# Chaitanya's Random Pages

## July 29, 2016

### Distribution of Melbourne’s length of day

Filed under: geography — ckrao @ 10:34 pm
According to timeanddate.com, Melbourne, Australia in 2016 has a minimum daylength of 9 hours 32 minutes and 32 seconds, and a maximum daylength of 14 hours 47 minutes and 19 seconds (the asymmetry is due to the way day length is calculated). Here is a look at the distribution of day length through the year.
 Duration of daylength (hrs) Dates Frequency < 10 19 May-24 July 67 10-10.5 25 July-10 August, 2-18 May 34 10.5-11 11-24 August, 18 April-1 May 28 11-11.5 25 August-6 September, 5-17 April 26 11.5-12 6-19 September, 24 March-4 April 25 12-12.5 20 September-1 October, 12-23 March 24 12.5-13 1-13 October, 28 February-11 March 25 13-13.5 14-26 October, 16-27 February 25 13.5-14 9 Nov, 2-15 Feb 28 14-14.5 10 Nov-27 Nov, 16 Jan-1Feb 35 >14.5 28 Nov-15 Jan 49

What surprised me the most about this was that only 100 days of the year have daylength between 11 and 13 hours and we have a good 84 days with light longer than 14 hours.

## June 26, 2016

### 2016 has many factors

Filed under: mathematics — ckrao @ 11:16 am

The number 2016 has at least as many factors (36) as any positive integer below it except $1680 = 2^4\times 3\times 5\times 7$ (which has 40 factors). The next time a year will have more factors is $2160 = 2^4\times 3^3\times 5$, also with 40 factors.

Here are the numbers below 2160 also with 36 factors:

• $1260 = 2^2 \times 3^2 \times 5 \times 7$
• $1440 = 2^5 \times 3^2 \times 5$
• $1800 = 2^3 \times 3^2 \times 5^2$
• $1980 = 2^2 \times 3^2 \times 5 \times 11$
• $2016 = 2^5 \times 3^2 \times 7$
• $2100 = 2^2 \times 3 \times 5^2 \times 7$

The first integer with more than 40 factors is $2520 = 2^3 \times 3^2 \times 5 \times 7$ (48 factors).

#### References

[1] N. J. A. Sloane and Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 10000 terms from N. J. A. Sloane), A000005 – OEIS.

## March 28, 2016

### Recent months of global warmth

Filed under: climate and weather — ckrao @ 5:02 am

According to both NASA’s Goddard Institute for Space Studies [1] and the NOAA National Centers for Environmental Information [2], February 2016 set another record of the highest deviation of global temperatures above the monthly mean. In fact NASA’s dataset has seen the past five months record the largest five monthly global warm anomalies [3]. Some plots of global temperatures from recent months can be seen at Makiko Sato’s page here. One case in point is Longyearbyen, Svalbard (78°N) whose temperatures have barely been below average for the past six months (data from [4-5]).

February set the record of greatest anomaly from mean monthly temperatures, beating the previous record (set only the previous month) by more than 0.2°C. The map here shows that the vast majority of the planet had above-average temperatures, with the greatest deviation in the arctic region. As an example, check out the temperatures of Salekhard, Russia on the arctic circle during this time (this is a place that registers temperatures below -40 during winters). Over the month its average was 12.5°C above the mean! Data is from [6].

[2] NOAA National Centers for Environmental Information, State of the Climate: Global Analysis for February 2016, published online March 2016, retrieved on March 27, 2016 from http://www.ncdc.noaa.gov/sotc/global/201602.

## March 26, 2016

### Applying AM-GM in the denominator after flipping the sign

Filed under: mathematics — ckrao @ 8:44 pm

There are times when solving inequalities that one has a sum of fractions in which applying the AM-GM inequality to each denominator results in the wrong sign for the resulting expression.

