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

Originally I thought to rephrase it in geometric terms, seeing one degree terms I think about equation of a line.

I also have that $\sqrt{1-x^{2}}-\frac{\sqrt{2}-1}{2} \leq p x+q \leq \sqrt{1-x^{2}}+\frac{\sqrt{2}-1}{2}$

I think that a geometric solution might be possible, please help me to proceed.


  • $\begingroup$ To get idea make a Taylor development of $ \sqrt{1-x^2}$ it could give you idea of the region where $(p,q)$ are because what you write is really close to a Taylor Lagrange inequality... $\endgroup$
    – EDX
    Dec 28, 2020 at 9:13
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    $\begingroup$ I think a geometric interpretation sounds good - you get a closed shape by taking the interior of the lines $x=0,1$ and $y=\sqrt{1-x^2} \pm \frac{\sqrt{2}-1}{2}$. The question is then which lines lie entirely in that shape. Does that help? $\endgroup$ Dec 28, 2020 at 10:24
  • $\begingroup$ In particular, if you fix an allowable value of $q$ (anything between $\frac{1+\sqrt{2}}{2}$ and $\frac{3-\sqrt{2}}{2}$), then you get an allowable range of $p$ by varying it between the two values at the tangent of the boundary. $\endgroup$ Dec 28, 2020 at 10:27
  • $\begingroup$ @EDX I don't really have much familiarity with calculus and I'm looking for an elementary solution, though calculus might help. $\endgroup$
    – user823576
    Dec 28, 2020 at 10:31
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    $\begingroup$ It's the optimal order 1 Chebyshev polynomial approximation to $\sqrt{1-x^2}$ on $[]0,1}$, so $p,q$ are uniquely determined. The criterion is here: en.wikipedia.org/wiki/Approximation_theory#Optimal_polynomials $\endgroup$
    – user436658
    Dec 28, 2020 at 13:11

3 Answers 3


Let $f(x) = \sqrt{1-x^{2}}-p x-q$, then the condition becomes $$|f(x)| \leq \frac{\sqrt{2}-1}{2}$$ Thus, the extrema of $f$ should lie in $\left[ -\frac{\sqrt{2}-1}{2}, \frac{\sqrt{2}-1}{2}\right]$. Now, let's see how its extrema look like: $$f'(x) = \frac{-x}{\sqrt{1-x^2}} - p = 0 $$ After bringing to a common denominator, numerator should be equal to zero: $$ -x - p\sqrt{1-x^2} = 0 \\[2mm] x = r\sqrt{1-x^2}, \ \ (r=-p) \\[2mm] x^2 = r^2(1-x^2) \\[2mm] (1+r^2)x^2 = r^2 \\[2mm] x_0 = \pm \frac{r}{1+r^2} = \mp\frac{p}{\sqrt{1+p^2}}$$ However, since $x \ge 0$, we get $$x_0 = -\frac{p}{\sqrt{1+p^2}} $$ Therefore, $$\begin{align}f(x_0) &= \sqrt{1-\frac{p^2}{1+p^2}} + \frac{p^2}{\sqrt{1+p^2}}-q \\[1mm]&= \frac{1 + p^2}{\sqrt{1+p^2}} -q \\[1mm]&= \sqrt{1+p^2}-q \end{align}$$ Hence, the condition seems to be satisfied for all $p$ and $q$ such that $$\left| \sqrt{1+p^2} -q \right| \le \frac{\sqrt2 - 1}{2} \tag{1}$$ Also, we need to see the edge cases when $x = 0$ and $x = 1$, which give $$|f(0)| = \left|1 - q\right| \le \frac{\sqrt{2}-1}{2} \tag{2}$$ and $$|f(1)| = \left|-p - q\right| \le \frac{\sqrt{2}-1}{2} \tag{3}$$ Finally, combining $(1)$, $(2)$ and $(3)$ we get the final answer.

I didn't solve the last three inequalities putting all together, but after graphing I noticed (but still needs to be justified rigorously by solving the inequalities) that only one pair satisfies the statement, namely, $$(p,q) = \left(-1, \frac{\sqrt 2 + 1}{2}\right)$$

enter image description here

Below is the graph with those $p$ and $q$:

enter image description here

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    $\begingroup$ You should also bring in the condition when $|f(x)|$ achieves its bounds when $x=0,x=1$ ie an additional condition $|f(0)|,|f(1)| \le \frac{\sqrt{2}-1}{2}$ $\endgroup$ Dec 28, 2020 at 11:28
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    $\begingroup$ @AlbusDumbledore Thanks, edited accordingly. $\endgroup$
    – VIVID
    Dec 28, 2020 at 11:38

Remark: @VIVID gave a nice solution. I rewrote it in elementary way (without using derivative, extrema). Note that while we used calculus to motivate our solution, we do not need to include any calculus in the solution!

