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1:43 AM
Is 'Tautness' in Algebraic topology an important concept?
In Spanier, it has a section named tautness and continuity but wikipedia doesnt have it
 
 
3 hours later…
4:27 AM
Pretty quiet in here recently
 
 
1 hour later…
5:43 AM
Afternoon all. Would like to ask for a feedback. Do you think the solution here is clear enough..? And engaging? Or something to be improved. Thanks
 
If we have x(x-6) + 1(x+6)=0. Then , how do I make x-6 and x+6 equal in sign. What I think is that if I put a minus sign on both sides. Then , we get -x(x-6)-1(x+6)=0 . Then , for -x^2. It doesn’t matter if x is +ve or -ve. Due to ^2 , it will be + ve always in end. So , x(x+6) - 1(x+6)=0. Is this right way ?
 
Some one flagged the comment above? That seems a strange thing to do. Was it a mistake?
 
5:59 AM
@SrijanM.T No, this is incorrect. I have no idea why you're trying to do this. Just expand and simplify and solve the quadratic: $x^2-6x+x+6 = x^2-5x+6=0$. Now solve in the usual way.
 
@TedShifrin There are two ways I know. One is completing the square which I know. 2nd is by x(x-6) + 1(x+6) = 0. If x+6 we’re both equal , then I would have just written x+6 = 0 or x+1=0. That’s what I wanted to do here. But I have one side x-6 and other side x+6 which is causing the problem.
 
@SrijanM.T if x(x-6) + 1(x+6)=0 then x(x-6) = - 1(x+6)
But this isn't going to help you solve the equation.
 
Ok. What else can I do then ? @JohnRennie
 
Like Ted says, multiply out the brackets and group terms together to turn it into a standard quadratic of the form ax² + bx + c = 0
Then you can solve that by completing the square or by just using the quadratic formula.
 
@JohnRennie ok.
Thank you @TedShifrin @JohnRennie sir.
 
6:11 AM
:-)
@SrijanM.T the roots are x = 2 and x = 3, as a few seconds work with the quadratic equation will show. I must admit I wonder if I'm missing something and the equation was written in the form x(x-6) + 1(x+6)=0 for a reason. I don't see it though ...
 
Just factor the quadratic I wrote. No completing square. No quadratic formula!
 
@TedShifrin Srijan is preparing for the JEE exam in India, and the JEE questions tend to be written in such a way as to give hints to the student, so it seems likely that writing the equation as x(x-6) + 1(x+6)=0 was supposed to give a hint. However I don't see it.
 
i don't either. ted and i are in the unfortunate position of agreeing.
 
@JohnRennie Ok. So , in many Q for similar like x(x-6)+1(x+6)=0. The sing for x-6 is changed to x+6 to make the Q easier to solve by some way. I am not getting is how to approach that way .
 
although between you and me, there was a point in my life where because i had the quadratic formula hammer in my hand, every quadratic equation looked like a nail.
 
6:25 AM
I know completing the square method. I just wanted to learn every way so that I expand my knowledge onto knowing everything about this topic
 
@leslietownes I am a physicist. Everything looks like a nail to me :-)
 
So , is it that there is no way to make x-6 change to x+6 or there is . That’s the entire Q now.
 
@JohnRennie there is an elegance in the simplicity of that approach.
 
@SrijanM.T Well I suppose you could write x-6 = x+6 - 12, but I don't see how that helps.
 
@leslietownes for 10 billionth time now
 
6:28 AM
@JohnRennie ok.
 
@leslietownes :-)
 
Well everyone. Thank you so much for your help. I will search about this way on internet as well and find if there is a way or not only. Then , Ill share it with you,
 
@leslietownes Worse still, I am (was) an experimental physicist so when I say everything looks like a nail I mean that quite literally :-)
 
@JohnRennie :)
 
just don't beam any lasers my way. i do not fluoresce.
 
6:30 AM
@leslietownes :)
 
But $x^2-5x+6=(x-2)(x-3)$, so how can we get that from $x(x-6)+1(x+6)$?
 
@robjohn x^2 - 6x+x+6
Then take brackets
 
I understand how to expand it, but how can we get it simpler than that?
 
@robjohn complete the square.
 
I am trying to see if the form $x(x-6)+1(x+6)$ helps at all
 
6:33 AM
@robjohn I was wondering if there was some cunning trick that was too obvious for me to see ...
 
