Leibniz formula using mathematical induction

Jkohn
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Homework Statement


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
 
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What they do is basically

\sum_{k=0}^n a_k = a_n+ \sum_{k=0}^{n-1} a_k = a_n + \sum_{k=1}^n a_{k-1}

Do you understand these steps?
 
micromass said:
What they do is basically

\sum_{k=0}^n a_k = a_n+ \sum_{k=0}^{n-1} a_k = a_n + \sum_{k=1}^n a_{k-1}

Do you understand these steps?

Interesting, I didnt know of this rule..whats it called??
 
Jkohn said:
Interesting, I didnt know of this rule..whats it called??

It's not really called anything.

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

\sum_{k\in B} a_k = \sum_{k\in A} a_{f(k)}

But most important: do you see why the equalities hold??
 
No I dont.
 
Jkohn said:
No I dont.

Perhaps try to write out the summations?? So in

\sum_{k=0}^n a_k = a_n + \sum_{k=1}^n a_{k-1}

replace all summation signs by the actual sums.
 
micromass said:
What they do is basically

\sum_{k=0}^n a_k = a_n+ \sum_{k=0}^{n-1} a_k = a_n + \sum_{k=1}^n a_{k-1}

Do you understand these steps?

Jkohn said:
No I dont.

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.
 
ohhhhh makes sense..thanks!
 
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