Difference between revisions of "Aufgaben:Problem 15"
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It says: "Auf beiden Seiten steht bei festem \(a\) eine elliptische Funktion mit denselben Nullstellen \((\pm a)\) und Polstellen (to prove 1). Daher stimmen sie bis auf einen konstanten Faktor überein (to prove 2). Für die Normierung benutzt man die Beziehung \(\lim_{z\rightarrow 0}z^2(\wp(z)-\wp(a))=1\). dass auf der rechten Seite dasselbe herauskommt, folgt aus den Relationen \(\sigma(a)=-\sigma(-a)\) und \(\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}=1\), welche unmittelbar aus der Definition folgen (better see the passages)." | It says: "Auf beiden Seiten steht bei festem \(a\) eine elliptische Funktion mit denselben Nullstellen \((\pm a)\) und Polstellen (to prove 1). Daher stimmen sie bis auf einen konstanten Faktor überein (to prove 2). Für die Normierung benutzt man die Beziehung \(\lim_{z\rightarrow 0}z^2(\wp(z)-\wp(a))=1\). dass auf der rechten Seite dasselbe herauskommt, folgt aus den Relationen \(\sigma(a)=-\sigma(-a)\) und \(\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}=1\), welche unmittelbar aus der Definition folgen (better see the passages)." | ||
| + | '''Claim''': \(-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\) is elliptic and has zeros first order in every \(z\in\left\{\pm a+\gamma | \gamma\in\Gamma_\omega\right\}\) and poles second order in every \(z\in\left\{\Gamma_\omega\right\}\) | ||
| + | |||
| + | ''Proof'': | ||
| + | ''i) Periodic in two directions'': Let \(\gamma\in\Gamma_\omega=\mathbb{Z}+\omega\mathbb{Z}\), then \(-\frac{\sigma(z+a+\gamma)\sigma(z-a+\gamma)}{\sigma(z+\gamma)^2\sigma(a)^2}\overset{prop\ \theta -fct}{=}-\frac{e^{a_\gamma(z+a)+b_\gamma}\sigma(z+a)e^{a_\gamma(z-a)+b_\gamma}\sigma(z-a)}{(e^{a_\gamma z+b_\gamma}\sigma(z))^2\sigma(a)^2}=-\frac{e^{2a_\gamma z+2b_\gamma}\sigma(z+a)\sigma(z-a)}{e^{2a_{\gamma} z+2b_\gamma}\sigma(z)^2\sigma(a)^2}=-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\) | ||
| + | ''ii) Meromorphic'' \(-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}=-\frac{(z+a)\prod[(1-\frac{z+a}{\gamma})e^{\frac{z+a}{\gamma}+\frac{(z+a)^2}{2\gamma ^2}}](z-a)\prod[(1-\frac{z-a}{\gamma})e^{\frac{z-a}{\gamma}+\frac{(z-a)^2}{2\gamma ^2}}]}{(z\prod(1-\frac{z}{\gamma})e^{\frac{z}{\gamma}+\frac{z^2}{2\gamma ^2}})^2(a\prod[(1-\frac{a}{\gamma})e^{\frac{a}{\gamma}+\frac{a^2}{2\gamma ^2}})^2}=-\frac{(z^2-a^2)\prod[(1-\frac{z+a}{\gamma})(1-\frac{z-a}{\gamma})e^{\frac{z+a}{\gamma}+\frac{(z+a)^2}{2\gamma ^2}+\frac{z-a}{\gamma}+\frac{(z-a)^2}{2\gamma ^2}}]}{z^2a^2\prod[(1-\frac{z}{\gamma})^2 (1-\frac{a}{\gamma})^2 e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{2a}{\gamma}+\frac{a^2}{\gamma ^2}}]}=-\frac{(z^2-a^2)\prod[(1-\frac{z+a}{\gamma})(1-\frac{z-a}{\gamma})e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{a^2}{\gamma ^2}}]}{z^2a^2\prod[(1-\frac{z}{\gamma})^2 (1-\frac{a}{\gamma})^2 e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{2a}{\gamma}+\frac{a^2}{\gamma ^2}}]}=-\frac{(z^2-a^2)\prod\gamma^2(\gamma-z-a)(\gamma-z+a)e^{\frac{-2a}{\gamma}}}{z^2a^2\prod(\gamma-z)^2 (\gamma-a)^2}\) | ||
'''Claim''': \(f,g\) elliptic with same pole, zeros then \(f=Ag\), with \(A\) const | '''Claim''': \(f,g\) elliptic with same pole, zeros then \(f=Ag\), with \(A\) const | ||