I don't see it
 
Just the examiners being mean then.
 
@robjohn The point is. I wanted to learn as many methods as possible. Especially , in this equation. The concept which I wanted to know was if we can change x-6 to x+6 somehow. So , the Q is not solving the quadratic equation and finding the roots only. It’s that I wanted to know if this concept is present or not.
 
Quadratic Formula and completing the square are the two assured ways of tackling that
if you can recognize the factorization, that's great, but there are a lot of polynomials to remember factorizations for.
 
@SrijanM.T: You should think more on this also and generalize: Suppose you want to solve $7x^2-157x+150=0$ then I claim that its roots will be 1/7 times the roots of $x^2-157x+1050=0$ (note that this is very easy to factorize).
and this will also help you save a lot of calculations!
 
6:42 AM
@Koro How did you do this ?
 
So we have a quadratic equation: $px^2+qx+r=0,p\ne 0$ and we are multiplying throughout by $p$ to get $(px)^2+q(px)+rp=0$ whence we solve for $px$ and divide the answer by $p$. :-)
cool right?
 
Yes
@Koro Pretty interesting. How did you get to know about this ? Teacher or book or self learned
 
@SrijanM.T Few months ago, I saw this type of question on this website only. That's when I observed it. So when I saw the discussion here on quadratic equations, I recalled that question..
 
 
3 hours later…
9:33 AM
@Koro that is not right. That equation will have the same roots as the other.
 
9:43 AM
@JohnRennie it wouldn't be the first time they were being mean, sir.
 
@user178758 :-)
 
:-)
perhaps it's not just being mean; but, mean squared :P
Oct 23 '11 at 13:51, by t.b.
@robjohn: you're mean squared :p
 
@robjohn ? I am solving second equation for $px$.
 
I just removed references to rape in this post and someone instantly rolled it back
 
Second equation: $y^2+qy+rp=0$, where $y=px$ (x is (are) the root(s) of the old equation), can be solved to get $y=y_1,y_2$ whence $px=y_1,y_2$ and we get x (the roots of the old equation).
 
9:58 AM
@robjohn thanks for mentioning the extended euclidian algrorithm
 
@Koro no, the second equation is just $p$ times the first. same roots.
 
@robjohn can you give some input on this?
the guy is also being sarcastic in comments about my edit "People are no longer raped. They are selected! – Rodrigo de Azevedo 1 min ago"
 
i would agree with rain1 that it is a very unfortunate question
 
@robjohn: It seems that I didn't paraphrase my sentence correctly and that there is a confusion. I request you to have a look at this please:
1 min ago, by Koro
Second equation: $y^2+qy+rp=0$, where $y=px$ (x is (are) the root(s) of the old equation), can be solved to get $y=y_1,y_2$ whence $px=y_1,y_2$ and we get x (the roots of the old equation).
 
the instant rollback and then acting like that is really rude
it seems like this person has some kind of "issue" and wants to take control of the question
 
10:00 AM
3 hours ago, by Koro
So we have a quadratic equation: $px^2+qx+r=0,p\ne 0$ and we are multiplying throughout by $p$ to get $(px)^2+q(px)+rp=0$ whence we solve for $px$ and divide the answer by $p$. :-)
@rain1 where is this? your comment does not link to it.
 
Yes. we are solving for $px$ in the second equation and not for $x$
 
he just deleted that comment
 
lots of people have control issues
 
after solving for $px$, we are dividing by $p$ :)
 
@robjohn do you think my edit was good? removed references to rape
can we undo his rollback and lock it maybe?
 
10:01 AM
it was also a very unfortunate sarcastic comment
 
let it go
 
@rain1 I don't know where this is
 
@robjohn Please note that I am not denying this.
 
here robjohn
 
10:02 AM
the mods will deal with it
2
Q: Probability that at least $52$ people are raped yearly in Greece

Γιάννης ΚαρλατήραςDisclaimer: I undestand this is a horrific subject to discuss. In Greece, rape has a prevalence of $1.1$ per $100,000$ people. Greece's population is $10.8$ million. To calculate the probability of at least $52$ yearly occurences, one would have to use the cumulative distribution function for $X...

