The Incomplete Gamma and Confluent Hypergeometric Functions (Part 6)

Calculus 2 Minutes

In the previous posts, I showed that the curious integral

\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \displaystyle e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s},

where the right-hand side is a special case of the confluent hypergeometric function when a is a positive integer, can be confirmed if I can show that

\displaystyle \sum_{s=0}^a (-1)^s \binom{n}{s} = (-1)^a \binom{n-1}{a},

where 0 \le a \le n-1. We now prove this combinatorical identity.

Case 1. If a=0, then

\displaystyle \sum_{s=0}^0 (-1)^s \binom{n}{s} = (-1)^0 \binom{n}{0} = 1 = (-1)^0 \binom{n-1}{0}.

Case 2. If 1 \le a \le n-1, then

\displaystyle \sum_{s=0}^a (-1)^s \binom{n}{s} = (-1)^0 \binom{n}{0} + \sum_{s=1}^{a} (-1)^s \binom{n}{s}

\displaystyle = 1 + \sum_{s=1}^{a} (-1)^s \binom{n}{s}

\displaystyle = 1 + \sum_{s=1}^a (-1)^s \left[\binom{n-1}{s-1} + \binom{n-1}{s}\right]

\displaystyle = 1 + \sum_{s=1}^a (-1)^s \binom{n-1}{s-1} + \sum_{s=1}^{a-1} (-1)^s \binom{n-1}{s}

\displaystyle = 1 + \sum_{s=0}^{a-1} (-1)^{s+1} \binom{n-1}{s} + \sum_{s=1}^a (-1)^s \binom{n-1}{s}

\displaystyle = 1 + (-1)^1 \binom{n-1}{0} + \sum_{s=1}^{a-1} (-1)^{s+1} \binom{n-1}{s} + \sum_{s=1}^{a-1} (-1)^s \binom{n-1}{s} + (-1)^a \binom{n-1}{a}

\displaystyle = 1 -1 + (-1)^a \binom{n-1}{a} + \sum_{s=1}^{a-1} (-1)^{s+1} \binom{n-1}{s} + \sum_{s=1}^{a-1} (-1)^s \binom{n-1}{s}

\displaystyle = (-1)^a \binom{n-1}{a} +\sum_{s=1}^{a-1} \left[(-1)^{s+1} \binom{n-1}{s} + (-1)^s \binom{n-1}{s} \right]

\displaystyle = (-1)^a \binom{n-1}{a} + \sum_{s=1}^{a-1} (-1)^s \binom{n-1}{s} \left[ -1 + 1 \right]

\displaystyle = (-1)^a \binom{n-1}{a} + \sum_{s=1}^{a-1} 0

\displaystyle = (-1)^a \binom{n-1}{a}

In retrospect, I’m really surprised that I don’t remember seeing this identity before. The proof uses many of the techniques that we teach in discrete mathematics, including peeling off a term from either the start or end of a series, reindexing a sum, and the identity for constructing the terms in Pascal’s triangle in terms of the binomial coefficients. So I would’ve thought that I would’ve come across this, either in my own studies as a student or else in a textbook while preparing to teach discrete mathematics.

The observant reader may have noted that the “proof” over the past few posts isn’t really a proof since I started with the equation that I wanted to prove and then ended up with a true statement. While working backwards helped me see what was going on, this logic is ultimately flawed since it’s possible for false statements to imply true statements (another principle from discrete mathematics). I’ll clean this up in the next post.

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Published by John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

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