Mathematics - 2016-01-05 (page 2 of 3)

Andrew Thompson

17:01

I'm unable to tell whether they are easier or harder than the Norwegian ones. Granted, had I prepared for a Norwegian calcexam and those integrals were thrown in my face I would likely not do too well, but if I knew the format beforehand I really don't know which is worse.

Andrew Thompson

17:17

Ah, just found the calcexam I took. Translation of the last question:

"Let $n \ge 2$ and let $P_n$ be the $n$th Taylorpolynomial for $f(x)$ about $x = 0$. Explain why $P_n''(x)$ is the $(n-2)$nd Taylorpolynomial for $f''(x)$, and explain why it follows from Taylor's formula with error term that there is a function $g(x)$ that is continuous at $x = 0$ and satisfies $P_n''(x) - P_n(x) = x^{n-1}g(x)$."

Mike Miller

Nobody in our classes would get that.

Andrew Thompson

To be sure I'm not creating a skewed perspective on our exams; that was the most difficult question on the exam.

The full exam in Norwegian: math.uib.no/adm/Eksamen/content/MAT111/…

Ted Shifrin

I expected this sort of thing in the Calc Theory class, which is now defunct. The top half of them could do that.

Owatch

Hello.

Andrew Thompson

Calc Theory? A substitute for calc1 for the good students?

Ted Shifrin

17:25

People wanting a hard, proof-based class — maybe 20 students max — as opposed to engineering calculus.

Andrew Thompson

Ah, I see. We only offer one calc1 course on the form provided in the link.

Mike Miller

Tip of my brand new G&H is soggy from the rain. Sigh

Ted Shifrin

You need special accommodations for books, Mike.

Mike Miller

Backpack sealed tight. But you have a point.

Andrew Thompson

Mike, are there admission fees for undegraduate applications?

Mike Miller

17:30

I think so.

Andrew Thompson

Hm. UCLA has an admissionrate of 17%, so one would expect that the admitted students are decent.

Superstar Monica

@robjohn Happy New Year!

Ted Shifrin

Yuppers.

Mike Miller

one would expect. I certainly did.

Andrew Thompson

Where did you do your undergrad?

Mike Miller

17:32

The onlh difference I see in these students and the ones at my undergrad are those in the upper echelon, the best people in each class. The average student as far as I can tell is pretty much the same.

robjohn

@SuperstarMonica Happy New Year!

Ted Shifrin

Off to meet a former student for lunch. He can teach me econ

robjohn

@TedShifrin are you back in the hood?

Mike Miller

In fact, because upper div math at my undergrad seemed kuch more self-selecting than here, the upper div students here seem worse.

@AndrewThompson: Santa Clara university

Ted Shifrin

hi/bye @robjohn. No! Ann Arbor, to ATL tomorrow

HNY

robjohn

17:34

@TedShifrin HNY and bon appetit!

Andrew Thompson

I only know Santa Clara exists because of Halmos.

Mike Miller

Same with everyone else. And the Putnam.

Andrew Thompson

That's the competition for undergrads?

Mike Miller

Yes. We run the grading.

Andrew Thompson

Ah, cool.

Mary Star

17:51

I found the following:

$$\kappa^2 \sin^2 v \left [((p^4+q^4)\sin^2 u\cos^2 u+p^2q^2(\cos^4 u+\sin^4 u))\cos^2 v+r^2\sin^2{v}(p^2\sin^2 u+q^2\cos^2 u)-(q^2-p^2)\sin^2\cos^2u \cos^2v\right ]-\kappa \frac{pqr\sin^2 v}{\sqrt{\sin^2 v(q^2r^2\cos^2 u+p^2r^2\sin^2 u)+p^2q^2\cos^2 v}} \left [(p^2\cos^2 u+q^2\sin^2 u)(\cos^2 v+1)+r^2\sin^2 v)\right ]+\frac{p^2q^2r^2\sin^2 v}{\sin^2 v(q^2r^2\cos^2 u+p^2r^2\sin^2 u)+p^2q^2\cos^2 v}=0 $$

Can this be correct? Can we simplify it? @TedShifrin

Andrew Thompson

18:22

Due to pollution my city is starting a system starting tomorrow where you can only drive half of the days - depending on the last number on your car registration (even/odd). The county is having an online Q&A regarding this, and one guy of course went: "Climate change is a liberal hoax and so I will drive regardless." The guy replied "I'm a liberal so I will fine you regardless."

