Because although $y$ obtained above satisfies the ODE but I'm confused as to show that $x\cos x$ also satisfies the ode.

But it's one of the solutions in your general solution family.

I mean the ode with indicial equation: $(r^2+1)^2=0$

The confusion is because: $xe^{ix}$ is one solution (I understand this) but $x\cos x$ is too?

a swarm of feral parrots have entered the yard.

Oh, have you never worked this out before? The real and complex parts are the solutions working over $\Bbb R$.

Just do the simple case of $y'' + y = 0$.

Leslie: why would you think that? Is background of this question not clear?

koro: i was mostly kidding. the background is clear, but without a specific theorem it is not clear from first principles that any value identified by finding an equation satisfied by x cos x is minimal. you need to leverage the full scope of some unspecified theorem to do that. or just check it by hand.

@leslietownes We used to have more of them around here than we do now. I think they were depleted by the same virus that depleted our crow population. The crows are coming back, but the parrots need people to release them into the "wild" here.

What are you doing, @Koro?

i'm sorry to hear that. my mom's local crow population was decimated. ours is bigger than ever.

the parrots are so cute.

Please ignore that, Ted. I messed up.

When I said to do the simple case, I meant that you should see how $\sin x$ and $\cos x$ come from $e^{\pm ix}$ as solutions of $y''+y=0$. You're totally muddling things.

To answer the question: $y= A\cos x+B\sin x$ is general solution of $y''+y=0$

And you get it from the complex solutions by allowing complex coefficients or by taking real and imaginary parts.

and yes, I understand that that comes from $c_1e^{ix}+c_2e^{-ix}$

using $e^{ix}=\cos x+i\sin x$

OK, so it works similarly, obviously.

We saw from this example that components $\cos, \sin$ work for $e^{ix}$ as solutions. But for $xe^{ix}, xe^{-ix}$ as solutions, can we say that $x\cos x$ also works as a solution?

hmm, i think we can.

Same algebra.

if A = 0 then Re(A) = Im(A) = 0.

Ah, so I got confused because I was applying it incorrectly on an equation which only had $xe^{ix}$ as a solution and not $xe^{-ix}$ as well.

:(

😃

Ah, that explains something.

this is all very important stuff, usually left out of books. i'm not sure i've seen a book that discusses it.

Thanks a lot, Ted. :-)

it's implicit in most textbook treatments but not actually there.

I think it was in the standard books way back when. I TAed the ODE course my first quarter at Berkeley, and I'm pretty sure we did this there.

Leslie: you mean this $a=0\iff \Re(z)=\Im(z)=0?$

Oh, that. Yes, it's there.

ted: i had an utterly horrible ODE/linear first course. it was gone by that point.

it may have been present in an appendix of a book.

I studied those different cases for ODE solutions at college.

@Koro That is correct. $c_1=y(0)$.

i remember learning it and thinking why didn't they just tell me that in the class

We did the usual Boyce-dePrima course. Not my favorite, but ... I took a year-long more challenging course my first year at MIT, integrating linear algebra and ODE. It was taught by a superb applied mathematician who did all sorts of amazing stuff.

@robjohn :(. But at $(0,0)$, the functions are not even defined.

I agree, @Koro. $(0,0)$ is not in the domain of that problem.

I think this is the question they asked in exam and everyone marked it wrong so they had to give bonus marks to everyone.

@Koro look near $(0,0)$. Don't look directly at the sun!

In fact, $x=0$ needs to be removed entirely. You can make a substitution $u=y/x$ and work on the blow-up at the origin :P

it has happened before and I won't be surprised if that happened for that question too.

@robjohn That's a nice way to say it. :-)

I think blow-ups should be part of the standard undergraduate curriculum. (Only semi-joking.)

today we are receiving shipment of a bed frame that we ordered in 2020.

Well, as I said, I'm still waiting for my masks, which were supposed to have arrived almost two weeks ago.

we'd actually forgotten that we ordered it, until my wife got an email saying it was out of stock. we phoned and said, that's funny, you charged us for this almost 2 years ago.

@TedShifrin Boom!

then suddenly it was in stock again and is arriving.

I speak in the mathematical venue, @robjohn.

hmm, why is the order delivery taking so long? Leslie and Ted.

the idea of having a bed frame seems ridiculously upper class to me.

COVID

@TedShifrin Blow ups transcend

koro: my guess would be pandemic-related shipping issues. hard to staff a port.

