Mathematics - 2017-01-08 (page 7 of 8)

TheGreatDuck

19:00

what's up man?

Hippalectryon

@TheGreatDuck o/ My studies are on a halt, resuming in three months

Astyx

Oh my, not this surrender joke again :p

TheGreatDuck

^^^

euclid

@arctictern but im sure it is same with your definition about $ o(\chi)=d$

SteamyRoot

Always works :D

arctic tern

19:00

@euclid no, $\chi^d=1$ and $o(\chi)=d$ are not the same. I have told you this many times now.

Ted Shifrin

Having Hippa around, I always have to make sure I'm not being lambasted.

arctic tern

There are $d$ different characters $\chi$ satisfying $\chi^d=1$ and there are only $\varphi(d)$ characters $\chi$ satisfying $o(\chi)=d$.

TheGreatDuck

@Hippalectryon remember that crazy floor function stuff we were discussing last year?

SteamyRoot

But, hey, cheer up. With all the radiation of the sun turning the American flags on the moon white, you can now claim your flag is up there :D

(okay, I'll stop now)

euclid

@arctictern so you mean it is not the correct answer.sorry if i am boring.i ask many questions

arctic tern

19:03

@euclid I mean the other answer does not apply to your question

the other answer writes "$o(\chi)=d$" when really the user means "$\chi^d=1$" because that is what they are talking about

user379685

can someone help me with lim x->inf (2+tan^2(x))*ln(1/x)

Hippalectryon

@TheGreatDuck I still have the google doc link ! Right in the middle of my bookmark bar !

Ted Shifrin

@Alessandro: I answered. I'm heading off to dim sum lunch now, but will return in a few hours.

Hippalectryon

TheGreatDuck

@Hippalectryon It extends to differential equations. Also, the jump series cannot be found in general.

Astyx

19:06

That doesn't converge, does it ? @user37

TheGreatDuck

see the answer here: math.stackexchange.com/questions/1683530/…

user379685

@astyx i don't know

TheGreatDuck

if there were an operator that worked overall it would solve the riemann hypothesis

SteamyRoot

the ln goes to $-\infty$, but the $\tan^2(x)$ should hop between $0$ and $\infty$ forever.

TheGreatDuck

i.e. it's a ridiculous long shot

SteamyRoot

19:08

Although, well, shouldn't it be $-\infty$ thanks to the $2$ term?

Astyx

$tan^2$ does not converge at $+\infty$ and neither does $\ln({1\over x}) = -\ln(x)$, which goes to $-\infty$

SteamyRoot

Like, $2 + \tan^2(x) \geq 2$ for all $x$

Astyx

So you thing cannot converge

SteamyRoot

and $\ln(1/x) \to -\infty$

Hippalectryon

@TheGreatDuck I didn't really have time to work on it since last time (exams etc), but it seems you've made some progress :D it's very interesting !

TheGreatDuck

19:08

anyone see all those recent "____ but when they do ____ it gets 5% faster videos recently"?

what's with that

:p

SteamyRoot

So I actually think it does go to $-\infty$, unless I messed up something.

TheGreatDuck

@Hippalectryon you've taken differential equations, right?

user379685

@Astyx product of two not converging things can't converge?

Astyx

So it does not converge @Steamy :p

Alessandro Codenotti

Thanks, and buon appetito @Ted! Of course I won't share the book

Astyx

19:09

Of course it can

SteamyRoot

@Astyx why not?

One factor is always $\geq 2$

the other gradually becomes $-\infty$

Astyx

So it diverges to $-\infty$

SteamyRoot

For me that's "converge to $-\infty$" :P

Astyx

Baah ..

:p

SteamyRoot

Either way, semantics. But the limit exists, that's the point :)

Hippalectryon

19:11

@TheGreatDuck bee movie meme

Astyx

Mmm sure

euclid

@arctictern in fact my question is a preposition in number theory that says :if $t$ is not primitive root modulo $p$ then $\sum\limits_{d|p - 1} {\frac{{\mu \left( d \right)}}{{\phi \left( d \right)}}\sum\limits_{o\left( \chi \right) = d} {\chi \left( t \right)} } = 0$

Hippalectryon

That's where it comes from

@TheGreatDuck yep

SteamyRoot

Well, looks like 34 hats on MSE is where this winter bash will end for me, unless a small miracle happens.

