Add content for numbers-to-indices/more-algebra

Add actual content to the skeleton of the
`numbers-to-indices/more-algebra` section.

Signed-off-by: Eggert Karl Hafsteinsson <[email protected]>
Signed-off-by: Teodor Dutu <[email protected]>
Signed-off-by: Razvan Deaconescu <[email protected]>

chapters/numbers-to-indices/more-algebra/reading/read.md

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+# More on Algebra
+
+## Some Squares
+
+If $a$ and $b$ are real numbers, then
+
+$$(a+b)^2=a^2+2ab+b^2$$
+
+### Details
+
+If $a$ and $b$ are real numbers, then:
+
+$$(a+b)^2=a^2+2ab+b^2$$
+
+This can be proven formally with the following argument:
+
+$$\begin{aligned} (a+b)^2 &=& (a+b)(a+b)\\ &=&(a+b)a+(a+b)b\\ &=& a^2+ba+ba+b^2\\ &=& a^2+2ab+b^2\end{aligned}$$
+
+## Pascal's Triangle
+
+Pascal's triangle is a geometric arrangement of the binomial coefficients in a triangle:
+
+$$\begin{array}{ccccc} & & 1 & &\\ & 1 & & 1&\\ 1 & & 2 & & 1\\ \vdots \quad \vdots && \vdots && \vdots \quad \vdots \end{array}$$
+
+### Details
+
+$$\begin{array}{ccccccccc} n=0: & & & & &1& & & \\ n=1: & & & &1& &1& & \\ n=2: & & &1& &2& &1& \\ n=3: & &1& &3& &3& &1 \end{array}$$
+
+To build Pascal's triangle, start with `1` at the top, and then continue placing numbers below it in a triangular pattern.
+Each number is just the two numbers above it added together (except for the edges, which are all `1`).
+
+### Examples
+
+:::info Example
+
+The following function in R gives you the Pascal's triangle for $n=0$ to $n=10$.
+
+```text
+> fN <- function(n) formatC(n, width=2)
+
+> for (n in 0:10) {
++ cat(fN(n),":", fN(choose(n, k = -2:max(3, n+2))))
++ cat("\n")
++ }
+
+ 0 :  0  0  1  0  0  0
+ 1 :  0  0  1  1  0  0
+ 2 :  0  0  1  2  1  0  0
+ 3 :  0  0  1  3  3  1  0  0
+ 4 :  0  0  1  4  6  4  1  0  0
+ 5 :  0  0  1  5 10 10  5  1  0  0
+ 6 :  0  0  1  6 15 20 15  6  1  0  0
+ 7 :  0  0  1  7 21 35 35 21  7  1  0  0
+ 8 :  0  0  1  8 28 56 70 56 28  8  1  0  0
+ 9 :  0  0  1  9 36 84 126 126 84 36  9  1  0  0
+10 :  0  0  1 10 45 120 210 252 210 120 45 10  1  0  0
+```
+
+Changing the numbers in the line `for(n in 0:10)` will give different portions of the triangle.
+
+:::
+
+## Factorials
+
+We define the factorial of an integer `n` as
+
+$$n!= n\cdot(n-1) \cdot(n-2)\cdot \ldots \cdot 3 \cdot 2 \cdot 1$$
+
+For convenience we define $0!$ to be `1`.
+
+### Details
+
+:::note Definition
+
+We define the factorial of an integer `n` as:
+
+$$n!= n\cdot(n-1) \cdot(n-2)\cdots \ldots \cdot 3 \cdot 2 \cdot 1$$
+
+:::
+
+### Examples
+
+:::info Example
+
+Suppose you have six apples: $\{a, b, c, d, e, f\}$ and you want to put each one into a different apple basket: $\{1,2,3,4,5,6\}$.
+
+For the first basket you can choose from `6` apples $\{a, b, c, d, e,f\}$, and for the second basket you have then `5` apples to choose from and so it goes for the rest of the baskets, so for the last one you only have `1` apple to choose from.
+
+The end result would then be $6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720$ possible allocations.
+
+This could also be calculated in R with the factorial function:
+
+```text
+> factorial(6)
+[1] 720
+```
+
+:::
+
+## Combinations
+
+The number of different ways one can choose a subset of size $x$ from a set of $n$ elements is determined using the following calculation:
+
+$$\displaystyle{n \choose x}= \displaystyle\frac{{n!}}{{x!\left( {n - x} \right)!}}$$
+
+### Details
+
+:::note Definition
+
+A **combination** is an un-ordered collection of distinct elements.
+
+:::
+
+Suppose we want to toss a coin $n$ times.
+In each toss we obtain head (`H`) or tail (`T`) resulting in a sequence of `H`, `T`, `T`, `H`, ..., `T`.
+
+How many of these possible sequences contain exactly $x$ tails?
+There are $n$ positions in the sequence, we can choose $x$ of these in $\displaystyle\binom{n}{x}$ ways and put our `T`s in those positions.
+If the probability of landing tails is $p$, then each one of these sequences with exactly $x$ tails has probability $p^x(1-p)^{n-x}$
+So the total probability of landing exactly $x$ tails in $n$ independent tosses is:
+
+$$\displaystyle{n \choose x}= \displaystyle\frac{{n!}}{{x!\left( {n - x} \right)!}}$$
+
+### Examples
+
+:::info Example
+
+Consider tossing a coin four times:
+
+(a) How many times will this experiment result in exactly `2` tails?
+There are a total of `16` possible sequences of heads and tails from `4` tosses.
+These can simply all be written down to answer a question like this
+
+We get two tails in `6` of these tosses.
+We can explicitly write the corresponding combinations of two tails as follows:
+
+```text
+HHTT HTHT HTTH THTH TTHH THHT
+```
+
+(b) How many times you will end up with `1` tail? The answer is `4` times and the output can be written as:
+
+```text
+HHHT HTHH THHH HHTH
+```
+
+The case of a single tail is easy: The single tail can come up in any one of four positions.
+
+:::
+
+## The Binomial Theorem
+
+$$(a+b)^n = \sum_{x=0}^n \displaystyle{n \choose x} a^xb^{n-x}$$
+
+### Details
+
+If $a$ and $b$ are real numbers and $n$ is an integer then the expression $(a+b)^n$ can be expanded as:
+
+$$(a+b)^n = a^n+ \displaystyle{n \choose 1}a^{n-1}b + \displaystyle{n \choose 2}a^{n-2}b^ + \ldots + \displaystyle{n \choose n-1}ab^{n-1}+b^n$$
+
+$$(a+b)^n = \sigma_{i=1}^n \displaystyle{n \choose x}a^xb^{n-x}$$
+
+This can be seen by looking at $(a+b)^n$ as a product of $n$ parentheses and multiply these by picking one item ($a$ or $b$) from each.
+If we picked $a$ from $x$ parentheses and $b$ from $(n-x)$, then the product is $a^x b^{n-x}$.
+We can choose the $x$ $a$'s in a total of $\displaystyle\binom{n}{x}$ ways so the coefficient of $a^x b^{n-x}$ is $\displaystyle\binom{n}{x}$.
+
+### Examples
+
+:::info Example
+
+Since
+
+$$(a+b)^n = \sum_{x=0}^n \displaystyle{n \choose x} a^xb^{n-x},$$
+
+it follows that
+
+$$2^n = (1+1)^n = \sum_{x=0}^n \displaystyle{n \choose x}$$
+
+i.e.
+
+$$2^n = \displaystyle{n \choose 0} + \displaystyle{n \choose 1} + \displaystyle{n \choose 2}\ldots + \displaystyle{n \choose n}$$
+
+:::