Leibniz formula using mathematical induction

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SUMMARY

The discussion centers on the Leibniz formula and its application through mathematical induction. Participants analyze the summation transformation, specifically how the expression ∑_{k=0}^n a_k = a_n + ∑_{k=0}^{n-1} a_k = a_n + ∑_{k=1}^n a_{k-1} is derived. The key takeaway is the understanding of separating terms in summations and adjusting indices, which maintains the integrity of the total number of terms. This method is likened to the substitution rule for integrals, emphasizing the bijection between finite sets.

PREREQUISITES
  • Understanding of summation notation and properties
  • Familiarity with mathematical induction techniques
  • Basic knowledge of bijections in set theory
  • Concept of substitution in integral calculus
NEXT STEPS
  • Study the principles of mathematical induction in depth
  • Learn about bijections and their applications in combinatorics
  • Explore the substitution rule in integral calculus
  • Practice transforming summation expressions and their implications
USEFUL FOR

Students of mathematics, educators teaching calculus and algebra, and anyone interested in deepening their understanding of summation techniques and 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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