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Math Help - Taylor/Maclaurin series

  1. #1
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    Taylor/Maclaurin series

    Hello, I was wondering if someone could explain this problem:

    Expand f(x)=x^3/(1+x^3) in powers of x for abs(x)<1.

    I think it's a maclaurin expansion, but I've evaluated up to the 9th derivative at f(0), and I come up with 2 terms. How long do I have to expand it? or am I just doing it wrong.

    Thanks.
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  2. #2
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    Quote Originally Posted by hasbro View Post
    Hello, I was wondering if someone could explain this problem:

    Expand f(x)=x^3/(1+x^3) in powers of x for abs(x)<1.

    I think it's a maclaurin expansion, but I've evaluated up to the 9th derivative at f(0), and I come up with 2 terms. How long do I have to expand it? or am I just doing it wrong.

    Thanks.
    There's an easier way to do this. First

     <br />
f(x) = \frac{x^3}{1+x^3} = \frac{1+x^3-1}{1+x^3} = 1 - \frac{1}{1+x^3}<br />

    Then from the geometric power series

     <br />
\frac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots<br />

    replace the x with x^3 giving


    \frac{1}{1 + x^3} = 1 - x^3 + x^6 - x^9 + \cdots

    so putting these together gives

     <br />
1 - \left( 1 - x^3 + x^6 - x^9 + \cdots\right) = x^3 - x^6 + x^9 - \cdots<br />
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  3. #3
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    Thank you that's a lot easier.

    What would you do if the function was (x^3)/(1+(x^4))?
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  4. #4
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    Quote Originally Posted by hasbro View Post
    Thank you that's a lot easier.

    What would you do if the function was (x^3)/(1+(x^4))?
    Use

    \frac{1}{1+x^4} = 1 - x^4 + x^8 - x^{12} + \cdots and multiply by x^3 (even easier).
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  5. #5
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    Indeed, you could have done \frac{x^3}{1+ x^3} in exactly that (easier) way:
    \frac{1}{1+ x^3}= \frac{1}{1- (-x^3)} and so is equal to the geometric series 1- x^3+ x^6- x^9+ \cdot\cdot\cdot= \sum_{n=0}^\infty (-1)^n x^{3n} and then multiply each term by x^3
    \frac{x^3}{1+ x^3}= x^3- x^6+ x^9+ \cdot\cdot\cdot = \sum_{n=0}^\infty (-1)^n x^{n+3}= \sum_{j=3}^\infty (-1)^j x^j where, in the last sum, j= n+3.
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