 
10:15 AM
@Koro the equation for $z=px$ is $z^2+qz+rp=0$ but the variable in the equation you give is $x$
 
> @shintuku Boohoo. – Rodrigo de Azevedo 1 min ago
I appreciated your comment shintuku!
thanks for contributing
now he has accused me of vandalism for changing the question from being about 'rape' to being about a generic random variable
 
@user178758 This is a crime statistic; however, it could be asked without reference to the crime. The mention of the crime is causing some people to be disturbed. Other people seem to think that the crime is being belittled by removing its mention.
 
nobody says that the crime is being belittled?
 
no one has said that anyone has been disturbed, either. I am trying to sort out why there is the rollback war.
 
There is no war, i did not do anything after he reverted my edit
 
10:25 AM
The system reported a rollback war.
 
that's very strange
 
Okay, the OP has accepted the change to remove the mention of the crime.
 
What's a decent source for the Lebesgue integral? (for a total beginner to it)
is baby Rudin's last chapter alright?
I have no measure theory tho
 
@robjohn yes so what? Both the equations have the same root (x). What I said earlier was that root of the 1st equation are $x=a,b$ then root of the second equation (considering $px$ as variable) are $px=pa, pb$ and dividing these by $p$ gives $x$ (roots of the first equation). I don't see what's causing confusion.
 
The equations $apx^2+aqx+ar=0$ and $px^2+qx+r=0$ have the same roots. You chose $a=p$
 
10:37 AM
@robjohn we solve this for z and get $z= -q\pm \sqrt {q^2-4pr}\implies px=-q\pm \sqrt {q^2-4pr}\implies x= \frac{-q\pm \sqrt {q^2-4pr}}{p}$. That's all.
@robjohn yes professor Rob, I agree. what do you want to say?
 
You have not written the equation for $z$
 
But I did say-solve for $px$.
@robjohn I have never objected to this professor Rob.
 
solve for $px$? you mean solve for $x$ and multiply by $p$?
 
No. I mean in $(px)^2+2(px)+3=0$, we solve to get $px=$ something.
 
4 hours ago, by Koro
@SrijanM.T: You should think more on this also and generalize: Suppose you want to solve $7x^2-157x+150=0$ then I claim that its roots will be 1/7 times the roots of $x^2-157x+1050=0$ (note that this is very easy to factorize).
that was a good statement
the following statements were confusing
@Koro that would be $z^2+2z+3=0$
solve that for $z$ and divide by $p$ to get $x$
 
10:44 AM
@robjohn Like I guessed, my symbol usage and paraphrasing of the statement!!
:-(
@robjohn yes so what?
 
I am just sorting out what is being solved for. That seems to be what is confusing.
 
professor Rob, when you make a statement I have to review even my 1+1=2 :) so I am typing a response to your "solve for z..." to clarify the confusion
 
reviewing 1+1=2 never hurt anyone :P
 
@robjohn so we get :$z= px=-1\pm i\sqrt 2$ and $x$ obtained from it satisfies the equation in $px$ (that is, $(px)^2+2(px)+3=0$)
By "solve for $px$ in $(px)^2+3(px)+4=0$", I mean get $px =a,b$ such that $a^2+3a+4=0$ and $b^2+3b+4=0$
 
here things are better, you are separating the usage of $p$, it is not also a coefficient in the original equation. You have also explicitly stated that $z=px$ and so the variable being solved for is clearer.
 
10:58 AM
Let me write a proof for the statement I made so that everything is clear. We have a quadratic equation: $px^2+qx+r=0, p\ne 0$, we multiply both sides by $p$ to get $(px)^2+q(px)+rp=0$ and we put $z=px$ to get $z^2+qz+rp=0$ which gives $z=px=-q\pm \sqrt {q^2-4rp}\implies x= \frac{-q\pm \sqrt {q^2-4rp}}p$(these roots are the same as those of the old equation).
@robjohn :)
since so many trailing messages are there, let me also state why multiplication by $p$ came up in the first place.
 
$z^2+qz+rp=0$ is clear as to what is being solved. much better
 
:) I see so the phrase "solve for $px$ in..." caused all the confusion.
 
Here is the way that might be clearer: start with $ax^2+bx+c=0$. Multiply by $p^2$ to get $ax^2p^2+bxp^2+cp^2=0$ now substitute $z=px$ to get $az^2+bpz+cp^2=0$
but your more recent description is good
 
@robjohn no. we want to multiply with $a$ in this case. Why $p$?
The idea is to knock off coefficient of square $x$.
 
why does it need to be the lead coefficient?
@Koro oh, was that the point?
I thought you were just trying to show how to adjust roots by adjusting the coefficients.
 
