@MarkDominus Okay, I added a conclusion and more about the effect of the phase shift than anyone probably wants to know :-)

I don't know if that helps, but I had to do it :-)

It's probably too late now to get a lot of points, but it is a beautiful answer .

I never knew that the frequency varies as well as the amplitude. I've only seen examples where the amplitudes are the same.

@MarkDominus I don't care so much about the points, but I definitely learned something writing it.

I was hoping you would say something like that.

I didn't know about the phase modulation either.

I wonder if you can observe a similar phenomenon with light waves? It seems like you shouldn't be able to, but it's hard to say.

@MarkDominus It would show up in interference patterns. The frequency of the light should vary slightly between valleys and peaks

I don't know how much. it might be too small to notice.

But light isn't really waves, it's photons, and as far as I know the photons don't interact that way.

Disclaimer: I am a physics ignoramus.

@MarkDominus It behaves as both. I wonder if there is something observed like this.

@MarkDominus I just measured the graph very closely and the change in frequency of the green wave matches with what I give due to the phase shift :-)

hola!

@MarkDominus Hmm, I think I was wrong about this showing up in interference patterns, since those are usually generated with the same frequency of light. I don't know if there are observed effects of light whose frequencies are close. It would take two coherent beams (lasers) with frequencies that are extremely close to notice anything like this.

Oh, well.

@leo Hey there!

Is there a version of the Arzèla bounded convergence theorem for Riemann integrals where the Riemann integrability is no longer assumed?

@leo are you talking about something like this?

@robjohn Let me see

thanks!

@robjohn yes. photons are both a particle and a wave depending on the observance.

@Eugene Hmm, slightly more subtile than that. (I worked and work on quantum optics).

@Eugene The phenomenon we were wondering about was if there were anything like what happens when two waves of almost equal frequency sound combine, where the frequency of the sum oscillates.

@JonasTeuwen You might know about this then. In this answer, I discovered that not only does the amplitude of the combined wave vary, but so does the frequency. I don't remember seeing that before.

@robjohn You are doing some kind of amplitude modulation right?

@JonasTeuwen No, I am simply adding two waves of slightly different frequencies and differing amplitudes.

@robjohn But the effect is apparently some amplitude modulation. So I would guess they should be equivalent for these particular wave. This is what I would expect. You have your "carrier" and you want to modulate the amplitude, then you can add the signal to your carrier wave to increase/decrease the amplitude. If the frequencies are far enough apart should be quite okay.

Of course, if you would do AM it is better to squeeze the amplitude using some transistor, easier.

Although, adding is not so bad either...

@robjohn Rob, I have settled the PV problem.

@PeterTamaroff PV?

The principal value problem? I hope it is not the one I thought about as well 8-).

@robjohn Pretty Videos.

JK, Principal Value.

Remember?

The one about $$\int\limits_{ - \infty }^{ + \infty } {\cos \left( {\varphi x} \right)dx} $$

@PeterTamaroff Yes, how did you settle it?

@robjohn I'm reading a part of Cauchy's Course D'Analyse

According to Agustin, it should be calculated as follows

$$\int\limits_{ - \infty }^{ + \infty } {\cos \left( {\varphi x} \right)dx} = \lim \left( {\int\limits_{ - \frac{1}{\varepsilon }}^{X - \varepsilon } {\cos \left( {\varphi x} \right)dx} + \int\limits_{X + \varepsilon }^{\frac{1}{\varepsilon }} {\cos \left( {\varphi x} \right)dx} } \right)$$

I think it would be OK to use $$\int\limits_{ - \infty }^{ + \infty } {\cos \left( {\varphi x} \right)dx} = \lim \left( {\int\limits_{ - \frac{1}{\varepsilon }}^{ - \varepsilon } {\cos \left( {\varphi x} \right)dx} + \int\limits_\varepsilon ^{\frac{1}{\varepsilon }} {\cos \left( {\varphi x} \right)dx} } \right)$$

Which gives $0$

No it doesn't

the limit doesn't exist

@robjohn Oh, darn, I wrote something the other way around, you're right.

@PeterTamaroff I'm sorry, I have to go for a while. I will be glad to talk later when I get back.

@robjohn Oh, it's OK. I have to study for my physics midterm tomorrow, and I guess the PV there is a lost case.

@robjohn i think there is wave interference in light particles if that's what you mean

@Eugene Sup?

