@Evinda Instead of 2 initial conditions, think of it like this: for each $x_0$, you have a 2nd-order ODE in $t$, and you need two initial values $u(0,x_0)$ and $u_t(0,x_0)$ to solve it. Since you need 2 initial values for every $x_0$, when you vary $x_0$ you get two functions as your initial conditions.

in a given function, f(1) = 4 and f(3) = -6. Does a zero exist in the interval [1,3]?

I was trying to set up an equation but aren't there like a lot of possibilities

@MATHASKER If it's a continuous function, yes. (Look up the Intermediate Value Theorem.)

If it's not required to be continuous, then one can come up with counterexamples easily enough.

i don't know if its required to be continous or not, i thought i had to like write down an equation cause they gave us like a points

Eh, there's not going to be a unique function unless you add some more info.

For instance, you could take f(x)=mx+b. That'd have a unique solution. (Two unknown parameters - two known coordinates.)

What kind of class is the question from, @MATHASKER?

what do you mean by class

But you could also do a quadratic, a cubic, a quartic, ... and all of those would have many different answers.

Where did you get this question?

math class pre calc

yes thats what i thought there are so many option

precalc ... what have you been studying?

And that's only if I stay within polynomials. Bottom line is that there's not anywhere near enough info as given to say where the root is. Nor am I guaranteed to even have a root without the assumption of continuity.

just how functions behave

have they talked about continuity? (that usually isn't discussed until calculus)

yea essential, removal and jump right

those are discontinuities

have they talked about the Intermediate Value Theorem if you have a continuous function? Your question doesn't mention continuous ... so you cannot assume it's continuous.

As an example of why you need continuity, take $f(x)=(x + 3)/(2 - x)$.

no i don't think we have discussed intermediate value theorem thing

That satisfies the conditions given (f(3)=-6, f(1)=4) but doesn't have a zero in [1,3] because of how it blows up at x=2.

OK, but you have talked about asymptotes and discontinuities of different kinds.

@Semiclassic: It's also not defined at 2. I wish we knew the exact question and the exact intention of the teacher.

True.

I wanted to give an example that didn't require a piecewise definition.

@MATHASKER: Does your question say that the function needs to be defined at every point of $[1,3]$?

This question annoys me because it's not clear what the teacher expects students to know.

Well, it may be clear in the class ... just not to us.

Which might be because of the teacher, or because the problem hasn't been stated precisely...

no all it says is In a given function, f(1) = 4 and f(3) = -6. Does a zero exists in the interval [1,3]?. Defend your answer

If that's really all that's being said, I'm definitely a bit annoyed.

OK, @MATHASKER. So the answer is: Not necessarily. Now you write a paragraph.

But then I'm also grumpy for other reasons ATM.

Did you just have that birthday, @Semiclassic?

Nah, still 6 weeks away.

i just said it depends on the function, if its continous yes its going to have zeros if its not then it might be a hole and not a zero

Oh, damn, good thing I'll be gone when it's 6 weeks. You'll be really grumpy then :D

But, the prof just sent out the draft for his final half an hour ago. Along with the remark: "For comment...sorry at the short notice, but it has to go [to the front office] Monday by noon"

@MATHASKER: OK. Or there might even be a vertical asymptote somewhere in there.

@Semiclassic: Can we assume the prof has worked the exam carefully?

I always wrote an answer key before I thought about duplicating

Especialllllly in large lecture classes.

We'll see once I've gone through it. I just saw the email.

Yeah, for that function you do, @MATHASKER. What if I give you x going to 1 and $$f(x)=\frac{x^2-1}{x-1}?$$

i i guess plug in 1 again

well, try.

oh undefined

so there's an asymptote there

so maybe dne?

well, maybe, maybe not.

you could try putting in numbers very close to 1 and see what happens ... or you could think about algebra for a moment

oh umm i guess thinking about algebra sounds good but idk lol

Try the first one, then.

note that you should usually thinking about that when you get 0/0

u can't divide by zero doe right

right

It's true that the function doesn't exist there.

What you need to find out is whether the limit exists there or not.

limit like if its still continous or not

not continuous because it's not even defined at $x=1$.

