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5:00 PM
Integration via differentiation
 
that's a special kind of crazy-looking
2
 
Physics typically calls for that.
 
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.
 
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:
$$L=\int_{\Bbb{R}}a(x)A^x dx$$ ?
 
Probably not.
At least not in that form.
I think you would see, though, stuff like $L=\int_{\mathbb{R}}a(x) e^{xA}\,dx$
 
5:06 PM
I see
 
Wait, what?
 
(I'm not positive i've seen that, but I think I have)
 
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...
I think those are examples, they are called nonlocal operators
 
5:25 PM
@Secret but, you can't.
That's the whole reason why spectral theory exists
One of the reasons anyway
 
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...
 
What's the intuition behind Open Set
 
0
Q: Intuition behind open set in topology

cr001I 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
 
i'm reviewing the definition on: brilliant.org/wiki/open-sets
 
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 $z\in\mathbb C,x\in\mathbb R$)

Simply Beautiful ArtMany 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)),&...

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.
 
5:58 PM
@Secret I have been pinged?
 
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
 
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.
 
Sounds like either dynamical systems related maths subjects or biomathematics
 
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?
 
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
 
6:06 PM
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 $a_y=-9.8$
 
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.
 
Draw a diagram as well
 
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?
 
What's the question, find the highest point?
That would be when $v_y=0$
But, yes.
 
ok guys, I am off to sleep, night
 
6:10 PM
I have to be honest, I am just reviewing this for fun :P
but trying to approach it from a calculus perspective.
this is algebra based physics.
 
[Unrelated] In the ring of formal power series $\Bbb R[[x]]$, there are unique square roots
 
Night secret
 
(Cont'd) $\sqrt{x+1}=1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac5{128}x^4+\dotsb$
 
So our $v_y = ~ 2.86$
hm.
 
Could someone explain to me me why I have a cold when it's been 35 degrees Celsius for the past 4 days ?
 
6:14 PM
now usually we'd use one of our kinematics here
 
How do I solve this question? If $3 tan (\theta -15^o) = tan (\theta +15^o); 0<\theta<90^o$ . Then prove that $\theta = 45^o$
 
@Astyx Because colds are not caused by the cold
 
Yeah but still, it feels pretty absurd
 
@LeakyNun ...
 
That's sad astyx..
I'm sorry to hear of your ailments.
 
6:15 PM
$x+1=\sqrt{(x^2+2x)+1}$
 
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?
 
That means $x+1=1+\frac12(x^2+2x)-\frac18(x^2+2x)^2+\dotsb$ in the ring of formal power series
 
anyone?
 
What's your definition of $\gcd$ ?
 
No convergence issues because each coefficient has finitely many terms contributing to it
 
6:16 PM
If $d|a$ and $d|b$ then $d| ax+yb$ for all $x,y$
 
Akiva
 
let me check @Astyx
 
so we can write this as $-4.9t^2+2.86t$ ?
 
${}+4$, since you start four units up, I think
 
6:17 PM
true.
 
But that's the right idea
 
@Astyx they just define it as the greatest number that divides both $a$ en $b$
 
Then see what I said above
 
I don't know how you got your velocity but I assume it has to do with trig
 
yeah
the simple trig
 
6:18 PM
oh shit
thanks @Astyx
 
Glad to help
 
that is indeed the easier approach I was looking for :P
 
my computer is being really "laggy"
 
$\dfrac1{\gcd(a,b)}=\dfrac{xa+yb}{\gcd(a,b)}=x\dfrac a{\gcd(a,b)}+y\dfrac b{\gcd(a,b)}$ must be an integer @ShaVuklia
so $\gcd(a,b)$ must equal $1$
 
oh wowwww
 
6:20 PM
(Essentially the same as what Astyx wrote)
 
nice thanks @Akiva
 
nice.
 
hahaha yea
 
Do you know the converse?
 
the shoe company?
 
6:21 PM
$\gcd(a,b)=1\implies\exists x,y:xa+yb=1$
 
well I use Euclid's extended algorithm to show the converse
 
Oh
There's a really quick way
 
oh tell me
 
shoes lol
 
Look at the smallest number that you can write in the form $xa+yb$ as $x$ and $y$ vary
 
6:22 PM
I used to have a pair of converse.
 
Well, that has $0$ and negative numbers in it. Look at the smallest positive number of that form.
 
