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12:03 AM
in quantum physics, each particle can be thought of as a wave function, which is a complex-valued function
so I think it's fair to say that complex numbers do exist in nature

Oh interesting, didn't know that @thor I'll have to learn QM before I can agree or disagree :P
I guess I just got annoyed that $\sqrt{-1}$ doesn't have a real number value
Like, why would we even consider complex numbers as a thing or spend time diddling with something like that :P
but mathematicians have no chill with their curiosity

the imaginary number $i$ appears in the Schrödinger equation
so I think it's pretty uncontested to say that complex numbers are indispensable for doing quantum physics

Right I just meant I never learned it so I can't say anything about it.

engineers also care a lot about complex numbers
something to do with Fourier transforms

12:21 AM
"Complex numbers" is just a bad term that we're stuck with imo. One could've chosen a name like "rotational numbers" or whatever. Anything that gives a hint about the structure

Yeah that would make more sense actually. Didn't hamilton develop quaternions or whatever right after anyway?
complex numbers on their own in like a college algebra class or whenever people learn them seem very unintuitive

12:53 AM
@Thorgott those regularity results on bounded subsets of $\mathbb R^n$ aren't trivial but are incredibly standard, I assume the move to manifolds isn't too hard
@user69608 if you want $A = |v+w|^2 + C$ and you want to know what $C$ is, its of course $C=A - |v+w|^2$? unless you assume $a,b,c$ are orthogonal the answer isn't clean. Since you understand the previous line, then you don't need me to open $|v+w|^2$ for you?

2 hours later…
2:35 AM
Can someone help me understand why an arithmetic series formula looks like the following?

So, we have the series {3, 5, 7...}

The put it into the form:

a(n) = a(n-1) + 2

But I'm not seeing why it shouldn't be written:

2(n-1) + a

Doing it the first way, and suppose you take the third term, you'd get the following:

3(3-1) + 2

3(2) + 2

6 + 2

= 8. Which is incorrect.

Doing it the second way, one gets:

2(3-1) + 3

2(2) + 3

4 + 3

= 7.

Which is correct.

@Beliod when you. write $2(n-1)+a$, what does $a$ mean?
I don't think you understand that $a(n)$ is not "$a$ times $n$"
If you have the sequence $\{ 3,5,7,9,11,\dots\}$, then $a(1)=3,a(2) = 5, a(3) = 7, a(4) = 9,$ and so on
$a(2) = a(1) + 2$, for example

3:31 AM
@CalvinKhor that "something" is $a^2 +b^2+c^2-((a+b+c)^2/4))$ i am interested in this value.

$|a|^2 + |b|^2 + |c|^2 - \frac14|a+b+c|^2 = \frac34 (|a|^2+|b|^2+|c|^2)-\frac12 (a\cdot b + a\cdot c +b\cdot c)$

@CalvinKhor yeah now what is summation a.b

I still don't understand. Summation a.b means? $\text{write something in \LaTeX}$?

dot product

$a\cdot b$ means dot product. There's no summation?

3:38 AM
i mean value of a.b +b.c+c.a

@user69608, I have no idea what you want, because that is my final answer. You asked me about something in the middle of a computation, and didn't provide any context. I can't read minds and I don't know what bigger problem you're trying to solve

3 hours later…
6:53 AM
Hello!! Does someone of you have an idea about my question:
0

Let $1\leq m,n\in \mathbb{N}$ and let $\mathbb{K}$ be a field. For $a\in M_m(\mathbb{K})$ we consider the map $\mu_a$ that is defined by $$\mu_a: \mathbb{K}^{m\times n}\rightarrow \mathbb{K}^{m\times n}, \ c\mapsto ac$$ We have that $trace(\mu_a)=n\cdot trace(a)$ and $\det (\mu_a)=\det (a)^n$. Ho...

?

2 hours later…
9:12 AM
Hello everyone, I was wondering, do you know of some extension of Minkowski inequality to the case of negative exponents?

9:32 AM
A sequence of real numbers converges to a number $s$ relative to the usual topology iff it is eventually in each neighborhood of $s$.

