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July 9, 2017

# A case for emptiness

It probably doesn't come as a surprise that I dislike being in a state of sadness, but I'm not very fond of being in a state of happiness either.

What do I mean by state? Something that lasts longer that a fleeting thought, that occupies your mind, that is with you for hours.

Imagine trying to read a book. If you're in a state of sadness, it might very hard to progress without dark thoughts popping in, disturbing you. This is probably the least surprising part of all this.

Imagine being in a state of happiness. You just can't clear that happy feeling, a feeling which is connected to something that has happened in your life. Maybe an event occurred, an event that aligned with some deeply rooted expectations. This made you very happy, and the feeling has conquered your whole being. Try reading a book in this state. It might be just as hard as in the sad state! You want to pace around, giggle, tell your friends about the happy thing, etc. You can't get anything done. Happiness is just another distraction, often created by the chance event that the coin toss that is life aligned with your expectations. The times when it doesn't align, you end up in sadness instead.

So, expectations cause happiness and sadness and I find that both are just a disturbance. So I grip the problem by the root and do away with expectations. When that is done, what state should I strive for? I strive for emptiness.

But what do I refer to by emptiness?

I do not refer to the feeling people have when they say "I feel empty inside". I think that what people actually mean when they say that they "feel empty inside" is that their mind is so saturated by impressions that they feel exhausted. This bad feeling can come to me when I have too much to do, or when a terrible social gathering has taken all the juice out of me. Often when people "feel empty inside", they try to push themselves, thinking they should be able to do something productive because they have an internal "emptiness". They try to fill it! But I think that people, whenever they feel like this, are actually completely saturated with impressions and should probably lie down on a couch and have tea for the rest of the day.

Now, what I do refer to by emptiness is the feeling of neither happiness or sadness. It's a completely empty head, like a blank canvas. There are very few thoughts randomly attacking your mind. You are at peace. Try reading a book in this state. It'll be like disappearing into the text. As long as there are no linguistic or logical hiccups in whatever you're reading, it will be like intuitive understanding of someone else's thoughts, put onto a piece of paper.

While I prefer happiness over sadness, I put emptiness on a pedestal high above both.

Note that the above is true for most activities in my life. I am after all a book-consuming student. An empty mind vastly improves almost everything I do. Now, there are of course exceptions. Sometimes euphoria is a very wanted thing, sometimes even sadness is sought, but the activities connected to those states are probably exceptions to my daily habits. The things I do every day is what I find interesting in life, and none of that improves by going into them with a mind bothered by other thoughts, be they happy, sad or of some other nature.

June 19, 2017

# Flight

In the empty darkness – no room, no walls, no sky, no stars – time did not pass. There was nothing that could tremble, nothing that could interact. Here, existence is undefinable and creation is both an event and an unobservable glitch.

There, the void suddenly disturbed. With the help of light, a flight of stairs would have hung in the black. But there was only a dark backdrop. On the top step, a head. It was not conscious, was not alive. No motion was possible in this realm, not because it could not happen, but because no one and nothing had ever tried. The head tried.

Slowly, almost adamant, over the edge, down a step. No thud, a loud scream:

"DEGELHE!"

Flash! The flight now visible, still unreadable. If a void contains something perfectly evenly red, it is just a smear. A new perspective required.

Lunge out into the void, see what the world looks like! A side-view. Like blood flowing down a pitch black volcano, the stairs were nothing but a red silhouette against the nullity.

"MUGHULEYA!"

A crack formed, the head split open. There, down in the slush, the most beautiful colors could be seen. Twisting and turning, the stairs – the world! – compacted itself by swirling into a maelstrom and disappearing, sucked into the colorful pit. Closing. Darkness again. Rumbling, shaking! Ever more violently, paving way for the final deed.

Exploding – not splitting, but exploding – into millions of colors, the head vanished. It left behind the most intricate palette of colored matter – but the matter was sad, and stayed that way.

May 26, 2017

# Everything binomial

Let's go on a mathematical adventure! We'll look at a couple of seemingly different things which are all actually related. First we'll look at calculating probabilities for slot machines, then we'll draw some spiffy triangles which make our lives easier. To finish off we'll look at a general and very nice rule for expanding $(x + y)^n$, where n can be any positive integer.

## Winning at the casino

Imagine playing a simple slot machine which has a 5% chance (probability) of you winning a single game. You have enough coins for 10 games. You wonder "what's the probability for me winning exactly 3 of these 10 games?". We can figure this out using some methods from probability theory and the information above.

