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I'm trying to develop the real function $f(x) = (1 + x)^{\alpha}$, $\alpha \in \mathbb{R}$, but I'm stuck in the middle.

What I did until now:

First, we have that the $f$ is defined for $\forall x \in \mathbb{R}$.

Then, I examined if the $f$ is infinitely diferentiable:

$$f'(x) = \alpha(1 + x)^{\alpha - 1},$$

$$f'(x) = \alpha(\alpha - 1)(1 + x)^{\alpha - 2},$$

$$f'''(x) = \alpha(\alpha - 1)(\alpha - 2)(1 + x)^{\alpha - 3},$$

$$\vdots$$

$$f^{(n)}(x) = \alpha(\alpha - 1)(\alpha - 2) \cdots (\alpha - n + 1)(1 + x)^{\alpha - n}.$$

So, $f$ is infinitely diferentiable on the whole set $\mathbb{R}$.

Now, I chose the point $x_0 = 0$ in which we are going to develop $f$.

Because, $f$ is infinitely diferentiable, that implies it's also continuous with all its derivatives up to the $n$-th order and in the $n$-th order, in the arbitrarily neighborhood $U$ of the point $x_0 = 0$, and it has derivative of the $n + 1$-st order in $x_0$. So, now we have that all conditions of the following theorem,

Checking my understanding of the process of developing function into power series,

are fulfilled. Now we have that we can write MacLaurien formula for the $f$ (because $x_0 = 0$):

$$f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + R_n(x) = $$

$$= 1 + \frac{\alpha x}{1!} + \frac{\alpha(\alpha - 1) x^2}{2!} + \frac{\alpha(\alpha - 1)(\alpha - 2) x^3}{3!} + \cdots + \frac{\alpha \cdots (\alpha - n + 1) x^n}{n!} + R_n(x) =$$

$$= 1 + \sum\limits_{n = 1}^{\infty} \frac{\alpha(\alpha - 1) \cdots (\alpha - n + 1) x^n}{n!} + R_n(x).$$

My question:

I think that is a mistake that I have $1$ in the last row of the last equation. I don't think that it can be added to the remainder $R_n(x)$.

Please, could you tell me where I made a mistake and how to fix it?

MathsLearner
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1 Answers1

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Tuck it into your series by making it $\sum_{n=0}^\infty$

Since your sum is effectively $$\sum_{n=1}^\infty \binom \alpha n x^n$$ setting $n=0$ gives the $1$ you took out.

Rhys Hughes
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  • So, $\n \choose k$ can use real numbers? I know it is defined using factorial and factorial is defined only in $\mathbb{N}$. I never saw this before. – MathsLearner Jun 27 '19 at 18:33
  • If I only use $\alpha \choose k$ as a mark for $\frac{\alpha(\alpha - 1)\cdots(\alpha - n + 1)}{n!}$ and not binomial coefficient, I think I would get $\alpha + 1$ for $n = 0$. – MathsLearner Jun 27 '19 at 18:41
  • They are the same thing: $$\alpha(\alpha-1)\ldots(\alpha-(n-1))=\alpha^{\underline{\alpha-n}}=\frac{\alpha!}{(\alpha-n)!}$$ Multiplying this by $\frac{1}{n!}$ gives $$\binom{\alpha}{n}=\frac{\alpha!}{n!(\alpha-n)!}$$ – Rhys Hughes Jun 27 '19 at 20:15
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    All I'm saying is that plugging $n=0$ into your sum would give $1$. Notice that you have a term being $$f^{(n)}(x)= (1+x)^{\alpha-n}\cdot\prod_{t=0}^{n-1}(\alpha-t)$$ When $n=0, x=0$, the product has no terms so you just get $(1+0)^\alpha=1$ – Rhys Hughes Jun 27 '19 at 20:25