Mathematics - 2017-06-22 (page 5 of 7)

0celo7

17:00

Integration via differentiation

Semiclassical

that's a special kind of crazy-looking

2

0celo7

Physics typically calls for that.

Semiclassical

Yeah.

And to be fair: sometimes, you really do need to keep track of that stuff.

but i'll leave that to braver workers than I.

Secret

I have another question:

Since we can have power series expansions of operators e.g. $e^D$

can we have a continuum version of it, for example:

Let $L,A$ be some linear operator. Then:

$$L=\int_{\Bbb{R}}a(x)A^x dx$$ ?

Semiclassical

Probably not.

At least not in that form.

I think you would see, though, stuff like $L=\int_{\mathbb{R}}a(x) e^{xA}\,dx$

Secret

17:06

I see

0celo7

Wait, what?

Semiclassical

(I'm not positive i've seen that, but I think I have)

Secret

would acting the hamiltonian on a state vector, and expand it in the position basis count?

o wait, nvm, that probably not what I am thinking about...

ma.utexas.edu/mediawiki/index.php/…

I think those are examples, they are called nonlocal operators

0celo7

17:25

@Secret but, you can't.

That's the whole reason why spectral theory exists

One of the reasons anyway

Secret

O wait, I forgot about the infinite case. That holds in finite cases however (because we use something similar to solve systems of ODE)

For finite dimensional vector spaces, $e^A$ is well defined because the taylor series converges pointwise via a proof involving operator norms if I recall my 2nd year linear algebra course...

Zophikel

What's the intuition behind Open Set

Secret

0

Q: Intuition behind open set in topology

I am reading Munkres Topology Chapter 13, in which some examples of bases of topologies are given. One of the examples compares the two possible bases (a, b) and [a, b) on the real line. I understand that both (a,b) and [a,b) satisfy the definition of basis (intersection between two element has ...

But basically, it generalises from open balls in metric spaces

Zophikel

i'm reviewing the definition on: brilliant.org/wiki/open-sets

Secret

Feb 24 '15 at 15:39, by Chris's sis

@Semiclassical $$\int_0^1 x^{x^x} \ dx$$

Let me just do a quick check of my database...

Jun 1 at 16:04, by Secret

[Random] Experiment No. 1
$$\int_0^1 {}^n x dx$$

Jun 1 at 16:40, by LegionMammal978

@Secret Found an explicit integral formula, turns out it converges to $0.6465031992...$

This is consistent to:

Based on all these findings, I suspect...

$$\lim_{n\to \infty} \int_0^1 {}^nx dx= \int_0^1 \lim_{n\to \infty} {}^nx dx = \int_0^1 e^{-W(-\ln x)}dx$$

Will check these later, or perhaps @Waiting can have a look at these results as she had explored the whole class of tetration integrands before

31

Q: A new interesting pattern to $i↑↑n$ that looks cool (and $z↑↑x$ for $z\in\mathbb C,x\in\mathbb R$)

Many of you may recall "An obvious pattern to $i↑↑n$ that is eluding us all?", an old question of mine, and just recently, I saw this new question that poses a simple extension to tetration at non-integer values: $$a↑↑b=\begin{cases}a^b,&b\in[0,1]\\a^{a↑↑(b-1)},&b\in(1,+\infty)\\\log_a(a↑↑(b+1)),&...

algebra-precalculus complex-numbers recreational-mathematics tetration power-towers

Meanwhile @SimplyBeautifulArt and co. have found the complex case

Now the question is. How far we can push the domain of tetration integrands...

$$\int_V {}^nxd^kx ?$$

or possibly...

$$\int_0^1 \mathop{{\Large E}}_{i=1}^n (x+i)dx$$

Definition of Tetration operator here:

Jun 6 at 16:53, by Secret

Another interesting thing is the following problem: The above feature of right associativity of hyperoperators suggest there exists a finite ascending sequence $(a_i)$ and a finite descending sequence $(b_i)$ such that the following will be true:

$$\mathop{\Large{E}}^n_{i=1}a_i=\mathop{\Large{E}}^n_{i=1}b_i$$

where e.g. $\mathop{\Large{E}}^n_{i=1}a_i=a_1^{a_2^{⋰^{a_n}}}$ and is understood to be right associative.