Don't get how it is eventually in *each* neighborhood of $s$.
$S_n$ is eventually in set $U$ if there is an $m \Bbb N$ s.t $n \geq m \Rightarrow \, S_n \in U$.

its eventually in $U$ for each nbd $U$, but there is no $n$ such that its in each (i.e. every) neighbourhood

@CalvinKhor It is not eventually in every nbd $U$ at the same time?

yes, as a rough first approximation, think about the sequence 1/n and all the balls around 0
for each $\epsilon$, $1/n$ is eventually in $B(0,\epsilon)$
but that doesnt imply that its also in $B(0,\epsilon/100000)$

Got it. Thanks

you're welcome :)

10:11 AM
The polar form of a complex number has magnitude and phase. Can I say the magnitude and phase are independent to each other? What is the condition for that?

define "independent"

Let $x=re^{j\theta}$. Can I say pdf $f(r,\theta)=f(r)f(\theta)$?

what's $f$

$f$ is the probability distribution function.

of which distribution?

10:22 AM
I am trying to find the integral of multiplication of two Gaussian Q function $Q(x)$. The conditional pdf is $y(y|r,\theta)=Q(rcos(\theta))*Q(rsin(\theta))$
I am trying to find the marginal pdf of $Z(y)$ where, $z(y) = \int\limits_0^\infinity {\int\limits_0^{2\pi } {Q(rcos(\theta ))*Q(rsin(\theta ))f(r,\theta )} } drd\theta$i
I was thinking can I write $f(r,\theta ) = f(r)f(\theta )$. If I can do that what is the condition?

11:15 AM
Can't theorem 3 be proved with just the isomorphism?
I mean since it needs to be bijection and that means it needs to be surjection
then both has to have finite order n?
I got bunch of question today

They're just being careful because the fact that $f(g)^n = e_H$ does not mean that the order of $f(g)$ is necessarily $n$---it just divides $n$.
But since an isomorphism gives an inverse which is also an isomorphism, you can apply the argument both ways and get equality.

@Fargle So I am right right?

I think I'm unclear on what you're asking.

I mean proving the theorem 3 by just bijection
which is faster

As opposed to...?

11:19 AM
and efficient
I mean can we proof the theorem with just saying since it is isomorphism then it means it needs to be bijective which needs to be subjective so □

No, because you're talking about a single element.
You're saying if $g \in G$ has order $n$, and $f$ is iso, then $f(g)$ has order $n$.
Surjectivity is not enough to require this.

But for this proof isn't it required to be cyclic?

No.
$G$ and $H$ are arbitrary isomorphic groups.
$g$ is an arbitrary finite-order element of $G$.

because I think only cyclic becomes neutral when they have their order as exponents

@NoseBleed What is the definition of order?

11:22 AM
A cyclic group is one where every element is a power of some specific element.

@AlessandroCodenotti size

size of what

@AlessandroCodenotti Size of group I guess or it's elements

The order of a group is its size, but what is the order of an element?

I am confused why elements have orders? It should be 1 for each

11:24 AM
Your book should have defined order of an element.
Go back and look

you mean for example elements of $D_3$ ?
I forgot how to spell it

@NoseBleed The theorem you're asking about talks about the order of an element $g\in G$, so your book should have defined what that means at some point before theorem 3

dihelral...
Oh now I remember
It is order of a element treated as generator
oh now I see I understand the proof
one more help in this book I got another too but talk it for later
in the first picture after the proof it talks about different between dihedral group and rotations of hexagon
how it theorem 2 is applied
so that they know they are not isomorphic
and there is alternative way which I kinda don't understand
I think they used theorem 2 (iii)
ok I understand it
I just need to understand the one begin with alternativey one

What's the alternative way, out of curiosity?
Oh---I see, it was in your last screenshot.

