# Taylor expansion of $\sqrt{n-k}$

I am reading a paper which casually assumes the asymptotic $$\sqrt{n-k} \simeq \sqrt{n}-\frac{k}{2\sqrt{n}}$$. This expression is what Wolfram calls Taylor expansion at infinity and from what I understand we work with $$\sqrt{x-k}=\sqrt{x}\cdot \sqrt{1-k/x}$$ and then we proceed on doing the Taylor expansion. My problem is that we also cannot do Taylor expansion near zero because said function is not differentiable there.
Another point is: do the Taylor expansion near point $$1$$ for $$\sqrt{x}$$ so that $$\sqrt{x}\sqrt{1-k/x} = \sqrt{x} - \frac{k}{2\sqrt{x}} - \frac{k^2}{8x^{3/2}}\frac{1}{\xi^{3/2}},$$ where $$\xi$$ as in Legendre residual form (between $$1-k/x$$ and $$1$$). We can then deduce that $$\sqrt{n-k}\leq \sqrt{n}-k/2\sqrt{n},$$ but what about the lower bound? Is there a way to fix an inequality relating the remainder and the second term of the above expansion?
My real question, is how to use the aforementioned fact in order to get a result of the form $$\sqrt{x-1/2}\geq \sqrt{x}\left(1-\frac{4}{x}\right).$$ The first two terms of $$\sqrt{x-1/2}$$ in the above expansion should be $$\sqrt{x}-\frac{1}{4\sqrt{x}},$$ so the remainder term should be $$\geq -\frac{15}{4\sqrt{x}}$$ or something.

• You want to show $\sqrt{x-\frac12}\ge\sqrt x(1-\frac4x)$? The left side must be positive, but for $x<4$, the right side is negative. Square both sides, and you'll see the left side is greater than the right side for $x>\frac{32}{15}$. QED – J. W. Tanner Jun 12 at 15:31
• When I wrote “positive”, I meant “non-negative” – J. W. Tanner Jun 12 at 16:10
• Which paper are you reading? – Jack Jun 18 at 13:54
• @Jack Both Ramirez, et al. in chapter $6$ and Edelman, Sutton (I believe in ch. 4) use the $\sqrt{n-k}$ estimate, while the explanation via Taylor can "almost surely" be inferred from Small Deviations for Beta Ensembles (Ledoux, Rider) (eq 2.8, 2.9) – separable ninja Jun 18 at 14:16
• So you "real" question is showing $\sqrt{n-k} \simeq \sqrt{n}-\frac{k}{2\sqrt{n}}$ for $k=1/4$, isn't it? That is, for large enough $n$, $$C_1\sqrt{n-k} \leq \sqrt{n}-\frac{k}{2\sqrt{n}}\leq C_2 \sqrt{n-k}$$ for some constants $C_1,C_2$. – Jack Jun 18 at 14:33

If you want to show that $$\sqrt{x-1/2}\geq \sqrt{x}\left(1-\frac{4}{x}\right),$$
(a) for $$\frac12\le x<4$$: $$\sqrt{x-1/2}\geq0\gt \sqrt{x}\left(1-\frac{4}{x}\right);$$
(b) for $$x>\frac{32}{15}, \frac{15}2>\frac{16}x,$$ so $$x-\frac12>x-8+\frac{16}x=x(1-\frac4x)^2,$$ so $$\sqrt{x-1/2}>\sqrt x(1-\frac4x);$$
so we have shown $$\sqrt{x-1/2}>\sqrt x\left(1-\frac4x\right)$$ for all $$x$$ where $$\sqrt{x-1/2}$$ is defined.
• That is indeed true. However, I still need to understand a way to do it via the Taylor expansion method, as I also want to apply it to $\sqrt{n-k}$. Thanks for your time. – separable ninja Jun 13 at 7:21