## ECE1505H Convex Optimization. Lecture 4: Sets and convexity. Taught by Prof. Stark Draper

January 25, 2017 ece1505 No comments , , , , , , ,

ECE1505H Convex Optimization. Lecture 4: Sets and convexity. Taught by Prof. Stark Draper

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1505H, Convex Optimization, taught by Prof. Stark Draper, covering [1] content.

### Today

• more on various sets: hyperplanes, half-spaces, polyhedra, balls, ellipses, norm balls, cone of PSD
• generalize inequalities
• operations that preserve convexity
• separating and supporting hyperplanes.

## Hyperplanes

Find some $$\Bx_0 \in \mathbb{R}^n$$ such that $$\Ba^\T \Bx_0 = \Bb$$, so

\label{eqn:convexOptimizationLecture4:20}
\begin{aligned}
\setlr{ \Bx | \Ba^\T \Bx = \Bb }
&=
\setlr{ \Bx | \Ba^\T \Bx = \Ba^\T \Bx_0 } \\
&=
\setlr{ \Bx | \Ba^\T (\Bx – \Bx_0) } \\
&=
\Bx_0 + \Ba^\perp,
\end{aligned}

where

\label{eqn:convexOptimizationLecture4:40}
\Ba^\perp = \setlr{ \Bv | \Ba^\T \Bv = 0 }.

fig. 1. Parallel hyperplanes.

Recall

\label{eqn:convexOptimizationLecture4:60}
\Norm{\Bz}_\conj = \sup_\Bx \setlr{ \Bz^\T \Bx | \Norm{\Bx} \le 1 }

Denote the optimizer of above as $$\Bx^\conj$$. By definition

\label{eqn:convexOptimizationLecture4:80}
\Bz^\T \Bx^\conj \ge \Bz^\T \Bx \quad \forall \Bx, \Norm{\Bx} \le 1

This defines a half space in which the unit ball

\label{eqn:convexOptimizationLecture4:100}
\setlr{ \Bx | \Bz^\T (\Bx – \Bx^\conj \le 0 }

Start with the $$l_1$$ norm, duals of $$l_1$$ is $$l_\infty$$

fig. 2. Half space containing unit ball.

Similar pic for $$l_\infty$$, for which the dual is the $$l_1$$ norm, as sketched in fig. 3.  Here the optimizer point is at $$(1,1)$$

fig. 3. Half space containing the unit ball for l_infinity

and a similar pic for $$l_2$$, which is sketched in fig. 4.

fig. 4. Half space containing for l_2 unit ball.

## Polyhedra

\label{eqn:convexOptimizationLecture4:120}
\begin{aligned}
\mathcal{P}
&= \setlr{ \Bx |
\Ba_j^\T \Bx \le \Bb_j, j \in [1,m],
\Bc_i^\T \Bx = \Bd_i, i \in [1,p]
} \\
&=
\setlr{ \Bx | A \Bx \le \Bb, C \Bx = d },
\end{aligned}

where the final inequality and equality are component wise.

Proving $$\mathcal{P}$$ is convex:

• Pick $$\Bx_1 \in \mathcal{P}$$, $$\Bx_2 \in \mathcal{P}$$
• Pick any $$\theta \in [0,1]$$
• Test $$\theta \Bx_1 + (1-\theta) \Bx_2$$. Is it in $$\mathcal{P}$$?

\label{eqn:convexOptimizationLecture4:140}
\begin{aligned}
A \lr{ \theta \Bx_1 + (1-\theta) \Bx_2 }
&=
\theta A \Bx_1 + (1-\theta) A \Bx_2 \\
&\le
\theta \Bb + (1-\theta) \Bb \\
&=
\Bb.
\end{aligned}

## Balls

Euclidean ball for $$\Bx_c \in \mathbb{R}^n, r \in \mathbb{R}$$

\label{eqn:convexOptimizationLecture4:160}
\mathcal{B}(\Bx_c, r)
= \setlr{ \Bx | \Norm{\Bx – \Bx_c}_2 \le r },

or
\label{eqn:convexOptimizationLecture4:180}
\mathcal{B}(\Bx_c, r)
= \setlr{ \Bx | \lr{\Bx – \Bx_c}^\T \lr{\Bx – \Bx_c} \le r^2 }.

