Compute Taylor Series

Question

For the question above I have done the first few Taylor series calculations; they are below. Now I am finding it difficult to transform these terms into a series. Every equation I come up with is inadequate. Any tips would be much appreciated.

Answer 1

So the easiest way I found to compute the Taylor series is to transform it to look like a binomial expansion. This will be $(1+(x-1))^{1/2}$, so the series is $$\sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}(x-1)^k$$

So, you need to find a way to write binomial coefficients for $n=1/2$.

Answer 2

The derivatives of $\sqrt{x}$ are:

$$\begin{align} x^{1/2}\\ \frac12x^{-1/2}\\ \frac12\left(-\frac12\right)x^{-3/2}\\ \frac12\left(-\frac12\right)\left(-\frac32\right)x^{-5/2}\\ \frac12\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)x^{-7/2}\\ \end{align}$$

There is a pattern. The $n$th derivative has $2^n$ in the denominator. Except for the $0$th derivative, they all have $(-1)^{n-1}$ for a factor. Again except for the $0$th derivative, the $n$th derivative has $$(2n-3)(2n-5)\cdots3\cdot1$$ in its numerator, which can be written in closed form as $\frac{(2n-2)!}{2^{n-1}(n-1)!}$. So excet for the $0$th derivative, $$f^{(n)}(1)=(-1)^{n-1}\frac{(2n-2)!}{2^{2n-1}(n-1)!}$$ And so $$T(x)=1+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2n-2)!}{2^{2n-1}(n-1)!n!}(x-1)^n$$ or, perhaps expressed in a simpler way $$T(x)=1+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2n)!}{4^{n}(n!)^2(2n-1)}(x-1)^n$$