And many truckers have quit, etc., etc., etc.

the port of LA/long beach, which i live next to, is one of the biggest links between the US and china. and every time we go out there we see literally dozens of container ships just sitting there. sometimes they honk at each other.

But ordering in 2020 and receiving in 2022 is just too much.

i'd be madder about it if i had remembered the original purchase.

@leslietownes I bet copper would staff a port, or any sweet red wine...

Anyhow, I think Koro should rewrite that ODE in coordinates $x$, $u$, with $y=ux$.

i think my wife bought it without asking me. i'd say, we don't need furniture for our furniture.

@TedShifrin that is how I did it

See, you blow up subconsciously.

indeed

In ODE courses they mumble homogeneity blah blah blah ...

my mind has been blown

@TedShifrin I think, you're right. I'll do it right away. Why didn't I recognise the equation as homogeneous? :)

The initial condition still makes no sense, though. We don't know what $u(0)$ will be.

My guess is that we can get different solutions with different values of $u(0)$ (all corresponding to $y=0$ at $x=0$).

there was a tsunami warning this morning here for 8:10am, a complete non event

yeah, this leads to the already obtained expression $|x\sin \frac yx|=c$.

cycled down to the water front. nothing other than a normal high tide.

But this is satisfied by $y=\pm n\pi x$ :)

So may be that's the answer.

But we know it's wrong due to the way the question has been posed.

copper unless accompanied by specific evacuation orders those things are meaningless. we got them too.

i guess my sand castle would have been f'ed up. i dunno.

@TedShifrin yeah, my comment about $y(0)$ is wrong. The condition $y(0)=0$ is confusing.

it does sound very serious for tonga. i don't mean to downplay the significance of the event.

but it is weird when we all get buzzing on our devices over nothing.

@copper.hat you cycled to the water front at the time of a tsunami warning?

@Koro Don't you get different solutions with different values of $u(0)$?

robjohn: that is exactly what he would do.

@robjohn the bay area has a long history of tsunami warnings with no adverse effects. i think the Monterey area had some pier damage decades ago, but that is on the coast

i used to windsurf & sail a lot, so one gets used to the reality after a while of being totally risk averse

in a real event people living near the water would be ordered by the authorities to leave their homes. there is a procedure for it.

in ireland when you get a weather warning it generally means something

@TedShifrin We obtain $|x\sin u|=c_1$ so $c_1=0$ at x=0. Hence the solution is: $x\sin u=0\implies u=\pm n\pi$

albany issued an eval alert 15 mins before the event even though it was known hours in advance

which is, $y/x=\pm n\pi$

@TedShifrin $x\sin(u)=c_1$. If there were a limit of $u$ at $0$, that would mean that $c_1=0$

city lawyers taking action i suppose

I think that's what the question had in its mind.

yeah, someone is trying to minimize potential liability.

@Koro the question is sentient?

I don't see why you have the absolute value in your solution, anyhow, @Koro. But we get $x=0$ as a solution and we get $x\sin u = c$. The point should be that $u(0)$ is not defined in general. The solution curves should not go through $x=0$.

the albany police would not let us cycle up the golden gate hill and sent us back along the half mile path on the shoreline. makes total sense.

i talked to the parks guy, he rolled his eyes and said "total non event"

we have a wind advisory. No tsunami warning here.

official notices can make all kinds of difference in terms of whether insurance policies do, or do not, enter into the equation. it's those investments that are being protected. it's a boring, dreadful part of the law.

i have never seen so many police out in force, however

don't get me started on overfunded cops.

maybe the legal ante was upped

they landed a helicopter and had several commandeered golf carts on a golf course nearby to stop, and i'm not joking, a theft of a pickup truck.

what a waste of money.

also they shoot people all the time without thinking about it.

we have had real issues (mostly wind related with high tide) but the police never show for these events wit real impact

@TedShifrin Ah, I put those because $\int \cot v\, dv=\log |\sin v|, \int \frac 1x\, dx=\log|x|$

would send a pic but my friend is in it

@robjohn haha.

@copper.hat we might track you through your friend.

beautiful day out there. sorry for the poor folks in Tonga

@Koro You should recognize from $\sin u + x\cos u\,u' = 0$ immediately that this is $(x\sin u)' = 0$.

@robjohn a moments sleuthing will find my yellow car in albany

for more on my extremely left wing FTP views, follow my podcast, leslie speaks.