Let $$\Bx_1, \Bx_2$$, $$\theta \in [0,1]$$

\label{eqn:convexOptimizationLecture4:200}
\begin{aligned}
\Norm{ \theta \Bx_1 + (1-\theta) \Bx_2 – \Bx_c }_2
&=
\Norm{ \theta (\Bx_1 – \Bx_c) + (1-\theta) (\Bx_2 – \Bx_c) }_2 \\
&\le
\Norm{ \theta (\Bx_1 – \Bx_c)}_2 + \Norm{(1-\theta) (\Bx_2 – \Bx_c) }_2 \\
&=
\Abs{\theta} \Norm{ \Bx_1 – \Bx_c}_2 + \Abs{1 -\theta} \Norm{ \Bx_2 – \Bx_c }_2 \\
&=
\theta \Norm{ \Bx_1 – \Bx_c}_2 + \lr{1 -\theta} \Norm{ \Bx_2 – \Bx_c }_2 \\
&\le
\theta r + (1 – \theta) r \\
&= r
\end{aligned}

## Ellipse

\label{eqn:convexOptimizationLecture4:220}
\mathcal{E}(\Bx_c, P)
=
\setlr{ \Bx | (\Bx – \Bx_c)^\T P^{-1} (\Bx – \Bx_c) \le 1 },

where $$P \in S^n_{++}$$.

• Euclidean ball is an ellipse with $$P = I r^2$$
• Ellipse is image of Euclidean ball $$\mathcal{B}(0,1)$$ under affine mapping.

fig. 5. Circle and ellipse.

Given

\label{eqn:convexOptimizationLecture4:240}
F(\Bu) = P^{1/2} \Bu + \Bx_c

\label{eqn:convexOptimizationLecture4:260}
\begin{aligned}
\setlr{ F(\Bu) | \Norm{\Bu}_2 \le r }
&=
\setlr{ P^{1/2} \Bu + \Bx_c | \Bu^\T \Bu \le r^2 } \\
&=
\setlr{ \Bx | \Bx = P^{1/2} \Bu + \Bx_c, \Bu^\T \Bu \le r^2 } \\
&=
\setlr{ \Bx | \Bu = P^{-1/2} (\Bx – \Bx_c), \Bu^\T \Bu \le r^2 } \\
&=
\setlr{ \Bx | (\Bx – \Bx_c)^\T P^{-1} (\Bx – \Bx_c) \le r^2 }
\end{aligned}

## Geometry of an ellipse

Decomposition of positive definite matrix $$P \in S^n_{++} \subset S^n$$ is:

\label{eqn:convexOptimizationLecture4:280}
\begin{aligned}
P &= Q \textrm{diag}(\lambda_i) Q^\T \\
Q^\T Q &= 1
\end{aligned},

where $$\lambda_i \in \mathbb{R}$$, and $$\lambda_i > 0$$. The ellipse is defined by

\label{eqn:convexOptimizationLecture4:300}
(\Bx – \Bx_c)^\T Q \textrm{diag}(1/\lambda_i) (\Bx – \Bx_c) Q \le r^2

The term $$(\Bx – \Bx_c)^\T Q$$ projects $$\Bx – \Bx_c$$ onto the columns of $$Q$$. Those columns are perpendicular since $$Q$$ is an orthogonal matrix. Let

\label{eqn:convexOptimizationLecture4:320}
\tilde{\Bx} = Q^\T (\Bx – \Bx_c),

this shifts the origin around $$\Bx_c$$ and $$Q$$ rotates into a new coordinate system. The ellipse is therefore

\label{eqn:convexOptimizationLecture4:340}
\tilde{\Bx}^\T
\begin{bmatrix}
\inv{\lambda_1} & & & \\
&\inv{\lambda_2} & & \\
& \ddots & \\
& & & \inv{\lambda_n}
\end{bmatrix}
\tilde{\Bx}
=
\sum_{i = 1}^n \frac{\tilde{x}_i^2}{\lambda_i} \le 1.

An example is sketched for $$\lambda_1 > \lambda_2$$ below.

Ellipse with $$\lambda_1 > \lambda_2$$.

• $$\lambda_i$$ tells us length of the semi-major axis.
• Larger $$\lambda_i$$ means $$\tilde{x}_i^2$$ can be bigger and still satisfy constraint $$\le 1$$.
• Volume of ellipse if proportional to $$\sqrt{ \det P } = \sqrt{ \prod_{i = 1}^n \lambda_i }$$.
• When any $$\lambda_i \rightarrow 0$$ a dimension is lost and the volume goes to zero. That removes the invertibility required.

