Summer of Math Exposition 4
First, the things on the left of the screen are not part of the submission, because it said in the rules "you can't make your submission ahead of time", this page exists, I'll be summarizing the website as of today, and finally, I'll only show original math that I did.
Calculus
\[ \frac{f(x + dx) - f(x)}{dx} = \frac{df}{dx} = f\prime (x) \]
sum rule
\[ \vdots \]
\[ (f + g)\prime = f\prime + g\prime \]
product rule
\[ \vdots \]
\[ (fg)\prime = fg\prime + f\prime g \]
chain rule
\[ \vdots \]
\[ (f(g))\prime = f\prime(g) g\prime \]
mbc rule
\[ \vdots \]
\[ (cf)\prime = c f\prime \]
exponent rule
\[ \vdots \]
\[ \frac{d(a^x)}{dx} = a^x \frac{a^{dx} - 1}{dx} \]
\[ \text{(lets figure this out later!)} \]
e
\[ \frac{de^x}{dx} = e^x \]
\[ \vdots \]
\[ e = (1 + dx)^{\frac{1}{dx}} \]
\[ log_e (x) = : ln(x) \]
logarithmic derivitave
\[ \vdots \]
\[ (ln(f))\prime = \frac{f\prime}{f} \]
\[ f(x) = a^x \]
\[ \vdots \]
\[ (a^x)\prime = a^x ln(a) \]
power rule
\[ f(x) = x^n \]
\[ \vdots \]
\[ (x^n)\prime = n x^{n - 1} \]
Polynomial
Skipping this one, it's just Mathologer
Calculus II
More about \(e\)
\[ e = \lim_{\Delta x \to 0} (1 + \Delta x)^{\frac{1}{\Delta x}} = \lim_{N \to \infty} (1 + \frac{1}{N})^N \]
\[ \vdots \]
\[ e^x = (1 + x dx)^{\frac{1}{dx}} \]
you will see why this is usefull here.
More about \(e\) to the \(x\)
\[ \sum\limits_{n=0}^{\infty} C_n x^n = e^x \]
\[ \vdots \]
\[ \sum\limits_{n=0}^{\infty} C_n n x^{n - 1} = \sum\limits_{n=0}^{\infty} C_n x^n \]
\[ \vdots \]
\[ C_n = \frac{1}{n!} \]
\[ e^x = \sum\limits_{n=0}^{\infty}\frac{x^n}{n!} \]
\[ e = \sum\limits_{n=0}^{\infty}\frac{1}{n!} \]
quoteint rule
\[ \vdots \]
\[ (\frac{f}{g})\prime = \frac{f\prime g - f g\prime}{g^2} \]
L'Hopital's rule
\[ f(c) = 0 \]
\[ g(c) = 0 \]
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = ? \]
\[ \vdots \]
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{\frac{d}{dx} f(x)}{\frac{d}{dx} g(x)} \]
Complex numbers
\[ i = \sqrt{-1} \]
\[ a + bi = c + di \iff c = a, d = b \]
\[ (a + bi)(c + di) = ac + adi + bci + bdii = ac + adi + bci - bd = (ac - bd) + (ad + bc)i \]
\[ (a + bi)^2 = (aa - bb) + (ab + ab)i = (a^2 - b^2) + 2abi \]
\[ (a + bi)r = ar + bri \]
\[ \text{ccong} (a + bi) = a - bi \]
the vertical line is the definition of \(sin(\theta)\)
the horizontal line is the definition of \(cos(\theta)\)
\(\alpha\) is the angle from the right of the \(x\) axis to the point.
I will define \(f(x)\) as a function that takes in a real number and outputs a complex number that is \(x\) radians around the unit circle. Because of how circles work, \(f(x) = cos(x) + isin(x)\). Fun fact! \(z\) equals \(rf(\theta)\) for any complex \(z\). Eccept pepole don't use \(f(\theta)\) because there are other complex functions \(f(z)\), and \(rcos(x) + irsin(x)\) dosen't reach the high bar for perfection set by mathematicians.
So, now the question is: solve for \(f(x)\).
I'll start by multiplying \(f(x) \cdot f(y)\)
\[ \vdots \]
\[ \vdots \]
\[ f(x)f(y) = f(x + y) \]
hmmm \(f(x)f(y) = f(x + y)\) sounds faniliar... Oh right! This is an exponential, but what's the base?
while, a common proof that I once used that if the derivitave of \(g(x)\) is \(g(x)\), then \(g(x) = ce^x\), it has a method of:
\[ \vdots \]
\[ g(x) = g(0)(1 + dx)^{\frac{x}{dx}} \]
and then, using facts from Calculus part 2, \(g(x) = g(0)e^x = ce^x\).
So I will use a proof like that, the proof will go like this:
proof
\[ f(x + dx) = f(x)f(dx) \]
\[ \vdots \]
\[ f(dx) = (cos(0) + dx \text{ } cos'(0)) + (sin(0) + dx \text{ } sin'(0))i \]
\(cos(x)\) reaches a peak at \(0\), so \(cos'(0) = 0\)
while as you zoom into the right of the unit circle, the hight is the distance travled and the sine of a small angle is that angle, so \(sin'(0) = 1\)
\[ \vdots \]
\[ f(x + ndx) = f(x)(1 + dxi)^n \]
and if you saw Calculus part 2, you know that...
