I don't remember in which topic I found it but I know it was there. And I still have not find a proof of this nice approximation.
Let $x$ be a non perfect square number. If $y$ is the closer perfect square to $x$ such that $y < x$ then
$$\sqrt{x}\approx \sqrt{y}+\frac{x-y}{2\cdot \sqrt{y}}$$
And it gives at the maximum two correct decimals after the decimal point.
My first reaction to try to find from where this formula goes, was to expand it, I found $$2\sqrt{xy}\approx y +\sqrt{2y}\cdot(x-y)$$ and I tried to find a remarkable identity but I failed and I'm still stuck there. Also I don't know how should I prove the maximum of two correct digits after the comma.
If we look for example to the sqare root time of $1000$, $961$ is the closer perfect square which verifies the condition. Then we have $\sqrt{1000} \approx 31 + \frac{39}{2*31} = 31.62903…$ and with a calculator we have $\sqrt{1000} = 31.6227766017…$ which is quite good.
Any hints would be appreciate, thank you in advance.
Best Answer
To understand why this is a good approximation note that $(\sqrt{x})' = \frac{1}{2\sqrt{x}}$.
So the formula is of the form $f(x) = f(y) + f'(y)(x-y)$. See Taylor's theorem for more details.