Ellipses will be seen a lot in this course, since we are interested in “bowl” like geometries (and the ellipse is the image of a Euclidean ball).

## Norm ball.

The norm ball

\label{eqn:convexOptimizationLecture4:360}
\mathcal{B} = \setlr{ \Bx | \Norm{\Bx} \le 1 },

is a convex set for all norms. Proof:

Take any $$\Bx, \By \in \mathcal{B}$$

\label{eqn:convexOptimizationLecture4:380}
\Norm{ \theta \Bx + (1 – \theta) \By }
\le
\Abs{\theta} \Norm{ \Bx } + \Abs{1 – \theta} \Norm{ \By }
=
\theta \Norm{ \Bx } + \lr{1 – \theta} \Norm{ \By }
\lr
\theta + \lr{1 – \theta}
=
1.

This is true for any p-norm $$1 \le p$$, $$\Norm{\Bx}_p = \lr{ \sum_{i = 1}^n \Abs{x_i}^p }^{1/p}$$.

Norm ball.

The shape of a $$p < 1$$ norm unit ball is sketched below (lines connecting points in such a region can exit the region).

## Cones

Recall that $$C$$ is a cone if $$\forall \Bx \in C, \theta \ge 0, \theta \Bx \in C$$.

Impt cone of PSD matrices

\label{eqn:convexOptimizationLecture4:400}
\begin{aligned}
S^n &= \setlr{ X \in \mathbb{R}^{n \times n} | X = X^\T } \\
S^n_{+} &= \setlr{ X \in S^n | \Bv^\T X \Bv \ge 0, \quad \forall v \in \mathbb{R}^n } \\
S^n_{++} &= \setlr{ X \in S^n_{+} | \Bv^\T X \Bv > 0, \quad \forall v \in \mathbb{R}^n } \\
\end{aligned}

These have respectively

• $$\lambda_i \in \mathbb{R}$$
• $$\lambda_i \in \mathbb{R}_{+}$$
• $$\lambda_i \in \mathbb{R}_{++}$$

$$S^n_{+}$$ is a cone if:

$$X \in S^n_{+}$$, then $$\theta X \in S^n_{+}, \quad \forall \theta \ge 0$$

\label{eqn:convexOptimizationLecture4:420}
\Bv^\T (\theta X) \Bv
= \theta \Bv^\T \Bv
\ge 0,

since $$\theta \ge 0$$ and because $$X \in S^n_{+}$$.

Shorthand:

\label{eqn:convexOptimizationLecture4:440}
\begin{aligned}
X &\in S^n_{+} \Rightarrow X \succeq 0
X &\in S^n_{++} \Rightarrow X \succ 0.
\end{aligned}

Further $$S^n_{+}$$ is a convex cone.

Let $$A \in S^n_{+}$$, $$B \in S^n_{+}$$, $$\theta_1, \theta_2 \ge 0, \theta_1 + \theta_2 = 1$$, or $$\theta_2 = 1 – \theta_1$$.

Show that $$\theta_1 A + \theta_2 B \in S^n_{+}$$ :

\label{eqn:convexOptimizationLecture4:460}
\Bv^\T \lr{ \theta_1 A + \theta_2 B } \Bv
=
\theta_1 \Bv^\T A \Bv
+\theta_2 \Bv^\T B \Bv
\ge 0,

since $$\theta_1 \ge 0, \theta_2 \ge 0, \Bv^\T A \Bv \ge 0, \Bv^\T B \Bv \ge 0$$.

fig. 8. Cone.

Inequalities:

Start with a proper cone $$K \subseteq \mathbb{R}^n$$

• closed, convex
• non-empty interior (“solid”)
• “pointed” (contains no lines)

The $$K$$ defines a generalized inequality in \R{n} defined as “$$\le_K$$”

Interpreting

\label{eqn:convexOptimizationLecture4:480}
\begin{aligned}
\Bx \le_K \By &\leftrightarrow \By – \Bx \in K
\Bx \end{aligned}

Why pointed? Want if $$\Bx \le_K \By$$ and $$\By \le_K \Bx$$ with this $$K$$ is a half space.

