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
$\Rightarrow \zeta(-1)(1-\frac{2}{2^{-1}}=1-2+3-4+5-6+...=\frac{1}{4}$
$\Rightarrow \zeta(-1)(-3)=\frac{1}{4}$
$\Rightarrow \zeta(-1)=\frac{-1}{12}$
$\Rightarrow 1+2+3+4+5+... =\frac{-1}{12}$