Simply Beautiful Art

17:58

@Secret I have been pinged?

Secret

yup

Seeing similar results from different people seemed to suggest to me that tetration expressions are easier to converge than lower hyoperations

at least within some interval. I am not sure if that's something in common with tetrations in general. Plan to investigate these later

MiKiDe

Hi everyone!
For a project in mathematics, I have to find a subject in maths (or computer science) related to these theme : "Environment: interface, interaction, homogeneity, break". I havn't find yet... What does that inspire you?
It has to be not too much complicated (I've just finished my first year in university).
It's a really important thing for me to find a subject, so if you can help me I will be forever grateful.

Secret

Sounds like either dynamical systems related maths subjects or biomathematics

Dodsy

is there a way to turn a physics question like "a child sitting in a tree 4 m up throws an apple at a veclocity of 5 m/s [35 degrees upwards]" into a calculus question?

Secret

first, you need to define your coordinates system (e.g. where you set x=0,t=0 etc.). Then you can express velocity in terms of the derivative of position. Using these you should get an ODE, which you can plug the IVP into it to solve

Akiva Weinberger

18:06

Let $x_x(t)$ be the $x$-coordinate and $x_y(t)$ the $y$-coordinate. $\frac d{dt}x_x=v_x$, and $\frac d{dt}v_x=a_x$

Note that $a_x=0$ and $a_y=-9.8$

Dodsy

right

so we know that the third derivative would be $f''(x)=-9.8$

so we can model the equation by $f(t)=-4.9t^2 + .... + ....$

right?

or am I looking at this wrong.

Akiva Weinberger

Draw a diagram as well

Dodsy

So you're suggesting breaking the velocity into a horizontal and vertical component.

and then finding at which t the vertex of the graph lies?

Akiva Weinberger

What's the question, find the highest point?

That would be when $v_y=0$

But, yes.

Secret

ok guys, I am off to sleep, night

Dodsy

18:10

I have to be honest, I am just reviewing this for fun :P

imgur.com/a/pPp0U

but trying to approach it from a calculus perspective.

this is algebra based physics.

Akiva Weinberger

[Unrelated] In the ring of formal power series $\Bbb R[[x]]$, there are unique square roots

Dodsy

Night secret

Akiva Weinberger

(Cont'd) $\sqrt{x+1}=1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac5{128}x^4+\dotsb$

Dodsy

So our $v_y = ~ 2.86$

hm.

Astyx

Could someone explain to me me why I have a cold when it's been 35 degrees Celsius for the past 4 days ?

Dodsy

18:14

now usually we'd use one of our kinematics here

Abcd

How do I solve this question? If $3 tan (\theta -15^o) = tan (\theta +15^o); 0<\theta<90^o$ . Then prove that $\theta = 45^o$

Akiva Weinberger

@Astyx Because colds are not caused by the cold

Astyx

Yeah but still, it feels pretty absurd

Abcd

@LeakyNun ...

Dodsy

That's sad astyx..

I'm sorry to hear of your ailments.

Akiva Weinberger

18:15

$x+1=\sqrt{(x^2+2x)+1}$

Sha Vuklia

Guys, how can I show that if $xa+yb=1$, then $\gcd(a,b)=1$, without resorting too much to Euclid's extended algorithm? If I have to use Euclid's algorithm, I would use the fact that Euclid's algorithm yields the smallest possible value such that $x'a+y'b=\gcd(a,b)$, but I haven't shown this yet (it's an exercise in our book, but I haven't done it). However, I'd rather avoid doing that exercise now, if there is a quicker way to show it?

Akiva Weinberger

That means $x+1=1+\frac12(x^2+2x)-\frac18(x^2+2x)^2+\dotsb$ in the ring of formal power series

Abcd

anyone?

