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Real Analysis : Series

  1. Mar 10, 2006 #1
    Consider the series

    1+ Σ((1/(2^k))coskx + (1/(2^k))sinkx)

    (a) Show that series converges for each x in R.
    (b) Call the sum of the series f(x) and show that f is continuous on R = real numbers

    My thoughts:

    From trig => cos + sin = 1. So, is it something like
    |coskx + sinkx| / |2^k| < or = (in particular = ) 1/2^k = M. Then since ΣM = Σ(1/2^k) converges (since 1/2^n approaches 0, even though it never attains it) Thus the given series converges uniformly and therefore converges. Adding 1 will not change the fact that it converges.
     
    Last edited: Mar 11, 2006
  2. jcsd
  3. Mar 11, 2006 #2

    HallsofIvy

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    This is your problem. What have you done on it?
     
  4. Mar 11, 2006 #3
    This is what I've done:
    My thoughts:

    From trig => cos + sin = 1. So, is it something like
    |coskx + sinkx| / |2^k| < or = (in particular = ) 1/2^k = M. Then since ΣM = Σ(1/2^k) converges (since 1/2^n approaches 0, even though it never attains it) Thus the given series converges uniformly and therefore converges. Adding 1 will not change the fact that it converges.
    Can you please correct me?
     
  5. Mar 11, 2006 #4
    Well, for starters, [tex]cos\theta+sin\theta \neq 1[/tex]. What you are looking for is [tex]cos^2\theta+sin^2\theta=1[/tex].

    -Dan
     
  6. Mar 12, 2006 #5

    matt grime

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    And nor is it even true that cos(x)+sin(x)<1 for all x, but one can do something with the abs value, or one can just appeal to any of the comparison tests.

    Further, when you say .

    "Then since ΣM = Σ(1/2^k) converges (since 1/2^n approaches 0, even though it never attains it)"

    please tell me that you don't think the terms tend to zero implies the sum converges. If there is exactly one result in analysis that is hammered home ad nauseum it is that this is is clearly false as can be seen just from trying to sum 1/n the simplest series whose terms tend to zero.
     
    Last edited: Mar 12, 2006
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