Leibniz formula using mathematical induction

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Homework Help Overview

The discussion revolves around the application of the Leibniz formula and the use of mathematical induction. Participants are examining a specific summation manipulation and its implications within the context of the formula.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the steps involved in manipulating summations, particularly how separating terms and shifting indices leads to equivalent expressions. Questions arise regarding the validity of these steps and the reasoning behind them.

Discussion Status

Some participants express confusion about the manipulations and seek clarification on the reasoning behind the equalities. Others provide insights into the nature of the summation and suggest writing out the terms explicitly to aid understanding. There is an ongoing exploration of the concepts without a clear consensus yet.

Contextual Notes

Participants are discussing a specific rule related to summations and its connection to the substitution rule for integrals, indicating a potential gap in familiarity with the underlying concepts.

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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