MATH 414 Lecture 36

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(homework questions)

Significance of Bonus: Singularity Detection

Bonus problem on homework was to show $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\psi (x)\mathrm{d}x=0$

Recall that when we project $f$ onto $V_j$: $f_j={\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_{V_j}(f)$

We had $a_k^j=f(2^{-j}k)$ (sampling) and $b_k^{j-1}=2^{j-1}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)\psi (2^{j-1}x-k)\mathrm{d}x$

Let's take a closer look at $b_k^{j-1}$:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)\psi (2^{j-1}x-k)\mathrm{d}x=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(2^{1-j}k+2^{1-j}u)\psi (u)\mathrm{d}u$

$\psi (u)$ has compact support, so it is bounded by some $a\le \psi (x)\le b$.

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(2^{1-j}k+2^{1-j}u)\psi (u)\mathrm{d}u=\underset{a}{\overset{b}{\int }}f(2^{1-j}k+2^{1-j}u)\psi (u)\mathrm{d}u$

Assuming $f$ is infinitely differentiable, we can expand a Taylor Series:

$f(2^{1-j}u+2^{1-j}k)=f(2^{1-j}k)+f^{\prime }(2^{1-j}k)(2^{1-j}u-2^{1-j}k)+O((2^{-j}{)}^2)$

Hence

$b_k^{j-1}=\text{stuff}\cancel{\underset{a}{\overset{b}{\int }}\psi (u)\mathrm{d}u}+\text{stuff}\cancel{\underset{a}{\overset{b}{\int }}u\psi (u)\mathrm{d}u}+\text{small}$

If $f(x)=ax+b$, then $b_k^{j-1}=0$

If the derivative condition fails at any point, then $b_k^j$ will be big: