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Analysis questions

This may sound like a stupid question, but if a function has a second derivative, does that mean it has a first derivative?

Also, if a function f : (0,1) --> R is 2 times differentiable, does that mean f : (x,x+h) --> R is 2 times differentiable? (x and x+h are in the interval (0,1)

Reply 1

Zacken

Original post by asdfyolo

Also, if a function f : (0,1) --> R is 2 times differentiable, does that mean f : (x,x+h) --> R is 2 times differentiable? (x and x+h are in the interval (0,1)

Are you saying that

(x, x+h) \subset (0,1)

Reply 2

asdfyolo

Original post by Zacken

Are you saying that

(x, x+h) \subset (0,1)

yes

Reply 3

Zacken

Original post by asdfyolo

yes

I can't see why that wouldn't be two times differentiable, but I'd gladly be corrected by someone else.

Reply 4

asdfyolo

Original post by Zacken

I can't see why that wouldn't be two times differentiable, but I'd gladly be corrected by someone else.

Thanks. How would you go about proving that the limit of (f(x+h)+f(x-h)-2f(x))/h^2 as h tends to zero is the second derivative, using taylor's theorem?

Reply 5

joostan

Original post by asdfyolo

Thanks. How would you go about proving that the limit of (f(x+h)+f(x-h)-2f(x))/h^2 as h tends to zero is the second derivative, using taylor's theorem?

If you're using Taylor's Theorem, this does however assume that the function is twice differentiable at

x

.
Write:

f(x+h)=f(x)+hf'(x)+\dfrac{h^2}{2}f''(x)+\mathcal{O}(h^3)

.
Expand

f(x-h)

similarly and the result follows.

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