# How is the approximation $\sqrt{k+1} - \sqrt{k}\approx \frac{1}{2\sqrt{k}}$ done?

How is the approximation $$\sqrt{k+1} - \sqrt{k}\approx \frac{1}{2\sqrt{k}}$$ done? (suppose $$k$$ is an integer)

Is this a Taylor expansion?

(P.S. I asked this on Physics Stack Exchange because I encountered this in a physics textbook and I think approximations like this are probably only done in physics/engineering instead of mathematics)

• Would Mathematics be a better home for this question? Oct 9, 2021 at 8:40
• 1. Yes, it is just the first order approximation for $\sqrt{x}$. 2. Qmechanic is correct Oct 9, 2021 at 8:43

This approximation is valid for large $$k\gg 1$$.

Consider the small quantity $$\frac{1}{k}\ll 1$$. Then, by the binomial expansion: $$\sqrt{1+\frac{1}{k}}-1=\left(1+\frac{1}{2k}+\mathcal{O}\left(\frac{1}{k}\right)^2\right)-1\approx\frac{1}{2k}$$ Multiplying both sides by $$\sqrt{k}$$, we have the desired result.

One can see this by simply rationalising the numerator: $$(\sqrt{k+1}-\sqrt{k})\times\dfrac{\sqrt{k+1}+\sqrt{k}}{\sqrt{k+1}+\sqrt{k}}=\dfrac{1}{\sqrt{k+1}+\sqrt{k}}~ \approx \dfrac{1}{2\sqrt{k}}$$

Since $$\sqrt{k+1} \approx \sqrt{k}$$ for sufficiently large $$k$$.

You can also use derivatives to obtain this approximation: $$f(x) = \sqrt x \rightarrow f'(x) = \frac{1}{2\sqrt x}$$ $$f(x+\delta x) - f(x) \approx f'(x).\delta x$$ $$f(x+1) - f(x) \approx \frac{1}{2\sqrt x}.1$$ $$\sqrt {x+1} - \sqrt x \approx \frac{1}{2\sqrt x}$$

provided $$x \gg 1$$

Hope this helps.

If you apply Lagrange's theorem to $$f(x)=\sqrt{x}$$ in the interval $$[k,k+1]$$, you get $$f(k+1)-f(k)=f'(\xi) (k+1-k),\quad \xi \in (k,k+1)$$

i.e.,

$$\sqrt{k+1}-\sqrt{k} = \frac{1}{2\sqrt{\xi}},\quad \xi \in (k,k+1)$$

From this relation, you deduce that $$\frac{1}{2\sqrt{k+1}} \leq \sqrt{k+1}-\sqrt{k}\leq \frac{1}{2\sqrt{k}}.$$ This inequality holds for all $$k\ge 1$$, large or small.