1
$\begingroup$

I've been reading Richard Stanley's proof of the upper bound conjecture, but I haven't been able to make sense of one statement he makes near the start while defining the h-vector of a simplicial complex. Section in question

He defines a function H

$$ H(m) = \begin{cases} 1 & m = 0 \newline \sum_{i=0}^{d-1} f_i {m-1 \choose i} & m > 0 \\ \end{cases} $$

And then claims the following equality to be elementary for some set of $h_i$s

$$(1-x)^d\sum_{m=0}^{\infty} H(m)x^m = h_0 + h_1x + \ldots + h_dx^d$$

It's unclear to me why this infinite sum would end up equal to a polynomial that only goes up to degree d when summing $x^m$ for $m$ going to infinity.

Here's a link to the full paper if more context is needed. https://math.mit.edu/~rstan/pubs/pubfiles/27.pdf

$\endgroup$

1 Answer 1

2
$\begingroup$

This is in general true for polynomials.

If $p(n)$ is a polynomial of degree $d-1,$ we can write it as a linear combination of the polynomials $g_0(n)=1,$ $g_{i+1}(n)=(n+i+1)g_i(n).$ So $$p(n)=\sum_{i=0}^{d-1} c_ig_i(n)$$

Now, for fixed $i,$ $$\sum_{n=0}^\infty g_{i}(n)x^n=\left(\frac{1}{1-x}\right)^{(i)}=\frac{i!}{(1-x)^{i+1}}$$

So:

$$(1-x)^d\sum_{n=0}^\infty p(n)x^n=\sum_{i=0}^{d-1}c_i\cdot i!\cdot (1-x)^{d-i-1},$$ which gives a polynomial of degree $d-1.$

If $p$ is an integer polynomial, we also get the $c_i$ are integers.

So you really only need to know how to show $H(n)$ is a polynomial of $n$ of degree $d-1.$


Example

So, for example, if $p(n)=n^2,$ then $$p(n)=1\cdot (n+1)(n+2)-3\cdot (n+1)+1=g_2(n)-3g_1(n)+g_0(n)$$ so:

$$\sum_{n =0}^\infty n^2x^n=1\cdot\frac{2!}{(1-x)^3}-3\cdot\frac{1!}{(1-x)^2}+\frac{0!}{1-x}$$

Multiplying both sides by $(1-x)^3,$ we get:

$$(1-x)^3\sum_{n=0}^\infty n^2x^n =2-3(1-x)+(1-x)^2=x+x^2$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .