# Chaitanya's Random Pages

## May 25, 2013

### Sine of half angles

Filed under: mathematics — ckrao @ 8:34 am

I found the following cool-looking formulas in [1].

$\displaystyle \sin \frac{45^{\circ}}{2^n} = \frac{\sqrt{2 - \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}}{2}$
(n + 1 nested roots)

$\displaystyle \sin \frac{15^{\circ}}{2^n} = \frac{\sqrt{2 - \sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}}{2}$
(n + 2 nested roots)

$\displaystyle \sin \frac{18^{\circ}}{2^n} = \frac{\sqrt{8 - 2\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}}{4}$
(n + 2 nested roots)

To prove them, we use the following form double angle formula for cosine.

$\displaystyle 4 \sin^2 x = 2(1 - \cos 2x) = 2 - \sqrt{4 - 4 \sin^2 2x }\quad \quad (1)$

Now we evaluate this for different values of $x$. Firstly $\sin^2 45^{\circ} = 1/2$ implying from (1) that

$4 \sin^2 22.5^{\circ} = 2 - \sqrt{2}.\quad \quad \quad \quad (2)$

Secondly,

\begin{aligned}4 \sin^2 15^{\circ} &= 2(1 - \cos 30^{\circ})\\ &= 2 - \sqrt{3},\end{aligned}

so that from (1),

$\displaystyle 4 \sin^2 7.5^{\circ} = 2 - \sqrt{4 - (2 - \sqrt{3}} = 2 - \sqrt{2 + \sqrt{3}}. \quad \quad (3)$

To find $\sin 18^{\circ}$ one could apply Ptolemy’s theorem to four points of a regular pentagon or alternatively set $y = 36^{\circ}$ and write

\begin{aligned} \cos 3y &= -\cos (180^{\circ} - 3y)\\&= - \cos 2y.\end{aligned}

Using the double and triple angle formulae for cosine this becomes

\begin{aligned} 4 \cos^3 y - 3\cos y = 1 - 2\cos^2 y \\ \hbox{i.e.}\ 4\cos^3 y + 2\cos^2 y - 3\cos y - 1 &= 0\\ \hbox{i.e.} \ (\cos y + 1) (4 \cos^2 y - 2 \cos y - 1) &= 0.\end{aligned}

From this the only valid (positive) solution is $\cos y = (2 + 2\sqrt{5})/8$, so from (1)

$16 \sin^2 18^{\circ} = 8(1 - \cos 36^{\circ}) = 6 - 2\sqrt{5}.$

Hence applying (1) again,

$\displaystyle 16 \sin^2 9^{\circ} = 2(4 - \sqrt{16 - 16 \sin^2 18^{\circ} }) = 8 - 2\sqrt{10 + 2\sqrt{5}}.\quad \quad (4)$

Now we set up a hypothesis for mathematical induction. Assume that

$\displaystyle 4 \sin^2 \frac{45^{\circ}}{2^k} = 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k nested roots}}$
$\displaystyle 4 \sin^2 \frac{15^{\circ}}{2^k} = 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+1 nested roots}}$
$\displaystyle 16 \sin^2 \frac{18^{\circ}}{2^k} = 8 - 2\underbrace{\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+1 nested roots}}$

Note that these are equivalent to our initial three formulas. In (2)-(4) we have seen that these results are true for $k = 1$. Then using (1) and the inductive hypothesis,

\begin{aligned} 4 \sin^2 \frac{45^{\circ}}{2^{k+1}} &= 2 - \sqrt{4 - \sin^2 \frac{45^{\circ}}{2^{k}}}\\&= 2 - \sqrt{4 - \left(2 - \underbrace{ \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k nested roots}}\right)} \\ &= 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k+1 nested roots}}\end{aligned}

Similarly,

\begin{aligned} 4 \sin^2 \frac{15^{\circ}}{2^{k+1}} &= 2 - \sqrt{4 - \sin^2 \frac{15^{\circ}}{2^{k}}}\\&= 2 - \sqrt{4 - \left(2 - \underbrace{ \sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+1 nested roots}}\right)} \\ &= 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+2 nested roots}}\end{aligned}

and

\begin{aligned} 16 \sin^2 \frac{18^{\circ}}{2^{k+1}} &= 8 - 2\sqrt{16 - 16 \sin^2 \frac{18^{\circ}}{2^{k}}}\\&= 8 - 2\sqrt{16 - \left(8 - 2\underbrace{ \sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+1 nested roots}}\right)} \\ &= 8 - 2\underbrace{\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+2 nested roots}}.\end{aligned}

In each case we have shown that if the result is true for $n = k$, it is true for $n = k+1$. By the principle of mathematical induction, the result is true for $n = 1,2, \ldots$ and we are done.

#### Reference

[1] E. Maor, Trigonometric Delights, Princeton University Press, 1998.

## April 28, 2013

### A few fast centuries

Filed under: cricket,sport — ckrao @ 8:58 am

Last week Chris Gayle scored a phenomenal 175* off 66 balls in an IPL cricket game for Bangalore featuring 17 6s. It was the best display of sustained hitting by a batsman I have seen. He didn’t give the fielders any chance of catching him and he barely played a false shot. Here is his ball-by-ball scoring off legal deliveries (not including leg byes).

0 0 1 | 4 4 0 4 4 0 4 | | 1 | 6 6 4 0 6 6 | | 4 6 1 6 0 | 6 6 4 6 6 | 0 1 6 1 | 0 1 1 0 | 1 1 | 0 0 6 1 0 | 1 0 1 6 | 4 1 | 4 6 6 4 6 | 1 1 | 0 1 | 4 1 4 1 |  1 0 | 6 1 1

Notice that he scored 28, 28, 26 and 20 off four of the overs… over a hundred runs just there! Interestingly he never hit 3 sixes in a row, though he managed 10 in the space of 16 balls (73 runs total)! The graph below shows how he took his time during the middle of his innings (102 in his first 30 balls, just 14 off his next 15) before surging again to 150.

Apparently, while playing for a Jamaican XI in 2006 he scored 196 off 68 balls and was dismissed in the 15th over!!

Below are a few more fast centuries of the past. More can be seen at the following ESPN CricInfo links.

Here are those centuries from the above lists scored off fewer than 40 balls.

 Name Balls Faced Teams Venue Season/scorecard Game type CH Gayle 30 Royal Challengers Bangalore v Pune Warriors Bangalore 2013 T20 A Symonds 34 Kent v Middlesex Maidstone 2004 T20 D Hookes 34 South Australia v Victoria Adelaide 1982/3 FC LP van der Westhuizen 35 Namibia v Kenya Windhoek 2011/12 T20 GD Rose 36 Somerset v Devon Torquay 1990 List A YK Pathan 37 Rajasthan Royals v Mumbai Indians Mumbai 2009/10 T20 SB Styris 37 Sussex v Gloucestershire Hove 2012 T20 Shahid Afridi 37 Pakistan v Sri Lanka Nairobi 1996 ODI

There is also a list here of the fastest centuries of each English county season of recent years. More on fast innings through declaration bowling here.

A few centuries worth mentioning: I still have a soft spot for Shahid Afridi’s 102 off 40 balls for Pakistan vs Sri Lanka with his century off 37 balls – amazingly it was his first ODI innings.

0 6 1 0 4 0 0 6 0 0 6 6 1 1 6 6 2 6 4 4 0 0 6 6 1 4 1 1 0 4 1 6 0 6 0 2 4 1 0 X

Yusuf Pathan’s IPL century off 37 balls featured 11 consecutive legal deliveries sent for 4 or 6.

1 1 | 1 2 1 | 6 . . . 1 | 1 1 . | . 6 6 6 | 6 4 4 6 | w 4 | 4 4 w 4 w 1 | . 4 | 1 1 | 2 4 1 | 1 6 4 6

Scott Styris’s century off 37 balls featured 38 in one over:

1 | 1 1 2 1 | 1 1 . . | 2 6 1 2 | 6 1 1 | 4 2 2 6 1 | 6 4 6 6 4 . 4 6 | 4 2 6 1 | 6 6 2 1

Finally, Don Bradman scored a century off just 22 balls in three eight-ball overs (not even one dot ball!) in a game for Blackheath XI vs Lithgow:

6 6 4 2 4 4 6 1 | 6 4 4 6 6 4 6 4 | 6 6 1 4 4 6

## April 22, 2013

### A few geometric inequalities proved using complex numbers

Filed under: mathematics — ckrao @ 11:33 am

The following inequalities all follow from the elementary triangle inequality for complex numbers:

$\displaystyle \left|z_1 \right| + \left| z_2 \right| \geq \left| z_1 + z_2 \right|.$

Equality holds if and only if $0, z_1, z_2$ are collinear or in other words $z_2/z_1$ is a real number when $z_1 \neq 0$.

\begin{aligned} (p-q)(r-s) + (p-s)(q-r) &= pr - qr - ps + qs + pq - qs - pr + rs\\ &= pq - qr - ps + rs\\ &= (p-r)(q-s) \end{aligned}

Hence

\begin{aligned}\left| p-r \right| \left| q-s\right| &= \left| (p-r)(q-s) \right| \\ &= \left| (p-q)(r-s) + (p-s)(q-r)\right| \\&\leq \left| (p-q)(r-s)\right| + \left| (p-s)(q-r)\right|\end{aligned}

Hence if $P, Q, R, S$ are four points in the plane,

$\displaystyle PR \times QS \leq PQ \times RS + PS \times QR, \quad \quad \quad (1)$

which is Ptolemy’s inequality. Equality holds if and only if

$\displaystyle \frac{(p-s)(q-r)}{(p-q)(r-s)} \in \mathbb{R}$. The ratio $(p-s)/(p-q)$ is a complex number with argument $\angle QPS$ while the ratio $(q-r)/(r-s)$ has argument $\pi - \angle SRQ$. Hence for the product of these ratios to be real means $\angle QPS + \angle SRQ$ is a multiple of $\pi$. In other words, the points $P, Q, R, S$ lie on a circle.

\begin{aligned}\frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)} &= \frac{bc(c-b) + ca(a-c) + ab(b-a)}{(a-b)(b-c)(c-a)} \\ &= 1 \end{aligned}

From this,

\begin{aligned}\frac{ |b| |c|}{|a-b| |a-c|} + \frac{|c| |a|}{|b-c| |b-a|} + \frac{|a| |b|}{|c-a| |c-b|} &\geq \left| \frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)}\right|\\ &= 1. \end{aligned}

In other words, if $P, A, B, C$ are four points in the plane, $(PB\times PC)/(AB \times AC) + (PC \times PA)/(BC \times AB) + (PA \times PB)/(AC \times BC) \geq 1$, or

$\displaystyle (BC \times PB \times PC) + (CA \times PC \times PA) + (AB \times PA \times PB) \geq AB \times BC \times CA.\quad (2)$

3. Similarly we have

\begin{aligned} a^2(b-c) + b^2(c-a) + c^2(a-b) &= -(a-b)(b-c)(c-a), \end{aligned}

from which

$\displaystyle PA^2 \times BC + PB^2 \times CA + PC^2 \times AB \geq AB \times BC \times CA.\quad \quad \quad (3)$

4. Finally, we have the equality

\begin{aligned} & a^3(b-c) + b^3(c-a) + c^3(a-b) \\&= (a + b + c)(a^2(b-c) + b^2(c-a) + c^2(a-b)) - \left[ (b+c) a^2(b-c) + (c + a) b^2 (c-a) + (a+b)c^2(a-b) \right]\\ &= -(a+b+c)(a-b)(b-c)(c-a) + \left[ a^2(b^2-c^2) + b^2(c^2 - a^2) + c^2(a^2-b^2) \right] \quad \text{(from the equality in 3.)}\\ &= -(a-b)(b-c)(c-a)(a + b + c)\end{aligned}

from which

$\displaystyle PA^3 \times BC + PB^3 \times CA + PC^3 \times AB \geq 3 AB \times BC \times CA \times PG, \quad \quad \quad (4)$

where $G$ is the centroid of $\triangle ABC$ (i.e. the vector sum $PA + PB + PC$ is $3PG$).

See [2] for more details of special cases of these inequalities.

#### References

[1] A. Bogomolny, Complex Numbers and Geometry from Interactive Mathematics Miscellany and Puzzles  http://www.cut-the-knot.org/arithmetic/algebra/ComplexNumbersGeometry.shtml#cycl, Accessed 22 April 2013.

[2] T. Andreescu and D. Andrica, Proving some geometric inequalities by using complex numbers, Educatia Matematica Vol. 1, Nr. 2 (2005), pp. 19–26.

## March 30, 2013

### Great slopes of the world

Filed under: geography — ckrao @ 9:40 am

Below are a few images collected via Google Earth of significant elevation gains.

In North America two of the most significant slopes Mt McKinley in Alaska and Mt St Elias on the Alaska-Yukon border. For the former, the base-to-peak rise is something like 5500m while the latter is one of the closest big mountains to the ocean.

Mt McKinley, USA

In Peru the Cotahuasi Canyon is one of the deepest canyons in the world. The following shows a rise of almost 5,500m up to Coropuna.

Cotahuasi Canyon to Coropuna, Peru

The remaining images are from Asia. For sheer rise it’s hard to go past Nanga Parbat in Pakistan. The following shows a gain of close to 7000 metres.

Nanga Parbat, Pakistan

The south-eastern slope of this mountain (Rupal face) has been called “the largest mountain face in the world” [1] rising some 4.5km in just 7km of horizontal distance!

Rupal face of Nanga Parbat, Pakistan

Finally we turn to the mighty Himalayas. There are numerous large elevation gains here – among the most notable are the area among the peaks of DhaulagiriAnnapurna I and Machhapuchhare. The Kali Gandaki River flows through this region.

Dhaulagiri, Nepal

Dhaulagiri to Annapurna I, Nepal

Muchhapuchhare, Nepal

Finally, one of the world’s deepest canyons is Yarlung Tsangpo Grand Canyon in Tibet, China. Near Namcha Barwa the drop to the Yarlung Tsangpo River is more than 5000m.

Namcha Barwa, China

Here is a world map of the mountains mentioned in this post.

### Map

Two oceanic rises that I could not gather images for are:

• Mauna Kea in the biggest island of Hawaii, which has the largest elevation gain anywhere on earth from the ocean floor – 10,200m, though this is over a distance of more than 100km.
• From Challenger Deep, the deepest point in the ocean, to a point about 65km north west there is a rise of some 9500m from 11900m below sea level to 2400m below sea level.

#### References

[1] World Top 25 by Reduced ORShttp://www.peaklist.org/spire/lists/world-top-25-angle.html

[2] summitpost.org discussion thread: Great Elevation Gains : General – http://www.summitpost.org/phpBB3/great-elevation-gains-t57242.html

## March 19, 2013

### Recovering ratios from two decimal place approximations

Filed under: mathematics — ckrao @ 11:52 am

This post is inspired by a question I asked myself during a recent one day international cricket game I watched on TV. It was pointed out that Pakistani batsman Nasir Jamshed had an impressive average of 50.26. I knew that he hadn’t played that many games so I was surprised that his average ended in .26 which suggests that he was dismissed more than just a few times. For example if his average were 50.22 I would have imagined that this was 50 and 2/9, so I could guess he had scored 452 runs and been dismissed 9 times.

So how many dismissals could he have had if his average were 50.26, assuming it is small? This was the way I thought about it mentally: the number 0.26 is just over 1/4, so it is probably of the form k/(4k-1), then it’s a question checking numbers of this form. To two decimal digits we have

• 2/7 = 0.29
• 3/11 = 0.27
• 4/15 = 0.27
• 5/19 = 0.26 (aha!)

Hence I guessed that he had been dismissed 19 times and had scored $50 \times 19 + 5 = 955$ runs. Later on I checked that this was in fact the case. I thought it was cool that from a single average to two decimal places one could make a good guess of both the number of dismissals and the runs scored by the player (i.e. both the dividend and divisor), just based on the assumption that the ratio is simple. Of course if the decimal had been something like 50.25, it would have been more difficult to guess whether the he had scored 201, 402, 603, … runs for 4, 8, 12, … dismissals.

Below is a table showing for each value from 0.01 to 0.50 (in 0.01 increments) the simplest fraction that rounds to that value correct to two decimal places. Interestingly apart from fractions close to 0 and 0.5, the only case where the denominator exceeds 20 is for 0.34.

 Decimal Simplest ratio Decimal Simplest ratio Decimal Simplest ratio Decimal Simplest ratio Decimal Simplest ratio 0.01 1/67 0.11 1/9 0.21 3/14 0.31 4/13 0.41 7/17 0.02 1/41 0.12 2/17 0.22 2/9 0.32 6/19 0.42 5/12 0.03 1/29 0.13 1/8 0.23 3/13 0.33 1/3 0.43 3/7 0.04 1/23 0.14 1/7 0.24 4/17 0.34 10/29 0.44 4/9 0.05 1/19 0.15 2/13 0.25 1/4 0.35 6/17 0.45 5/11 0.06 1/16 0.16 3/19 0.26 5/19 0.36 4/11 0.46 6/13 0.07 1/14 0.17 1/6 0.27 3/11 0.37 7/19 0.47 7/15 0.08 1/12 0.18 2/11 0.28 5/18 0.38 3/8 0.48 10/21 0.09 1/11 0.19 3/16 0.29 2/7 0.39 7/18 0.49 17/35 0.10 1/10 0.20 1/5 0.30 3/10 0.40 2/5 0.50 1/2

To show how one of the above ratios is calculated (apart from trial and error), we will show as an example how to find fractions m/n whose decimal equivalent to two places is 0.26. In other words, $0.255 \leq m/n < 0.265$ or equivalently $51n \leq 200m < 53n$. We first find a fraction that is close to m/n, in this example 1/4 is appropriate. Using this, we set $n = 4m - d$ (more generally for the approximation p/q choose $pn = qm \pm d$) and find that $204m - 51d \leq 200m < 212m - 53d$ or equivalently,

$\displaystyle \frac{53d}{12} < m \leq \frac{51d}{4}.$

Hence for $d = 1$, $m$ ranges from 5 to 12, corresponding to the fractions $5/19, 6/23, \ldots, 12/47$ (note that $12/47 \approx 0.2553$ which rounds to 0.26 while $13/51 \approx 0.2549$ which rounds down to 0.25).

For $d = 2$, $m$ ranges from 9 to 25, corresponding to the fractions $9/34, 10/38, \ldots, 25/98$. We can continue to find all fractions satisfying the above inequality but it is apparent that the simplest will correspond to $d=1$ and $m = \lceil 53/12 \rceil = 5$.

## 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 $A=1, \phi=\pi/3$).

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 $A = 2$).

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:

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

## January 31, 2013

### Some noteworthy time zones

Filed under: geography — ckrao @ 11:10 am

I was recently learning about time zones and thought I would collect some notes about some of the more interesting ones here. In total there are 40 time zones spanning the globe.

• UTC – 12:00: Nobody lives here (it covers the uninhabited US islands of Baker Island and Howland Island in the Pacific)
• UTC -11:00: Samoa used to be in this time zone but it switched to UTC+13, skipping 30 December 2011 in the process.
• UTC -9:30: Only covered by the Marquesas Islands (French Polynesia), population ~9,000.
• UTC -4:30: Used in Venezuela since December 9 2007 (previously used UTC -4:00)
• UTC -1:00: Used in Cape Verde, eastern Greenland (in winter) and the Azores (Portugal, in winter)
• UTC +3:30: Used by Iran in winter
• UTC +4:30: Used by Afghanistan and Iran (in summer)
• UTC +5:30: Used in India and Sri Lanka, the world’s second most populated time zone
• UTC +5:45: Used in Nepal since 1986
• UTC +6:30: Used in Myanmar and Cocos Islands (Australia)
• UTC +8:45: Unofficial time zone used in Eucla and surrounds, Western Australia, population ~200
• UTC +9:30: Used in Northern Territory, South Australia (winter), western New South Wales around Broken Hill (winter) (all in Australia)
• UTC +10:30: Used in Lord Howe Island, South Australia (summer), western New South Wales around Broken Hill (summer) (all in Australia)
• UTC +11:30: Used in Norfolk Island (Australia)
• UTC +12:45: Used in Chatham Islands (New Zealand)
• UTC +14:00: The highest time zone (used by the Line Islands of Kiribati)

Here are the time zones used in the past but not at present.

• UTC -10:30: Used by Hawaii from 1900 to 1947 (now UTC -10:00)
• UTC -8:30: Used in Pitcairn Islands (UK) until 1998 (now UTC -8:00)
• UTC -0:44: Used in Liberia from 1919 to 1972 (now UTC 0:00)
• UTC -0:25: Used in Ireland from 1880 to 1916 (now UTC 0:00 in winter)
• UTC +0:20: Used in Netherlands from 1909 to 1940 (now UTC +1:00 in winter)
• UTC +0:30: Sandringham time used in the British royal household, 1901-1936
• UTC +1:30: Was used in present-day Nambia and parts of South Africa, 1892-1903
• UTC +2:30: Was used in Moscow in late 19th century (now UTC +4:00 in winter)
• UTC +4:51: Used as Bombay time (India) until 1951
• UTC +7:20: Was used as daylight saving time in Singapore between 1933 and 1940
• UTC +7:30: Used in Singapore before 1970 (now UTC +8:00)
• UTC +8:30: Formerly used in north eastern China and South Korea
• UTC +9:45: Unofficially used in Eucla and surrounds (Australia, see above) in summer when Western Australia had daylight saving

#### References

[1] Time zone (Wikipedia)

[2] List of time zones by UTC offset (Wikipedia)

## January 30, 2013

### Similar triangles fitting together in two ways

Filed under: mathematics — ckrao @ 11:58 am

In my previous mathematical post we observed that if three triangles fit together to form a triangle or quadrilateral, then they also can fit together in a second way. I thought this point was noteworthy enough to form a blog post of its own.

One of the examples from the previous post is reproduced below.

More generally, we have the following result.

Given points $A, B, C, D$ in the plane there exist points $P, Q, R, S$ in the plane so that

• $\displaystyle \triangle ABD \sim \triangle PQS$
• $\displaystyle \triangle ACD \sim \triangle RQS$
• $\displaystyle \triangle BCD \sim \triangle RPS$

To prove this, first construct $\triangle PQS \sim ABD$ and choose $R$ so that $\triangle ACD \sim \triangle RQS$ and the triangles have the same orientation (i.e. vertices labelled in the same direction clockwise or anticlockwise). This can be done in one way only. We now wish to show that in this configuration we have the third condition $\triangle BCD \sim \triangle RPS$. We shall use complex number geometry to show this.

Let $z_1$ represent the spiral similarity that maps $DA$ to $DB$.

Let $z_2$ represent the spiral similarity that maps $DB$ to $DC$.

Let $z_3$ represent the spiral similarity that maps $DC$ to $DA$.

Then the composition of these three maps is the identity so we have $z_3 z_2 z_1 = 1$. But since $\triangle ABD \sim \triangle PQS$, $z_1$ also maps $SP$ to $SQ$. Similarly $z_3$ maps $SQ$ to $SR$.

Therefore $z_3z_1 = 1/z_2$ maps $SP$ to $SR$, or in other words $z_2$ maps $SR$ to $SP$. This shows that $\triangle BCD \sim \triangle RPS$, as we wished to show.

It’s fascinating to me that non-trivial geometrical ideas can follow from the arithmetic of complex numbers (in this case, the simple fact that $z_1 z_2 z_3 = z_2 z_3 z_1$).

## January 16, 2013

### Australia’s recent heat wave

Filed under: climate and weather — ckrao @ 11:09 am

Australia has started 2013 with something like 10 of its first 11 days among the hottest 20 previously determined over some 100 years of records! The previous record area-averaged maximum of 40.17°C (set in December 1972) was broken on January 7 with 40.33°C. The two striking features of this heat wave have been its duration and wide-reaching nature. It included Hobart’s highest recorded maximum of 41.8°C and Australia’s hottest day for 15 years (49.6°C at Moomba, SA). Other very high temperatures included 49.0 at both Leonora (WA) and Birdsville (QLD).

The following graph based on data in [1] shows the maximum temperatures over the states and territories over this period. Note that the southern states of VIC and TAS are the smallest and have the least impact on the average temperature for Australia. For information on how area averages are calculated based on the weather station locations, refer to [2].

Birdsville probably was probably the hottest town in Australia over this period with the following maximum temperatures recorded (an average maximum of 45.6°C over the 13 days!). (Data from here).

 Date (Jan 2013) Maximum Temp (°C) 1 43.9 2 45.2 3 45.5 4 47.3 5 46.7 6 46.3 7 45.1 8 44.6 9 44.2 10 40.8 11 46.2 12 48.6 13 49.0

#### References

[1] Extreme January heat: Special Climate Statement Issued 7th of January 2013, updated 25th January 2013, Bureau of Meterology, Australia.

[2] D. Jones, W. Wang and R. Fawcett, High-quality spatial climate data-sets for Australia, Australian Meteorological and Oceanographic Journal 58 (2009) 233-248.

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