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Cool Math Tricks: Pulling Linear Factors out of Binomial Sums

While doing my statistical mechanics homework today I arrived at a sum that looked like


Sums like this come up frequently when you're working with random walks, or flipping coins, or counting states of quantum mechanical systems where only two energies are allowed, or in any other number of contrived situations.

It's almost a good looking sum because everything to the right of the factor of m looks like the sum for a binomial expansion:


which simply evaluates to:


It turns out that there's an easy way to get the factor of m out of the sum and get on with your life! First notice that:

so that the sum


can be re-written as:


So, we rather handily got rid of the factor of m. The extra factor of p can be taken outside of the sum since it has nothing to do with the summation index m. Furthermore, the order of summation and differentiation can be interchanged to arrive at:


Now the sum actually is a binomial expansion and after simplifying and performing the derivative that was introduce we arrive at:

I didn't come up with this trick, I'm merely passing it along. I found it in a Statistical Mechanics text that I'm very impressed with:


NOTE: I hope these little pointers are helping folks out, because they're definitely helping me. While preparing this post, I initially put in a summation index of i and found myself wondering what m had to do with anything. It turns out that I didn't have the concept firmly in my head yet, and upon re-investigating I corrected the summation index and got a much better understanding of what's going on. It just goes to show that what my childhood piano teacher said is true:

"The best way to learn something is to prepare yourself to be able to teach it to someone else."

Comments

Anonymous said…
Wow that is a nifty trick and I'll keep it in mind when I do stat mech. I've noticed that all the time a complex problem can be simplified by realizing that some expression is really the derivative of some other expression and it makes the problem trivial.
Hamilton said…
Thanks for the pointer! I always thought of derivatives for finding the rate of change. It's very cool that they can be used for simplifications as well!

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