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Breaking this expressions into areas.

  1. Dec 26, 2009 #1
    [tex]\sum_{n=-\infty}^{\infty}r^{|n|}e^{inx}=[/tex][tex]\sum_{n=1}^{\infty}e^{inx}[/tex]+ [tex]r^{0}e^{0}[/tex] +[tex]\sum_{n=-\infty}^{\infty}r^{n}e^{-inx}[/tex]

    if we look at the left side
    i have been told that it was broken by intervals
    the left most is for n=1.. infinity the central is for n=0
    the right most for n=-1 ..-infinity

    i know the definition of the |x| function
    for some values its x for other its -x

    but here the power of e changes too
    why?
     
    Last edited: Dec 26, 2009
  2. jcsd
  3. Dec 26, 2009 #2

    tiny-tim

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    Hi nhrock3! :smile:

    I suspect you've copied it wrong, is it …

    [tex]\sum_{n=-\infty}^{\infty}r^{|n|}e^{inx}\ =\ [/tex] [tex]\sum_{n=1}^{\infty}r^{n}e^{inx}[/tex]+ [tex]r^{0}e^{0}[/tex] +[tex]\sum_{n=1}^{\infty}}r^{n}e^{-inx}[/tex]
    ?​
     
  4. Dec 26, 2009 #3
    why we always have [tex]r^n[/tex] why in the minus case we dont have [tex]r^{-n}[/tex]

    and why the power of the exponent changes to minus in the negative case
    ?
     
  5. Dec 26, 2009 #4

    tiny-tim

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    Because it depends what the limits are.

    Perhaps it's clearer if I write it …

    [tex]\sum_{n=-\infty}^{\infty}r^{|n|}e^{inx}\ =\ [/tex] [tex]\sum_{n=1}^{\infty}r^{n}e^{inx}[/tex]+ [tex]r^{0}e^{0}[/tex] +[tex]\sum_{m=1}^{\infty}}r^{m}e^{-imx}[/tex]

    where m = -n, so |n| = m, and inx = -imx.

    (and m and n are "dummy" indices, so you can then change m to n :wink:)
     
  6. Dec 26, 2009 #5
    i cant see how the minus indexes changes the original expression

    ?

    why the power of the exponent changes to minus in the negative case

    they should be negative
    and we dont need to change it to minus
     
  7. Dec 26, 2009 #6

    HallsofIvy

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    Do you see that both sums are for positive n?
    Your original sum is
    [tex]\sum_{n=-\infty}^\infty r^{|n|}e^{inx}[/tex]

    You can break that into three parts, as you say:
    [tex]\sum_{n=-\infty}^{-1} r^{|n|}e^{inx}+ r^0e^{i0x}+ \sum_{n=1}^\infty r^{|x|}e^{inx}[/tex]

    Now look at each of that first sum separately. For n< 0, |n|= -n so
    [tex]\sum_{n=-\infty}^{-1} r^{|n|}e^{inx}= \sum_{n=-\infty}^{-1} r^{-n}e^{inx}[/tex]
    Change the index: let k= -n so n= -k:
    [tex]= \sum_{-k= -\infty}^{-1} r^{k}e^{i(-k)x}= \sum_{k=1}^\infty r^k e^{-ikx}[/tex]

    Since the index here is "dummy" we can use any letter we like, say "n", in place of k:
    [tex]= \sum_{n= 1}^\infty r^n e^{-inx}[/tex]

    Of course, [itex]r^0e^{i(0)x}= 1[/itex], and, for n> 0, |n|= n, so the entire sum becomes
    [itex]\sum_{n= 1}^\infty r^n e^{-inx}+ 1+ \sum_{n=1}^\infty r^n e^{inx}[/tex].
     
  8. Dec 26, 2009 #7

    tiny-tim

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    Try changing m to -n in the right-hand term in my equation …

    what do you get? :smile:
     
  9. Dec 26, 2009 #8
    no you cant use is as dummy
    if n=-k
    and k=n
    then n=-n

    which works only for n=0
     
  10. Dec 26, 2009 #9
    no you cant use it as dummy
    if n=-k
    and k=n
    then n=-n

    which works only for n=0
    and for n= -1 to -infinity
     
  11. Dec 26, 2009 #10

    HallsofIvy

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    What that tells me is that you don't know what a "dummy" index is. If I had an equation that said k+ n= 3, then I could not replace k by either n or -n- they mean different things. But if my equation is [itex]\sum_{k=1}^5 \frac{1}{k}+ \sum{n=1}^5 \frac{(-1)^n}{n}[/itex] I could replace the "k" in the first sum by "n" because [itex]\sum_{k=1}^5 \frac{1}{k}= 1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}[/itex] and [itex]\sum_{n=1}^5 \frac{1}{n}= 1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}[/itex] are both numbers and do not depend upon either "n" or "k". Having changed from "n" to "k", I could now make the change from k to -n as well and have [itex]\sum_{n=-1}^{-5}\frac{1}{-n}= \frac{1}{-(-1)}+ \frac{1}{-(-2)}+ \frac{1}{-(-3)}+ \frac{1}{-(4)}+ \frac{1}{-(-5)}[/itex] which is exactly the same thing. This "n" has nothing to do with the original n. That does NOT imply "k= n" and "k= -n" for the same n!
     
  12. Dec 26, 2009 #11
    aahhh thanks :)
     
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