MATH 414 Lecture 36

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(homework questions)

Significance of Bonus: Singularity Detection

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

Recall that when we project $f$ onto $V_{j}$: $f_{j}=\mathrm {proj} _{V_{j}}{(f)}$

We had $a_{k}^{j}=f(2^{-j}\,k)$ (sampling) and $b_{k}^{j-1}=2^{j-1}\,\int _{-\infty }^{\infty }f(x)\,\psi (2^{j-1}\,x-k)\,\mathrm {d} x$

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

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

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

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

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

$f(2^{1-j}\,u+2^{1-j}\,k)=f(2^{1-j}\,k)+f'(2^{1-j}\,k)\left(2^{1-j}\,u-2^{1-j}\,k\right)+O((2^{-j})^{2})$

Hence

$b_{k}^{j-1}={\mbox{stuff}}\,{\cancel {\int _{a}^{b}\psi (u)\,\mathrm {d} u}}+{\mbox{stuff}}\,{\cancel {\int _{a}^{b}u\,\psi (u)\,\mathrm {d} u}}+{\mbox{small}}$

If $f(x)=a\,x+b$, then $b_{k}^{j-1}=0$

If the derivative condition fails at any point, then $b_{k}^{j}$ will be big: