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.

  • $\begingroup$ What do you mean by: This expression uses $\sqrt {x+y}$ and $y$ as required? $\endgroup$ – Enthusiastic Engineer Mar 17 '15 at 21:20
  • $\begingroup$ @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. $\endgroup$ – Mark Bennet Mar 17 '15 at 21:25
  • $\begingroup$ Series adapted from WA. $\endgroup$ – lhf Mar 18 '15 at 2:51
  • 1
    $\begingroup$ 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? $\endgroup$ – 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 $$\sqrt{x+y}-\sqrt{x-y}\approx \frac{y}{x}\sqrt{x+y}.$$

  • $\begingroup$ You could have $\sqrt {x+y}$ in the denominator, since the question seems to want that expression retained. $\endgroup$ – 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

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

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

  • $\begingroup$ not as fast as you were :-) $\endgroup$ – Math-fun Mar 17 '15 at 19:21
  • $\begingroup$ 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 $r=y/x$. $\endgroup$ – ja72 Mar 17 '15 at 19:22
  • $\begingroup$ what is a CAS? (if I may ask) $\endgroup$ – Math-fun Mar 17 '15 at 19:23
  • $\begingroup$ Computer Algebra System : en.wikipedia.org/wiki/Computer_algebra_system $\endgroup$ – ja72 Mar 17 '15 at 19:25
  • $\begingroup$ I see, many thanks. $\endgroup$ – 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:

$$ f(y) \approx \sqrt{x} \left( \arcsin\left(\frac{y}{x}\right) \right) $$


$$ f(y) \approx \sqrt{x} \left( \arcsin\left(\frac{y}{x}\right)-\frac{1}{24}\left(\arcsin\left(\frac{y}{x}\right)\right)^3 \right) $$

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



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.