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Prove that the power series defined as: $$ S(x) = \sum_{n = 0}^{+ \infty} \frac{n!}{1.3....(2n + 1)} x^{2n + 1} $$ is solution to the differential equation: $$ (E): (x^2 - 2)y' + xy + 2 = 0 $$ and find its expression.

So far I found:

$ S'(x) = \sum_{n = 0}^{+ \infty} \frac{n! (2n + 1)}{1.3....(2n + 1)} x^{2n} $

$$ \implies S'(x) = \sum_{n = 0}^{+ \infty} \frac{n! (2n)}{1.3....(2n + 1)} x^{2n} + \sum_{n = 0}^{+ \infty} \frac{n!}{1.3....(2n + 1)} x^{2n} $$

$$ \implies x^2 S'(x) = x \sum_{n = 0}^{+ \infty} \frac{n! (2n)}{1.3....(2n + 1)} x^{2n + 1} + x\sum_{n = 0}^{+ \infty} \frac{n!}{1.3....(2n + 1)} x^{2n + 1} $$

$$ \implies x^2 S'(x) = x \sum_{n = 0}^{+ \infty} \frac{n! (2n)}{1.3....(2n + 1)} x^{2n + 1} + xS(x) $$

I do not see how to proceed to the other terms.

Zouhair El Yaagoubi
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1 Answers1

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$$S(x) = \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)\cdot(2n+1)} x^{2n+1} \\ \implies S'(x) = \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n}$$

The left side of the ODE is then

$$\sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n+2} - 2 \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n} + \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)\cdot(2n+1)} x^{2n+2} + 2$$

Combine the series with the same-degree terms.

$$2 \sum_{n=0}^\infty \frac{(n+1)!}{1\cdot3\cdot\cdots\cdot(2n-1)\cdot(2n+1)} x^{2n+2} - 2 \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n} + 2$$

Shift the index on the first sum.

$$2 \sum_{n=1}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-3)\cdot(2n-1)} x^{2n} - 2 \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n} + 2$$

Join the constant term to the first sum and we're done.

$$2 \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n} - 2 \sum_{n=0}^\infty \frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} x^{2n} = 0$$

To get a closed form for $S$, observe that

$$\frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)} = \frac{n!\cdot2\cdot4\cdot\cdots\cdot2n}{1\cdot2\cdot3\cdot4\cdot\cdots\cdot(2n-1)\cdot2n} = \frac{2^n\cdot(n!)^2}{(2n)!} = \frac{2^n}{\binom{2n}n}$$

Compare this to the series expansion of the square of $\arcsin$,

$$\arcsin^2(x) = \sum_{n=0}^\infty \frac{2^{2n-1}}{n^2\binom{2n}n} x^{2n}$$

We can then find $S'$ in terms of derivatives of $\arcsin^2(x)$, and recover $S$ by integrating.

Differentiate $\arcsin^2$ twice and multiply by $2x$ as needed to obtain $$\frac{2x\arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=0}^\infty \frac{2^{2n}}{n\binom{2n}n} x^{2n}$$ $$\frac{x^2\sqrt{1-x^2}+x\arcsin(x)}{(1-x^2)^{3/2}} = \sum_{n=0}^\infty \frac{2^{2n}}{\binom{2n}n} x^{2n} = S'\left(\sqrt2\,x\right)$$ and upon replacing $x\mapsto \frac x{\sqrt2}$, $$S'(x) = \frac{x^2\sqrt{2-x^2}+2x\arcsin\left(\frac x{\sqrt2}\right)}{(2-x^2)^{3/2}}$$ By the fundamental theorem of calculus, noting that $S(0)=0$, we have $$S(x) = \int_0^x \frac{y^2\sqrt{2-y^2}+2y\arcsin\left(\frac y{\sqrt2}\right)}{(2-y^2)^{3/2}} \, dy = \frac{2\arcsin\left(\frac x{\sqrt2}\right)}{\sqrt{2-x^2}} - x$$ as long as $|x|<\sqrt2$.

user170231
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  • After combining the series with the same degree terms, I don't see why on the nominator we have $(n + 1)!$. Can you tell me what I am missing? Thank you. – Zouhair El Yaagoubi Feb 01 '23 at 19:35
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    In the first sum, rewrite the coefficient as$$\frac{n!}{1\cdot3\cdot\cdots\cdot(2n-1)}=\frac{(2n+1)n!}{1\cdot3\cdot\cdots\cdot(2n-1)\cdot(2n+1)}$$Then adding the two $x^{2n+2}$ sums together, we get a coefficient of$$\frac{(2n+1)n!+n!}{\cdots}=\frac{2(n+1)n!}{\cdots}=\frac{2(n+1)!}{\cdots}$$ – user170231 Feb 01 '23 at 19:38
  • Ah, I see. How can I get the explicit expression of $S$? I am stuck in that part too, thank you for your help. – Zouhair El Yaagoubi Feb 01 '23 at 19:45
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    I've added a few more details to my answer – user170231 Feb 01 '23 at 19:53
  • Accepted answer, thank you so much for your help. – Zouhair El Yaagoubi Feb 01 '23 at 19:57