0
$\begingroup$

I defined the support function $h_A:R^n→R$ of a non-empty closed convex set $A\subseteq \mathbb{R}^n$ as $$h_A(x)= \sup\left\{x⋅a |\ a \in A\right\}$$

Everything I know about this topic I found it. I have to calculate the support function of an ellipse $$\text{E=$\left\{(x,y) \in \mathbb{R}^2 \quad|\quad \frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1 \right\}$}$$

For $x=a\cos\theta ,\ y=b\sin\theta, \ 0\le\theta\lt2\pi\ $, how can calculate $h_E(\theta)$?

The result should be $h_E(\theta)=\left(a^2\cos^2\theta+b^2\sin^2\theta \right)^\frac12$ but my calculations lead me to have $h_E(\theta)=\sup\{(a\cos\theta,b\sin\theta)⋅(\cos\theta,\sin\theta)\}=a\cos^2\theta+b\sin^2\theta$.

Where is the error?

$\endgroup$
  • $\begingroup$ Please change the variable over which the supremum is taken so that it's different from the function's variable. At present the definition of $h_A$ isn't clear. $\endgroup$ – Micapps Jun 5 '16 at 10:22
  • $\begingroup$ I've done. Thank you for remark. Is clearer now? $\endgroup$ – J.Doe Jun 5 '16 at 10:33
1
$\begingroup$

The main problem you have is giving the same name to different things: $\theta$ means two different things, and so does $a$.

The parametric equation $x=a\cos t$, $y=b\sin t$, leads to $$ h_E(\theta)=\sup_t (a\cos t,b\sin t)\cdot (\cos\theta, \sin\theta) $$ Observe that the dot product can just as well be written as $$ (\cos t,\sin t)\cdot (a\cos\theta, b\sin\theta) $$ which is simply the projection of $(a\cos\theta, b\sin\theta)$ onto the direction determined by $t$. The maximal possible value of scalar projection is the length of the vector, hence $$ h_E(\theta)=|(a\cos\theta, b\sin\theta)| = \sqrt{a^2\cos^2\theta+b^2\sin^2\theta} $$

$\endgroup$
1
$\begingroup$

A general hyper-ellipoid is the affine image of the unit-ball. This can be written as $\mathcal E := \{Ax + c \text{ s.t } x \in \mathbb B_n\}$, where $\mathbb B_n := \{x \in \mathbb R^n \text{ s.t } \|x\|_2 \le 1\}$ is the unit-ball in $\mathbb R^n$, $A : \mathbb R^n \rightarrow \mathbb R^n$ is a linear transformation and $c \in \mathbb R^n$, is the center of the ellipsoid. Now, one computes the support function $\sigma_{\mathcal E}$ of $\mathcal E$ as

\begin{equation} \begin{split} \sigma_{\mathcal E}(z) &= \sup\{\langle z, y\rangle \text{ s.t } y \in \mathcal E\} = \sup\{\langle z, Ax + c\rangle \text{ s.t } x \in \mathbb B_n\} \\ &= \langle z, c\rangle + \sup\{\langle A^Tz, x\rangle \text{ s.t } x \in \mathbb B_n\} = \langle z, c\rangle + \sigma_{\mathbb B_n}(A^Tz) = \langle z, c\rangle + \|A^Tz\|_{(2^*)}\\ &= \langle z, c\rangle + \|A^Tz\|_2. \end{split} \end{equation}

In your particular case $c = 0 \in \mathbb R^n$, and $A$ is an orthonormal matrix with entries $\pm \sin(\theta), \pm \cos(\theta)$. Figure out the the precise values for these entries, plug them into the formula i derived above, and you're done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.