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Abel's Lemma

  1. Oct 12, 2004 #1

    Integral

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    Abel’s Lemma,
    Let [itex]a_0,a_1,a_2,\cdots [/itex] and [itex]b_0,b_1,b_2,\cdots [/itex] be elements of a field;
    let[itex] s_k = a_0 + a_1 + a_2 + \cdots + a_k [/itex] k= 0,1,2,… And s-1 =0.

    Then for any positive real integer n and for m= 0,1,2,…,n-1,

    [tex] \sum^n _{k=m} a_k b_k = \sum ^{n-1}_{k=m} (b_k - b_{k+1}) s_k + b_n s_n - b_m s_{m-1}[/tex]



    Proof:

    Expanding the terms of the sum gives

    [tex] (b_m - b_{m+1})s_m + (b_{m+1} - b_{m+2} )s_{m+1} + \cdots + (b_{n-1} - b_n) s_{n-1} + b_n s_n - b_m s_{m-1}[/tex]

    By the definition of sk we have sk+1 = sk + ak+1

    Now for the expansion of the sum we can write:

    [tex] (b_m s_m - b_{m+1} s_m + b_{m+1} s_{m+1} - b_{m+2} s_{m+1} + \cdots + b_{n-1} s_n - b_n s_{n-1} + b_n s_n - b_m s_{m-1}[/tex]


    [tex] b_m a_m + b_m s_{m-1} - b_{m+1} s_m + b_{m+1} s_m + b_{m+1} a_{m+1} - b_{m+2} s_{m+1} + \cdots + b_{n-1} s_n - b_n s_{n-1} + b_n s_n - b_m s_{m-1}[/tex]

    Now group canceling terms:

    [tex] b_m a_m + ( b_m s_{m-1} -b_m s_{m-1}) + ( - b_{m+1} s_m + b_{m+1} s_{m}) + b_{m+1} a_{m+1} + \cdots + b_{n-1} a_{n-1} + b_{n-1} s_n - b_n s_{n-1} + b_n s_n [/tex]

    All that remains is

    [tex] \sum _{k=m} ^ n a_k b_k [/tex]

    Is this proof sufficient or do I need do an induction proof?
     
    Last edited: Oct 12, 2004
  2. jcsd
  3. Oct 12, 2004 #2

    Galileo

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    It's sufficient. But my teacher would say it's too suggestive with those dots.
    Use induction. It's much neater.
     
  4. Oct 13, 2004 #3

    Integral

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    OK, let me see if I can do the induction proof.
    The part that has bothered me is the m, not clear to me what this buys? I guess it it a bit more generalized, don't have to start at zero.

    For the induction proof I need to show that the relatationship is ture for n=1 and m=0

    [tex]\sum _ {k=0}^1 a_k b_k = a_0 b_0 + a_1 b_1 [/tex]

    for the RHS I have

    [tex] \sum ^0 _ {k=0} (b_k - b_{k+1})s_k + b_n s_n - b_m s_{m-1}= (b_0 - b_1)s_0 + b_0 s_0 + b_1 a_0 + b_1 a_1 - b_0 s_{-1} [/tex]
    [tex] = b_0 a_0 - b_1 a_0 + b_0 a_0 + b_1 a_0 + b_1 a_1 - 0 [/tex]
    [tex] = b_0 a_0 + a_1 b_1 [/tex]

    So the statement is valid for n=1, now assume that the general statement for is true n= N.

    Let n= N+1

    [tex] \sum^{N+1} _{k=m} a_k b_k = \sum ^N_{k=m} (b_k - b_{k+1}) s_k + b_{N+1} s_{N+1} - b_m s_{m-1}[/tex]
    For the RHS we have

    [tex] \sum^{N+1} _{k=m} a_k b_k = \sum ^{N-1}_{k=m} (b_k - b_{k+1}) s_k + b_{N+1} s_{N+1} - b_m s_{m-1}+( b_{N} - b_{N+1})s_N[/tex]


    [tex] = \sum ^{N-1}_{k=m} (b_k - b_{k+1}) s_k + b_{N+1} s_{N} + b_{N+1}a_{N+1} - b_m s_{m-1}+ b_{N}s_N - b_{N+1}s_N[/tex]

    [tex] = \sum ^{N-1}_{k=m} (b_k - b_{k+1}) s_k + b_Na_N - b_m s_{m-1}+ b_{N+1}a_{N+1} [/tex]

    [tex] = \sum ^N_{k=m} a_k b_k + a_{N+1}b_{N+1} [/tex]

    [tex] = \sum^{N+1}_{k+m} a_k b_k [/tex]
    QED
     
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