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Radius of convergence (Power series)

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi there,
    I have just started taylor series for my course.. most seems arlgiht so far, however when it comes to validating a given series( tayor or maclaruin), I have an idea on how to find out the x value.. but I don't know what I am doing wrong.


    Take for example: The following function (1+x)^-1 . Expand the maclaurin series of it and give values of x for which the above series is valid:


    2. Relevant equations
    F(x) = f(0) + f'(0)x /1! + f'' (0)x^2/ 2! + f'''(0)x^3 /3! + f^n (0)x^n/n!


    3. The attempt at a solution

    Here's my working:

    Let f(x) = (1+x)^-1 so f(0) = (1+0)^-1 = 1
    f'(x) = -1. (1+x)^-2 f' (0) = -1 . (1)^-2

    f''(x) = 2.1 (1+x)^-3 = > f''(0) = 2.1 (1)^-3

    f'''(x) = -3.2.1. (1+x)^-4 => f'''(0) = -3.2.1 (1)^-4

    Now initially i was puzzled on how to predict (next term or the general pattern) but I think next term will be f'4 = 4.3.2.1 (1x)^-5 so f'4 (0) = 4.3.2.1 (1)^-5

    Now I know test ratio can be done to find the radius of convergence i.e for a given function say anx^n ratio would be => (an+1)x^(n+1) / anx^n....

    I tried doing the same for the above series.. My general formula is : f(n) : (-1)^n . n! (1+x)^ -(n+1) .... by manipulation i got : 1.(n+1) x [ 1^-n-2] / 1^-n-1]

    = > 1. (n+1) but since n => infinity , this function diverges ?
    Any constructive input will be appreciated. Thanks!
     
    Last edited: Oct 29, 2011
  2. jcsd
  3. Oct 29, 2011 #2
    Hint: think of geometric series.
     
  4. Oct 29, 2011 #3
    I know geometric series goes like : 1+ q+ q^2 +q^3 ... and is in the form of a/1-r but how do I approach this problem ? I am confused .. can't I use ratio method to get value/s of x which converge.
     
  5. Oct 29, 2011 #4
    Ok.. so the general form of my series is :

    [itex]F^{n}[/itex] = [tex] (-1)^n (n!) (1+x)^{-(n+1)} [/tex]

    So we can ignore [tex] (-1)^n[/tex] as n -> infinity.

    I am left with : [tex] (n!) (1+x)^{-(n+1)} [/tex]


    [itex]F^{n+1}[/itex] / [itex]F^{n}[/itex]= [tex] (n+1!) (1+x)^{-(n+2)} / n! (1+x)^{-(n+1)} [/tex]

    I end up with the expression 1/ (1+x)....

    EDIT: I don't mean to spam or brea any forum rules, I can find series and predict nth term.. it's only the convergence bit confusing me :(
     
    Last edited: Oct 29, 2011
  6. Oct 29, 2011 #5
    How can you express [itex] \frac{1}{1+x} [/itex] as a geometric series? If what you've given above is what you think the nth term of the series is, then it's incorrect. It looks more like [itex] f^{(n)}(x)[/itex], but you can avoid calculating this by using my hint. If you're having trouble, look up geometric series in Wikipedia.

    For what values of x does this series converge? I don't think you need to use the ratio test--facts about geometric series should be sufficient.
     
  7. Oct 29, 2011 #6
    Well... a /1-r = 1/ x+1 ... I am totally lost...
    hm..
     
  8. Oct 29, 2011 #7
    If my hint isn't helping you, maybe you should go back to your original approach. Afterward I can show you the faster method I was hinting at if you're interested.

    What do you get for [itex] \displaystyle \frac{f^{(n)}(0)}{n!}[/itex]?
     
  9. Oct 29, 2011 #8
    Coming back to my original approach.. Is the general form of the series wrong ? hm.. I am determined to finish ( and understand it) this problem , tonight.
    Thanks for your patience with me.

    EDIT: Oh i think i see where i may have gone wrong.
    Previous

    [itex]F^{n}[/itex] = [tex] (-1)^n (n!) (1+x)^{-(n+1)} [/tex]

    Shouldn't the expression [tex](1+x)^{-(n+1)} [/tex] actually be [tex] (1+x)^{-n-1} [/tex] ?
    The form which you have posted looks like a general form of maclaurin or taylor series hm...

    EDIT: Actually i will give a little read(google) over the 'geometric approach' hm..
     
