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Math Help - Taylor approximation

  1. #1
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    Taylor approximation

    Find the second Taylor approximation for the following

    g(x)=ln(cos)(x))

    I think I have to use the chain rule, but i'm not sure how. I really need help.
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  2. #2
    Member Miss's Avatar
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    Quote Originally Posted by tim_mannire View Post
    Find the second Taylor approximation for the following

    g(x)=ln(cos)(x))

    I think I have to use the chain rule, but i'm not sure how. I really need help.
    Do you mean taylor series?
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  3. #3
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    yes, i believe taylor series and taylor approximation mean the same thing.
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  4. #4
    Member Miss's Avatar
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    centered at what?
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    Quote Originally Posted by Miss View Post
    centered at what?
    Assume MacLaurin (Taylor centred at 0)

    CB
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  6. #6
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    Wouldn't it just be -ln(x^2/2!)?

    How I got my answer was just by identifying that cos(x) = 1 - x^2/2!, then just take the ln of each nth. This left me with 0 - ln(x^2/2!).

    Please keep in my that I could be wrong, fore I am learning Taylor/MacLaurin series at this moment.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tim_mannire View Post
    Find the second Taylor approximation for the following

    g(x)=ln(cos)(x))

    I think I have to use the chain rule, but i'm not sure how. I really need help.
    T_2(x;0)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2.

    So, f(0)=\ln(1)=0

    f'(0)\frac{-\sin(0)}{\cos(0)}=0

    f''(0)=\frac{-1}{\cos^2(0)}=-1.

    So T_2(x;0)=\frac{-1}{2}x^2
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Brandong954 View Post
    Wouldn't it just be -ln(x^2/2!)?

    How I got my answer was just by identifying that cos(x) = 1 - x^2/2!, then just take the ln of each nth. This left me with 0 - ln(x^2/2!).

    Please keep in my that I could be wrong, fore I am learning Taylor/MacLaurin series at this moment.
    There is a way to doing this. For a small enough neighborhood around zero \ln(\cos(x))=\ln\left(\sqrt{1-\sin^2(x)}\right)=\frac{1}{2}\ln\left(1-\sin^2(x)\right) and remembering that \ln(1-x)=-x-\frac{x^2}{2}-\cdots we see that \frac{1}{2}\ln\left(1-\sin^2(x)\right)=\frac{-1}{2}\left(\sin^2(x)+\frac{\sin^4(x)}{2}+\cdots\ri  ght).


    But, as to your question, since when does \ln(a+b)=\ln(a)+\ln(b)?
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