Hey all, quick question regarding derivative quotient rule for calculus. I often see the rule written in different ways. Whether its gf'-fg' / g^2, g'f - f'g /g^2, etc. Is one way the correct way to do it? Or will you always get the same answer each time? I personally like saying it as f'g - fg' / g^2.

but that's different from asking whether it has a limit

whoa ... one is right, one isn't.

Which one is right?

Check thy signs.

the way to remember it is that the g' term has to be negative (because the derivative of 1/x is -1/x^2)

so i its 0/0 the limit doesn't exist?

0/0 is not enough to conclude that.

@MATHASKER: But the limit does exist for the one I gave you.

oh..how

I suggested you look at algebra. What can you do with $x^2-1$?

(x-1)(x+1)?

Good, @MATHASKER. So now look at the question again.

so thers a hole?

if theres a hole thers a limit?

Hey there chat!

Hi, Demonark.

Slow down, @MATHASKER. So how can you rewrite $f(x)$ when $x\ne 1$?

because x-1 in numerator and denominator canceled

and you get ...

umm x does not equal 1?

no, the function made no sense when $x=1$.

yes but i don't know how i could re-write it

You just told me you canceled $x-1$. What do you have then?

just (x+1)

so now its a linear function? with holes at positive 1?

OK ... and what happens to that when x goes to 1?

a hole maybe?

wait let me go to desmos

Right. So the graph of $f(x)$ will be the graph of $x+1$ with a hole at $x=1$. But what is the LIMIT?

You can draw a darn straight line, @MATHASKER:)

oh ya lol desmos is faster haha

limit...

you better learn how to do basic graphs without computers

i still donn't know cause there is no value for 1

hahaha ya i still need to practice a bit

there is no value, @MATHASKER, but when you say there's a "hole" in the graph, where is that hole drawn? That's telling you what the limit is!

at the x value of 1?

so the limit is y = 1...?

Look closer.

What does x+1 do when x gets close to 1?

nothing..

Oh come on.

like a value is just not there

For x+1?

i mean it still keeps on going after x is 2 and 3 and so on

Reminds my of the unit where I sketched rational functions

We're looking microscopically close to x=1. Forget all the rest.

i still can't figure it out even after zooming in

As x gets closer and closer to 1, does x+1 get closer and closer to anything?

Okay, now I'm officially annoyed.

One of the topics which we didn't cover in lab/discussion/HW was geometric optics.

closer to 1?

Say x goes from 0.9 to 0.99 to 0.999.

Definitely no stuff with lenses and mirrors. (I see it in his lecture notes, but only just the formulas.)

Then what is x+1 in these cases?

to 1.9, 1.99, 1.999

and 2 out of the 20 multiple choice problems are geometric optics. @ted

@MATHASKER Now what value does that seem to get very close to?

2

so is the limit 2

Exactly, so the limit of x+1 as x goes to 1 is 2. QED.

Good, @MATHASKER.

Lovely, @Semiclassic. Did he cover it in lecture in detail?

Good for you too, @jamesbond.

Somehow, Mr. Bond, I'm not too worried about your understanding of the limit concept.

But thanks for helping. Actually, I need to head out. So ... bye, all.

Bye.

@Ted Based on his notes, I don't believe so.

I'm emailing him about it now.

Perhaps he'll have more family emergencies this time, too, @Semiclassic. Good luck.

ohhh ok so for (3x+4)(3x-4)/(3x+4) as x goes close to -4/3 the limit is negative 8?

hnnnnggh

He probably just stole someone else's exam, @Semiclassic.

Right, @MATHASKER.

@Semiclassical What class is this for?

Considering that the problems I'm seeing look awfully similar to the ones that were in the 1302 exam last semester...yeeeaahhh.

ohhh ok thanks for helping @TedShifrin and @JamesBond

(this is 1202, so it really shouldn't be the same.)

Bye all. 'Til next time.

second-semester intro physics for pre-med and bio majors

Gotta love lazy ass professors.

To be clear, I'm not objecting about geometric optics being in a course like that. It's not an unreasonable subject.

It's OK to use other's notes to teach as long as you read through it yourself first, which is not the case it seems here.

But I do object if it shows up on the final when it wasn't presented in class.

So yeah lazy instructor.

I agree on both counts, @Semiclassic.

Hi, if ln(A) - ln(B) = ln(A/B) can this still be done if they have different coefficients? eg: 1/2*ln(A) - 1/3*ln(B)

Soooo email time.

