Math Thread

ITT let's talk about math!

0.999...!=1
The concept of two different numbers being equal is sophistry. 0.999... can't get to 1.

What loungramming language is this?

\(\huge Exponential \; growth \; and \; decay\)
\(\large \frac{dy}{dt} = ky\)
Rate of change of y with respect to t.
k is a constant and y is y itself.

When k > 0, it's exponential growth. When g < 0, it's exponential decay.

\(\Large Theorem\)
The only solutions of the differential equation \(\frac{dy}{dt} = ky\) are the exponential functions of \(\large y(t)=y(0)e^{kt}\)

Sometimes, you might need to convert t from hours to minutes, or vice versa. Depends on the problem and what starting information you're given.

y(0) is the starting amount, such as number of bacteria or mass of a radioactive substance.

k determines the rate of growth.

For these types of problems, remember that \(\ln e = 1\).

Also remember these things:
\(\ln (1) = 0\)
\(\ln (<1) < 0\)
\(\ln (>1) > 0\)

So you can easily figure out if something is growth or decay.

Use base e to make things easier. For example, \(e^{\ln 2} = 2\).

\(\huge Related \; Rates\)
Sometimes rates can be related to one another.
Let's say we're talking about a related rate problem involving volume.

\(V\) stands for volume that is changing in time.
Then \(\large \frac{dV}{dt}\) is the rate for how fast the volume is changing with respect to time \(t\).

If you have a \(\frac{dV}{dt}\) value of \(10cm^{3}/s\), then that means the volume is increasing by \(10cm^{3}\) per second.

Other examples of related rates: walking away from a light source and the length of the shadow changing.

Sometimes, you will need to use implicit differentiation for related rate problems.

\(A=\pi r^{2}\)
Then differentiate it to get the change in area.
\(\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}\)

Don't forget the \(\frac{dr}{dt}\) (because of the chain rule when differentiating).

A is a function of r, but r is a function of t.

It can help to draw pictures for these problems. They are often word problems and it's easier to figure out what to do when you have a diagram that visualizes the problem.

Rate of change is negative when shrinking, or positive when increasing.

If you have a problem asking about the change in surface area, the function for finding the surface area is the function itself, and the derivative of said function is the rate of change in the surface area.

For some of these kinds of problems, they might have similar triangles. Recall that \(a^{2} + b^{2} = c^{2}\) and the ratios of the sides can be the same, if that's the kind of problem it is.

Remember how implicit differentiation works for these. And review ratios of similar triangles.

\(\huge Random \; stuff\)
Product rule
Chain rule
Quotient rule
Laws of logarithms
Implicit differentiation
etc.
\(\frac{d}{dx}(a^{x}) = a^{x}\ln a\)
Of course, there is one special case for that:
\(\frac{d}{dx}(e^{x}) = e^{x}\)
Because \(\ln e\) = 1.

\(\Large Logarithmic \; differentiation\)
\(f(x)=x^{\tan x}\)
\(\ln (f(x)) = \ln (x^{\tan x})\)
\(\ln (f(x)) = \tan (x) \ln (x)\)
\(\frac{1}{f(x)} f'(x) = \tan x \frac{1}{x} + \sec^{2} x \ln x\)
Product rule invoked above.

Remember that just because something looks like a special character (such as \(\pi\) or \(e\) doesn't make it a variable. These are still just constants.

The derivative of a constant is zero.
\(\frac{d}{dx} \ln \pi = 0\)

If\(f'(a)\) exists, then \(\displaystyle{\lim_{x \to a}} f(x) = f(a)\)

how do i Big Text?

\(\huge Linear \; approximations \; and \; differentials\)
\(y=f(x)\)
Point \((a, f(a))\)
tangent line
\(y-f(a)=f'(a)(x-a)\)
\(y=f'(a)(x-a)+f(a)\)

Linear approximation
\(\large L(x) = f(a) + f'(a)(x-a)\)
L can approximate f.