Vedant Chandra

@AndrewThompson They started that exact scheme in my city just three days ago. It's barely made a difference though, and I've seen hundreds of people simply ignore it.

My city is New Delhi by the way, with the proud badge of being the most polluted city in the world.

Superstar Monica

How would you say in English that a term can be absorbed in/by/into a series? Absorbed by the series? In(to) the series?

Huy

gets eaten

can you give an example?

Superstar Monica

18:37

@Huy got absorbed?

@Huy 'The term may be absorbed by the series.' My example is very complicated, but a simple example is this

$$\frac{1}{n}+\sum_{k=1}^{n-1} \frac{1}{k}=\sum_{k=1}^{n} \frac{1}{k}$$

In this case, absorbing a term in/into a sum (not sure how is this formulated in English).

Huy

hm, not sure how I'd say that, maybe someone else has a good idea

Superstar Monica

OK

Hippalectryon

19:01

@SuperstarMonica o/

Superstar Monica

19:21

@Hippalectryon o/ how is it going?

Hippalectryon

@SuperstarMonica Fine :-) and you ?

Superstar Monica

@Hippalectryon Great news then. Working on some fantastic stuff here. :-)

Hippalectryon

I bet :-)

Superstar Monica

@Hippalectryon :D

Adda - Lupii (Clanker Jones Remix)

►

@Hippalectryon ^^^ (not sure if this is for your taste)

Owatch

Don't think IBP is going to get me far with $\int ln(t)cos(t) dt$...

I can't eliminate any of the t's, or make a substitution.

Any hints to how to solve it. .

Hippalectryon

19:34

@SuperstarMonica nah, not really for me :P

Superstar Monica

Integrating by parts you get the sine integral.

@Hippalectryon hehe, OK. :-)

Semiclassical

mathematica gives as antiderivative Log[t] Sin[t] - SinIntegral[t]

Owatch

Well, I get an integral of $\int \frac{sin(t)}{t}dt$

Hippalectryon

@Semiclassical Yeah, that's what @SuperstarMonica said :-)

Semiclassical

so yeah, not much more one can do with that re: indefinite integration

Owatch

19:36

Which is where I was earlier.

Semiclassical

yes, well, i figured i'd include the boundary term :P

in any case, there's really no reason to think one can get a nice answer. best one can do is show that it's equivalent to another non-elementary integral

Owatch

So I'd be going back. Or, I'll get this using the other IBP option: $\int -sin(t) * (tln(t) - t) dt$

Which makes it even worse.

Semiclassical

well, what are you trying to show in the first place?

Balarka Sen

@AndrewThompson I'd think snow and smoke will cancel off. Aren't you in Norway?

Owatch

Oh, you were talking to me?

I was just trying to solve it.

Semiclassical

19:40

because if your intention is to get something elementary, my point is that you probably can't

Owatch

Aw..

Andrew Thompson

@BalarkaSen No snow at the moment, even though its cold. The combination good weather + cold + a lot of traffic really ruins the air.

Semiclassical

any more than you'd hope to get an elementary answer for $\int e^{-x^2}\,dx$

Owatch

Is there no elementary solution for $\int \frac{1}{t^{2}} cost(t) dt$?

Balarka Sen

@AndrewThompson Ah.

Owatch

19:41

I started out with that.

Semiclassical

i don't know off the top of my head, but i doubt it

i'll ask mathematica to be sure, though

Owatch

:<

Semiclassical

and it returns -Cos[t]/t - SinIntegral[t]

which is to say, nope

Andrew Thompson

Owatch

Well, I don't know why a differential equations problem would give me that. So I guess I must have done something wrong earlier.

Andrew Thompson

19:42

My city recently, the fog is due to pollution.

