Continuous function

  • Thread starter akoska
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  • #1
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How do I show that f(x)=sum (from n=0 to infinity) cos(nx)e^-nx is a continuous function? x is from (0, infinity)

So, I need to show that the series converges uniformly. I'm trying to say that |cos nx e^-nx| <= |e^-nx| and use Weierstrass comparison, but I can't find a function M_n to use for Weierstrass.

I feel like I'm missing something really simple...
 
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Answers and Replies

  • #2
1,789
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For positive [itex]n[/itex] what is [itex]e^{-nx}[/itex] always less than?
 
  • #3
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For positive [itex]n[/itex] what is [itex]e^{-nx}[/itex] always less than?

1, but the sum of 1s is obviously not convergent.
 
  • #4
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1, but the sum of 1s is obviously not convergent.

Okay, lets see:

[tex]f(x) = \sum_{n = 0}^{\infty} e^{-nx}\cos(nx)[/tex]

Clearly,

[tex]|e^{-nx}\cos(nx)| \leq e^{-nx}[/tex]

Now, as you said, [itex]e^{-nx} < 1[/itex] for all [itex]x \in (0,\infty)[/itex] and nonnegative n. So, clearly, your "[itex]M_{n}[/itex]" is simply the (convergent) geometric sum

[tex]\sum_{n = 0}^{\infty} e^{-nx}[/tex]
 

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