Question

If \(\alpha\) is real and \(-10\).

Short answer

Question: Prove Newton's binomial theorem for a real number α and -1 < x < 1 by showing the convergence of the series, proving the given differential equation, and showing the desired expression for the inverse function. Answer: For a real number α and -1 < x < 1, we have shown that the series converges absolutely using the ratio test. We then proved the given differential equation holds true by differentiating the function f(x) and manipulating the equation. Finally, we showed the desired expression for the inverse function g(x) by taking its derivative and using the fact that g(x) is the inverse function of f(x). Thus, we have proven Newton's binomial theorem for the given conditions.

Step-by-step solution

Define the function and series

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

Prove convergence of the series

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.

Prove the given differential equation

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.

Show the desired expression for the inverse function

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.