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Then we write $$f(z)=h(z)\frac{\prod_{a_i zero}(z-a_i)}{\prod_{b_i pole}(z-b_i)},\ \ g(z)=l(z)\frac{\prod_{a_i\ zero}(z-a_i)}{\prod_{b_i\ pole}(z-b_i)}$$ with \(h,l\) analytic and with prop. of above. | Then we write $$f(z)=h(z)\frac{\prod_{a_i zero}(z-a_i)}{\prod_{b_i pole}(z-b_i)},\ \ g(z)=l(z)\frac{\prod_{a_i\ zero}(z-a_i)}{\prod_{b_i\ pole}(z-b_i)}$$ with \(h,l\) analytic and with prop. of above. | ||
| − | If \(f,g\) elliptic, then \(\frac{f}{g}\) is elliptic too: \(\frac{f}{g}=\frac{h(z)}{l(z)}\) analitic as a quotient of analitic functions, then \(\frac{h(z)}{l(z)}\) has no singularities and then no pole. But an elliptic function without pole is constant \(q(z):=\frac{h(z)}{l(z)},\ q(\mathbb{C}=q(\Gamma_\omega)\) compact, so bounded and after Liouville is \(q(z)\) constant) \(\Rightarrow \ f(z)=\frac{h(z)}{l(z)}g(z)=Ag\ \square\) | + | If \(f,g\) elliptic, then \(\frac{f}{g}\) is elliptic too: \(\frac{f}{g}=\frac{h(z)}{l(z)}\) analitic as a quotient of analitic functions, then \(\frac{h(z)}{l(z)}\) has no singularities and then no pole. But an elliptic function without pole is constant \(q(z):=\frac{h(z)}{l(z)},\ q(\mathbb{C})=q(\Gamma_\omega)\) compact, so bounded and after Liouville is \(q(z)\) constant) \(\Rightarrow \ f(z)=\frac{h(z)}{l(z)}g(z)=Ag\ \square\) |
Revision as of 11:47, 28 December 2014
Problem
Recall, \(\vartheta\) is a theta function relative to a lattice \(\Gamma_{\omega}=\mathbb{Z}+\mathbb{Z}\omega\), \(\omega\in\mathbb{H}\), if \(\vartheta\) is entire and \(\forall\gamma\in\Gamma_{\omega}\ \exists a_{\gamma},b_{\gamma}\in\mathbb{C}\) such that \(\vartheta\left(z+\gamma\right)=e^{a_{\gamma}z+b_{\gamma}}\vartheta(z),\ \forall z\in\mathbb{C}\). Consider the following theta function relative to \(\Gamma_{\omega}\) (you may assume that the product converges and that \(\sigma\) satisfies the above definition): $$\sigma(z)=z\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\left(1-\frac{z}{\gamma}\right )e^{\frac{z}{\gamma}+\frac{z^{2}}{2\gamma^{2}}}$$
a) Show that \(\forall a\notin\Gamma_{\omega}\), $$\wp(z)-\wp(a)=-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^{2}\sigma(a)^{2}}$$ where \(\wp=-\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2}+c_1)\) is the Weiertrass \(\wp\)-function, and \(\theta(z)=\sum_{n=-\infty}^{\infty}e^{2\pi inz}e^{\pi i\omega z^2}\) is the Riemann theta function and \(c_1\) is the coefficient in the expansion \(\frac{d}{dz}\log\theta(z+\frac{1+\omega}{2})=\frac{1}{z}+c_0+c_1z+\cdots\)
Hint: show that both sides of the equality are elliptic funtions with the same poles and zeros.
b) Show that $$\wp'(a)=-\frac{\sigma(2a)}{\sigma(a)^4},\ \forall a\notin\Gamma_{\omega}$$
Solutions
Coming soon, ab 28 erste Skizze.
a) First "idea": [1], to develop... or pages 351-2 of [2] (to see if possible to use it)
From Freitag, Busam - Funktionentheorie [Springer], Problem V.6.4 is exactly the same problem (the hint is related too): only to add some proofs.
It says: "Auf beiden Seiten steht bei festem \(a\) eine elliptische Funktion mit denselben Nullstellen \((\pm a)\) und Polstellen (to prove 1). Daher stimmen sie bis auf einen konstanten Faktor überein (to prove 2). Für die Normierung benutzt man die Beziehung \(\lim_{z\rightarrow 0}z^2(\wp(z)-\wp(a))=1\). dass auf der rechten Seite dasselbe herauskommt, folgt aus den Relationen \(\sigma(a)=-\sigma(-a)\) und \(\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}=1\), welche unmittelbar aus der Definition folgen (better see the passages)."
