Mathematics - 2019-03-25 (page 2 of 3)

@PolineSandra Is that an interval?

Or coordinates

an open interval

I would recommend drawing a graph

It's gonna involve $\sin^{-1}(\sin(x)-\varepsilon)$ and $\sin^{-1}(\sin(x)+\varepsilon)$

which can't be simplified as far as I know

I think that we have to determine if $\sin(x)-\varepsilon,\sin(x)+\varepsilon[\subset [-1,1]$ no?

1:15 PM

Seems like that gets tricky when sin(x) is nearly 1 (or minus 1)

1:40 PM

Graphs will be useful here, as one is basically mapping an infinitesimal interval of values of sin x back to the corresponding interval(s) in the x axis

Hi, Could someone take a look on this: math.stackexchange.com/questions/3160866/…

Hi

hi

Hello

could please someone take a look on the following: math.stackexchange.com/questions/3160866/…

?

user21820

← 1 message moved from Logic

1:45 PM

what is <x,y>, inner product?

Ordered pair

$ARB \implies A\overline{\sim}M=B$

$BRA \implies B\overline{\sim}J=A$

$ARBRA \implies (A \overline{\sim} M) \overline{\sim} J$

$ARBRC \implies (A \overline{\sim} T) \overline{\sim} S$

$(A \overline{\sim} M) \overline{\sim} J = A$

Thus $(A \overline{\sim} M)RA$

you are used associative in your proof.

nope I did not, noticed I have not dropped the brackets

and I have not finished writing

sorry

2:01 PM

So the next step is to prove that $J$ has to be some subset of $A$ (since we already have the set $A \overline{\sim} M$ thus we have all the element of $A$)

In fact, $M$ also have to be a subset of $A$ else you end up with elements not in $A$

I am not sure how to prove that rigorously, but $M,J$ basically behave like zero elements in this particular case

That is, my suspicion is that you need the result of $ARA$ first before you can prove transitivity and reflexivity

$ARA \implies A \overline{\sim} X = A$

Thus $X$ has to be some subset of $\Bbb{N}$ such that it only shift elements of $A$ (assuming $A$ is infinite)

Hmm...

$ARB \implies A \overline{\sim} M = B$
$BRC \implies B \overline{\sim} P = C$

$(ARB)RC \implies (A \overline{\sim} M) \overline{\sim} P = C$

$AR(BRC) \implies A \overline{\sim} Q = (B \overline{\sim} P = C)$

$\implies A \overline{\sim} Q = (B \overline{\sim} P = (A \overline{\sim} M) \overline{\sim} P)$

Now since $=$ is transitive and associative (this is from the definition in logic), thus we have:

$\implies A \overline{\sim} Q = (A \overline{\sim} M) \overline{\sim} P$

$\implies A \overline{\sim} Q = C$

$\implies ARC$

Thus $AR(BRC) = ARC$

The above also showed that $(ARB)RC=AR(BRC)$ and hence proving associativity for $\overline{\sim}$

Thus the associativity of $\overline{\sim}$ is guaranteed by the transitivity, associativity of $=$ and the existence of some subset of natural numbers defined by the relation $R$

With transitivity proved, ARBRA=ARA should be straightforward

2:24 PM

Thanks a lot for your help. I am going write it on a paper to understand it right now.

Note, you have to write those in set notation. I tend to have a bad habit of skipping the let $a \in A$ step because I don't want to write many words

@Secret hello, how to do the graphe please?

I cannot think of any way to formalise it, I just drew a sine curve , coloured some small sections of it, and check its projection on the x axis (because that is what it means by $\sin ^{-1} (I)$ where $I$ is an interval)

My guess is you might want to calculate the derivative of $\sin x$ and then project that onto the x axis

That is, let $Y = \sin^{-1}(\sin x - \epsilon , \sin x + \epsilon)$ thus you want to find intervals $Y$ such that $\sin (Y) = (\sin x - \epsilon , \sin x + \epsilon)$

Thus this basically means you need to project the tangent line at $\sin x$ onto the x axis

what I must draw ?

sin(x) ? or arcsin(y)?

