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Expansion of series and possible induction

  1. Sep 21, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Consider the infinite series $$\frac{x}{e^x - 1} = A_o + A_1 x + \frac{A_2}{2!}x^2 + ... + \frac{A_n}{n!}x^n + ...$$ Determine that ##1 = A_o,\,\,\,\,\,0 = A_o/2! + A_1,\,\,\,\,\,0 = A_o/3! + A_1/2! + A_2/2!##.
    Show that for ##n > 1##, one can write the relations as $$(A+1)^n - A^n,$$ where ##A^n \rightarrow A_n##.

    2. Relevant equations

    Taylor expansions and induction

    3. The attempt at a solution
    First part is fine. I thought about using induction on the second part since ##n## is confined to an integer. I understand ##A^n \rightarrow A_n## as meaning the powers of the series converge to the actual terms in the series, which is why I thought ##n## was an integer, labeling the terms in the series. However, I can't seem to make sense of the base case (n=2). It is $$(A+1)^2 - A^2 = 2A + 1$$ What is ##A##?

    thanks
     
  2. jcsd
  3. Sep 23, 2014 #2

    CAF123

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    Anyone any comments or help? thanks.
     
  4. Sep 23, 2014 #3

    pasmith

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    What this is saying (I think) is that you can expand the left hand side of [itex](A + 1)^n - A^n = 0[/itex], and if you then replace [itex]A^k[/itex] with [itex]A_k[/itex] where [itex]0 \leq k < n[/itex], you can find [itex]A_{n-1}[/itex] in terms of [itex]A_0, \dots, A_{n-2}[/itex].

    Thus: [itex](A + 1)^2 - A^2 = 2A + 1[/itex] and indeed [itex]2A_1 + A_0 = 0[/itex].
     
  5. Sep 23, 2014 #4

    CAF123

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    I see, is there a reason why you neglected the case where ##k=n##?
    I nearly see what you did there except how did you get that ##A=A_1##? Is this just a notation ##A = A^1 = A_1##?

    Would you say induction is the way to go? I can try further given I now understand what the base case means, thanks.
     
  6. Sep 23, 2014 #5

    pasmith

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    The [itex]A^n[/itex] terms cancel.

    [itex]x^1 = x[/itex] for every [itex]x[/itex].

    It is the obvious method.
     
  7. Sep 23, 2014 #6

    haruspex

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    It's not that ##A^1 = A_1##, but that it represents it in the analogy between the two sets of equations.
     
  8. Sep 24, 2014 #7

    CAF123

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    I tried the induction argument, but seemed to run into a mistake. Assume ##P(n) = (A + 1)^n - A^n = 0 ## is true for ##n##. Then $$(A+1)^{n+1} - A^{n+1} = (A+1)^n (A+1) - A^n A $$$$= ((A+1)^n - A^n)A + (A+1)^n = A^n \neq 0$$ The ##A^n## are just numbers so I am not sure where the mistake is.
     
  9. Sep 24, 2014 #8

    pasmith

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    I think you want to be using induction to establish that [tex]
    \sum_{k=0}^{n-1} \binom{n}{k} A_k = 0
    [/tex] since [tex]
    (A+ 1)^n - A^n = \sum_{k=0}^{n-1} \binom{n}{k} A^k.
    [/tex]
     
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