@PeterTamaroff nothing much. reading books on algebraic geometry, algorithmic number theory, and homological algebra

@Eugene Which one is harder?

@robjohn $O\left(\frac 1 {\log n} \right)$ is $o(1)$ right?

@PeterTamaroff wish i knew.

@PeterTamaroff Yes, but not the other way around

@robjohn Right.

@PeterTamaroff, can you help me?

or some one?

there's not a discussion going on, just throw your question into the arena. you don't need to request permission or someone's availability.

it seems that I can't download this (I forgot the password provided by my institution) so I'll be glad if someone can download it and send me a link or something

@leo What up?

@JasperLoy that î

is there a pw recovery process? or is your emergency question "what is my pw?" or somesuch?

@JasperLoy ♫♪ Yo-ho Yo-ho, pirate always be! ♫♪

At any rate, @PeterTamaroff here's a problem: Given a natural number $k$, what is the largest $m$ such that $m\mid(n^k-n)$ for all $n\in\Bbb Z$?

@anon there is a recovery procces. In order to recover my pw I have to call to an office which isn't working at this time of day

@anon Interesting.

@JasperLoy Yes I can. But right now it is closed.

@JasperLoy :-S

anyone here familiar with pell equations

i am trying to figure out the best solution for 10 x + 84 y + 16 = 0 but different solvers give me different results; which result makes the most sense?

@AgainstASicilian that's a linear equation with multiple solutions. not a pell equation.

Unless you're missing some ^2's..

it's linear, sorry

but how do I express the fundamental solution?

@leo Try here

Solve for y in terms of x. Then the fundamental solution is (x,y(x)) (the second component as a function of x)

@anon Should Fermat's Theorem help there?

so the fundamental solution will have y = 84 something?

@PeterTamaroff No. It's elementary. Hint: use CRT to investigate the prime factorization of $m$.

One solution seems to be x = -136 + 42 t' and y = 16 - 5 t'. Another x = -10 + 42 t and y = 1 - 5 t. Another x = 42t + 32 and y = -5t - 4

@anon Yes, I was thinking about $\mod m$ (of course) and using Fermat's $a^{p-1} \equiv 1 \mod p$ on every prime divisor of $m$. Is that ok?

@PeterTamaroff Oh, that Fermat.

@anon What other do you know?

Apart from FLT.

That works, but is $p-1$ the only number for which $a^x\equiv1$ for all $p\not\mid a$? :) (yes)

@PeterTamaroff thanks I'll see

well, the smallest positive one anyway

@anon Well, that follows from the Theorem, doesn't it? Wasn't it an $\iff$ theorem?

oh, maybe. hold on, fireworks are scaring the shit out of my dog..

@anon Ah, yeah, it's a trying week for dogs.

@anon Why fireworks?

@leo Here's the main page

Dunno if this helps, too.

I'll go to sleep in a while. Wish me luck in my Physics second mid-term tomorrow!

Good luck.

@PeterTamaroff Sleep well and good luck man!

@PeterTamaroff converse

@PeterTamaroff Sleep well and good luck man!

@JasperLoy his name is shadow, named him after sonic adventure 2 came out :)

@PeterTamaroff 4th of July is coming up

(Independence Day in the US)

@anon Oh! Our indep. day is on the $9^{th}$ (And by bday on the $6^{th}$)

@JasperLoy You don't?

@anon Ah! Forgot about Carmichael numbers.

@JasperLoy Why not?

@JasperLoy Maybe you're allergic or something....

Or Korean....

(btw the question was asked by lhf this morning)

can anyone explain what a fundamental solution would be for the equation above?

Ok, bye all!

a fundamental solution is one that contains every particular solution

(the particulars obtainable from the fundamental one by plugging in values)

so what would it be for that equation?

10 x + 84 y + 16 = 0

any of the ones you gave, or the fundamental solution obtained from solving for y in terms of x (as I said). not all expressions for the fundamental solution look the same: take one such expression, and then replace $t$ with $t+1$ for instance; the appearance of the expression will change but it will still be the same fund solution

so are ALL of them I posted fundamental?

yes

interestig

assuming whatever systems you were using don't make arithmetic errors

To go from the first to the second, replace $t$ with $t+3$. To go from the second to the third replace $t$ with $t+1$.

the solver i used gave me x = -136 + 42 t' and y = 16 - 5 t' but then arbitrarily decides to substitute t=3+t'

how did you know that? the t+3 and t+1 thing? i don't understand what on earth is going on here, haha

I just used arithmetic. The difference between -136 and -10 divided by 42 is?

potatoe

@anon Oh sorry. misunderstood

I am just trying to understand why the program might wish to make the substitution

that, I have no idea about

it's not like it's an opinion -- clearly something in the algorithm instructed it to do this

Remember to set your clocks back one second on June 30 just before midnight! The minute of 23:59 on June 30 will contain 61 seconds.

