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# Initial value problems, arcsin power series expansion and Euler’s zeta(2)

In this post we shall discuss an ODE with initial conditions to derive a power series expansion for $\arcsin^2(x)$ near $x=0$. Let us consider the following initial value problem

$\displaystyle (1-x^2)y^{{\prime}{\prime}}-xy^{\prime}-2=0, y(0)=y^{\prime}(0)=0$.

It is quite clear that the function $g:(-1, 1)\to\mathbb{R}$, $\displaystyle g(x)=\arcsin^2(x)$ satisfies the ODE above and since the coefficients are variable it is natural to look for a power series solution $\displaystyle y(x)=\sum_{n=0}^{\infty}a_{n}x^n$. Differentiation term by term yields

$\displaystyle y^{\prime}(x)=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n$ and $\displaystyle y^{{\prime}{\prime}}(x)=\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$.

Since $a_{0}=y(0)=0$, $a_{1}=y^{\prime}(0)=0$, our initial value problem will be equivalent to

$\displaystyle 2a_{2}+6a_{3}x+\sum_{n=2}^{\infty}((n+2)(n+1)a_{n+2}-n^2a_{n})x^n=2$.

Clearly, $a_{2}=1$, $a_{3}=0$, and for $n\geq 2$, $\displaystyle a_{2n+1}=0, a_{n+2}=\frac{n^2}{(n+2)(n+1)}a_{n}$. We easily obtain that for $n\geq 2$,

$\displaystyle a_{2n+1}=0$ and $\displaystyle a_{2n}=\frac{(2^{n-1}(n-1)!)^2}{(2n)(2n-1)\ldots 4\cdot 3}=\frac{1}{2}\frac{2^{2n}}{n^2\binom{2n}{n}}$.

Therefore, this implies

$y(x)=\boxed{\displaystyle \arcsin^2(x)=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}, |x|\leq 1}$.

The above formula serves as a good ingredient in evaluating series involving the central binomial coefficient like

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2\binom{2n}{n}}=\frac{\pi^2}{18}, \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^3\binom{2n}{n}}=\frac{2}{5}\zeta(3), \sum_{n=1}^{\infty}\frac{1}{n^4\binom{2n}{n}}=\frac{17}{3456}\pi^4.$

This series converges at $x=1$ by Raabe’s test, and for $x\in (-1, 1)$ it is uniformly convergent by the Weierstrass M-test.

Now, substitute $x=\sin t$, $0, and we get

$\displaystyle t^2=\sum_{n=1}^{\infty}\frac{2^{2n-1}}{n^2\binom{2n}{n}}\sin^{2n}t$.

Integrating from $0$ to $\frac{\pi}{2}$, we have

$\displaystyle \frac{\pi^3}{24}=\sum_{n=1}^{\infty}\frac{2^{2n-1}}{n^2\binom{2n}{n}}\int_0^{\frac{\pi}{2}}\sin^{2n}t$.

On the other hand, Wallis’ formula (integral form) tells us that $\displaystyle\int_0^{\frac{\pi}{2}}\sin^{2n}tdt=\frac{\pi}{2^{2n+1}}\binom{2n}{n}$, and thus we finally obtain Euler’s celebrated

$\displaystyle\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$.

Remark. The ideas presented above have been generalized in this short paper.