My friend thinks the this is equal to $ f"(x_0) $, but I don’t see just how to prove that this is true.
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Thanks because that the help!
The easiest method to prove your friend dorn is to use this come a function that we really understand, choose $f(x) = x$. We"ll do it in ~ $x_0 = 1$. Then you are calculating
$$ fracf(1 + ah) - f(1 + bh)h = frac1 + ah - 1 - bhh = (a - b).$$
This is not $1$ and depends critically top top $a$ and $b$ (exactly choose derivatives do not).
More generally, we have the right to write (but whereby I omit all border signs)$$ eginalignfracf(x_0 + ah) - f(x_0 + bh)h &= fracf(x_0 + ah) - f(x_0)h - fracf(x_0 + bh) - f(x_0)h \& o af"(x_0) - bf"(x_0) = (a-b)f"(x_0).endalign$$
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