For example (from [1], p18), if we wish to show that for real numbers $x_1, x_2, \ldots, x_n$ with sum $n$ that

$\displaystyle \sum_{i = 1}^n \frac{1}{x_i^2 + 1}\geq \frac{n}{2},$

we may write $x_i^2 + 1 \geq 2x_i$ (equivalent to $(x_i-1)^2 \geq 0$), but this implies $\frac{1}{x_i^2 + 1} \leq \frac{1}{2x_i}$ and so the sign goes the wrong way.

A way around this is to write

\begin{aligned} \frac{1}{x_i^2 + 1} &= 1 - \frac{x_i^2}{x_i^2 + 1}\\ &\geq 1 - \frac{x_i^2}{2x_i}\\ &= 1 - \frac{x_i}{2}. \end{aligned}

Summing this over $i$ then gives $\sum_{i=1}^n \frac{1}{x_i^2 + 1} \geq n - \sum_{i=1}^n (x_i/2) = n/2$ as desired.

Here are a few more examples demonstrating this technique.

2. (p9 of [2]) If $a,b,c$ are positive real numbers with $a + b + c = 3$, then

$\dfrac{a}{1 +b^2} + \dfrac{b}{1 +c^2} + \dfrac{c}{1 +a^2} \geq \dfrac{3}{2}.$

To prove this we write

\begin{aligned} \frac{a}{1 + b^2} &= a\left(1 - \frac{b^2}{1 + b^2}\right)\\ &\geq a\left(1 - \frac{b}{2}\right) \quad \text{(using the same argument as before)}\\ &=a - \frac{ab}{2}. \end{aligned}

Next we have $3(ab + bc + ca) \leq (a + b + c)^2 = 9$ as this is equivalent to $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$. This means $ab + bc + ca \leq 3$. Putting everything together,

\begin{aligned} \frac{a}{1 + b^2} + \frac{b}{1 + c^2} + \frac{c}{1 + a^2}&\geq \left( a - \frac{ab}{2} \right) + \left( b - \frac{bc}{2} \right) + \left( c - \frac{ca}{2} \right)\\ &= (a + b + c) - (ab + bc + ca)/2\\ &\geq 3 - 3/2\\ &=\frac{3}{2}, \end{aligned}

as required.

3. (based on p8 of [2]) If $x_i > 0$ for $i= 1, 2, \ldots, n$ and $\sum_{i = 1}^n x_i^2 = n$ then

$\displaystyle \sum_{i=1}^n \frac{1}{x_i^3 + 2} \geq \frac{n}{3}.$

By the AM-GM inequality, $x_i^3 + 2 = x_i^3 + 1 + 1 \geq 3x_i$, so

\begin{aligned} \frac{1}{x_i^3 + 2} &= \frac{1}{2}\left( 1 - \frac{x_i^3}{x_i^3 + 2} \right)\\ &\geq \frac{1}{2}\left( 1 - \frac{x_i^3}{3x_i} \right)\\ &= \frac{1}{2}\left( 1 - \frac{x_i^2}{3} \right). \end{aligned}

Summing this over $i$ gives

\begin{aligned} \sum_{i=1}^n \frac{1}{x_i^3 + 2} &\geq \frac{1}{2} \sum_{i=1}^n \left( 1 - \frac{x_i^2}{3} \right)\\ &= \frac{1}{2}\left( n - \frac{n}{3} \right)\\ &= \frac{n}{3}. \end{aligned}

4. (from [3]) If $x, y, z$ are positive, then

$\dfrac {x ^ 3}{x ^ 2 + y ^ 2} + \dfrac {y ^ 3}{y ^ 2 + z ^ 2} + \dfrac {z ^ 3}{z ^ 2 + x ^ 2 } \geq \dfrac {x + y + z}{2}.$

Once again, focusing on the denominator,

\begin{aligned} \dfrac {x ^ 3}{x ^ 2 + y ^ 2} &= x\left(1 - \dfrac {y ^ 2} {x ^ 2 + y ^ 2} \right)\\ &\geq x \left(1 -\dfrac{xy^2}{2xy} \right)\\ &= x-\dfrac{y}{2}. \end{aligned}

Hence,

\begin{aligned} \dfrac {x ^ 3}{x ^ 2 + y ^ 2} + \dfrac {y ^ 3}{y ^ 2 + z ^ 2} + \dfrac {z ^ 3}{z ^ 2 + x ^ 2 } &\geq x-\dfrac{y}{2} + y-\dfrac{z}{2} + z-\dfrac{x}{2}\\ &= \dfrac {x + y + z}{2}, \end{aligned}

as desired.

