# 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?

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.

• What do you mean by: This expression uses $\sqrt {x+y}$ and $y$ as required? – Enthusiastic Engineer Mar 17 '15 at 21:20
• @EnthusiasticStudent The original question says that it is crucial that the factor $\sqrt{x+y}$ is included. Some of the suggested answers use the variable $x$ rather than $x+y$. Changing the variable to $w$ means that this part of the question is answered. Note also that $y$ is said to be small, so that is the best choice in the expansion to help identify small terms. – Mark Bennet Mar 17 '15 at 21:25
• Series adapted from WA. – lhf Mar 18 '15 at 2:51
• Not that three terms of this series is by far much worse than 3 terms of a taylor series as presented in other answers. I wonder why this was accepted when its more complex and less accurate? – ja72 Mar 18 '15 at 12:16

$$\sqrt{x+y}-\sqrt{x-y}=\frac{(\sqrt{x+y}-\sqrt{x-y})(\sqrt{x+y}+\sqrt{x-y})}{\sqrt{x+y}+\sqrt{x-y}}=\frac{2y}{\sqrt{x+y}+\sqrt{x-y}}\approx\frac{y}{\sqrt{x}}$$ Or, if you want a factor of $\sqrt{x+y}$, you can write it as $$\sqrt{x+y}-\sqrt{x-y}\approx \frac{y}{x}\sqrt{x+y}.$$

• You could have $\sqrt {x+y}$ in the denominator, since the question seems to want that expression retained. – Mark Bennet Mar 17 '15 at 19:15

You can convert $f(y) = \sqrt{x+y}=\sqrt{x-y}$ into a 3rd order taylor series

$$f(y) \approx f(0) + (y-0)\,\left. \dfrac{\partial f}{\partial y} \right|_{y=0} + \dfrac{(y-0)^2}{2}\,\left. \dfrac{\partial^2 f}{\partial y^2} \right|_{y=0} + \dfrac{(y-0)^2}{6}\,\left. \dfrac{\partial^3 f}{\partial y^3} \right|_{y=0}$$

$$f(y) \approx \dfrac{y}{\sqrt{x}} + \dfrac{y^3}{8 \sqrt{x^5}}$$

• not as fast as you were :-) – Math-fun Mar 17 '15 at 19:21
• I used a CAS for speed and accuracy. I see you went one up on terms. Well $f \approx \sqrt{x} \left(r+\frac{r^3}{8}+\frac{7r^5}{128}+\frac{33 r^7}{1024}+\frac{715 r^9}{32768} \right)$ with $r=y/x$. – ja72 Mar 17 '15 at 19:22
• what is a CAS? (if I may ask) – Math-fun Mar 17 '15 at 19:23
• Computer Algebra System : en.wikipedia.org/wiki/Computer_algebra_system – ja72 Mar 17 '15 at 19:25
• I see, many thanks. – Math-fun Mar 17 '15 at 19:27

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}

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