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Math Help - Taylor Series

  1. #1
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    Taylor Series

    1. Find the first five terms in the Taylor series about x = 0 for f(x) = \frac{1}{1-2x} and find the interval of convergence for the series.

    2. Then, using partial fractions, find the first five terms in the Taylor series about x = 0 for g(x) = \frac{1}{(1-2x)(1-x)}.


    Originally I took the first four derivatives of f(x), plugged in x=0 for them, then used those 5 results (four derivatives + original) to get the first five terms. However, my teacher told me that was incorrect, so I instead just plugged in x=0, x=1, up to x=4 into f(x) to get 1, -1, -1/3, -1/5, -1/7. Is that right?

    (I've got issues with later parts too but I want to clear up the first part.)
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  2. #2
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    Quote Originally Posted by Lord Darkin View Post
    1. Find the first five terms in the Taylor series about x = 0 for f(x) = \frac{1}{1-2x} and find the interval of convergence for the series.

    2. Then, using partial fractions, find the first five terms in the Taylor series about x = 0 for g(x) = \frac{1}{(1-2x)(1-x)}.


    Originally I took the first four derivatives of f(x), plugged in x=0 for them, then used those 5 results (four derivatives + original) to get the first five terms. However, my teacher told me that was incorrect, so I instead just plugged in x=0, x=1, up to x=4 into f(x) to get 1, -1, -1/3, -1/5, -1/7. Is that right?

    (I've got issues with later parts too but I want to clear up the first part.)

    If your teacher told you that was wrong perhaps she/he meant that you didn't derivate or substitute or something properly, because taking the first four derivatives and plugging in x = 0 is what must be done.

    The series' first 5 elements are  f(0)+\frac{f'(0)x}{1!}+\frac{f''(0)x^2}{2!}+\frac{  f'''(0)x^3}{3!}+\frac{f^{(iv)}(0)x^4}{4!} =1+2x+4x^2+8x^3+16x^4 , which is exactly the same as plugging in 2x instead of x in the well-known Taylor series \frac{1}{1-x}=1+x+x^2+x^3+x^4+\ldots , which is true for |x|<1 . In your case it'd be for |2x|<1\Longleftrightarrow |x|<\frac{1}{2}

    Tonio
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  3. #3
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    Ah, I understand part 1 now. It was a little misunderstanding between us.

    My first four terms come out to be:

    1, 2, -4, 12, 48

    Now for the other part you said ... are you saying the interval of convergence is from -1/2 to 1/2? The way I was taught to do it was to put the series in series/sigma notation, such as this:

    \sum (2x)^n

    Then do a ratio test or limit comparision test to get 2 possibilities for the endpoints of x, then plug the 2 endpoints back in the original series and determine whether it convergers/diverges to determine if I need a bracket or a paranthesis (includes the endpoint or does not include the endpoint).

    If I plug in -1/2 and 1/2 in for the series, I get:

    \sum (-1)^n

    and

    \sum (1)^n

    Which ultimately causes both series to diverge, so my final interval of convergence is:

    (-1/2,1/2)

    With paranthesis to indicate that the halves are NOT INCLUDED.
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  4. #4
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    Okay, I understand the first part. But I"m not sure how to use partial fractions for the last part. I know how to aply them to get an integrand, but not a derivative. And it's hard to manually find the derivatives because taking derivatives of g(x) results in messy equations due to the chain rule.
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  5. #5
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    Anyone?
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  6. #6
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    Partial fractions:
    \frac{1}{(1- 2x)(1- x)}= \frac{A}{1- 2x}+ \frac{B}{1- x}

    Multiply on both sides by (1- 2x)(1- x) to get
    1= A(1- x)+ B(1- 2x) for all x.

    Taking x= 1 and x= 1/2 gives you A and B very easily.

    Now use the result before to find the Taylor series for this function.
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