Largest circle between $y=x^n$ and $y=\sqrt[n]{x}$

Something I have been wondering about for a while. Let us look at the area between $x^n$ and $\sqrt[n]{x}$ when $x\in [0,1]$. Where $n$ is a positive integer. Below is an image.

With a given n, how large of a circle can you fit between $x^n$ and $\sqrt[n]x$?

I know that the circle must lie on the line y=x, but more than that I have not been able to figure out. Will update with my research and images. soon =)

Cheers for all help and solutions

-
For each point $p$ on the line $y = x$, consider the minimum distance $m(p)$ from this point to any of the graphs $y = x^n$ or $y = \sqrt[n]{x}$ (due to symmetry it doesn't matter which). You can easily compute explicit expression for $m(p)$. Then maximize $m(p)$ as a function of $p$, and you'll have your answer. – user12918723509187 Mar 1 '12 at 10:54
I am having difficulties here. I keep obtaining -2p + x for a line perpendicular to the point (p,p). Then I need to solve the equation x^n + 2p - x = 0. To find the interesection point between per line and x^n. – N3buchadnezzar Mar 1 '12 at 11:49
No no no! The minimum of $m(p)$ is NOT along a perpendicular! What you need to do is this: for fixed $p$ on the line $y = x$, let $M_p(u) = \sqrt{(p - u)^2 + (p - u^n)^2}$. Now let $m(p) = \min_{u\in [0,1]}\{ M_p(u) \}$. Now you want to maximize $m(p)$ as a function of $p$. All of this optimization of course involves taking derivatives, finding critical points, etc... – user12918723509187 Mar 1 '12 at 20:15
Yes, yes, yes! Your method of course works, but the circle has to be tangent to the power curves whenever they meet. For $p\in\mathbb{N}$, this is proven in Jacobson's Basic Algebra (see 1st paragraph of my post for exact reference). If the power curve's tangent is not also tangent to the circle, then it contains a point interior to the circle whence, by the lemma, so does the power curve itself, contradicting that the curve bounds the circle. Yes, of course, I mean the maximum of $m(p)$. I am just trying to see how far I can get without numerical methods. – bgins Mar 3 '12 at 23:51

Let's say we have the curves $y=x^{p^{\pm1}}$ on $[0,1]$ for $p>1$, which are symmetric about $y=x$ the case $p=1$. Let $Q=(q,q)$ ($0 < q < 1$) be the center of the desired circle $C$. Let $r$ be the minimum distance from Q to the curves. Let $P=(x,y=x^p)$ be a closest point to $Q$ (fix $y=x^p$). Then $C$ is tangent to $x^p$ at $P$ (proof for $p\in\mathbb{N}$: Lemma 2, p. 330 in Jacobson, Basic Algebra 1, 2nd ed.). This amounts to an equation of slopes. The slope, $\beta$, of $x^p$ at $P$ is found from the derivative. The slope of the tangent to the circle at $P$ is the negative reciprocal (perpendicular) slope of the radial line $\overline{QP}$. Thus $$\frac{x^p-q}{x-q}=-\frac{1}{px^{p-1}}=-\beta^{-1} \tag{1a}$$ implicitly relates $x$ and $q$ for a given $p$, which can be written: $$px^{p-1}(x^p-q)+(x-q)=0 \tag{1b}$$ $$q=\frac{x(1+px^{2p-2})}{1+px^{p-1}} \tag{1c}$$ $$q'=\frac{p^3x^{3(p-1)}+p(2p-1)x^{2(p-1)}-p(p-2)x^{p-1}+1}{p^2x^{2(p-1)}+2px^{p-1}+1} =\frac{\beta^3+(2-\frac1p)\beta^2-(p-2)\beta+1}{(\beta+1)^2} ~\tag{1d}$$ Now the distance $r$ from $P$ to $Q$ has square $$r^2=(x^p-q)^2+(x-q)^2 \tag{2}$$ and we wish to maximize (2) on the interval $(0,1)$ subject to the constraint (1) and to the constraint that the resulting circle given by $q$ and $r$ never crosses the dual power curves elsewhere. However, in order for the $r$ to be maximal, (I think) this means that each $x$ at which $C$ meets $x^p$ must be a local or global maximum of $r$. In fact, there may be either one or two such points; for a given $p>1$, $q$ and $r$ are uniquely determined, but $P$ will have one or two solutions, depending upon whether $p$ is past a certain threshold, around $5.2$.

Here is an animated graph of the solutions for $p$ varying from $1$ to $10$. The solution circle $C$ is shown in green. The dotted blue & purple (cyan/violet) circles show perturbations of $q$. The dashed grey lines show the critical points of $q$, or roots of $q'$. Between these two lines is the area where $q$ displays "retrograde motion" as $x$ increases. The purple and blue curves show how $q$ and $r$ depend on $x$. The green radial line represents $\overline{QP}$, or an arbitrary choice of one of the two possibilities for $p>5.2$.

