Mathematics - 2017-01-22 (page 1 of 5)

TheGreatDuck

12:00 AM

@SimplyBeautifulArt been looking for impulse problems regarding differential equations. Notice any around here?

brb

Simply Beautiful Art

No thank you

XD

TheGreatDuck

gotta alter my rep

Simply Beautiful Art

Euler's identity

In mathematics, Euler's identity (also known as Euler's equation) is the equality e i π + 1 = 0 {\displaystyle e^{i\pi }+1=0} where e is Euler's number, the base of natural logarithms, i is the imaginary unit, which satisfies i2 = −1, and π is pi, the ratio of the circumference of a circle to its diameter. Euler's identity is named after the Swiss mathematician Leonhard Euler. It is considered to be an example of mathematical beauty. == Explanation == Euler...

@Socrates

TheGreatDuck

no... thank you? I'm asking you if you've seen any...

Simply Beautiful Art

@TheGreatDuck No and thank you

XD

TheGreatDuck

12:02 AM

why are you thanking me?!?

XD

Simply Beautiful Art

Say, can we define the exponential function as follows?

$$e^z=\lim_{|n|\to\infty}\left(1+\frac zn\right)^n$$

TheGreatDuck

idk

not sure

frankly, it always seemed weird to "define" e. I mean we can find it via a limit.

but it's just a constant

i see no such thing for pi.

XD

Simply Beautiful Art

You sure?

TheGreatDuck

pi is the ratio of diameter to circumference

but i mean there;s no limit or equation denoting pi that i know of

Simply Beautiful Art

@TheGreatDuck Could you define those concepts?

TheGreatDuck

12:04 AM

it's just... 3.1415926...

Simply Beautiful Art

Define ratio of diameter to circumference for me

TheGreatDuck

huh?

Simply Beautiful Art

Define that stuff

TheGreatDuck

im busy. I was just saying equations seemed weird

Simply Beautiful Art

Remember, no limits

TheGreatDuck

12:05 AM

and i have no clue

Simply Beautiful Art

D:

TheGreatDuck

i don't think that can even be expressed with or without limits

What?!

can it...?

How so?

12:06 AM

how does one find the circumference without knowing the ratio?

Socrates

a rope

Simply Beautiful Art

@TheGreatDuck Exactly

@Socrates PFT Wtf man

@TheGreatDuck But you can define it in two different ways

TheGreatDuck

no

it's pi

Simply Beautiful Art

As the "arc length" with an integral...

TheGreatDuck

oooh

actually

if you use arc length

you don't need to know the ratio.

Simply Beautiful Art

12:07 AM

Or you could consider the limit of regular polygons inscribed and outscribed and apply squeeze theorem

@TheGreatDuck AHA!

But how to define arc length...without limits?

TheGreatDuck

dude, you're acting like you're telling me something ultra special..

Simply Beautiful Art

YES

TheGreatDuck

All I was saying is that pi is pi regardless of a formula for it

Socrates

a is a

Simply Beautiful Art

:D

@TheGreatDuck No...

TheGreatDuck

12:09 AM

i mean that e is the irrational number 2.76 and whatever goes on.

it's a number, not an expression.

1+1 =2, but which defines 2?

Socrates

1+1=0 in $\mathbb{F}_2$

there is a set approach to natural numbers

Simply Beautiful Art

@TheGreatDuck Simple

Akiva Weinberger

In $\Bbb Z$, everything is the sum of four squares (this is Lagrange's four-square theorem). Is this true in $\Bbb Z[x]$?

Simply Beautiful Art

$$2\equiv\text{the number after }1$$

Akiva Weinberger

Well, clearly not, actually, since $x$ isn't

TheGreatDuck

12:11 AM

@SimplyBeautifulArt hahaha no lol

wanna know the official definition of 2?

Simply Beautiful Art

$$n+1\equiv\text{the number after }n$$

Nah, mine is better

TheGreatDuck

$\{\emptyset, \{ \emptyset\} \}$

Simply Beautiful Art

$${\frac {2}{\pi }}={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdots $$

Socrates

@SimplyBeautifulArt I think the word you want to use might be successor?

Simply Beautiful Art

@Socrates Yup

Akiva Weinberger

12:12 AM

@TheGreatDuck $\{\emptyset\}$ is $1$

TheGreatDuck

im editing

Simply Beautiful Art

And I thought that $\pi$ formula was pretty interesting

Akiva Weinberger

$2$ is $\{0,1\}=\{\emptyset,\{\emptyset\}\}$

Simply Beautiful Art

Lol

Akiva Weinberger

And this is arbitrary, actually. @TheGreatDuck We just need some convention of writing numbers in terms of sets, for the purposes of ZFC, and Von Neumann's definition turns out to be convenient

TheGreatDuck

12:13 AM

@AkivaWeinberger I KNOW. I was trying to see what empty set was in latex. Geez.