5. (from the 1991 Asian Pacific Maths Olympiad, see [4] for other solutions) Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be positive numbers with $\sum_{i = 1}^n a_i = \sum_{i = 1}^n b_i$. Then

$\displaystyle\sum_{i=1}^n\frac{a_i^2}{a_i + b_i} \geq \frac{1}{2}\sum_{i=1}^n a_i.$

Here we write

\begin{aligned} \sum_{i=1}^n\frac{a_i^2}{a_i + b_i} &= \sum_{i=1}^n a_i \left(1 - \frac{b_i}{a_i + b_i} \right)\\ &\geq \sum_{i=1}^n a_i \left(1 - \frac{b_i}{2\sqrt{a_i b_i}} \right) \\ &= \frac{1}{2} \sum_{i=1}^n \left( 2a_i - \sqrt{a_i b_i} \right) \\ &= \frac{1}{4} \sum_{i=1}^n \left( 4a_i - 2\sqrt{a_i b_i} \right)\\ &= \frac{1}{4} \sum_{i=1}^n \left( 2a_i +a_i - 2\sqrt{a_i b_i} + b_i \right) \quad \text{(as } \sum_{i=1}^n a_i = \sum_{i=1}^n b_i\text{)}\\ &= \frac{1}{4} \sum_{i=1}^n \left( 2a_i + \left(\sqrt{a_i} - \sqrt{b_i}\right)^2 \right)\\ &\geq \frac{1}{4} \sum_{i=1}^n 2a_i\\ &= \frac{1}{2} \sum_{i=1}^n a_i, \end{aligned}

as required.

#### References

[1] Zdravko Cvetkovski, Inequalities: Theorems, Techniques and Selected Problems, Springer, 2012.

## February 28, 2016

### The race up the charts for two recent movies

Filed under: movies and TV — ckrao @ 7:59 pm

In recent times Jurassic World and Star Wars VII (The Force Awakens) have respectively become the fourth and third biggest movies of all time worldwide (behind Avatar and Titanic). Here is how they ranked in the all-time US/Canada charts day by day (using data from boxofficemojo.com).

 Day Jurassic World Star Wars: The Force Awakens 1 786 464 2 275 183 3 143 96 4 108 68 5 79 40 6 69 29 7 54 22 8 37 11 9 28 6 10 18 5 11 15 5 12 11 5 13 10 4 14 9 3 15 7 2 16-19 5 2 20-21 5 1 22-44 4 1 45+ 3 1

I was amazed by Jurassic World’s summer run and then that of The Force Awakens simply blew my mind.🙂

## February 27, 2016

### Cutting a triangle in half

Filed under: mathematics — ckrao @ 9:40 pm

Here is a cute triangle result that I’m surprised I had not known previously. If we are given a point on one of the sides of a triangle, how do we find a line through the triangle that cuts its area in half?

Clearly if that point is either a midpoint or one of the vertices, the answer is a median of the triangle. A median cuts a triangle in half since the two pieces have the same length side and equal height.

So what if the point is not a midpoint or a vertex? Referring to the diagram below, if $P$ is our desired point closer to $A$ than $B$, the end point $Q$ of the area-bisecting segment would need to be on side $BC$ so that area(BPQ) = area(ABC)/2.

In other words, we require area(BPQ) = area(BDQ), or, subtracting the areas of triangle BDQ from both sides,

$\displaystyle |DPQ| = |DCQ|.$

Since these two triangles share the common base $DQ$, this tells us that we require them to have the same height. In other words, we require $CP$ to be parallel to $DQ$. This tells us how to construct the point $Q$ given $P$ on $AB$:

1. Construct the midpoint D of $AB$.
2. Draw $DQ$ parallel to $AP$.

See [1] for an animation of this construction.

