Double Sum Identity Problem Solution

In summary, the problem is to prove that the given statement is true, which involves manipulating the double sum and using concrete examples. The result can also be obtained by manipulating the sums, such as shifting indices and switching summation orders.
  • #1
CompuChip
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Hi. The problem is as follows:

Homework Statement


Let m and n be integers, we may assume that (if they are not equal), m is the smallest. Then
[tex]\sum_{i=0}^m \sum _{j=0}^n f((m+n)-2 (i+j)) = \sum_{i=0}^m \sum _{j=0}^{-2 i+m+n} f((m+n)-2 (i+j))[/tex]
for some sequence [itex]f(k)_k[/itex].

Homework Equations


Anything you can think of, but probably just the standard manipulations on sums and some smart rewriting and/or separating different cases.

The Attempt at a Solution


I have reduced from a given question to the above statement, and I'm quite positive that I haven't made a mistake in this -- so the given statement should be true. It doesn't really look true to me though, so I plugged this into Mathematica: it seems to hold for any two [itex]m \le n[/itex] I choose, but I just don't see how to prove this. I think it involves a manipulation with the double sum, because if I remove the outer sum and consider i a fixed number in the rest of the expression, it doesn't seem to hold anymore (I always need one or more of the f(n) to be zero, which they aren't in general).
So I think the given information now is necessary and sufficient (e.g. no extra information on f(n) or m and n should be needed).
 
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  • #2
I'd start with concrete examples (m = 1, n = 2, etc.) to see how exactly it is working.
 
  • #3
I already did, and I also explicitly worked out the multiplicities of the various f(m + n - 2 i) of the left hand side, I found it to be equal to
[tex]
\sum_{i = 0}^{m - 1} (i + 1) f(m + n - 2i) +
{\sum_{k = m - n}^{n - m}}' (m + 1) f(k) +
\sum_{i = 0}^{m - 1} (i + 1) f(-(m + n - 2i)),
[/tex]
where the prime in the second sum denotes summation over odd values only (so m - n, m - n + 2, ...).So I could just do the same for the right hand side and show that the multiplicity of each term is equal, but I was sort of hoping that the result can also be obtained by just manipulating the sums (e.g. operations like shifting indices, switching summation orders, etc.).
 
Last edited:

1. What is the Double Sum Identity Problem Solution?

The Double Sum Identity Problem Solution is a mathematical concept used to simplify and solve equations involving the summation of two terms. It is commonly used in algebra and calculus to reduce complex expressions into simpler forms.

2. How does the Double Sum Identity work?

The Double Sum Identity states that the sum of two consecutive terms in a sequence can be expressed as the sum of two terms, one of which is the sum of the first and last term, and the other is the original sum of the two consecutive terms. In other words, the sum of (a + b) and (b + c) can be written as (a + c) + (2b).

3. Why is the Double Sum Identity important?

The Double Sum Identity is important because it allows for the simplification of complex equations, making them easier to solve. It also helps to identify patterns and relationships between terms in a sequence.

4. How is the Double Sum Identity used in real-life applications?

The Double Sum Identity has many real-life applications, particularly in fields such as physics, engineering, and finance. It is used to model and solve problems involving sequences, series, and mathematical patterns.

5. Are there any limitations to the Double Sum Identity?

While the Double Sum Identity is a powerful tool in mathematics, it is not applicable to all types of sequences. It can only be applied to sequences where the difference between consecutive terms remains constant. It also does not work for infinite series or divergent sequences.

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