Ted: yes, of course. I had missed the product. :)

the city won't take action on a dying camphor tree (3 outside our house) from which huge branches drop (homeowner is responsible for maintaining city trees on their property) but i can't cycle up a hill.

this is really neat:

copper: seems like a good allocation of resources to me.

using SO(3) and graph theory to design a puzzle

Time to start multiple integrals/Green theorem/Surface integrals now.

what is the holonomy part of all that? the 'constraints' on movement?

copper: the only party that might give a f- about any of that would be the insurer of the homeowner's property. if you wanna snitch, that's how i'd do it.

i've done this in the past and weirdly got immediate results.

@copper.hat no, the fact that you can move the 'rook' (as he calls it) around the sphere to change its orientation

@leslietownes i am getting old and losing my combative edge, but i may try that

@Semiclassical how is that holonomic?

what makes it a maze is that some of the moves are blocked

if they get an email about something potentially creating liability, all kinds of alarm bells go off. because it would be the first thing that would be discoverable in litigation brought by someone crushed by a branch.

i usually think of holonomic constraints as limiting versions of more complicated realities

maybe he should've called it a parallel transport maze but that doesn't sound as good

holonomy can mean the change of orientation when vector are transported along a path on a manifold

it has several meaning in maths

@leslietownes you mean write to the insurer, or write to the city indicating that i will contact the insurer?

i would contact the insurer directly.

if they are doing something illegal for the city, like renting out an apartment or something, go to them too. they get right on that.

another one being solution to a linear differential equation with polynomial coefficients

the problem is, you probably have no idea who the insurer is.

just got an update, the berkeley evac order is still in effect

you mean the city insurer? i can find out. i have people inside :-)

oh, and he made an dodecahedral version: youtube.com/watch?v=wzjUTPCAF4Y

driveway connections

whoever owns the policy that would be drawn upon if someone gets hurt by falling branches. it's probably not a municipal entity. you could maybe find it in a records search for the property, as they may have touched documents relating to financing the purchase of the house.

@leslietownes good suggestion, i never thought of that. thanks. the city of albany will be out looking for you

:D i should put together a one hour lecture on this and sell it to people.

i've done this before. an inspector for hte insurance company comes out and demands that a homeowner fix things. it's humiliating.

Does anyone here know about NV Centers?

that's more of a physics question

OK

what do you want to know, though?

If nitrogen-vacancy centers can be biologically produced (i.e., by living organisms)

the usual examples of NV centers are in diamond, so that seems unlikely

Yes, I know -- it's typically used a magnetometer or for other applications, but I was just wondering about biological applications.

Well, thank you anyways @Semiclassical

google does give some biological applications, but they seem to be in the vein of introducing NV centers to a host

not that the host would naturally contain diamonds

Yup, it has been used to detect tumorous areas in the human body, for instance

still waiting for our tsunami here

Get out your scuba gear.

i'm hiding in my campor trees, sorry Albany's camphor trees

Sorta like the beginning of Winnie the Pooh. You'll be attacked by a swarm of bees and Christopher Robin will need to lend you his umbrella for you to escape.

since participating in this chat room, whenever i type 'x' in normal conversation, i instinctively add \$ signs.

one of the camphor trees has a deep hole as such, and i used to check it every now and then. once a friend hid a bottle of wine for me, but someone else got to it first

Everyone always climbs up trees looking for spare bottles of wine.

in albany apparently

such is the tsunami risk, i suppose

@TedShifrin wine doesn't grow on trees?!

Well, vines.

our neighbor has grape vines in their back yard. Rats keep them from getting all but a handful each year.

@robjohn by any chance do you have a reference for the Riemann Stieltjes integration by parts theorem used in math.stackexchange.com/a/3551095/27978

Is that not in Rudin when he does RS integrals?

my library is full of holes.

@TedShifrin in PMA?

i think i have a pdf version, lemme see.

No, he does IBP after RS integrals. Maybe an exercise?

i will check. i spend a few unsuccessful minutes trying to prove the result without the RS IBP

Nope, doesn't look like it.

discontinuities always make me nervous :-) perhaps that explains my discrete fears

Oh wait. It's exercise 17 in chapter 6.

@copper.hat not really that goes beyond the Wikipedia article. I don't like ending the intervals on points of discontinuity, so I limit the endpoints from one side or the other.

Will check, i need to restore my pdf collection from disk unfortunately

See also this post.

Now I'm copying leslie's tricks.

@robjohn its a lovely application, but i like to be able to do it without such tools as well.

