$\langle x \rangle $ = the expected value

$\bar{x}$ = vector

$\overrightarrow{X}$ = matrix vector?

n! = Factorial: Denoted by the exclamation mark (!). Factorial means to multiply by decreasing positive integers. For example, 5! = 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 = 120

Convolution = a mathematical operation on two functions (f and g) that produces a third function (f*g) that expresses how the shape of one is modified by the other. The term convolution refers to both the result function and to the process of computing it (source: wikipedia).

A cumulant of a probability distribution are a set of quantities that provide an alternative to the moments of the distribution. According to wikipedia, the first cumulant is the *mean*, the second the *variance*, and the third is the same as the *third central moment*.

Visually, this comes down to:

Some more sources in an attempt to understand cumulants:

- http://www.scholarpedia.org/article/Cumulants (perhaps the best)
- https://academic.oup.com/book/24921/chapter-abstract/188767614?redirectedFrom=fulltext
- http://www.stat.uchicago.edu/~pmcc/courses/stat306/2013/cumulants.pdf
- https://www.youtube.com/watch?v=OYhfMRAf_zQ (calm explaination)

The characteristic function:

$$ \log \langle e^{ikx} \rangle = \sum _{n=1}^{\infty} \frac{(ik)^n}{n!}C_n $$Two equations are given, with the characteristic function (7.25):

$$ exp \left(\sum _{n=1}^{\infty} \frac{(ik)^n}{n!}\right)C_n = \sum _{n=0}^{\infty} \frac{(ik)^n}{n!} \langle x^n \rangle $$and expanding the exponential as (7.26):

$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots $$The goal is to rework the equation to find $C_n$. We do this by looking at the righthand side and lefthand side independently.

The equation on the right hand side is as follows:

$$ \sum _{n=0}^{\infty} \frac{(ik)^n}{n!} \langle x^n \rangle $$As stated above, $\langle x^n \rangle $ is the expected value. The equation goes until infinity, but we're only looking for three cumulants. Therefore, we plug in $n=0, n=1, n=2$ and $n=3$.

n | plugged in equation | result |
---|---|---|

n = 0 | $$\frac{(ik)^0}{0!} \langle x^0 \rangle $$ | $$1$$ |

n = 1 | $$\frac{(ik)^1}{1!} \langle x^1 \rangle $$ | $$ik\langle x \rangle $$ |

n = 2 | $$\frac{(ik)^2}{2!} \langle x^2 \rangle $$ | $$\frac{-k^2}{2} \langle x^2 \rangle $$ |

n = 3 | $$\frac{(ik)^3}{3!} \langle x^3 \rangle $$ | $$\frac{-ik^3}{6} \langle x^3 \rangle $$ |

Now let's move on to the left hand side.

$$ exp \left(\sum _{n=1}^{\infty} \frac{(ik)^n}{n!}\right)C_n $$Add m as an intermediate step to eventually remove the exponential.

$$ \sum _{m=0}^{\infty} \frac{1}{m} \left(\sum _{n=1}^{\infty} \frac{(ik)^n}{n!}C_n \right)^m $$Next, plug in the results from the right hand side.

$$ \sum _{m=0}^{\infty} \frac{1}{m!} \left( \frac{ik}{1}C_1 + \frac{-k^2}{2}C_2 + \frac{-ik^3}{6}C_3 \right)^m $$The next step is to plug in values for m.

m | plugged in equation | result |
---|---|---|

m = 0 | $$1$$ | |

m = 1 | $$\sum _{m=1}^{\infty} \frac{1}{1} \left( \frac{ik}{1}C_1 + \frac{-k^2}{2}C_2 + \frac{-ik^3}{6}C_3 \right)^1 $$ | $$ik C_1 + \frac{-k^2}{2}C_2 + \frac{-ik^3}{6}C_3 + \dots $$ |

m = 2 | $$\sum _{m=2}^{\infty} \frac{1}{2} \left( \frac{ik}{1}C_1 + \frac{-k^2}{2}C_2 + \frac{-ik^3}{6}C_3 \right)^2 $$ | $$\frac{1}{2}\left[-k^2C_1^2 + \frac{-ik^3}{2}C_1C_2 + \dots \right]^2 $$ |

m = 3 | $$\sum _{m=3}^{\infty} \frac{1}{6} \left( \frac{ik}{1}C_1 + \frac{-k^2}{2}C_2 + \frac{-ik^3}{6}C_3 \right)^3 $$ | $$\frac{1}{6} \left[ -ik^3C_1^3 + \dots \right]^3 $$ |

The next step is to fill in k to find the three cumulants.

We start with putting the left and right side together.

$$ikC_1 = ik \langle x \rangle $$Next, fill out k = 1

$$i1C_1 = i1 \langle x \rangle $$Resulting in:

$$C_1 = \langle x \rangle $$Thus, the first cumulant $C1$ is $\langle x \rangle$, the expected value of $x$.