11:08 AM
@robjohn The point was to simply calculation: if we have $7x^2-179x+100=0$ then solving it by usual formula may be complicated so if we multiply both sides by $7$ and solve for $7x=y$ in $y^2-179y+700=0$ we get $(y-175)(y-4)=0\implies y=7x =175,4$.
Something like that :)
now we just have to divide by $7$ to recover $x$. :)
 
Greetings
I have a question to ask
 
@Koro Another question: how did you factor $x^2-157x+1050$? it does not seem evident.
 
If anyone is free to help?
@robjohn Use Quadratic fomula
 
@SamyakMarathe I know how to solve it. I was asking about a statement made earlier that seemed to preclude the quadratic equation.
 
@robjohn professor Rob, I created that example like that : $150 \times 7=1050$ and $150+7=157$
 
11:16 AM
Oh
Would u mind if u answer this question.
-1
Q: Randomness of 3n+1 with other odds

Samyak MaratheI recently found out that the randomness followed by f(n)=3n+1 and g(n)=n/2 based on some rules mentioned below. 1: If the output f(n) is even we insert it to g(n). 2:If the output is odd, we insert it to f(n) again. Now here is my question, does this work for any odd in place of 1, for example f...

 
@Koro it is simple to factor in hindsight, I see.
 
:)
 
Can u answer the question I asked
I m super curious to know
 
@robjohn professor Rob, recently I learned to handle non-homogeneous recursive sequences a bit differently
For example: $a_{n+1}=2a_n+9$ then we can convert this to homogeneous recursive sequence: $a_{n+1}-2a_n=a_{n+2}-2a_{n+1}$ and then solve this as usual. :)
 
@Koro I have done that in several answers.
@SamyakMarathe it is unclear as to what you are asking in that question.
probably the reason for the downvotes (I have not downvoted).
 
11:22 AM
@robjohn I'll check those out, thanks professor Rob :)
 
@Koro another way is to use $b_n=a_n+9$. then the equation is $b_{n+1}=2b_n$. This does not increase the degree of the equation.
 
@robjohn that's perfect :)
 
Yet another way is to set $c_n=2^{-n}a_n$, then the equation becomes $c_{n+1}=c_n+9\cdot2^{-n-1}$
 
but it's not making the original equation homogeneous :(
professor Rob, I want to ask one question that's been bothering me
I am given $a_{n+1}=\frac 1{4-3a_n}$ for $n\ge 1$ and I want to find $a_1$ such that $a_n$ converges.
 
@SamyakMarathe I see the veritasium video has started taking effect :)
 
11:35 AM
When I look at the above question, I would want to find $n$th term by guessing the pattern of first few terms
Is that the only way?
 
@Koro it makes it easy to figure out how to solve it, it's just a sum.
 
Another way that comes to mind is: if $f(x)=\frac 1{4-3x}$ then the given recursive sequence is $a_{n+1}=f(a_n)$ and then using an exercise problem from Rudin's derivative chapter if $|f'(x)|\le c\lt 1$ for some constant $c$ then $(a_n)$ is a Cauchy sequence. But here ofcourse the difficulty is to find such bound $c$.
 
@Koro Can you find the value to which it would converge?
 
@robjohn yes right!! It gives us a G.P. That's great :)
@robjohn yes, to that end suppose that the given sequence converges to $l$ then we must have $4l-3l^2=1$ and then we solve for $l$.
 
yes
 
11:41 AM
But finding $a_1$ seems tricky to me because what if I don't see pattern from first few terms then what is the way to get $a_1$?
 
$3x^2-4x+1=0$ so $x\in\left\{1,\frac13\right\}$
@Koro there must be a sequence of values for $a_1$
 
I observe that if $a_1=1$ then we get a constant sequence.
 
the way to find the values would be to follow Rudin
also for $a_1=\frac13$
one of those is stable and one is not.
$\frac{\mathrm{d}}{\mathrm{d}x}\frac1{4-3x}=\frac3{(4-3x)^2}$
 
but then what would be $c$? Why I ask for $c$ is because $f$ does not seem to be bounded in the appropriate interval and hence supremum of range of f will not exist
if it I did, I would have said $c=sup$ of range of $f$ on that interval.
also, $c$ can't be $1$.
 
