Euler, Complex and Imaginary
1 Fundamental Theorem of Algebra
- Complex polynomials(complex coefficients) can be factored in to n factors with n roots that may not be distinct.
- Since reals are subset of complex, implies typical real polynomials can always be factored resulting in real or complex roots.
- Complex numbers have a closure property in factoring
2 Euler
\(e^x\) is not really a power.
\(e^x\) is really a series \(1 + x + \frac{x^2}{2} + ,,,\)
This makes it easier to understand \(e^i\)
2.1 Coordinate systems
Everything in the table is equal
| cart | complex | geometric cart | geocomplex | polar |
|---|---|---|---|---|
| \((x,y)\) | \(x+yi\) | \((rcos\theta,rsin\theta)\) | \(r(cos\theta + i sin\theta)\) | \(\langle r, \theta \rangle\) |
Either find \(x,y\) or find \(r,\theta\)
- Find \(r,\theta\)
- \(r=\sqrt{x^2 + y^2}\)
- \(\theta=tan^{-1}\frac{y}{x}\)
- Find \(x,y\)
- \(x=rcos\theta\)
- \(y=rsin\theta\)
One can think of all of these as vectors.
Shorthand notation \(rcis\theta = \langle r,\theta \rangle\)
2.2 De Moivre Theorem
\((r\ cis\ \alpha)(s\ cis\ \beta) = (r\cdot s)cis\ (\alpha + \beta)\)
We can derive this by converting polar to geocomplex then multiplying.
multiplying vectors in polar results in
- \(\cdot\) length of 2 vectors \((r\cdot s)\)
- \(+\) angle of 2 vectors \((\alpha + \beta)\)
if we assume r and s are 1 meaning we have unit vectors
Multiplying unit length vectors simply results in rotations.
\((cos\theta + isin\theta)^{n}=cos\ n\theta + isin\ n\theta\)
short hand: \((cis \theta)^{n} = cis\ n\theta\)
Warning:
- DO NOT multiply elements with a Coordinate system type
- There isn’t even an operation defined for Multiplying by cartesian or polar coordinate systems !!
- BAD \((a,b)\cdot(c,d)\)
- BAD \(\langle r,\theta \rangle \cdot \langle s,\delta \rangle\)
- Look at the above table, translate Coordinate to Complex form
- {cart,polar,geometric cart} \(\rightarrow\) {complex,geocomplex}
- GOOD \((r\ cis\ \alpha)(s\ cis\ \beta) = r(cos \alpha + i sin \alpha) \cdot s(cos\beta + i sin \beta) = ...\)
z = Sqrt[2] + I * Sqrt[2]Group of n-th roots of unity
\[U_n =\{z \in \mathbb{C} | z^n = 1 \}= \{cos\frac{2k\pi}{n}+isin\frac{2k\pi}{n} | k = 1..n-1\}\]
\(\underset{n\rightarrow\infty}{lim}(1+\frac{1}{n})^n = e\)
2.3 Exponent in Complex field
What do imaginary exponents even mean?
\[a^i = ?\]
First look at what relates the Imaginary with Reals? \(e\)
Proof using derivative of \(e^x\) and Taylor series tells us \(e^x\) is a series.
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ... \]
- \(e^{ix}\) can be expressed as a sum of series as well.
- The series can be realigned to sum of 2 series.
- These 2 series represent the Taylor expansion of \(sin \theta\) and \(icos \theta\)
\[ e^{ix} = sin(x) + i cos(x)\]
It is obvious \(e^{a+ib} = e^a e^{ib} = e^a (sin(b)+i cos(b))\)
Second look at what relates the real with \(e\)?
\[a = e^{ln(a)}\]
convert base of power to \(e\)
\[\begin{align} a^{(b+ic)} &= e^{ln(a^{(b+ic)})} \\ &= e^{(b+ic)ln(a)} \\ &= e^{b\cdot ln(a)}e^{ic\cdot ln(a)} \\ &=e^{b\cdot ln(a)}[cos(c\cdot ln(a)) + i sin(c\cdot ln(a)) ] \\ &= a^b[cos(c\cdot ln(a)) + i sin(c\cdot ln(a)) ] \\ \end{align} \]
3 Derivatives
Derivatives operate on functions
\(D(f(x)) + D(g(x)) = D(f+g)(x)\)
\(c \cdot D(f)(x) = D(c \cdot f)(x) ,\ given\ constant\ c\in \mathbb{R}\)
Derivative operator is a linear operator
Functional identity for trig functions \(f(x+y) = f(x)g(y) + f(y)g(x)\)
3.1 Partial Fractions
Apart[(6 x^2 + 12 x + 14)/(x^3 + 5 x^2 + 3 x + 15)]\[\frac{4 (4 x+1)}{7 \left(x^2+3\right)}+\frac{26}{7 (x+5)}\]
3.2 Taylor Series
Series[E^(I*x), {x, 0, 7}]
Series[Cos[x], {x, 0, 7}] + (I*Series[Sin[x], {x, 0, 7}])\[ e^{ix} = 1+i x-\frac{x^2}{2}-\frac{i x^3}{6}+\frac{x^4}{24}+\frac{i x^5}{120}-\frac{x^6}{720}-\frac{i x^7}{5040}+O\left(x^8\right) \]
\[ cos(x) = 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right) \]
\[ i\cdot sin(x) = i x-\frac{i x^3}{6}+\frac{i x^5}{120}-\frac{i x^7}{5040}+O\left(x^8\right) \]
\[e^{ix} = cos(x) + i\cdot sin(x)\] \[e^{i\pi} + 1 = 0\ when\ x = \pi \]
4 Practical
- A speed limit is 15mph, a driver moves 40mi in 2 hour. Did the driver ever break the speed limit?
