MathB.in
New
Demo
Tutorial
About
Prove that if $f(x)$ is a function such that $f(x) = \overline{f}(-x)$ then it's Fourier Transform is real. Try of proof: $F(w)$ will be the Fourier tranform of $f(x)$. We will try to prove that $F(w) = \overline{F}(w)$, and in doing so we prove that $F(w)$ is real. So: $\overline{F}(w) = \overline{\int_{-\infty}^\infty \! f(x)e^{-iwx} \, \mathrm{d}x} = \int_{-\infty}^\infty \! \overline{f(x)e^{-iwx}} \, \mathrm{d}x =$ $ \int_{-\infty}^\infty \! \overline{f(x)}*\overline{e^{-iwx}} \, \mathrm{d}x = \int_{-\infty}^\infty \! f(-x)e^{iwx} \, \mathrm{d}x$. Lets set a new variable $x' = -x$, so ${d}x' = -{d}x$. Continuing the equalities: $\int_{-\infty}^\infty \! f(-x)e^{iwx} \, \mathrm{d}x = -\int_{-\infty}^\infty \! f(x')e^{-iwx'} \, \mathrm{d}x' = -F(w)$. So we get $\overline{F}(w) = -F(w)$, and not $\overline{F}(w) = F(w)$. What did I do wrong?
ERROR: JavaScript must be enabled to render input!
Thu, 07 Nov 2013 23:38 GMT