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Taylor series radius converge

  1. Jul 15, 2012 #1
    Hi everybody,
    Firstly sorry for my bad English . I have a question related to taylor series . I did not find easy way to solve it .Derivatives are becoming more and more complex . Please help me.

    question : Work out the taylor series of the function x/(1+x^2) at x =0 .Find the radius of converge and interval of converge of the series Integrating the series obtained estimate the value of ln2 ?

    You don't have to find answer .I need only easy way to solve this problem . Thanks a lot
  2. jcsd
  3. Jul 15, 2012 #2
    If you know the following then it'll be very easy:

    [tex]\text{For}\,\,|x|<1\,\,,\,\frac{1}{1+x}=1-x+x^2-x^3+...=\sum_{n=0}^\infty(-1)^nx^n\Longrightarrow \frac{1}{1+x^2}=1-x^2+x^4-x^6+...=\sum_{n=0}^\infty(-1)^nx^{2n}[/tex]

    Well, now just multiply the above by [itex]\,x\,[/itex] and you're done.

  4. Jul 15, 2012 #3
    wow thanks a lot . Can I multiply series and x directly ? (-1)^n (x)^2n+1 ?
  5. Jul 15, 2012 #4
    Another way lies in the Taylor series for the arctangent function. We know that
    [tex]\int \frac{1}{1+x^2}dx=\arctan(x)+C[/tex]
    so we first start by assuming that the power series expansion exists. If so, we can write this for some sequence a:
    Integrating both sides, we obtain
    [tex]\arctan(x)+C=\sum_{k=0}^{\infty}\frac{a_{k} x^{k+1}}{k+1}[/tex]
    Evaluation at x=0 easily yields C as 0. Hence, we can strike that out. Now setting x=1, we obtain
    [tex]\pi/4 = \sum_{k=0}^{\infty}\frac{a_{k}}{k+1}[/tex]
    For this equality to hold, we use the Leibniz formula for pi. This formula requires all odd k, so all even k+1 to vanish, which implies for odd k the sequence [itex]a_k[/itex] equals zero. For even k, the sequence equals [itex](-1)^{k/2}[/itex]. This leaves us with the formula:
    [tex]\frac{1}{1+x^2}=\sum_{k=0}(-1)^k x^{2k}[/tex]

    Indeed complicated, but it is still legitimate.
  6. Jul 15, 2012 #5


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    a0=pi/4, ak=0 otherwise would satisfy the formula for x=1, too, and give a wrong answer.

    You cannot derive the whole taylor series by evaluating it at a single point. It works in this case, but it is pure luck.
  7. Jul 15, 2012 #6
    It is not pure luck. This also holds true. Take arctangent and differentiate it, along with its power series, to obtain the very same formula I gave. I just showed another way of working it out. You are right, but it is not a coincidence.
  8. Jul 15, 2012 #7


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    Yes you can! :smile:
  9. Jul 15, 2012 #8


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    You picked a specific series to get the sum pi/4. This choice is arbitrary, and you just picked the one which gives the correct taylor series (by luck, by "good eye", whatever). Any other choice would be valid in this step, but lead to a wrong taylor series. You have no way to check whether you picked the right series or not in your post.

    You can differentiate arctan n times (and find a general formula), but you did not use this.
  10. Jul 16, 2012 #9
    mfb, if I started with arctan, it would be a valid proof but it would have no sense of intuition. How did I get arctan in the first place? You need to do what I did before you can actually think of differentiating arctan.
  11. Jul 17, 2012 #10


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    As far as the "radius of convergence" is concerned, the simplest thing to do is to note that the denominator of the fraction, [itex]x^2+ 1[/itex], will be 0 when x= i or -i, both at distance 1 from 0 in the complex plane. A Taylor's series will converge until it "hits a singularity".
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