In turns out that the set of all area-bisecting lines are tangent to three hyperbolas and enclose a deltoid of area $(3/4)\ln(2) - 1/2 \approx 0.01986$ times the original triangle. [2,3,4]

#### References

[1] Jaime Rangel-Mondragon, “Bisecting a Triangle” http://demonstrations.wolfram.com/BisectingATriangle/ from the Wolfram Demonstrations Project Published: July 10, 2013

[4] Henry Bottomley, Area bisectors of a triangle, January 2002

## January 31, 2016

### Most wickets after n test matches from debut

Filed under: cricket,sport — ckrao @ 12:18 am

Here is a list of the leading test cricket wicket takers after playing n matches. A few current-day players feature and I hope to update this list over time (last updated January 31 2016). It was created largely manually using ESPNCricinfo’s Statsguru starting from [1] and [2], and using lists of the fastest to multiples of 50 wickets here.  Corrections are more than welcome (updated: August 8 2016).

#### References

[1] Most wkts in consec Tests from debut – Google Groups

[2] Top 10 bowlers with most wickets in 10 Tests or less (crictracker.com)

## January 30, 2016

### Patterns early in the digits of pi

Filed under: mathematics — ckrao @ 8:59 pm

Recently when taking a look at the early decimal digits of $\pi$ I made the following observations:

3.141592653589793238462643383279502884197169399375105820974944592307816406286…

• The first run of seven distinct digits (8327950, shown underlined) appears in the 26th to 32nd decimal place. Curiously the third such run (5923078, also underlined) in decimal places 61 to 67 contains the same seven digits. (There is also a run of seven distinct digits in places 51 to 57 with 5820974.)
• Decimal digits 60 to 69 (shown in bold) are distinct (i.e. all digits are represented once in this streak). The same is true for digits 61 to 70 as both digits 60 and 70 are ‘4’.

Assume the digits of $\pi$ are generated independently from a uniform distribution. Firstly, how often would we expect to see a run of 7 distinct digits? Places $k$ to $k+6$ are distinct with probability

$\displaystyle \frac{9}{10} \times \frac{8}{10} \times \frac{7}{10} \times \frac{6}{10} \times \frac{5}{10} \times \frac{4}{10} = \frac{9!}{3.10^6} = \frac{189}{3125}.$

Hence we expect runs of 7 distinct digits to appear $3125/189 \approx 16.5$ places apart. This includes the possibility of runs such as 12345678 which contain two runs of 7 distinct digits that are only 1 place apart.

How often would we expect to see the same 7 digits appearing in a run as we did in places 26-32 and 61-67? Furthermore let’s assume the two runs have no overlap, so we discount possibilities such as 12345678 which have a six-digit overlap. We expect a given sequence (e.g. 1234567, in that order) to appear $1/10^7$ of the time. There are $7! = 5040$ permutations of such a sequence, but of these $1$ has overlap 6 (2345671), $2!$ has overlap 5 (3456712 or 3456721), $3!$ has overlap 4, …, $6!$ has overlap 1 with the original sequence. This leaves us with $5040 - (1 + 2 + 6 + 24 + 120 + 720) = 4167$ possible choices of the next run to have the same 7 digits but non-overlapping (or to appear in precisely the same order – overlap 7). Hence we expect the same 7 digits to recur (no overlap with the original run) after approximately $10^7/4167 \approx 2400$ places apart so what we saw in the first 100 places was remarkable.

Now let us turn to runs of all ten distinct digits. Repeating the argument above, such runs occur every $10^{10}/10! \approx 2756$ places. According to [1] the next time we see ten distinct digits is in decimal places 5470 to 5479.

To answer the question of when we would expect to see the first occurrence of ten distinct digits, we adopt an argument from renewal-reward theory based on Sec 7.9.2 of [2] (there also exist approaches based on setting up recurrence relations, or martingale theory (modelling a fair casino), see [3]-[4]). Firstly we let $T$ be the first time we get 10 consecutive distinct values – we wish to find $E[T]$ where $E$ denotes the expected value operator. Note that this will be more than the 2756 answer we obtained above since we make no assumption of starting with a run of ten distinct digits – there is no way $T$ could be 1 for example, but we could have two runs of ten distinct digits that are 1 apart.