2:39 PM

For each $(\sin x - \epsilon , \sin x + \epsilon)$ (the sloped line segment), there is a corresponding $Y$ (the horizontal line segment)

Thus what that question is asking is to find the representation of the tangent line of $\sin x$ when it is projected to the x axis

So one possible way to do is is to find the equation of the tangent line, project that onto the x axis, and then use a limiting procedure to get rid of the $\epsilon$

ok nvm

that does not make sense

0.0

Actually, since sin x and arcsin are monotonic and continuous, that means mapping from $Y$ to $(\sin x - \epsilon, \sin x + \epsilon)$ is also bijective

but only in [-1,1]

this means $f(Y)$ where $f$ is monotonic should only depend on the endpoints

sorry on $[-\pi/2,\pi/2]$

2:48 PM

Hi, secret, can you explain from where come variable Q?
I think there`s missing line(s) before it...

Thus $\sin^{-1} (\sin x - \epsilon, \sin x + \epsilon) = (\sin^{-1}(\sin x - \epsilon), \sin^{-1}(\sin x + \epsilon))$ for $x \in [-\frac{\pi}{2},\frac{\pi}{2}]$. I don't know of any further simplification

@JohnD $AR(BRC)$ means the set $A$ is related by $\overline{\sim}$ to a set corresponding to $BRC$, thus there exists some $Q \in \mathscr{P}(\Bbb{N})$ such that $A \overline{\sim} Q = G$ where $G$ is the set resulted from $BRC$

4:16 PM

22 secs ago, by Akiva Weinberger

16 secs ago, by Akiva Weinberger

10 secs ago, by Akiva Weinberger

Recursion

rolls $\pi^2$ eyes

BREAK

Oh high

Ah, programming humor.

@TedShifrin hey

4:17 PM

I'm leaning towards not correcting the spelling mistake

Saw this earlier: math.columbia.edu/~woit/wordpress

hi @Leaky

@TedShifrin would you have any idea how to find $x_0, \cdots, x_n \in [-1, 1]$ that minimizes $\displaystyle \sup_{x \in [-1, 1]} \prod_{j=0}^n |x-x_j|$

Nope.

ok thanks

4:20 PM

I still think that substituting x=cos(t) is an idea

(And similarly for the subscripted x’s)

What about looking at $\prod (x-x_j)^2$?

the answer does contain a lot of cosines

That was another suggestion I made, heh

Also

That product looks like the discriminant?

It’s also therefore related to the Vandermonde determinant

Oh, derp

No. Discriminant would involve all pairs of roots

Ignore meeee

Isn't it clear that we can do the question with $f^2$ instead of $f$? And then take $\log$?

Yeah. It’s just not clear to what extent that’s tractable

Also there’s the restriction to x-values between -1 and 1, which is a bit tedious

@LeakyNun something I’m noticing: Suppose that the the polynomial $\prod_{j=0}^n (x-x_j)^2$ is, up to an irrelevant positive factor, the square of the (n+1)-th Chebyshev poly of the first kind

Then it seems to be the case that the max value possible for that square is 1

(well. Up to said positive constant)

4:55 PM

I've noticed that the (monic) polynomial of degree $n$ that minimizes $\int_{-1}^1f^2dx$ necessarily has all its roots in $(-1,1)$

and I'm fairly certain that that is a constant times the Chebyshev polynomials, though I'd want someone to check that

If they're orthogonal under the $\int_{-1}^1fg~dx$ inner product then they are

And you end up with the supremum being $1/2^{n-1}$ or something similar

The question is, how do we turn that into something about $\sup|f|$

One thing to note: $dT_n/dx=U_{n-1}(x)$ where Tn, Un are Chebyshev polys of the first and second kind

Maybe consider the inner product $\sum_{k=0}^nf(k\pi/n)g(k\pi/n)$ or something

which is where $T_n$ has its maximums and minimums I think

Another useful point: if n>0, then $2T_{n}(x/2)$ is monic

@Semiclassical No longer has its roots in $[-1,1]$ though

You'd want $2^{1-n}T_n(x)$ I think

Hrm, yeah

Note that every f in this problem has degree at least 1

Which is suggestive to me somehow

I sorta think one should be looking at $f(x)-2^{-n}T_{n+1}(x)$

(f is degree n+1)

5:06 PM

What is $\sum_{k=0}^n\cos(k\pi)\cos(\frac{n-1}nk\pi)$

If it's zero then that would be amazing

Well, it’s an alternating sum

Not convinced it’s zero tho

Well it's the real part of $\sum_{k=0}^ne^{k\pi}e^{\frac{n-1}nk\pi}$

Or, more usefully,

the real part of $\sum_{k=0}^ne^{k\pi}e^{\frac{1-n}nk\pi}$

which is $\sum_{k=0}^ne^{k\pi/n}$

One think one can say is that $f(x)-2^{-n}T_{n+1}(x)$ has degree n, one lower than $f(x)$