    Last edited: Oct 29, 2011
  10. Oct 30, 2011 #9
    I think what you had for [itex] f^{(n)}(x) [/itex] was right, but then you tried to do the ratio test using it. The ratio test examines [itex] \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n} \right| [/itex] for a power series [itex] \sum_{n=0}^\infty a_n x^n [/itex]. So, now that you have [itex] f^{(n)}(x) [/itex], compute [itex] \frac{f^{(n)}(0)}{n!}[/itex], which will give you the coefficient [itex] a_n [/itex].

    These are the same thing, so both are correct.

    You can try looking at http://en.wikipedia.org/wiki/Geometric_series" [Broken] link for geometric series. The idea is to find a way to write [itex] \frac{1}{1-x} [/itex] as a geometric series, and then use this to do the same for [itex] \frac{1}{1+x} [/itex].
     
    Last edited by a moderator: May 5, 2017
  11. Oct 31, 2011 #10
    Did you get the answer? If you did and still want to see the shortcut using geometric series, I'll post it.
     
  12. Oct 31, 2011 #11
    Sorry other assignments have kept me busy.. i will type down my approach in a bit.If everything fails you can post geometric method.Thanks for your concern =]
     
  13. Oct 31, 2011 #12
    Can you show me how to get the co-efficient [itex]a_{n}[/itex] ? (as above).. I am confused.. do I plug in n = 1 ?

    I have used the following link as my reference... : http://www.sosmath.com/calculus/radcon/radcon02/radcon02.html

    Oh do I basically find out my power series and then find out it's general form ? the maclaurin series in this case is : 4

    [tex]1-x+x^2 -x^3 +x^4 -x^5 [/tex]
     
    Last edited: Oct 31, 2011
  14. Oct 31, 2011 #13
    It's just as you had it above, [itex] a_n = \frac{f^{(n)}(0)}{n!} [/itex].
    That looks like a good start, but you need an infinite series. What is the general pattern? You said before that [itex]f^{(n)}(x) = (-1)^n (n!) (1+x)^{-(n+1)} [/itex]. So what is [itex] f^{(n)}(0)[/itex]?
     
  15. Nov 2, 2011 #14
    I believe I have finally got the solution.. it took me less than a minute to figure it out... the thing which I was over looking at was the fact that I wasn't trying to find out radius of the derivates rather than the series itself. ( As you have pointed out as well).
    My series is : [tex]1-x+x^2 -x^3 +x^4[/tex]
    In the general form : [tex] (-1)^n * x^n[/tex]
    This is An.
    So the ratio which's in the form of : [itex]a_{n+1}/a{n} [/itex] as n -> infinity becomes:
    [tex] ((-1)^n+1) *(x^n+1)/-1^n * x^n [/tex]

    after I manipulate them I end up with -x only... which's weird since the limit is given.. how am I suppose to plug in n -> infinity , if n's cancel... hmm
     
  16. Nov 2, 2011 #15

    HallsofIvy

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    In general, the power series for a function will converge "until there is a problem". Here, the function is 1/(1+ x) which "has a problem" at x= -1. So there is no reason for the series not to converge between 0 and -1 but it obviously cannot converge to a value at x= -1.

    Or, as others have pointed out, 1/(1+ x)= 1/(1- (-x)) is the sum of a goemetric series with "common ratio", r= -x. For what "r" does a geometric series converge?
     
  17. Nov 2, 2011 #16
    Looks good. (You can add the LaTeX markup \cdots to make it [itex]1-x+x^2 -x^3 +x^4 - \cdots[/itex].)
    That's okay. If [itex] \left| \frac{a_{n+1}}{a_n} \right| [/itex] is the same thing for every n (i.e., doesn't depend on n), then taking the limit shouldn't change anything, right? For example, [itex] \lim_{n \to \infty} 2 = 2 [/itex].

    Read what the ratio test has to say a bit more carefully. First of all, it examines [itex] L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| [/itex], so don't forget the absolute values. Now, for which values of L does the series converge? For which does it diverge? And for which is the test inconclusive?
     
  18. Nov 2, 2011 #17
    Thanks for the reply.
    Would the r value be r>1 ? since if that's plugged into the equation... i get negative number. I still however am confused since typical ratio test problems have some value of x turning out to be either negative or positive value which tells of divergence,convergence of the function.
     
    Last edited: Nov 2, 2011
  19. Nov 2, 2011 #18
    How do you get a negative number since you're taking the absolute value?
    Like I said, read carefully what the ratio test tells you: http://en.wikipedia.org/wiki/Ratio_test
     
  20. Nov 2, 2011 #19
    I see. So in my case L = -x, right ?
     
  21. Nov 2, 2011 #20
    Almost, but don't forget to take the absolute value!
     
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