Which one is more conventional?

$\{z|P(z)\}$ or $\{z:P(z)\}$?

@FrankScience I use the second, but I know that the first is also used.

@FrankScience some also use $\{z\ni P(z)\}$

@robjohn The first one is taught in China. I wonder what the high-school education in US uses.

the second appears more conventional to me.

@FrankScience I was taught the second and the one that I mentioned in high school. I didn't see the first until college or grad school

@FrankScience Of course that was 35 years ago.

Get it.

you mean "got it"

What's difference?

Get it? Got it. Good! from The Court Jester

It's abbreviation for I've got it?

@FrankScience Yes. "Get it" is in the imperative voice; you are telling someone to understand :-)

Well, but I'm taught that past participle is usually used to express the passive voice, for example, spilt milk.

@FrankScience But the past participle of "get" is "gotten"

get (third-person singular simple present gets, present participle getting, simple past got, past participle (chiefly British) got, (North American or British archaic) gotten)

See Wikitionary

@FrankScience I've got to say that "got" as a past participle just sounds bad to me.

Ooh! I just noticed that I've written 666 answers!

Sometimes awful. For instance, I have got a pen in AmE only means I have a pen in BrE.

Here's proof that I'm bathing in brimstone

Two recent gold specialist badges: André Nicolas in calculus and Brian M. Scott in general topology.

claps

Should any question which does not have single correct answer be CW?

Do we have some guidelines about which questions should be community-wiki? I don't see such thing in faq.

Ok, maybe this can be considered as guideline: What kind of questions should be asked as CW?

How to compute $\lim_{n\to\infty}(\sum_{k=1}^n1/\sqrt k-2\sqrt n)$?

euler-maclaurin

What's your answer?

I don't think it can be solved out through that formula.

I'm looking for links, this is def EM on $H_{n,s}$ and $\zeta(s)$, $s=1/2$.

According to Wikipedia the answer should be $$\gamma = \lim_{a\to1}\left[ \zeta(a) - \frac{1}{a-1}\right]$$ in your case $a=2$.

wat

Wait for a moment. I suddenly remember that there's something asymptotics on CMath about $\zeta(z)$ where $\Re z\le1$.

Sorry about my bad memory

It's just a special case of CMath EXERCISE 9.27.

wat = west Africa time?

Sorry, I copied incorrect formula from wiki.

wat = how do you plug a=2 in a lim a->1

and also how that formula is relevant

$$\gamma_{f_a} = \frac{(a-1)\zeta(a)-1}{a-1}$$

and $a=2$

For $H_{n,s}$ the generalized harmonic numbers, we have $H_{n,s}\sim n^{1-s}/(1-s)$ (via a simple Riemann summation argument). The constant term involves $\zeta(s)$, and the lower terms involve the Euler-Maclaurin formula.

$\zeta(1/2)$ is right.

aha, here is the link

very first formula in the paper

I'll read it in many years.

@MartinSleziak your formula is right, it just needs to be $a=1/2$ rather than $a=2$

the extra 2 comes from the lower bound of 1 in the integral given on wikipedia

Wikipedia claims this is formula for $$c_f = \lim_{n \to \infty} \left[ \sum_{k=1}^n f(k) - \int_1^n f(x) \, dx \right]$$ in the special case $$f_a(x) = x^{-a}.$$

So they should write $x^a$ instead?

Arbitrary choice.

I'm so stupid.

Of course, we have $1/\sqrt{k}=k^{-1/2}$, so $a=1/2$.

Embarrassing that I somehow came up with $a=2$.

@MartinSleziak It looks as if you are leading up to the Euler-Maclaurin Sum Formula.

I was just looking for online reference for the result Frank was asking about.

@FrankScience That limit is $\zeta(1/2)$

@anon Ah, I just saw that. I was looking back at what Frank Science had said.

$$a_n=\sqrt{1+2\sqrt{1+3\sqrt{\cdots+n\sqrt{1+n}}}}$$

@anon This is closely related to this answer

$\lim_{n\to\infty}a_n$.

