SNCA:Calculus

Calculus is currently a work in progress page!

You will NOT edit parts of this page until they are properly finished.

reeeeeeee i hate math

Calculus is a subject of mathematics many 'teens will have to take in school when they are cacas.

It primarily refers to infinitesimal calculus, which is the study of continuous change. It is primarily split between differential calculus, which is related to the instantaneous rates of changes and slopes of curves, and integral calculus, which is related to the accumulation of quantities and areas under curves.

These two topics are connected by the fundamental theorem of calculus: $\underset{a}{\overset{b}{\int }}{f}^{\prime }(t)dt=f(b)-f(a)$

The Real Numbers

The reason why we put so much importance on the concept of the real numbers (denoted $\mathbb{ℝ}$ ) in calculus, is due to the way they are measured. Neither the naturals, integers, nor the rationals, can make a bijection to the real numbers- this proved by German Mathematician Georg Cantor, leading to:

$\left|\mathbb{ℝ}\right|>\left|\mathbb{\mathbb{Q}}\right|=\left|\mathbb{\mathbb{Z}}\right|=\left|\mathbb{\mathbb{N}}\right|$

Roughly speaking, this is proven by the following:

• There is a bijection from the natural numbers to the integers. Intuitively, fix the point $0$, then use the following: $1\mapsto 1$, $2\mapsto -1$, $3\mapsto 2$, $4\mapsto -2$, etc.
• There is a bijection from the natural numbers to the rational numbers. Intuitively, we do something like: $0\mapsto \frac{0}{1}$, $1\mapsto \frac{1}{1}$, $2\mapsto \frac{2}{1}$, $3\mapsto \frac{1}{2}$, $4\mapsto \frac{-1}{2}$, $5\mapsto \frac{-2}{1}$, etc., skipping all fractions that are not lowest terms.
• However, there is no bijection from the natural numbers to the real numbers. Suppose for contradiction, that there was. Then, we could list all real numbers $r_1=0.a_{1,1}{a}_{1,2}{a}_{1,3}\dots ,r_2=0.a_{2,1}{a}_{2,2}{a}_{2,3}\dots ,\dots$. Then, we construct a new number $r$ by taking all of the diagonal digits $a_{1,1},a_{2,2},a_{3,3},\dots$, and adding $1mod10$. This is guaranteed to not be in the list, because $r$ differs from $r_1$ in the first digit, $r$ differs from $r_2$ in the second digit, etc, which implies that such a number was not in the list, contradicting the earlier assumption, and therefore there is no bijection from the naturals to the reals.

This result is very important, because all of these sets have a size of infinity. You will commonly see pop-soyence influencers make the spurious claim that "some infinities are bigger than others, or something", which is technically true, but lacks context. The difference in infinities here is in the ability to make bijections from countably infinite sets, which is the 'smaller' infinity. Intuitively, we can see how there are infinite numbers on the unit interval $[0,1]$, yet the measure of this interval is finite. So, when measuring sets on the real numbers, we use a measure called the Lebesgue measure. This measure assigns a value of zero to individual points in a domain, hence, while the rationals $\mathbb{\mathbb{Q}}$ are dense on any subset of $\mathbb{ℝ}$, the measure of the rationals is zero.

Limits and Bounded Sequences

A sequence of real numbers is a function $f:\mathbb{\mathbb{N}}\to \mathbb{ℝ}$. A sequence $\{f_n{\}}_{n=1}^{\mathrm{\infty }}$ is said to be bounded iff there exists a $M\in \mathbb{ℝ}$ such that:

$\left|f_n\right|⩽M\mathrm{\forall }n\in \mathbb{\mathbb{N}}$

The limit of a bounded sequence is said to converge to a number $L\in \mathbb{ℝ}$ iff for every $ϵ>0$, there exists some $N\in \mathbb{\mathbb{N}}$ such that $\mathrm{\forall }N⩽n$:

$|f_n-L|<ϵ$

Thus, we say say the limit of sequence $\{f_n\}$ converges to $L$. In notation:

$\underset{n\to \mathrm{\infty }}{\lim }\{f_n\}$ = L

Extrapolating to functions, we get the important result:

$\underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)$

Differentiation

The derivative of a function $f\left(x\right)$ denoted $f^{\prime }\left(x\right)$, $\frac{df}{dx}$, or $\frac{d}{dx}\left[f\left(x\right)\right]$ quantifies the local behavior of $f\left(x\right)$ near the point $x=c$. The derivative is given by:

$f^{\prime }\left(c\right)=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$

There are many differentiation rules for different kinds of elementary functions, but we will not cover these, as they are fairly simple to prove using the limit definition of the derivative and require only basic algebraic manipulation.