From a sequence of digits we first define a renewal process in which after we get 10 consecutive distinct values (at time $T$) we start over and wait for the next run of 10 consecutive distinct values without using any of the values  up to time $T$. Such a process will then have an expected length of cycle of $E[T]$.

Next, suppose we earn a reward of $1 every time the last 10 digits of the sequence are distinct (so we would have obtained$1 at each of decimal places 69 and 70 in the $\pi$ example). By an important result in renewal-reward theory, the long run average reward is equal to the expected reward in a cycle divided by the expected length of a cycle.

In a cycle we will obtain

• $1 at the end •$1 at time 1 in the cycle with probability $1/10$ (if that digit is the same as ten digits before it)
• $1 at time 2 in the cycle with probability $2/100$ (if the last two digits match those ten places before it) •$1 at time 9 in the cycle with probability $9!/10^9$ (if the last nine digits match those ten places before it)

Hence the expected reward in a cycle is given by

$\displaystyle 1 + \sum_{i=1}^9 \frac{i!}{10^i} = \sum_{i=0}^9 \frac{i!}{10^i}.$

We have already seen that the long run average reward is $10!/10^{10}$ at each decimal place. Hence the expected length of a cycle $E[T]$ (i.e. the expected number of digits before we expect the first run of ten consecutive digits) is given by

$\frac{10^{10}}{10!}\sum_{i=0}^9 \frac{i!}{10^i} \approx 3118.$

Hence it is pretty cool that we see it so early in the decimal digits of $\pi$.🙂

#### References

[2] S. Ross, Introduction to Probability Models, Academic Press, 2014.

[4] A Collection of Dice Problemsmadandmoonly.com

## December 30, 2015

### Novak Djokovic against younger players

Filed under: sport — ckrao @ 9:11 pm

Novak Djokovic had a stellar tennis record in 2015, winning three out of the four grand slams (runner-up in the other), the ATP World Tour finals for the fifth time, a record six Masters titles and reached 15 consecutive finals (winning 11 of them) in attaining an 82-6 record (31 wins against top 10 players!!) and US\$21.6m for the year.

I noticed that all his losses for the year were to players older than him and I looked up his record against younger players. Here is a list of the players younger than Djokovic who have beaten him in his entire ATP career. The month refers to when the associated tournament started.

• Ernests Gulbis (Jan 2009)
• Filip Krajinovic (May 2010 – Djokovic retired after losing the first set)
• Juan Martin del Potro (Sep 2011, Jul 2012 and Mar 2013)
• Kei Nishikori (Oct 2011 and Aug 2014)
• Sam Querry (Oct 2012)
• Grigor Dimitrov (May 2013)

That’s right, only six players have had this honour for a total of nine losses. After his loss to Dimitrov at the Madrid Masters he has won 71 of his last 72 matches against younger players, losing only to Nishikori in the 2014 US Open semi-finals. He is currently on a 40-match winning streak and to the end of 2015 his record is 130-9 against younger players (career: 686-146). As a comparison here is the record of Murray, Nadal and Federer against younger players not named Novak Djokovic, Andy Murray or Rafael Nadal.

• Murray: 118-19
• Nadal: 177-17 (including 41 in a row from Nov 2009 to the end of 2011)
• Federer: 517-61 (the higher numbers show the clear generation gap)

#### Reference

Tennis Abstract: ATP and WTA Match Results, Splits, and Analysis

## December 23, 2015

### Areas of sections of a triangle from distances to its sides

Filed under: mathematics — ckrao @ 12:35 pm

If a point $P$ is in the interior of triangle $ABC$ distance $x, y$ and $z$ from the sides, what is the ratio of the area of quadrilateral $BXPZ$ to that of $ABC$?

