Prove or disprove that the ellipse of largest area (centered at origin) inscribed in $y=\pm e^{-x^2}$ has the equation $x^2+y^2=\frac12(1+\log2)$.

Question

I can show that $x^2+y^2=\frac12(1+\log2)$ is the equation of the circle of largest area inscribed in $y=\pm e^{-x^2}$:

The minimum distance $r$ (which will be the radius of the circle) between the origin and $y=e^{-x^2}$ can be found by finding the critical numbers of the derivative of the distance function. \begin{align} r&=\sqrt{x^2+(e^{-x^2})^2}\\ \frac{dr}{dx}&=\frac{2x-4xe^{-2x^2}}{2\sqrt{x^2+e^{-2x^2}}}\\ 0&=2x(1-2e^{-2x^2})\\ x&=0\quad\text{(obviously inadmissible, or)}\\ x&=\sqrt{\frac12\log2}\\ \\ r&=\sqrt{\frac12\log2+e^{-\log2}}\\ r^2&=\frac12\log2+\frac12\\ \implies\quad x^2+y^2&=\frac12\left(1+\log2\right) \end{align} as desired.

The relationship between a circle and an ellipse is like that of a square and a rectangle: given a set perimeter, the more square the rectangle, the larger the area. But since this question is based on the curve $y=e^{-x^2}$ and not on any set perimeter, it does not seem like the same conclusion can be drawn. I would need something stronger (perhaps based on concavity?) to show whether $x^2+y^2=\frac12(1+\log2)$ is the largest ellipse or not.

Answer 1

We can assume by symmetry and without loss of generality that the ellipse can be parametrized by $$(x,y) = (a \cos \theta, b \sin \theta), \quad a, b > 0, \quad \theta \in [0,2\pi).$$ We require tangency to the curve $y = e^{-x^2}$ as well as a single point of intersection in the first quadrant. That is to say, $$b \sin \theta = e^{-(a \cos \theta)^2}$$ has a unique solution for $\theta \in (0, \pi/2)$, and at this point, $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{b}{a} \cot \theta = -2x(\theta)e^{-x(\theta)^2} = -2a (\cos \theta )e^{-(a \cos \theta)^2}.$$ Consequently, $$-\frac{b}{a} \cot \theta = -2ab \cos \theta \sin \theta,$$ or $$\sin \theta = \frac{1}{a \sqrt{2}}.$$ Note if $a < 1/\sqrt{2}$, no such angle exists. The ellipse is "too narrow"--this is due to the fact that the point of tangency is at $(x,y) = (0,1)$. At the point of tangency, we also have $$\cos \theta = \sqrt{1 - (2a^2)^{-1}}$$ so that we now have $$\frac{b}{a \sqrt{2}} = e^{1/2 - a^2}$$ or $$b = a e^{1/2 - a^2} \sqrt{2}.$$ Finally, we seek to maximize the area of this family of ellipses parametrized by $a$. Since the area is proportional to $ab$, we need to maximize $$f(a) = a^2 e^{1/2-a^2}.$$ Computing the derivative with respect to $a$ and solving for critical values, we get $$0 = \frac{df}{da} = 2(a-1)a(a+1)e^{1/2-a^2},$$ hence $$a = 1$$ is the unique solution, with $b = \sqrt{2/e}$ and the ellipse has equation $$\frac{x^2}{1} + \frac{y^2}{2/e} = 1.$$ The area of this ellipse is simply $$\pi a b = \pi \sqrt{\frac{2}{e}}.$$ This is obviously not a circle.

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For your enjoyment, I have included an animation of the family of ellipses $$\frac{x^2}{a^2} + \frac{y^2}{2a^2 e^{1-2a^2}} = 1,$$ for $a \in [1/\sqrt{2},2]$:

Answer 2

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \implies \frac{dy}{dx}=-\frac{b^2}{a^2}\frac xy=-\frac{b}{a^2}\frac x{\sqrt{1-\frac{x^2}{a^2}}}. $$

Hence we have the following system to find the tangent point: $$\begin{cases} e^{-x^2}=b\sqrt{1-\frac{x^2}{a^2}}\\ 2xe^{-x^2}=\frac{b}{a^2}\frac x{\sqrt{1-\frac{x^2}{a^2}}} \end{cases}\implies 1-\frac{x^2}{a^2}=\frac{1}{2a^2}\implies x^2=a^2-\frac12.\tag1 $$

Substituting this back into the equation of the tangent point one obtains equation to determine $b$: $$ e^{\frac12-a^2}=\frac ba\sqrt{\frac12}\implies b=\sqrt{2e}\,ae^{-a^2}. $$ The area is respectively: $$ A=\pi ab=\pi \sqrt{2e}\,a^2e^{-a^2}.\tag2 $$ To find the extremum of the area we differentiate over $a$ to obtain: $$ \frac1{\pi\sqrt{2e}}\frac{dA}{da}=2ae^{-a^2}-2a^3e^{-a^2}=2ae^{-a^2}(1-a^2), $$ meaning that the largest value of area is achieved at $a=1$: $$ A_\text{max}=\pi\sqrt{\frac2e}. $$

Since $a\ne b$ the ellipse of the largest area is not a circle.

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PS. In fact the system (1) has another solution: $x=0, b=1$. It can be however shown that the largest possible area of the inscribed ellipse in this case is $\frac\pi{\sqrt2}$ which corresponds to $a=\frac1{\sqrt2}$ in equation (2).

Answer 3

I think what you say holds good if the Bell curve has a single constant in its parametrization. But there are two constants in: $y= y_{max} e^{- x^2/(2 \sigma^2)} $

The touching curve is an ellipse, not a circle.

This is vague/intuitive at present,trying to prove that ellipse area is maximized when constrained on an envelope with symmetry of one axis with single inflection. We can proceed with intersections and differentiation etc., but for the present I proceed on this basis as given and may be shall ask a question separately about such an elegant general possibility.

Taking classical Bell curve as ( in present case $\sigma = \frac{1}{\sqrt 2})$

$$y= y_{max} e^{- x^2/(2 \sigma^2)} $$

Bell Curve by differentiation:

$$ x_I= \sigma; y_I=y_m /\sqrt{e};\;y_I'= \dfrac{-xy}{\sigma^2} \rightarrow y_I'= -\dfrac{y_m}{\sigma \sqrt{e}}\tag1$$

Ellipse and derivatives: $$x_I^2/a^2+y_I^2/b^2=1; \;y_I'=-\dfrac{x_Ib^2}{y_Ia^2} \tag2$$

Eliminate $y'$ between (1),(2) simplifying and solving for $(a^2,b^2)$ we get $$ (a,b)= (\sqrt 2 \sigma,y_m \sqrt {2/e} )\tag3 $$