One less reason to procrastinate, damnit :(

Astyx

I for myself have 6

TheGreatDuck

19:15

@Hippalectryon take the differential equation y'' = h(x) where h is heavyside

arctic tern

@euclid sure. look inside the factorization, the factors $1-f(d)/\varphi(d)$ where $f(d)=\sum_{o(\chi)=d}\chi(g^k)=\sum_{o(\zeta)=d}\zeta^k$ and $g$ is a primitive root. if $t=g^k$ is not also a primitive root, then it shares a prime factor with $p-1$, i.e. $q\mid k$ for some $q\mid(p-1)$, in which case $\zeta^k=1$ for all $\zeta$ with $o(\zeta)=q$, so $f(q)=\sum_{o(\zeta)=q}1=\varphi(q)$, so $1-f(q)/\varphi(q)=0$

SteamyRoot

If you quickly award someone a bounty you'll get another hat...

TheGreatDuck

y = 1/2x^2h(x) + xC(x) + D(x) where D and C are piecewise constants such that y is continuous

SteamyRoot

Totally worth it for those 5 hours left amirite?

TheGreatDuck

wait no...

Astyx

19:16

Totally

Let's ask a silly question

TheGreatDuck

such that the individual terms are continuous

cause they are individual solutions

@Hippalectryon

anyway finding the continuous solution to solving a differential equation by treating piecewise constants as constants results in the actual solution to the differential equation

(that's a mouthful)

@Hippalectryon also, assuming these are consistent the formal rules of that system of differentiation are here

0

Q: Are these alternate axioms of differential algebra (on real number functions) consistent?

In another question I asked, I asked about using a particular axiom to define the derivative so that I could negate it in general. However, I realize that there are two alternative systems in particular that I find interesting*, and that they might not be inconsistent with the pre-existing differ...

Hippalectryon

that's great :-)

TheGreatDuck

did you read it? :-)

Hippalectryon

I did

TheGreatDuck

does it make sense to you?

euclid

19:25

@arctictern Shapiro has a book that he has proof it. id like we talk about it if it is possible.

Hippalectryon

@TheGreatDuck Are those axions supposed to be enough to make derivatives ?

arctic tern

@euclid well, we just proved it too, but okay

Hippalectryon

Because it seems to be that for instance they don't tell us how to derive simple functions like $f:x\to x$

Astyx

Can an algebra morphism be defined on a basis ?

TheGreatDuck

@Hippalectryon supposedly they all span differential algebra and define it fully. Apparently some of the rules are enough to prove the derivative of x is 1.

shrugs

take number 3

let f(x) = x

(f(f(x)))' = f'(f(x))*f'(x)

but then this is the same as saying f'(x) = f'(x)^2

which means f'(x) = 1

assuming it exists of course

therefore the chain rule proves the identity derivation

multiplication rule is also implied somehow

from that everything else of normal derivation falls into place

:-)

it's only the first statement that jams up the system potentially

cause the rules might conflict

Hippalectryon

19:33

@TheGreatDuck Maybe I'm missing something very obvious here >.> (haven't done maths in a few months), but how do you get that ?

euclid

@arctictern is there a way for share pics here?

Hippalectryon

@euclid paste the link image.extension, like so

euclid

i dont have any link.it is a page of a book

Astyx

upload it to imgur and share the link

That's what most people here do

TheGreatDuck

imgur.com

go there

upload

Hippalectryon

19:38

you can also upload a pic from your computer @euclid, unless you're on mobile

euclid

thanks

TheGreatDuck

@Hippalectryon not quite sure what you mean. What were you having trouble following?

(sorry got distracted by people who just wouldn't stop talking)

Hippalectryon

@TheGreatDuck how (f(f(x)))' = f'(f(x))*f'(x) => f'(x) = f'(x)^2

TheGreatDuck

f(x) = x

euclid

the link needs to registration.there is not other way here?

Hippalectryon

19:44

ah true @TheGreatDuck forgot >.>

TheGreatDuck

so (f(f(x)))' = (f(x))' = f'(x)

Martin Sleziak

@euclid chat.stackexchange.com/faq#talk You also need 100 reputation points to upload pictures in chat Chat requirement to post images should be made explicit.

