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Homework Help: 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
    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
    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
    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

    Staff: Mentor

    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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