I need help with a problem. For context, the section of the textbook the problem is in is about power series. Note that the textbook uses the convention that $f^{(n)}$ represents the $n$th derivative of $f$, and $f^{(0)}(x) = f(x).$ I'll now state the problem exactly as stated in the textbook:
Consider the function $f$ defined by
$f(x) = \arcsin x$, for $\lvert x \rvert \leq 1$.
The derivatives of $f(x)$ satisfy the equation
$
(1 – x^2)f^{(n + 2)}(x) – (2n + 1)xf^{(n + 1)}(x) – n^2 f^{(n)}(x) = 0$, for $n \geq 1.
$
The coefficient of $x^n$ in the Maclaurin series for $f(x)$ is denoted by $a_n$. You may assume that the series only contains odd powers of $x$.
$\textbf{a.1)}$ Show that, for $n \geq 1, (n+1)(n+2)a_{n+2} = n^2 a_n.$
$\textbf{a.2})$ Given that $a_1 = 1$, find an expression for $a_n$ in terms of $n$, valid for odd $n \geq 3.$
$\textbf{b})$ Find the radius of convergence of this Maclaurin series.
$\textbf{c})$ Find an approximate value for $\pi$ by putting $x = \frac{1}{2}$ and summing the first three non-zero terms of this series. Give your answer to $\textbf{four}$ significant figures.
I'm stuck on $\textbf{a.1}$. The way the question is formulated makes me think you're not supposed to use the actual derivatives of $\arcsin$ to solve it, but I can't figure out how to do it. I know that $a_n = \frac{f^{(n)}(0)}{n!}$,so I was thinking that if I can find a formula for the nth derivative of $f(x)$, I should be good to go.
I know the derivative of $f(x)$:
$f^\prime(x) = \frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1- x^2}}$. From here, I can easily also find the second, third, etc. derivatives. However, when I try to come up with a formla for the $\textit{nth}$ derivative, I have a problem. I came up with the following formula:
$\frac{d^n}{dx^n}\arcsin x = (-1)^n \prod\limits_{k = 0}^n \left(\frac{1}{2} – k\right)$.
Unfortunately, I have no idea how to proveed from here, as I don't know how to evaluate the product $\prod\limits_{k = 0}^n \left(\frac{1}{2} – k\right)$. Anyway, I don't think this is the right approrach, as my textbook hasn't dealt with products yet, only sums. Can anyone help with $\textbf{a.1}$?
Best Answer
The given differential equation shows you that
$$ f^{(n + 2)}(0) - n^2 f^{(n)}(0) = 0. $$
Then using Taylor,
$$(n+2)!\,a_{n+2}-n^2\,n!\,a_n=0.$$
Simplify.