@robjohn I've already upvoted that one :)

@anon I wasn't trolling, just mentioning that you can show Frank's result using it.

@FrankScience look on Ramanujan's wikipedia page or something

@anon Thanks

that reminds me of a funny incident in middle school. I made up a story about a nonexistent unlockable in a video game me and a friend played in order to screw with him. he spent months trying to get it, having no idea how (to this day I don't think information about it is online). however one day he called me over and lo and behold, the trophy I had made up was a real thing.

@anon Cool story, bro!

indeed

it is probably funnier if you're me

@JasperLoy I have switched to Arch!

@JasperLoy Took too long.

@JasperLoy Excellent!

@JasperLoy Pretty hardcore, this Buddha.

@JM Are you there? I have a question relating special functions. Usually we can have a generating function for stuff like $H_m(x) H_m(y) \dots$ right? What if we say have some sum over $m^k H_m(x) H_m(y)$ instead of the usual generating function. What I did now is reduce the problem to a case where I have $H_{m_1} H_{m_2}$ with $m_1 + m_2$ constant. I was now wondering... if I do it this way, wouldn't it actually be easier to just get it directly from the generating function...?

Or is this not something true in general... I wonder. Maybe it is way easier than I think.

Others can answer as well of course, just do not know people here with special functions knowledge.

the Hermite polynomials?

if you mean $\sum_m m^k H_m(x)H_m(y)z^m$, then notice that $(z\frac{d}{dz})^k z^m=m^kz^m$...

@JM Jonas has a question if you're around.

@JonasTeuwen you have to use an old conversation to ping JM :-)

@robjohn Oh! How so?

@JonasTeuwen Two character names are not recognized by chat.

@JonasTeuwen his name is only two letters, so just writing @JM won't work.

@anon Slightly more complicated 8-).

@robjohn @anon Thanks.

Yeah, so Mario and Luigi are not recognized. Being two character names.

@anon That is exactly what I used to obtain the $m^k$.

oh, well I guess I'm working backwards then

@anon Like here.

But maybe I mixing things up, let me see...

@anon Well, yes. So you mean that you get the Mehler kernel with that operator applied to it? I am trying to compute the Mehler kernel with that operator applied to it.

If you see what I mean...

dunno what the mehler kernel is

@anon Well, the thing you wrote down with $k = 0$.

z^m?

:)

not sure if that has a simple answer

It does not, I can assure you.

Except if your name is Doron.

Then you argue about hetero- and homosexual relationships between your polynomials.

Usually when I look for papers, I first check out the ones written by people I have never heard off. Then at least I can feel good about myself for a couple of minutes before finding that someone else proved it in much greater generality 8-).

Hm... Zorn's Lemma is equivalent to Axiom of Choice. Does this mean that we should tag questions about ZL by axiom-of-choice? (Probably not.) Would perhaps Zorn's Lemma deserve a tag of its own. (ZL is a frequently used proof technique.)

But I guess the things about tagging ZL questions are ok as they are now.

@MartinSleziak Hmm... good question. If we would do this, could we go as far as tagging everything that is correct as "ZFC"?

lol

with the exception of some large cardinals questions

Exactly.

Theorem: Tagging on MSE is trivial.

Hmm, usually to the things I write about some problem I have a section: "Review of relevant and irrelevant literature". Would that be accepted in a journal? 8-). With irrelevant I mean that I explain why a certain paper is not applicable here while it might be so. Not that I put in a review of Dan Browns whatever.

Creating tag for transfinite induction was discussed here in chat. Both transfinite induction and Zorn's lemma are proof techniques that are frequently employed. (But tag for transfinite induction was not created.)

But I'll stop blabbing about tags and leave some space for more important discussion.

@MartinSleziak It is much more important than the other (non-existent) discussions here! 8-).

"2.1.13 How the Da Vinci Code will not help with transfinite recursion"

Maybe we can start a discussion on comparing the importance of two non-existent discussion...

@anon Or you try to be Doronish and prove $7 + 2 = 3 \cdot 3$ and then remark the proof also works with other numbers 8-).

Hello hello helo

anyone in here

hey

I want to ask one question which isn't strictly mathematical but I am too much bothered about it

k

this question is a doozy

Generally in physics we assume functions to be twice differentiable, and leave it there. For example if a function is representing the position of a particle (or any other physical quantity), then we want it to be twice differentiable so that we can define velocity and acceleration (or Force).