As I send first picture it says:"Alternatively, bal bla bla"
@Fargle it's first

11:35 AM
The punchline is that the rotations of the hexagon contain an element of order 6.

but $d_3$ also has 6 orders
but $d_3$ also has 6 as order

If there was an isomorphism with $D_3$, then that would imply $D_3$ has an element of order $6$---by Theorem 3.
Not just order 6 on the group---order 6 on an element.

you need to differentiate between the order of the group and the order of an element in the group

oh now I get it

fun fact: these two groups are the only groups of order 6
2

11:39 AM
Good morning from Italy everybody into chat. Why my question is yet closed?

@Thorgott r u sure? I think we can create one

0

Supposing to have these commons integrals with $x\in\Bbb R$, for example I have these: $$\int\limits_{-\infty}^{+\infty}\frac{dx}{(x^2+1)^2}=\frac12\left(\arctan x+\frac{x}{1+x^2}\right)\tag 1$$ $$\int\limits_{0}^{+\infty}\frac{x^2}{x^4+1}dx=\sqrt{2}\pi/4 \tag 2$$ $$\int\limits_{0}^{2\pi}\frac{1}... I am sure the only groups of order 6 up to isomorphism, I should specify ok now we move to fibonacci number Thank you very much for your suggestions. Best regards. 11:40 AM @Thorgott I hope to find another one that is to say, any group of order 6 is either isomorphic to D_3 or the group of rotations of a hexagon you can try to find another one it's oftentimes instructive to fail @Thorgott but they are not isomorphic lol Thank you very much for reopen my question. THank you with my heart. @Sebastiano you have 4 reopen votes already, you can probably just wait for another. But also, I don't know why it should be reopened. I don't understand the question @CalvinKhor What is not clear? 11:42 AM Lol well it has now been opened. But I still don't understand I'm not saying they are, though "Whether an integral cannot be solved by classical methods when I will have to use to complex analysis and what needs to be done to solve them?" this sentence, I can't understand it I'm saying that there are two groups of order 6, D_3 and the groups of rotations of a hexagon, which I shall call C_6, that these two are not isomorphic, but that any other group of order 6 is isomorphic to one and exactly one of D_3 and C_6 @CalvinKhor I never dealt with complex analysis at university. I did some private tutorials in 2003 but I don't remember anything. The question is: when do I have to use complex analysis if I can't solve it by substitution or by parts, or by artifice? @Thorgott paulsible let's move to geometry of numbers 11:45 AM @CalvinKhor Is there a criterion, a clue that makes me think that certain integrals can also be solved through complex analysis and how to solve them? @Sebastiano OK, I understand now, and I don't think you can get a simple answer besides learn more and examples and practice more examples @Thorgott The statement I saw was: \Omega^k = \Delta\Omega^k \oplus \mathcal{H}^k. Why is \Delta\Omega^k = \delta\Omega^{k+1} \oplus d\Omega^{k-1}? I just came across an interesting topic, Relation between Kirchhoff's circuital laws and Matrix tree theorem, can someone provide any refrence or insights why is path from 1 to n is f_n? I cannot see the pattern what is the reason behind it which is not mentioned @CalvinKhor Can you help me, please? Would you kindly edit my question? I don't know how many times I've changed it. There's three wholemeal, two impropres and one definite. 11:46 AM I tried drawing the rabbit breeding chart still can't get it 0 I'm not a professional mathematician, just an undergraduate student. I was reading Introduction to Graph Theory by West, I came over the topic which discuses the methods to find the spanning trees in a general graph. Kirchhoff's Matrix tree theorem was there. Out of curiosity, I thought, could th... Oh okay. It may not be. But \Omega^k = \Delta\Omega^k \oplus \mathcal{H}^k implies \Omega^k = \delta\Omega^{k+1} \oplus d\Omega^{k-1} \oplus \mathcal{H}^k. Its just set theoretic It there any way we can see the pattern? @CalvinKhor I hope I haven't disturbed anyone. Thank you so much for your comments. @feynhat It is, but it's not entirely trivial. In fact, you can go like \Delta\Omega^k=d\delta\Omega^k\oplus\delta d\Omega^k=d\Omega^{k-1}\oplus\delta\Omega^{k+1}. It