The Incomplete Gamma and Confluent Hypergeometric Functions (Part 5)

Calculus 1 Minute

In the previous two posts, I show 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}.

When I first arrived at this equation, I noticed that this was really a property about the numbers in Pascal’s triangle: if we alternate adding and subtracting the binomial coefficients \displaystyle \binom{n}{s} on the nth row of Pascal’s triangle, the result is supposed to be an entry in the previous row (perhaps multiplied by -1).

For example, using the 10th row of Pascal’s triangle:

  • 1 - 10 = -9, which is negative the number to the “northeast” of 10.
  • 1 - 10 + 45 = 36, which is the number northeast of 45.
  • 1 - 10 + 45 - 120 = -84, which is negative the number northeast of 120.
  • 1 - 10 + 45 -120 + 210 = 126, which is the number northeast of 210.

And so on.

These computations suggest a proof of this identity. Any term in the interior of Pascal’s triangle can be found by adding the two terms immediately above it. Indeed, we can see this working out in the sequence of sums listed above.

I’ll write up the formal proof of the identity 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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