Use the linear approximation to approximate the value of \( \sqrt[3]{8.05}\) and \(\sqrt[3]{25}\).
Convert it into a function
\(f(x) = \sqrt[3]{x}\)
Rewrite as a fractional exponent to make it easier to derive.
\(f(x) = x^{\frac{1}{3}}\)
Differentiate.
\(f'(x)=\frac{1}{3}x^{\frac{-2}{3}}\)
a = 8
x = 8
(8,2)
\(m = \frac{1}{3 \sqrt[3]8^{2}} = \frac{1}{3(\sqrt[3]{8})^{2}} = \frac{1}{3 \times 2^{2}} = \frac{1}{12}\)
\(y-2=frac{1}{12}(x-8)\)
\(y = \frac{1}{12}x - \frac{8}{12} + \frac{24}{12}\)
\(y = \frac{1}{12}x + \frac{4}{3}\)
The linear approximation is \(L(x)=\frac{1}{12}x + \frac{4}{3}\)
Now plug in the 8.05 and 25 to linearly approximate them. It's not 100% accurate, but it's close enough, and shouldn't require a calculator.
\(L(8.05) = \frac{1}{12}(8.05) + \frac{4}{3}\)
\(\sqrt[3]{25} \approx L(25) = \frac{1}{12}(25)+\frac{4}{3}\)


\(\Large Differentials\)
\(y=f(x)\)
\(\frac{dy}{dx}=f'(x)\)
\(\large dy=f'(x)dx\)
\(\Delta x\) is an arbitrary increment in the independent variable of \(x\).
\(\Delta y\) is the actual change in the dependent variable\(y\) as \(x\) changes from \(x\) to \(x+ \Delta x\), that is:
\(\large \Delta y = f( \Delta x + x) - f(x)\)


\(dy\) is the differential of the dependent variable\(y\) and is defined by:
\(\large dy = f'(x)dx\)

\(\displaystyle{ \Large f'(x) = \lim_{\Delta x \to 0} \Big( \frac{f(x+\Delta x) - f(x)}{\Delta x} \Big) }\)

If \(\Delta x\) is very small \((\Delta x \approx 0)\), then
\(\large f'(x) \approx \frac{f(x+\Delta x)-f(x)}{\Delta x}\), \(\Delta x \neq 0\)

\(\large f'(x)\Delta x \approx f(x+\Delta x)-f(x)\)

\(y=e^{x}\)
x = 0
\(\Delta x = 0.5\)
\(\Delta y = f(x+\Delta x)-f(x) = e^{0.5}-e{0}=e^{0.5}-1\)
\(dy=e^{x}dx\)
x=0
dx=0.5
\(\Delta x = dx = 0.5\)

\(\Delta y = f(x+\Delta x)-f(x) \approx f'(x)dx = dy\)
---------------------------------------

\(\frac{d}{dx}\sin x = \cos x\)
\(\frac{d}{dx}\cos x = - \sin x\)
\(\frac{d}{dx}\tan x = sec^{2} x\)
\(\frac{d}{dx}\csc x = -\csc x \cot x\)
\(\frac{d}{dx}\sec x = \sec x \tan x\)
\(\frac{d}{dx}\cot x = - \csc^{2} x\)

\(\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^{2}}}\)
\(\frac{d}{dx}\cos^{-1}x=\frac{-1}{\sqrt{1-x^{2}}}\)
\(\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^{2}}\)
\(\frac{d}{dx}\cot^{-1}x=\frac{-1}{1+x^{2}}\)
\(\frac{d}{dx}\sec^{-1}x=\frac{1}{x\sqrt{x^{2}-1}}\)
\(\frac{d}{dx}\csc^{-1}x=\frac{-1}{x\sqrt{x^{2}-1}}\)

Distance is the function, velocity is the first derivative, and acceleration is the second derivative.

>>7
If I told you, you'd probably misuse this feature. I am only using it for academic posts, not random shitposting. I taught myself \(\large \LaTeX\) in college even though it wasn't taught in any of my classes. Very useful though.
\(\huge Hyperbolic \; functions\)
\(\Large Definition:\)
Hyperbolic sine of x:
\(\large \sin h x = \frac{e^{x}-e^{-x}}{2}\)
Hyperbolic cosine of x:
\(\large \cos h x = \frac{e^{x}+e^{-x}}{2}\)
Hyperbolic tangent:
\(\large \tan h x = \frac{\sin h x}{\cos h x}\)
Hyperbolic cotangent:
\(\large \cot h x = \frac{\cos h x}{\sin h x}\)
Hyperbolic secant:
\(\large \sec h x = \frac{1}{\cos h x}\)
Hyperbolic cosecant:
\(\large \csc h x = \frac{1}{\sin h x}\)

