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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)

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

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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.

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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 $k$.

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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.

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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.

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