Mathematics - 2021-07-31

love_sodam

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

robjohn

4:27 AM

Pretty quiet in here recently

Arief

5:43 AM

IMO solution that everyone can understand! simple math.

►

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

Srijan M.T

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 ?

John Rennie

Some one flagged the comment above? That seems a strange thing to do. Was it a mistake?

Ted Shifrin

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.

Srijan M.T

@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.

John Rennie

@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.

Srijan M.T

Ok. What else can I do then ? @JohnRennie

John Rennie

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.

Srijan M.T

@JohnRennie ok.

Thank you @TedShifrin @JohnRennie sir.

John Rennie

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 ...

Ted Shifrin

Just factor the quadratic I wrote. No completing square. No quadratic formula!

John Rennie

@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.

leslie townes

i don't either. ted and i are in the unfortunate position of agreeing.

Srijan M.T

@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 .

leslie townes

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.

Srijan M.T

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

John Rennie

@leslietownes I am a physicist. Everything looks like a nail to me :-)

Srijan M.T

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.

leslie townes

@JohnRennie there is an elegance in the simplicity of that approach.

John Rennie

@SrijanM.T Well I suppose you could write x-6 = x+6 - 12, but I don't see how that helps.

Koro

@leslietownes for 10 billionth time now

Srijan M.T

6:28 AM

@JohnRennie ok.

John Rennie

@leslietownes :-)

Srijan M.T

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,

John Rennie

@leslietownes Worse still, I am (was) an experimental physicist so when I say everything looks like a nail I mean that quite literally :-)

Srijan M.T

@JohnRennie :)

leslie townes

just don't beam any lasers my way. i do not fluoresce.

Srijan M.T

6:30 AM

@leslietownes :)

robjohn

But $x^2-5x+6=(x-2)(x-3)$, so how can we get that from $x(x-6)+1(x+6)$?

Srijan M.T

@robjohn x^2 - 6x+x+6

Then take brackets

robjohn

I understand how to expand it, but how can we get it simpler than that?

Srijan M.T

@robjohn complete the square.

robjohn

I am trying to see if the form $x(x-6)+1(x+6)$ helps at all

John Rennie

6:33 AM

@robjohn I was wondering if there was some cunning trick that was too obvious for me to see ...

robjohn

I don't see it

John Rennie

Just the examiners being mean then.

Srijan M.T

@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.

robjohn

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.

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).

and this will also help you save a lot of calculations!

Srijan M.T

6:42 AM

@Koro How did you do this ?

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 $p$. :-)

cool right?

Srijan M.T

Yes

@Koro Pretty interesting. How did you get to know about this ? Teacher or book or self learned

Koro

@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..

robjohn

9:33 AM

@Koro that is not right. That equation will have the same roots as the other.

user178758

9:43 AM

@JohnRennie it wouldn't be the first time they were being mean, sir.

John Rennie

@user178758 :-)

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

Koro

@robjohn ? I am solving second equation for $px$.

rain1

math.stackexchange.com/posts/4213488/revisions

I just removed references to rape in this post and someone instantly rolled it back

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).

SAJW

9:58 AM

@robjohn thanks for mentioning the extended euclidian algrorithm

robjohn

@Koro no, the second equation is just $p$ times the first. same roots.

rain1

@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"

shintuku

i would agree with rain1 that it is a very unfortunate question

Koro

@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).

rain1

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

robjohn

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.

Koro

Yes. we are solving for $px$ in the second equation and not for $x$

rain1

he just deleted that comment

user178758

lots of people have control issues

Koro

after solving for $px$, we are dividing by $p$ :)

rain1

@robjohn do you think my edit was good? removed references to rape

can we undo his rollback and lock it maybe?

shintuku

10:01 AM

it was also a very unfortunate sarcastic comment

user178758

let it go

robjohn

@rain1 I don't know where this is

rain1

math.stackexchange.com/posts/4213488/revisions

Koro

@robjohn Please note that I am not denying this.

