# calculating taylor series and investigating a statement

So i have a function: $$f:(-1,1) \rightarrow \mathbb{R} , f(x)= \frac{1}{\sqrt{(1-x)^3}}$$

I need to calculate it's taylor series(Maclaurin to be exact, because it's to be centered around x=0). And use that to either prove either disprove the following inequality(i presume it will be of help, i could be wrong though): $$\sqrt{1-x} \sum_{n=0}^{\infty} \frac{x^n}{(2n)!!} \leq \frac{1}{1-x}$$

double factorial meaning, so there is no confusion

Any help would be appreciated, thank you in advance.

• The function is not defined for $x=1$. But you could write it as $(1-x)^{-3/2}$ to get the Maclaurin series – H. H. Rugh Aug 3 '16 at 8:09
• i made a mistake, sorry, will edit – MathIsTheWayOfLife Aug 3 '16 at 8:15
• @H.H.Rugh i made an edit, now it's defined correctly – MathIsTheWayOfLife Aug 3 '16 at 8:16

## 3 Answers

You may use the formula: $$(1-x)^p = 1 + \sum_{k\geq 1} \frac{p \times \cdots \times (p-k+1)}{k \times \cdots \times 1} (-x)^k$$ valid for all $p$ and $|x|<1$ (or better)

• will it help the inequality, or was it a long shot? – MathIsTheWayOfLife Aug 3 '16 at 8:29
• Inequality doesn't hold, as you can see from my answer. – alans Aug 3 '16 at 8:42
• As $(1-x)^{-3/2}=1-3/2x +...$, your sum starting with one should necessarily have the same linear term in order for the inequality to be valid for $x$ small. And this is not the case. – H. H. Rugh Aug 3 '16 at 9:10

Let $f(x) = (1-x)^{-3/2}$, then $$f'(x) = \frac{3}{2}(1-x)^{-5/2} \\ f''(x) = \frac{5}{2}\frac{3}{2}(1-x)^{-7/2} \\ f'''(x) = \frac{7}{2}\frac{5}{2}\frac{3}{2}(1-x)^{-9/2} \\$$ Now for $f^{(n)}$, we focus on the fraction in front : $$Numerator = \frac{(2n+1)!}{2^n \cdot n!} \\ Denominator = 2^n$$ Therefore $$f^{(n)}(x) = \frac{(2n+1)!}{4^n\cdot n!}(1-x)^{-n/2}$$ Now you can plug $x := 0$ and get your MacLaurin series !

Series expansion: $$(1-x)^{-\frac{3}{2}}=\sum_{n=0}^\infty (-x)^n\binom{-\frac{3}{2}}{n}$$ for $|x|<1$.

Since $$(2n)!!=2n(2n-2)\dots2=2^nn!,$$ inequality $$\sqrt{1-x} \sum_{n=0}^{\infty} \frac{x^n}{(2n)!!} \leq \frac{1}{1-x}$$ becomes $$\sqrt{1-x} \sum_{n=0}^{\infty} \frac{(\frac{x}{2})^n}{n!} \leq \frac{1}{1-x},$$ now use series expansion $e^t=\sum_{n=0}^\infty \frac{t^n}{n!}$, in order to get $$\sqrt{1-x} e^{\frac{x}{2}} \leq \frac{1}{1-x}.$$ But, this doesn't hold (check: for $x=-\frac{1}{2}$, it is equivalent with $e>9$, which doesn't hold), so we can conclude that initial inequality doesn't hold.