Advanced Calculus/Newton's general binomial theorem

Newton's generalized binomial theorem
We shall now describe a generalized binomial theorem, which uses generalized binomial coefficients.

Definition:

Let $$k \in \mathbb N$$ and $$r \in \mathbb R$$. Then we define
 * $$\binom{r}{k} := \frac{r (r - 1) \cdots (r - (k-1))}{k!}$$.

For these generalized binomial coefficients, we have the following formula, which we need for the proof of the general binomial theorem that is to follow:

Lemma:


 * $$(r-k) \binom{r}{k} + (r-(k-1)) \binom{r}{k-1} = r \binom{r}{k}$$.

Proof:


 * $$\begin{align}

(r-k) \binom{r}{k} + (r-(k-1)) \binom{r}{k-1} & = (r - k) \frac{\prod\limits_{j=0}^{k-1} (r - j)}{k!} + \frac{\prod\limits_{j=0}^{k-1} (r - j)}{(k-1)!} \\ & = \frac{(r-k+k) \prod\limits_{j=0}^{k-1} (r - j)}{k!} \\ & = r \binom{r}{k} \end{align}$$

Now we are ready for the theorem:

Theorem (generalized binomial theorem; Newton): If $$r \in \mathbb R$$ and $$|x| < |y|$$, then
 * $$(x + y)^r = \sum_{k=0}^\infty \binom{r}{k} x^k y^{r-k}$$,

where the latter series does converge.

Proof:

We begin with the special case $$y = 1$$. First we prove that whenever $$|x| < 1$$, the latter series converges; this we do by employing the quotient formula for the radius of convergence of power series. Since continuity of the absolute value allows us to compute the limit inside the absolute value first, we have
 * $$\lim_{k \to \infty} \frac{|a_k|}{|a_{k+1}|} = \lim_{k \to \infty} \left| \frac{k+1}{r-k} \right| = |-1| = 1$$;

thus we indeed have a convergence radius of $$1$$.

This convergence allows us to apply term-by-term differentiation inside the convergence area of $$|x| < 1$$, which gives
 * $$\frac{d}{dx} \sum_{k=0}^\infty \binom{r}{k} x^k = \sum_{k=1}^\infty (r-(k-1)) \binom{r}{k-1} x^{k-1}$$.

If we denote the function defined from the series we are considering by $$g(x)$$, we thus get
 * $$\begin{align}

(1 + x) \frac{d}{dx} g(x) & = \sum_{k=1}^\infty (r-(k-1)) \binom{r}{k-1} x^{k-1} + \sum_{k=1}^\infty (r-(k-1)) \binom{r}{k-1} x^k \\ & = r + \sum_{k=1}^\infty \left( (r-k) \binom{r}{k} + (r-(k-1)) \binom{r}{k-1} \right) x^k \\ & = r + r \sum_{k=1}^\infty \binom{r}{k} x^k \\ & = r g(x), \end{align}$$ where we used the preceding lemma in transposing from line 2 to 3. Defining now $$f(x) = (1 + x)^r$$, we get by the usual differentiation rules:
 * $$\frac{d}{dx} \left( \frac{g(x)}{f(x)} \right) = \frac{g'(x) f(x) - f'(x) g(x)}{f(x)^2} = \frac{r\frac{g(x)}{x+1}(1+x)^r - rg(x)(1 + x)^{r-1}}{f(x)^2} = 0$$

$$|x| < 1$$ implies $$f(x) \neq 0$$, which is why every fraction above is defined. We have thus proven that $$g/f$$ is constant and the special case $$y=1$$ follows from $$g(0) = 1 = f(0)$$ (as the empty product is defined to equal $$1$$).

For general $$x, y \in \mathbb R$$ with $$|x| < |y|$$, we have
 * $$\begin{align}

\frac{(x + y)^r}{y^r} & = (x/y + 1)^r \\ & = \sum_{k=0}^\infty \binom{r}{k} (x/y)^k; \end{align}$$ the convergence is ensured by $$|x/y| < 1$$ by assumption. All there is to do to obtain the claim is to multiply by $$y^r$$.