Hippalectryon

@euclid are you on mobile ? if not there's an "upload" button in the chat

euclid

no

TheGreatDuck

and the RHS of the final equation is the chain rule which is axiom 3

Ilan Aizelman WS

19:45

@TedShifrin Hey Ted. how are you?

TheGreatDuck

product rule is foreign to me though. Would need proper proof of that. :p

granted, since differential algebra is a rigorous subject less things need to be proven as they are already in existence

all that would change is how piecewise constants are manipulated

Martin Sleziak

@euclid After you gain 100 reputation points, in the same window where you are typing messages you will also see upload button. Like here: Chat room image instead of link.

TheGreatDuck

also notice that the integral is never mentioned

it doesn't change

the antiderivative is what changes

and the second fundamental theorem of calculus also changes

Hippalectryon

@TheGreatDuck Well the first set of axioms doesn't shock me, but the second seems really weird

TheGreatDuck

see this question: math.stackexchange.com/questions/1992503/…

Hippalectryon

19:48

for instance $f(x)=\{x\}$ would have a derivative of $0$ on $[0,2]$ but $1$ on $[0,1[$ ?

TheGreatDuck

what no

x is not periodic is it?

x mod 1 has a derivative of 0, essentially.

Hippalectryon

$\{x\}=x-\lfloor x\rfloor$

TheGreatDuck

i see

Hippalectryon

yeah so x mod 1

It's not periodic on [0,1[ but it is on [0,n>1]

TheGreatDuck

no

it is periodic with period 1 everywhere

anyway... it's just something that works. Probably useless. Worth mentioning. Worth knowing about.

:p

it has "nice qualities"

Hippalectryon

19:52

@TheGreatDuck Everywhere ? I don't really see how the function defined on [0,1[ by f(x)=x is periodic everywhere

TheGreatDuck

what...

x mod 1 is defined everywhere

@Hippalectryon do you not agree that (x+n) mod 1 = x mod 1 where n is an integer?

euclid

@arctictern look these please imgur.com/GkVaCJr imgur.com/QM4Q0Wm imgur.com/0FOAWls

Hippalectryon

@TheGreatDuck that's not what I mean. f(x)=x is defined everywhere, but I'm creating the function g(x)=f(x) defined only on [0,1[, aka the restriction of f to [0,1[

Akiva Weinberger

Hola qué 'ta pasando

TheGreatDuck

@Hippalectryon that's not what we are discussing. It's not rationally periodic so how does it at all say anything relevant?

Hippalectryon

19:56

@TheGreatDuck Exactly, it's not rationally periodic, thus its derivative is not 0

TheGreatDuck

but it isn't x mod 1 so how is that relevant?

x mod 1 does not equal what you proposed.

Hippalectryon

What I mean is that using the second set of axioms, a function's derivative changes depending on the interval on which it's defined, and that bothers me

SteamyRoot

It says alot about my day when the most exciting thing that has happened, is starting up TeXstudio and being informed a new version is available.

TheGreatDuck

@Hippalectryon all functions are defined for all real numbers...

wolframalpha.com/input/?i=x+mod+1

Ramanujan

@TheGreatDuck good morning?

TheGreatDuck

19:59

@Ramanujan good morning to you too?

Ramanujan

One night over since we last time chat?

Hippalectryon

@TheGreatDuck Uh ? How so ? By definition, the restriction of $f:x\to1$ on $[0,1]$ is only defined on $[0,1]$ for instance

TheGreatDuck

well then it's not periodic

Hippalectryon

I agree, but why does that matter ? Does the second set of actions not apply to any function ?

TheGreatDuck

it does

take x mod 1/2 on domain [0,1]

it's periodic

is it not?

x + 1/2 mod 2 = x mod 1/2

well... assuming it's defined

Hippalectryon

20:02

It is. But it's not on [0,1/2], therefore when restricted to that domain its derivative is nonzero according to your axioms (at least that's what I understand from what they say)

TheGreatDuck

when we restrict it to that domain it equals x

why would we restrict a domain anyway?

aren't functions always assumed to exist for all real numbers?