@copper.hat Do what?

what happens is i prove it at some point in my life but then promptly forget the details of the assumptions. sometimes texts have implicit assumptions mentioned earlier in the book, and i prove stuff from the stated result not remembering the implicit assumptions. so i like to have a known, reliable reference for a result that i will add to my little repertoire.

@robjohn the equivalent of line 2 -> line 3 with the RS IBP in the Euler constant expansion result. I have not spent much time (yet) on it.

@TedShifrin thanks for the reference.

Yeah, it's hard to remember technical details in things that we do not use frequently.

Thank goodness I thought to scan my more important lecture notes before I shredded everything upon my retirement.

I also got rid of the Baby Rudin that I bought in 1971 as the text for my real analysis class. It was an early edition.

It was the only book I ever owned in which I made notes in pencil (because Rudin's exercises were so tricky). He added a number of hints in the subsequent edition.

i am not sure at what point i moved from treating text books like royalty to actually pencilling in notes. i suspect Moe's Books helped.

@copper.hat That step is the one key line, and I never felt comfortable just ending the intervals on points of discontinuity. If we are using $\lfloor x\rfloor$, The RS integral includes the upper bound, but not the lower bound, if I remember correctly. I want to include both endpoints.

@robjohn yeah, i get the $n^+,1^-$ parts, i just thought it would be a trivial exercise to do it another way.

@copper.hat without RS or without the limits?

without RS

@Semiclassical the graph theory aspect is very cool indeed, I hadn't thought about it

that is the first time i encountered a RS IBP that added something non trivial (to me), hence my preoccupation.

ikr

especially for the dodecahedral one

the prices of these is ridiculous tho

i hadn't looked lol

@copper.hat Apostol's mathematical analysis

@Koro Thanks!

The theorem is this: If $f\in RS(g)$, then $g\in RS(f)$ and the following holds: $\int_a^b f d(g)+\int_a^b g d(f)=(fg)|_a^b$

RHS is: $(fg)(b)-(fg)(a)$

Are they the only conditions on $f,g$?

You need parentheses in that, @Koro.

any right/left continuity, bdd variation etc sort of thing?

what does RS mean?

copper: no BV, no continuity.

only RS.

just seems too good to be true

Astyx: Riemann Stieltjes

copper: IBP stated above is true.

@copper But saying $f\in RS(g)$ means that $f$ has to be continuous where $g$ jumps and maybe more?

@TedShifrin yes, i suppose i am looking for a result that spells that out. but i can live with this version.

just like i avoid any proof that uses nets

I have never netted, either. But Apostol surely explains earlier what it means for $f\in RS(g)$.

My Apostol is likewise long gone.

@TedShifrin yes, i am sure, but when i want to apply the result i don't want to have to recurse down to some other level because i am basically lazy.

@Koro do you not need some limits there for the $a,b$ stuff?

Note: Apostol doesn't use the symbol RS(function).

a<b and f is defined on [a,b].

does $f$ have continuity properties at $a,b$?

surely it must

You mean: is f continuous at a and b?

did you mean $a< n \le b$ or $a \le n \le b$, that might be relevant

Certainly right-continuous at $a$ and left-continuous at $b$?

@copper.hat $a<n\le b$

@TedShifrin i would have assumed something along those lines

but want to see it in the text :-)

ok, i will have a cup of tea now

It is almost lunchtime.

No continuity has been mentioned in the hypothesis.

@copper

Using that identity, we can re-write Robjohn's How to Answer (math.stackexchange.com/questions/3550990/…) differently, I think.

the $a<n$ bit bothers me, are they assumed to be integers and it so why not write it as $a+1\le n\le b$.

Copper: that means $n$ starts from $[a]+1$, where [.] is floor function.

it is the details that cause me loss of sleep.

if $f$ assumed differentiable?

or ac?

Definitely not assumed to be integers!

For the summation formula, yes and in fact f' is assumed continuous there.

For IBP, no.

@Koro I posted a link to Robjohn's answer but God knows how it changed to "How to answer..."

You mean math.stackexchange.com/a/3551095/27978 ?

I mean this: math.stackexchange.com/questions/3550990/…

Using the sum formula above, $1-$ and $n+$ also make sense. :)

So $\int_a^{b^+}$ means $\lim_{c\to b^+} \int_a^c$, I presume.

I would imagine.

This is complex enough as it is. Don't bring in imaginary numbers.

hamilton's infuence

then again, Boole taught at my alma mater, so one would imagine a simplifying influence there

I prefer boules to boole.