rain1

here robjohn

user178758

10:02 AM

the mods will deal with it

2

Q: Probability that at least $52$ people are raped yearly in Greece

Disclaimer: 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...

robjohn

10:15 AM

@Koro the equation for $z=px$ is $z^2+qz+rp=0$ but the variable in the equation you give is $x$

rain1

> @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

robjohn

@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.

rain1

nobody says that the crime is being belittled?

robjohn

no one has said that anyone has been disturbed, either. I am trying to sort out why there is the rollback war.

rain1

There is no war, i did not do anything after he reverted my edit

robjohn

10:25 AM

The system reported a rollback war.

rain1

that's very strange

robjohn

Okay, the OP has accepted the change to remove the mention of the crime.

shintuku

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

Koro

@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.

robjohn

The equations $apx^2+aqx+ar=0$ and $px^2+qx+r=0$ have the same roots. You chose $a=p$

Koro

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?

robjohn

You have not written the equation for $z$

Koro

But I did say-solve for $px$.

@robjohn I have never objected to this professor Rob.

robjohn

solve for $px$? you mean solve for $x$ and multiply by $p$?

Koro

No. I mean in $(px)^2+2(px)+3=0$, we solve to get $px=$ something.

robjohn

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 $x$

Koro

10:44 AM

@robjohn Like I guessed, my symbol usage and paraphrasing of the statement!!

:-(

@robjohn yes so what?

robjohn

I am just sorting out what is being solved for. That seems to be what is confusing.

Koro

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

user178758

reviewing 1+1=2 never hurt anyone :P

Koro

@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 $b^2+3b+4=0$

robjohn

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.

Koro

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.

robjohn

$z^2+qz+rp=0$ is clear as to what is being solved. much better

Koro

:) I see so the phrase "solve for $px$ in..." caused all the confusion.

robjohn

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

Koro

@robjohn no. we want to multiply with $a$ in this case. Why $p$?

The idea is to knock off coefficient of square $x$.

robjohn

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.

Koro

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 $x$. :)

Samyak Marathe

Greetings

I have a question to ask

robjohn

@Koro Another question: how did you factor $x^2-157x+1050$? it does not seem evident.

Samyak Marathe

If anyone is free to help?

@robjohn Use Quadratic fomula

robjohn

@SamyakMarathe I know how to solve it. I was asking about a statement made earlier that seemed to preclude the quadratic equation.

Koro

@robjohn professor Rob, I created that example like that : $150 \times 7=1050$ and $150+7=157$

Samyak Marathe

11:16 AM

Oh

Would u mind if u answer this question.

-1

Q: Randomness of 3n+1 with other odds

I 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...

education random

robjohn

@Koro it is simple to factor in hindsight, I see.

Koro

:)

Samyak Marathe

Can u answer the question I asked

I m super curious to know

Koro

@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. :)

robjohn

@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).

Koro

11:22 AM

@robjohn I'll check those out, thanks professor Rob :)

robjohn

@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.

Koro

@robjohn that's perfect :)

robjohn

Yet another way is to set $c_n=2^{-n}a_n$, then the equation becomes $c_{n+1}=c_n+9\cdot2^{-n-1}$

Koro

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.

satan 29

@SamyakMarathe I see the veritasium video has started taking effect :)

Koro

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?

robjohn

@Koro it makes it easy to figure out how to solve it, it's just a sum.

Koro

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 $c$.

robjohn

@Koro Can you find the value to which it would converge?

Koro

@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 $l$.

robjohn

yes

Koro

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 $a_1$?

robjohn

$3x^2-4x+1=0$ so $x\in\left\{1,\frac13\right\}$

@Koro there must be a sequence of values for $a_1$

Koro

I observe that if $a_1=1$ then we get a constant sequence.

robjohn

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.