boils down to a handful of adjointness computations. I'm in a seminar rn, but I can try telling you the proof later if you want. 11:50 AM @Sebastiano In my opinion, there can't be a good answer, but I would remove most of the information, especially about the integrals you can already solve, and add this which you said to me: "Is there a criterion, a clue that makes me think that certain integrals can also be solved through complex analysis and how to solve them?" also you don't need to say Riemann integrals, "solve the integrals" is fine I am doubtful about the second equality. is it called fibonacci zigzag path counting or there is any resource online to see this You guys seem busy so I will question it later 12:16 PM @CalvinKhor Done!! Thank you very muchhhhhhhhhhhhhh. @Sebastiano you're welcome. Hope I'm wrong and someone gives a great answer 12:47 PM no worries I found the pattern by matching with points going through (n-1) and (n-2) then I found that it is f_n and is the same one as fibonacci number I should had done it before I appreciate your helps guys! Thanks for involving in discussion! 1:28 PM quick question again :p why does \int_a^{x}(f'(a)+\int_a^{x_1}f''(x_2)dx_2)dx_1=f'(a)(x-a)+\int_a^{x}\int_a^{x_1}f''(x_2)dx_2dx_1? It says it is integration by parts but I think this is mean value theorem used here but instead of c it is a trying to review Taylor theorem derivation huh? Its just linearity of \int. feynhat is right What does it means [f(a)]_a^b? it means ur overthinking it, theres no fancy calculation, just \int 3f+g dx = 3\int f dx+\int g dx 1:38 PM I know it but I actually mean why wait lemme type how to integrate \frac{1}{(a+bcosx)^2} where a greater than b why does \int_a^{x}f'(a)dx_1=f'(a)(x-a)? I think I knew it in past but now I forgot why does \int_a^x 5 dx = 5(x-a)? Yeah, the ticket here is that a is constant @CalvinKhor ? 1:43 PM @CalvinKhor hmm... \int a dx=ax+c @user69608 dont know I don't know hehe @CalvinKhor u must be joking :) @CalvinKhor Why. No riddles please ;_; its not a riddle idk what else to say if you can't integrate a constant then review that before taylor's theorem 1:45 PM any hint? you r right may be need a serious review of integration lol no I don't need review my first step is correct then I perform |_a^b lmao wth did I ask... really embarrassing so it means that f'(@) is constant in order to be treated like this Yep---f'(a) doesn't vary as you vary x_1. It might in principle vary if you varied a, but integration only cares about whether the inside looks constant to the variable you're integrating on. oh cr@p u r right can anyone help in that integral? thank u guys ok enough for today 2:26 PM Hello everyone! I got this question hi @user69608 which one? Find all real solutions (x,y) such that$$sin^4 ⁡x=y^4+x^2 y^2−4y^2+4 \\cos^4 ⁡x=x^4+x^2y^2−4x^2+1$$Here is my attempt:$$
\cos^4 x - \sin^4 x = x^4 - y^4 + 4y^2 -4x^2 - 3 \\
(cos^2 x - \sin^2 x )(\cos^2 + \sin^2 x) = x^4 - y^4 + 4y^2 -4x^2 - 3 \\
\cos 2x = x^4 - y^4 + 4y^2 -4x^2 - 3
$$thats what i did too. now you might wanna try: -1<= cos(2x) <= 1$$
-1 \leq \cos 2x \leq 1 \\
-1 \leq x^4 -y^4 + 4y^2 - 4x^2 -3 \leq 1 \\
2 \leq x^4 - 4x^2 +4 -(y^4 -4y^2 +4) \leq 4 \\
2 \leq (x^2 -2)^2 - (y^2 -2)^2 \leq 4
$$@Knight I want to ask for clarifications. 2:31 PM You can Getting that last inequality it is possible to find as many real x and y as we wish. @Knight If your equation for \sin^4 x, LHS is a function of x only and RHS is a function of x,y i.e. 2 variables, did you notice that? 5 mins ago, by Knight Find all real solutions (x,y) such that$$sin^4 ⁡x=y^4+x^2 y^2−4y^2+4 \\cos^4 ⁡x=x^4+x^2y^2−4x^2+1$$Yes Abhas why I wouldn't notice that Do you know what that means? no um, why is that an issue? 2:33 PM it means nothing substantial It means, take all x to one side and y to other, the thing you get on both sides, corresponds to a constant. Now you can easily find all such (x,y) pairs Moving on we have @Knight i also noticed something else Solution for such system of equation we have$$
S = \{ (x,y) | 2 \leq (x^2 -2)^2 - (y^2 -2)^2 \leq 4 \}
$$@satan29 YES? @Knight No, that's silly thing. You say \cos 2x = x^4 - y^4 + 4y^2 - 4x^2 - 3 Actually solve it like this:$$ cos 2x - x^4 + 4x^2 = -y^4 + 4y^2 - 3 = C 