Claim: \(-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\) is elliptic and has zeros first order in every \(z\in\left\{\pm a+\gamma | \gamma\in\Gamma_\omega\right\}\) and poles second order in every \(z\in\left\{\Gamma_\omega\right\}\)
Proof: i) Periodic in two directions: Let \(\gamma\in\Gamma_\omega=\mathbb{Z}+\omega\mathbb{Z}\), then \(-\frac{\sigma(z+a+\gamma)\sigma(z-a+\gamma)}{\sigma(z+\gamma)^2\sigma(a)^2}\overset{prop\ \theta -fct}{=}-\frac{e^{a_\gamma(z+a)+b_\gamma}\sigma(z+a)e^{a_\gamma(z-a)+b_\gamma}\sigma(z-a)}{(e^{a_\gamma z+b_\gamma}\sigma(z))^2\sigma(a)^2}=-\frac{e^{2a_\gamma z+2b_\gamma}\sigma(z+a)\sigma(z-a)}{e^{2a_{\gamma} z+2b_\gamma}\sigma(z)^2\sigma(a)^2}=-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\) ii) Meromorphic \(-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}=-\frac{(z+a)\prod[(1-\frac{z+a}{\gamma})e^{\frac{z+a}{\gamma}+\frac{(z+a)^2}{2\gamma ^2}}](z-a)\prod[(1-\frac{z-a}{\gamma})e^{\frac{z-a}{\gamma}+\frac{(z-a)^2}{2\gamma ^2}}]}{(z\prod(1-\frac{z}{\gamma})e^{\frac{z}{\gamma}+\frac{z^2}{2\gamma ^2}})^2(a\prod[(1-\frac{a}{\gamma})e^{\frac{a}{\gamma}+\frac{a^2}{2\gamma ^2}})^2}=-\frac{(z^2-a^2)\prod[(1-\frac{z+a}{\gamma})(1-\frac{z-a}{\gamma})e^{\frac{z+a}{\gamma}+\frac{(z+a)^2}{2\gamma ^2}+\frac{z-a}{\gamma}+\frac{(z-a)^2}{2\gamma ^2}}]}{z^2a^2\prod[(1-\frac{z}{\gamma})^2 (1-\frac{a}{\gamma})^2 e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{2a}{\gamma}+\frac{a^2}{\gamma ^2}}]}=-\frac{(z^2-a^2)\prod[(1-\frac{z+a}{\gamma})(1-\frac{z-a}{\gamma})e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{a^2}{\gamma ^2}}]}{z^2a^2\prod[(1-\frac{z}{\gamma})^2 (1-\frac{a}{\gamma})^2 e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{2a}{\gamma}+\frac{a^2}{\gamma ^2}}]}=-\frac{(z^2-a^2)\prod\gamma^2(\gamma-z-a)(\gamma-z+a)e^{\frac{-2a}{\gamma}}}{z^2a^2\prod(\gamma-z)^2 (\gamma-a)^2}\)
Claim: \(f,g\) elliptic with same pole, zeros then \(f=Ag\), with \(A\) const
Proof: We know that a function \(k(z)\) with zeros \(a_1 ,\cdots ,a_m\), pole \(b_1, \cdots , b_r\) can be written as \(k(z)=q(z)\frac{\prod_{a_i\ zero}(z-a_i)}{\prod_{b_i\ pole}(z-b_i)}\) with \(q(z)\) analitic and \(q(a_i)\neq 0\ \forall a_i, q(b_i)\neq 0\ \forall b_i\).
Then we write $$f(z)=h(z)\frac{\prod_{a_i zero}(z-a_i)}{\prod_{b_i pole}(z-b_i)},\ \ g(z)=l(z)\frac{\prod_{a_i\ zero}(z-a_i)}{\prod_{b_i\ pole}(z-b_i)}$$ with \(h,l\) analytic and with prop. of above.
If \(f,g\) elliptic, then \(\frac{f}{g}\) is elliptic too: \(\frac{f}{g}=\frac{h(z)}{l(z)}\) analitic as a quotient of analitic functions, then \(\frac{h(z)}{l(z)}\) has no singularities and then no pole. But an elliptic function without pole is constant \(q(z):=\frac{h(z)}{l(z)},\ q(\mathbb{C})=q(\Gamma_\omega)\) compact, so bounded and after Liouville is \(q(z)\) constant) \(\Rightarrow \ f(z)=\frac{h(z)}{l(z)}g(z)=Ag\ \square\)
