## Saturday, May 07, 2016

### Nil Battey Sannata Review

I just finished Nil Battey Sannata. What a wonderful movie about a single mom who is trying to raise her daughter. The film was amazing and gives insight into what many of us are not even aware of the plight of poor people. For example growing up I was always told that or kind of knew that I will be an engineer. So in a way I always had confidence about myself and the future. The film hit me hard when you realize that not everyone is so fortunate.

Arvind Kejriwal and Manish Sisodia has made this movie tax free in Delhi and that speaks volume about the quality of the movie and their intention is to make education free for all. It's an uplifting story that should be seen by everyone. Few weeks back there was a story about the drought in Bundelkhand district of India and how the people were fleeing their villages and migrating to big cities. They hardly had enough money to pay for the train tickets. Most of these families will end up with menial jobs in the big city and that will totally wreck the future of their families.
I am totally convinced that good quality education is the most important tool to uplift the economy of the country. Take for instance the Ecuador example the president there has been proactive in implementing quality education through the IB program. They have been aiming to have over 500 IB schools by 2017. Working in IB school has taught me its one of the best program for schools as it prepares them in multi facet way. Especially I like the IB learner Profile where they are trained to be better communicative, risk-takers, inquirers, knowledgeable, open-minded, thinkers, principled, caring, balanced and reflective. Because once you have defined it then its easy to focus on each of those ideas and that's where I think IB shines.
So to sum up my review of Nil Battey Sannanta. Please go watch it. It's a great uplifting movie that should be shown in each government school in India. It helps one to think and have dreams!

## Sunday, May 01, 2016

### Bringing Moocs to solve India's higher education problem

One of the big problem in India right now is to provide quality higher education to every one. One of the good step in that direction is the NPTEL program and they have done a good job of providing certification for their online courses. However this year just in Delhi there will be over 150,000 kids who will have a rough time getting admission in university as there are not enough number of seats. What government can do is to open a mooc university which will help not only in reskilling but can also provide everybody the opportunity to gain a certification and degree. Considering the success of Georgia Universities tie up with udacity to provide a legitimate degree and now Coursera jumping into that its high time we do this in India.  Edx is also providing the global freshman curriculum through Arizona State University. American Universities will be very reluctant to provide degree as the end product as education is a big business and this is where India can come in and fill the gap. Several factors work in our favor. We have a pressing need to educate large number of people and we can do it cheaply as we are doing with NPTEL certification.

Take for instant the MITs free OCW program. Its been 10 years and still they haven't been able to provide all the video lectures for their courses and they are way behind India's NPTEL in terms of sharing their resources inspite of the massive funding they enjoy!

Coursera when started initially had big and lofty goal of making the content available for free but inspite of thier best intention have now been struggling to provide the upper level courses.
India will have to take a lead by providing the way to certification. The model could be where you can take exam and get the degree. The exams should be very hard and test your mastery and should be affordable. The goal is that anyone who has passed those exams has demonstrated the mastery of the subject.

## Saturday, April 16, 2016

### Angry Indian Goddesses

I think this is one of the best movie to have come highlight the plight of women in India. It covers many areas that women in Indian find themselves at the receiving end and anyone who has lived there can readily identify with that. It's truly a wonderful one directed by Pan Nalin. Finally a movie that should be watched by every Indian. I hope Kejriwal sees it and tweets about it and more people get to see this really touching movie and I do hope that it inspires new generation of directors to take up such issues and bring positive change in the society. Kudos Pan Nalin, cast and everyone involved for the amazing work.