Example:1: $$K = \mathbb{R}^n_{+}, \Bx \in \mathbb{R}^n, \By \in \mathbb{R}^n$$

fig. 12. K is non-negative “orthant”

\label{eqn:convexOptimizationLecture4:500}
\Bx \le_K \By \Rightarrow \By – \Bx \in K

say:

\label{eqn:convexOptimizationLecture4:520}
\begin{bmatrix}
y_1 – x_1
y_2 – x_2
\end{bmatrix}
\in R^2_{+}

Also:

\label{eqn:convexOptimizationLecture4:540}
K = R^1_{+}

(pointed, since it contains no rays)

\label{eqn:convexOptimizationLecture4:560}
\Bx \le_K \By ,

with respect to $$K = \mathbb{R}^n_{+}$$ means that $$x_i \le y_i$$ for all $$i \in [1,n]$$.

Example:2: For $$K = PSD \subseteq S^n$$,

\label{eqn:convexOptimizationLecture4:580}
\Bx \le_K \By ,

means that

\label{eqn:convexOptimizationLecture4:600}
\By – \Bx \in K = S^n_{+}.

• Difference $$\By – \Bx$$ is always in $$S$$
• check if in $$K$$ by checking if all eigenvalues $$\ge 0$$.
• $$S^n_{++}$$ is the interior of $$S^n_{+}$$.

Interpretation:

\label{eqn:convexOptimizationLecture4:620}
\begin{aligned}
\Bx \le_K \By &\leftrightarrow \By – \Bx \in K \\
\Bx \end{aligned}

We’ll use these with vectors and matrices so often the $$K$$ subscript will often be dropped, writing instead (for vectors)

\label{eqn:convexOptimizationLecture4:640}
\begin{aligned}
\Bx \le \By &\leftrightarrow \By – \Bx \in \mathbb{R}^n_{+} \\
\Bx < \By &\leftrightarrow \By – \Bx \in \textrm{int} \mathbb{R}^n_{++}
\end{aligned}

and for matrices

\label{eqn:convexOptimizationLecture4:660}
\begin{aligned}
\Bx \le \By &\leftrightarrow \By – \Bx \in S^n_{+} \\
\Bx < \By &\leftrightarrow \By – \Bx \in \textrm{int} S^n_{++}.
\end{aligned}

## Intersection

Take the intersection of (perhaps infinitely many) sets $$S_\alpha$$:

If $$S_\alpha$$ is (affine,convex, conic) for all $$\alpha \in A$$ then

\label{eqn:convexOptimizationLecture4:680}
\cap_\alpha S_\alpha

is (affine,convex, conic). To prove in homework:

\label{eqn:convexOptimizationLecture4:700}
\mathcal{P} = \setlr{ \Bx | \Ba_i^\T \Bx \le \Bb_i, \Bc_j^\T \Bx = \Bd_j, \quad \forall i \cdots j }

This is convex since the intersection of a bunch of hyperplane and half space constraints.

1. If $$S \subseteq \mathbb{R}^n$$ is convex then\label{eqn:convexOptimizationLecture4:720}
F(S) = \setlr{ F(\Bx) | \Bx \in S }
is convex.
2. If $$S \subseteq \mathbb{R}^m$$ then\label{eqn:convexOptimizationLecture4:740}
F^{-1}(S) = \setlr{ \Bx | F(\Bx) \in S }
is convex. Such a mapping is sketched in fig. 14.

fig. 14. Mapping functions of sets.

# References

[1] Stephen Boyd and Lieven Vandenberghe. Convex optimization. Cambridge university press, 2004.

## Motivation

Lance told me they’ve been covering the circumference of a circle in school this week. This made me think of the generalization of a circle, the ellipse, but I couldn’t recall what the circumference of an ellipse was. Sofia guessed $$\pi ( a + b )$$. Her reasoning was that this goes to $$2 \pi r$$ when the ellipse is circular, just like the area of an ellipse $$\pi a b$$, goes to $$\pi a^2$$ in the circular limit. That seemed reasonable to me, but also strange since I didn’t recall any $$\pi ( a + b )$$ formula.

It turns out that there’s no closed form expression for the circumference of an ellipse, unless you count infinite series or special functions. Here I’ll calculate one expression for this circumference.

## Geometry recap

There’s two ways that I think of ellipses. One is the shape that you get when you put a couple tacks in a paper, and use a string and pencil to trace it out, as sketched in fig. 1.
The other is the basic vector parameterization of that same path

\label{eqn:elipticCircumference:20}
\Br = ( a \cos\theta, b \sin\theta ).

fig. 1. Ellipse, showing tracing curve from the foci.

