Taylor series of $\sinh{(x)}$ at $\ln{(2)}$.

Question

Determine the Taylor series of $\sinh{(x)}$ about $x = \ln{(2)}$.
Equating each derivative at $x = \ln{(2)}$ gives: \begin{equation*} \begin{split} f'(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\ f''(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\ f^{(3)}(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\ f^{(4)}(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\ f^{(5)}(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \end{split} \end{equation*} and so on. This means the Taylor series is \begin{equation*} \begin{split} \sum_{n=0}^{\infty} \frac{f^{(n)}(\ln{(2)})}{n!}(x-\ln{(2)})^n &= f(\ln{(2)}) + \frac{f'(\ln{(2)})}{1!}(x-\ln{(2)}) + \frac{f''(\ln{(2)})}{2!}(x-\ln{(2)})^2 \\ &+ \frac{f^{(3)}(\ln{(2)})}{3!}(x-\ln{(2)})^3 + \frac{f^{(4)}(\ln{(2)})}{4!}(x-\ln{(2)})^4 + \frac{f^{(5)}(\ln{(2)})}{5!}(x-\ln{(2)})^5 + \ldots \\ &= \frac{3}{4}+\frac{5}{4\cdot1!}(x-\ln{(2)})^1+\frac{3}{4\cdot 2!}(x-\ln{(2)})^2 \\ &+\frac{5}{4\cdot 3!}(x-\ln{(2)})^3+\frac{3}{4\cdot 4!}(x-\ln{(2)})^4+\frac{5}{4\cdot 5!}(x-\ln{(2)})^5+\ldots \end{split} \end{equation*} How do I find the general sum of this? Has it got something to do with the Maclaurin series which is $\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$.

Answer 1

Your computation leads naturally to:

$$\sinh x=\frac54\sum_{n=0}^{\infty} \dfrac{(x-\log2)^{2n+1}}{(2n+1)!} +\frac34\sum_{n=0}^{\infty} \dfrac{(x-\log2)^{2n}}{(2n)!}\\ =\sum_{n=0}^{\infty} \left(1-\dfrac{(-1)^n}4\right)\dfrac{(x-\log2)^{n}}{n!}$$

The radius of convergence is infinite.

Answer 2

$$\sinh(x)=\dfrac{e^x-e^{-x}}2=\sum_{r=0}^\infty\dfrac{x^{2r+1}}{(2r+1)!}$$

Set $x=\ln2$

Answer 3

$$\sinh (\ln 2)= \sum_{k=0}^{\infty} \left(\frac{2-(-1)^k\frac{1}{2}}{k!}\right) (x-\ln 2)^k$$