Semiclassical

well, could be that the solution is in terms of the sine integral function, but it's probably worth rechecking what youv'e done

Balarka Sen

@AndrewThompson Yikes. Looks polluted indeed.

Owatch

What does Mathematica give you for $\int \frac{cos(t)}{t^{4}}dt$?

Semiclassical

((-2 + t^2) Cos[t] + t Sin[t] + t^3 SinIntegral[t])/(6 t^3)

Huy

@AndrewThompson: Bergen?

Andrew Thompson

19:52

Yup.

Owatch

That is very unpleasant.

Sigh..

Semiclassical

i think any indefinite integral of that kind won't be nice

Owatch

Well, I don't know why this question I'm doing would lead me to solve it.

Semiclassical

if it were $t^n$ instead of $t^{-n}$ it'd be fine

well, what's the question in the first place?

Owatch

I'm trying to solve $t^{3}\frac{dy}{dt} + 3t^{2}y = cost$, where $y(\pi) = 0$.

Semiclassical

19:54

okay. i'll plug it into mathematica just to see if it gives an elementary solution

actually. can't the left side of that be written as a total derivative?

Owatch

I put it into form: $\frac{dy}{dt} + \frac{3}{t}y = \frac{cos(t)}{t^{3}}$/

Which was, as the book indicates, the form I want it in: $\frac{dy}{dx} + P(x) = Q(x)$.

Semiclassical

ummm

your second term isn't of the form $P(x)$

it's of the form $y P(x)$

Owatch

Well. I am supposed to also find an Integration factor.

I don't know if that addresses that or not.

Semiclassical

well

in the ODE at hand, you actually started out with something that doesn't need an integrating factor

hence my remark about it being a total derivative

Owatch

I am also watched a youtube video, which has an example in which the same 'point' at which I am solving it has: $y' - \frac{1}{x}y = xsinx$. So I assumed it was okay for me to have a y there too.

Semiclassical

19:58

maybe so, but looking for an integrating factor doesn't mean putting it into that form and trying to integrate the RHS

Owatch

No, I did not do that for an integrating factor.

I used the equation: $I(t) = e^{\int P(t)dt}$.

Semiclassical

okay? but the integral you'd get would involve P, not Q, i.e. the 3/t not the cos(t)/t^3

Owatch

Which I then solved as: $e^{ \int \frac{3}{t} dt }$.

(To get 3t)

Semiclassical

that's not equal to 3t

Owatch

$e^{3 * \int{\frac{1}{t}} dt }$

$ 3e^{\int \frac{1}{t} dt }$

Semiclassical

20:01

false!

the second line, i mean

Owatch

:(

Semiclassical

i mean, is 2^3=3*2?

Owatch

Doesn't work like it does with ln

Semiclassical

to what are you referring? the identity is $e^{\ln t}=t$ not $e^t = te$

Owatch

I'm saying, I guess it doesn't work like it would with ln()

Where you are allowed to move coefficients to exponent spots.

Semiclassical

20:03

can you give an example of what you mean by that?

like, $n \ln t = \ln(t^n)$?

Owatch

$2ln(6) = ln(6^{2})$

Semiclassical

sure, that's fine. and if you exponentiate both sides you get $6^2 = e^{2\ln 6}$ which is valid

Owatch

Somehow though I could do that with $e$.

Semiclassical

ah.

Owatch

Probably because I don't know why it's allowed. So I mixed it up.

Anyways, given that's what was wrong there, it should actually be $e^{3ln(t)}$.

Semiclassical

20:07

right. and then the 'movement' you pointed out does work within the exponent, i.e. $3\ln t = \ln(t^3)$

Owatch

Ah. I didn't see that, so I get $t^{3}$!

Semiclassical

right. so your integrating factor is $t^3$...exactly what you divided out by

which is to say, the initial form you started with was already in the right form to be integrated :/

Owatch

oh..

