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Math Help - Taylor expansion questions

  1. #1
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    Taylor expansion questions

    Hello im new to this forum so im not sure if this is on the right forum. Im having trouble with taylors when 2 functions are combined. E.g. Expand ln(sinx) up to x^3 with a=pi/6. Could someone show me the steps to do this, thank you.
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  2. #2
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    Do you know the formula?
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  3. #3
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    [Fixed post based on SammyS's corrections below. Thanks SammyS.]

    Just start from the beginning. It is the same as any function.
    Take derivatives:
    f(x) = \ln(\sin(x))
    f'(x) = \frac{1}{\sin(x)}\cos(x) = \cot(x)
    f''(x) = -\csc^2(x)
    f'''(x) = 2\csc^2(x)\cot(x)

    Evaluate derivatives:
    f(\frac{\pi}{6}) = \ln(\frac{1}{2})
    f'(\frac{\pi}{6}) = \sqrt{3}
    f''(\frac{\pi}{6}) = -4
    f'''(\frac{\pi}{6}) = 8\sqrt{3}

    Now just write the Taylor expansion:
    <br />
f(x)=f(\frac{\pi}{6}) + f'(\frac{\pi}{6})(x - a) +\displaystyle {{f''(\frac{\pi}{6})}\over{2!}}(x - a)^2 + {{f'''(\frac{\pi}{6})}\over{3!}}(x - a)^3 + ...<br />
    <br />
=<br />
-\ln(2) + \sqrt{3}(x - \frac{\pi}{6}) - 2(x - \frac{\pi}{6})^2 + \displaystyle{{{4\sqrt{3}}\over{3}}}\textstyle (x - \frac{\pi}{6})^3 + ...<br />

    Disclaimer: Note this is just to give an idea (there may be mistakes).
    Last edited by snowtea; December 27th 2010 at 09:09 PM.
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  4. #4
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    Thanks i understand that method, but is there any way you can use the separate expansions of lnx and sinx and combine them to get the answer?
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  5. #5
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    In this case, I don't think there is an "easy" way.

    If it was ln(x)sin(x), then you can multiply two Taylor expansions, but ln(sin(x)) is nested.
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  6. #6
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    Ah ok i see, thank you
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  7. #7
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    Quote Originally Posted by snowtea View Post
    Just start from the beginning. It is the same as any function.
    Take derivatives:
    f(x) = \ln(\sin(x))
    f'(x) = \frac{1}{\sin(x)}\cos(x) = \cot(x)
    f''(x) = -\csc^2(x)
    f'''(x) = 2\csc^2(x)\cot(x)

    Evaluate derivatives:
    f(\frac{\pi}{6}) = \ln(\frac{1}{2})
    f'(\frac{\pi}{6}) = \sqrt{3}
    f''(\frac{\pi}{6}) = -4
    f'''(\frac{\pi}{6}) = 8\sqrt{3}

    Now just write the Taylor expansion:
    <br />
f(\frac{\pi}{6}) + f'(\frac{\pi}{6})(x - a) + f''(\frac{\pi}{6})(x - a)^2 + f'''(\frac{\pi}{6})(x - a)^3 + ...<br />
    <br />
=<br />
\ln(\frac{1}{2}) + \sqrt{3}(x - \frac{\pi}{6}) - 4(x - \frac{\pi}{6})^2 + 8(x - \frac{\pi}{6})^3 + ...<br />

    Disclaimer: Note this is just to give an idea (there may be mistakes).
    I expect you're supposed to evaluate the derivatives at \displaystyle \frac{\pi}{2} in order to get rational values for the coefficients. The whole point of a Taylor Series is to be able to evaluate the function at any value, it's pretty hard to do that when your coefficients can't even be evaluated...
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  8. #8
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    Quote Originally Posted by snowtea View Post
    ...

    Now just write the Taylor expansion:
    <br />
f(\frac{\pi}{6}) + f'(\frac{\pi}{6})(x - a) + f''(\frac{\pi}{6})(x - a)^2 + f'''(\frac{\pi}{6})(x - a)^3 + ...<br />
    <br />
=<br />
\ln(\frac{1}{2}) + \sqrt{3}(x - \frac{\pi}{6}) - 4(x - \frac{\pi}{6})^2 + 8(x - \frac{\pi}{6})^3 + ...<br />

    Disclaimer: Note this is just to give an idea (there may be mistakes).

    You forgot the factorials !

    Also dropped the \displaystyle \sqrt3 from the cubic term.

    <br />
f(x)=f(\frac{\pi}{6}) + f'(\frac{\pi}{6})(x - a) +\displaystyle  {{f''(\frac{\pi}{6})}\over{2!}}(x - a)^2 + {{f'''(\frac{\pi}{6})}\over{3!}}(x - a)^3 + ...<br />

     <br />
=<br />
-\ln(2) + \sqrt{3}(x - \frac{\pi}{6}) - 2(x - \frac{\pi}{6})^2 + \displaystyle{{{4\sqrt{3}}\over{3}}}\textstyle (x - \frac{\pi}{6})^3 + ...<br />

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