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## Sunday, January 10, 2016

### Proof Cauchy Riemann Equations

Cauchy Riemann equations are necessary conditions but not sufficient conditions to check if a function is analytic or not. According to the definition of derivative we have
$f'(z_0)=\lim\limits_{\Delta z\to 0}\dfrac{f(z_0+\Delta z)-f(z_0)}{\Delta z}$
One important fact to take notice is $f(z)=f(x+iy)=f(x,y)$
$\Rightarrow f'(z_0)=\lim\limits_{\Delta x+i \Delta y \to 0}\dfrac{f(z_0+\Delta z)-f(z_0)}{\Delta x+ i\Delta y}$
For the derivative to exists at point $z_0$ it should be same no matter what path we take to reach $z_0$
$\Rightarrow f'(z_0)=\lim\limits_{x_0\to 0}\dfrac{f(z_0+\Delta z)-f(z_0)}{\Delta x}$
Lets go around only x axis which means $\Delta y=0$
$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta x \to 0}\dfrac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}$
$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta x\to 0}\dfrac{u(x_0+\Delta x,y_0)+iv(x_0+\Delta x,y_0)-(u(x_0,y_0)+iv(x_0,y_0))}{\Delta x}$
$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta x\to 0}\dfrac{u(x_0+\Delta x,y_0)-u(x_0,y_0)+iv(x_0+\Delta x,y_0)-iv(x_0,y_0)}{\Delta x}$
$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta x\to 0}\dfrac{u(x_0+\Delta x,y_0)-u(x_0,y_0)}{\Delta x}+\lim\limits_{\Delta x\to 0}\dfrac{iv(x_0+\Delta x,y_0)-iv(x_0,y_0)}{\Delta x}$
$\Rightarrow f'(z_o)=u_x+iv_x$
Now let’s go along only y axis i.e $\Delta x = 0$
$\Rightarrow f'(z_0)=\lim\limits_{\Delta y\to 0}\dfrac{f(z_0+\Delta z)-f(z_0)}{i\Delta y}$
$\Rightarrow f'(x_0+iy_0)=\lim\limits_{\Delta y\to 0}\dfrac{f(z_0+\Delta z)-f(z_0)}{i\Delta y}$
$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta y \to 0}\dfrac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{i\Delta y}$
$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta y\to 0}\dfrac{u(x_0,y_0+\Delta y)+iv(x_0,y_0+\Delta y)-(u(x_0,y_0)+iv(x_0,y_0))}{i\Delta y}$
$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta y\to 0}\dfrac{u(x_0,y_0+\Delta y)-u(x_0,y_0)+iv(x_0,y_0+\Delta y)-iv(x_0,y_0)}{i\Delta y}$
$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta y\to 0}\dfrac{u(x_0,y_0+\Delta y)-u(x_0,y_0)}{i\Delta y}+\lim\limits_{\Delta x\to 0}\dfrac{iv(x_0,y_0+\Delta y)-iv(x_0,y_0)}{i\Delta y}$
$\Rightarrow f'(x_0,y_0)=\lim\limits_{\Delta y\to 0}-i\dfrac{u(x_0,y_0+\Delta y)-u(x_0,y_0)}{\Delta y}+\lim\limits_{\Delta x\to 0}\dfrac{v(x_0,y_0+\Delta y)-v(x_0,y_0)}{\Delta y}$
$\Rightarrow f'(z_o)=-iu_y+v_y$
$\Rightarrow f'(z_o)=v_y-iu_y$
Now if the derivative has to exist both of these should be same i.e
$\Rightarrow f'(z_o)=u_x+iv_x=v_y-iu_y$
Equating real and imaginary parts we get
$\Rightarrow u_x=v_y$ and $\Rightarrow u_y=-v_x$

### -1/12

Let’s define three different series
$S = 1+2+3+4+...$
$S_1 = 1-1+1-1+1-1+... =\frac{1}{2}$
$S_2 = 1-2+3-4+5-...$
To find $S_2$ Let’s place a shifted copy of itself and add the corresponding terms $2S_2=S_2 +S_2 = (1-2+3-4+5-...)+(1-2+3-4+5-...)=1-1+1-1+1-1+.. = \frac{1}{2} \Rightarrow S_2= \frac{1}{4}$
Now lets find $S-S_2=(1+2+3+4+...)-(1-2+3-4+5-...)=(4+8+12+...)=4(1+2+3+...)=4S$

$\Rightarrow -S_2=3S$

$\Rightarrow -\frac{1}{4}=3S$
$\Rightarrow -\frac{1}{12}=S$
The second proof also makes use of the idea we learned that
$S_2 = 1-2+3-4+5-... =\frac{1}{4}$
The way he goes to proof is using the idea of power series and differentiating it
$\frac{1}{1-x}=1+x+x^2+x^3+...$
Differentiating both sides we get
$\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+5x^4+...$
Now plug in $-1$ on both sides we get
$\Rightarrow \frac{1}{4}=1-2+3-4+5-...$ and we have the same result.
The next thing he does is invoke the Reimann zeta function $\zeta (s) = \frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+...$
Notice that $\zeta(-1) = 1+2+3+4+5+...$
Now multiply both sides of $\zeta(s)$ by $\frac{2}{2^s}$ we get
$\Rightarrow \zeta(s)\frac{2}{2^s}=\frac{2}{2^s}(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+...)=(\frac{2}{2^s}+\frac{2}{4^s}+\frac{2}{6^s}+\frac{2}{8^s}+...)$
Now subtracting this equation from the original $\zeta(s)$ we get $\Rightarrow \zeta(s)(1-\frac{2}{2^s})=\frac{1}{1^s}-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+...$
Plugging back $s= -1$ we get
$\Rightarrow \zeta(-1)(1-\frac{2}{2^{-1}}=1-2+3-4+5-6+...=\frac{1}{4}$
$\Rightarrow \zeta(-1)(-3)=\frac{1}{4}$
$\Rightarrow \zeta(-1)=\frac{-1}{12}$
$\Rightarrow 1+2+3+4+5+... =\frac{-1}{12}$

