Approximation of $\sqrt{ x + y } - \sqrt{ x - y }$ I've been struggling to try and find a way to approximate the function:
$\sqrt{ x + y } - \sqrt{ x - y }$
I should mention that $y$ is positive and a small number, so that $0<y<<1$.
What I'm hoping for is to approximate this in such a way that, we have roughly:
$\sqrt{ x + y } - \sqrt{ x - y } \approx (1-y)\sqrt{ x + y }$
There may be some numerical factor in front of this. This could very well be absurd, I'm wondering if this can be done at all. It is crucial that I have this factor $\sqrt{ x + y }$ in the approximation.
I've thought about defining a function $\ f(r)=\sqrt{r}$. Then I could write:
$f(x+y)-f(x-y)$ = $f(x+y)-f(x+y-2y)$
i've tried taking a Taylor expansion but my result isn't working out. Does anyone have some advice?
 A: You can convert $f(y) = \sqrt{x+y}=\sqrt{x-y}$ into a 3rd order taylor series
$$f(y) \approx f(0) + (y-0)\,\left. \dfrac{\partial f}{\partial y} \right|_{y=0} + 
\dfrac{(y-0)^2}{2}\,\left. \dfrac{\partial^2 f}{\partial y^2} \right|_{y=0} + 
\dfrac{(y-0)^2}{6}\,\left. \dfrac{\partial^3 f}{\partial y^3} \right|_{y=0}$$
$$ f(y) \approx \dfrac{y}{\sqrt{x}} + \dfrac{y^3}{8 \sqrt{x^5}} $$
A: Use a Taylor's series expansion for $f(y)=\sqrt{ x + y } - \sqrt{ x - y }$:
\begin{align}
f(y)&=\frac{y}{\sqrt x}+\frac{y^3}{8x^{5/2}}+\frac{7y^5}{128x^{9/2}}+O(y^7)\\
\end{align}
A: Try this one for size:
$$ f(y) \approx \sqrt{x} \left( \arcsin\left(\frac{y}{x}\right) \right) $$
or 
$$ f(y) \approx \sqrt{x} \left( \arcsin\left(\frac{y}{x}\right)-\frac{1}{24}\left(\arcsin\left(\frac{y}{x}\right)\right)^3 \right) $$
The last one is almost an exact fit throughtout the range of $y$. I got it by setting $y=x \sin(r)$ and perfoming a taylor expansion on $$f(r)=\sqrt{x} \left( \sqrt{1+\sin(r)}-\sqrt{1-\sin(r)} \right) \\ = \sqrt{x} \left(r-\frac{r^3}{24}\right)$$
See the comparison for yourself at Wolfram Alpha
Alternative link to Wolfram, http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ef6ds3l1c74

A: $$\sqrt{x+y}-\sqrt{x-y}=\frac{(\sqrt{x+y}-\sqrt{x-y})(\sqrt{x+y}+\sqrt{x-y})}{\sqrt{x+y}+\sqrt{x-y}}=\frac{2y}{\sqrt{x+y}+\sqrt{x-y}}\approx\frac{y}{\sqrt{x}}$$
Or, if you want a factor of $\sqrt{x+y}$, you can write it as
$$\sqrt{x+y}-\sqrt{x-y}\approx \frac{y}{x}\sqrt{x+y}.$$
A: Let $z=x+y$ and $w=\sqrt{z}=\sqrt{ x + y }$. Then 
$$
\def\F#1/#2;{\frac{#1}{#2}}
\sqrt{ x + y } - \sqrt{ x - y }
=
\sqrt{z } - \sqrt{ z - 2y }
=
\F y/w;+
\F y^2/2 w^3;+
\F y^3/2 w^5;+
\F 5 y^4/8 w^7;+
\F 7 y^5/8 w^9;+
\F 21 y^6/16 w^{11};+O(y^7)
$$
This expression uses $\sqrt{ x + y }$ and $y$ as required.