XD

Socrates

1,2,3 could easily be a,b,c

or I,II,III

Simply Beautiful Art

Or yi er san

uno dos tres

Akiva Weinberger

3Q

Sān-kyu

Thank you

Simply Beautiful Art

;D

TheGreatDuck

I'm just trying to tell simple art that the definition of a number cannot be an expression. That's silly. They're just equal.

Simply Beautiful Art

12:15 AM

:D

Socrates

define $e^x$ non circlic with only expressions.

and get me a coffee please while you do

Simply Beautiful Art

:-/ Using set's to describe transcendental numbers sounds unnatural

Akiva Weinberger

I feel like numbers in a tonal language could easily have been "ǎ, ǎǎ, ǎǎǎ, ..."

@Socrates $\lim_{n\to\infty}(1+\frac xn)^n$

where $n$ ranges over $\Bbb Z$, and the exponentiation there is repeated multiplication.

Simply Beautiful Art

@AkivaWeinberger But what of $z\in\mathbb C$ and $n\in\mathbb C$, and the limit $|n|\to\infty$?

Hm, might post it as a question

Daminark

@Simply Natural numbers are not even irrational, much less transcendental

Simply Beautiful Art

12:19 AM

@Daminark I wasn't talking about natural numbers. Just wondering about transcendental ones

Daminark

You said it sounds "unnatural"

:P

Socrates

is there "more trancendential"?

Akiva Weinberger

I like the Dedekind cuts version.

Simply Beautiful Art

:P

Daminark

@Akiva No but like equivalence classes of Cauchy sequences

Simply Beautiful Art

12:20 AM

@Socrates What about a number that is not roots to any polynomial with algebraic coefficients?

Akiva Weinberger

I prefer Dedekind cuts to equivalence classes of Cauchy sequences.

It's simpler.

@Socrates Relevant

Daminark

But like, Cauchy sequences are more general, and somehow (for me at least) they feel like they correspond better to some intuition on the real numbers

Simply Beautiful Art

@Socrates Then we can define "more transcendental" :)

Akiva Weinberger

Then $T_1$ is the set of transcendentals, and $T_2=\emptyset$

Simply Beautiful Art

@AkivaWeinberger Oh wait, lemme fix that up XD

Socrates

12:23 AM

I can do $e^{x+y}=e^xe^y$, but I can't understand why

Akiva Weinberger

@Daminark They're only "more general" if you plan on generalizing to metric spaces. Dedekind cuts also generalize, to linear orders.

Socrates

and I am watching something which explains it, but don't understand it.

Akiva Weinberger

Using the limit definition?

Daminark

True, but metric spaces are (at least in my experience) far more common.

Socrates

could be any number, $2^{x+y}$ being the same in this regard

Simply Beautiful Art

12:24 AM

Let $A_n$ to be defined as the set of numbers not included in $A_k$ for $k<n$ and are also roots to polynomials with coefficients that are in $A_k$.

@Socrates Then we have "more algebraic", and likewise, more transcendental

And $A_0=\mathbb N$.

Akiva Weinberger

Then $A_n$ are all algebraics for $n\ge1$? @Simply

Oh, wait

Simply Beautiful Art

Haha, I got you ;)

Akiva Weinberger

$A_1$ is algebraics that are not integers, and $A_2$ is the empty set (since you defined them to be disjoint)

Socrates

what do you imagine if you imagine $e$?

Simply Beautiful Art

@Socrates That's easy

That interest problem

which is equal to $(1+1/n)^n$

@AkivaWeinberger So there exists no number that is not algebraic and also not the root to a polynomial with algebraic coefficients?

Akiva Weinberger

12:29 AM

Yeah, using algebraic coefficients doesn't give you any new numbers. Sorry.

Socrates

But why is $e$ correlated with $\pi$, talking about the trig-functions. I can't seem to connect the two.

Simply Beautiful Art

:O Darn, sucks

@Socrates Because...

$$y=\frac{\cos(x)+i\sin(x)}{e^x}$$

What is $\frac{dy}{dx}$?

Akiva Weinberger

@Socrates My thinking is: $e^x$ is defined such that, around $x=0$, we have $e^x\approx x+1$.

That's the closest linear approximation at that point (because the slope at $0$ is $1$).

So, we have $e^{0.00001i}\approx 1+0.00001i$, if we want this property to carry to the complex numbers.