I met lots of lazy instructors in undergrad. Maybe they only care about their research

@WDUK You could use A*ln(B) = ln(B^A) for that

Eh, this is a prof who is going into staged retirement.

So I think he just doesn't care at this point in general.

By the way how many people actually self-study from Fundamentals Of Physics 1 and 2 by Shankar? It's the Yale Open Courses.

@LegionMammal978 not sure i understand how that equates to A*ln(B)?

@WDUK In your case, 1/2*ln(A) - 1/3*ln(B) = ln(A^(1/2)) - ln(B^(1/3)) = ln(A^(1/2)/B^(1/3))

@blue How does it compare with say Feynman lectures, in your opinion?

@LegionMammal978 ohhh that makes sense ! thank you :)

@blue Yeah.

@blue Do you have the Shankar books too?

i like the MIT video lectures

I watched the MIT calculus ones, good but nothing really special.

@blue Ah I see. Well they are now published as books!

3blue1brown just started a calculus series perfect timing for me

I've got Shankar's quantum mech book. That's more a grad student book, though.

Well, as many know, I always study from real books. =)

PDFs and videos don't work for me.

@Semiclassical He has 4 books, the 2 basic physics, this one and one on math.

I have now watched all Bond movies except for the Daniel Craig ones.

I think the movies just go downhill from 1962 till 2002.

I have a question about end behavior for lets say (x-2)(x+1)/(x-1) the end behavior i put down was limf(x) as x goes to negative infinity = pos infinity and limf(x) as x goes to postivie infinity = what..because as the function goes to the positive side it goes near 1 and then jumps to go to like the positive infinity

how can i write it in interval notaion?

@JamesBond Makes sense. One thing I like about Shankar's QM is that he includes some stuff in the closing chapter on path integral stuff e.g. coherent states.

Plus, no terrible philosophy. (My undergrad quantum had that and I hate hate hate that part of it.)

I just learned that MathOverflow calls us "math underflow" :(

pfffffft

Hi! How many orthonormal basis $\mathbb{R}^n$ has?

uncountably many

You can get a orthonormal basis by rotating the standard one, right?

or reflecting, yes

But what about R? Just {1}, right?

and {-1}

Ohhh, yes! Thank you :)

indeed the space of all orthonormal frames of R^n is itself an n(n-1)/2 dimensional space

o.O

with two connected components (corresponding to the two possible orientations of space)

I have another question :$ if V is orthogonal to W, is $V^{\perp}$ orthogonal to $W^{\perp}$? I think not, but I can't find a counter example :(

it's clearly true in 2D. is it true in 3D?

The orthogonal complement of a line is a plane, so it seems to me that it is true :/

I can't find a pair that does not work

besides {0} and all of R^3, the two types of subspaces of R^3 are lines and planes. a plane cannot be orthogonal to another plane, so there are two types of pairs of orthogonal (nontrivial, proper) subspaces: two lines or a line and a plane. as you just said, if a line and a plane are parallel, then their orthogonal complements (a plane and a line respectively) will also be orthogonal. but what about if two lines are orthogonal - are their orthogonal complements still orthogonal?

OHHHHH

THanks so much, that was sooo clear.

Hello. Is the spectral projection of a point the same thing as a point spectrum?

@IcedPalmer what context are these from?

there are psychological consequences to adapting the scientific worldview

mathematicians are very adept at stripping meaning away from symbols. I think this is why there are so many schizophrenic (finding connections of meaning where there are none) mathematicians.

An amusing thought:

Since I know basically nothing about schizophrenia beyond what I know culturally, the word itself is basically a symbol for me.

but you interpret meanings all the time

schizophrenia is where that goes haywire

leading to paranoia, etc

My point is merely that trying to talk about schizophrenia in that way risks itself being a stripping away of meaning from it.

i.e. of attempting to interpret a meaning onto something which has its own definition.

i think I see

good, because hell if I can actually explain it.

@BalarkaSen anything to add here: mathoverflow.net/questions/269250/connectedness-in-the-plane

?

Hey there everybody!

@Forever Not off the top of my head at this moment but I'll add something if I do

Yo @Eric

RIP France

Hey @Soumyob, @Mike, and @s.harp!

hey @Daminark!