For the second cumulant, the left and right hand side are:

$$ \frac{-k^2}{2}C_2 - \frac{1}{2}k^2C_1^2 = \frac{-k^2}{2}\langle x^2 \rangle $$Simply filling out k = 2

$$ \frac{-2^2}{2}C_2 - \frac{1}{2}2^2C_1^2 = \frac{-2^2}{2}\langle x^2 \rangle $$We know what $C_1$ is from the previous step, and plug this in:

$$ \frac{-C_2}{2}- \frac{\langle x \rangle ^2}{2} = -2 \langle x^2 \rangle $$Algebraically solving:

$$C_2= \langle x^2 \rangle - \langle x \rangle^2 $$Lastly, the left and right hand side of the third cumulant are:

$$ \frac{-ik^3}{6}C_3 - \frac{ik^3}{2}C_1 C_2 - \frac{-ik^3}{6}C_1^3 = \frac{-ik^3}{6}\langle x^3 \rangle $$Next, dividing everything by $ik^3/6$ gives:

$$ -C_3 -3C_1C_2 - C_1^3 = - \langle x^3 \rangle$$We know that $C_1 = \langle x \rangle$, and plug this in. Then divide everything by -1 to improve readability.

$$ C_3 + 3 \langle x \rangle C_2 + \langle x \rangle^3 = \langle x^3 \rangle$$Next, plug in $C_2= \langle x^2 \rangle - \langle x \rangle^2 $ and bring everything but $C_3$ to the right side.

$$ C_3 = \langle x^3 \rangle - \langle x \rangle^3 - 3 \langle x \rangle \left( \langle x^2 \rangle - \langle x \rangle^2 \right) $$Expanding out gets us closer to the third cumulant:

$$ C_3 = \langle x^3 \rangle - \langle x \rangle^3 - 3 \langle x \rangle \langle x^2 \rangle + 3 \langle x \rangle^3 $$The final solution for the third cumulant is:

$$ C_3 = \langle x^3 \rangle - 3 \langle x \rangle \langle x^2 \rangle + 2 \langle x \rangle^3 $$Let's start with visualizing the differential element:

Find the expected value of the transformation.

The area of the paralellogram can be calculated by multiplying the length of its base by its height.

looks like a derivative. (dy_1/dx_1)dx_1 (dy_2

Cross-product of this is the determinant

Make a diagram to visualize.

I assume that $p$ is probability.

Plugging in $y_1$ and $y_2$ gives:

$\overrightarrow{x} = \sqrt{-2 \ln x_1} \sin(x_2)(x_1, x_2), \sqrt{-2 \ln x_1} \cos(x_2)(x_1, x_2))$

Jacobian's inverse is the same as the Jacobian of the derivative (Miana):

$J_f^{-1} = J_{f'}$

Notes:

- Square them, add them up.

Transform by equation

Random number generator

In [33]:

```
import numpy as np
rng = np.random.default_rng(seed=42) # Input
arr2 = rng.random((3, 3)) # Place it in a 3x3 matrix
arr2 # Print results
```

Out[33]:

array([[0.77395605, 0.43887844, 0.85859792], [0.69736803, 0.09417735, 0.97562235], [0.7611397 , 0.78606431, 0.12811363]])

An LFSR is a Linear Feedback Shift Register. It is a method to generate random numbers, as truly pseudo-random bits. In table 7.1 we can find the tap values $i = 1,4$ at order $M=4$. So the recursion becomes:

$$x_n=x_{n-1} + x_{n-4}$$Next, compute the values using Python, given the The recursion relation in (7.75):

$$ x_n = \sum _{i=1}^{M}a_ix_{n-i} $$With the help of a short how-to on Python random by GeeksforGeeks, a good blogpost on LFSR, a W3Schools list with Python arithmetic operators and the Python Random Module on W3Schools, here's a bit sequence:

In [32]:

```
import random
def randombit(): # Define the random number.
return random.randint(0,1) # Return a random integer between 0 and 1.
LFSR = [randombit(), randombit(), randombit(), randombit()] # Create the register with 4 values on each line.
for i in range(len(LFSR)**2): # Adding a for loop, range() and len() function to return the number of items in an object. https://www.w3schools.com/python/python_for_loops.asp & https://www.w3schools.com/python/ref_func_len.asp
print(LFSR)
x = (LFSR[0] + LFSR[3]) %2
LFSR.pop(3) # Removes the element at the specified position 3 (note: 0 is the first element). https://www.w3schools.com/python/ref_list_pop.asp
LFSR.insert(0,x) # Inserts specified value at position x.
```

In the text, it is given that the repeat time is $2^n-1$ for a register with $n$ steps.

A frequency of 1 Hz is equivalent to one cycle per second. A clock rate of 1 GHz is $10^9$ Hz.

There are $60 \cdot 60 \cdot 24 = 86,400$ seconds in a day. Multiplied by 365 we have $31,536,000$ seconds in one year.

Therefore, the age of the universe is: $3.1536 \cdot 10^7 \cdot 10^{10} = 3.1536 \cdot 10^{17}$ in seconds.

In class, it was mentioned that $2^{10}$ is approximately $10^3$.

Bringing it together, and calculating the last step with Wolfram Alpha:

$$2^n - 1 = 10^{9}\cdot 3.1536 \cdot 10^{17}$$$$2^n - 1 = 3.1536 \cdot 10^{26}$$$$n = \log_2 (3.1536 \cdot 10^{26}+1)$$$$n \approx 88 $$The diffusion equation (7.57) is:

$$ \frac{\partial p}{\partial t} = \frac{\langle \delta^2 \rangle}{2 \tau} \frac{\partial^2 p}{\partial x^2} $$Resources used for answering this question include: the explanation on a Fourier transform by 3Blue1Brown, a wikipedia page on Fourier transform (not so helpful), and the textbook *A First Course in Differential Equations with Applications* by Dennis G. Zill.

On page 526 of the textbook, it is explained how you can do a Fourier transform. Integral transforms appear in transform pairs.