I just saw a post on instagram which said that

$$\sqrt{x}\approx \frac{x+y}{2\sqrt{y}}$$

I tried it out on a few values, and surprisingly, it came within 1 decimal point of the actual answer. Is there a reason for this or is it coincidental?

y is the closest perfect square to x

  • 4
    $\begingroup$ What is $y$?.... $\endgroup$ – Shubham Johri Mar 2 at 14:16
  • 1
    $\begingroup$ observe $x+y-2\sqrt{xy} = (\sqrt{x} - \sqrt{y})^2$ and thus the error is very small if y is close to x $\endgroup$ – Aditya Dwivedi Mar 2 at 14:17
  • $\begingroup$ @ShubhamJohri y is the closest perfect square, probably should add that in $\endgroup$ – Brien Lim Mar 4 at 15:15

This is because, for $x\approx y$:

$$\sqrt{xy}\approx \frac{x+y}{2}$$

This is a consequence of the AM-GM Inequality, which states that:


For all $x, y\geq 0$, with equality occurring if and only if $x=y$.

  • $\begingroup$ so does this means that it'll only work for relatively small values, and the bigger the number, the more inaccurate the answer will be? $\endgroup$ – Brien Lim Mar 4 at 15:19

Using Taylor series around $y=x$ (assuming them to be positive), we have $$\sqrt{x\,y}=x+\frac{y-x}{2}-\frac{(y-x)^2}{8 x}+\frac{(y-x)^3}{16 x^2}+O\left((y-x)^4\right)$$ and,sice it is an alternating series, you can build a lot of ineaqualities.


Google AM–GM inequality.

If we square any real number $z$, which can be expressed as a difference of two other real numbers $x$ and $y$, the result is always greater or equal $0$, this means $\color{blue}{z^2 \geq 0}$, assume $z = x-y$, it follows $$\color{blue}{z^2} = (x-y)^2 = x^2 \color{red}{-2xy} + y^2 = x^2 \color{red}{+ 2xy} + y^2 \color{red}{- 4xy} =(x+y)^2-4xy\color{blue}{\geq 0} $$

$$\Rightarrow (x+y)^2-4xy\geq 0 \Leftrightarrow(x+y)^2\geq 4xy \Rightarrow x+y \geq \sqrt{4xy} \Leftrightarrow \frac{x+y}{2\sqrt{y}} \geq \sqrt{x}$$

The result is:

$$\sqrt{x} \leq \frac{x+y}{2\sqrt{y}} $$

If $x = y$, then

$$\sqrt{x} =\frac{x+y}{2\sqrt{y}} $$

Example: $$\sqrt{4} =\frac{4+4}{2\sqrt{4}} = 2$$


Your approximation is for lower disadvantages, there's a better approximation for the Square root function, check here for it's derivation $$\sqrt{x} \approx \frac{ x+k^2+k}{2k+1}$$ Say $k^2+k = y$, $k^2+k-y = 0$ $$k = \frac{ -1 \pm \sqrt{1+4y} }{2}$$ $$\sqrt{x} = \approx \frac{x+y}{2k+1}$$ $$2k+1 = \sqrt{1+4y}$$ $$2k+1 = 2\sqrt{y+\frac{1}{4}}$$ $$2k+1 \approx 2\sqrt{y}$$ $$\sqrt{x} = \approx \frac{ x+y}{2\sqrt{y}}$$


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.