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We consider the following function.

$$u(x) = \int_{0}^{x} \frac{dt}{\sqrt{1 - t^4}}$$

$u(x)$ is defined on $[-1, 1]$. Since $u'(x) = \frac{1}{\sqrt{1 - x^4}} > 0$ on $(-1, 1)$, $u(x)$ is strctly increasing on $[-1, 1]$. Hence there exists the inverse function of $u(x)$. We denote the inverse function of $u(x)$ by $s(u)$. We call $s(u)$ lemniscate sine(see this question).

By the binomial series expansion,

$\frac{1}{\sqrt{1 - x^4}} = \sum_{k=0}^{\infty} {k-\frac{1}{2} \choose k}x^{4k}$

This series converges uniformly on $|x| \le \rho$, where $\rho$ is any real number such that $0 < \rho < 1$.

Hence when $|x| \le \rho$,

$u(x) = \int_{0}^{x} \frac{dx}{\sqrt{1 - x^4}} = \sum_{k=0}^{\infty} \int_{0}^{x} {k-\frac{1}{2} \choose k}x^{4k} dx = \sum_{k=0}^{\infty} \frac{1}{4k+1}{k-\frac{1}{2} \choose k}x^{4k+1}$

Then the power series expansion of $s(u)$ can be calculated as a formal power series.

$s(u) = u -\frac{1}{10}u^5 + \frac{1}{120}u^9 + \frac{11}{15600}u^{13}+ \frac{211}{353600}u^{17}+\cdots$

My question How do we prove that the power series expansion of $s(u)$ is of the following form?

$s(u) = \sum_{k=0}^{\infty} b_{4k+1}u^{4k+1}$

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Makoto Kato
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1 Answers1

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An analytic function $f(z)$ has Maclaurin series of the form $\sum_k b_{4k+1} z^{4k+1}$ iff $f(iz) = i f(z)$ for all $z$ in a neighbourhood of $0$. If $g(z) = 1/(1 - z^4)^{1/4}$ we have $g(iz) = g(z)$. Now $u(z)$ is an antiderivative of $g(z)$ in a neighbourhood of $z=0$ with $u(0) = 0$; since $v(z) = u(iz)/i$ also has $v(0) = 0$ and $v'(z) = u'(iz) = g(iz) = g(z)$, we must have $v(z) = u(z)$ in a neighbourhood of $0$, i.e. $u(iz) = i u(z)$ there. And then $s(z)$ is the unique inverse function of $u(z)$ on a neighbourhood of $0$: $s(iz) = w$ where $u(w) = iz$, but since $u(is(z)) = u(s(iz)) = iz$ we have $w = i s(z)$, i.e. $s(iz) = i s(z)$ in some neighbourhood of $0$.

Robert Israel
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