Let's give the $5\%$ probability a name. Let's call it $p$. We say that $p = 5/100 = 0.05$. In probability theory, probabilities range between $0$ and $1$, not $0$ and $100$, so we make sure to divide percentages by $100$. Furthermore, we assign the number of games we can play to $n = 10$. The number of games we hope to win is assigned to $k = 3.$ Finally we also calculate the probability of loosing a single game. Since we had $p = 0.05$, the probability of loosing is one minus $p$, let's call this $q = 1 - p = 1 - 0.05 = 0.95$. You can think of this as subtracting the probability of winning a single game, $5\%$, from the maximum probability possible, $100\%$.

The probability of winning 3 out of 10 games can be calculated using what's known as the binomial probability mass function, the bin pmf. We'll explore it in detail below:

${n \choose{k}} \cdot p^k \cdot q^{n - k} = {10 \choose{3}} \cdot 0.05^3 \cdot 0.95^7$

If you're unfamiliar with the $n \choose{k}$ bit, don't fear. This bit (factor) has the dry name binomial coeffecient, which simply states that it is a number in the binomial pmf. This coefficient is very nifty and can be used to figure out how many ways you can order $k$ things among a total of $n$ things. For us it represents how many ways we can arrange winning $k=3$ games out of a total of $n=10$ games. We do this because we do not know which of the 10 games we will win, so we must take into account that any 3 games of the total 10 can be won, in any order. You can think of it like this: Let W denote winning a single game and L loosing it. Then one possible order of winning 3 out of 10 games is: WWWLLLLLLL. But another one is WWLWLLLLLL. Yet another one is WWLLWLLLLL. And WLWLWLLLLL. And LLLLWWLLLW. The total number of ways to arrange the Ws and Ls is denoted $n \choose{k}$, or more specifically $10 \choose{3}$. This is equivalent to the number of ways 3 out of 10 games can be won.

Now, ${n \choose{k}}$ is just mathematical notation, it doesn't tell us much unless we know how to calculate it. One way is to list all possible WWWLLLLLLL-combinations and count them, but this is tedious and error-prone. A purely algebraic way is

\begin{aligned} {n \choose{k}} &= \frac{n!}{k!(n - k)!} \\ \\ {10 \choose{3}} &= \frac{10!}{3!7!} = \frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\cdot(7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)} = \frac{3628800}{30240} = 120 \end{aligned}

You might not be familiar with the ! in the expression above. It's called the factorial and is simply the number in front of the ! multiplied by all numbers smaller than itself, all the way down to 1. For example $8! = 8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$.

So, now we can go back to our probability and swap ${10 \choose{3}}$ with $120$:

${n \choose{k}} \cdot p^k \cdot q^{n - k} = {10 \choose{3}} \cdot 0.05^3 \cdot 0.95^7$ $= 120 \cdot 0.05^3 \cdot 0.95^7 = 120 \cdot 0.000125 \cdot 0.6983 \approx 0.0105$

This means that if we have a slot machine with a 5% chance of winning and we want to win 3 out of 10 games, then this will occur with a 0.0105 probability, which corresponds to 1.05%. Not a great outlook in other words!

You probably wonder why we take $0.05$ to the power of $3$ ($0.05^3$) and $0.95$ to the power of $7$. Recall that $0.05^3$ means that we take $0.05$ times itself three times, i.e. $0.05^3 = 0.05\cdot0.05\cdot0.05$. In probability theory you can take two or more unrelated events and multiply their probabilities together in order to find the probability of all of them occurring. This is exactly what we're doing here, the probability of winning 3 games is the probability of winning 1 game multiplied by itself 3 times. We can do this since the events are unrelated; winning one game does not affect the outcome of the next. Using the same idea we multiply the probability of loosing a game, $0.95$, with itself 7 times in order to find the probability of loosing 7 games. We then multiply the probability of winning 3 games with the probability of loosing 7 games, $0.05^3\cdot0.95^7 = 0.00008729$. This is the probability of winning 3 and loosing 7 games, but this can be done in ${10 \choose{3}} = 120$ different ways! So we just multiply the probability of the wins and losses with the number of ways it can happen, i.e. $120\cdot0.05^3\cdot0.95^7 = 0.0105$, which is our final probability.

Please note that we asked for the probability of winning exactly 3 of 10 games, not 3 or more. If we wish to calculate the probability of winning 3 or more out of 10 games, we'd have to calculate the probability of winning exactly 3 games, and the probability of winning exactly 4 games and 5 and 6 and so on up to 10. We'd then have to add these probabilities up. If you do this you'll end up with a probability of $0.0115$, which is only slightly higher than the one for winning exactly 3 games. This is because the probability of winning 4 games is very small, winning 5 even smaller. Adding them up doesn't do much in this case.

