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I consider the the ODE $ -u''(x)=f(x),\quad x\in \left]0,1\right[ $ with $ u(0)=u(1)=0 $. So I search a function $ u:\ [0,1]\rightarrow \mathbb{R} $ which solves this problem.

First step is to discretize the domain of $ u $ with $ x_i:=\frac{i}{n},\ i=0,\ldots,n $ and I search approximated values $ u_i\approx u(x_i) $. Further define $ f_i=f(x_i),\ i=1,\ldots,n-1 $.

So far so good, but now here comes my problem: The second derivative is approximated by the difference quotient which is reached by taylor approximation. Lets claim $ u $ is in $ \mathcal{C}^4([0,1]) $. Then we get:

$$ u(x\pm h)=u(x)\pm u'(x)\cdot h\pm\frac{1}{2!}\cdot u''(x)\cdot h^2\pm \frac{1}{3!} \cdot u^{(3)}(x)\cdot h^3+\underbrace{\frac{1}{4!}\cdot u^{(4)}(x\pm \xi_{\pm} \cdot h)\cdot h^4}_{(*)} $$ with $ \xi_{\pm}\in [0,1] $

I cannot comprehend how to get to this formula. And which remainder term is $(*)$? When I try to get it I only get this:

At first I do taylor approximation for fourth order at $ x_0\in [0,1] $: $ u(x)\\=u(x_0)+u'(x_0)\cdot (x-x_0)+\frac{1}{2!}\cdot u''(x_0)\cdot (x-x_0)^2+\frac{1}{3!}\cdot u^{(3)}(x_0)\cdot (x-x_0)^3+\underbrace{R_4(x;x_0)}_{(**)} $

From here I get stuck.

(**) Some remainder term... . I'm only familiar with the lagrange remainder term.

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hallo007
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    It is the Lagrange remainder, but instead of writing $$\frac{u^{(4)}(c)}{4!}h^4$$ where $c$ is between $x$ and $x \pm h$, they made a substitution $\xi = \frac{c - x}{h}$. – Mason Nov 28 '20 at 04:52
  • Ahh ok. Thanks! But one thing is still confusing me. Why they plugged in $ u $ on the left side the argument $ x\pm h $ and on the right side the argument $ x $ in $ u $ ? When I do normally taylor approximation then I would get $ u(x)=\sum\limits_{k=0}^n \frac{u^{(k)}(x_0)}{k!}\cdot (x-x_0)^k +R_n(x;x_0)$. But they came out with the upper expression instead. – hallo007 Nov 28 '20 at 20:43
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    They expanded $u(x \pm h)$ with a Taylor series centered at $x$. Their formula holds for any real $x$. To make your formulation look like theirs, replace "$x_0$" with "$x$" and "$x$" with "$x \pm h$". – Mason Nov 29 '20 at 00:30

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