I used WNY's suggestion to find the solution, but had to first overcome a numerical instability issue. The red arcs display crossover outside the power curves, which results from numerical inaccuracies. To overcome these, I used symmetry properties and the inequalities $0 < r < \min(r,1-q)$ and $y < q < x < 1$. If there is interest, I could post an animation or sage code demonstrating this as well. When time permits, I will update and revise the below with some further observations, and hopefully find an analytic value for the threshold value of $p$ around $5.2$, unless someone beats me to it (or knows where the problem is already solved)!

Let $f(x)=\frac12r^2$ with $r$ as in (2) and $q$ as in (1). Its local extrema occur when its derivative vanishes. \eqalign{f\,'(x) &=(x^p-q)(px^{p-1}-q')+(x-q)(1-q') \\&=\left(px^{p-1}(x^p-q)+(x-q)\right)-q'\big(x^p+x-2q\big) \tag{3a} } Since the circle is tangent to the curve at all local extrema, we can use (1b) to eliminate the first parenthesized term, $$f\;'(x)=r\;r'=-q'\cdot\big(x^p+x-2q\big) \tag{3b=1\wedge3a}$$ $$\matrix{ r' &=& \frac{2}{r} \left(q-\frac{x+y}{2}\right) &\cdot& q' \cr \frac{dr}{dx} &=& \frac{dr}{dq} &\cdot& \frac{dq}{dx} } \tag{3c}$$ resulting in two factors, where $q'$ is given in (1d). Thus, $f$ & $r$ have the same critical points as $q$, plus one more, representing the difference in coordinate values between two points on $y=x$, namely, between $Q=(q,q)$ and the midpoint of $P=(x,y)$ and its "dual" point $(y,x)$, with coordinate $\frac{x+y}{2}$. Let's assume for now that $q'\ne0$. This implies $x^p+x-2q=0$, i.e., $q=\frac{x^p+x}{2}=\frac{x+y}{2}$. With the slope equation (1b) above, then, we have \eqalign{ 0 &=px^{p-1}\left(x^p-\frac{x^p+x}{2}\right)+\left(x-\frac{x^p+x}{2}\right) \\&=\left(px^{p-1}-1\right)\left(\frac{x^p-x}{2}\right) \tag{3.1a} } which has one real root in $(0,1)$, at $x=p^{-1/(p-1)}=p^{1/(1-p)}$. (The other roots are all $(p-1)^\text{th}$ roots of $1$ and $\frac1p$, which, aside from a negative root of each for odd $p$, lie off the real axis). Thus $$x=p^{1/(1-p)} \qquad y=x^p=p^{p/(1-p)} \qquad q=\frac{x+y}{2}, \tag{3.1b}$$ the distance and area are $$r=\frac{|x^p-x|}{\sqrt{2}} \qquad A=\frac{\pi}{2}|x^p-x|^2, \tag{3.1c}$$ and $$y-q=x^p-q=\frac{x^p-x}{2}=\frac{y-x}{2}=-(x-q) \,. \tag{3.1d}$$ In fact, (3.1) gives the global maximum of $f$ for $1<p\le p^*\approx 5.72$. Using L'Hopital's rule as $p\rightarrow\infty$, \eqalign{ &\log x= -\frac{\log p}{p-1} \rightarrow -\frac1p \rightarrow 0, \quad \log y= -\frac{\log p}{1-\frac1p} \rightarrow -\infty \\&\implies \quad x\rightarrow1, \quad y\rightarrow0, \quad q\rightarrow\frac12, \text{ and } \quad r\rightarrow\frac{1}{\sqrt2} \,. \tag{3.1e} } Similar analysis show that $x\rightarrow y\rightarrow q\rightarrow\frac1e$ as $p\rightarrow 1$. Equations (3.1a-d) must maximize $f$ for all $p$ for which $q'\ne0$ on $(0,1)$. The limiting behavior, however, shows that this cannot hold for $p$ sufficiently large. This is because the circle meets $(x,y=x^p)$ at a tangent, but must also have a smaller radius of curvature to stay above the curve. This rules out the limit point at $(1,0)$ as $p\rightarrow\infty$.

The problem is that, for $p$ large enough (the threshold is around $5.72$), the other term, $q'$, also has real roots in $(0,1)$, and the minimum crosses over to one of these; the root of $x+x^p-2q$ becomes a maximum at the crossover. The graph below shows the approximate threshold case. The extrema of the green circle's area occur when the product (i.e. either) of the blue ($2q-x-x^p$) and/or red-brown ($q'$) curves has a root. For $1 < p < 5.72$, $q'$ is strictly positive on $(0,1)$, while for $p$ above the threshold, two new roots are introduced.

