91Ó°ÊÓ

Let's denote the right side of the given theorem by \(f(x)\). So, we have: $$ f(x) = 1 + \sum_{n=1}^{\infty} \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n!} x^n $$

We can use the ratio test to check the convergence of the series. Consider the n-th term of the series: $$ a_n = \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n!} x^n $$ Now, compute the ratio between consecutive terms: $$ \frac{a_{n+1}}{a_n} = \frac{\alpha-n}{n+1} x $$ As x tends to one from the left, the absolute value of the ratio tends to \(\mid \alpha-n \mid / (n+1)\). This converges to 0 as n goes to infinity, which means that the series converges absolutely for -1 < x ≤ 1.

We differentiate f(x) with respect to x and get: $$ f'(x) = \sum_{n=1}^{\infty} \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{(n-1)!} x^{n-1} $$ Now, we will multiply both sides of the differential equation by (1 + x) and see if it holds true: $$ (1+x) f'(x) = \sum_{n=1}^{\infty} \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{(n-1)!} x^n + \sum_{n=1}^{\infty} \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{(n-1)!} x^{n-1} $$ The above expression simplifies to: $$ (1+x) f'(x) = \alpha \sum_{n=1}^{\infty} \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n!} x^n = \alpha f(x) $$ Hence, the given differential equation holds true.

If we define the inverse function as: $$ g(x) = (1-x)^{-\alpha} $$ We need to show that: $$ g(x) = \sum_{k=0}^{\infty} \frac{\Gamma(k+\alpha)}{k!\Gamma(\alpha)} x^k $$ Taking the derivative of g(x) and using the chain rule, we get: $$ g'(x) = \alpha (1-x)^{-\alpha - 1} $$ Now, multiplying both sides of the original differential equation by g'(x) and using the fact that g(x) is the inverse function of f(x), we get: $$ \alpha g(x) = g'(x) (1-x) $$ Finally, we let n = k and change the notation of the sum to match the given form. Thus, we have shown the desired expression for the inverse function as: $$ (1-x)^{-\alpha} = \sum_{k=0}^{\infty} \frac{\Gamma(k+\alpha)}{k!\Gamma(\alpha)} x^k $$ The proof is complete, and we have proven Newton's binomial theorem for the given conditions.