Lol seems like you preferred Le Pencil? :P

yussir

I see

people always say that Le Pen appeals to the fears of the masses, but nobody minds it when Merkel tells Britain that the EU will end all trading with it if it exits. I mean that's totally not blackmailing right?

with politics we get what we deserve

I try not to worry about it

I have never heard of Merkel saying anything of the sort to Britain, I have heard them warn Britain that it stands to lose more than the EU but that's it

I mean tbh I don't often engage in politics, on general principle I roll centrist in the absence of strong beliefs to the contrary (which aren't terribly common)

@Daminark when hoisting the British flag in the capital of Britain itself offends people of a certain background then I think there's reason to have a strong belief against centrism

I mean what you described there is pretty non-center

@Daminark centrism supports the immigration of the people of that certain background with open doors, so it is centrist

at least according to the centrist version of Macron

but anyway, I'll leave it here. Time will tell what the reality is. Unfortunately it might be too late by then

To be fair American center is somewhat more conservative

Also, offense of that sort seems to be less... serious? Like yeah, sure, oh well, such is life

The main thing I'm iffy about with this Le Pen figure is that I generally don't think emotions/feelings should play much of a role in policy, so using fear seems... I dunno, out of place?

"Somewhat" lol

This being said I have an exam to go to - bye !

Anyway, things always work themselves out somehow so I'll just roll with it and get back to my stuff

Hi @Astyx

Bye @Astyx

deletes Paris from bucket list of places to visit

either I will continue to practice mathematics, or I'll wake up in a forced labor camp. meh

Hey @Alessandro!

Hi

think about it

Huh

so if you employ 10,000 people, about 100 will do over half of the work

that's what happens

like if you play monopoly, eventually one person ends up with all the money, while everyone else goes broke

doesn't matter who plays it or how many times

I heard monopoly was invented by a socialist (or communist?) to try and get people to dislike capitalism

@ForeverMozart It's actually a circle, so overflow=underflow, lol.

@SoumyoB First, news writers often paraphrase what people actually say or try to sensationalise things, and we don't know what actually happened. Second, we can't judge a person's intentions simply from a single sentence or so.

@Daminark Then bored people must have invented monotonic functions.

@JohnJiang functional analysis

Is there an example of polynomial in $\Bbb Z[X}$ that factors in $\Bbb Q[X]$?

Most polynomials that factor in $\Bbb Z[X]$?

*$\Bbb Z[X]$ hello eric

$X^2+X$ is a polynomial in $\Bbb Z[X]$ that factors in $\Bbb Q[X]$ as $X(X+1)$, I don't understand your question

it's demonstration of Gauss lemma

Gauss lemma says that polynomials irreducible in $\Bbb Z[X]$ are irreducible in $\Bbb Q[X]$ (the converse is false)

Oh!!!

facepalm :S

@AlessandroCodenotti The converse is false?

$2X-2$ is irreducible in $\Bbb Q[X]$ but not in $\Bbb Z[X]$ @steamy

Ummm

As far as I know, irreducible means you can't write it as the product of nonconstant polynomials

Doesn't $a$ irreducible mean that if $a=bc$ then one of $b$ and $c$ is a unit? ($a$ nonzero non-unit)

hi

In general, yes; but for polynomials I always used non-constant...

Although, well, I may just be thinking of polynomial rings over fields, in which both definitions agree

And non-monic polynomials are ugly anyway

hi

I think you can get an iff in Gauss lemma if you consider only primitive polynomials in $\Bbb Z[X]$

@SteamyRoot good point

I walked a kilometer and it's burning hot outside

feel ill man

Hello @BalarkaSen, LOL.

@BalarkaSen I believe you use British spelling, so it should be kilometre.

@BalarkaSen What you need to do now is drink lots of water, like a whole litre of it.

@JasonBourne hi. nah I use american spelling (although i should use british spelling)

@BalarkaSen I miss Chris's Sis. I pinged her yesterday after discovering her new username... I also hope that you become the next Ramanujan of India. =)

i'm fine w being a nobody

For me, I am still trying to get better, still hanging in there.

You might be as sickly as Ramanujan @BalarkaSen

nah i don't try to survive on vegetables

Haha

When I answer a simple question, I get no upvotes. When a high rep user does so, he gets five upvotes.

Well why do you think they have a high reputation, they get five votes

Nice proof @paul, lol.

@JasonBourne We meet again.

@ParthKohli Hello. I remember you used a very bad dictionary to correct my spelling and your dictionary was wrong, a long time ago.

