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Is it true that Pascal's Rule holds for binomial coefficients with a negative upper index?

With $n = -1$ and $k = 3$, for example, it appears not to hold.

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3 Answers 3

up vote 2 down vote accepted

Substituting your values of $n,k$ into $${n-1 \choose k }+{n-1 \choose k-1} = {n \choose k}$$ we get $${-2 \choose 3 }+{-2 \choose 2} = {-1 \choose 3}$$ but this is simply $$-4+3=-1$$ which of course is true.

To see why this works, consider that there's nothing stopping us from putting negative numbers in there. Expanding out the proof of the identity everything will work the same. In fact, we can use it to extend Pascal's triangle into negative rows by formally rearranging Pascal's rule into $${n-1 \choose m}={n \choose m }-{n-1 \choose m-1}.$$ (If we really want to get crazy, we can even extend Pascal's rule to $q$-binomials!)

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The binomial coefficients may actually be viewed as rational-coefficient polynomials:

$${X\choose k}:=\frac{X(X-1)\cdots\big(X-(k-1)\big)}{k\cdot(k-1)\cdots3\cdot2\cdot1}.$$

This is useful/interesting for a number of reasons. These polynomials are a basis for the space of all rational-coefficient polynomials which take integer values at integer arguments. The binomial theorem generalizes to powers that need not be natural numbers:

$$(1+T)^s=\sum_{k\ge0}{s\choose k}T^k.$$

This allows us to extend the notion of taking-powers to complete topological rings of characteristic zero (which is to say, we can do it in abstract algebraic structures equipped with limits where a priori it is not defined), although there are caveats as to regions of convergence. It may also be useful in the contexts of generating functions and special functions.

Anyway, Pascal's rule holds on the more fundamental level of polynomials:

$${X-1\choose k}+{X-1\choose k-1}=\frac{\color{Green}{(X-1)\cdots}\color{Blue}{\big(X-1-(k-1)\big)}}{k!}+\frac{\color{Purple}{k}\color{Green}{(X-1)\cdots\big((X-1)-(k-2)\big)}}{k!}$$

$$=\frac{\color{Green}{(X-1)\cdots (X-(k-1))\,}[\color{Blue}{(X-k)}+\color{Purple}{k}]}{k!} ={X\choose k}.$$

Since the identity holds for polynomials, it holds for the polynomials evaluated at any arguments; they can be negative, fractional, irrational, complex, $p$-adic, whatever floats your boat.

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+1 Nice answer. While what you say is entirely correct, the final sentence might be read over-optimistically by some. One should not forget that we have rational coefficients here, so we must stay in characteristic $0$; one cannot evaluate binomial coefficients in $\Bbb Z/n\Bbb Z$ (even if $n$ is prime). One might naively think so since at integer arguments binomial coefficients are always integral. –  Marc van Leeuwen Apr 5 '13 at 5:11

$$\dbinom{n}r = \dfrac{\Gamma(n+1)}{\Gamma(r+1) \Gamma(n-r+1)}$$ wherever the quantity on the right hand side is defined. Now make use of the fact that $\Gamma(z+1) = z \Gamma(z)$. \begin{align} \dbinom{n}{r} + \dbinom{n}{r+1} & = \dfrac{\Gamma(n+1)}{\Gamma(r+1) \Gamma(n-r+1)} + \dfrac{\Gamma(n+1)}{\Gamma(r+2) \Gamma(n-r)}\\ & = \dfrac{\Gamma(n+1)}{(n-r) \Gamma(r+1) \Gamma(n-r)} + \dfrac{\Gamma(n+1)}{(r+1) \Gamma(r+1) \Gamma(n-r)}\\ & = \dfrac{\Gamma(n+1)}{\Gamma(r+1) \Gamma(n-r)} \left(\dfrac1{n-r} + \dfrac1{r+1}\right)\\ & = \dfrac{\Gamma(n+1)}{\Gamma(r+1) \Gamma(n-r)} \left(\dfrac{n-r+r+1}{(n-r)(r+1)}\right)\\ & = \dfrac{\Gamma(n+1)}{\Gamma(r+1) \Gamma(n-r)} \left(\dfrac{n+1}{(n-r)(r+1)}\right)\\ & = \dfrac{\Gamma(n+2)}{\Gamma(r+2) \Gamma(n-r+1)}\\ & = \dbinom{n+1}{r+1} \end{align}

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