Hippalectryon

Well, a lot of functions don't exist for all reals

like log(x)

TheGreatDuck

unless they have an undefined result due to an illegal operation?

log(-x) is a complex number actually

SteamyRoot

is many complex numbers actually

Hippalectryon

Well it's a generalization, depends how you define it :P

TheGreatDuck

20:04

it exists. We just don't like to talk about it.

Hippalectryon

But anyway, reducing function's intervals come in handy in several proofs. But it might indeed seem "artificial" at first.

TheGreatDuck

@SteamyRoot log(-x) for some specific value of x only gives one complex number. It's not a vector valued function.

@Hippalectryon I would assume we're not talking about restricted domains and even if we were it doesn't imply a contradiction. It just implies a weird property.

unless you actually have a contradiction with the other axioms?

SteamyRoot

No, it's a multi-valued function. Until you choose a branch cut and a branch to work on, it's not even a function.

$\log(-1) = \pi i = - \pi i = 3 \pi i = -3 \pi i = \dots$

TheGreatDuck

@SteamyRoot fair enough. I don't do complex number stuff. I just know it gives a complex number. Probably the principal logarithm or something

also pi i \dne -pi i

SteamyRoot

Well, most calculators choose the principal branch... which is defining $\log(re^{i\theta})$ as $\log(r) + i\theta$ with $\theta \in (-\pi,\pi]$

TheGreatDuck

20:09

most calculators dont support complex numbers

SteamyRoot

I know they don't equal eachother, but there's no real good way to express the image of a multi-valued function I know of

TheGreatDuck

@Hippalectryon to be fair, the periodic thing is a random afterthought. It's probably the less useful half-sibling.

SteamyRoot

4

A: Graphing error found

You're right that $$f(x) = e^x|\log(x)|$$ is only defined on $(0,+\infty$), so the "problem" lies with the graphing software. Below is a plausible explanation that seems to match the graph drawn. Many calculators and software follow the steps listed elow, e.g. my TI-83, WolframAlpha and GeoGeb...

A recent question you reminded me of.

Hippalectryon

@TheGreatDuck let me think for a bit :P

SteamyRoot

Many calculators or software that, at first sight, don't have anything to do with complex numbers, do have them programmed. Not always with good results, as the question demonstrates.

TheGreatDuck

20:12

@Hippalectryon fair enough but try using the rules to derive a solution to the differential equation y'' + 2floor(x)y' + floor(x)^2y = 0 in the piecewise constant alternate system. It has a really clever trick. ;) Don't use laplace transforms. Ugliest possible way to do it.

@SteamyRoot im talking about a handheld arithmetic calculator

usually they just say "error"

I am thinking of starting a 500 rep bounty on this

-1

Q: Is there a set of axioms governing the properties of derivatives in calculus that include this particular axiom and what would it be?

Moved from Math Overflow due to not being regarded as a high degree of research Note: I am looking in particular at real valued/real input functions at all values regardless of differentiability. In this question a series of axioms or postulates governing calculus are proposed. Granted, that is...

calculus derivatives axioms research

is it worth doing or am I being rash?

Is there any likelihood of it being answered as a result?

Hippalectryon

@TheGreatDuck What is $f,g$ are non rationally periodic but $f+g$ is ? how does that work out ?

TheGreatDuck

oh...

you're right

you just found a contradiction

wait no

not neccessarily

Hippalectryon

Indeed, we could have f'=-g'

TheGreatDuck

^^^

Hippalectryon

but it's not always the case

Let me draw a quick example

TheGreatDuck

20:20

let f(x) = x

and g(x) = floor(x)

(f - g)' = 0

therefore f' + (-g)' = 0

which in turn implies that f' = -(-g)'

a contradiction only occurs if we show that either g or x have a derivative of 0 everywhere

however x' = 1

so therefore

1 = -(-g)'

-1 = (-g)'

we lose the ability to pull out constant multiplication

otherwise

1 = g'

meaning floor(x) is an antiderivate as well as x

but

this makes sense

they vary by a periodic function

x - {x} = floor(x)

@Hippalectryon make sense? I don't see a contradiction. Just a weird rule.

piecewise constant function may vary well have derivative of 1 in this system

weird but not a contradiction

Hippalectryon

@TheGreatDuck I've got an example :P wait I'm busy rn

Kari

I can't see why the second equality from $(2.3.4)$ follows.

