Question

Use binomial series to expand the function as a power series. State the radius of convergence.

Answer 1

\[\sqrt[4]{1-x}\]

Answer 2

recall the binomial formula:$$(a+b)^n=\sum_{k=0}^n\binom{n}ka^kb^{n-k}$$so$$(1+x)^n=\sum_{k=0}^n\binom{n}kx^k=\sum_{k=0}^n\frac{n!}{k!(n-k)!}x^k=\sum_{k=0}^n\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}x^k$$it turns out this generalizes for arbitrary \(n\):$$(1+x)^n=\sum_{k=0}^\infty\frac{n(n-1)(n-2)\dots(n-k+1)}{k!}x^k$$this is known as the binomial series and it is a special case of the generalized binomial theorem

Answer 3

in this case we have $$\sqrt[4]{1-x}=(1+(-x))^{1/4}$$so consider:$$(1+(-x))^{1/4}=\sum_{k=0}^\infty\frac{1/4\cdot(1/4-1)\cdot(1/4-2)\cdots(1/4-k+1)}{k!}(-x)^k$$

Answer 4

$$\begin{align*}\sqrt[4]{1-x}&=\sum_{k=0}^\infty(-1)^k\prod_{n=0}^k\left(\frac{1/4-n}n\right)x^k\\&=\sum_{k=0}^\infty(-1)^k\prod_{n=0}^k(-1)\left(1-\frac1{4n}\right)x^k\\&=\sum_{k=0}^\infty(-1)^{2k+1}\prod_{k=0}^n\left(1-\frac1{4n}\right)x^k\\&=-\sum_{k=0}^\infty\prod_{n=0}^k\left(1-\frac1{4n}\right)x^k\end{align*}$$

Answer 5

to determine the radius of convergence, consider the ratio test:$$\lim_{k\to\infty}\left|\frac{-x^{k+1}\prod_{n=0}^{k+1}(1-1/(4n))}{-x^k\prod_{n=0}^k(1-1/(4n))}\right|=\lim_{k\to\infty}\left(1-\frac1{4(k+1)}\right)|x|=|x|<1$$