We need to ensure that we are always choosing the solution with minimal $r$. We must choose the least positive root $x\in(0,1)$ of $q'(x)$, when it exists.

To check these statements, we can take the second derivative of $f$: \eqalign{ f\;''(x)&= -q''\left(x^p+x-2q\right) -q'\left(px^{p-1}+1-2q'\right)\\&= 2(q')^2 -q''\left(x^p+x-2q\right) -q'\left(px^{p-1}+1\right) \tag{4} } Its sign tells us whether each critical point is a local minimum or maximum. Or, as already mentioned, we can check that the radius $r$ is less than the radius of curvature, $R$, of $y=x^p$ at $P$, given by \eqalign{ R^2 &= \frac{\left(1+(y')^2\right)^3}{(y'')^2} &= \frac{\left(1+(px^{p-1})^2\right)^3}{\left(p(p-1)x^{p-2}\right)^2} &= \frac{\left(1+p^2x^{2(p-1)}\right)^3}{p^2(p-1)^2x^{2(p-2)}} } \tag{5} which should lead us to the cutoff we desire when we equate $R^2$ with $r^2$ above (if it is not equivalent to $q'=0$).

Here is an animation of the result of using (3.1) for $p\in[1,11)$ with increments of $\frac1{10}$. You can clearly see how the root becomes a local maximum after the crossover threshold.

Before evaluating critical points using (4) or (5), we will also need to find the critical points of $q$, using the numerator of $q'$ from (1d): $$0=p^3x^{3(p-1)}+p(2p-1)x^{2(p-1)}-p(p-2)x^{p-1}+1 \tag{3.2a}$$ which I call equation (3.2) since it represents a second branch of (3) corresponding to the first, previously discounted, factor of (3b). With $\beta=px^{p-1}\in(0,p)$ (the common slope at $P$), we have the cubic $$0=\beta^3+(2-\tfrac1p)\beta^2-(p-2)\beta+1 \tag{3.2b}$$ which for $p$ sufficiently large has roots near $$px^{p-1} = \beta \approx \frac1{p-2},~ -1\pm\sqrt{p-1-\frac1{p}}+\frac1{2p}, \tag{3.2\approx}$$ one of which is in the interval $(0,1)$ and must therefore be our solution. Its precise location can be expressed as (I suspect) $$\beta = \gamma - \frac{2p-1}{3p} + \frac{3p^3-2p^2-4p+1}{9p^2\gamma} \tag{3.2\beta}$$ for $\gamma$ given by (the cube root of) $$\gamma^3 = \left(p + 1\right) \sqrt{\frac{-p^4+7p^3-11p^2+7p-1}{27p^3}} - \frac{9p^4-p^3-3p^2+6p-1}{27p^3}. \tag{3.2\gamma}$$ Interestingly, the complex roots of (3.2a) for integral $p>2$ all seem to lie in the annulus $\frac12<|x|<1$.

To be continued...

-
This almost works alas, when p tends to infinity all hell breaks lose. When $p\to \infty$ the radi should be $1/2$, but the limit of $r$ is not. – N3buchadnezzar Mar 1 '12 at 15:30
@N3buchadnezzar The reason is that bgins, assumed that $q'\neq 0$. – Norbert Mar 1 '12 at 19:17
There is a mistake in evaluation of $q'$ – Norbert Mar 2 '12 at 7:26
I've checked all solutions of this equation and previous one. All of them gives circles with radius tending to 1. So neither yours nor my approaches doesn't give an answer. The problem turn up to be much more complicated then I expected. – Norbert Mar 2 '12 at 13:29
Very nice work! – N3buchadnezzar Mar 9 '12 at 0:46

What is at stake here is an optimization problem with a parameter, namely $n$. In such a case it is easily possible that the "maximum" (or whatever) not only depends numerically on the value of the parameter, but that the extremal situation changes drastically at certain values of the parameter. This is the case in this example: For $n\leq5$ the largest circle touches the two curves $\gamma:\ y=x^n$, $\gamma':\ y=x^{1/n}$ in the two points where $y'(x)=1$; but for $n\geq6$ the curvature of $\gamma$ and $\gamma'$ in these two points is too large to make this possible. This implies that for $n\geq6$ the extremal situation is globally completely different: The extremal circle will now touch the curves $\gamma$ and $\gamma'$ in totally 4 points. This will make the computing of the extremal circle much more difficult in the case $n\geq6$.

-
I've note this at the very begining. The problem is that Lagrange method doesn't give this 4 points – Norbert Mar 2 '12 at 20:26
I'm still not quite far enough along to see if I can get both points on each curve. – bgins Mar 3 '12 at 1:15