Mathematics - 2017-07-07 (page 3 of 8)

Typhon

5:00 AM

or rather a = \bar a

Akiva Weinberger

@Typhon $x$ can be complex, it's $a$ that we know is real

Typhon

i know. Said it wrong

Akiva Weinberger

Subtracting $x\bar x=a\bar a$ and subtracting $1$, we get $x+\bar x=a+\bar a$

Typhon

4

Q: Proving that if $a^n$ and $(a+1)^n$ are both elements of a polynomial field, then $a$ is also an element of that polynomial field?

In the main MSE chat @AkivaWeinberger proposed the following conjecture they found from an old question. If $a$ is a real number and $a^n$ and $(a+1)^n$ are both rational, then $a$ is a rational number. I have tried several times to prove it and I just wish to know how to prove it. Since I...

Akiva Weinberger

and dividing by $2$, we get ${\rm Re}(x)={\rm Re}(a)$

er, uh, $=a$

Typhon

5:01 AM

@LasVegasRaiders see this meme please

XD

Akiva Weinberger

I'm having trouble remembering how to finish

Las Vegas Raiders

True dat

Akiva Weinberger

We have $|x|^2={\rm Re}(x)^2+{\rm Im}(x)^2$, I guess

chrismc

Hi!

Las Vegas Raiders

Hi

Akiva Weinberger

5:03 AM

or $a^2=a^2+{\rm Im}(x)^2$, so $x$ is real

chrismc

How hard is it to get into University of VA if im in state

Virginia?

yes

What's your gpa?

So $x$ is either $a$ or $-a$. Only one of those also satisfies $|x+1|=|a+1|$, though.

Typhon

5:05 AM

@AkivaWeinberger slow clap

cumulative?

Yup

about a 4.4

So, $x=a$, QED. And that was probably more messy than it had to be

chrismc

weighted

Akiva Weinberger

5:06 AM

but whatever

Also, note that we had to use that $a$ was real

As noted earlier, there are counterexamples otherwise. For example, the cube root of unity with $n=6$.

Typhon

hmmm

actually there is an easier one

Akiva Weinberger

The circles $|x|=|a|$ and $|x+1|=|a+1|$ are only tangent when $a$ is real, essentially.

Typhon

i^2 and (i+1)^2 are both real methinks

wait wait wait nope

Akiva Weinberger

No, $(i+1)^2$ is $2i$

Typhon

no

i^2 + 2i +1

oh nevermind

but i + 1 has a length of...

\sqrt{2}!

Akiva Weinberger

5:09 AM

…$\sqrt2$

Yup

Note, by the way, that the two circles can only intersect at at most 2 places (when $a$ is allowed to be complex). This means that, when $a$ is any complex counterexample, it can have at most one (Galois) conjugate.

This means that the only complex counterexamples are quadratic irrationals.

Typhon

@AkivaWeinberger hmmm

so solutions to the polynomial x^2 + ax + b like earlier?

Akiva Weinberger

Thus the examples of the cube root of unity (which is $-\frac12+\frac{\sqrt{-3}}2$) and the example of $-.5+.5i$ (which is $-.5+.5\sqrt{-1}$)

Typhon

heh

golden ratio

x^2 - x - 1 = 0

Leaky Nun

@AkivaWeinberger for $n=2$, $2a+1$ is rational, so $a$ is rational...

Typhon

$-.5+.5i$ that's the golden ratio, right?

Akiva Weinberger

5:12 AM

No @Typhon

Typhon

wait no

Akiva Weinberger

It's $\frac{1+\sqrt5}2$, which is like 1.618 I think

Typhon

i was thinking of $\frac{1}{2} + \frac {\sqrt{5}}{2}$

Akiva Weinberger

I see, sure

Typhon

stop. sniping. MEEEEEEE

Akiva Weinberger

5:12 AM

That is not a complex counterexample.

Typhon

thank you

Akiva Weinberger

Nor is it complex.

Oh, speaking of

Typhon

thanks, sherlock

yaw?

oh welll...

Las Vegas Raiders

They are listed as extremely competitive @chrismc

Akiva Weinberger

@LeakyNun It's important that $a$ is real. There are counterexamples if it's allpwed to be complex

Typhon

5:13 AM

they arent tangent for all complex numbers

Akiva Weinberger

@LeakyNun Is it?

