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| -rw-r--r-- | notes/build/FourierOnS2.pdf | bin | 54623 -> 54623 bytes |
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diff --git a/notes/FourierOnS2.tex b/notes/FourierOnS2.tex index de15f01..24b972f 100644 --- a/notes/FourierOnS2.tex +++ b/notes/FourierOnS2.tex @@ -40,7 +40,7 @@ \section{Fourier on a flat surface} -%% TODO: replace everywhere "nice" with this definition +%% FIXME First, we need to discuss about the class of functions on which the Fourier works at all. We will denote the set of such functions with \(C(\Omega; \mathbb{C})\), that means: continuous bounded functions from \(\Omega\) to the complex numbers. Informally, we will refer to these as ``nice'' functions. %% TODO: Check that R^2 / Z^2 = (R / Z) \times (R / Z) @@ -134,46 +134,46 @@ is easy to compute. This is shown in the next lemma. \] \end{lemma} \begin{proof} -To start, we first expand the left side of the statement: -\begin{gather} - \nonumber - \langle \nabla^2 f, B_{m, n} \rangle + To start, we first expand the left side of the statement: + \begin{gather} + \nonumber + \langle \nabla^2 f, B_{m, n} \rangle = \iint_{[0, 1]^2} \nabla^2 f B_{m,n} d\mu d\nu \\ = \iint_{[0, 1]^2} \left( - \frac{\partial^2 f}{\partial \mu^2} - + \frac{\partial^2 f}{\partial \nu^2} - \right) e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu. + \frac{\partial^2 f}{\partial \mu^2} + + \frac{\partial^2 f}{\partial \nu^2} + \right) e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu. \label{eqn:laplacian-coeffs-expanded} -\end{gather} -Since the integrand is a sum of partial derivatives, we now have 2 integrals. Notice that inside each integral we have an expressions of the form: -\begin{equation} \label{eqn:inner-by-parts} - \int_{[0, 1]} \frac{\partial^2 f}{\partial \xi^2} e^{-i2\pi x\xi} d\xi, -\end{equation} -once with \(x = m, \xi = \mu\) and the second time with \(x = n, \xi = \nu\). -The integral \eqref{eqn:inner-by-parts} can be integrated by parts twice resulting in this ugly expression: -\[ - e^{-i2\pi x\xi} \left( + \end{gather} + Since the integrand is a sum of partial derivatives, we now have 2 integrals. Notice that inside each integral we have an expressions of the form: + \begin{equation} \label{eqn:inner-by-parts} + \int_{[0, 1]} \frac{\partial^2 f}{\partial \xi^2} e^{-i2\pi x\xi} d\xi, + \end{equation} + once with \(x = m, \xi = \mu\) and the second time with \(x = n, \xi = \nu\). + The integral \eqref{eqn:inner-by-parts} can be integrated by parts twice resulting in this ugly expression: + \[ + e^{-i2\pi x\xi} \left( \frac{\partial f}{\partial \xi} - i2\pi f - \right)\Bigg|_0^1 - + (i2\pi x)^2 \int_{[0, 1]} f e^{-i2\pi x\xi} d\xi. -\] -However, actually this is not too bad. That is because once we substitute the bounds two things happen: the exponential in the front always equals 1 and what is inside of the parenthesis can be rewritten as -\[ - \frac{\partial f}{\partial\xi}(1) - \frac{\partial f}{\partial\xi}(0) - + i2\pi \left[ f(1) - f(0) \right], -\] -which equals zero, since \(f\) and its derivative are continuous and periodic. Hence, we are left with two integrals, that when substituted back into \eqref{eqn:laplacian-coeffs-expanded} give: -\begin{align*} - \langle \nabla^2 & f, B_{m, n} \rangle = \\ + \right)\Bigg|_0^1 + + (i2\pi x)^2 \int_{[0, 1]} f e^{-i2\pi x\xi} d\xi. + \] + However, actually this is not too bad. That is because once we substitute the bounds two things happen: the exponential in the front always equals 1 and what is inside of the parenthesis can be rewritten as + \[ + \frac{\partial f}{\partial\xi}(1) - \frac{\partial f}{\partial\xi}(0) + + i2\pi \left[ f(1) - f(0) \right], + \] + which equals zero, since \(f\) and its derivative are continuous and periodic. Hence, we are left with two integrals, that when substituted back into \eqref{eqn:laplacian-coeffs-expanded} give: + \begin{align*} + \langle \nabla^2 & f, B_{m, n} \rangle = \\ &(i2\pi m)^2 \iint_{[0, 1]^2} f e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu \\ &\quad + (i2\pi n)^2 \iint_{[0, 1]^2} f e^{-i2\pi m\mu} e^{-i2\pi n\nu} d\mu d\nu. -\end{align*} -Finally, to complete the proof we rewrite the right side using the compact notation: -\[ - (i2\pi m)^2 \langle f, B_{m, n} \rangle - + (i2\pi n)^2 \langle f, B_{m, n} \rangle. - \qedhere -\] + \end{align*} + Finally, to complete the proof we rewrite the right side using the compact notation: + \[ + (i2\pi m)^2 \langle f, B_{m, n} \rangle + + (i2\pi n)^2 \langle f, B_{m, n} \rangle. + \qedhere + \] \end{proof} % TODO: closing words: this is why fourier is useful @@ -244,10 +244,10 @@ Thus we first need examine the solutions to this equation before constructing th \end{proposition} \begin{lemma} The expression -\begin{equation} \label{eqn:legendre-rodrigues} - P_n(x) = \frac{1}{n!2^n}\frac{d^n}{dx^n}(x^2-1)^n. -\end{equation} -is equivalent to \eqref{eqn:legendre-poly}. + \begin{equation} \label{eqn:legendre-rodrigues} + P_n(x) = \frac{1}{n!2^n}\frac{d^n}{dx^n}(x^2-1)^n. + \end{equation} + is equivalent to \eqref{eqn:legendre-poly}. \end{lemma} \begin{proof} diff --git a/notes/build/FourierOnS2.pdf b/notes/build/FourierOnS2.pdf Binary files differindex c327e27..887616d 100644 --- a/notes/build/FourierOnS2.pdf +++ b/notes/build/FourierOnS2.pdf |