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Math Help - Limit Formula For Second Derivative

  1. #1
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    Limit Formula For Second Derivative

    Show that f''(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}.

    OK, so I'm trying to think through this but getting something completely different. I'm starting with knowing that:

    f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x+h)-f'(x)}{h}

    But also that:

    f'(x+h)=\displaystyle\lim_{h\to 0}\frac{f(x+2h)-f(x+h)}{h}

    f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

    So when I substitute these in, I'm instead getting the following:

    \displaystyle\lim_{h\to 0}\frac{\frac{f(x+2h)-f(x+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}

    =\displaystyle\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}

    Which I know is just off by an h. It seems to me that they get the second derivative formula like this:

    f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x)-f'(x-h)}{h}

    But I don't really understand why it's that way. That seems like f''(x-h) to me. Why is it done that way?
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  2. #2
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    Look here,

    Limit Formula For Second Derivative-img.jpg
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by mathematicalbagpiper View Post
    Show that f''(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}.

    OK, so I'm trying to think through this but getting something completely different. I'm starting with knowing that:

    f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x+h)-f'(x)}{h}



    But also that:

    f'(x+h)=\displaystyle\lim_{h\to 0}\frac{f(x+2h)-f(x+h)}{h}

    f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

    So when I substitute these in, I'm instead getting the following:

    \displaystyle\lim_{h\to 0}\frac{\frac{f(x+2h)-f(x+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}

    =\displaystyle\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}

    Which I know is just off by an h. It seems to me that they get the second derivative formula like this:

    f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x)-f'(x-h)}{h}

    But I don't really understand why it's that way. That seems like f''(x-h) to me. Why is it done that way?
    It is all true! Now just do substitution: x=y-h. (Why you allowed to do this kind substitution?)
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  4. #4
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    Quote Originally Posted by Also sprach Zarathustra View Post
    It is all true! Now just do substitution: x=y-h. (Why you allowed to do this kind substitution?)
    You mean x=y+h...
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Pandevil1990 View Post
    You mean x=y+h...


    No... x=y-h will work.
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  6. #6
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    where exactly are you going to do this substitution?
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Pandevil1990 View Post
    where exactly are you going to do this substitution?
    Here:

    \displaystyle\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}
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  8. #8
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    It's ok...But i think this is a more difficult way than L'hospital 's rule.Thanks...
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