\[ f(0 + \frac{x}{dx}dx) = f(x) = f(0)((1 + dxi)^{\frac{1}{dx}})^x = f(0)(e^i)^x \]
so, surprisingly, the base of the exponential is \(e^i\), eccept many of you would of guessed that because of eueler's identity, which I was trying to derive, anyways there is one more step (more like three but you get the point)
\[ f(0) = 1 \]
\[ f(x) = e^{ix} \]
thus:
\[ cos(x) + isin(x) = e^{ix} \]
proof complete!
the Holy Grail of complex numbers (I forgot to do this \(3\) months ago.)
\[ cos(\pi) + isin(\pi) = e^{i \pi} \]
\[ e^{i \pi} = -1 + i0 \]
\[ e^{i \pi} = -1 + i0 = -1 \]
\[ e^{i \pi} + 1 = 0 \]
Jacobian matrices
Learn more here, and here.
\[ \begin{bmatrix} a \\
b\end{bmatrix} = a + bi \]
\[ \hat{I} = \begin{bmatrix} 1 \\
0\end{bmatrix} = 1 \]
\[ \hat{j} = \begin{bmatrix} 0 \\
1\end{bmatrix} = i \]
\[ a + bi = \begin{bmatrix} ? \quad ? \\
? \quad ?\end{bmatrix} \]
\[ \vdots \]
\[ a + bi = \begin{bmatrix} a & -b \\
b & a\end{bmatrix} \]
complex functions
\[ z = x + yi =\begin{bmatrix} x \\
y\end{bmatrix} \]
\[ f(z) = u + vi = \begin{bmatrix} u \\
v\end{bmatrix} \]
\[ dz = dx + dyi = \begin{bmatrix} dx \\
dy\end{bmatrix} \]
\[ df = du + dvi = \begin{bmatrix} du \\
dv\end{bmatrix} \]
\[ \vdots \]
\[ \begin{bmatrix} du \\
dv\end{bmatrix} = f\prime (z) \begin{bmatrix} dx \\
dy\end{bmatrix} \]
\[ \text{And remember that!} \]
\[ df = f(z + dz) - f(z) \]
\[ \frac{\partial}{\partial x} f(x, t) = : \frac{f(x + dx, t) - f(x, t)}{dx} \]
\[ \partial f(x, t)_x = : f(x + dx, t) - f(x, t) \]
\[ f(z + dx \hat{I}) = f(z) + \begin{bmatrix} \partial u_x \\
\partial v_x\end{bmatrix} \]
\[ f(z + dy \hat{j}) = f(z) + \begin{bmatrix} \partial u_y \\
\partial v_y\end{bmatrix} \]
\[ f(z + dx + dyi) = f(z + dz) = f(z) + \begin{bmatrix} \partial u_x \\
\partial v_x\end{bmatrix} + \begin{bmatrix} \partial u_y \\
\partial v_y\end{bmatrix} \]
\[ f(z + dz) - f(z) = df = \begin{bmatrix} \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy \\
\frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy\end{bmatrix} = \begin{bmatrix} \frac{\partial u}{\partial x} \quad \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} \quad \frac{\partial v}{\partial y}\end{bmatrix} \begin{bmatrix} dx \\
dy\end{bmatrix} \]
\[ \begin{bmatrix} du \\
dv\end{bmatrix} = \begin{bmatrix} \frac{\partial u}{\partial x} \quad \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} \quad \frac{\partial v}{\partial y}\end{bmatrix} \begin{bmatrix} dx \\
dy\end{bmatrix} \]
\[ f\prime (z) = \begin{bmatrix} \frac{\partial u}{\partial x} \quad \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} \quad \frac{\partial v}{\partial y}\end{bmatrix} \]
cauchy-reimann equations
the equation or jacobian matrix is a matrix, but as you know, evry complex number has a corrasponding matrix, but not every matrix has a corrasponding complex number, so to find out if the jacobian matrix is a complex number or just a matrix, or said another way, if complex function \(f(z)\) has a derivitave, we need the cauchy-reimann equations, lets go derive them!
so, if the jacobian matrix is a complex number \(a + bi\) (which I have been saving for something like this), than the corrasponding matrix is:
\[ \begin{bmatrix} a & -b \\
b & a\end{bmatrix} \]
so, that means that
\[ \begin{bmatrix} \frac{\partial u}{\partial x} \quad \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} \quad \frac{\partial v}{\partial y}\end{bmatrix} = \begin{bmatrix} a & -b \\
b & a\end{bmatrix} \]
well, what makes a matrix of that form?
for one, the top left equals the bottom right equals the real part
and for another, the top right equals the negatave of the bottom left equals the negatave of the imaginary part
in conclusion...
\[ f\prime (z) = \begin{bmatrix} \frac{\partial u}{\partial x} \quad \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} \quad \frac{\partial v}{\partial y}\end{bmatrix} \]
and to test if this is a matrix or a complex number
\[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \]
\[ -\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} \]
examples
\[ f(z) = z^2 \]
\[ f(x + yi) = (x^2 - y^2) + (2xy)i \]
\[ u = x^2 - y^2 \]
\[ \vdots \]
\[ 2x = 2x \]
\[ 2y = 2y \]
\[ f\prime (z) = 2x + 2yi = 2z \]
another one!
\[ f(z) = \text{ccong} (z) \]
\[ f(x + yi) = x - yi \]
\[ \vdots \]
\[ 1 \ne -1 \]
\[ 0 = 0 \]
\[ f\prime (z) = \begin{bmatrix} 1 & 0 \\
0 & -1 \end{bmatrix} \]
in conclution, \(z^2\) has a derivitave, and \(\text{ccong} (z)\) does not.
Gamma
Skipping this one, it's just BriTheMathGuy