And I realize that I'm getting a little handwavy, but bear with me.

Similarly, $\cos(x)\approx1$ and $\sin(x)\approx x$ near $0$.

Simply Beautiful Art

Lol, so $e^x\approx\sin(x)+\cos(x)$?

Akiva Weinberger

Which means that $e^{0.00001i}\approx\cos(0.00001)+i\sin(0.00001i)$.

Now, raise to the $100000x$-eth power, using De Moivre on the right-hand side.

Try to think of the geometric intuition here:

Socrates

12:34 AM

@SimplyBeautifulArt we didn't prove deriviatives, so..

well, I can accept deriviative of $e^x$ is $e^x$ by definition, if you want to go that route

Akiva Weinberger

I'm saying that $e$ to a very small imaginary power is roughly equal to the place on the unit circle whose angle is the power over $i$.

(And, in fact, it's exactly equal to it.)

Simply Beautiful Art

$$\frac d{dx}\frac{\cos(x)+i\sin(x)}{e^{ix}}=0$$

With some basic trig

Thus, I leave you with proving only constants can have a derivative equal to $0$.

And likewise, at $x=0$...

Akiva Weinberger

As an example, @Socrates, let's see what happens if I ask Google what (1+0.00001*pi*i)^(100000) is.

-1.00004935 + 1.0335934 × 10^-9 i

Socrates

@AkivaWeinberger geometry is no course at my university, where could I learn all those neat things?

Akiva Weinberger

Well, you surely learn linear approximations at points in a calculus course

Socrates

12:37 AM

we even learn it in analysis

oh

"linear"

not sure ;)

Akiva Weinberger

$f(x_0)\approx f(x_0)+f'(x_0)(x-x_0)$

@Socrates It's just the equation of the tangent line.

The tangent line is the linear approximation.

Socrates

Ah ok

why only approximation?

Akiva Weinberger

Because $f(x)$ isn't a line, so you can only approximate it by a line

Like, $e^x$ isn't equal to $x+1$

Simply Beautiful Art

:P That ought to be obvious

Akiva Weinberger

but, for small $x$ (a.k.a., for $x$ near to $0$), they're very close together.

Simply Beautiful Art

12:40 AM

cya guys, gotta eat

Akiva Weinberger

And no other line is closer (near $x=0$).

Socrates

meh, i just realized using trigfunctions with a textbook with plug and chuck was not benefitial

Simply Beautiful Art

12:56 AM

I wonder if we can prove there is a unique continuous function with $z\in\mathbb C$ such that $(f(z))^n=f(nz)$ and $f'(0)=1$.

Akiva Weinberger

$n\in\Bbb Z$ or $n\in\Bbb C$

Actually, it wouldn't matter. The answer is, yes.

Simply Beautiful Art

Cool then

So DeMoivre's is all that's needed then?

Akiva Weinberger

I guess

Simply Beautiful Art

Lol. Ok then

Wow

I hit max rep yesterday? And I was only on for a few hours

Socrates

given enough answers, one might collect 200 rep without even being online that day

Simply Beautiful Art

1:03 AM

@Socrates I hit 200 rep with 10 minutes of activity one day

but it still amazes me

Socrates

not bad

i have +150 rep for a 3 liner, it's not even particularly beautiful

Akiva Weinberger

I like it when I get reputation from questions

It's nice because it's like I didn't do any actual work, I just was curious

Simply Beautiful Art

Yeah, I do too, but it also feels...easy

Yeah, same XD

Akiva Weinberger

1:26 AM

So there's this old game, which was an $n\times n$ array of buttons, some of which were lit up red

and if you pressed a button, it and its four neighbors would toggle the lighting (from on to off or vice versa)

Ah, it was called "Light's Out" and it looked like this.

Simply Beautiful Art

Mhm...

Akiva Weinberger

And the question is, can every starting configuration eventually be turned completely off (which is the goal of the game)

and I believe the answer is yes

and I think I have a proof

but I wonder what the simplest proof would be.

Note that it does not matter in which order the buttons are pressed.

Clearly, we only need to show that the configurations with a single light on can be turned off, since for a general configuration we can then turn off each light one at a time.

I'm sure if I look it up I'd find lots of people writing about it.

Also, I don't think it's possible to turn off just a single light on an infinite grid.

The finiteness comes into play.

sarcopsy

Well in an infinite grid you just make a parity argument

i think

Akiva Weinberger

How? Each push changes the parity.

Five things toggle; the button and its four neighbors.

sarcopsy

Doesn't it toggle the four neighbors?

