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    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)
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    (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)?
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    (Original post by Zacken)
    Are you saying that (x, x+h) \subset (0,1)?
    yes
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    (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.
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    (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?
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    (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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