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Stuck on power series, need a refreshment! Diff EQ! converge?

  1. Mar 9, 2006 #1
    Hello everyone! I remember doing these in calc II, but forgot 99% of it. Here is the question:
    Find the interval of convergence for the given power series.
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/5e/119a4cd07ce5d3f6f673477ed169641.png [Broken]
    The series is convergent
    from x = , left end included (enter Y or N):
    to x = , right end included (enter Y or N):

    Here is my work:
    http://suprfile.com/src/1/4fiy34/lastscan.jpg [Broken]

    Now i'm confused on what i do now, also u see where i have
    |x-6|? Is that right or should it be |x+6| ? Alos is it suppose to be the limit as x-> or n->? Thanks!
    I'm using the ratio test btw.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 9, 2006 #2
    Well you have the ratio test wrong there is no |x - x0| tacked on to the fron of the limit, and the limit should be as "n" goes to infinity, then in order for the series to converge the limit must be less than or equal to 1.
  4. Mar 9, 2006 #3
    thanks for the responce d_leet i'm looking at this professors website and following his example, and for some reason he tacked on a |x-xo|.

    Scroll down to example one if u want to know where I got that |x-xo|
    http://tutorial.math.lamar.edu/AllBrowsers/3401/PowerSeries.asp [Broken]
    Last edited by a moderator: May 2, 2017
  5. Mar 9, 2006 #4
    Grrrr... I lost what I just typed.. Ok well I see what he did. What you did is not quite the same, you tried to doboth methods simultaneously, if you had left out that factor then you would get the right answer doing what I suggested. The reason that it is like that on the other website is that he knows that factor will come out of the limit if he takes it as part of the a_n so he just takes it out right away. Tell me if any of that made any sense at all, I can try to exxplain better if it didn't.
  6. Mar 9, 2006 #5
    I really didn't understand but I think i'll just ignore that factor of |x-6| and just leave it as

    okay if i take this as n goes to infinity, wouldn't i get infinity/infinity if i plug in infinity for n?
    I remeber there was a rule for this but I forgot. Do i divide everything by n then plug in infity and see waht i'm left with? kinda sounds familiar?
  7. Mar 9, 2006 #6
    Okay i still didn't figure out how to take the limit of that, but the TI-89 did, and it got:
    -6(x-6), now i know the rule is:
    http://tutorial.math.lamar.edu/AllBrowsers/3401/PowerSeries_files/eq0013M.gif [Broken]
    So do i set -6(x-6) = 1?
    so i'm getting x = 5.833 which is > 1, so this means the series diverges. But how do i know what intervals? Because it wants:
    The series is convergent
    from x = , left end included (enter Y or N):
    to x = , right end included (enter Y or N):

    So I know the radius of convergence is 35/6, but what do i do with that? I entered it in, +/- and it didn't like it
    Last edited by a moderator: May 2, 2017
  8. Mar 9, 2006 #7
    Okay I redid this problem, and i think i did it right, but not, lies!
    But i'm way closer to it then i was before, check my work out, it looks legit hah:
    Note: ignore the First series up there, where i have the 4 scribbled out.
    http://img233.imageshack.us/img233/7356/lastscan4ge.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  9. Mar 10, 2006 #8


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    Homework Helper

    Note that since [tex]|x-6|\geq 0,[/tex] for all x by definition of absolute value, it follows that [tex]|x-6|\geq 0 >-6[/tex] for all x (so you can also include x=0 above.)
  10. Mar 10, 2006 #9


    User Avatar
    Science Advisor

    You have, at one point,
    and then immediately after
    What happened to the absolute value in the denominator? You are aware that |-6|= 6 aren't you? Shouldn't it be
    |x- 6|< 6?
  11. Mar 10, 2006 #10


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    Homework Helper

    [tex]\sum_{n=1}^{\infty} \frac{(x-6)^n}{n(-6)^n}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\left( \frac{x-6}{6}\right) ^n = -\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\left( \frac{x}{6}-1\right) ^n = -\log \left| \frac{x}{6}\right| [/tex]

    which converges for [itex]-1<\frac{x}{6}-1\leq 1[/itex] or [itex]0<x\leq 12[/itex] per the usual Taylor series

    [tex]\log |1+x| = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}\mbox{, valid for }-1<x\leq 1[/tex]

    and which agrees with HallsofIvy's post (and that is not bad for the confidence.)
    Last edited: Mar 10, 2006
  12. Mar 10, 2006 #11
    Oo my bad, i do'nt know why i just said f the abs sign in the denominator, that made it work out perfect. And thanks for the explanation benorin! I'm e-mailing u both milk and cookies. Enjoy. :biggrin:
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