# finding the taylor series for a function

so im trying to find the taylor series representation of $\frac{1}{\sqrt{x}}$ around $1$. i managed to find this so far: $$\sum_{n=0}^\infty \frac{(x-1)^n \cdot (-1)^n }{ 2^n \cdot n!}$$ the last thing im missing is in the numerator- but i cant figure out what exactly and how to represent it. Any ideas?

• addition: i basically need to find a way to represent the series 1,1,3,5,15,105,1575.... – dvd280 May 1 '17 at 18:00

By using the binomial theorem, one has, for $|t|<1$, $$(t+1)^{-1/2}=1-\tfrac12t+\tfrac12\tfrac32\cdot t^2-\tfrac12\tfrac32\tfrac52\cdot t^3+\dots+(-1)^n\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\cdot t^n+\cdots$$ or, by putting $t=x-1$, as $x \to 1$ \begin{align} \frac1{\sqrt{x}}&=\frac1{\sqrt{t+1}} \\\\&=1+\sum_{n=1}^\infty\:(-1)^n\cdot\frac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\cdot(x-1)^{n} \\\\&=1+\sum_{n=1}^\infty\:(-1)^n\cdot\frac{(2n)!}{2^{2n} (n!)^2}\cdot(x-1)^{n}. \end{align}
Why not just apply the binomial theorem? We note that $$\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{1+(x-1)}} = [1+(x-1)]^{-1/2} =\sum_{n=0}^\infty(x-1)^n{-1/2 \choose n}$$ Note that we can turn this into your form by using the definition of Binomial Coefficients
If you're just looking for some compact notation, let me introduce to you the double factorial: for $n$ odd, $$n!! = n \cdot (n-2) \cdot (n - 4) \cdots 1.$$ So, if you take $a_n = (2n - 1)!!$, you have $a_1 = 1, a_2 = 3, a_3 = 15, a_4 = 105, a_5 = 1575$, and so forth; this is exactly the sequence of odd coefficients that appears in the power series coefficients.