<s>[INST] Bounding the difference between square roots I want to compute the value of $\frac{1}{\sqrt{a + b + c}}$.  Say I can observe a and b, but not c.  Instead, I can observe d which is a good approximation for c in the sense that $P( |c-d| \leq 0.001 )$ is large (say 95%), and both c and d are known to have $|c| \leq 1, |d| \leq 1$ so a difference of 0.001 is actually small.
I want to argue that $\frac{1}{\sqrt{a + b + d}}$ is a good approximation because d is a good approximation for c.  Is there anything I can say about the difference
$\frac{1}{\sqrt{a + b + c}} - \frac{1}{\sqrt{a + b + d}}$?
Maybe I could say something like $P(\frac{1}{\sqrt{a + b + c}} - \frac{1}{\sqrt{a + b + d}} \leq 0.001) = x?$  I'm worried that being under a square root might mess things up.
Would I need to find out the exact distribution of $c - d$, or anything else before I can make such claims?
 [/INST] Use a Taylor series (or equivalently the Binomial Theorem) to expand around $c$.  This is valid provided $|d-c| \lt |a+b+c|$:
$$\eqalign{
&\frac{1}{\sqrt{a+b+c}} - \frac{1}{\sqrt{a+b+d}}\\
&= (a+b+c)^{-1/2} - (a+b+c + (d-c))^{-1/2} \\
&= (a+b+c)^{-1/2} - (a+b+c)^{-1/2}\left(1 + \frac{d-c}{a+b+c}\right)^{-1/2} \\
&= (a+b+c)^{-1/2} - (a+b+c)^{-1/2}\sum_{j=1}^{\infty}\binom{-1/2}{j}\left(\frac{d-c}{a+b+c}\right) ^j\\
&= \frac{1}{2}(a+b+c)^{-3/2}(d-c) - \frac{3}{8}(a+b+c)^{-5/2}O(d-c)^2
}
$$
The difference therefore is approximately  $\frac{1}{2}(a+b+c)^{-3/2}$ times $(d-c)$ and the error is (a) negative (because this is an alternating series when $d-c$ is positive and has all negative terms when $d-c$ is negative), (b) proportional to  $\frac{3}{8}(a+b+c)^{-5/2}$, and (c) of second order in $d-c$.  That should be enough to complete your analysis.  (This leads essentially to the delta method.)
</s>