Astyx

What's your definition of $\gcd$ ?

Akiva Weinberger

No convergence issues because each coefficient has finitely many terms contributing to it

Astyx

18:16

If $d|a$ and $d|b$ then $d| ax+yb$ for all $x,y$

Dodsy

Akiva

Akiva Weinberger

Yeah?

Sha Vuklia

let me check @Astyx

Dodsy

so we can write this as $-4.9t^2+2.86t$ ?

Akiva Weinberger

${}+4$, since you start four units up, I think

Dodsy

18:17

true.

Akiva Weinberger

But that's the right idea

Sha Vuklia

@Astyx they just define it as the greatest number that divides both $a$ en $b$

Astyx

Then see what I said above

Akiva Weinberger

I don't know how you got your velocity but I assume it has to do with trig

Dodsy

yeah

the simple trig

Sha Vuklia

18:18

oh shit

thanks @Astyx

Astyx

Glad to help

Sha Vuklia

that is indeed the easier approach I was looking for :P

Dodsy

my computer is being really "laggy"

Akiva Weinberger

$\dfrac1{\gcd(a,b)}=\dfrac{xa+yb}{\gcd(a,b)}=x\dfrac a{\gcd(a,b)}+y\dfrac b{\gcd(a,b)}$ must be an integer @ShaVuklia

so $\gcd(a,b)$ must equal $1$

Sha Vuklia

oh wowwww

Akiva Weinberger

18:20

(Essentially the same as what Astyx wrote)

Sha Vuklia

nice thanks @Akiva

Dodsy

nice.

Sha Vuklia

hahaha yea

Akiva Weinberger

Do you know the converse?

Dodsy

the shoe company?

Akiva Weinberger

18:21

$\gcd(a,b)=1\implies\exists x,y:xa+yb=1$

Sha Vuklia

well I use Euclid's extended algorithm to show the converse

Akiva Weinberger

Oh

There's a really quick way

Sha Vuklia

oh tell me

Avantgarde

shoes lol

Akiva Weinberger

Look at the smallest number that you can write in the form $xa+yb$ as $x$ and $y$ vary

Dodsy

18:22

I used to have a pair of converse.

Akiva Weinberger

Well, that has $0$ and negative numbers in it. Look at the smallest positive number of that form.

Astyx

Setting $\gcd(a,b)$ to be the unique positive integer such that $a\Bbb Z + b\Bbb Z = \gcd(a,b)\Bbb Z$ is the quickest way I know :p

Akiva Weinberger

(I have never heard of this company)

Sha Vuklia

@AkivaWeinberger I'm not sure I see why that would be the gcd then

Astyx

Converse ? They're a famous brand of shoes here (in France)

Akiva Weinberger

18:23

Oh, wait, hold on

(Where's "here")

Dodsy

France

Akiva Weinberger

Oh right duh

Dodsy

for astyx

Canada for me.

Astyx

I'm ze silly frenchman

Sha Vuklia

you don't recognise these?

Akiva Weinberger

18:25

To be honest, I've never really paid attention to people's shoes

Sha Vuklia

anyhow I'll just stick to Euclid's algorithm. but I'm glad the other way is easy

Dodsy

World famous sneakers

Akiva Weinberger

@ShaVuklia Yeah, I forget how it goes

but I think it ends up being the same as Euclid more or less

Dodsy

Euclid's algorithm is taught in number theory, I believe.

Akiva Weinberger

Do you know what an ideal is?

Sha Vuklia

18:26

not yet but will soon

Dodsy

Strawberry Fields Forever - Restored HD Video

►

Akiva Weinberger

An ideal is a set of numbers such that adding any two keeps you in the set, and multiplying by any number keeps you in the set

Sha Vuklia

I have to leave :P kinda time-pressured 'n stuff

Akiva Weinberger

Oh, OK, bye

Astyx

Bye

Sha Vuklia

18:27

oh right, well I'll get to that soon! thanks

Dodsy

Hey Danu

Astyx

You also have to be a group

So you'd have to change "adding" to "substracting"

SteamyRoot

Ideals are the normal subgroups of rings :P

Akiva Weinberger

Ideals are what you can quotient by.