Typhon

i understood

we're speaking colloquially

@AkivaWeinberger who?

Leaky Nun

wait, you're right @AkivaWeinberger

Typhon

If $a^n$ and $(a+1)^n$ are both elements of a field $Q[x]$ where $x$ is the solution to a second order polynomial with integer coefficients and a leading coefficient of $1$, then $a$ is an element of $Q[x]$.

im trying to think of the correct way to make this one work

i dont want to just say "real"

that might not work and it might limit the possible ways to look at it since elements of Q[x] need not be real always

Akiva Weinberger

Is $x$ specifically real there?

Typhon

5:17 AM

@AkivaWeinberger no

it can be ANY solution

Akiva Weinberger

$n$ needn't be $2$?

Leaky Nun

@AkivaWeinberger I've been having too many brain farts

Typhon

i was wondering there

Akiva Weinberger

Leaky, take a nap

Typhon

@AkivaWeinberger n can be any positive power

Akiva Weinberger

5:19 AM

or eat

or go to the bathroom, or something

Typhon

it is the quadratic ring equivalent of your statement

only problem is that the integers are a quadratic ring

Akiva Weinberger

@Typhon I was talking to Leaky there

Typhon

and a need not be real here

oh

hahaha

Akiva Weinberger

I'm half-regretting using the letter $a$ for this, it's looking to much like the actual word

What $a$ confusing variable

5

Typhon

im trying to build a legit conjecture out of the one you posted

Akiva Weinberger

5:21 AM

Hm, do there exist complex counterexamples if $x$ is a complex thingy?

Typhon

and I think that the restriction need not be that a is "real" but rather something more "exotic"

@AkivaWeinberger dont know

try?

append i

Akiva Weinberger

OK, I appended $i$

It made a 'pop' sound

Typhon

look for a counter example

-_-1

omg

looks like a cell phone

Akiva Weinberger

$\Bbb Q[i]$ 'pop'

Typhon

-_-1 <== im on the phone, see

XD

best... typo... evah

@AkivaWeinberger that's alpha-braic number theory? eh? eh? gettit?

Akiva Weinberger

5:25 AM

I'm kinda confused how my last comment got three stars, since there's only three of us here and I can't star my own thing…

Is there someone else here?

Hello? 'echo-y reverb'

Typhon

@LasVegasRaiders quit lurking

Las Vegas Raiders

That's my specialty.

Akiva Weinberger

Aah! jumpscare'd

Typhon

no, your specialty is being annoying in minecraft

Las Vegas Raiders

};-)

Typhon

5:26 AM

@AkivaWeinberger see the meme

Akiva Weinberger

I should take all this as an indication of how I need to sleep

@Typhon ?

Typhon

@LasVegasRaiders raiders are the most annoying of players in that game

pillaging is one thing

but when they steal my fucking math theorems

Akiva Weinberger

…Isn't that rainbow backwards?

Typhon

they're dead to me

Akiva Weinberger

Unless it's the second of a double rainbow

Typhon

5:29 AM

idk

@AkivaWeinberger you know how you can have written books in minecraft? Well anyways, I had a written book full of proofs and someone stole it and I was like:

Akiva Weinberger

…Probably hard to typeset that, I imagine

Typhon

they were geometry stuff

no need to typeset too much

Akiva Weinberger

Ah OK

Fun fact, De Moivre correctly calculated the day he was gonna die

Apparently he got chronically lethargic when he got older

He noticed he needed 15 more minutes of sleep each night

so he calculated when that would accumulate to 24 hours of sleeping time

and, as it turned out, he was correct—he did die that day

Typhon

huh

creepy

how did he die?

Akiva Weinberger

I think they put something like "lethargy" as his cause of death

Typhon

5:34 AM

heh

Akiva Weinberger

Let me look it up, though

Typhon

how old was he?

Las Vegas Raiders

Was he accurate to within 15 minutes of his time of death?

Akiva Weinberger

87, apparently @Typhon

Typhon

sounds to me like his prediction actually caused him to die out of him believing it

@AkivaWeinberger oh well then ok.

