Exercise 15.4.14

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
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The function $\varphi(z)$ is analytic at $z=0$ and hence has a power series expansion.
\be
\item[(a)] In Exercise 3 of Section 15.2, you used $\varphi'^2(z) = 1 - \varphi^4(z)$ to show that $\varphi''(z) = -2 \varphi^3(z)$. Use these two identities to prove by induction that for every $n\geq 1$, there is a polynomial $G_n(u) \in \Z[u]$ such that $\varphi^{(n)}(z)$ equals $G_n(\varphi(z))$ if $n$ is even and $G_n(\varphi(z)) \varphi'(z)$ if $n$ is odd.
\item[(b)] Use part (a) to prove that the coefficients of the power series expansion of $\varphi(z)$ at $z=0$ lie in $\Q$.
\item[(c)] Use part (b) and $\varphi(iz) = i \varphi(z)$ to show that $\varphi(z) = \sum_{j=0}^\infty c_j z^{4j+1}, c_j \in \Q$.
\item[(c)] Show that $c_0 = 1, c_1 = -\frac{1}{10}$, and $c_2 = \frac{1}{120}$.
\ee

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\be} {\begin{enumerate}} ewcommand{\ee} {\end{enumerate}} ewcommand{\deb}{\begin{eqnarray*}} ewcommand{\fin}{\end{eqnarray*}} ewcommand{\ssi} {si et seulement si } ewcommand{\D}{\mathrm{d}} ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{\U}{\mathbb{U}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\im}{\,\mathrm{Im}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{\Gal}{\mathrm{Gal}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \begin{proof} \item[(a)] $\varphi^{(0)}(z) = \varphi(z)$, thus $\varphi^{(0)}(z) = G_0(\varphi(z))$, where $G_0(z) = z$. Now assume that the property holds for all integers $k \leq n$, where $n\geq 2$. $\bullet$ If $n$ is even, there is a polynomial $G_n \in \Z[u]$ such that $$ \varphi^{(n)}(z) = G_n(\varphi(z)).$$ Then $n+1$ is odd, and, for all $z\in \C$ such that both members are defined, \begin{align*} \varphi^{(n+1)}(z) &= G_n'(\varphi(z)) \varphi'(z)\ &=G_{n+1}(\varphi(z)) \varphi'(z), \end{align*} where $G_{n+1}(u) = G'_n(u) \in \Z[u]$, so that the property holds for $n+1$ if $n$ is odd. $\bullet$ If $n$ is odd, there is a polynomial $G_n\in \Z[u]$ such that $$\varphi^{(n)}(z) = G_n(\varphi(z))\varphi'(z).$$ Then $n+1$ is even. Using $\varphi'^2(z) = 1 - \varphi^4(z)$ and $\varphi''(z) = -2 \varphi^3(z)$, we obtain \begin{align*} \varphi^{(n+1)}(z) &= G'_n(\varphi(z)) \varphi'^2(z) + G_n(\varphi(z)) \varphi''(z)\ &=G'_n(\varphi(z)) (1- \varphi^4(z)) -2 \varphi^3(z) G_n(\varphi(z))\ &=G_{n+1}(\varphi(z)), \end{align*} where $G_{n+1}(u) = G'_n(u)(1-u^4) - 2 u^3 G_n(u) \in \Z[u]$. The induction is done. To resume, for each integer $n \geq 0$, there is a polynomial $G_n(u) \in \Z[u]$ such that $$\begin{array}{lll} \varphi^{(n)}(z) &= G_n(\varphi(z)) &(n \equiv 0 \pmod 2)\ &= G_n(\varphi(z))\varphi'(z) & (n \equiv 1 \pmod 2) \end{array}$$ where $$\begin{array}{ll} G_0(u) = u,&\ G_{n+1}(u) = G'_n(u) &\qquad (n \equiv 0 \pmod 2),\ G_{n+1}(u) = G'_n(u)(1-u^4) - 2 u^3 G_n(u) &\qquad (n \equiv 1 \pmod 2). \end{array}$$ \item[(b)] The power series expansion of $\varphi(z)$ at $z=0$ is $$\varphi(z) = \sum_{k=0}^\infty \frac{\varphi^{(k)}(0)} {k!} z^k, \qquad (|z| < r,\ r>0),$$ where $\varphi^{(k)}(0) = G_k(\varphi(0)) = G_k(0) \in \Z$ if $k$ is even, and $\varphi^{(k)}(0) = G_k(\varphi(0)) \varphi'(0) = G_k(0) \in \Z$ if $k$ is odd ($r$ is the radius of convergence, which is positive). Thus $\varphi_k(0) \in \Z$ for all $k\geq 0$, and $\frac{\varphi^{k}(0)}{k!} \in \Q$. The coefficients of the power series expansion of $\varphi(z)$ at $z=0$ lie in $\Q$. \item[(c)]We write the power series expansion of $\varphi(z)$ at $z = 0$ under the form $$\varphi(z) = \sum_{k=0}^\infty a_k z^k.$$ Then $$\varphi(iz) = \sum_{k=0}^\infty a_k i^k z^k,\qquad (|z| = ZZ['u'] def G(n): G0 = u for k in range(n): if k % 2 == 0: G0 = G0.diff() else: G0 = G0.diff()*(1-u^4)-2*u^3*G0 return G0 G(5)(0)/5.factorial(), G(9)(0)/9.factorial() \end{verbatim} $$ -\frac{1}{10}, \frac{1}{120} $$ So $$c_0 = 1, c_1 = -\frac{1}{10}, c_2 = \frac{1}{120}.$$ With more effort, $$\varphi(z) = z -\frac{1}{10} z^5 + \frac{1}{120} z^9 -\frac{11}{15600}z^{13} + \frac{211}{3536000} z^{17} -\frac{1607}{318240000} z^{21} + \frac{1511}{3536000000}z^{25} + \cdots. $$ \end{proof}

Exercise 15.4.14

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Exercise 15.4.14

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