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)Let $t = \frac{x - 4}{4}$. Then form above, we have that:
(2)We have that the Maclaurin series for the natural logarithm is given by:
(3)Substituting back $t = \frac{x - 4}{4}$ into the series above gives us that:
(4)Example 2
Find the Taylor series of the function $f(x) = \cos x$ in powers of $(x - \pi)$.
We have that:
(5)We have that the Maclaurin series for $\cos x$ is:
(6)Substituting $x - \pi$ for $x$ and negating the result gives us:
(7)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)The Maclaurin series for $e^{t}$ is:
(9)Therefore the Maclaurin series for $e^{2t}$ is:
(10)Multiplying both sides by $e$ gives us:
(11)Substituting $t = x + 1$ back in and we have that:
(12)