Akiva Weinberger

5:35 AM

Wikipedia has more words

Typhon

so... 4*16 is 64 days

Akiva Weinberger

https://en.wikipedia.org/wiki/Abraham_de_Moivre

Typhon

wow

2 months to live

geez

aah

1667

i was thinking this was a modern man

heh

Akiva Weinberger

Hm, apparently it's probably apocryphal

> "The story isn't completely false. It is true that De Moivre was, towards the end of his life, in an increasingly frail state. He required in excess of 20 hours per day of sleep, and his condition only worsened. However, the story about predicting his own death has no reliable source, and can reasonably be assumed to be at least partly apocryphal."

Hm

Typhon

heh

Akiva Weinberger

5:39 AM

Seems he did die in his sleep, at least

Las Vegas Raiders

13

Q: Did Abraham De Moivre really predict his own death?

At the last point of his life, he noted that he slept 15 minutes extra every night then he calculated that he would die on the day that the extra 15 minutes a night accumulated to 24 hours. That day was November 27, 1754, the actual day of his death. He died in London and was buried at St. Mar...

biographical-details death mathematicians

Typhon

well anyways'

@AkivaWeinberger can you find a counter example?

im afraid to star that yet i want to

Las Vegas Raiders

Don't they'll ban me.

2

Typhon

xD

no they wont

D:

Las Vegas Raiders

We're not ready allowed to talk about that

Typhon

5:48 AM

...

dude, we are

lighten up

relax

Las Vegas Raiders

It's not "nice."

Typhon

yes it is

it is fine in context

and pretty funny

Las Vegas Raiders

We should talk about rainbows and unicorns

Some might flag it.

and then taken out of context

BAM!!!

Akiva Weinberger

This is all very interesting: snopes.com/college/homework/unsolvable.asp

Las Vegas Raiders

I've been there pal

Akiva Weinberger

5:51 AM

George Dantzig mistakes open problems for homework, solves them

(Also math.stackexchange.com/questions/533146/…)

Las Vegas Raiders

Possible, but highly unlikely...

Akiva Weinberger

Snopes has a quote from the man himself @LasVegasRaiders

And the second link says what the problems were

Las Vegas Raiders

...yup, saw that.

Secret

6:09 AM

NB: I screwed up. Will fix this later

\begin{align}
C_{\alpha} & : \{\gamma,\delta,\eta\}\mapsto \{\gamma+\delta,\gamma\delta,\gamma^\delta, I(\gamma,\delta),C(\eta),\psi_\gamma(\eta)|\eta < \alpha\}\\
C(\alpha)_0 & =\{0,1\}\\
C(\alpha)_{n+1} & =C_{\alpha}C(\alpha)_n\\
C(\alpha) & =\sup (C(\alpha)_n|n\in\Bbb{N})\\
\psi_\beta(\alpha) & =\sup\{\max\{\gamma_n|\gamma_n\in C(\alpha)\land \gamma_n<I(0,\beta)\} |n\in\mathbb N\}
\end{align}

----
All cross terms after $C(\alpha)_1,\alpha > 0$ were omitted from print. $n,m \in \Bbb{N}$

NB Nonstandard notation for subscript towers:

Jul 2 at 21:13, by Simply Beautiful Art

Let $I(0,\alpha)$ be the $\alpha$th ordinal that is either admissible or a limit or admissibles.

An ordinal $\alpha$ is said to be $\beta$-inaccessible if it is $\gamma$-inaccessible and $\alpha=\sup\{I(\gamma,\delta):\delta<\alpha\}\forall\gamma<\beta$.

Let $I(\beta,\alpha)$ be the $\alpha$th ordinal that is either $\beta$-inaccessible or a limit of $\beta$-inaccessibles.
$$C(\alpha)_0=\{0,1\}\\ C(\alpha)_{n+1}=C(\alpha)_n\cup\{ \gamma+\delta,\gamma\delta,\gamma^\delta, I(\gamma,\delta),\psi_\gamma(\eta)| \gamma,\delta,\eta\in C(\alpha)_n\land \eta<\alpha\}\\ \psi_\beta(\alpha)=\sup\{\ma…

Daminark

@LasVegas finally figured out the Bassel problem

Well, that one was easier, the one we /just/ figured out was proving that $\sum \frac{1}{n^4} = \frac{\pi^4}{90}$

Alessandro Codenotti

6:39 AM

How did you do it? With a lucky fourier series?

