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Leibniz formula using mathematical induction

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Here is this problem: IMG_20121206_202225.jpg

    IMG_20121206_202342.jpg

    I have the solution http://www.proofwiki.org/wiki/Leibniz%27s_Rule/One_Variable

    This is where I get stuck..
    Where it says: 'For the first summation, we separate the case k=n and then shift the indices up by 1.'

    Why does this lead to the conclusion??
    thanks
     
  2. jcsd
  3. Dec 6, 2012 #2

    micromass

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    What they do is basically

    [tex]\sum_{k=0}^n a_k = a_n+ \sum_{k=0}^{n-1} a_k = a_n + \sum_{k=1}^n a_{k-1}[/tex]

    Do you understand these steps?
     
  4. Dec 6, 2012 #3
    Interesting, I didnt know of this rule..whats it called??
     
  5. Dec 6, 2012 #4

    micromass

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    It's not really called anything.

    If you're familiar to the substitution rule for integrals, then this is pretty similar.
    In general: if [itex]f:A\rightarrow B[/itex] is a bijection between the finite sets A and B, then

    [tex]\sum_{k\in B} a_k = \sum_{k\in A} a_{f(k)}[/tex]

    But most important: do you see why the equalities hold??
     
  6. Dec 6, 2012 #5
    No I dont.
     
  7. Dec 6, 2012 #6

    micromass

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    Perhaps try to write out the summations?? So in

    [tex]\sum_{k=0}^n a_k = a_n + \sum_{k=1}^n a_{k-1}[/tex]

    replace all summation signs by the actual sums.
     
  8. Dec 6, 2012 #7

    Mark44

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    It's really not very complicated. From the first sum to the next step, all that has happened is that they have separated out an from the summation. The first summation has n + 1 terms, a0, a1, ..., an. The expression in the middle has an and a summation with n terms, a0, a1, ..., an-1. In all, it's the same (n + 1) terms as in the first summation.

    Going from the expression in the middle to the last one, all they are doing is fiddling with (i.e., adding 1 to) the index, where the index ranges between 1 and n instead of between 0 and n - 1 as before. To adjust for this, the subscript on a is adjusted correspondingly.

    In all three expressions, they are adding n + 1 terms, a0, a1, ..., an.
     
  9. Dec 6, 2012 #8
    ohhhhh makes sense..thanks!
     
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