Theorems

Mean Value Theorem

Let $f\left(x\right)$ be a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$, then for some $c\in [a,b]$:

$\frac{f\left(b\right)-f\left(a\right)}{b-a}=f^{\prime }\left(c\right)$

This theorem is an extrapolation from Rolle's theorem, which states if $f\left(b\right)=f\left(a\right)$ then there exists some $c\in [a,b]$ such that $f^{\prime }\left(c\right)=0$.

L'Hôpital's rule

Let $f\left(x\right),g\left(x\right)$ be functions on the open interval $X$, such that $g\left(x\right)\ne 0,f\left(x\right)\ne \pm \mathrm{\infty }$, for $x\in X\setminus \{c\}$ for a number $c\in X$ Then if either:

$\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }g\left(x\right)=0$

Or

$\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }g\left(x\right)=\pm \mathrm{\infty }$

Then:

$\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

Taylor's Theorem

Let $f\left(x\right)$ be a function satisfying $f\left(x\right)\in C^{\mathrm{\infty }}\left(I\right)$ (this means $f\left(x\right)$ is infinitely differentiable on interval $I$) Then there exists a polynomial $P\left(x\right)$ of degree $n$, such that for some $a\in I$:

$P^{\left(n\right)}\left(a\right)=n!f^{\left(n\right)}\left(a\right)$

This polynomial is called the Taylor polynomial of $f\left(x\right)$ centered at $a$, and has the property of uniqueness, since the derivative of a function is unique. The Taylor expansion of $f\left(x\right)$ about $x=a$ is:

$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{\left(n\right)}\left(a\right)}{n!}{(x-a)}^n$

Inverse Function Theorem

Let $f\left(x\right)$ be a bijective function for $x\in X$, then there exists a function $f^{-1}\left(x\right)$ such that $f\left(f^{-1}\left(x\right)\right)=x,\mathrm{\forall }x\in X$. Then, we can differentiate both sides leading to:

$f^{\prime }\left(f^{-1}\left(x\right)\right)f^{{-}^{\prime }1}\left(x\right)=1$

$\frac{d}{dx}\left[f^{-1}\left(x\right)\right]=\frac{1}{f^{\prime }\left(f^{-1}\left(x\right)\right)}$

Integration

The integral, or as some people call it, the 'anti-derivative' is the measure of a function over a certain domain. There are a few types of integrals (mostly appearing in analysis), but the most common is known as the Riemann integral; if a function $f\left(x\right)$ is continuous over a domain $X=[a,b]$, then the Riemann integral is defined as:

$\underset{X}{\int }f\left(x\right)dx=\underset{N\to \mathrm{\infty }}{\lim }\frac{b-a}{N}\sum _{k=0}^{N}f(a+\frac{b-a}{N}k)$

If one of the endpoints of the interval is infinite, and $\left|f\left(a\right)\right|<\mathrm{\infty }$:

$\underset{b\to \mathrm{\infty }}{\lim }\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{b\to \mathrm{\infty }}{\lim }F\left(b\right)-F\left(a\right)$

For example, integrating $\frac{1}{x^2}$ over $[1,\mathrm{\infty })$:

$\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2}=\underset{b\to \mathrm{\infty }}{\lim }\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2}=\underset{b\to \mathrm{\infty }}{\lim }-\frac{1}{x}\underset{1}{\overset{b}{\mid }}=\underset{b\to \mathrm{\infty }}{\lim }1-\frac{1}{b}=1$

The methods of integration are much more difficult than differentiation, as most functions do not have elementary primatives.

Methods of Integration

Substitution

$\underset{a}{\overset{b}{\int }}f\left(g\left(x\right)\right)g^{\prime }\left(x\right)dx=\underset{g\left(a\right)}{\overset{g\left(b\right)}{\int }}f\left(u\right)du$

Integration by Parts

$\int f\left(x\right)dg\left(x\right)=f\left(x\right)g\left(x\right)-\int g\left(x\right)df\left(x\right)$

Leibniz Method

$\frac{d}{dt}\left[\underset{a}{\overset{b}{\int }}f(x,t)dx\right]=\underset{a}{\overset{b}{\int }}\frac{\partial }{\partial t}f(x,t)dx$

For example:

$\underset{0}{\overset{1}{\int }}\frac{x^{\alpha }-1}{\log \left(x\right)}dx=\underset{0}{\overset{1}{\int }}\frac{d}{d\alpha }\left[\underset{0}{\overset{1}{\int }}\frac{x^{\alpha }-1}{\log \left(x\right)}dx\right]d\alpha =\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{x}^{\alpha }dxd\alpha =\underset{0}{\overset{1}{\int }}\frac{1}{\alpha +1}d\alpha =\log \left(2\right)$

Fundamental Theorem of Calculus

$\underset{a}{\overset{b}{\int }}{f}^{\prime }\left(x\right)dx=f\left(b\right)-f\left(a\right)$