About first derivative approximation with third order.

Let $$f'(t)=\frac{(2t+h)\cdot{f(h)}-4t\cdot{f(0)}+(2t-h)\cdot{f(-h)}}{2h^2}+R.$$ Show that $$R=\frac{f'''(\xi)\cdot{(3t^2-h^2)}}{6}$$ and $$\frac{-h}{2}≤t≤\frac{h}{2}.$$

The Taylor-expansion $h\cdot{f(h)}$ and $-h\cdot{f(-h)}$ is common knowledge. However, what about $2t\cdot{f(h)}$ and $2t\cdot{f(-h)}$? This looks like a more variables Taylor expansion. How to solve? Thanks in advance!

  • 1
    $\begingroup$ I am curious what your Taylor expansion for $f(h)$, $f(-h)$ and $f(0)$ look like since the result should follow from writing down the correct Taylor expansion and then doing the algebra $\endgroup$ – user47693 Dec 8 '13 at 22:58

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