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In a forward difference formula, $D_-$ refers to the backward difference operator. Thus, $(D_- u)(x) = u(x)-u(x-h)$.

In the answer key of a problem that I was working on, one of the steps is:

$[D_- + \frac{1}{2}D_-^2+\frac{1}{3}D_-^3+O(h^4)]^s$

=$D_-^s + {s \choose 1}\frac{1}{2}D_-^{s+1}+ ({s \choose 1}\frac{1}{3}+{s \choose 2}\frac{1}{2^2})D_-^{s+2}+O(h^{s+3})$

Unfortunately, I don't know how they got the second expression from the first.

Aside from my own experimentation, I've checked my textbook which has a similar expression. Unfortunately, it doesn't show the derivation.

Twilight Sparkle
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1 Answers1

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In the $s$ factors, you need to pick $D$ all $s$ times in $$\binom{s}{s}=1$$ ways to get $D^s$.

To get $D^{s+1}$, you need to pick a $D^2$ once in $$\binom{s}{1}$$ ways and the $1/2$ will appear once in the term.

To get $D^{s+2}$ you either pick two $D^2$ in $$\binom{s}{2}$$ ways with a factor of $(1/2)^2$ or you pick one $D^3$ in $$\binom{s}{1}$$ ways with one factor of $1/3$.

All other terms involve $D^{s+3}$ or greater so the equality follows.

  • Hi! Thanks a ton for your help, why is it that we don't pick higher powers than that? e.g. pick $\frac{1}{3}D_-^3$, $s$ times to get $\frac{1}{3^s}D_-^{3s}$ – Twilight Sparkle Apr 14 '14 at 15:55
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    @TwilightSparkle I haven't the slightest idea. Apparently this is an "approximation" where only the lower order terms matter, but without any context it's hard to say. –  Apr 14 '14 at 16:01
  • I see, much appreciated! – Twilight Sparkle Apr 14 '14 at 16:01