But what about singularities of function in the form of 3rd or higher derivatives jumping? Why we never care about them, Why we never bring them inside a physical theory, even in mathematice (to my limited knowledge) they are neglected?

@anon : If you do not find this question interesting then no problem

pretty sure we don't neglect singularities in mathematics

But i couldn't find anything on a internet search

ok we braodly say whether a function is analytic/ smooth or not so that we can use Taylor series, but i didn't find anything more

in physics they normally assume that there simply aren't any singularities in the relevant functions, or if there are they can be modeled accurately with impulse functions (the mathematical theory justifying these ideas would be distribution theory and dirac deltas in particular) alone, in my experience.

In Physics they need functions to be twice differentiable generally and they do not care what is happening to the higher derivatives

if you say so bro

well, wait, is that necessarily even a problem?

@anon : its generally said I heard it on physics.stackexchange

if singularities in higher-order derivatives don't affect anything and the calculations done are legitimate regardless of their existence, I don't see why they should matter

I think that is literally true, that when you start driving a car the fourth derivative of position is discontinuous at the moment you start to push down the gas pedal.

They do not matter, I completely agree, but sometimes muse myself that any mathematical idea does have a role in physics/applications. So I wonder where such a concept of singularity (higher derivatives) has a role. Poor things, they too want to involved in the real world

:-)

@MarkDominus : What you mean to say, i couldn't get

Which fractal looks like a cloud or a water drop?

(I am looking for some reading about fractals and environment, just curious)

@RajeshD What did you not get?

What is literally true?

That singularities in higher-order derivatives don't affect anything.

ok

Not just in a mathematical sense, but in a physical sense, that we deal every day with physical systems with discontinuous higher-order derivatives.

ok

The regularity of higher-order derivatives is usually not a problem. In PDE we usually want the regularity to be at least equal to the highest derivative in your PDE. If this fails you might only have weak solutions. If it does not, everything is okay and we don't worry about the higher order derivatives as we solved our equation.

Sometimes physics omits math details.

But there is of course a whole theory about dealing with "derivatives" of strange functions, check out distributions and weak derivatives (Sobolev spaces). They have been developed independently by Schwartz and Sobolev, but they are practically equivalent.

For example, they usually assume that functions are analytic at $z=0$.

@FrankScience Yes, but that is obvious. In physics we try to solve a real life problem. We do not worry about the math until we actually know that our model is the right one.

@JonasTeuwen Especially here.

Many good mathematicians which I know actually have a similar approach to mathematics. They somewhat use their intuition to justify steps, go on, if it gives a good solution they go back to see why you can do this.

@JonasTeuwen Yeah but I find that they're different.

Mathematicians' intuition is an intuition of the whole problem, but physics sometimes makes assumptions which are not always true.

Nah.

This physicist/mathematician dichotomy is driving me monkeys! 8-).

(I do both)

Inexactly, mathematicians' intuition is just like this: something can be proved, and this way seems accessible, but physics' is just like this: something is true indeed, whether it's proved or not.

I might misunderstand the physics theorists.

aka physicists

(the more theoretical ones being called "theoretical physicists")

I don't know how computer science is.

Maybe it's rigorous, maybe not.

Depends... what do you call computer science? Same thing as with the physics.

I know some physicists which are much better at mathematics than many mathematicians I know. So, it is quite some bold statement, this dichotomy. You will be surprised how much they actually do understand, the difference is they don't really care about filling in all the details when they know it will work. Mathematicians will fill in the details (not always) when they know something is correct.

@JonasTeuwen For example, The Art of Computer Programming is a book of CS.

@FrankScience Way too narrow if you ask me.

A computer scientist is generally pure as long as you don't ask him to solve your real world problem, but when you do, he doesn't hesitate to do anything, to make you happy.

computer science is about more than just programming

I made it up just now

but true from my experience

@JasperLoy xkcd.com/435 :-)

I agree that some important math has come out while trying to solve a physical problem. For example Fourier series was first discovered when Fourier was trying to solve a heat transfer problem.

@JasperLoy Really? I'm a machine, bro. I need oil.

@JonasTeuwen For you, I think we need to tweak the "coffee into theorems" adage...

yes @Jasper

thats what i intended to say too