1
$\begingroup$

Show that $u_0-u_1+u_2-u_3+...=\frac{1}{2}u_0-\frac{1}{4}\Delta u_0+\frac{1}{8}\Delta^2 u_0-\frac{1}{16}\Delta^3 u_0+...$, where $\Delta$ is the forward difference operator.

My attempt:

$(u_0+u_2+u_4+...)-(u_1+u_3+u_5+...)=(u_0+E^2u_0+E^4u_0+...)-(Eu_0+E^3u_0+E^5u_0+...)$

$E$ being shift operator.

$\endgroup$
  • $\begingroup$ $1-0+0-0+-\ldots=\frac{1}{2}-\frac{1}{4}+0-0+-\ldots$? $\endgroup$ – Leonhardt von M Apr 7 '15 at 9:19
  • $\begingroup$ $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+-\ldots = \frac{2}{3} $ but $\frac{1}{2} -\frac{1}{4}\frac{-1}{2}+\frac{1}{8}\frac{-1}{4}-\frac{1}{16}\frac{-1}{8}+-$ $\ldots =\frac{1}{2}+\frac{1}{8}-\frac{1}{32}+\frac{1}{128}\ldots=\frac{3}{5}$. $\endgroup$ – Leonhardt von M Apr 7 '15 at 10:24
1
$\begingroup$

Let $T$ be the unit translation operator, $(T^ku)_0=u_k$. Then $Δ=T-1$, $T=1+Δ$, $$ \sum_{k=0}^\infty (-T)^k=\frac1{1+T}=\frac1{2+Δ}=\frac12·\frac1{1-(-\frac12Δ)}=\frac12\sum_{k=0}^\infty \left(-\frac12Δ\right)^k. $$ Which is all rather questionable since the norm and spectral radius of the translation operator is $1$ for the standard norms and thus the geometric or Neumann series does not converge. But these transformations may serve as general guideline for a correct proof in a distributional sense.


Considering the modified supremum norm norm $\|u\|_r=\sup_k r^k|u_k|$ for some $r>1$ one finds that in operator norm $\|T\|=r^{-1}<1$ and $\frac12\|Δ\|\le\frac{1+r^{-1}}2<1$, so that in this topology on the sequence space the identities are also correct as converging sequences.

$\endgroup$
  • $\begingroup$ I like your answer but I can not find a good condition which guarantees that the equality holds -- numerically. $\endgroup$ – Leonhardt von M Apr 7 '15 at 10:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy