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Understanding series?

  1. Nov 8, 2004 #1
    Understanding series?????

    The question in the book states Find the values of x for which the series converges, Find the sum of the series for those valuese of x

    this is the series [tex]\sum_{n=1}^{\infty}\frac{x^n}{3^n}[/tex]

    first of all I dont even really understand what its saying, is it saying find the sum and set x equal to it.? The first thing I did was write out the first few terms [tex][\frac{x}{3}],[\frac{x^2}{9}],[\frac{x^3}{27}][/tex] so this thing is geometric and [tex]a=\frac{x}{3}[/tex] with the ratio [tex]r=\frac{x}{3}[/tex] am I assuming that that the ratio is less than one b/c if not you cant even go anyfurther can you? well anyway after that by using the [tex]\frac{a}{1-r}[/tex] theorm I get [tex]\frac{x}{3-x}[/tex] where do I go from here? Im pretty confused :confused: If this was a series with real numerical values I would have been done with the question b/c I have already fournd the sum...but what next...should there be another function that I set this equal to to get the "values of x" that the book wants
     
    Last edited: Nov 8, 2004
  2. jcsd
  3. Nov 8, 2004 #2
    I unfortunately haven't done series for a couple years, but I definitely know that r<1 in order for the series to converge, which would lead me to believe that 0<x<3 maybe? or 0 < or = x < 3 ?

    I guess from there you would have x^n/3^n = (x/3)^n, where you could then apply your summation of geometric series rule, where r = x/3 (given any x in the interval defined above), and a_0 = x/3. So,

    (x/3)/ [1-(x/3)] = x/(3-x), for x >/= 0, x<3. Maybe?
     
  4. Nov 8, 2004 #3

    Galileo

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    You correctly recognized it as a geometric series.

    For which values of r does [itex]\sum_{n=0}^{\infty}r^n[/itex] converge?
     
  5. Nov 8, 2004 #4
    those whose absolute values are less then one...should i set my equation equal to 1?
     
  6. Nov 8, 2004 #5

    Galileo

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    You know a geometric series converges for |r|<1.
    You also identified your series as a geometric series with [itex]r=\frac{x}{3}[/itex].

    If the series converges for [itex]|r|<1[/itex] and [itex]r=\frac{x}{3}[/itex] for what values of x does the series converge?
     
  7. Nov 8, 2004 #6
    for [tex]\frac{x}{3}<1[/tex] i dont know how to put in the absoulute value signs so imagine that fraction has an abs sign :redface: and that comes out to be [tex]x<3[/tex] :rofl: Yippy! is that my answer? Seems to be correct :rolleyes:
     
  8. Nov 9, 2004 #7

    Galileo

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    Yep :smile:
    Since [itex]|r|=|\frac{x}{3}|[/itex]
    [tex]|r|<1 \iff |x/3|<1 \iff |x|<3[/tex]

    I just use pipelines for absolute values. (shift+backslash)
     
  9. Nov 9, 2004 #8

    HallsofIvy

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    It's relatively easy to show that any power series (i.e. anything of the form [itex]\Sigma a_n x^n [/itex]) has a "radius of convergence": a number r such that the series converges absolutely for |x| < r, diverges for |x|> r and may or may not converge for |x|= r. Often the simplest way to find the radius of convergence is to use the "ratio test"- a series [itex]\Sigma a_n[/itex] converges if [itex]lim\frac{a_{n+1}}{a_n}< 1[/itex] and diverges if that limit is larger than 1.

    In this particular example, [itex]\sum_{n=1}^{\infty}\frac{x^n}{3^n}[/itex], the ratio becomes [itex]\frac{|x^{n+1}|}{3^{n+1}}\frac{3^n}{|x^n|}= \frac{|x|}{3}[/itex]. The series converges if [itex]\frac{|x|}{3}< 1[/itex] or |x|< 3. Of course, as was pointed out before, this is really a geometric series.

    Another example might be [itex]\Sigma\frac{x^n}{n}[/itex]. Now the ratio becomes [itex]\frac{|x^{n+1}|}{n+1}\frac{n}{|x^n|}= \frac{n||x}{n+1}[/itex]. The limit of that as n goes to infinity is just |x| so we must have |x|< 1. The radius of convergence is 1.
     
    Last edited: Nov 9, 2004
  10. Nov 9, 2004 #9
    [tex]\frac{|x^{n+1}|}{n+1}\frac{n}{|x^n|}= \frac{n|x|}{n+1}[/tex]

    is that what you ment? I think your latex was wrong
     
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