$1$ is an unstable equilibrium
$\frac13$ is stable
So there is a range around $\frac13$ where $a_1$ will lead to convergence
 
11:47 AM
I think I read those terms in differential equations but never in sequences
 
anything less than $\frac{4-\sqrt3}3$ should work
 
12:08 PM
anything except $a_1=\frac{3^{n+1}-1}{3^{n+1}-3}$ works as per the answer.
And for $a_1=1$ the sequence converges to 1 but for other allowed values of $a_1$, the sequence converges to 1/3
if we write first few terms we get this pattern: $a_n=\frac{(3^{n-1}-1)-(3^{n-1}-3)a_1}{3^n-1-(3^n-3)a_1}$ and we get the above answer so it seems that the answer is correct but the problem is what if we are not able to guess this nth term
 
Don't guess the $n^\text{th}$ term, compute it.
 
In such an event, what should be done ?@professor Rob
@robjohn I don't think that's possible but I'll try this.
 
@satan29 Yes Indeed, but i guess i have to make a program and try first part of the question by myself.
 
12:31 PM
Given that you don't say in your question what "randomness" you're referring to, there's not really any way you're going to get a useful answer.
If you're thinking of a specific portion of the Veritasium video on Collatz, that's a start to providing context. But you'll need to actually say what you mean.
(Having watched that video as well, I would guess it's the appearance of Benford's law. But there's no way anyone reading your question can read your mind and know that's what you mean.)
 
Veritasium has a tendency to generate a lot of heat without light.
in The h Bar, yesterday, by Nihar Karve
I think Veritasium's one-way speed of light video has resulted in 100+ questions on this site
Not to mention his vid on learning styles.
 
Well, the appearance of Benford's law at least is genuine math. The Veratisium video includes comments from Alex Kontorovich, who has published papers on Collatz including one with proofs regarding Benford's law showing up in it (arxiv.org/abs/math/0412003).
i haven't watched the one-way speed of light video, though. seeing it in my feed did motivate me to learn about the Fizeau experiment, though.
 
12:47 PM
Classic.
 
@SamyakMarathe If you're simply asking whether there's other $x\mapsto ax+b$ maps for which their version of the Collatz conjecture is known, there are certainly some examples where the conjecture is definitely false
Take $x\mapsto 3x-1$. Then 5 -> 14 -> 7 -> 20 -> 10 ->5, i.e., a cycle
so iterating starting from 5 will never reach 1
 
is anyone familiar with spivak's appendix on riemann sums?
 
However, I would strongly suspect that there are other maps of that kind for which no cycles have been discovered
 
i'm wondering about a specific case that isn't mentioned in the proof on page 282
 
and for those, they presumably have exactly the same status as the original Collatz conjecture
 
1:17 PM
@shintuku for Riemann integration?
or rearrangement theorem for series?
 
identity between integral and riemann sum
I mean, the limit of the riemann sum
I'm just wondering about the following case:
$t_{i-2} < u_{j-3} < t_{i-1} < u_{j-2} < u_{j-1} < t_i=u_j$
oh I'm beginning to realize this one is covered too it seems
 
@Shin: that's not possible. $a<a$ is not true for any real $a$
 
my bad
 
1:35 PM
so, actually, the above case is covered under case #2 (there are at most K-1 of them)
 
 
1 hour later…
2:59 PM
@robjohn I tried to compute $a_n$ in terms of $a_1$ but it's getting inordinately lengthy so I think writing the first few terms and recognizing the pattern to write $a_n$ (general term) and then verifying it by induction, is the only way I have right now to solve the question :-(
 
 
1 hour later…
4:21 PM
Are there non-trivial examples of objects for which every homomorphism is an isomorphism?
 
every homomorphism is too much to expect. but if you have a field F, every nonzero homomorphism F -> F is an isomorphism
eh, i guess not an isomorphism to the range, but an isomorphism onto image :)
 
interesting ty, was just curious
 
you can of course always make up some category in which all arrows are isomorphisms
 
That seems like cheating :P
 
well, maybe. categories in which all arrows are isomorphisms are called groupoids, there's plenty of examples of those in nature.
a group G is a groupoid with one object, namely G with all arrows G -> G being "multiplication by g"
this seems made up but actually a useful way of thinking of a group
 
4:57 PM
in this language, the natural automorphisms of the identity functor are canonically identified with the center of the group
this explains why the center is called center
 
@LeakyNun hey sorry for the random question, did you take differential topology this year?
 
