# Find the Mclaurin series of $f(x)=(1+x)^\alpha$ then find the radius of the convergence by ratio test.

Find the Mclaurin series of $f(x)=(1+x)^\alpha$ then find the radius of the convergence by ratio test. I'm told that i should not assume n is an integer.

I think to find the Mclaurin series i need to differentiate the function a couple of times. And then find the Taylor series about 0.

$$f(x) = (1+x)^\alpha \quad f(0) = 1^\alpha = 1$$ $$f'(x) = \alpha(1+x)^{\alpha-1} \quad f'(0) = \alpha1^{\alpha-1}=\alpha$$ $$f''(x) = \alpha(\alpha-1)(1+x)^{\alpha-2} \quad f''(0) = \alpha(\alpha-1)$$ $$f'''(x) = \alpha(\alpha-1)(\alpha-2)(1+x)^{\alpha-3} \quad f'''(0) = \alpha(\alpha-1)(\alpha-2)$$ $$f^n(x) = \alpha(\alpha-1)(\alpha-2)...(\alpha-n)(1+x)^{\alpha-n} \quad f^n(0) = \alpha(\alpha-1)(\alpha-2)....(\alpha-(n-1))$$

So Mclaurin series is... $\alpha\frac{1}{0!}x^0+\alpha(\alpha-1)\frac{1}{1!}x^1+\alpha(\alpha-1)(\alpha-2)\frac{1}{2!}x^2...+\alpha(\alpha-1)(\alpha-2)....(\alpha-k)\frac{1}{k!}x^k$

Hence the power series is $$\sum^\infty _{k=o }(\alpha-k)\frac{1}{k!}x^k$$

I've kinda forgotten how to do ratio tests since it's not covered in this class.

I think it goes $$C_n = (x-\alpha)^n$$ if.... $$lim_{n \to \infty}\frac{|a_{n+1}|}{|a_n|} \quad \text{ is infinite, then R=0}$$ $$lim_{n \to \infty}\frac{|a_{n+1}|}{|a_n|} \quad \text{ =0, then R=\infty}$$ $$lim_{n \to \infty}\frac{|a_{n+1}|}{|a_n|} \quad \text{ =K|x- \alpha |, where K is finite and nonzero, then R = 1/K}$$

So by this logic $$lim_{n \to \infty}\frac{|(\alpha-(n+1))\frac{1}{(n+1)!}x^{n+1}|}{|(\alpha-n)\frac{1}{n!}x^n|} = lim_{n \to \infty}\frac{|(\alpha-(n+1))\frac{n!}{(n+1)(n)!}x^{n+1}|}{|(\alpha-n)|} = lim_{n \to \infty}\frac{|(\alpha-(n+1))\frac{1}{(n+1)}x|}{|(\alpha-n)|} = |x|lim_{n \to \infty}\frac{|(\alpha-(n+1))\frac{1}{(n+1)}|}{|(\alpha-n)|}$$

Not sure if i'm going the right direct and with this limit. I'm guessing because $\frac{1}{(n+1)}$ factor the limit equals 0 and hence r = infinity.

Find the Mclaurin series of $$f(x)=(1+x)^\alpha$$ then find the radius of the convergence by ratio test. I'm told that i should not assume n is an integer.

to find the Mclaurin series differentiate the function a couple of times. And then find the Taylor series about 0.

$$f(x) = (1+x)^\alpha \quad f(0) = 1^\alpha = 1$$ $$f'(x) = \alpha(1+x)^{\alpha-1} \quad f'(0) = \alpha1^{\alpha-1}=\alpha$$ $$f''(x) = \alpha(\alpha-1)(1+x)^{\alpha-2} \quad f''(0) = \alpha(\alpha-1)$$ $$f'''(x) = \alpha(\alpha-1)(\alpha-2)(1+x)^{\alpha-3} \quad f'''(0) = \alpha(\alpha-1)(\alpha-2)$$ $$f^n(x) = \alpha(\alpha-1)(\alpha-2)...(\alpha-n)(1+x)^{\alpha-n} \quad f^n(0) = \alpha(\alpha-1)(\alpha-2)....(\alpha-(n-1))$$

So Mclaurin series is... $$1+\alpha\frac{1}{1!}x^1+\alpha(\alpha-1)\frac{1}{2!}x^2+...+\alpha(\alpha-1)(\alpha-2)....(\alpha-k+1)\frac{1}{k!}x^k$$

Hence the power series is $$r = 1+\sum^\infty _{k=1}\alpha(\alpha-1)(\alpha-2)....(\alpha-k+1)\frac{1}{k!}x^k$$

Radius of the convergence is.... $$lim_{n \to \infty}\frac{|a_{n}|}{|a_n+1|}<1 \quad \text{ excluding x}$$

So by this logic $$lim_{k \to \infty}\frac{|\alpha(\alpha-1)(\alpha-2)....(\alpha-k)\frac{1}{k!}|}{|(\alpha(\alpha-1)(\alpha-2)....(\alpha-k)(\alpha-k-1)\frac{1}{(k+1)!}|}$$

$$lim_{k \to \infty}\frac{|(k+1)|}{|((\alpha-k-1)|} \quad \text{cancelling like terms and redistributing the fraction}$$ $$lim_{k \to \infty}\frac{|(1+1/k)|}{|((\alpha/k-1-1/k)|} \quad \text{simply dividing everything by k and taking the limit}$$

$$\frac{|(1)|}{|((-1)|} = 1$$

Hence radius of convergence is 1. Not sure of the significance of the less than 1 definition.