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$F(x) = \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$. Find $G(x) = \sum_{n=0}^{\infty}g_nx^{n}$ if $F(G(x)) = x$. Specifically find a formula for $g_n$

I am learning about formal power series, so please don't use $\sin x = F(x)$ which I think is true here, hence we can easily get $G(x)$ using this way.

I tried $F(G(x)) = \sum_{n=0}^{\infty}(-1)^{n}\frac{G(x)^{2n+1}}{(2n+1)!}$ and somehow getting the coefficient of x as 1 and all other cofficients of $x^j$ as $0$. I am unable to do so. If I need to provide the tedious computation which is not leading me anywhere, I am happy to do so.

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    $\begingroup$ This is the power series for $\sin $. $\endgroup$ – Andres Mejia Sep 23 '16 at 20:43
  • $\begingroup$ g(x) = inverse of sine function. I don't want to approach this like that. I mentioned that already in the question. That teaches me nothing about proofs of formal power series $\endgroup$ – Amrita Sep 23 '16 at 20:44
  • $\begingroup$ I'm sorry, that was my fault for rashly typing a comment. I didn't read your full question, obviously. $\endgroup$ – Andres Mejia Sep 23 '16 at 20:46
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    $\begingroup$ What you have here is a series reversion problem. The Mathworld page just linked has some discussion, as does Wikipedia's page on the Lagrange inversion theorem. (Also, while you may not want to use the fact of $G(x)=\sin^{-1}(x)$, you can at least use it to check your work.) $\endgroup$ – Semiclassical Sep 23 '16 at 20:47
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Hint: Differentiate to get $F'(G(x))G'(x) = 1$. Then differentiate again to get

$F''(G(x))G'(x)+F'(G(x))G''(x) = 0$

but note $F''(G(x)) = -F(G(x))$ (this secretly is true because $F$ is sine, but of course can be verified by differentiating the power series), so

$F'(G(x))G''(x) = -xG'(x)$ and multiplying both sides by $G'(x)$ gives

$G''(x) = -x(G'(x))^2$

which we can use to solve for the coefficients of $G(x)$.

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  • $\begingroup$ How do you prove product rule, chain rule for formal series? It looks so similar like differentiation of functions. $\endgroup$ – Amrita Sep 24 '16 at 2:54
  • $\begingroup$ it's well known that these rules hold. you can look up the proofs online if you wish but I don't think they're that exciting $\endgroup$ – mathworker21 Sep 24 '16 at 3:23
  • $\begingroup$ I am not finding anything easy to understand. By just using coefficients or something, not convergence and all. Any resource you can provide $\endgroup$ – Amrita Sep 24 '16 at 3:30
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    $\begingroup$ so "formal series" means we can ignore questions of convergence. they aren't technically polynomials. you can think of them as infinite sequences where $(a_0,a_1,a_2,\dots)$ corresponds to $\sum_{n=0}^{\infty} a_nx^n$. so all you have to do is compare coefficients. math.stackexchange.com/questions/570542/… $\endgroup$ – mathworker21 Sep 24 '16 at 3:38
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    $\begingroup$ Thanks that looks easy to understand. I will try to do for chain rule following that. $\endgroup$ – Amrita Sep 24 '16 at 3:41
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Your $G(x)=\sum_{n\geq 1} g_n x^n$ should start from $n=1$ and not 0. Then e.g. $$ x = (g_1 x+g_2 x^2 + g_3 x^3+ \cdots) - (g_1 x + g_2 x^2+ \cdots)^3/3! + (g_1 x +\cdots)^5/5!+ \cdots$$ comes with a natural grading and allows you to read off $g_1=1$, $g_2=0$, $g_3=1/6$,...

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In this answer, it is shown that for a formal power series $$ x+\sum_{n=2}^\infty a_nx^n $$ the inverse series is $$ x+\sum_{n=2}^\infty b_nx^n $$ where the $b_n$ can be computed inductively using the composition formula $$ 0=a_n+b_n+\left[x^n\right]\sum_{k=2}^{n-1}b_k\left(x+a_2x^2+a_3x^3+\dots+a_{n-1}x^{n-1}\right)^k $$ This may be arduous, but it works.

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The method for computing the coefficients of the inverse function is known as Lagrange inversion. In this particular case, we are inverting the sine function, hence we are interested in the power series around $x=0$ of $G(x)=\arcsin(x)$, and we may exploit the fact that the Taylor series of $$ G'(x) = \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-1/2} $$ can be computed from the extended binomial theorem: $$ G'(x) = \sum_{n\geq 0}\binom{-1/2}{n}(-1)^n x^{2n}=\sum_{n\geq 0}\binom{2n}{n}\frac{x^{2n}}{4^n} $$ and by termwise integration: $$ G(x) = \sum_{n\geq 0}\binom{2n}{n}\frac{x^{2n+1}}{(2n+1)4^n}.$$ The radius of convergence of such a power series is $1$, by Weierstrass M-test or the fact that the closest singularities to the origin of $G'(x)=\frac{1}{\sqrt{1-x^2}}$ lie at $x=\pm 1$.

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