## Making the binomial coefficient easier to calculate

${n \choose{k}} = \frac{n!}{k!(n - k)!} = {10 \choose{3}} = \frac{10!}{3!7!} = \frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1\cdot(7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)}$

For big values of $n$ or $k$ the fraction becomes very lenghty. We can make our lives easier by canceling factors in the fraction:

${10 \choose{3}} = \frac{10\cdot9\cdot8\cdot\sout{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}{3\cdot2\cdot1\cdot\sout{(7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)}} = \frac{10\cdot9\cdot8}{3\cdot2\cdot1} = \frac{720}{6} = 120$

The general result of this cancellation is usually given as an explicit formula, which is handy for doing the calculation by hand (har har):

${n \choose{k}} = \frac{n\cdot(n-1)\cdot(n-2)\cdot...\cdot(n-k+1)}{k!}$

Try to see how ${10 \choose{3}} = \frac{10\cdot9\cdot8}{3\cdot2\cdot1}$ would fall out of this formula. The numerator in the formula says that we should start at $10$ and then multiply it with all numbers smaller than itself all the way down to $n - k + 1 = 10 - 3 + 1 = 8$. It is especially useful when $n$ is big and $k$ is small, since we get way fewer factors than with our previous formula. For situations where $n$ is big and $k$ is close to $n$, it gives little help since just a few factors will cancel. Note that $k$ can never be bigger than $n$ nor can any of $n$ or $k$ be less than zero.

There is another, more graphic and mnemonic-y way of finding the value of a specific binomial coefficient. Let's draw Pascal's triangle!

\begin{aligned} & & & & & & & & & & & 1 & & & & & & & & & &\\ & & & & & & & & & & 1 & & 1 & & & & & & & & &\\ & & & & & & & & & 1 & & 2 & & 1 & & & & & & & &\\ & & & & & & & & 1 & & 3 & & 3 & & 1 & & & & & & &\\ & & & & & & & 1 & & 4 & & 6 & & 4 & & 1 & & & & & &\\ & & & & & & 1 & & 5 & & 10 & & 10 & & 5 & & 1 & & & & &\\ & & & & & 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1 & & & &\\ & & & & 1 & & 7 & & 21 & & 35 & & 35 & & 21 & & 7 & & 1 & & &\\ & & & 1 & & 8 & & 28 & & 56 & & 70 & & 56 & & 28 & & 8 & & 1 & &\\ & & 1 & & 9 & & 36 & & 84 & & 126& & 126& & 84 & & 36 & & 9 & & 1 &\\ & 1 & & 10 & & 45 & & 120& & 210& & 252& & 210& & 120& & 45 & & 10 & & 1 \end{aligned}

Finding $10 \choose {3}$ using Pascal's triangle goes as follows: Count the rows from the top, starting at $0$ for the first row, $1$ for the second and so on until you reach $10$. For the specific triangle above, this will be the last row. On that row, start counting the columns from left or right, again starting at $0$. Do this until you reach $3$. You should now be staring at the value of ${10 \choose {3}} = 120$.

Drawing your own Pascal's triangle is very easy. Start with the top triangle consisting only of 1:s:

\begin{aligned} & & 1 &\\ & 1 & & 1 \end{aligned}

Add a new row at the bottom with 3 numbers. Let the numbers in this new row be the sum of the numbers diagonally above them. Numbers on the edges (furthest to the left or right) should be 1.

\begin{aligned} & & & 1 & &\\ & & 1 & & 1 &\\ & 1 & & 2 & & 1 \end{aligned}

Repeat.

\begin{aligned} & & & & 1 & & &\\ & & & 1 & & 1 & &\\ & & 1 & & 2 & & 1 &\\ & 1 & & 3 & & 3 & &1 \end{aligned}

Doing this 7 more times should produce the big triangle we used to calculate $10 \choose {3}$ above. Huzzah!

Pascal's triangle is named after the French mathematician and physicist Blaise Pascal. He's known for Pascal's law, which describes transmission of fluid pressure. The SI-unit for pressure, Pascal, is named after him.

## The binomial theorem

Have you ever memorized that $(x + y)^2 = x^2 + 2xy + y^2$? Maybe $(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$? How about memorizing $(x + y)^4$? Have no fear, the binomial theorem is here (also called binomial expansion). It is a theorem which can be used to calculate a binomial to any power. A binomial is just a polynomial with exactly two terms, hence the bi in the name. Examples are $(x + y)$, $(3x + 1)$, $(a + z)$ or whatever you like.

