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Another try at a difficult proof

  1. Feb 14, 2005 #1
    I found something very strange working with Riemann sums and expansion series.
    Can anyone tell me why
    [tex]\frac{x^n}{\sum_{i=1}^n x^{n-i}}=\frac{(x-1)x^n}{x^n-1}[/tex]

    Excuse the profanity, but I thought in the beginning this was a disgusting joke! :rofl: :yuck: :confused:
     
  2. jcsd
  3. Feb 14, 2005 #2

    dextercioby

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    Define a new variable of summation
    [tex] j=:n-i [/tex]

    Transform the sum [itex] \sum_{i=1}^{n} x^{n-i} [/itex] into a sum over "j" and the result will come out immediately.

    Daniel.

    P.S.What profanity?:confused:
     
  4. Feb 14, 2005 #3
    Sorry, but I don't quite understand how to do what you just wrote.
     
  5. Feb 14, 2005 #4

    learningphysics

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    The way I did, just multiply the numerator and denominator by x-1. So in the denominator you get 2 summations (one minus the other). Simplify the denominator.
     
  6. Feb 14, 2005 #5

    dextercioby

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    It doesn't make any sense to me.
    [tex] j=:n-i [/tex] (1)
    [tex] i=1 \Rightarrow j=n-1 [/tex] (2)
    [tex] i=n \Rightarrow j=0 [/tex] (3)

    The new sum (over "j") becomes
    [tex] \sum_{j=n-1}^{j=0} x^{j}=\sum_{j=0}^{j=n-1} x^{j}=\frac{x^{n}-1}{x-1} [/tex] (4)

    Now plug it in the initial expression and u'll find the desired result...

    Daniel.
     
  7. Feb 14, 2005 #6

    learningphysics

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    If you multiply the denominator through by x-1 you get:

    [tex] \sum_{i=1}^{n} x^{n-i+1} - \sum_{i=1}^{n}x^{n-i} [/tex]

    You can simplify the above (try to make the two sums look the same) and get [tex]x^n-1[/tex]
     
  8. Feb 14, 2005 #7

    dextercioby

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    How do you make that simplification...?

    Daniel.
     
  9. Feb 14, 2005 #8

    learningphysics

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    Similar to your method:

    [tex] \sum_{i=1}^{n} x^{n-i+1} - \sum_{i=1}^{n}x^{n-i} =[/tex]

    [tex] \sum_{i=1}^{n} x^{n-(i-1)} - \sum_{i=1}^{n}x^{n-i}=[/tex]

    For the first sum, let j=i-1

    [tex] \sum_{j=0}^{n-1} x^{n-j} - \sum_{i=1}^{n}x^{n-i} =[/tex]

    [tex] (x^{n-0} + \sum_{j=1}^{n-1} x^{n-j}) - (\sum_{i=1}^{n-1}x^{n-i} + x^{n-n}) =[/tex]

    I take out the first term from the first sum and last term from the second sum.

    The two sums that are left cancel each other out.

    [tex] x^n-1[/tex]
     
  10. Feb 14, 2005 #9

    dextercioby

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    Wow,that's so simple!*admirative*...Gosh,why didn't i think of that...?*envy*

    Daniel.

    P.S.We should let the OP decide which solution he likes best.
     
  11. Feb 14, 2005 #10
    I think they are both equally creative, well, thanks a lot for your help.
     
  12. Feb 14, 2005 #11

    learningphysics

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    :smile: well, yours is the faster solution.
     
  13. Feb 15, 2005 #12

    Curious3141

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    Why can't one just use the sum for Geometric series ?

    The denominator is [tex]x^{n-1} + x^{n-2} + ... + x^0 = \frac{x^n - 1}{x - 1}[/tex]

    using the well worn sum for GPs, and the answer immediately follows.
     
  14. Feb 15, 2005 #13

    Galileo

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    Indeed, you can just cancel the [itex]x^n[/itex] on both sides. Invert to get:
    [tex]\frac{x^n-1}{x-1}=\sum_{i=1}^n x^{n-i}}[/tex]
    A well known fact.

    PS: [itex]\sum_{i=1}^n x^{n-i}=\sum_{i=0}^{n-1} x^{i}[/itex]
     
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