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Stream: deprecated: mathematics

Topic: Taylor expansion of y'(x-h)

Jan Pax (Dec 03 2022 at 18:25):

How can I derive this approximate equation via the Taylor expansion: $\frac{y(x+h)-y(x-h)}{2}-\frac{2h}{12}(y'(x+h)+4y'(x)+y'(x-h))=-\frac{1}{80}h^5y^{(5)}(x)+O(h^7)$ ? Do we have to assume that $y(x+h)-y(x-h)=2hy'(x)+\frac{1}{3}h^3y^{'''}(x))+\frac{1}{360}h^5y^{(5)}(x))+O(h^7)$ or does this follow ?

Jean-Baptiste Vienney (Dec 03 2022 at 23:18):

You should use the Taylor expansions of $y(x+h)$ and $y(x-h)$ up to $O(h^7)$ to have an expression of $\frac{y(x+h)-y(x-h)}{2}$, then obtain in the same way an expression of $\frac{y'(x+h)+y'(x-h)}{2}$ that you're gonna use to make disappear some terms of low degree:

• Start with:
  $y(x+h)=y(x)+hy'(x)+\frac{h^2}{2}y''(x)+\frac{h^3}{3!}y'''(x)+\frac{h^4}{4!}y^{(4)}(x)+\frac{h^5}{5!}y^{(5)}(x)+\frac{h^6}{6!}y^{(6)}(x)+O(h^7)$
  $y(x-h)=y(x)-hy'(x)+\frac{h^2}{2}y''(x)-\frac{h^3}{3!}y'''(x)+\frac{h^4}{4!}y^{(4)}(x)-\frac{h^5}{5!}y^{(5)}(x)+\frac{h^6}{6!}y^{(6)}(x)+O(h^7)$
  It gives that:
  $\frac{y(x+h)-y(x-h)}{2}=hy'(x)+\frac{h^3}{3!}y'''(x)+\frac{h^5}{5!}y^{(5)}(x)+O(h^7)$
  and then:
  $\frac{y(x+h)-y(x-h)}{2}-hy'(x)=\frac{h^3}{3!}y'''(x)+\frac{h^5}{5!}y^{(5)}(x)+O(h^7)\qquad(1)$

• We now use:
  $y'(x+h)=y'(x)+hy''(x)+\frac{h^2}{2}y'''(x)+\frac{h^3}{3!}y^{(4)}(x)+\frac{h^4}{4!}y^{(5)}(x)+\frac{h^5}{5!}y^{(6)}(x)+\frac{h^6}{6!}y^{(7)}(x)+O(h^7)$
  $y(x-h)=y'(x)-hy''(x)+\frac{h^2}{2}y'''(x)-\frac{h^3}{3!}y^{(4)}(x)+\frac{h^4}{4!}y^{(5)}(x)-\frac{h^5}{5!}y^{(6)}(x)+\frac{h^6}{6!}y^{(7)}(x)+O(h^7)$
  It gives that:
  $\frac{y'(x+h)+y'(x-h)}{2}=y'(x)+\frac{h^2}{2}y'''(x)+\frac{h^4}{4!}y^{(5)}(x)+\frac{h^6}{6!}y^{(7)}(x)+O(h^7)$
  then:
  $\frac{y'(x+h)+y'(x-h)}{2}-y'(x)=\frac{h^2}{2}y'''(x)+\frac{h^4}{4!}y^{(5)}(x)+\frac{h^6}{6!}y^{(7)}(x)+O(h^7)$
  then:
  $h(\frac{y'(x+h)+y'(x-h)}{2}-y'(x))=\frac{h^3}{2}y'''(x)+\frac{h^5}{4!}y^{(5)}(x)+O(h^7)$
  then:
  $\frac{2h}{6}(\frac{y'(x+h)+y'(x-h)}{2}-y'(x))=\frac{h^3}{3!}y'''(x)+\frac{2h^{5}}{4!6}y^{(5)}(x)+O(h^7)$
  ie. :
  $\frac{2h}{12}(y'(x+h)+y'(x-h)-2y'(x))=\frac{h^3}{3!}y'''(x)+\frac{2h^{5}}{4!6}y^{(5)}(x)+O(h^7)\qquad(2)$

• $(1) - (2)$ gives:
  $\frac{y(x+h)-y(x-h)}{2}-hy'(x)-\frac{2h}{12}(y'(x+h)+y'(x-h)-2y'(x) = (\frac{h^5}{5!}-\frac{2h^5}{4!6})y^{(5)}(x)+O(h^7)$
  ie. :
  $\frac{y(x+h)-y(x-h)}{2}-\frac{2h}{12}(y'(x+h)+4y'(x)+y'(x-h)) = (\frac{h^5}{5!}-\frac{2h^5}{4!6})y^{(5)}(x)+O(h^7)$
  but we compute:
  $\frac{1}{5!} - \frac{2}{4!6} = \frac{1}{5!}-\frac{1}{3.4.6} = \frac{1}{2.3.4.5}-\frac{1}{3.4.6} = \frac{1}{12}(\frac{1}{10}-\frac{1}{6})=\frac{1}{12}(\frac{3}{30}-\frac{5}{30}) = \frac{1}{12}(\frac{-2}{30})=-\frac{2}{360}=-\frac{1}{180}$
  and thus:
  $\frac{y(x+h)-y(x-h)}{2}-\frac{2h}{12}(y'(x+h)+4y'(x)+y'(x-h)) =-\frac{1}{180}h^{5}y^{(5)}(x)+O(h^7)\qquad(*)$
  which is what you wanted (modulo a little typo a priori).

Jean-Baptiste Vienney (Dec 03 2022 at 23:26):

There are some tricks to use to solve such questions. Your goal is to obtain an approximation of $f^{5}(x)$. You must do things like combining Taylor expansions of $f(x+h)$ and $f(x-h)$ to eliminate the terms of low degree ie. the terms $f''(x),f'''(x)...$ before $f^{5}(x)$, by using parities. You can try first to compute an approximation of $f'(x),f''(x)...$ and you're gonna get a feeling of the tricks.

Jean-Baptiste Vienney (Dec 03 2022 at 23:30):

As far as I know, these expressions are useful to compute solutions of differential equations numerically for instance. It's nice to have access to the fifth derivative without computing the first, then the second, then the third... and it's a more secured and efficient way to compute this fifth derivative numerically.

Jan Pax (Dec 04 2022 at 12:22):

Originally, my main goal was to obtain (5.3-10) especially the number $24$ as below. I obviously need to integrate $G(s)$ but I do not know from where to where. The source is here on page 173. 111.jpg

Notification Bot (Dec 05 2022 at 10:10):

This topic was moved here from #learning: questions > Taylor expansion of y'(x-h) by Matteo Capucci (he/him).