I should go eat lunch and take a pun time-out.

@TedShifrin that is what is meant in my post

We figured, @robjohn. I figured it was analogous to the notation $f(b^+) = \lim_{x\to b^+} f(x)$, etc. Just not altogether commonplace.

yeah, it's not too often that inclusion/exclusion of an endpoint matters in an integral

in this case, it does

it is a beautiful computation, but the details are a bit of a highwire act for someone like myself with limited memory

With RS integrals, of course it can :)

The other posted answer there looks even crazier.

@copper But, still, more your bailiwick than algebra :)

@TedShifrin indeed, hence my slow moving desire to up my algebra game.

once past $(3)$ the integrals no longer have point masses, and all returns to normal.

robjohn is the only one I know who annotates every line of his proof. I think I've done mild annotation only a few times in all my posts.

certainly justified for that answer

Yup.

Unless you use \intertext and put the annotations in-line ... but I don't think MathJax allows that.

his mathjax-fu is way beyond my yellow belt level

Well, my LaTeX skills are pretty solid, since I made it through typesetting four books, but a lot of the LaTeX schemes don't work in MathJax.

Does it have any commutative diagram implementation yet? I recall it didn't have tikz-cd and people would use arrays with arrows for that purpose.

mine are copy & paste skills. while miles better than troff, i did not like the xml like way that Knuth setup the 'blocks'.

copper: i think we can proceed from here and write a complete answer.

@Koro it takes me longer than that to justify that result

since $f$ is smooth the limits do not matter so much.

it's similar to what Robjohn gets. But I'm just trying to use the summation formula.

my experience in life is that wherever there is a discontinuity there is a lawyer waiting.

But here, we have no discontinuity except at $0$ and that's why we are taking $a$ away from $0$. :)

dang, i seem to have deleted the vm with my pdfs :-(

Guys, do your eyes also hurt if you spent lot of time reading from your screen?

I am getting used to buy all my books in ebook format instead of paper and damn... not sure if it was a good idea at all

Also, i guess it doesn't help to have two monitors with different resolutions

i prefer paper

@anakhro Spivak created AMS-LaTeX decades ago to have fancy CD implementations. It never caught on too much. Some CD stuff in LaTeX, but for diagonal arrows one has to import to something else to draw the arrows. I never learned tikz.

Someday I have to leave lyx and jump to real tex :D

Tikz is fantastic, Ted. You should fiddle around with it some day if you enjoy that sort of thing (fiddling around with LaTeX packages...I guess it's not everyone's thing :P)

I found it easier to draw my diagrams in Illustrator and import as eps. Sadly, I no longer have Illustrator to work with, but I haven't had to learn anything new yet.

LaTeXIt is a wonderful little app that generates LaTeX in whatever font to paste into such documents.

Inkscape is a good free vector drawing program that works well with LaTeX. You can output the svg_latex file or whatever to ensure the math gets rendered as such.

A little bit of a learning curve, though, as with any program.

Yes, I have Inkscape. But things I could do without thinking in Illustrator I have to learn all from scratch.

I told robjohn about it months ago, too.

I suppose you could buy illustrator if you still wanted to use it. :P

there's a nice one online for making diagrams which can be converted to Tikz

right, mathcha.io

I bought it long ago. Now they make you pay monthly rental costs, something like $30 a month. No way in hell I'm doing that.

The MacOS improvements over the years rendered all Adobe and Microsoft products I actually owned worthless.

I had to buy a new version of Mathematica for the same reason.

Yeah, I am not too fond of the subscription style systems companies are using these days...

I would state that categorically with profanity inserted.

You mean you don't like the AMS magazines very much?

@TedShifrin +1 to that sentiment

My AMS membership ended many years ago. I stayed in the MAA, but stopped paying AMS dues.

But that is not software. That's different. I do subscribe to (on-line) newspapers and several (paper) magazines.

Do you do a lot of mathematical reading these days, Ted? If so, about what lately?

No.

Now it's just cooking.

AMS had a huge jump in fees a few years ago and that meant no hard copy Monthly so I quit.

I think I had electronic subscriptions to journals the last number of years, anyhow.

Which means that we don't sit down and browse through them so much ...

for me nothing will replace libraries or hardcopies for browsing and getting overviews

i used to to go evans library and just browse maths books at random

stretching a bit here, but i am taking weak closure psqs to be convex problems