2:37 PM
@Knight The first equation gives sin^4(x)= y^2( x^2 + y^2-4) +4
and the next one gives cos^4(x)= x^2(x^2 + y^2-4) + 1

That expression gives bounds for $y$, solve it, then solve possible values of $x$

after Adding, On the RHS, let x^2 + y^2=t

2:39 PM

so you get sin^4(x) + cos^4(x) = t(t-4)+5
$sin^4(x)+cos^4(x)=t(t-4)+5$

okay
keep going

@satan29 integrate$\frac {1}{(a+bcosx)^2}$ where a greater than b

This might make things simpler. Since The LHS can be maximized/minimized easily
@Knight we can then

@satan29 Brother we got to find all the real solutions

2:43 PM
write the LHS as $(sin^2(x) + cos^2(x))^2 - 2sin^2(x)cos^2(x)$
$=1-(sin^2(2x)/2)$

okay keep going

now, we can write $sin^(2x)= (1-cos(4x))/2$

Brother can you take a moment to read the question?

which simplifies to $1/4 (3+cos(4x))$

okay

2:47 PM
which is bounded: it has a maximum value of 1 and a minimum value of 1/2
which means the RHS must be between 1/2 and 1

did you check with calculus?

uh cos(4x) can be 1 at max
and -1 at min
and the value of that expression depends only on cos(4x)
The RHS was $t^2-4t+5$

okay

when anyone gets any hint to the integral please ping me.

so have $1/2 <= t^2-4t+5<=1$
@user69608 i have a nice idea, ill ping in a moment
@Knight
subtracting one from both sides, we get
-1/2 <= t^2-4t+4<=0
-1/2<= (t-2)^2 <=0

2:51 PM
yes

which means the only solution is t=2

yes, but we have $t=x^2 +y^2 = 2$ and hence infinite solutions

thats not the only equation though,brother

(And I really just accepted your calculations without a check)
@satan29 Thats why I said read the question, we are asked to find all solutions

remember, we had $1/4(3+cos(4x))= t^2-4t+5$
when we put t=2

2:54 PM

cos(4x) has to be 1
i.e $4x=0, 2\pi$
i.e x=0 or pi/2

@satan29 Do you see this step?

yes

You got a square less than 0, that means no real solutions

kar di bejjati

2:57 PM
less than or EQUAL to zero @knight

OK
meri galti thi

@Knight and thats exactly why t=2 is the only solution

okay

so 0, sqrt(2) is a solution for x,y
but we need to check once
and yes, it indeed is a solution

Thanks

3:00 PM
correkt!

and we need to do it for x=pi/2 and y=sqrt(2-pi^2/4) also
but its unlikely that thats a solution.
and so i believe we have found all the solutions!
(x,y)=(0,sqrt(2))

okay

@Knight :)
@user69608
I have got an idea
we need to integrate $1/ (a+bcos(x))^2$
equivalently stated, we need to find a function whose derivative is $1/(a+bcosx)^2$
the square in the denominator reminds me of the quotient rule.
so i guess its reasonable to assume that the required function is of the form $f(x)/(a+bcos(x))$

Just find $p(x)$ for which $p'(x)(a+b\cos x) - (a + b\cos x)'p(x) = 1$

when we differentiate this, we get f'(x)(a+bcos(x)) - (-bsin(x))(f(x)) / (a+bcos(x))^2
and we want the numerator to be 1

3:06 PM
But, that's impossible (I guess, not sure), because that's unintelligible class of function that's why limits of integration is given... (Not sure, I'm damn weak with integration)

and its not hard to see that f(x) = sin(x) does the job.