Semiclassical

hence my remark about it being a total derivative: $\dfrac{d}{dt}(t^3 y)=?$

Owatch

genius

Semiclassical

20:12

one thing i should point out, though, is the presence of the integrating factor $t^3$ inside that derivative. the point of the method is that you don't need to be able to 'spot' the integrating factor, but rather can reduce it to taking an antiderivative

so showing that the integrating factor was $t^3$ didn't just tell you that your initial form was a total derivative, it also tells you that it's of the form I said above

if you can spot it by inspection, great; otherwise, you just need to set up an indefinite integral

Owatch

I am not sure I know why I am looking for an integrating factor to begin with. The book only explains that I want to use it to modify the LHS of the equation to ensure that it is the derivative of the product I(x)y.

Semiclassical

well, I(x) is itself the integrating factor

relative to the form $y'+P(x)y=Q(x)$ at any rate

bad typo right there

since then $\dfrac{d}{dx}(I(x)y)=I(x)y'+I'(x)y$ can match the LHS of that ODE

Owatch

The entire explanation for linear equations makes sense at first. I can see that by definition, a first order linear differential equation has to be in the form: $\frac{dy}{dx} + P(x)y = Q(x)$. But the example they use: $xy' + y = 2x$ loses me a bit. They say it can be written in the form $y' + \frac{1}{x}y = 2$. That is okay. But saying we can solve it by the product rule: $xy` + y = (xy)` $ confuses me. And so does the part saying we can re-write the equation as $(xy)` = 2x$.

Semiclassical

well, first let's check: $\dfrac{d}{dx}(xy)=x\dfrac{dy}{dx}+\dfrac{dx}{dx}y=x y'+y$

so the LHS is, using the term i gave above, a total derivative (i.e. it's the derivative of some definite function)

Owatch

Yeah.

Semiclassical

20:25

so your ODE is equivalent to $\dfrac{d}{dx}(xy)=2x$

Owatch

Okay. Yes.

Ah.

I see. It is just to isolate y.

Semiclassical

right.

Owatch

So they want a derivate that can be eliminated when integrating both sides, that modifies LHS so that it can be solved in that way.

Okay. That makes sense.

So I need to see, that $xy' + y$ is $(xy)` $ ? Or else I use integrating factor to make it so if it can be put in such a form?

Semiclassical

i'd put it a little differently: if you recognize that the product rule implies that, then you can immediately use it to find the general solution by integration

if you don't spot it, you can rearrange the equation into the "derivative + stuff" form and look for an integrating factor

Owatch

So this is only a specific case. I must look in general to see if the product rules implies that.

Balarka Sen

20:31

@MikeMiller Ok, here's a better more explicit example. $f_n(x) = x^n$ in $[0, 1]$ converges in the $L^2$ norm to the function which is $0$ on $[0, 1)$ and $1$ on $1$ (so, the delta function), which is not continuous. However, if I haven't mucked up basic calculus, this is Cauchy.

Motivated from @Semiclassical's nascent delta stuff.

Semiclassical

right. what you're hoping to spot is something of the form $I(x)y'+I'(x)y=(I(x)y)'$

(might as well stick with the same symbols as before)

Owatch

Okay. Thank you!

Semiclassical

if you can spot that by inspection, great; if you can't, then you can rearrange to the $y'+P$ form and see if an integrating factor exists

no problem

one thing i'd lastly point out. suppose they gave you, say, $x^{11} y'+x^{10} y$ as the LHS

Balarka Sen

Ok, fairly sure that works.

Semiclassical

in that case, it might well not be obvious that the integrating factor approach would work. but the moment you divide out by $x^11$, you get back the same form $y'+y/x$ as you did in the case before. so you'd still multiply that last equation by $x$ and have something that can be integrated

Owatch

20:35

Ah.

Balarka Sen

@Semiclassical: Your comments are really helping me solving the homework. You should chat even more from now on. :P

Owatch

I was thinking of dividing by $x^{10}$.

Semiclassical

hah

Huy

@BalarkaSen: you call $$f(x) = \begin{cases} 0 & x \in [0,1)\\ 1 & x = 1 \end{cases}$$ the delta function?

Semiclassical

and that'd work too, if you spot the product rule

Balarka Sen

20:36

it is, sort of.

Semiclassical

but you might not notice it in practice, especially if the example was complicated. my point is that you'd still be able to fall back on the $y'+Py$ approach

Owatch

Yes.

