Radius of a largest circle inscribed under $y=\frac{1}{(1+x^2)^n}$, closed form

Question

The curve $y=\frac{1}{1+x^2}$ has an obvious connection to circles, because it's the derivative of the arctangent function.

Besides, if we inscribe a circle under it, its radius is exactly $R=\frac{1}{2}$, so it takes exactly one fourth of the full area under the curve.

Actually, I don't know an easy way to find $R$ even in this simple case. So, let's consider how I solve the more general problem.

Find the largest circle fitting between the curve $y=y (x)$ and the line $y=0$

$$y(x)=\frac{1}{(1+x^2)^n}~~~~~~ n=1,2,3,\dots$$

The distance from the circle origin $(0,R)$ to the curve is the minimum of a function:

$$s(x)=\sqrt{x^2+\left(R-\frac{1}{(1+x^2)^n} \right)^2}$$

$$s(x)'=0$$

The condition for the inscribed circle is $s(x)=R$.

Let's denote:

$$t=1+x^2$$

Then from the above we obtain the system of equations:

$$t^{2n+1}+2n~R~t^n-2n=0$$

$$t^{2n+1}-t^{2n}-2~R~t^n+1=0$$

This is how I got the solution for $n=1$ (using Mathematica). Other solutions do not have obvious closed forms. Here are the numerical roots I've got with Mathematica:

$$R_2=0.4735710971151933$$

$$R_3=0.4401444298014721$$

$$R_4=0.41216506385826285$$

And so on.

The questions I ask:

Can there be closed forms for $R$ for $n \neq 1$? How to find them?

Is there an easier way to find $R$? At least for some $n$?

Edit

The most simple equation for $t$, as far as I can see, is:

$$(n+1)~t^{2n+1}-n~t^{2n}-n=0$$

I think I might ask a separate question about this equation. It's easy to solve by Newton-Raphson, but can it have closed form solutions for any $n$?

Answer 1

It is possible to find tangent points and the circle of contact for given $n$ using the conventional and classical Lagrangian multiplier method. The circle can be the object, and given curve the constraint function or vice-versa. A sort of non-linear programming in Operations Research.

Obtained the same contact radii values for each $n$ as the OP.

$R = \lambda $ parameter circles with x-axis tangent at origin are

$$ F = x^2 + (y-\lambda)^2 = x^2 +y^2 -2 y \lambda = 0 \tag{1} $$

$$ y = 1/(1+x^2)^n ;\, \rightarrow G = y (1+x^2)^n - 1 =0 \tag{2} $$

We have a constant Lagrange multiplier $$ \frac{F_x}{F_y}= \frac{G_x}{G_y} = C, \; \tag{3} $$

which after algebraic simplification yields the required link relation:

$$ 2 n y (y-x) = ( 1 + x^2) \; \tag{4} $$

Solving together (1), (2) and (4) we get $ x,y, \lambda = R $ that allow the curves and maximum radius contacts to be sketched as follows for $n = 1,3,6,20.$

The radius cannot be fully expressed in closed form , but only as an algebraic number in a field given from Mathematica for $ n\ne 1 $ example:

    n=3, \; \lambda = R =  AlgebraicNumber[
    Root[-8192 + 10240 #1^2 + 9984 #1^4 + 5120 #1^6 + 1520 #1^8 + 
    264 #1^10 + 25 #1^12 + #1^14 &, 2], {1/6, 0, 5/12, 0, 1/4, 0, 11/
    192, 0, 7/1536, 0, 0, 0, 0, 0}]

EDIT 1 (deleted)

Answer 2

The radius of curvature of a function $f(x)$ is

$$ \rho =- \frac{ \left( 1 + \left(\frac{{\rm d}}{{\rm d}x}f(x)\right)^2 \right)^\frac{3}{2} } { \frac{{\rm d}^2}{{\rm d}x^2} f(x) } $$

_{NOTE: If the curve is given implicitly then $$\rho = \frac{ (x'^2+y'^2)^\frac{3}{2}}{y' x'' - y'' x'}$$}

Case 1

Consider $f(x) = \frac{1}{1+x^2}$

$$ \rho = -\frac{ \left(1+\frac{4 x^2}{(1+x^2)^4} \right)^\frac{3}{2}} { \frac{2(3 x^2-1)}{(1+x^2)^3} } $$

and for $x=0$ the circle is $\rho =\frac{1}{2}$ as you expect.

Case 2

Consider $f(x) = \frac{1}{(1+x^2)^n}$

The radius of curvature is

$$ \rho = \frac{ \left( 4 n^2 x^2 + (1+x^2)^{2(n+1)}\right)^\frac{3}{2} }{ \tfrac{2 n (1-x^2 (2n+1))}{(1+x^2)^{-(2n+1)}}} $$

and for $x=0$ the curvature is simply ${\rho = \frac{1}{2n}}$

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Now the circle is going to be tangent to the curve (one of the curvature circles) and its center is going to be located at

$$ x_c = x - \frac{ y' (1+y'^2)}{y''} \\ y_c = y + \frac{1+y'^2}{y''} $$

_{here $y' = \frac{{\rm d}}{{\rm d}x} f(x)$ and $y'' = \frac{{\rm d}^2}{{\rm d}x^2} f(x)$}

For the circle to be tangent to $y=0$, its center has to be at $y_c = \rho$ or

$$ y + \frac{1+y'^2}{y''} = - \frac{(1+y'^2)^\frac{3}{2}}{y''} $$

For $y=\tfrac{1}{1+x^2}$ my CAS says the solution (besides $x=0$) is the roots of

$$ 2 x^{16} + 15 x^{14} + 48 x^{12} + 97 x^{10} + 126 x^8+81 x^6-20 x^4 -33 x^2 = 12$$

$ x = 0.759119999241623 $ or $\rho = 4.32460278011942$, but that ends up being a circle over the curve.

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For the general case of $y = \frac{1}{(1+x^2)^n}$ the equation to be solved is

$$ \left( (1+x^2)^{2(n+1)} + 4 n^2 x^2\right)^\frac{3}{2} + (1+x^2)^{3(n+1)} + 2 n ( 1+x^2)^{n+1} \left( 4 n x^2 + x^2 -1 \right) = 0 $$