Typhon

@AkivaWeinberger if I were to prove some conjecture in Z/1 and show that it being true in Z/n implied it were true in Z/{n+1} could I then conclude it is true for the integers?

Daminark

6:51 AM

Nope, the idea is to use $\frac{\sin(x)}{x} = \prod_{n=1}^{\infty} (1-\frac{x^2}{\pi^2 n^2})$

I mean, that's what the problem said to use

You compute the coefficient of $x^2$, which becomes $\frac{1}{\pi^2}\sum_{n=1}^{\infty} \frac{1}{n^2}$

But the coefficient of $x^2$ in the Taylor series of $\frac{\sin(x)}{x}$ is $\frac{1}{6}$

For the other case, the coefficient of $x^4$ is $$\frac{1}{\pi^4}\sum_{n,m = 1, n < m}^{\infty} \frac{1}{n^2 m^2} = \frac{1}{120}$$

But then $$\sum_{1 \le n < m}^{\infty} \frac{1}{n^2m^2} = \frac{1}{2}((\sum_{n=1}^{\infty} \frac{1}{n^2})^2 - \sum_{n=1}^{\infty} \frac{1}{n^4})$$

The reason why you know this is because you can express the first term on the right to be the sum over all possible $\frac{1}{n^2}\frac{1}{m^2}$

But then, in the left, $n < m$

The way you make this true is by eliminating $n=m$, thereby knocking off the sum of quartic reciprocals, and then you divide by 2 since the $m < n$ and $n< m$ cases are symmetric. Then you have it, and use the last result to get what we're looking for

@Alessandro

Alessandro Codenotti

7:06 AM

Interesting

Daminark

Of course, the question you'd likely have is how we know the product for $\sin(x)/x$

To my understanding, the basic idea is somehow, it's sorta like factoring a polynomial by roots

Alessandro Codenotti

I'm trying to obtain it from the other infinite product for $\frac{\sin x}{x}$ that I know (the one I posted yesterday or the day before) with no success

Daminark

$x = n\pi$ are precisely the roots of both the product and $\sin(x)$, so by some analogy (and maybe actually so) this sorta gives you what you want up to a constant. And then well, you let $x = 0$, the left side tends to $1$, and the right side equals $1$

I don't know for sure if analytic functions are determined by zeroes or not, but this is at least a reasonable heuristic

Alessandro Codenotti

Hm, makes sense

Daminark

7:27 AM

Actually it's not occurring to me yet for some reason whether or not analytic functions are determined by their zeroes up to a constant

Do you know?

PVAL-inactive

@Daminark e^z has no roots.

As does e^(e^z)

Daminark

Rip

PVAL-inactive

There's a sort of characterization of entire functions like this en.wikipedia.org/wiki/Weierstrass_factorization_theorem

and there's other things you can do on other domains

Daminark

I see

Daniel Cortild

Hi

parvin

7:50 AM

hi

can any one help me with finding out how the venn diagram of two independent events is like?

Emolga

8:46 AM

If a continuous function $f$ is subharmonic (i.e. smaller than its average on spheres around any point) then $\frac 1f$ is superharmonic? (meaning same with opposite signs.)
I can prove this in the smooth case, because there it is equivalent to checking the sign of the laplacian.

I also assume the function is negative.

(My aim is tackling this question: math.stackexchange.com/questions/2348511/…)

Liad

9:06 AM

im trying to calculate $\int \int \int_D 6x +8 $ where $D=\{(x,y,z) : x,y \ge 0 , z = 4 - x \ ^ 2 - y \ ^ 2 $. i use cylindrical coordinates to get $x = r cos(\theta) , y = r sin(\theta) z =z $

now im trying to find the boundaries of $r , \theta$: $\theta \in [0,2\pi]$

how can i find the boundaries of $r$ ?

we have $z = 4 - r \ ^ 2 $

huh, $r \in [0,2]$ , right?

Albas

@SohamChowdhury Hey soham, did you get IISER Pune?

s.harp

9:28 AM

What is the etymology of the word homology?

> One should realize that the homology groups describe what man does in his home; in French, l’homme au logis. The cohomology groups describe what co-man does in his home; in French, le co-homme au logis, that is, la femme au logis. Obviously this is not politically correct, so cohomology should be renamed. The big question is: what is a better name for cohomology?