Normal subgroups are what you can quotient by.

Astyx

That would be a terrible textbook definition :p

SteamyRoot

18:31

Indeed it would :P

Akiva Weinberger

But it's the motivation of the definition!

Astyx

"A set is said to be an ideal if you can quotient by it."

Akiva Weinberger

At least add a note saying, "In Chapter Blah, when we learn quotient groups, you'll learn that normal subgroups are precisely the ones we can quotient by."

After the "proper" definition or whatever.

Astyx

Algebra : an obscure approach

SteamyRoot

I'm not sure I agree. These are the kind of things you may want students to find out themselves.

Astyx

18:34

Defining quotienting before that might be tedious though

Akiva Weinberger

It's the reason the definition exists

I mean, intuitively, modular arithmetic is just "What happens to the integers if we decide seven should equal zero"

and ideals are kind of like "If seven is zero, what else is zero"

…Don't put that into a textbook

Or anywhere

Dodsy

holy poop

defnitely need to restart my computer

user193319

18:47

Suppose that $T : V \to W$ is a linear operator between finite dimensional vector spaces. If $T$ is not injective, does that mean that it maps one of the basis vectors of $V$ to zero?

Astyx

No, it means it maps one nonzero vector to 0

Danu

@user193319 To see that this couldn't possibly be true in general, perform a change of basis that changes all basis vectors.

Astyx

For instance if you take $V=W=\Bbb R^2$ with its canonical basis $(e_1, e_2)$, then the linear operator defined by $T(e_1) = e_1$ and $T(e_2)=e_1$ is not injective

Danu

On the other hand, changing bases also shows that you can always find a basis where a non-injective map sends a basis element to 0

Astyx

And $T(e_1-e_2) = 0$ but none of the basis vectors are sent to $0$

Sha Vuklia

18:50

Guys, say we have that $G$ is a monoid. I have to show that
$$
G^*=\{a\in G\vert \exists x\in G:xa=ax=e\}
$$
is a group. I don’t see how this is not trivial. Since $G$ already is a monoid, all we have to do is show that $G^*$ has an inverse for each $a\in G^*$, however that’s exactly what $G^*$ says?

Danu

@ShaVuklia You could check the other group axioms... But yeah.

Sha Vuklia

yea but they are already checked

by $G$ being a monoid

I guess there really is nothing more to it then, and that I'm not missing anything

SteamyRoot

Technically, they are not.

You're taking a subset of $G$.

Danu

That's a good point

Sha Vuklia

well yea we could maybe lose $e$

right?

SteamyRoot

18:52

You can't claim that $G^*$ is a group until you show that $G^*$ is non-empty.

Danu

you need to check that things still work inside the subset

Sha Vuklia

right

so I have to state that $ee=e$

SteamyRoot

"Consider $e \in G$. Then clearly some $x \in G$ exists such that $xe = ex = e$, namely $x = e$. Hence $e \in G^*$."

Sha Vuklia

wow geez

don't be so.. what's the word

"pompeus" :P

alright I've written it down

thanks

SteamyRoot

This is pretty much the bare minimum, though :P

Sha Vuklia

18:55

hahaha true I guess :P

Astyx

You also have to show it's stable under the operation

That is, if $a$ and $b$ have left and right inverses, so does $ab$

SteamyRoot

Yup

Sha Vuklia

oh right

let me think

SteamyRoot

Also, the definition kind of says that $a$ has an inverse $x$ in $G$.

You want an inverse in $G^*$.

Which is again, pretty trivial, but not completely :P

Astyx

So you really have to check all group axioms, none are inherited from $G$ (except the existence of a neutral element)

Sha Vuklia

18:58

the associativity is inherited tho?