@satan29 What?
cool observation!
@satan29 great dude :)

no wait, does it?

@satan29 don't confuse me!

@user69608 were you sure it was one in the numerator?
@abhas_RewCie lol sorry

3:10 PM
lol
@user69608 Are you sure denominator is squared? otherwise, the integration is simple and straightforward!
that square hurts

bruh
Yeah @user69608 i dont think the numerator is 1

@satan29 why?
WRA solved it with one...

if its some f(sin(x),cos(x)) , then the comparing thing works out quite nicely i suppose
in the numerator

WRA solved it with one...
I still don't understand why the limit of the integration was 1? It doesn't appear good even after putting the limits $a$ and $b$... or does it?
Solve this integration $\int \frac {\sin x} x dx$
and this $\int e^{-x^2} dx$
bye

@abhas_RewCie limit?
idk man
maybe @robjohn has an insight...

3:38 PM
@feynhat Ok, let's do this. First of all, for any two $k$-forms, $\omega_1,\omega_2$, we have $\langle d\delta\omega_1,\delta d\omega_2\rangle=\langle\delta\omega_1,\delta\delta d\omega_2\rangle=0$, so $d\delta\Omega^k$ and $\delta d\Omega^k$ are orthogonal. Next assume $\omega\in d\delta\Omega^k\cap\delta d\Omega^k$, so that there are $\omega_1,\omega_2\in\Omega^k$ with $\omega=d\delta\omega_1=\delta d\omega_2$. On one hand, $d\omega=dd\delta\omega_1=0$ and $\delta\omega=\delta\delta d\omega_2=0$, so $\omega\in\mathcal{H}^k$. On the other hand, $\langle\omega,\gamma\rangle=\langle d\delta\ Onto the second part: For$\omega_1\in\Omega^{k-1},\omega_2\in\Omega^{k+1}$, we have$\langle d\omega_1,\delta\omega_2\rangle=\langle\omega_1,\delta\delta\omega_2\rangle=0$, so$d\Omega^{k-1}$and$\delta\Omega^{k+1}$are orthogonal. Assume$\omega\in d\Omega^{k-1}\cap\delta\Omega^{k+1}$, so that there are$\omega_1\in\Omega^{k-1},\omega_2\in\Omega^{k+1}$, such that$\omega=d\omega_1=\delta\omega_2$. We have$d\omega=dd\omega_1=0$and$\delta\omega=\delta\delta\omega_2=0$, so$\omega\in\mathcal{H}^k$. We also have$\langle\omega,\gamma\rangle=\langle d\omega_1,\gamma\rangle=\langle\omega_1,
you could probably structure this better, but it's not like this is sophisticated argument

@Thorgott What. The first part is the definition of $\Delta$, right?

no
the reverse inclusion is non-obvious
you need to show that something of the form $d\delta\omega_1+\delta d\omega_2$ is actually of the form $d\delta\omega+\delta d\omega$

@satan29 c'mon man wolfram alfa most of the times gives complex wierd answers .
@satan29 this question is from "resonance high level problems"

3:53 PM
@user69608 see what i tried to do earlier.
It has to be along the lines of this only....and i say this because I have previously solved a similar question before, (very similar) using the same procedure.

I mean, in general you won't have $(f+g)V=fV+gV$ for linear operators on vector spaces

Oh ofcourse lol. take $f = I, g= -I$.

@satan29 should i send the answer?

So yeah. Its not that obvious.

ye
it looks tautological but isn't
still, the actual work is only to show that $\left(\mathcal{H}^k\right)^{\perp}=\Delta\Omega^k$ and the rest is more of a linear algebra exercise

3:59 PM
I mean, for me the statement is what I wrote. So I wouldn't need that LA exercise.

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