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
 
(I have never heard of this company)
 
@AkivaWeinberger I'm not sure I see why that would be the gcd then
 
Converse ? They're a famous brand of shoes here (in France)
 
6:23 PM
Oh, wait, hold on
(Where's "here")
 
France
 
Oh right duh
 
for astyx
Canada for me.
 
I'm ze silly frenchman
 
you don't recognise these?
 
6:25 PM
To be honest, I've never really paid attention to people's shoes
 
anyhow I'll just stick to Euclid's algorithm. but I'm glad the other way is easy
 
World famous sneakers
 
@ShaVuklia Yeah, I forget how it goes
but I think it ends up being the same as Euclid more or less
 
Euclid's algorithm is taught in number theory, I believe.
 
Do you know what an ideal is?
 
6:26 PM
not yet but will soon
 
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
 
I have to leave :P kinda time-pressured 'n stuff
 
Oh, OK, bye
 
Bye
 
6:27 PM
oh right, well I'll get to that soon! thanks
 
Hey Danu
 
You also have to be a group
So you'd have to change "adding" to "substracting"
 
Ideals are the normal subgroups of rings :P
 
Ideals are what you can quotient by.
Normal subgroups are what you can quotient by.
 
That would be a terrible textbook definition :p
 
6:31 PM
Indeed it would :P
 
But it's the motivation of the definition!
 
"A set is said to be an ideal if you can quotient by it."
 
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.
 
Algebra : an obscure approach
 
I'm not sure I agree. These are the kind of things you may want students to find out themselves.
 
6:34 PM
Defining quotienting before that might be tedious though
 
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
 
holy poop
defnitely need to restart my computer
 
6:47 PM
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?
 
No, it means it maps one nonzero vector to 0
 
@user193319 To see that this couldn't possibly be true in general, perform a change of basis that changes all basis vectors.
 
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
 
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
 
And $T(e_1-e_2) = 0$ but none of the basis vectors are sent to $0$
 
6:50 PM
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?
 
@ShaVuklia You could check the other group axioms... But yeah.
 
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
 
Technically, they are not.
You're taking a subset of $G$.
 
That's a good point
 
well yea we could maybe lose $e$
right?
 
6:52 PM
You can't claim that $G^*$ is a group until you show that $G^*$ is non-empty.
 
you need to check that things still work inside the subset
 
right
so I have to state that $ee=e$
 
"Consider $e \in G$. Then clearly some $x \in G$ exists such that $xe = ex = e$, namely $x = e$. Hence $e \in G^*$."
 
wow geez
don't be so.. what's the word
"pompeus" :P
alright I've written it down
thanks
 
This is pretty much the bare minimum, though :P
 
6:55 PM
hahaha true I guess :P
 
You also have to show it's stable under the operation
That is, if $a$ and $b$ have left and right inverses, so does $ab$
 
Yup
 
oh right
let me think
 
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
 
So you really have to check all group axioms, none are inherited from $G$ (except the existence of a neutral element)
 
6:58 PM
the associativity is inherited tho?
 
Oh yeah, true
 
Psh, the existence of the neutral is not the inherited one :P
 
lol true
that was the whole point in the beginning :P
okay cool I think I got it
 
@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.
 
Oh I have a cool problem
Find which integers $n$ verify the existence of integers $c_1, \dots, c_{n-1}$ such that $$\arctan(n) = \sum_{k=1}^{n-1}c_k \arctan(k)$$
 
7:01 PM
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
 
Why not both
 
Zee
Well, I am doing both, but I mean for $$$
 
That's not a decision we can take for you :p
 
And unless you pay yourself, it's not even a decision you might be able to make for yourself :/
 
Zee
7:16 PM
@SteamyRoot 100% true
Unless your Wittgenstein, in which case you give all your money away and become a gardner
 
SteamyRoot, the frustrated gardner
 
Haha :P
I'm getting paid plenty, though
 
7:42 PM
Oh wow
 
hello
 
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.
 
@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
 
Note that $n<\frac{3n+1}2$
 
exactly.
 
7:53 PM
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
 
So going from, say, $5$ to $8$ increases it
($5\mapsto16\mapsto8$)
 
@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.
 
Well, for Collatz, you also expect to "eventually" decrease the number.
 
yes but you can't prove it
and the "eventually" can take you a while
 
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
 
7:59 PM
mathematics isn't built on chances
 

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