It’s been a long time since grade 11 when I would have taken it for granted that these two representations are identical. To do so, we’d have to know where the foci of the ellipse sit. Cheating a bit I find in [1] that the foci are located at

\label{eqn:elipticCircumference:40}
\Bf_{\pm} = \pm \sqrt{ a^2 – b^2 } (1, 0).

This and the equivalence of the pencil and tack representation of the ellipse can be verified by checking that the “length of the string” equals $$2 a$$ as expected.

That string length is

\label{eqn:elipticCircumference:60}
\begin{aligned}
\Abs{ \Br – \Bf_{+} } + \Abs{ \Br – \Bf_{-} }
&=
\sqrt{ (a \cos\theta – f)^2 + b^2 \sin^2\theta }
+
\sqrt{ (a \cos\theta + f)^2 + b^2 \sin^2\theta } \\
&=
\sqrt{ a^2 \cos^2 \theta + f^2 – 2 a f \cos\theta + b^2 \lr{ 1 – \cos^2\theta } } \\
\sqrt{ a^2 \cos^2 \theta + f^2 + 2 a f \cos\theta + b^2 \lr{ 1 – \cos^2\theta } }.
\end{aligned}

These square roots simplify nicely

\label{eqn:elipticCircumference:80}
\begin{aligned}
\sqrt{ a^2 \cos^2 \theta + f^2 \pm 2 a f \cos\theta + b^2 \lr{ 1 – \cos^2\theta } }
&=
\sqrt{ (a^2 – b^2) \cos^2 \theta + a^2 – b^2 \pm 2 a f \cos\theta + b^2 } \\
&=
\sqrt{ f^2 \cos^2 \theta + a^2 \pm 2 a f \cos\theta } \\
&=
\sqrt{ (a \pm f \cos\theta)^2 } \\
&=
a \pm f \cos\theta.
\end{aligned}

So the total length from one focus to a point on the ellipse, back to the other focus, is

\label{eqn:elipticCircumference:100}
\Abs{ \Br – \Bf_{+} } + \Abs{ \Br – \Bf_{-} }
=
a + f \cos\theta + a – f \cos\theta = 2 a,

as expected. That verifies that the trigonometric parameterization matches with the pencil and tacks representation of an ellipse (provided the foci are placed at the points \ref{eqn:elipticCircumference:40}).

## Calculating the circumference

The circumference expression can almost be written by inspection. An element of the tangent vector along the curve is

\label{eqn:elipticCircumference:340}
\frac{d\Br}{d\theta} = ( -a \sin\theta, b \cos\theta ),

so the circumference is just a one liner

\label{eqn:elipticCircumference:280}
C = 4 \int_0^{\pi/2} \sqrt{ a^2 \sin^2\theta + b^2 \cos^2 \theta} d\theta.

The problem is that this one liner isn’t easy to evaluate. The square root can be put in a slightly simpler form in terms of the eccentricity, which is defined by

\label{eqn:elipticCircumference:300}
e = \frac{f}{a} = \frac{\sqrt{a^2 – b^2}}{a} = \sqrt{ 1 – \frac{b^2}{a^2} }.

Factoring out $$a$$ and writing the sine as a cosine gives

\label{eqn:elipticCircumference:320}
\begin{aligned}
C
&=
4 a \int_0^{\pi/2} \sqrt{ 1 – \cos^2\theta + \frac{b^2}{a^2} \cos^2 \theta} d\theta \\
&=
4 a \int_0^{\pi/2} \sqrt{ 1 + \lr{ \frac{b^2}{a^2} -1} \cos^2 \theta} d\theta \\
&=
4 a \int_0^{\pi/2} \sqrt{ 1 – e^2 \cos^2 \theta} d\theta.
\end{aligned}

For the square root, it’s not hard to show that the fractional binomial expansion is

\label{eqn:elipticCircumference:360}
\sqrt{1 + a}
=
1 – \sum_{k=1}^\infty \frac{(-a)^k}{2k – 1}\frac{ (2k – 1)!!}{(2k)!!},

so the circumference is

\label{eqn:elipticCircumference:380}
C
=
4 a \int_0^{\pi/2} d\theta
\lr{ 1 – \sum_{k=1}^\infty \frac{(e \cos\theta)^{2k}}{2k – 1}\frac{ (2k – 1)!!}{(2k)!!}}.

