First I need the derivatives of the function. I’ll take the first few derivatives.
\begin{align*}
f(x) \amp = \cosh x \\
\frac{d}{dx} f(x) \amp = \sinh x \\
\frac{d^2}{dx^2} f(x) \amp = \cosh x\\
\frac{d^3}{dx^3} f(x) \amp = \sinh x \\
\frac{d^4}{dx^4} f(x) \amp = \cosh x\\
\frac{d^5}{dx^5} f(x) \amp = \sinh x \\
\frac{d^6}{dx^6} f(x) \amp = \cosh x
\end{align*}
Then I’ll evaluate at the centre point.
\begin{align*}
f(0) \amp = 1\\
\frac{d}{dx} f(x) \Bigg|_{x=0} \amp = 0\\
\frac{d^2}{dx^2} f(x) \Bigg|_{x=0} \amp = 1\\
\frac{d^3}{dx^3} f(x) \Bigg|_{x=0} \amp = 0\\
\frac{d^4}{dx^4} f(x) \Bigg|_{x=0} \amp = 1\\
\frac{d^5}{dx^5} f(x) \Bigg|_{x=0} \amp = 0\\
\frac{d^6}{dx^6} f(x) \Bigg|_{x=0} \amp = 1
\end{align*}
Then I need to find a pattern. Here, the pattern is relatively obvious: a sequences of alternating zeros and ones. Only the even terms are non-zero, so I’ll only write the even terms in the final Taylor series. To write the even terms, I use \((2n)\) as \(n \in \NN\text{.}\)
\begin{equation*}
\frac{d^{2n}}{dx^{2n}} f(x) \Bigg|_{x=0} = 1
\end{equation*}
After describing a pattern for the derivatives at the centre point, I put that pattern into the standard form of the Taylor series.
\begin{equation*}
f(x) = \sum_{n=0}^\infty \frac{1}{(2n)!} x^{2n}
\end{equation*}
I take the limit of the ratio of the coefficients to determine the radius of convergence. (In this case, since there are only even terms, the resulting radius will be a bound on \(x^2\) instead of \(x\text{.}\))
\begin{equation*}
R = \lim_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}}
\right| = \lim_{n \rightarrow \infty} \left|
\frac{\frac{1}{2n!}}{\frac{1}{(2n+2)!}} \right| = \lim_{n
\rightarrow \infty} \frac{(2n+2)!}{2n!} = \lim_{n \rightarrow
\infty} (2n+1)(2n+2) = \infty
\end{equation*}
The radius of convergence for \(x^2\) is \(R =
\infty\text{.}\) Since \(x^2\) can be any real number, \(x\) can also be any real number, to the radius of convergence for the series is, indeed, infinity.