## Tuesday, December 29, 2015

### Viete's Pi

$\sin(x) = 2\sin \left( \frac{x}{2}\right)\cos \left( \frac{x}{2}\right)$
$\Rightarrow \sin(x) = 2^2\sin \left( \frac{x}{4}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{2}\right)$
$\Rightarrow \sin(x) = 2^3\sin \left( \frac{x}{8}\right)\cos \left( \frac{x}{8}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{2}\right)$
$\ldots$
$\Rightarrow \sin(x) = 2^n\sin \left( \frac{x}{2^n}\right)\cos \left( \frac{x}{2^n}\right)\ldots\cos \left( \frac{x}{8}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{2}\right)$
$\Rightarrow \sin(x) = \cos \left( \frac{x}{2}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{8}\right)\ldots\cos \left( \frac{x}{2^n}\right)\sin \left( \frac{x}{2^n}\right)2^n$
$\Rightarrow \dfrac{\sin(x)}{x} = \cos \left( \frac{x}{2}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{8}\right)\ldots\cos \left( \frac{x}{2^n}\right)\dfrac{\sin \left( \frac{x}{2^n}\right)}{\frac{x}{2^n}}$
as $n \to \infty$
$\Rightarrow \dfrac{\sin(x)}{x} = \cos \left( \frac{x}{2}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{8}\right)\ldots\cos \left( \frac{x}{2^n}\right)$
$\Rightarrow \dfrac{\sin(x)}{x} =\displaystyle{\Pi_{k=1}^{\infty} \cos \left( \frac{x}{2^k}\right)}$
Plug $x=\frac{\pi}{2}$ we get

$\Rightarrow \dfrac{\sin(\frac{\pi}{2})}{\frac{\pi}{2}} =\displaystyle{\Pi_{k=1}^{\infty} \cos \left( \frac{\frac{\pi}{2}}{2^k}\right)}$

$\Rightarrow \dfrac{2}{\pi}=\displaystyle{\Pi_{k=1}^{\infty} \cos \left( \frac{\pi}{2^{k+1}}\right)}$

$\displaystyle{\Rightarrow \dfrac{2}{\pi}=\Pi_{k=2}^{\infty} \cos \left( \frac{\pi}{2^{k}}\right)}$

$\displaystyle{\Rightarrow \dfrac{2}{\pi}=\cos \left( \frac{\pi}{4}\right)\cos \left( \frac{\pi}{8}\right)\cos \left( \frac{\pi}{16}\right)}\ldots$
$\displaystyle{\Rightarrow \dfrac{2}{\pi}=\dfrac{\sqrt{2}}{2}\dfrac{\sqrt{2+\sqrt{2}}}{2}\dfrac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}}\ldots$

### Gammow's Problem

This is from the book One, Two infinity by George Gammow but I came across this in Paul Nahin’s “An Imaginary Tale”
A young and adventurous man who discovers an ancient parchment among his great grandfather’s paper listing about the great riches of golds and diamond necklaces he had stashed in an island up in the sea.
Sail to this island and you will find a deserted island. You will see two trees there. One is a Pine tree and other is an oak tree. There you will also see an old gallow on which we used to hang out traitors. Start from the gallow and walk to the oak tree counting your steps. At the oak tree you must turn right by a right angle and take the same number of steps. Put here a spike in the ground. Now you must return to the gallows and walk to the pine counting the steps. At the pine you must turn left by a right angle and take the same number of steps and put another spike into the ground. Dig halfway between the spikes; the treasure is there.
The young man follows the instructions, at least to the point of locating the island, where he sees the oak and pine trees. But, alas there is no gallow! Unlike the living trees, the gallows has long since disintegrated in the weather and not a trace of it or its location remains. Unable to carry out the rest of the instructions, the young man sails back without a gold coin or a diamond necklace to show for his troubles.
If he knew complex analysis he could have found the treasure.
To solve this let the coordinate of oak tree is -1 and pine tree is 1 on some scale. Let $a+ib$be the coordinate of the gallows. Then shifting the origin to oak $-1+0i$ it becomes $(a+1)+ib$. Moving it by $90^o$ it becomes $-b+i(a+1)$. Shifting back we get $-b-1+i(a+1)$. Now for the oak tree. We shift the origin to $1+0i$ it becomes $(a-1)+ib$rotating it by $90^o$ in the clockwise direction is same as multiplying it by $-i$ we get $b-i(a-1)$. Shifting back to the origin this is $b+1-i(a-1)$. Now the treasure is buried between these two spikes.So the mid point is $\dfrac{(-b-1+i(a+1))+(b+1-i(a-1))}{2}=\dfrac{2i}{2} =i$. Thus to find the treasure all he had to do was go the mid point between the two trees and then walk $90^o$ the same midway distance and dig there!