Wait

That's zero isn't it

It has to be

@AkivaWeinberger need i in there somewhere

5:10 PM

Replace $e$ with $e^i$ everywhere

Sorry

But the real part of that is zero 'cause the sum is symmetric about the imaginary axis

which means

Oh but that's not quite what I want I think

:) darn complex multiplication

Lol

$\sup$

find the flattest monic polynomial of degree n+1 on [-1,1] with all roots in [-1,1]. idk.

Oh good that works

@Semiclassical OK so here's what I have now

Let $S$ be the set of local maxima of $|T_n|$, including $-1$ and $1$

Then, if $m<n$, then $\sum_ST_nT_m=0$ if $n-m$ is odd and $1$ if it's even

5:15 PM

My line of thinking right now is $$\sup_x |f(x)|\leq \sup_x |f(x)-2^{-n}T_n(x)|+\sup_x |2^{-n}T_{n+1}(x)|$$

In the case where it's $0$

That means $T_n$ and $T_m$ sometimes have the same sign and sometimes have the opposite sign

which means $\sup|T_n+cT_m|$ will always be greater than $\sup|T_n|$ for any $c$, positive or negative

In the case where it's 1 I dunno

but this is an inner product

OH I have an idea

Count $-1$ and $1$ as half

so it's $\sum_{S\setminus\{-1,1\}}+\frac12\sum_{\{-1,1\}}$

where again $S$ is the set of local maxima of $|T_n|$ on $[-1,1]$

Conjecture: under that, $\langle T_n,T_m\rangle=0$ for all $m<n$

Note that Chebyshev’s can only lie between -1 and 1, so the second supremum is $2^{-n}$

I should write that as $\langle T_n,T_m\rangle_n$ to emphasize that the inner product depends on $n$

@Semiclassical Reciprocal of that

Fixed

Main thing is that it doesn’t depend on f(x)

$2^{1-n}$ except for $n=0$

5:21 PM

No?

Coefficient of $T_n$ is $2^{n-1}$ for $n>0$

Sure, but I’m using $T_{n+1}$

Since f has degree n+1

Oh whoops sorry

Erk

I see a typo above

I had n+1 in my last supremum but not my first

Anyways. What i’d like to argue is that this bound on the supremum will be as small as possible when the supremum of the difference vanishes

In any case I think I've solved it

We want to show that $\sup|T_n+cf|>\sup|T_n|$ for $\deg f<n$, yeah?

And I'm fairly certain that $\langle T_n,f\rangle_n=0$

5:27 PM

That’s do it yeah

Well

Not when f=0. :P

So it should be $\geq$ with equality only if $f=0$

Or just write $f\ne0$

So in any case the only way for that to be zero is if, (a), $f$ is zero at all local maxima of $|T_n|$ on $[-1,1]$, or (b) $f$ is positive at some of them and negative at others

In the former case $f$ has to be zero

In the latter case we get our inequality

I don't mean positive at some and negative at others. I mean it has the same sign as $T$ and some and opposite sign at others

@Semiclassical This is a bit weird though

because if I did this right, $2^{1-n}T_n$ has the lowest $\sup_{[-1,1]}|f|$ of all monic polynomials of degree $n$, even if they don't have roots in $[-1,1]$

@Semiclassical Oh also

I think I have a much much simpler proof

We don't need any inner product business at all

What makes $T_n$ special?

The fact that all its local maxima in $[-1,1]$ have value $1$ and all its local minima have value $-1$

And that it has $n$ roots

What’s confusing me right now is that my argument seems to give $$\sup |f(x)|\leq 2^{-n}+\sup_x |f(x)-2^{-n}T_{n+1}(x)|$$

Which is true but also seems arbitrary

Beside from the fact that the second supremum is (the absolute value of) a degree n polynomial

OK but Mike it's that property (of having all its maxima and minima at the same value, modulo sign) that makes the $\sup|T_n+cf|>\sup|T_n|$, $-1<\deg f<n$ *iff $f$ has the opposite sign of $T_n$ at each local extremum thing true

hey semiclassical

Oh hey Antonio!