@JasonBourne Something along the lines of practice vs. practise.

@ParthKohli In AmE, practice is both noun and verb. In BrE, practice is noun and practise is verb. QED.

@Waiting I am waiting for you to chat! I changed my username again!

@Daminark I prefer pencil to pen for doing math. I also prefer not to use mechanical pencils.

If I rotate $(u_1,u_2,u_3)$ an angle $\pi/2$ about the $x_3$-axis, what is the resulting vector?

Should be $(\pm u_2,\mp u_1,u_3)$ with the sign depending on the sense of rotation

eg (1,0,0) rotates to either (0,1,0) or (0,-1,0) @Lozansky

How did you come up with that, @Semiclassical ?

Multiply the rotation matrix with the vector?

Draw your vector in the $x_1x_2$-plane and then draw the vector which is perpendicular to it

If you are semiclassical, then you are semimodern.

Sure.

I have no idea what classical and modern mean though in the context of mathematics.

Well, the word is borrowed from physics, where the distinction is really between Newtonian vs quantum mechanics

I see. I need to study some physics.

@Semiclassical @SteamyRoot Thanks, I'm an idiot

More broadly though I tend to think that how I think of problems is a mix of classical and modern mindsets

Is there a book covering mechanics all the way from high school physics onwards?

Penrose's book "The Road to Reality" comes to mind, though that also includes the relevant math

It doesn't sound like a mechanics book.

It's not a textbook, if that's what you mean

There are many calculus books now with no colourful pictures, and I am still looking for a basic physics book with the same specs.

I know about Shankar and Feynman, but somehow neither is satisfactory for me.

Do you know of any?

Hmm, not really. Physics needs some level of diagrams

Diagrams can be simple and not extremely colourful, and the photos are quite useless.

O'Hanion, maybe? (Not sure of spelling, will check)

Like photos showing people running in the mechanics chapter, nuts...

No apostrophe in there

I think you'd want to look for an older edition, tgen

Ohanian

@JasonBourne Many people may disagree with me, but, honestly, if you want to have a deep understanding of fundamental theoretical physics, read the works of mathematicians working in these fields. The amount of mathematics needed to understand fundamental theoretical physics is extraordinary. Learn as much mathematics as you can--in fact, just get a degree in mathematics and pick up the physics along the way.

Right

amazon.com/Physics-Hans-C-Ohanian/dp/0393957489/…

@user193319 I see. Are you interested in physics too?

@JasonBourne For example, here is a huge 40-year old open problem in the field of von Neumann Algebras: en.wikipedia.org/wiki/Connes_embedding_problem This is not only a huge problem in operator algebras, but an extremely important problem in quantum field and information theory.

Not entirely in agreement with that, but I'm not awake enough to debate it

@JasonBourne Reaad about Connes' Embedding Conjecture and Tsirelson's problem to get a feel for what fundamental theoretical physics is like. It has to do with quantum correlations and quantum entanglement. To answer your question, yes! The reason I study math is because I want to better understand physics.

QFT is weird in general

For one because how a physicist uses QFT relies so much on intuition built up from other physics courses

And the overall attitude of doing whatever 'works' to get meaningful calculations

Here is a nice little paper of Tsirelon's problem: arxiv.org/pdf/0812.4305.pdf Note that Connes' Embedding Conjecture is logically equivalent to Tsirelson's problem, which I believe was shown by Ozawa: arxiv.org/pdf/1212.1700.pdf These are really nice papers.

Lots of wickedly complicated math involved in this physics problem.

My boy is wicked smart, lol.

@JasonBourne Are you referring to me? No! I am very far from it! I am just very dedicated. Quite frankly, most of the stuff in those papers I linked in my previous post I don't quite understand yet. I suppose I just love being confused; that confusion serves as great motivation. Whenever I get confused by some intellectual puzzle, I think to myself, "I have to understand this--I must." It seems that most people are put-off by very hard problems, but I drawn to them.

@user193319 Yes, I was referring to you, but only in a half-serious manner. That line also appears in my favourite movie.

@user193319 Good Will Hunting, played by the same actor who played Jason Bourne, that is, the handsome and kind Matt Damon.

I know what a bounded function is. abs(U(k,y))<=C, where C is smaller than 1 but why here U (k, y) → 0.