\(\Large Hyperbolic \; identities\)
\(\sin h (-x) = - \sin h x\)
\(\cos h (-x) = \cos h x\)
\(\cos h^{2} (x) - \sin h^{2} (x) = 1\)

\(\large Derivatives \; of \; hyperbolic \; functions\)
\(\frac{d}{dx}\sin hx =\cos hx\)
\(\frac{d}{dx}\cos hx =\sin hx\)
\(\frac{d}{dx}\tan hx =\sec h^{2}x\)
\(\frac{d}{dx}\sec hx =- \sec hx \tan hx\)
\(\frac{d}{dx}\cot hx =- \csc h^{2}hx\)
\(\frac{d}{dx}\csc hx =- \csc hx \cot hx\)
Note that some of the signs here are different compared to regular functions and their derivatives.

>>9
Too bad you didn't learn vertical space.

Simply typed lambda calculus
- Abstraction rule:
\((\lambda x : t_x.\ y) : t_x \to t_y\) for \(y : t_y\)

- Application rule:
let \(f : t_x \to t_y\) and \(a : t_x\)
\(f\ a : t_y\)

- Syntax:

term(f, y, a) ::= \(\lambda x : type.\ y\)             Abstraction
                | \(f\ a\)                    Application

type(x, y)    ::= symbol
                | \(x \to y\)

Edited on 12/07/2018 14:02.

Simply typed lambda calculus
- Abstraction rule:
\((\lambda x : t_x. y) : t_x \to t_y\) for \(y : t_y\)
\((\lambda x : t_x.\ y) : t_x \to t_y\) for \(y : t_y\)

- Application rule:
let \(f : t_x \to t_y\) and \(a : t_x\)
\(f\ a : t_y\)
↵
- Syntax:↵
term(f, y, a) ::= \(\lambda x : type.\ y\)             Abstraction↵
                | \(f\ a\)                    Application↵
↵
type(x, y)    ::= symbol↵
                | \(x \to y\)

>>7
\huge

[b]Simply typed lambda calculus with polymorphic types (System F)[\b]
- Syntax:

term(f, y, a) ::= \(\lambda x : type.\ y\)             Abstraction
                | \(\lambda x : kind.\ y\)             Polymorphic Abstraction
                | \(f\ a\)                    Application

type(y)       ::= symbol
                | \(\forall x : type.\ y\)
                | \(\forall x : kind.\ y\)

kind          ::= \(\star\)

\(t_0 \to t_1\) is the same as \(\forall x : t_0.\ t_1\) for \(t_0 \not\in t_1\)

- Example
\(id = \lambda a : \star.\ \lambda x : a.\ x\)
\(id\ int\ 1 = 1\)
Most programming languages based on System F insert the type argument automatically.

\(id : (\forall t_0 : \star.\ (\forall t_1 : t_0.\ t_0))\)
\(id\ int : (\forall t_1 : int.\ int)\)
\(id\ int = \lambda x : int.\ x\)
\(id\ int\ 1 : int\)

\(\forall x : type. y\)

This is actually not needed for non-dependant calculi. You can just use \(type \to type\) in this case (the kind version is needed however).

>>12
but enough about my dick, this is a math thread

Did anyone save the pack of wild set theorists kopipe from /prog/rider's /math/? I'd like to read it again.

>>17
I don't know about that one, but I made the original ``consider this'' post. I am significantly less edgy now though. I have never used an alias on this board, but some of my posts, mainly on 4chan, have become widely-known memes. I don't consider the ``consider this'' one to be widely known outside of textboard circles, but I mean like one of them became pretty mainstream quite a few years ago.

>>18
I developed and introduced DRIVE LIKE JEHU.

>>18

one of them became pretty mainstream quite a few years ago

What was it?

>>18
Did you also invent faggot quotes?

>>21
Those have been around longer than http.

>>20
3.4 million views on this article alone:
https://knowyourmeme.com/memes/has-anyone-really-been-far-even-as-decided-to-use-even-go-want-to-do-look-more-like

4.9 million views for an old post I made on my old blog and /g/:
https://www.youtube.com/watch?v=HDDLTwS4zgs

>>21
of course not