$\frac{\mathrm{d}}{\mathrm{d}x}\frac1{4-3x}=\frac3{(4-3x)^2}$

Koro

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 $1$.

robjohn

$1$ is an unstable equilibrium

$\frac13$ is stable

So there is a range around $\frac13$ where $a_1$ will lead to convergence

Koro

11:47 AM

I think I read those terms in differential equations but never in sequences

I think you meant this: en.wikipedia.org/wiki/Recurrence_relation#Stability

robjohn

anything less than $\frac{4-\sqrt3}3$ should work

Koro

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

robjohn

Don't guess the $n^\text{th}$ term, compute it.

Koro

In such an event, what should be done ?@professor Rob

@robjohn I don't think that's possible but I'll try this.

Samyak Marathe

@satan29 Yes Indeed, but i guess i have to make a program and try first part of the question by myself.

Semiclassical

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.)

user178758

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.

Semiclassical

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.

user178758

12:47 PM

Classic.

Semiclassical

@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

shintuku

is anyone familiar with spivak's appendix on riemann sums?

Semiclassical

However, I would strongly suspect that there are other maps of that kind for which no cycles have been discovered

shintuku

i'm wondering about a specific case that isn't mentioned in the proof on page 282

Semiclassical

and for those, they presumably have exactly the same status as the original Collatz conjecture

Koro

1:17 PM

@shintuku for Riemann integration?

or rearrangement theorem for series?

shintuku

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

Koro

@Shin: that's not possible. $a<a$ is not true for any real $a$

shintuku

my bad

shintuku

1:35 PM

so, actually, the above case is covered under case #2 (there are at most K-1 of them)

Koro

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 :-(

Charlie

4:21 PM

Are there non-trivial examples of objects for which every homomorphism is an isomorphism?

Balarka Sen

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 :)

Charlie

interesting ty, was just curious

Balarka Sen

you can of course always make up some category in which all arrows are isomorphisms

Charlie

That seems like cheating :P

Balarka Sen

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

Thorgott

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

porridgemathematics

@LeakyNun hey sorry for the random question, did you take differential topology this year?

Balarka Sen

@Thorgott how so

Thorgott

cause it's the elements from which the group looks the same

Balarka Sen

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)

Wolgwang

> 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?

Thorgott

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

Wolgwang

Why? I am unable to find a reason... :-/

Leaky Nun

@porridgemathematics no

porridgemathematics

oh never mind then

Thorgott

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

Wolgwang

Some high level thing, but I at least know it has a unique solution, and I believe you. Thanks :-)

Ted Shifrin

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.

user178758

Howdy professor @TedShifrin

Thorgott

6:26 PM

hi @Ted

geocalc33

6:37 PM

I need an August math project/problem to mull over. Any topic suggestions?

shintuku

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?

satan 29

@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

@shintuku oh it's just a less obvious triangle inequality

Semiclassical

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 $f(x)$)

copper.hat

@shintuku the triangle inequality surely?

shintuku

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

copper.hat

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???

shintuku

oh

you're right

I could just always pick $\delta_2$

whew

thanks

Addem

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"

Thorgott

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

Balarka Sen

thorgott, sent you something elsewhere

robjohn

@Koro I will post something, but I have a 5 hour drive ahead, so it will be a while.

Addem

@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?

Thorgott

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

Addem

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!

asthasr

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...

fido9dido

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

copper.hat

8:52 PM

@Addem it is fubini's theorem. for example web.math.ucsb.edu/~cmart07/fubini.pdf

Thorgott

@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 $B$

asthasr

@Thorgott Cool, thanks. :)

fido9dido

@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

Thorgott

indeed

fido9dido

Thanks alot

Addem

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.

copper.hat

@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 $B$.

Thorgott

@Addem it is a special case of the general measure-theoretic version of Fubini's theorem

Addem

@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).

Thorgott

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.

Addem

Ah I see. Right, all the ideas sort of swim together for sure.

fido9dido

9:46 PM

@copper.hat I don't understand this formula

copper.hat

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.

fido9dido

ah thanks

Shiranai

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

geocalc33

10:08 PM

if you assume the harmonic numbers are a discrete subset of an envelope function, f, can f be expressed elementarily?

copper.hat

10:32 PM

@Shiranai take the points $x^* \pm L e_k$ for suitably large $L$.

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