Expected Difference Surds

Let $y=\sqrt{x-k}+k$ or sine real $k$ . We rearrange to get $y-k=\sqrt{x-k}$ . This is the equation of the $y=\sqrt{x}$ graph translated $k$ units right and up.

This means we can assume WLOG that $m=0$ . We thus have to solve the equation $\sqrt{x}=\sqrt{x-n}+n$

First off, note that as long as $n\in [-1,1]$ then there will be a solution $x_1$ . This can be easily seen by sliding the graph $\sqrt{x}$ along the line $y=x$ .

Then, we can proceed to solve the equation:

Squaring both sides: $x=x-n+n^2+2n\sqrt{x-n}$ $1-n=2\sqrt{x-n}$

Squaring both sides again: $n^2-2n+1=4(x-n)$ $x=\dfrac{n^2+2n+1}{4}=\dfrac{(n+1)^2}{4}$

Thus, $\boxed{x_1=\dfrac{(n+1)^2}{4}}$ . This means $P=\sqrt{x_1}=\dfrac{n+1}{2}$ .

We want to find the expected value of $P-x_1=\dfrac{n+1}{2}-\dfrac{(n+1)^2}{4}$ in the range $n\in [-1,1]$ . Thus we want to find the value of the constant $c$ such that $\int_{-1}^1\dfrac{n+1}{2}-\dfrac{(n+1)^2}{4}-c\text{ d}n=0$

It is easy to compute that $\int_{-1}^1\dfrac{n+1}{2}-\dfrac{(n+1)^2}{4}\text{ d}n=\dfrac{1}{3}$ and $\int_{-1}^1c\text{ d}n=2c$ so $\dfrac{1}{3}-2c=0\implies c=\dfrac{1}{6}$

Thus, the expected value of $P-x_1=\dfrac{1}{6}$ and our answer is $1+6=\boxed{7}$ .

$y=\frac{\sqrt{x-m}-\sqrt{x-n}}{n-m}=\frac{1}{\sqrt{x-m}+\sqrt{x-n}}$

The right-hand-side is continuous and decreasing on $x\geq max(n,m)$ , and tends to 0 as $x$ goes to infinity. So by Intermediate Value Theorem we have a unique solution for $y=1$ at $x=x_{1}$ provided $|n-m|<1$

At $x_{1}$ we have $\sqrt{x-m}+\sqrt{x-n}=1$ so adding the two expressions for $P$ gives $2P=2+m+n$ . Combining this with $x_{1}=(P-m)^{2}+m$ gives with very little pain that: $P-x_{1}=\frac{1-(n-m)^{2}}{4}$

Taking Expectations gives that $\frac{p}{q}=\frac{1-Var(n-m)}{4}$

It is not completely clear that everything does not depend on the distribution of $n$ and $m$ and technically there is no "Uniform" distribution on the unbounded region $|n-m|<1$ but assuming $(n-m)$ is uniformly distributed on $[-1,1]$ gives the variance as $\frac{1}{3}$ and the required value as $\boxed{\frac{1}{6}}$ .