Long time no see

5:41 PM

:)

yeah I "died" as Erdos would say

(Edited last comment)

We’re talking about Chebyshev polynomials, which as a species of orthogonal polys seems like the kind of thing that’s summon you :P

And because of the intermediate value theorem, that would imply $f$ has at least $n$ roots

Well, my point is more that I don’t have anything like that in my argument yet :P

@AntonioVargas a permanent death or merely temporary?

@Semiclassical permanent. I'm trying to do stats now, outside of academia.

tbh I wish I had learned more stats earlier. Bayesian stuff is actually awesome.

6:27 PM

Logic is a deductive system where each rule has a semantic meaning

Therefore, illogic is everything that is not logic

Types of illogic:

1. Mental shortcuts and common sense

2. Emotional guided reasoning (ha only some basic deduction rules, with quite unpredictable outcomes

3. Intuition based reasoning: Often operates on the level of similarities and congruences

4. Spiritual reasoning: Involves rules that cannot be written down

5. Dynamic inconsistency: Deduction rules changes unpredictably as it goes

6. Total randomness: From any step to another step, there is no framework or rules that can describe them

6:56 PM

So, a ring with the property that every $a$ has a unique $b$ such that $aba=a$. This property seems to imply that $ab=1$, but how to prove that there must be a multiplicative identity there in the first place?

So, proving that $abk=kab=k$ for all $k\in R$ seems untenable. Perhaps there is another approach I should be considering.

You can ignore the additive structure entirely @Rithaniel. This is actually known, but is far from trivial.

Spoiler: mse q'n

7:14 PM

you want to say "for every nonzero $a$", or $R$ will be boring

Ah, right, I always forget to safeguard against lines that end up collapsing the object into a trivial one.

Also, the proofs on this question you linked are intense, Karl.

@MatheinBoulomenos I was just going to prove it is a group under multiplication then observe the ring must be trivial. :P

@Rithaniel The ideas are natural, but getting them to work is a little painful. I remember filling up at least a page on that.

The $aba=a\implies bab=b$ is an excellent starting point, at least, and the proof on that part is so intuitive that I feel dumb for not having pieced it together myself. (I'm blaming that one on a lack of sleep)

@KarlKronenfeld did you systematically study semigroup theory?

@MatheinBoulomenos did you read the discussion above?

3 hours ago, by Leaky Nun

@TedShifrin would you have any idea how to find $x_0, \cdots, x_n \in [-1, 1]$ that minimizes $\displaystyle \sup_{x \in [-1, 1]} \prod_{j=0}^n |x-x_j|$

7:25 PM

I read it but I don't know anything about optimization

ok thanks

@AntonioVargas gotcha. I'm sorta half-dead right now, insofar as I'm still doing research but staying in academia just seems like a death spiral

@Semiclassical could you conclude the discussion above?

not really, I'm missing the last piece that @AkivaWeinberger put together

@MatheinBoulomenos No.

7:28 PM

[awakens from my slumber]

yo

@MatheinBoulomenos I'm not very sure how Stone-Weierstrass shows that the Fourier series sums to itself

@LeakyNun Oh yeah the key is that $T_n$ has all its maxima have the same value

and all its minima are the negative of that value

After that all you needs is IVT and the fact that something with $n$ roots has degree at least $n$

something something interlacing

In fact, $T_n$ (and constant multiples of it) is the only polynomial of degree $n$ with that property

*local maxima and minima

in $[-1,1]$

7:30 PM

yeah, that's neat

(which means $-1$ and $1$ count as local maxima/minima themselves even though they don't have zero derivative)

That uniqueness follows from the same proof.

@LeakyNun if $(f_i)_{i \in I}$ is an orthonormal basis in a Hilber space $V$, then for any $x \in V$, $x=\sum_{i \in I} \langle x,f_i\rangle f_i$

sure, but why is it a basis? in particular, why does it span?

@LeakyNun Suppose that $\sup|T_n+cf|>\sup|T_n|$ for all $c\ne0$. What condition should we have on $f$?

