{n=0}^{\infty} e \frac{2^n (1 + x)^n}{n!} \end{align}