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I’m having trouble parsing the solution for the 1995 Putnam, question A2. The proof proceeds:

The easiest proof uses ``big-O'' notation and the fact that $(1+x)^{1/2} = 1 + x/2 + O(x^{2})$ for $|x|<1$. (Here $O(x^{2})$ means bounded by a constant times $x^{2}$.) Me: That’s all well and good, and I can arrive at that by the generalized binomial theorem.

The proof continues: \begin{align*} \sqrt{x+a}-\sqrt{x} &= x^{1/2}(\sqrt{1+a/x} - 1) \\ &= x^{1/2}(1 + a/2x + O(x^{-2})), \end{align*}

It would appear that’s an error; it would appear the proof writer dropped the minus 1.

But it continues, hence

$\sqrt{\sqrt{x+a} - \sqrt{x}} = x^{1/4} (a/4x + O(x^{-2}))$

Which I’m at a loss how to arrive at.

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  • $\begingroup$ For what it’s worth, you aren’t the first person to have asked about this. math.stackexchange.com/questions/2020837/… $\endgroup$ – Steve Kass Dec 30 '18 at 0:44
  • $\begingroup$ I am not sure if you have copied things properly. Why do we have $O(x^{2})$ in one place and $O(x^{-2})$ at others? $\endgroup$ – Kavi Rama Murthy Dec 30 '18 at 0:45
  • $\begingroup$ Note that on the right side of the first equation, x is in the denominator, whereas at the top, x is in the numerator. $\endgroup$ – 7l-l04 Dec 30 '18 at 0:46
  • $\begingroup$ @KaviRamaMurthy Not OP, but that is because the new "$x$" is now $a/x$, so when you substitute that in $O(x^2)$ becomes $O(a^2/x^2)$, which is the same as $O(x^{-2})$ $\endgroup$ – Noble Mushtak Dec 30 '18 at 0:46
  • $\begingroup$ You should show the question. I looked up the reference and it doesn't appear to be what you want. Also, where is the proof that you refer to? $\endgroup$ – herb steinberg Dec 30 '18 at 0:46
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OK, so the first step is clearly a mistake with them dropping the $-1$. However, I'm going to try to explain what they may have done, even though it doesn't make sense. First, let's start with their (albeit wrong) equation:

$$\sqrt{x+a}-\sqrt{x}=x^{1/2}(1+\frac{a}{2x}+O(x^{-2}))$$ Now, take the square root: $$\sqrt{\sqrt{x+a}-\sqrt{x}}=[x^{1/2}(1+\frac{a}{2x}+O(x^{-2}))]^{1/2}$$

Distribute the exponent: $$\sqrt{\sqrt{x+a}-\sqrt{x}}=x^{1/4}[(1+\frac{a}{2x}+O(x^{-2}))]^{1/2}$$

Now, the second expression on the right is in the form of $(1+z)^{1/2}$, so I will use the following formula:

$$(1+z)^{1/2}=1+\frac z 2+O(z^2)\rightarrow \\ (1+\frac a{2x}+O(x^{-2}))^{1/2}=1+\frac{a}{4x}+\frac{1}{2}O(x^{-2})+O(\frac{a^2}{4x^2})=1+\frac{a}{4x}+O(x^{-2})$$

Thus, we get:

$$\sqrt{\sqrt{x+a}-\sqrt{x}}=x^{1/4}(1+\frac{a}{4x}+O(x^{-2}))$$

Oh, and remember that $-1$ we dropped in the first step? Let's try adding that back in here, because...why not?

$$\sqrt{\sqrt{x+a}-\sqrt{x}}=x^{1/4}(1+\frac{a}{4x}+O(x^{-2})-1)$$ $$\sqrt{\sqrt{x+a}-\sqrt{x}}=x^{1/4}(\frac{a}{4x}+O(x^{-2}))$$ Thus, we get the same equation they did, and we only had to make two mathematical errors to get there!

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  • $\begingroup$ I like that you were able to decipher what the original proof writer most likely meant. Thank you for that. $\endgroup$ – 7l-l04 Dec 30 '18 at 0:56

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