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Homework Help: 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 = (1x)^-5 so f'4 (0) = (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...
  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.

    [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


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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!
  22. Nov 2, 2011 #21
    Ah... I see. Thank you very much for your help. I really appreciate it.

    So I have another expression: ln (x+a) for which I did the above process all over again.. I end up with L = |x/a|, am I right ?
    Last edited: Nov 2, 2011
  23. Nov 2, 2011 #22
    So what did you get for the radius of convergence?

    I don't understand... Are you trying to find a series for f(x) = ln(x + a)?
  24. Nov 2, 2011 #23
    Wouldn't that be simply L = |x| ?

    f(x) = ln(x + a)
    Series and the x value/s for which it's valid.
  25. Nov 2, 2011 #24
    Right, and now the link from post #18 tells you for what values of L the series converges. Post what you get for the region of convergence and I'll show you the geometric series trick.

    Once again, you can either do this problem by calculating successive derivatives, or by noting a nice geometric series trick. What is the derivative of f(x) = ln(x+a) and can you see its connection with the previous problem?
  26. Nov 3, 2011 #25
    Hi there, I have just had my test on this series.... for which i have got 9/10 .. not bad.. for the previous question since nth term's cancel out.. and we're left with -x... for a series to be valid it must converge so I take absolute value of -x i.e |x| .. and the condition under which this series converges is when x<1.

    EDIT: In case I am wrong.. I will rethink this again.. I have lecture in few minutes.
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