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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?

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  • $\begingroup$ addition: i basically need to find a way to represent the series 1,1,3,5,15,105,1575.... $\endgroup$ – dvd280 May 1 '17 at 18:00
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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 $x \to 1$ $$ \begin{align} \frac1{\sqrt{x}}&=\frac1{\sqrt{t+1}} \\\\&=1+\sum_{n=1}^\infty\:(-1)^n\cdot\frac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\cdot(x-1)^{n} \\\\&=1+\sum_{n=1}^\infty\:(-1)^n\cdot\frac{(2n)!}{2^{2n} (n!)^2}\cdot(x-1)^{n}. \end{align} $$

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    $\begingroup$ beat me by 28 seconds! Dang :) +1! $\endgroup$ – Brevan Ellefsen May 1 '17 at 18:01
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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

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  • $\begingroup$ Nice answer. (+1) $\endgroup$ – Olivier Oloa May 1 '17 at 18:04
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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.

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