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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<t<\frac{\pi}{2}, 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.

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