Astyx

Oh yeah, true

SteamyRoot

Psh, the existence of the neutral is not the inherited one :P

Sha Vuklia

lol true

that was the whole point in the beginning :P

okay cool I think I got it

Leaky Nun

@Abcd Let $\alpha = \theta-15^\circ$ and $t=\tan\alpha$. $\begin{array}{rcl} 3\tan(\theta-15^\circ) &=& \tan(\theta+15^\circ) \\ 3\tan(\alpha) &=& \tan(\alpha+30^\circ) \\ 3t &=& \dfrac{t+\frac1{\sqrt3}}{1-\frac t{\sqrt3}} \\ 3t - \sqrt3t^2 &=& t+\frac1{\sqrt3} \\ 3t^2 - 2\sqrt3t + 1 &=& 0 \\ t&=& \frac1{\sqrt3} \\ \alpha &=& 30^\circ \\ \theta &=& 45^\circ \end{array}$

Oh, and RIP formatting.

Astyx

Oh I have a cool problem

Find which integers $n$ verify the existence of integers $c_1, \dots, c_{n-1}$ such that $$\arctan(n) = \sum_{k=1}^{n-1}c_k \arctan(k)$$

SteamyRoot

19:01

You'll want to use some \text in there, probably :P

Meh, cheater!

Zee

@Daminark Am gonna become a mathematician or a Gardner, still deciding

Astyx

Why not both

Zee

Well, I am doing both, but I mean for $$$

Astyx

That's not a decision we can take for you :p

SteamyRoot

And unless you pay yourself, it's not even a decision you might be able to make for yourself :/

Zee

19:16

@SteamyRoot 100% true

Unless your Wittgenstein, in which case you give all your money away and become a gardner

Astyx

SteamyRoot, the frustrated gardner

SteamyRoot

Haha :P

I'm getting paid plenty, though

Akiva Weinberger

19:42

Oh wow

olympiad math

hello

Dodsy

WHy do you think the collatz conjecture hasn't been proven yet?

for instance, it makes sense that if you divide even numbers by 2 over and over you'll eventually reach 2.

since the ld of any even number is 2.

and then 2/2 = 1

so, it shouldn't be that hard to prove

I think it's the 3n + 1 business that is making it harder to define.

but even just adding 1 to n and then dividing is the same.

but also note that any odd integer multiplied by an odd integer is equal to an odd number.

and an odd number plus 1 is equal to an even number

then the ld of any even number is 2.

maybe I don't understand the conjecture.

Leaky Nun

@Dodsy if you replace "3n+1" with "n+1", then for odd numbers you are effectively doing (n+1)/2, so in each step you are guaranteed to decrease the number (unless you are at 1)

it isn't hard to prove that (n+1)/2 < n for n > 1

Akiva Weinberger

Note that $n<\frac{3n+1}2$

Leaky Nun

exactly.

olympiad math

19:53

Well if its a power of 2 you are good to go, but as soon as you end up with a odd number trouble starts. This might be because for an given odd number we dont know how many steps it would take to reach 1. Few will reach quickly but others might take up time

Akiva Weinberger

So going from, say, $5$ to $8$ increases it

($5\mapsto16\mapsto8$)

Leaky Nun

@Dodsy also, nothing you said has anything to do with collatz conjecture; the conjecture says that you eventually go to 1

5 mins ago, by Dodsy

for instance, it makes sense that if you divide even numbers by 2 over and over you'll eventually reach 2.

This is demonstrably false: 6 is even, yet you can only reach 3.

Maybe you have confused even numbers with powers of 2.

SteamyRoot

Well, for Collatz, you also expect to "eventually" decrease the number.

Leaky Nun

yes but you can't prove it

and the "eventually" can take you a while

SteamyRoot

If you define $T(n)$ as $n/2$ for even $n$ and $(3n+1)/2$ for odd $n$, you can verify that, by considering values mod $4$, you expect $T(n)$ to be even with $50$% chance and odd with $50$% chance

Leaky Nun

19:59

mathematics isn't built on chances

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$Mathematics$ Mathematics