Chaitanya's Random Pages

February 21, 2013

The non-winning streak of the West Indies in Australia

Filed under: cricket,sport — ckrao @ 11:32 am

It surprised me to learn that the West Indies recently won their first match in Australia against Australia since early 1997. Over 32 matches (11 tests, 19 one day internationals, 2 Twenty 20 internationals) prior to the win they had two no results in ODIs and one draw in the tests. In the first of the no-result games they were in a winning position, reducing Australia to 5/43 chasing 264 before rain washed out the game.

A list of the matches can be found via ESPN CricInfo’s Statsguru here. The best batting and bowling performances by the Windies during that time were as follows.

  • Brian Lara scored two test centuries (including a 226 in Adelaide) and one ODI century.
  • Chris Gayle scored two test centuries including 165* in the drawn game and the fifth fastest ever test century (in balls faced).
  • Dwayne Bravo also scored two test centuries and had the best test bowling figures of 6/84.
  • Pedro Collins played two ODIs, getting figures of 5/43 and 3/8!

I’ll take this chance to link to a clip of highlights of Lara’s 226.

February 20, 2013

Summing two sinusoids where one has double the frequency of the other

Filed under: mathematics — ckrao @ 11:10 am

In this post we look at the sum two sinusoids of different amplitude and phase but where one has twice the frequency of the other. How should we choose the phase so that the sum has minimal peak?

Mathematically, given a constant A > 0 we wish to find

\displaystyle \min_{\phi} \max_x \sin 2x + A \sin (x + \phi ).

The sum can take a variety of forms, two of which are shown in red below (A=2, \phi = 0 and A=1, \phi=\pi/3).

2sinusoids_phi=02sinusoids_phi=pion3

We claim that a value of \phi for which the peak value of \sin 2x + A \sin (x + \phi ) is maximised is \phi = 3 \pi/4 as illustrated below (again A = 2).

2sinusoids_phi=3pion4

We see here the two sinusoids reach their minimum at x = 3\pi /4, resulting in a minimum value of -A -1 for the sum. However the maximum value is only 1.5 for the case A=2. Hence we make the observation that finding the lowest peak is different from finding the minimum deviation from zero.

I initially approached this problem using the method of Lagrange multipliers. Consider the problem of finding the maximum value of f(x) = \sin 2x + A \sin (x + \phi ) where A and \phi are fixed. We write

\begin{aligned} f(x) &= 2 \sin x \cos x + A (\sin x \cos \phi + \cos x \sin \phi )\\ &= 2(uv + ru + sv),\end{aligned}

where u = \cos x, v = \sin x are variables and r = \frac{A}{2} \cos \phi, s = \frac{A}{2} \sin \phi are constants. Hence we wish to maximise uv + ru + sv subject to u^2 + v^2 = 1. By the method of Lagrange multipliers, at the maximum the gradient vector of the objective function is parallel to the gradient of the constraint. In other words, (v + r, u + s) \parallel (u, v). Hence there exists a constant \lambda so that

\displaystyle v + r = \lambda u \ \ \ \ \ (1)
\displaystyle u + s = \lambda v \ \ \ \ \ (2)

Solving these two equations for u and v gives u = (\lambda r + s)/(\lambda^2 -1) and v = (r + \lambda s)/(\lambda^2 -1). Then the condition u^2 + v^2 = 1 gives us the quartic equation

\displaystyle (\lambda r + s)^2 + (r + \lambda s)^2 = (\lambda^2 - 1)^2. \ \ \ \ \ (3)

By multiplying (1) by u and (2) by v and adding, we obtain

\begin{aligned} 2(uv + ru + sv) &= 2(\lambda - uv)\\ &= 2 \left( \lambda - (\lambda r + s)(r + \lambda s) / (\lambda^2 -1)^2 \right), \end{aligned}

where \lambda is one of the solutions of (3).

Unfortunately only for a few special cases does the solution in x have a “nice” form. So this approach to finding the best \phi was not appealing.

After conjecturing that the best \phi is 3\pi/4, one alternative approach is to proceed as follows. We firstly note that for \phi = 3 \pi /4,

\begin{aligned} \sin 2x + A \sin (x + \phi) &= 2 \sin x \cos x + A \frac{\sqrt{2}}{2} \left(\cos x - \sin x \right)\\ &= 1 - (\cos x - \sin x)^2 + A \frac{\sqrt{2}}{2} \left(\cos x - \sin x \right)\\ &= -(\cos x - \sin x - A \frac{\sqrt{2}}{4})^2 + 1 + \frac{A^2}{8}\\ &\leq \begin{cases} 1 + \frac{A^2}{8} & \text{if } A \leq 4\\ -(-\sqrt{2} - A \frac{\sqrt{2}}{4})^2 + 1 + \frac{A^2}{8} = A-1 & \text{if } A > 4. \end{cases}\end{aligned}

Hence the maximum value is 1 + A^2/8 or A - 1 depending on how large A is.