Letting $x = 0$, we have $|1 - q| \le \frac{\sqrt{2}-1}{2}$ or $$1 - \frac{\sqrt{2}-1}{2} \le q \le 1 + \frac{\sqrt{2}-1}{2}. \tag{1}$$

Letting $x = 1$, we have $|-p - q| \le \frac{\sqrt{2}-1}{2}$ or $$-p - \frac{\sqrt{2}-1}{2} \le q \le -p + \frac{\sqrt{2}-1}{2}. \tag{2}$$

From (1) and (2), we know that $p < 0$ (reason: if $p\ge 0$ then $-p + \frac{\sqrt{2}-1}{2} < 1 - \frac{\sqrt{2}-1}{2}$).

Since $-\frac{p}{\sqrt{1+p^2}} \in [0, 1]$, by letting $x = -\frac{p}{\sqrt{1+p^2}}$, we have $|\sqrt{p^2+1} - q| \le \frac{\sqrt{2}-1}{2}$ or $$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le q \le \sqrt{p^2 + 1} + \frac{\sqrt{2}-1}{2}. \tag{3}$$

From (2) and (3), since $\sqrt{p^2 + 1} > -p$, we have $$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le -p + \frac{\sqrt{2}-1}{2}$$ that is $\sqrt{p^2 + 1} + p - \sqrt{2} + 1 \le 0$ or $\frac{(p^2+1) - 2}{\sqrt{p^2+1} + \sqrt{2}} + (p+1) \le 0$ or $\frac{(p - 1 + \sqrt{p^2+1} + \sqrt{2})(p+1)}{\sqrt{p^2+1} + \sqrt{2}} \le 0$ which results in $p \le -1$.

From (1) and (3), since $1 + \frac{\sqrt{2}-1}{2} \le \sqrt{p^2 + 1} + \frac{\sqrt{2}-1}{2}$, we have $$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le 1 + \frac{\sqrt{2}-1}{2}$$ that is $p^2 \le 1$. Thus, we have $p = -1$.

From (1) and (3), we have $q = 1 + \frac{\sqrt{2}-1}{2}$.

Finally, when $p = -1, q = 1 + \frac{\sqrt{2}-1}{2}$, it is easy to prove that $|\sqrt{1-x^2} + x - 1 - \frac{\sqrt{2}-1}{2}| \le \frac{\sqrt{2}-1}{2}, \forall x \in [0, 1]$ that is $1\le \sqrt{1-x^2} + x \le \sqrt{2}, \forall x\in [0, 1]$.

Thus, $p = -1, q = 1 + \frac{\sqrt{2}-1}{2}$ is the only one pair satisfies the statement.


I went ahead and drew an example in Geogebra. Let $\alpha = \frac{\sqrt{2}-1}{2}$. It was clearly chosen specially, since it looks like almost no choices of $q$ (labelled $E$ in the plot) allow any solutions.

Plot of the region bounded by the lines in the question, showing an unallowed value of p and q

With that in mind, I'll consider a necessary condition for there to be any solutions $p$ for a given value of $q$.

Condition: The line joining $(0,q)$ to $(1,\alpha)$ must not cross the lower circle.

When does that happen? Well, experimenting suggests it never happens, and the chord is tangent when $q=1+\alpha$. So maybe we try to prove that the line $AB$ is tangent to the circle $DC$.

The line $AB$ has equation $y=1+\alpha - x$. Does this intersect the circle $DC$? That is equivalent to asking for solutions of $1+\alpha-x = \sqrt{1-x^2} - \alpha$, or:

\begin{align} 1-x^2 &= (1+2\alpha-x)^2 \\ &= (\sqrt{2}-x)^2 \\ &= 2-2\sqrt{2}x + x^2 \\ 2x^2 - 2\sqrt{2} x + 1 &= 0 \\ (\sqrt{2}x - 1)^2 &= 0. \end{align}

So we have a point of tangency. Clearly, therefore, if $q < 1+\alpha$ then there is no value of $p$ such that the line is contained in the figure $ABCD$. This means the only solution is $(-1, 1+\alpha)$.


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