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Math Help - Calculate the taylor series expansion

  1. #1
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    Calculate the taylor series expansion

    Calculate the taylor series expansion about x=0 as far as the term in x^2 for the function

    f(x)= { \displaystyle \frac{sinx}{1-e^{-x}}   x not = 0,
    f(x)= 1 x=0.

    so at x=0, the function blow up to 0, which is a discontinuous at 0, how do we find the expansion?
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  2. #2
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    Quote Originally Posted by wopashui View Post
    Calculate the taylor series expansion about x=0 as far as the term in x^2 for the function

    f(x)= { \displaystyle \frac{\sin x}{1-e^{-x}}  \quad x \ne 0,
    f(x)= 1 \quad x=0.

    so at x=0, the function blow up to 0, which is a discontinuous at 0, how do we find the expansion?
    I don't know why you say that the function is discontinuous at 0. In fact, it is continuous at 0, because \displaystyle\lim_{x\to0}\frac{\sin x}{1-e^{-x}} = 1 (as you can check by using l'H˘pital's rule, for example).

    To find the first few terms of the Taylor series, first write the first few terms for the series of the numerator and denominator:

     \displaystyle \frac{\sin x}{1-e^{-x}} = \frac{x - x^3/6 + \ldots}{x - x^2/2 + x^3/6 - \ldots} = \frac{1 - x^2/6 + \ldots}{1 - (x/2 - x^2/6 + \ldots)}.

    For the denominator, use the series \frac1{1-z} = 1+z+z^2+\ldots, with z = x/2-x^2/6+\ldots. Then, ignoring all powers greater than x^2, we get

    \displaystyle \frac{\sin x}{1-e^{-x}} = (1 - x^2/6 + \ldots)(1 + (x/2-x^2/6 + \ldots) + (x/2 - \ldots)^2).

    Now you just have to pick out the coefficients of the first few powers of x on the right side of that equation.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    I don't know why you say that the function is discontinuous at 0. In fact, it is continuous at 0, because \displaystyle\lim_{x\to0}\frac{\sin x}{1-e^{-x}} = 1 (as you can check by using l'H˘pital's rule, for example).

    To find the first few terms of the Taylor series, first write the first few terms for the series of the numerator and denominator:

     \displaystyle \frac{\sin x}{1-e^{-x}} = \frac{x - x^3/6 + \ldots}{x - x^2/2 + x^3/6 - \ldots} = \frac{1 - x^2/6 + \ldots}{1 - (x/2 - x^2/6 + \ldots)}.

    For the denominator, use the series \frac1{1-z} = 1+z+z^2+\ldots, with z = x/2-x^2/6+\ldots. Then, ignoring all powers greater than x^2, we get

    \displaystyle \frac{\sin x}{1-e^{-x}} = (1 - x^2/6 + \ldots)(1 + (x/2-x^2/6 + \ldots) + (x/2 - \ldots)^2).

    Now you just have to pick out the coefficients of the first few powers of x on the right side of that equation.
    sorry, I dun see what you are really doing here, what is the substituion of z for, can you complete the last step to show what happen?
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  4. #4
    Super Member TheChaz's Avatar
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    Why aren't we just taking the first and second derivatives and putting them into
    f(a) + f'(a)x + f''(a)x^2/2 ???
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