@Thorgott how so
 
cause it's the elements from which the group looks the same
 
i agree now but i didnt understand that from your Aut(id) interpretation at all.
seems its better to understand what ghg^-1 does (translation of POV) to a group and then realize g-g^-1 is trivial for g in Z(G)
 
> If $f(x)$ is a polynomial of degree 5 with leading coefficient one and $f(1) =1^2,f(2)=2^2, f(3)=3^2,f(4)=4^2,f(5)=5^2$, then find $f(6)$ .
This question has a unique solution?
 
5:08 PM
Aut(id) are something like universal symmetries of the category
I think of them as homogeneities of some sort
it's effectively the same thing as saying that conjugation is a base change
just categorified
@Wolgwang yes
 
Why? I am unable to find a reason... :-/
 
@porridgemathematics no
 
oh never mind then
 
5:30 PM
cause a monic polynomial of degree 5 is uniquely determined by its value at 5 different points
more precisely, $P\mapsto(P(1),...,P(5))$ is a linear map from monic polynomials of degree 5 to $\mathbb{R}^5$ and its represented by a Vandermonde matrix relative to the standard bases, hence is invertible
with a bit of work, you can write down the inverse explicitly - this is known as Lagrange interpolation
 
Some high level thing, but I at least know it has a unique solution, and I believe you. Thanks :-)
 
6:07 PM
@Wolgwang Where is this question coming from? It's either a system of linear equations or it's an application of Lagrange interpolation polynomials.
Howdy, a @Balarka and @Thor.
 
Howdy professor @TedShifrin
 
6:26 PM
hi @Ted
 
6:37 PM
I need an August math project/problem to mull over. Any topic suggestions?
 
7:01 PM
I'm convinced $\frac{\epsilon}{2} > |a-b|$ and $\frac{\epsilon}{2} > |b-c|$ implies $\epsilon > |a-c|$. isn't there a way to show this algebraically?
 
@Wolgwang the polynomial behaves as x^2 for x=1,2,3,4,5...so I would guess its something like P(x)=x^2 + (x-1)(x-2)(x-3)(x-4)(x-5)
 
@shintuku oh it's just a less obvious triangle inequality
 
7:28 PM
Another way to go about it is to approach it via finite differences
If f(x) is a degree-n polynomial, then f_1(x)=f(x+1)-f(x) is one degree lower
and now satisfies f_1(1) =f(2)-f(1) = 3, f_1(2) = 5, .., f_1(5)=f(6)-25
so now the problem is reduced by one degree. this can be repeated again, i.e., f_2(x)=f_1(x+1)-f_1(x)
repeat that five times and you've got f_5(1) in terms of f(6). but f_5(x) is degree zero
and then it's basically done
(the nice thing about this approach is that you don't actually need to know $f(x)$)
 
@shintuku the triangle inequality surely?
 
yeah, it was the triangle inequality, but I actually now have a more complicated problem
$\forall \epsilon, \exists \delta_1, \delta_2:$

$0<|x-x_0|<\delta_1 \implies |f(x) - L| < \epsilon/2$

$0<|x-x_0|<\delta_2 <\delta_1 \implies |f(x) - G| < \epsilon/2$

$\therefore\epsilon/2 + \epsilon/2=\epsilon>|L-f(x)|+|f(x)-G|$
I can't use it here, right? to say $|L-G|<\epsilon$ since the two $f(x)$ on the last line aren't actually the same and have different bounds on their domain
 
7:43 PM
that is a peculiar statement. you can always pick $\delta_2$ so that the second line is vacuous. are you sure that is what you meant???
 
oh
you're right
I could just always pick $\delta_2$
whew
thanks
 
8:03 PM
Trying to find a proof of the following: If {a_n,k} is a doubly-indexed sequence of non-negative real numbers, and if {b_m} is an enumeration of {a_n,k}, then (sum_n sum_k a_n,k) = sum_m b_m.
Seems like it should be true but can't find a fully satisfying proof.
It's not quite the same as a "rearrangement"
 
8:20 PM
you can use that $\sum_{n=0}^{\infty}a_n=\sup_{I\subseteq\mathbb{N},I\text{ finite}}(\sum_{i\in I}a_i)$ for any non-negative sequence $(a_n)_n$
then it comes down to comparing a joint supremum with an iterated supremum, which is straightforward
 
thorgott, sent you something elsewhere
 
@Koro I will post something, but I have a 5 hour drive ahead, so it will be a while.
 