SteamyRoot

Because you're working on a sphere?

Kari

Oops, I meant the first equality

How does being on a sphere help to explain the second one anyway, @SteamyRoot?

SteamyRoot

20:35

By using the fact that the sum of squares is $1$

Seems like there's a square missing, though.

Kari

Oh right

Yea, on the very left right?

SteamyRoot

Yup...

Kari

$|P-P'|^2$ is what it should read

Sorry, I'm burnt out right now. May be time to stop math for the day.

SteamyRoot

Also, the notation seems to suddenly jump from lowercase to uppercase $p$.

Kari

It's not a very accurate set of notes that I'm working from

Sadly, it's the one the lecturer wrote and follows so I wanna get it down first

Then I'll probably jump at a book

SteamyRoot

20:42

Well, on the one hand, finding and correcting the mistakes is a good exercise.

On the other hand, it's not fun to make exercises if the notes you're using aren't reliable.

Kari

I guess so!

euclid

20:57

hi @SteamyRoot do you know primitive roots?

SteamyRoot

Maybe... In a skype call though, so can't really answer any questions I'm afraid

Hippalectryon

@TheGreatDuck

f,g don't have opposite deriatives

Akiva Weinberger

5. Show that the tangent plane of the cone $z^2=x^2+y^2$ at $\begin{bmatrix}a\\b\\c\end{bmatrix}\ne0$ intersects the cone in a line.

@TedShifrin I'm not sure I did it the intended method. Noting that we must have $c^2=a^2+b^2$ for it to line on the surface, I found that the tangent plane is $cz=ax+by$. I want to find the intersection of it with $z^2=x^2+y^2$. Using the formula for the product of the sums of two squares, we get:
$$(a^2+b^2)(x^2+y^2)=(ax+by)^2+(ay-bx)^2$$
That's $(c^2)(z^2)=(cz)^2+(ay-bx)^2$, which implies $0=ay-bx$ and thus $x/a=y/b$. Thus, defining $t:=x/a$, we see that al…

I don't think it's the intended method because I used the product of sums of squares formula.

Krijn

21:13

It's hard to study mathematics when your flatmate puts on The Sound of Music just a little bit too loud

TheGreatDuck

@Hippalectryon but do the other rules of the periodic system imply in any way that the two functions DONT have opposite derivatives. The purpose isn't to find differences between systems. I'm asking whether the rules contradict each other.

Jonathan Richard Lombardy

Does anybody knows a book which has a collection of papers by mathematicians before 20th century, not any historical or philosophical stuff just the works translated to english. Thanks a lot.

TheGreatDuck

@JonathanRichardLombardy ask it as a question rather than here. More likely to be answered. :)

SteamyRoot

Whelp, surprisingly just got a 35th hat :O

TheGreatDuck

21:28

i have 5 or 6

Jonathan Richard Lombardy

@TheGreatDuck Thanks for the suggestion.

TheGreatDuck

looking for an opinion

math.stackexchange.com/questions/2078995/… would it be productive to offer a very large bounty on this?

Akiva Weinberger

(@TedShifrin I meant *lie on the surface, not line)

TheGreatDuck

@robjohn nice hat placement. me like.

hi @AkivaWeinberger

math.stackexchange.com/questions/2045828/…

Hippalectryon

21:57

@TheGreatDuck I'm not sure. It doesn't appeal to many people I suppose

TheGreatDuck

no it just hasn't been answered yet

:p

unless you mean the closed one

...that's because I went a little close-heavy on my own questions.

I thought inactive questions had to be closed

and deleted.

Hippalectryon

@TheGreatDuck I mean that it's not something people are familiar with, so they're less likely to spend much time on it given the sheer amount of questions out there

There are more "accessible" high bounties

TheGreatDuck

that's why I asked if a very large bounty (on the level of 500 rep) might entice people or if it won't work.

Hippalectryon

then again, I offered 500rep bounties several times and usually got at least interesting comments

Even on weird questions

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$Mathematics$ Mathematics