Proof of the Binomial theorem; does ${n-1}\choose{n}$ make sense?

Question

I wanted to read a proof for the Binomial theorem, so I googled "proof of the binomial theorem".

My question is about the proof from the top link of that search. In the sixth line of the induction step, $$\sum_{k=0}^{n-1}{n-1 \choose k}x^{n-k}y^{k}$$ becomes $$\sum_{k=0}^{n}{n-1 \choose k}x^{n-k}y^{k}-{n-1 \choose n}x^{0}y^{n}$$ This is in order to get the indexes of the sums to match up, so the two sums can be combined.

So, I don't know what to make of ${n-1}\choose{n}$. WolframAlpha says it's zero. When using the formula ${{n}\choose{k}}=\frac{n!}{k!(n-k)!}$ with $n-1$ and $k=n$ I get $\frac{1}{n(-1)!}$

WolframAlpha says $(-1)!$ is complex infinity.

Also, ${n-1 \choose -1}$ is used, which I don't understand.

I'd be happy if someone could explain what is going on here.

Thanks.

Answer 1

Consider the following definitions/formulas for binomial coefficients $\binom{n}{m}$:

• The coefficient of $x^m$ in the expansion of $(1+x)^n$. Hence "binomial coefficient."
• The number of subsets of $\{1,\cdots,n\}$ of size $m$. Nice in combinatorial arguments.
• The factorial formula $n!/(m!(n-m)!)$. Note there is a canonical generalization of the factorial function known as the gamma function $\Gamma$.
• The formula $n(n-1)\cdots(n-(m-1))/m!$ (numerator and denominator are both products of $n$ terms). Efficient for hand computation, defines $\binom{x}{m}$ as a degree $m$ polynomial in $x$ so that it can be evaluated in any unital ring for which $m!$ is invertible. Indeed, allowing ourselves to evaluate at noninteger argument allows us to write down the Newton-binomial series expansion of $(1+x)^n$ when $n$ is nonintegral. These polynomials also form a basis for the subring of $\Bbb Q[x]$ of polynomials $p(x)$ for which $p(\Bbb Z)\subseteq\Bbb Z$.

All four definitions/formulas above would prescribe for us $\binom{n}{m}=0$ if $n<m$. The coefficient of $x^m$ in the expansion of $(1+x)^n$ would be $0$; there will be no subsets of $\{1,\cdots,n\}$ of size $m$, the gamma function has poles at nonpositive integers so its reciprocal should have zeros; and evaluating the polynomial form will result in zero. Thus the convention $\binom{n}{m}=0$ when $n<m$ harmonizes all four standard definitions/formulas we have for the binomial coefficient.

Moreoever, if one adopts this convention once and for all, then all future instances of sums that would involve such terms will not require breaking apart individual terms as David Wheeler mentions in the comments. Bookkeeping is made much more efficient in this way. A similar thing happens with $0^0$ - not in terms of limits and indeterminate forms, but in terms of how to evaluate the expression that appears in summation notation. Nobody frets about writing down $\sum_{n=0}^\infty a_nx^n$ instead of $a_0+\sum_{n=1}^\infty a_nx^n$ for instance.

The combinatorial interpretation is most aligned with the summation bookkeeping purpose, as one can often interpret addition and individual summands themselves combinatorially, as in the case of the binomial theorem. Indeed the same occurs with $0^0$: in general $Y^X$ denotes the set of functions $X\to Y$ so that $|Y^X|=|Y|^{|X|}$, and $\varnothing^\varnothing$ has one element, the empty function, so $0^0=1$.

All of these reasons imply this convention is the "morally correct" choice of definition.

Answer 2

You can define the binomial coefficient for an arbitrary (real) $\alpha$ and nonnegative integer k as ${\alpha \choose k} = \prod \limits_{j = 1}^k \frac{\alpha - j + 1}{j}$

If $\alpha$ is a positive integer smaller than k, one of the factors is 0 and the whole product becomes 0.

Answer 3

With the use of the beta function or the gamma function, you can also define the binomial coefficient $\binom{x}{y}$ for almost all $x,y\in\mathbb{C}$. That is, $$\binom{x}{y}=\frac{1}{(x+1)\cdot B(y+1,x-y+1)}=\frac{\Gamma(x+1)}{\Gamma(x-y+1)\cdot \Gamma(y+1)}\,.$$ In particular, if $x \neq -1$, $$\binom{x}{x+1}=\frac{1}{(x+1)\cdot B(x+2,0)}=\frac{\Gamma(x+1)}{\Gamma(0)\cdot\Gamma(x+2)}=0\,,$$ whereas $\binom{-1}{0}=1$.

Warning: The equality above should be taken in the limit sense if we encounter singularities. There are pairs $(x,y) \in \mathbb{C}\times\mathbb{C}$ such that the binomial coefficient $\binom{x}{y}$ is undefined (as the limit does not exist), namely, $(x,y)\in\mathbb{Z}_{<0}\times\mathbb{Z}_{<0}$.

Answer 4

Let's define $\binom mn = \dfrac{m(m-1)(m-2) \cdots (m-(n-1))}{n!}$

Note that this definition works even when $m$ is a real number and it can be shown that $\displaystyle(1+x)^m = \sum_{i=0}^{\infty}\binom mi x^i$

Now , for example, $\binom 46 = \dfrac{4 \cdot 3 \cdot 2 \cdot 1 \cdot 0 \cdot(-1)}{6!} = 0$

And, in general, $\binom mn = 0 $ if $m \lt n$.

This makes sense because, then, $ \sum_{i=0}^{\infty}\binom mi x^i$ becomes a finite series when $m$ is a positive integer and an infinite series when $m$ is a not an integer.

We can justify this definition.

Let $f(x) = (1+x)^m$. Then $f(0) = 1$ and $m f(x) = (1+x)f'(x)$.

Let $g(x) = 1 + \alpha_1 x + \alpha_2 x^2 + \alpha_3 x^3 + \cdots$

Then $g(0) = 1$.

$m g(x) = (1+x)g'(x) \implies$

$m + m\alpha_1 x + m\alpha_2 x^2 + m\alpha_3 x^3 + \cdots = (1+x)(\alpha_1 + 2\alpha_2 x + 3\alpha_3 x^2 + \cdots) \implies$

$\alpha_1 = m$
$\alpha_1 + 2\alpha_2 = m\alpha_1$
$2\alpha_2 + 3\alpha_3 = m\alpha_2$
$\cdots$
$i\alpha_i + (i+1)\alpha_{i+1} = m\alpha_i$
$\cdots \implies$

$\alpha_1 = m$
$\alpha_{i+1} = \dfrac{m-i}{i+1}\alpha_i \implies$

$\alpha_i = \dfrac{m(m-1)(m-2) \cdots (m-(n-1))}{n!}$

Since $f(0)=0$ and $f'(x) = g'(x)$, we conclude that $f(x)=g(x)$