0

My Combinatorics notes mention the theorem Orthogonality of Binomial Coefficients which states: $$\sum_0^n{{n \choose k}{(-1)}^kp(k)} = 0$$ where $n \in \mathbb{N}$ and $p(k) \in \mathbb{R}_{n-1}[k]$

The proof given is to differentiate $$(1-t)^n = \sum_0^n{{n \choose k}{(-1)}^kt^k}$$ $r$ times, where $r<n$ and then evaluate at $t=1$, which is where I get stuck, since all terms in the sum where $k\geq r$, the inside of the sum is $\frac{k!}{(k-r)!}$ inside the sum, while for $k<r$, the $t^k$ term becomes $0$, so I'm left with $$0 = \sum_r^n{{n \choose k}{(-1)}^k\frac{k!}{(k-r)!}}$$ which would be enough to finish the proof if the sum indices ran from $0$ as I could just use linear combinations of this sum with different values of $r$ to get any polynomial of degree at most n-1

What I want to know is how am I meant to take this derivative to keep the sum index from $0$ to $n$, or is there another step after having taken the derivative to show the equality is true for all $n-1$ degree polynomials?

Comma
  • 53
  • Should it be n! rather than k! in the numerator ? – Simon Oct 16 '20 at 10:02
  • @Simon I don't believe so, the fraction comes from differentiating $t^k;$ r times, which is completely independent of n – Comma Oct 16 '20 at 10:09
  • I'm very sorry for my stupid comment ! I shouldn't comment on here when not fully awake. I got confused because one can cancel the k! with the k! in the denominator of the binomial coefficient. – Simon Oct 16 '20 at 10:15
  • What is $r$ in the initial statement? – Gary Oct 16 '20 at 10:54
  • @Gary sorry about the confusion with r, just a bit of formatting gone wrong. As for that question, the second answer does indeed use differentiating, but doesn't explain how we can take the $r$th derivative inside the sum, or at least not In a way I understand – Comma Oct 16 '20 at 12:02

1 Answers1

1

So my result of the differentiation was right and $$0 = \sum_r^n{{n \choose k}{(-1)}^k\frac{k!}{(k-r)!}}$$ is indeed the desired result. The missing step was that we define $p_r(k):=k(k-1)\ldots(k-r+1)$ which is slightly different from $\frac{k!}{(k-r)!}$. We can then say $$\sum_r^n{{n \choose k}{(-1)}^k\frac{k!}{(k-r)!}} = \sum_0^n{{n \choose k}{(-1)}^kp_r(k)=0}$$ Since $\{p_r(k)|r=0,1,2,\ldots,n-1\}$ forms a basis for $\mathbb{R}_{n-1}[k]$, we can write any $p(k) \in \mathbb{R}_{n-1}[k]$ as a linear combination of these $p_r(k)$'s to get the desired result

Comma
  • 53