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


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
Differentiating both sides we get
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)
\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+ibbe 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)+ibrotating 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!
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