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I'm using the following book as reference:

** Walter G. Kelly and Allan C. Peterson Difference Equations Second Edition, Harcourt Academic Press, 2001. **

I gather from the book and from research that it's generally difficult to find sums involving fractions with complicated denominators.

I'd like to know what is known about these sums. One thing that particularly springs up is what techniques generally work well for finding these sums. The book lists some general properties of indefinite sums, and some properties seem to go well with products, but don't seem to work well with fractions. I wonder if there is some general way to organize complicated fractions so that they are easy to sum.

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I'm wondering about this now because I'm examining the possibility of combining discrete multiplicative integrals with these fractional sums. It seems that using the discrete multiplicative techniques may allow one to calculate complicated fractions with ease, by finding a common denominator. –  Generator Nov 30 '10 at 3:59
    
Voting to close - not a real question. Fractions are products, $x/y=x{\rm\ times\ }y^{-1}$, so premise seems false. If there were a "general way" to "organize complex fractions" to make them "easy to sum," don't you think it would be in that book? On the other hand, if you have an example of using "discrete multiplicative techniques" to "calculate complicated fractions with ease," maybe you could give an example and ask for generalizations. –  Gerry Myerson Nov 30 '10 at 5:50
    
@Gerry: Here's my example...take $\sum_{i=1}^n{1/i}$ = $\left(\sum_{i=1}^n{(\prod_{j=1}^n{j})/i}\right) / \sum_{j=1}^n{j}$. I take the product of all denominators and multiply each term by this product. So I essentially establish a common divisor, which is the product of all denominators. Thusly, I divide each term by its divisor, after having multiplied by the common divisor. When I'm done, I divide by this common divisor, thus finding the correct sum. I believe that this can be generalized. We multiply each term by the product of all denominators. –  Generator Nov 30 '10 at 6:10
    
...We have the understanding that we'll later divide by this same product. We then only have to divide each term by its denominator. I think that this may have an extension to the general problem. –  Generator Nov 30 '10 at 6:11
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I think you meant that sum on $j$ to be a product. If you are claiming that the left side is a complicated fraction which the right side allows you to calculate with ease, I'm afraid I'm going to need some convincing. More generally, putting a bunch of quotients over a common denominator is a bit too high school to be a significant breakthrough this late in the 21st century, don't you think? –  Gerry Myerson Nov 30 '10 at 9:38
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