I think this is not a correct etymology^^

Daminark

@s.harp nah this has gotta be correct

I mean, this sounds Greek

homo-logos?

s.harp

yes, the study of sameness

Daminark

Hmm, that does a better job at explaining what the word entails to a biologist

Dattier

Hi, $$\forall n \in \mathbb{N^*},\forall a \in \mathbb{N^*}, a^n(a^{n!}-1) \mod n!=0\text{ ?}$$

Secret

10:21 AM

@LeakyNun I have not read these in detail yet, but these are your best bet for somehting very different from ZFC. I stumbled upon them today when trying to make sense of admissible ordinals

Alternative set theory

Generically, an alternative set theory is an alternative mathematical approach to the concept of set. It is a proposed alternative to the standard set theory. Some of the alternative set theories are: the theory of semisets the set theory New Foundations Positive set theory Internal set theory Specifically, Alternative Set Theory (or AST) refers to a particular set theory developed in the 1970s and 1980s by Petr Vopěnka and his students. It builds on some ideas of the theory of semisets, but also introduces more radical changes: for example, all sets are "formally" finite, which means that sets...

Fuzzy set

In mathematics, fuzzy sets are sets whose elements have degrees of membership. Fuzzy sets were introduced by Lotfi A. Zadeh and Dieter Klaua in 1965 as an extension of the classical notion of set. At the same time, Salii (1965) defined a more general kind of structure called an L-relation, which he studied in an abstract algebraic context. Fuzzy relations, which are used now in different areas, such as linguistics (De Cock, Bodenhofer & Kerre 2000) decision-making (Kuzmin 1982) and clustering (Bezdek 1978), are special cases of L-relations when L is the unit interval [0, 1]. In classical set theory...

Evinda

10:43 AM

Hello!!! We have $2^n$ possible n-tuples in $\{0,1\}^n$. How can we find the number of n-tuples in $\{0,1\}^n$ for which two coordinates are 1 and the rest 0?

Liad

you have $n$ choose $2$ options to pick the coordinates that are $1$ , and the rest is zero. @Evinda

Evinda

I see, thank you :) @Liad

Liad

No problem :)@Evinda

Pritt Balagopal

11:22 AM

Hi

I want to learn multivariable integration, do you know any sources that can explain me how to?

Sha Vuklia

11:49 AM

@Steamy I think I made it!! And also: I'm FREE NOW!!!!!! After an entire year of suffering (and amazement), I am finally free! I think I'm going to revise on all the subjects I didn't have enough time for this summer! And I can do that without any time pressure!! And it won't be difficult, because I've already read everything, so it's really placing the dots on the i's :P It feels so good!!!! The year is over!

SteamyRoot

Haha, congratulations :)

Secret

Congratulations. Just don't follow my footsteps in the past of procrastinating e.g. in facebook while what I really want to do on that holiday before PhD is to devour Munkres, and i am sure your maths background will be massively beefed up without any pressure

Sha Vuklia

@Secret yes I'm afraid of that.. I can be extremely lazy if I don't feel any time pressure :(

I wish someone could just force me to do it this summer :P but then it wouldn't be a holiday anymore, but just school at a lower pace I guess

Secret

I love academic related learning, but my issue is laziness. It is one of those few things that can override my passion on something. For me it took a 30 mins warmup period in order to achieve flow and not distracted anymore. Perhaps you and other mathers might be better than me at concentration

user84215

DIFFERENTIAL CALCULUS APOSTOL

SteamyRoot

11:58 AM

For PhD students, summer is that time when you can finally concentrate on research ;)

Secret

Indeed, and supervisors like to take sabbaticals during this period

SteamyRoot

o.O

Sha Vuklia

Oh I have that too, I need a warm-up period. But I also like to work on stuff obsessively or just not at all, so my warm-up period is often 1-2 days, and then I can do it for maybe an entire week obsessively, and then it's over all of a sudden. But getting yourself to start is the most difficult part :( especially without time pressure. Ugh! I'll have to figure something out to really force myself.

Maybe I could make small youtube videos on the bits I want to revise, and then I have something to work towards

SteamyRoot

Sabbatical is at least half a year at my university

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$Mathematics$ Mathematics