Akiva Weinberger

1:36 AM

And the button itself.

sarcopsy

oops

Akiva Weinberger

Ah: Suppose you could. Draw the smallest rectangle around the set of buttons you'd push.

Everything outside that rectangle starts off dark,

sarcopsy

That seems right I guess

Simply Beautiful Art

Lol

Akiva Weinberger

so consider the buttons that you'd push that are right on the edge. They'd have to turn on a light outside the rectangle. And no two buttons inside the rectangle affect the same button outside it, so at the end you're left with something outside the rectangle on. QED.

So it's only possible in the finite case.

If you guys want my proof (that it's possible to turn off any combination), let me know

(though I wonder if the internet has better ones)

Simply Beautiful Art

1:41 AM

@AkivaWeinberger What if you were given a hexagonal grid?

Akiva Weinberger

Hm. Interesting. I'd have to think

You know what, I think my proof (for finite grids) still works.

Actually, for infinite grids, here's a point: Pushing infinitely many buttons is well-defined.

Assuming each individual button is only pressed finitely many times.

So I think that we might be able to solve it on an infinite grid, theoretically, using an infinite "push set"

Simply Beautiful Art

Lol

TheGreatDuck

math.stackexchange.com/questions/2104702/… my final solution seems slightly different than the other answer there. Did I do something wrong in finding the roots of the auxiliary equation or in finding the particular solution?

I know my answer is slightly different than most but I think I just made a simple math error I'm blind to.

@AkivaWeinberger also, remember that whole heavyside sin cos continuity thing?

Akiva Weinberger

No

I remember a sin cos thing but no Heaviside thing

TheGreatDuck

y = C(x)sin(x) + D(x)

oh

sometimes I would say D was heavyside.

XD

for triviality

Akiva Weinberger

1:50 AM

OK, so remind me what the question was, exactly

TheGreatDuck

find a piecewise constant C(x) such that y is continuous

and D(x) is piecewise constant

we'll say it's heavyside

for triviality

Ironically, that doesn't work as I realized there is no solution when D(x) jumps at a point where sin(x) = 0

cause no matter what

the left and right hand limits of C(x)sin(x) at x will be 0

Akiva Weinberger

But at other points just set the left lim and the right lim equal to each other

Socrates

@AkivaWeinberger can we work sinus out in private? I can't emphazise the need for it. Why did someone gave a name to it. Same is then analogous to cos and tan.

I know it's very useful, but I can't seem to see what the inventor(s) saw.

Akiva Weinberger

Say $D(x)=\begin{cases}0,&x<a\\1,&x\ge a\end{cases}$ @TheGreatDuck

TheGreatDuck

@AkivaWeinberger no. We want to make a discontinuity at a to cancel out the discontinuity at D

but if let's say a = 0

Akiva Weinberger

1:54 AM

@Socrates It was very useful in navigation

TheGreatDuck

no matter what C is

C(a)sin(a) = 0

and both sides i think will also approach 0

Akiva Weinberger

Sailing used a lot of spherical trigonometry (trigonometry on a sphere like the Earth)

TheGreatDuck

so there's no way to produce a discontinuity at those points

Akiva Weinberger

@Socrates Also astronomy.

TheGreatDuck

in fact, since that comes about from a diff eq.

I'm gonna go use laplace transform and see the result. It's possible there's a major hole in my logic.

Socrates

1:55 AM

@AkivaWeinberger astronomy and navigation go hand in hand or?

Akiva Weinberger

Like, if the North Star is such-and-such angle above the horizon, what latitude am I at

Socrates

i mean for the middle ages

Akiva Weinberger

@Socrates Yeah, they could, if you're navigating by the stars

TheGreatDuck

sine cos and tangent are good for trigonometry

Akiva Weinberger

though you could also do astronomy by itself

TheGreatDuck

1:56 AM

and working with angles

engineering

trajectory

etc.

Akiva Weinberger

Wait, my above example wasn't good, you don't need sine. Hm.

Socrates

where navigators well educated people?

Akiva Weinberger

I dunno

TheGreatDuck

YES

most of them were princes

or wealthy people

Socrates

I wonder how pirates got access to good navigators^^

Akiva Weinberger

1:58 AM

I'm pretty sure that, during the Age of Sail, they carried tables of trig values

TheGreatDuck

well...

pirates I'm sure learned by practice

not everyone must be well educated to look at the stars

Akiva Weinberger

That's why secant and cosecant are a thing. If they show up in a formula, it's easier to look up the secant in your table than to look up the cosine and take the reciprocal.

I'm sure if you Google "sine history" you'd get stuff.

goes to Google "sine history"

Transcript for

$Mathematics$ Mathematics