The binomial theorem looks very similar to the binomial probability mass function, in fact it looks just like a sum of several bin pmf:s. Here are two examples for different powers:

\begin{aligned}\\ (x + y)^2 &= {2 \choose{0}}x^0y^2 + {2 \choose{1}}x^{1}y^{1} + {2 \choose{2}}x^{2}y^0 \\ \\ &= 1y^2 + 2xy + 1x^2 = x^2 + 2xy + y^2 \end{aligned}

\begin{aligned}\\ (x + y)^3 &= {3 \choose{0}}x^0y^3 + {3 \choose{1}}x^1y^2 + {3 \choose{2}}x^2y^1 + {3 \choose{3}}x^3y^0\\ \\ &= 1y^3 + 3xy^2 + 3x^2y + 1x^3 = x^3 + 3x^2y + 3xy^2 + y^3 \end{aligned}

See the pattern? For the sake of consistency I start with the terms in the same order as in the bin pmf, but I flip everything around at the last equals sign to make them look like the rules you might have memorized. Also, recall that any number taken to the power of $0$ equals $1$, that's why $x^0 = 1$ and $y^0 = 1$ above vanish, multiplying with $1$ does nothing. Together with Pascal's triangle, binomial expansions to high powers can easily be "remembered". Let's do one to the power of $5$. Here's a Pascal's triangle with 6 rows:

\begin{aligned} & & & & & & 1 & & & & &\\ & & & & & 1 & & 1 & & & &\\ & & & & 1 & & 2 & & 1 & & &\\ & & & 1 & & 3 & & 3 & & 1 & &\\ & & 1 & & 4 & & 6 & & 4 & & 1 &\\ & 1 & & 5 & & 10 & & 10 & & 5 & &1 \end{aligned}

Let's write out the binomial expansion using the pattern above, looking up binomial coefficients in the triangle

\begin{aligned} (x + y)^5 &= {5 \choose{0}}x^0y^5 + {5 \choose{1}}x^1y^4 + {5 \choose{2}}x^2y^3 + {5 \choose{3}}x^3y^2 \\ \\ &+ {5 \choose{4}}x^4y^1 + {5 \choose{5}}x^5y^0 \end{aligned}

Note that all the binomial coefficients are on the form $5 \choose{n}$, meaning that all of them are on the 6:th, the last, row in the triangle. In fact, the coefficients can be read straight off this row:

\begin{aligned} (x + y)^5 &= 1x^0y^5 + 5x^1y^4 + 10x^2y^3 + 10x^3y^2 + 5x^4y^1 + 1x^5y^0 \\ \\ &= 1y^5 + 5xy^4 + 10x^2y^3 + 10x^3y^2 + 5x^4y + 1x^5 \\ \\ &= x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5 \end{aligned}

Ta-da! You can now more easily remember how to write crazy-big binomial expansions. Drawing Pascal's triangle and doing all this might not be fast, but it's an easy enough way to remember it all. It's better to use dumb, slow methods and be correct than to use fast, smart methods and make silly mistakes.

One last thing. There's a shorter, more general way of writing the binomial theorem, using a sum-sign:

$(x + y)^n = \sum\limits_{k=0}^n{n \choose{k}}x^ky^{n - k}$

The $\sum$ says that we should start with $k = 0$ and calculate the stuff to the right of the $\sum$ using $k = 0$ and then set $k = 1$ and calculate the stuff again using that and do this again and again all the way until $k = n$. We then add all the calculated parts together. Try doing this for $n = 2$ and $n = 3$, the examples at the start of this section should fall out.

May 25, 2017

# Primary directive

On the edge,
but still far off.
Internal monologues
that could easily take lives.
This chatter, these screams
everlastingly frightening.
Forces, strong, keep these
propagating ideas at bay.Simply put, these forces are that
all this madness is secondary to
the primary directive:
onto those around you
an everlasting supply of kindness,
to always be aware when others hurt
and find the courage

May 24, 2017

# Internal tea luggage

I have a huge internal energy cache and
the need to share words with a friend.
Long past dusk, but the path still visible,
a thermos my only luggage.Across the stream a bridge was built.
Rocks thereunder suffice as chair.
The trees are green, my shoes are red,
shrouded in darkness all around me
these colors live their nightly life.I brought no companion,
only this hot beverage
and my thoughts.Minutes turn into an hour.
A mug a mug a mug.
Earl's reserves are low.They'll never catch the wacko
sitting by the stream
alone at night
drinking tea
pondering life.