Hot answers tagged

3

A basis of convex functions is a tricky concept. Convexity is not closed under operations over vector spaces: a linear superposition of convex functions does not have to be convex (in fact, it most likely isn't). So... the entire formalism of vector spaces (Hilbert spaces actually) goes through the window and the concept of a basis doesn't make much sense. ...


2

Yes. By Cauchy's inequality, \begin{align*} (S+\#A)^2 = \bigg( \sum_{h=0}^{N-1} |\hat A(h)| \bigg)^2 &\le \bigg( \sum_{h=0}^{N-1} |\hat A(h)|^2 \bigg) \bigg( \sum_{h=0}^{N-1} 1^2 \bigg) \\ &= N \sum_{h=0}^{N-1} |\hat A(h)|^2 \\ &= N^2 \sum_{a\in A} 1^2 = N^2 \#A, \end{align*} where the last line follows by Plancherel's theorem applied to $x_a = 1$...


2

This is valid, so long as $|f(t)|=O(e^{b_-t})$ as $t\to\infty$ and $|f(t)|=O(e^{b_+t})$ as $t\to-\infty$ with $b_-<b_+$. Then $f(t)e^{bt}\in L^1(-\infty,\infty), \forall b\in(b_-,b_+)$ and the complex Fourier transform converges. Under this condition, the Fourier transform is analytic in $\omega=a+ib,\, a,b\in\mathbb R$ if $f(t)$ has finite number of ...


1

First, note that \begin{align} \|\sin-\cos\|^2&=\|\sin(x)\|^2+2\langle\sin,\cos\rangle+\|\cos\|^2 \\ & =\|\sin\|^2+\|\cos\|^2 \\ & = \|1\|^2=\pi. \end{align} You want $$ \|f-\sin\|+\|f-\cos\| \le \frac{2}{3}\sqrt{\pi}+\frac{1}{3}\sqrt{\pi}=\sqrt{\pi} $$ If you have the above, then $$ \sqrt{\pi}=\|\sin-\cos\| \le \|\sin-f\|+\|...


1

If $f(x)$ is real-valued, take $\dfrac{1}{3}$ times the first inequality plus $\dfrac{2}{3}$ times the second inequality to get: $$\int_{0}^{\pi}\dfrac{1}{3}(f(x)-\sin x)^2+\dfrac{2}{3}(f(x)-\cos x)^2\,dx \le \dfrac{1}{3} \cdot \dfrac{4\pi}{9} + \dfrac{2}{3}\cdot\dfrac{\pi}{9}$$ $$\int_{0}^{\pi}\left[f(x)^2-2f(x)\left(\dfrac{1}{3}\sin x - \dfrac{2}{3}\cos ...


1

I saw this proof in an extract of the College Mathematics Journal. Consider the Integeral : $I$ = $\int_0^{\pi/2}ln(2cosx)dx$ From $2\cos(x)$ = $e^{ix}$ + $e^{-ix}$ , we have: $\int_0^{\pi/2}In(e^{ix}$ + $e^{-ix})dx$ = $\int_0^{\pi/2}In(e^{ix}(1 + e^{-2ix}))dx$ =$\int_0^{\pi/2}ixdx$ + $\int_0^{\pi/2}In(1 + e^{-2ix})dx$ The Taylor series expansion of ...



Only top voted, non community-wiki answers of a minimum length are eligible