@Thorgott: Interesting. Can the conditions for this result be broadened to something like the conditions for swapping the order of summation? Or is non-negativity the broadest interesting condition that you know of?
Also, know of any book that discusses this identification with the sup over finite sums?
 
non-negativity is the best possible condition in order to write down the series as a sup. the result you want is also true under the assumption of absolute convergence, but that requires a more subtle argument
I don't know a reference off-hand, but this should no doubt be standard. if not in real analysis, it will be in measure theory textbooks
 
8:37 PM
Hm, I've searched tons of real analysis and measure theory texts. I did find basically this in one of them but not in any of the others. I'll study the one that I did find though.
Thank you!
 
Is there a term for the concept of treating different prime factors as the differentiators between "spaces?" e.g. the numbers that are multiples of 1 are the natural numbers, the numbers that are multiples of 2 are another, the numbers that are multiples of 3 are another; with intersections between the "spaces" at 6, and so on
For context, I was thinking idly about the concept of 3, 6, 9, ... as 1, 2, 3 in the "space" of 3; 5, 10, 15 as 1, 2, 3 in the "space" of 5; and so on...
 
regarding dot product, I am a bit confused of it's actual meaning. I know it to be |A|*|B|*cos Theta, I read somewhere that we use it to get the angle between two vectors but I think they have to be unit vectors, so i used to think of it to be the angle between A and B. recently I read that A dot B is the projection of A into B, does this means that the dot product return the length of the projection?
another thing is I read that the scalar projection is |A|*cos(Theta),so is the it different from the projection of the dot product? it looks like the projection of A into B from the dot product en.wikipedia.org/wiki/Scalar_projection
 
8:52 PM
@Addem it is fubini's theorem. for example web.math.ucsb.edu/~cmart07/fubini.pdf
 
@asthasr these are known as "ideals" (see en.wikipedia.org/wiki/Ideal_(ring_theory))
@fido9dido it's the length of the projection of $A$ onto the line spanned by $B$ times the length of $B$
 
@Thorgott Cool, thanks. :)
 
@Thorgott Thanks, so to be sure, can we say that the scalar projection is the length of the projection of $A$ onto the line spanned by $B$ and the dot product is the scalar projection times the length of $B$, right? assuming the definition I said for the scalar projection is true
 
indeed
 
Thanks alot
 
9:10 PM
@copp
@copper.hat Problem is that Fubini's theorem talks about changing the order of summation but not in the sense of enumerating the terms in an arbitrary enumeration.
In effect, you don't get a single-indexed series where every term is represented uniquely by an index.
 
@fido9dido if $B \neq 0$ you can always write $A = A \cdot ({B \over \|B\|}) {B \over \|B\|} + (A- A \cdot ({B \over \|B\|}) {B \over \|B\|})$. it is easy to check that the term in parentheses is perpendicular to $B$.
 
@Addem it is a special case of the general measure-theoretic version of Fubini's theorem
 
@Thorgott True, but as far as I know, the measure-theoretic version relies on sub-additivity of Lebesgue outer-measure, the proof of which relies on taking sums in any order (at least as proved in Royden and Axler).
 
this can all be non-circularly, but it's not like there's any need to invoke Fubini to prove this much more elementary result. it's really just a remark that it does follow from there.
 
Ah I see. Right, all the ideas sort of swim together for sure.
 
9:46 PM
@copper.hat I don't understand this formula
 
it shows that $A$ can be written as a component along the direction of $B$ and a component orthogonal to $B$. The scalar $(A \cdot {B \over \|B\|})$ gives the length along a unit vector in the $B$ direction.
 
ah thanks
 
Hello, I'm reading Misha's proof here https://math.stackexchange.com/questions/2961783/is-a-convex-function-always-continuous
And I don't know how to prove that the convex hull of the hypercube w circumscribed aroud S contains S
pictorially it seems right, at least in three dimensions, but I don't know how to prove it
 
10:08 PM
if you assume the harmonic numbers are a discrete subset of an envelope function, f, can f be expressed elementarily?
 
10:32 PM
@Shiranai take the points $x^* \pm L e_k$ for suitably large $L$.
 

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