Real Number Pairs $(p,\,q)$ Satisfying Inequality

[anemone]

Find all pairs of real numbers $(p,\,q)$ such that the inequality $|\sqrt{1-x^2}-px-q|\le \dfrac{\sqrt{2}-1}{2}$ holds for every $x\in [0,\,1]$.

 

[lfdahl]

Spoiler: My attempt

Let $f(x) = \sqrt{1-x^2}-(px+q)$ and $\delta_0 = \frac{\sqrt{2}-1}{2} \approx 0,2071$.

With this notation, the given problem reads: $|f(x)| \leq \delta_0, \;\; x \in \left [ 0;1 \right ]$.

As seen from the expression, the problem can be interpreted graphically as a constraint

on the absolute difference between the two functions: $\sqrt{1-x^2}$ and $px + q$.

Note, that the pair $(x,y) = (x, \sqrt{1-x^2})$, for $x \in \left [ 0;1 \right ]$ is the arc of a unit circle in the 1st quadrant of the coordinate system.

Hence, a symmetry argument intuitively suggests $p = -1$ as a solution. Another way to see this, is to use the endpoint constraints:

$x = 0$: $|1-q| \leq \delta_0 \Rightarrow 1-\delta_0 \leq q \leq 1 + \delta_0\;\;\;\;\;(1)$

$x = 1$: $|-p-q| \leq \delta_0\;\;\;\;\;(2)$

Obviously, putting $p = -1$ makes the two conditions identically valid. In addition, we have the function maximum constraint:

$0 < x < 1$: $|\sqrt{1+p^2}-q| \leq \delta_0 \;\;\;\;\;(3)$

The inequality follows from two observations:

(a). $f’’(x) = \frac{-1}{(1-x^2)^{\frac{3}{2}}} \leq 0$ (i.e. $f$ is truly concave on the open interval $]0;1[$, so if an extremum is detected, it must be a maximum).

(b). $ f’(x_0) = 0 \Rightarrow x_0 = -\frac{p}{\sqrt{1+p^2}}$.

Note the minus sign (we know from $(2)$, that $p < 0$). The condition: $|f(x_0)| < \delta_0$ yields $(3)$.

The presumed solution: $p = -1$ also satisfies $(3)$, because:

\[ \left | \sqrt{1+(-1)^2}-q \right | \leq \delta_0 \Rightarrow q \geq \sqrt{2}-\delta_0 = 1+\delta_0 \Rightarrow q = 1+\delta_0\] – which is the upper limit of $q$. We can conclude, that $(p,q) = ( -1, 1+\delta_0) = \left (-1, \frac{\sqrt{2}+1}{2} \right )$ is a solution to the problem.

We need to check for other possibilities:

The conditions $(2)$ and $(3)$ imply:

$-\delta_0 -p \leq q \leq \delta_0 -p \;\;\;\;(2a)$

$\sqrt{1+p^2}-\delta_0 \leq q \leq \sqrt{1+p^2}+\delta_0 \;\;\;\; (3a)$

Note from $(3a)$ and $(1)$, that $p$ is restricted to the interval $-1 \leq p \leq 0$. We can graphically illustrate the set of possible solutions:

Only $(p,q)$-pairs in the intersection of the yellow and green sets are allowed. Thus only the o-marked corner point at $(-1,1+\delta_0)$, is a valid solution. Indeed, the lower limit in $(3a)$ intersects the upper limit in $(2a)$ in $p = -1$:

\[\sqrt{1+(-1)^2}-\delta_0 = \delta_0 - (-1) \Rightarrow \sqrt{2}-\delta_0 = 1+\delta_0\Rightarrow 2\delta_0 = \sqrt{2}-1 \Rightarrow \delta_0 = \frac{\sqrt{2}-1}{2}\;\;\;\;TRUE\]

We can conclude, that $(p,q) = \left (-1, \frac{\sqrt{2}+1}{2} \right )$ is the only solution.

 

[Opalg]

Spoiler: A geometric approach

[TIKZ] [scale=5]
\draw (-0.25,0) -- (1.25,0) ;
\draw (0,-0.25) -- (0,1.25) ;
\draw [dashed] (1,0) arc (0:90:1) ;
\draw [thick,green] (1,0.2071) arc (0:90:1) ;
\draw [thick,red] (1,-0.2071) arc (0:90:1) ;
\draw [very thick,dotted] (0,1.2071) -- (1,0.2071) ;[/TIKZ]
The dashed curve is the quadrant $y= \sqrt{1-x^2}$. Raise it by a distance $\frac{\sqrt2-1}2$ to get the green quadrant, and lower it by a distance $\frac{\sqrt2-1}2$ to get the red quadrant. The dotted line is the unique line that can be fitted between the green and red quadrants. It has slope $-1$ and $y$-intercept $\frac{\sqrt2+1}2$, and that gives the values for $p$ and $q$ that lfdahl obtained. But I never got round to writing up the details.