From 5b477ea41aa43f5e8b6a02ac686fe310a4b245f0 Mon Sep 17 00:00:00 2001 From: joshuab <> Date: Thu, 15 Jul 2010 18:30:12 +0000 Subject: finished FT --- ClassJune28.page | 12 +++++++++--- 1 file changed, 9 insertions(+), 3 deletions(-) diff --git a/ClassJune28.page b/ClassJune28.page index 6359933..8b71664 100644 --- a/ClassJune28.page +++ b/ClassJune28.page @@ -217,11 +217,17 @@ It is important to note at this point that we have simply expressed the periodic $$\sum_{n=-\infty}^\infty a_n e^{inx}$$ converges for all $x$. Then -$$f(x) = \sum_{n=-\infty}^\infty a_n e^{inx}.$$ +$$f(x) = \frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^\infty a_n e^{inx}.$$ *Proof.* -Consider the difference $g:=f-\sum_{n=-\infty}^\infty a_n e^{inx}$. -****inner product with delta functions +Consider the difference $g:=f-\sum_{n=-\infty}^\infty a_n e^{inx}$. We will show that $g=0$. Since inner products are linear in each variable, +$$(g,f_m)=(f,f_m)-\sum{a_n (f_n,f_m)}=a_n-a_n=0.$$ + +In other words, $g$ is orthogonal to all of the complex exponentials $e^{imx}$. In a previous section, we showed that the function $\delta_n(x):=\cos^{2n}(x)+\cos^{2n+1}(x)$ has a peak at 0, and is very small everywhere else. Moreover, since sums and products and shifts of complex exponentials, and thus sinusoids, are complex exponentials, we can write $\delta_n(x-x_0)$ as a sum of complex exponentials, for any real shift $x_0$. This gives a function with a peak wherever we want it. + +Since the inner product of $g$ with every complex exponential vanishes, its inner product with all the $\delta_n$'s also vanishes. This implies that $g$ is everywhere $0$! For suppose otherwise. Since $g$ is continuous, there is some point $x_0$ on the interval where $|g|$ attains its maximum $M>0$. We have +$$0=(g,\delta_n(x-x_0))=\int g \delta_n(x-x_0) dx.$$ +but if $n$ is sufficiently small, we can prove that almost all of the mass of $\delta_n$ is contained in a narrow window (width $O(1/\sqrt{n}) of the peak. Since the peak is centered at the maximum of $g$, and $g$ is continuous, we can choose $n$ large enough so that more than 80% of the mass of $\delta_n$ is contained in a window around $x_0$ where $g$ still attains 90% of its own peak value. Thus the integral of their inner product is at least, say, $M/2$ times the total mass of $\delta_n$, which is, to wit, nonzero. Contradiction. #Why the Fourier decomposition is useful? Applications will be covered on Monday July 5, 2010. \ No newline at end of file -- cgit v1.2.3