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Is there a differnt way to write this expression?

  • Thread starter Firepanda
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  • #1
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f(z) = SUM [(-1)n

Where SUM = sum from n=0 to infinity

Looks to me like there isn't but there must be a simpler way to write that.

Thanks
 
Last edited:

Answers and Replies

  • #2
Dick
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It's divergent, isn't? The nth term doesn't approach 0.
 
  • #3
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It's divergent, isn't? The nth term doesn't approach 0.
The question is to write this power series as a function

2hn01gp.jpg
 
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  • #4
Dick
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z^(2n)/n! isn't equal to e^(z^2). The SUM of the series z^(2n)/n! is e^(z^2). That would be fine if the (-1)^n weren't there. Can't you think of a function that's a lot like e^(z^2) that will give you the alternating sign in the series?
 
  • #5
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I just keep getting into the routine of trying to simply it like before. Do you mean replace e^(z^2) or multiply in some function f(z)/f(z)?
 
  • #6
Dick
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I just keep getting into the routine of trying to simply it like before. Do you mean replace e^(z^2) or multiply in some function f(z)/f(z)?
(-1)^n*z^(2n). Can you think of a way to write that as r^n for some number r? What would r be?
 
  • #7
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ahh yes
 
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  • #8
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For the 2nd part I have it down to

z-2 SUM (-1)n.(z2/4)n+1

Is there anyway I can progress from here or should I go back and try a different way?
 
  • #9
Dick
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For the 2nd part I have it down to

z-2 SUM (-1)n.(z2/4)n+1

Is there anyway I can progress from here or should I go back and try a different way?
It's a geometric series.
 
  • #10
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It's a geometric series.
(z2/4)n+1 starting at n=0

= 1/(1-z2/4) - 1

=> Series = z-2 SUM (-1)n/(1-z2/4) + (-1)n+1
 
  • #11
Dick
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(z2/4)n+1 starting at n=0

= 1/(1-z2/4) - 1

=> Series = z-2 SUM (-1)n/(1-z2/4) + (-1)n+1
You just did the same wrong thing you did in the first problem. You replaced a term is the series with the sum of a series. And you missed the cue to do the same thing you did with the (-1)^n in the first problem. You series has the form a*r^n. r just might have a (-1) in it.
 
  • #12
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ah yes, i have my r now thanks, was simple after that!

Thanks alot
 

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