Leibniz formula using mathematical induction

[Jkohn]

Homework Statement

Here is this problem:

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

 

[micromass]

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]

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??

 

[micromass]

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??

 

[Jkohn]

No I dont.

 

[micromass]

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.

 

[Mark44]

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?

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 a_n from the summation. The first summation has n + 1 terms, a₀, a₁, ..., a_n. The expression in the middle has a_n and a summation with n terms, a₀, a₁, ..., a_n-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, ..., a_n.

 

[Jkohn]

ohhhhh makes sense..thanks!