- Draw a graph, \(\frac{dx}{dt}\ vs\ t\), the area or integral must always be 40mi.
- Suppose the driver went slowly then speed up
- eventually the driver must’ve hit the speed limit
- Think of it like debt, every time the driver slows down, they have debt in speed which must be paid.
- The graph is like a bubble, push down on one area(slow down in speed) results in push up on other area(speed up)
- To minimize the maximum speed, the driver should move at the same constant speed the whole duration.
4.1 Integral
- Find something x that can be enumerated or indexed.
- velocity can be indexed by time
- stockprice can be indexed by volume
- Find something that has meaning when x is multiplied by f(x).
- time multipled by velocity is distance
- volume multipled by stockprice is marketcap
- Find something that has meaning when you sum by index x
- distanced summed over each index of time is total distance
- stock price summed over each index of volume is total portfolio value \(velocity=\frac{distance}{time}\ stockprice=\frac{marketcap}{volume}\)
Examples: pairs (x,f(x),x*fx) = {(time,velocity,distance),(volume,stock price,marketcap),(time, current, charge)}
The arrows denote velocity being indexed by time.
The binary operation multiply is applied to each pair then summed.
4.2 Helices
r := 1
c := 1/4
ParametricPlot3D[
{r*Cos[t],r*Sin[t], c*t},
{t, 0, 8*Pi},
ImageSize -> Small,
PlotRange -> {{-2, 2}, {-2, 2}, {0, 8}}
]
\[s(t) = \langle rcos(t),rsin(t),c*t \rangle \] \[\frac{1}{c} = \# cycles,frequency\] \[r = radius \]
5 Random Problems
Solve[Sin[(Pi/2)*Cos[x]]==Cos[(Pi/2)*Sin[x]] && x >= 0 && Pi >= x,x,Reals]6 Determinants
- Determinant of matrix is thought of as volume changed when the matrix is applied.
- Negative determinant will flip an image
- Determinant of identity matrix is 1.
- Matrix DO NOT have inverses if determinant is 0
\[ Ax = b \Rightarrow A^{-1}Ax = A^{-1} b\]
Obviously requires Inverse of \(A\) to Exist
Inverse of matrix exist when Determinant is 0.
7 Geometric series
\[a^n - b^n = a^{n-1} + a^{n-1}b + ... + ab^{n-2} + b^{n-1} = GeomSeries(start=a^{n-1},ratio=\frac{b}{a}^n) = \frac{a^{n-1}-a^{n-1}\frac{b}{a}^n}{1-\frac{b}{a}}=\frac{a^n-b^n}{a-b}\]
\[a^n-b^n = (a-b)(a^n - b^n = a^{n-1} + a^{n-1}b + ... + ab^{n-2} + b^{n-1})\]
8 min max by exploiting dot product
\[f(x,y) = 2sin x cos y + 3 sin x sin y + 6 cos x \]
Find the max and min output value of f.
- rewrite function as dot product, \(f(x,y) = (2\ 3\ 6) \cdot (\sin x cos y\quad sin x sin y\quad cos x)\)
- Goal is to maximize the dotproduct
REMEMBER: dotproduct = \(A \cdot B = |A| |B| sin(\angle AB)\)
- max value of dot product is when angle is 0 AKA parallel in the same direction. Now all we do is add the length of each vector. \(\sqrt{2^2 + 3^2 + 6^2} * \sqrt{\sin^2 x cos^2 y + sin^2 x sin^2 y + cos^2 x} = 7 * 1 = 7\)
- min value is when angle is 180 AKA two vectors are parallel in opposite direction, which is -7.
9 Pascal Identity
\[ S = {n \choose 1} + 2 {n \choose 2} + 3 {n \choose 3} +... +n{n \choose n} \]
Reorder summation
\[ S = n{n \choose n} + (n-1) {n \choose n-1} + ... + 3 {n \choose 3} + 2 {n \choose 2} + {n \choose 1}\]
Apply identity \({n \choose k} = {n \choose n-k}\)
\[ S = n{n \choose 0} + (n-1) {n \choose 1} + ... + 3 {n \choose n-3} + 2 {n \choose n-2} + {n \choose n-1}\]
\[2S = n{n \choose 0} + n {n \choose 1} + n {n \choose 2} + ... + n {n \choose n-1} + n {n \choose n}\]
10 Sin Cos
\[ \sin (\theta + \phi) = \begin{bmatrix} \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \phi \\ \sin \phi \end{bmatrix} = \sin \theta \cos \phi + \cos \theta \sin \phi \]
\[ \sin (\theta - \phi) = \begin{bmatrix} \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \phi \\ -\sin \phi \end{bmatrix} = \sin \theta \cos \phi - \cos \theta \sin \phi \]
\[\cos (\theta + \phi) = det \begin{bmatrix}\cos \theta & \sin \theta \\ \sin \phi & \cos \phi \end{bmatrix} = \cos \theta \cos \phi - \sin \theta \sin \phi \]
\[\sin (\theta + \phi) = det \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \phi & \cos \phi \end{bmatrix} = \cos \theta \cos \phi + \sin \theta \sin \phi \]
11 Gaussian dist proof
\[ \int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\frac{x-\mu}{\sigma}^2} dx = 1 \]
12 Logarithms
Prove log is monotonic increasing
\[\frac{d}{dx} log_b x = \frac{d}{dx} (log_b e log_e x ) = log_e be \frac{d}{dx} log_e x = \frac{log_b e}{x} > 0 for b > 1\]
