From 6959541d2c9f7c4d92f4f632f5ec30d75c90f526 Mon Sep 17 00:00:00 2001 From: Razvan Deaconescu Date: Wed, 13 Sep 2023 05:44:26 +0300 Subject: [PATCH] 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 Signed-off-by: Teodor Dutu Signed-off-by: Razvan Deaconescu --- .../more-algebra/reading/README.md | 1 - .../more-algebra/reading/read.md | 183 ++++++++++++++++++ 2 files changed, 183 insertions(+), 1 deletion(-) delete mode 100644 chapters/numbers-to-indices/more-algebra/reading/README.md create mode 100644 chapters/numbers-to-indices/more-algebra/reading/read.md diff --git a/chapters/numbers-to-indices/more-algebra/reading/README.md b/chapters/numbers-to-indices/more-algebra/reading/README.md deleted file mode 100644 index 0f1ade1..0000000 --- a/chapters/numbers-to-indices/more-algebra/reading/README.md +++ /dev/null @@ -1 +0,0 @@ -# More on Algebra diff --git a/chapters/numbers-to-indices/more-algebra/reading/read.md b/chapters/numbers-to-indices/more-algebra/reading/read.md new file mode 100644 index 0000000..551c859 --- /dev/null +++ b/chapters/numbers-to-indices/more-algebra/reading/read.md @@ -0,0 +1,183 @@ +# 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. + +$$2^n = \displaystyle{n \choose 0} + \displaystyle{n \choose 1} + \displaystyle{n \choose 2}\ldots + \displaystyle{n \choose n}$$ + +:::