Hi,long time no see.I am back again ^_^.
One question on numerical analysis,

GIVES:
To approximate \(f'(x)\),suppose first that \(x_0 \in(a,b)\),where \(f \in C^2[a,b]\),and that \(x_1 = x_0 + h \) for some \( h \neq 0 \) that is suffficiently small to ensure that \( x_1 \in [a,b]\).We construct the first Lagrange polynomical \(P_{0,1}(x)\) for f determined by \(x_0, x_1 \),with its error term:
\(f(x) = P_{0,1}(x) + \frac{(x-x_0)(x-x_1)}{2!}f''(\xi(x))
= \frac{f(x_0)(x-x_0-h)}{-h} + \frac{f(x_0+h)(x-x_0)}{h} + \frac{(x-x_0)(x-x_0-h)}{2}f''(\xi(x)),\)
for some \(\xi(x)\) in [a,b],Differenttiating gives
\(f'(x) = \frac{f(x_0+h)-f(x_0)}{h} + \frac{2(x-x_0)-h}{2}f''(\xi(x)) + \frac{(x-x_0)(x-x_0-h)}{2}D_x(f''(\xi(x))),\)
so
\( f'(x) \simeq \frac{f(x_0+h) - f(x_0)}{h} \)

Why we could ignore the last two part of the formula above, knowing nothing about \( f''(\xi(x)) \).

Thanks.

 

I can't say for sure, but it looks like your formula is using a Taylor's series error estimate for your Lagrange interpolating polynomial. To be able to discard the last two terms, your text must be assuming something about the magnitude of the 2nd derivative, perhaps that it is some constant multiple of h^2, which for small h would be very much smaller. Do you see any discussion like this in your text?