$c$ positive or negative

the span of $e^{2\pi i n z}$ is a subalgebra of $C(S^1, \Bbb C)$ that separtes points, and hence it is dense by Stone-Weierstraß

7:34 PM

Remember that all of $|T_n|$'s local maxima have the same value

what does it mean to separate points?

and $C(S^1, \Bbb C)$ is dense in $L^2(S^1)$

@MatheinBoulomenos yeah that's the content I think

you can't escape from analysis :P

@LeakyNun For all pairs of points, there's a function in the algebra with a different value at each point

oh ok

7:35 PM

Saw this problem in the March issue of the AMM

"Suppose that the centroid of a triangle with semiperimeter $s$ and inradius $r$ lies on its incircle. Prove that $s\geq 3\sqrt{6}r$ and determine conditions for equality."

now, my geometry skills aren't enough to actually solve that myself and I'm not that interested in trying

Which one's the centroid again

Angle bisectors?

median

average value of the vertices (in Cartesian coordinates)

@LeakyNun the proof I've given to you is only almost correct

Oh

7:38 PM

I was cheating

Barycenter

Center of gravity

yeah

Center of mass

The center to end all centers

@MatheinBoulomenos how were you cheating?

I wanna see the center of antigravity, personally

7:38 PM

But I did play around with it in Mathematica enough that I think I know the condition for equality

So the area is $sr$ if I'm not too mistaken

yes

By the way, the fact that the angle bisectors intersect is such a cool fact,

because it basically follows directly from the transitive property of equality

@AkivaWeinberger what condition?

@LeakyNun let $A$ be the span of $e^{2 \pi i nz}$, then we can't just say $A$ is dense in $C(S^1,\Bbb C)$ and $C(S^1,\Bbb C)$ is dense in $L^2(S^1)$ and hence $A$ is dense in $L^2(S^1)$ because there are two different topologies involved!
But this is not an actual problem because $\|f\|_{L^2(S^1)}^2 \leq \mathrm{vol}(S^1) \sup_{x \in S^1} |f(x)|^2$

7:41 PM

If $d(p,A)=d(p,B)$ and $d(p,B)=d(p,C)$ then $d(p,A)=d(p,C)$

and then let $A$, $B$, and $C$ be the sides of the triangle

If you do the same with the vertices you get the perpendicular bisectores intersecting

@LeakyNun That is the question I asked you, yes

When is that inequality true for $f$

Alternatively

When could it not be true

I don't know lol

8 mins ago, by Akiva Weinberger

Remember that all of $|T_n|$'s local maxima have the same value

Important^

hrm. now I'm confusing myself

You're basically "nudging" $T_n$ by a little bit and trying to get the supremum to be lower

I'm not having any idea

7:44 PM

@Semiclassical come join me in oregon, we'll do consulting :D

lol

I feel like I"m heading that way albeit slowly

it's a tough decision to make

amusingly, the stuff I've been working on lately has had a statistical/probability flavor to it

nothing bayesian

oh neat

@LeakyNun Draw a graph

7:45 PM

just fun stuff with correlation matrices

Put $\cos(5\cos^{-1}(x))$ on Desmos

and $\cosh(5\cosh^{-1}(x))$

and $\pm\cosh(5\cosh^{-1}(-x))$ (forget if it's plus or minus)

okay, the interpretation I had of the case of equality was wrong

so that's annoying

Looks a bit like that

note the interlacing of critical points and roots

What happens if you add $\epsilon f$ for some function $f$ and tiny $\epsilon$

7:49 PM

that be important

I changed $c$ to $\epsilon$ to make it clearer

that's the unique triangle (up to similarity) for which the centroid is on the incircle and the inequality is saturated

Hm

it's...not that remarkable, really. The base angle of the isosceles is $\cos^{-1}(1/5)$ but that's about it

What's the center of the incircle of the triangle bounded by $(a,0,0)$, $(0,b,0)$, and $(0,0,c)$

7:54 PM

Mind using upper case instead of lower case?

Most sources I see use lower case for side lengths

> The coordinates of the incenter are the weighted average of the coordinates of the vertices, where the weights are the lengths of the corresponding sides.

Wat

Why

@AkivaWeinberger why is it the weighted average, you mean?

shrug

Ya

this is why I'm not trying to actually prove it

@Semiclassical Those aren't side lengths

7:55 PM

did you know that there is only one general acute-angled triangle?

@AkivaWeinberger ya, but that's my point

@akiva most sources do use $a,b,c$ for the side lenghts

Oh I misread

The (very very tedious) way I did it in Mathematica

Pick vertices (0,0), (1,0), (R cos theta, R sin theta)

Oh I see why the formula is like that now