Opiskelijan muistiinpanot

Geogebra applet with 4 datapoints, fitting line with best square distance (from point to line)

square here, as well as in variance concept – is because distance can well be negative, but we need to be able to sum it neatly somehow, therefore square is rational with MMSE and variance

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Have met a concept of Bayes theorem, and would pretty much like to visualize it here. As per purpose of this blog, its studying diary, so pretty much – visualizations, and making of those – would give me myself broader understanding.

Not to mention, that keeping notes publicly does make wanders to being neat in thoughtful way, so – lets begin.

Suppose we hear the description of a man: “He is strong and with a loud strong but assuring voice”.

Who is this man, do we want to guess – is he a librarian or a military drill sergeant?

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Suppose we have two countably infinite sets, which means we can find injection (one-to-one) relation to natural numbers. 

In other words, we could literally assign natural number indexes to only one of each member of the set.

So, we have with cardinality , and with cardinality

Then cartersian product (set of all possible pairs from A and from B) – is also infinite, but countable. I mean, for , we can assign single natural number for each member of P.

Lets prove it. Suppose we injections f and g, so that and

Lets define function this way:

Then for and , because of  unique division into primes and f,g injectivity (a,b)=(c,d).

Therefore there is injection from p(a,b) to natural numbers, and cartesian product is infinitely countable.

– Let us construct natural numbers (0,1,2,3..)
– Then from naturals, lets construct integer numbers (…,-2,-1,0,1,2…)
– Then from integers, lets construct rationals (…,-0.3,…,0,…,1/2,..,0.75,..)
– And finally from rationals – real numbers (…,-0.3,…,0,…,1/π,…,1/2,..,0.75,..,e,…,π,…)
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Lets consider linear cellular automata with Wolfram rule 90

Elementary cellular automata is essentially string of two symbols, lets call them 0 and 1, equipped with rules (and starting configuration, of course) 

Rule takes symbol in the position, as well as left and right neighbors. Ie – when calculating symbol of the cell, we should consider 3 cells, cell to the left, cell itself and cell to the right.

Ie – suppose we have set of rules(*):
111 - 0
110 - 1
101 - 0
100 - 1
011 - 1
010 - 0
001 - 1
000 - 0

Or graphically:

So if we have 10011 for starting configuration, calculation goes like this(from top to bottom):

(italics) means that we dont know which symbol is to the left and right to the starting configuration, for simplicity I consider here absent symbols to be 0s (ie …00000100110000…)

Now, what is 90, which is called Wolfram rule.

It is obvious, that  3 cells define state, and since we have two symbols, all possible combinations are , and since we have two possible symbols for each combination – rule of CA can be expressed as (0,2⁸=256) number.

90 in binary would be (*), which is enough to describe elementary ca behavior.

There were question in exam, which I understood very close to correct, thus absolutely wrong.

Roughly translated: “Define real axis ℝ Lebesque outermeasure on ℝ² surface“

Lets get into it. Bit intuitively first.. Suppose we have segment of a line (a,b)

segment (s) itself can be expressed as

Actually, lets forget about s, lets consider without segments, ie

Then outer measure can be expressed as union of open sets:

for all

Where intervals are:
Thus length of intervals are from one integer to another – ,.. all integers
Height of intervals are

Ie , where biggest height is , which is quater of epsilon.

Therefore

or shorter: and for all

It is obvious from here

This article was inspired by Margarita.