One way of determining this is to draw parallels to the sides of the triangles through $P$. Let $X_1$ and $X_2$ be where these parallels meet side $BC$ as shown below.

Let the sides of the triangles have lengths $a, b, c$ with corresponding altitudes $h_a, h_b, h_c$.

Then as $\triangle PX_1 X_2$ and $\triangle ACB$ are similar,

\begin{aligned}|PX_1X| &= |PX_1X_2| \frac{X_1X}{X_1X_2}\\ &= |PX_1X_2| \frac{b\cos C}{a}\\ &= |PX_1X_2| \frac{b (a^2 + b^2 - c^2)}{2a^2b} \quad \text{ (cosine rule)}\\ &= \left(\frac{x}{h_a} \right)^2|ABC|\frac{(a^2 + b^2 - c^2)}{2a^2}\\ &= |ABC|\left(\frac{ax}{ax+by+cz}\right)^2\frac{(a^2 + b^2 - c^2)}{2a^2}\\&= \frac{|ABC|x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2},\quad\quad (1) \end{aligned}

where the second last line follows from twice the area of |ABC| being $ah_a = ax + by + cz$.

Similarly,

$\displaystyle |PY_1Z| = \frac{|ABC|z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}.\quad \quad (2)$

Finally,

\begin{aligned}|X_1Y_1B| &= \left(\frac{h_b-y}{h_b}\right)^2|ABC|\\ &= \left(1-\frac{by}{bh_b}\right)^2 |ABC|\\ &= \left(1-\frac{by}{2|ABC|}\right)^2|ABC|\\ &= \left(1-\frac{by}{ax+by+cz}\right)^2|ABC|\\ &= \left(\frac{ax +cz}{ax+by+cz}\right)^2|ABC|. \quad\quad(3)\end{aligned}

Combining (1), (2) and (3), we obtain our desired answer as

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{|X_1Y_1B|-|PX_1X|-|PY_1Z|}{|ABC|}\\&= \left(\frac{ax +cz}{ax+by+cz}\right)^2-\frac{x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2}-\frac{z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}\\&=\frac{(ax+cz)^2 - x^2(a^2 +b^2-c^2)/2 - z^2(b^2+c^2-a^2)/2}{(ax+by+cz)^2}\\ &= \frac{2axcz + x^2(a^2 - b^2 + c^2) + z^2(c^2 +a^2-b^2)}{(ax+by+cz)^2}\\&= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}.\quad\quad(4)\end{aligned}

Similar formulas can be found for quadrilaterals $XPYC$ and $YPZA$ by permuting variables. Note that if $P$ is outside the triangle or if the triangle is obtuse-angled, care must be taken in the signs of the areas (the quadrilaterals may not be convex) and variables $x, y, z$.

Note that (4) may also be written as

$\displaystyle \frac{|BXPZ|}{|ABC|} = \frac{ac(2xz + (x^2 + z^2)\cos B)}{(ax+by+cz)^2}.\quad\quad(5)$

### Special cases

1) If $\triangle ABC$ is equilateral, $a=b=c$ and from (4) we obtain

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\ &= \frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\quad\quad(6)\end{aligned}

2) If $P$ is at the incentre of $\triangle ABC$, then $x = y = z = r$ (the inradius) and from (4) we have

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\ &= \frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\quad\quad(7)\end{aligned}

3) If $\triangle P$ is right-angled at $B$, then quadrilateral $BXPZ$ is a rectangle with area $xz$ and $\triangle ABC$ has area $ac/2$ and from (5),

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{2acxz )}{(ax+by+cz)^2}\\ &= \frac{2acxz )}{(ac)^2}\\ &= \frac{2xz}{ac}.\quad \quad (8)\end{aligned}

as expected.

4) If $a=c$ and $x=z$ (symmetric isosceles triangle case) then from (4),

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2x^2 + 2x^2(2a^2-b^2)}{2(2ax+by)^2}\\ &= \frac{x^2(4a^2 -b^2)}{(2ax+by)^2}.\quad\quad(9)\end{aligned}

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