Using \ref{eqn:elipticCircumference:260}, this is

\label{eqn:elipticCircumference:400}
C
=
2 \pi a

4 a
\sum_{k=1}^\infty \frac{e^{2k}}{2k – 1}\frac{ (2k – 1)!!}{(2k)!!}
\frac{(2k – 1)!!}{(2k)!!} \frac{\pi}{2}

\label{eqn:elipticCircumference:420}
\boxed{
C
=
2 \pi a \lr{ 1 –
\sum_{k=1}^\infty \frac{e^{2k}}{2k – 1} \lr{ \frac{ (2k – 1)!!}{(2k)!!} }^2
}.
}

Observe that this does reduce to $$2 \pi r$$ for the circle (where $$e = 0$$), and certainly isn’t as nice as $$\pi (a + b)$$.

## Appendix. Integral of even cosine powers.

The integral

\label{eqn:elipticCircumference:120}
\int_0^{\pi/2} \cos^{2k} \theta d\theta

can be evaluated using integration by parts.

\label{eqn:elipticCircumference:140}
\begin{aligned}
\int_0^{\pi/2} \cos^{2k} \theta d\theta
&=
\int_0^{\pi/2} \cos^{2k-1} \theta \frac{d \sin\theta}{d\theta} d\theta \\
&=
\evalrange{
\cos^{2k-1} \theta \sin\theta
}{0}{\pi/2}

(2 k -1)
\int_0^{\pi/2} \cos^{2k-2} \theta (-\sin\theta) \sin\theta d\theta \\
&=
(2 k -1)
\int_0^{\pi/2} \cos^{2k-2} \theta (1 – \cos^2\theta) d\theta \\
&=
(2 k -1)
\int_0^{\pi/2} \cos^{2k-2} \theta d\theta

(2 k -1)
\int_0^{\pi/2} \cos^{2k} \theta d\theta.
\end{aligned}

Bringing the $$2k$$ power integral to the other side and solving for the original integral gives a recurrence relation

\label{eqn:elipticCircumference:160}
\begin{aligned}
\int_0^{\pi/2} \cos^{2k} \theta d\theta
&= \frac{2 k – 1}{2 k}
\int_0^{\pi/2} \cos^{2k -2} \theta d\theta \\
&=
\frac{2 k – 1}{2 k}
\frac{2 k – 3}{2 k – 2}
\int_0^{\pi/2} \cos^{2k -4} \theta d\theta \\
&=
\frac{2 k – 1}{2 k}
\frac{2 k – 3}{2 k – 2}
\cdots
\frac{3}{4}
\int_0^{\pi/2} \cos^{2} \theta d\theta.
\end{aligned}

This last can also be solved using integration by parts

\label{eqn:elipticCircumference:180}
\begin{aligned}
\int_0^{\pi/2} \cos^{2} \theta d\theta
&=
\int_0^{\pi/2} \cos \theta \frac{d \sin\theta}{d\theta} d\theta \\
&=

\int_0^{\pi/2} (-\sin \theta) \sin\theta d\theta \\
&=
\int_0^{\pi/2} \lr{ 1 – \cos^2 \theta } d\theta,
\end{aligned}

or

\label{eqn:elipticCircumference:200}
\begin{aligned}
\Abs{ \Br – \Bf_{+} } + \Abs{ \Br – \Bf_{-} }
&=
\int_0^{\pi/2} \cos^{2} \theta d\theta \\
&=
\inv{2} \frac{\pi}{2}.
\end{aligned}

This gives

\label{eqn:elipticCircumference:220}
\int_0^{\pi/2} \cos^{2k} \theta d\theta
=
\frac{2 k – 1}{2 k}
\frac{2 k – 3}{2 k – 2}
\cdots
\frac{3}{4}
\frac{1}{2}
\frac{\pi}{2}.

Using the double factorial notation (factorial that skips every other value), this is

\label{eqn:elipticCircumference:260}
\boxed{
\int_0^{\pi/2} \cos^{2k} \theta d\theta
=
\frac{(2k – 1)!!}{(2k)!!} \frac{\pi}{2}
}

# References

[1] Wikipedia. Ellipse — wikipedia, the free encyclopedia, 2015. URL http://en.wikipedia.org/w/index.php?title=Ellipse&oldid=650116160. [Online; accessed 9-March-2015].