Now we wish to show that for other values of \phi there exists x for which this maximum value is exceeded. Firstly note that the more interesting case is A < 4 since when A \geq 4, we can achieve a minmax value of A-1 by aligning the peak of A \sin (x + \phi ) with the trough of \sin 2x (e.g. for \phi = 3 \pi/4), obtaining a sum value of A - 1 at x = -\pi/4. This is at the local maximum since here,

f'( -\pi/4) = 2 \cos (-\pi/2) + A \cos (-\pi/4 + 3\pi/4) = 0 +0 = 0

and

f''(-\pi /4) = -2 \sin \pi /2 - A \sin (-\pi / 4 + 3\pi /4) = -2 - A < 0.

From now on we consider A < 4.

Choose x such that \sin (x + \phi) = \frac{A}{4} (i.e. x + \phi is kept constant). Then \cos (x + \phi) = \pm \sqrt{1 - A^2/16},

\sin 2(x + \phi ) = 2 \sin (x + \phi ) \cos (x + \phi ) = \pm \frac{A}{2} \sqrt{1 - A^2/16} and

\cos 2(x + \phi ) = 2 \cos^2 (x + \phi ) - 1 = 2(1 - A^2/16) - 1 = 1 - A^2/8. Then

\begin{aligned} \sin 2x &= \sin(2(x + \phi ) - 2\phi )\\&= \sin 2(x+\phi ) \cos 2\phi - \cos 2(x + \phi ) \sin 2\phi \\ &= \pm \frac{A}{2} \sqrt{1 - A^2/16} \cos 2\phi - (1 - A^2/8) \sin 2\phi. \end{aligned}

Write this as g(\phi) = C_1 \cos 2\phi + C_2 \sin 2\phi, and note that C_1 \neq 0. We wish to show if \sin 2\phi \neq -1 there exists \phi close to 3\pi/4 such that g(\phi ) > -C_2. We use the fact that near \phi = 3\pi/4, \cos 2\phi = 2(\phi-3\pi/4) + o((\phi-3\pi/4)^2) while \sin 2\phi = -1 + (2(\phi-3\pi/4))^2/2 + o((\phi-3\pi/4)^3). In particular for all \phi sufficiently close (but not equal) to 3\pi/4,

\displaystyle \left| \cos 2\phi \right| > 2 \epsilon - \frac{(2\epsilon)^3}{3!}

\displaystyle \left| 1 + \sin 2\phi \right| < (2 \epsilon^2),

where |\phi - 3\pi/4| := \epsilon > 0. Furthermore choose \epsilon so that 4\epsilon^2 < 3 (i.e. (2 \epsilon)^3/6 < \epsilon) and \frac{1}{2\epsilon} \geq \left| C_2/C_1 \right|. We then find

\begin{aligned} \left| \frac{\cos 2\phi}{1 + \sin 2\phi} \right| &> \frac{2\epsilon - (2 \epsilon)^3/3!}{2 \epsilon^2} \\ &> \frac{\epsilon}{2\epsilon^2}\\ &= \frac{1}{2 \epsilon}\\ &\geq \left| \frac{C_2}{C_1} \right|. \end{aligned}

It follows that

-C_2(1 + \sin 2\phi ) \leq \left| C_2(1 + \sin 2\phi ) \right| < \left| C_1 \cos 2\phi \right| = C_1 \cos 2\phi ,

or C_2 (1 + \sin 2\phi) + C_1 \cos 2\phi > 0, where C_1 \cos 2\phi = \pm \frac{A}{2} \sqrt{1 - A^2/16} \cos 2\phi can be chosen to be positive for some \phi, x. Hence we have \sin 2x = C_1 \sin 2\phi + C_2 \cos 2\phi > -C_2 as desired and

\begin{aligned}\sin 2x + A \sin (x + \phi ) &> -C_2 + \frac{A^2}{4}\\ &= ( 1 - A^2/8) + \frac{A^2}{4} \\ &= 1 + \frac{A^2}{8}. \end{aligned}

Hence we would have an x for which the claimed maximum value of (1 + A^2/8) has been exceeded. We conclude that we must have \cos 2 \phi = 0 and \sin 2\phi = -1 (e.g. when \phi = 3\pi/4), leading to the minmax value of 1 + A^2/8.

Recapping, we have shown that

\displaystyle \min_{\phi} \max_x \sin 2x + A \sin (x + \phi ) = \begin{cases} 1 + A^2/8 & \text{ if } A \leq 4\\ A - 1 & \text{ if } A > 4.\end{cases}

It is interesting to see that the behaviour of the minmax changes from being quadratic to linear in A as A exceeds 4. For the examples plotted above, A = 2, leading to a minmax value of 1 + 2^2/8 = 3/2 as we found in the third graph.

If we wish to find \min_{\phi} \max_x A_1 \sin 2x + A_2 \sin (x + \phi ) where A_1, A_2 > 0, we simply write this as \min_{\phi} \max_x A_1 ( \sin 2x + A \sin (x + \phi ) where A = A_2/A_1 and use the above result to obtain the following:

\displaystyle \min_{\phi} \max_x A_1 \sin 2x + A_2 \sin (x + \phi ) = \begin{cases} A_1 + A_2^2/8A_1 & \text{ if } A_2 \leq 4A_1\\ A_2 - A_1 & \text{ if } A_2 > 4A_1 \end{cases}

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