Balarka Sen

fine, characteristic function of $1$ then.

happy?

Huy

^_____________^

Semiclassical

that either means yes or "i'm going to devour you om nom nom"

Huy

20:38

om nom nom ^_________________________________________^

Balarka Sen

Huy is famous for his very long version of -_-

this is just a minor modification

Huy

famous? I'd say I'm well-known for it

Semiclassical

> ^_____^
> /_____/

sigh, that doesn't quite work

Balarka Sen

lol at lopsided mouth

Owatch

Misaligned jaw.

Balarka Sen

20:40

change the |'s to /

Owatch

I am currently trying another problem. I will report back. This one uses y``

Semiclassical

better, i think

Balarka Sen

oh now it looks better actually

but looks like one of the ad guys advertising for toothpastes.

Huy

4/5 dentists recommend Trident

Balarka Sen

always makes me wonder what the 5th dentist said

Owatch

20:43

4/5 dentists recommend trois dents (minimum)

Semiclassical

4/5 ascii monsters prefer the taste of humans

Owatch

Si vous n'avez pas le choix, quelles trois dents garderez vous?

I think I'd pick my front two, and maybe one of the incisors.

Balarka Sen

@Huy Oh, by delta function I was talking about Kronecker delta. It turns out I never knew Dirac delta existed, or at least, convinced myself it was Kronecker delta.

Huy

-________________________________-

Semiclassical

...

Balarka Sen

20:48

what is this monstrosity. nonzero at only 0 and integral is 1???

Huy

@BalarkaSen: it's not a function. it's a distribution

Balarka Sen

what is a distribution

Huy

a generalization

Balarka Sen

what do we need it for

Huy

Distribution (mathematics)

Distributions (or generalized functions) are objects that generalize the classical notion of functions in mathematical analysis. Distributions make it possible to differentiate functions whose derivatives do not exist in the classical sense. In particular, any locally integrable function has a distributional derivative. Distributions are widely used in the theory of partial differential equations, where it may be easier to establish the existence of distributional solutions than classical solutions, or appropriate classical solutions may not exist. Distributions are also important in physics and...

distributional derivative / weak derivative

is where I've encountered them

Semiclassical

20:51

in physics you encounter them a lot as either ways of representing point charges (where you want their charge density to vanish everywhere but zero, and yet to have a finite charge) or in the context of greens functions

Balarka Sen

can you elaborate on the charge thing?

oh, nevermind, I see what you mean.

Huy

:P

Balarka Sen

@Huy I no physics.

Semiclassical

bah, i hate when mathematica starts being dumb

tell it to enter a command, it thinks about it, then beeps and clears the kernal

Balarka Sen

this is a weird creature.

Owatch

20:57

Tea time! Today's flavor is: "Full English black tea"

Huy

@BalarkaSen You talkin' to me? - Travis

Balarka Sen

Erm. Don't remember any Travis.

Semiclassical

not sure who else you'd be talking to :P

Owatch

I am attempting to solve $xy'' + 2y' = 12x^{2}$. I have made the substitution $u = y'$, and so: $xu' + 2u = 12x^{2}$. And so.. $x\frac{d}{du} + 2u = 12x^{2}$. So: $\int \frac{d}{du} x*u = \int{12x^{2}}dx$. And sooooo.. $u = 4x^{2} + \frac{1}{x}c$. But now what.

Superstar Monica

I didn't know there is a handbook of the differential equation (by Daniel Zwillinger).

Semiclassical

21:05

you mean $x \frac{du}{dx}$ for the first term, i imagine

Owatch

oops

Semiclassical

in which case your integrating factor computation looks to be nonsense, alas

Owatch

sigh.

Semiclassical

i'll say first off that $xu'+2u$ isn't a total derivative, so you do need an integrating factor

and if we want to get that factor mechanically, we should rewrite it to $u'+(2/x)u=12x$

can you get the integrating factor from that?

Owatch

I have re-written it.

Actually, I am doing something right!

Okay, hold on. I am getting the factor.

$\frac{du}{dx}(x^{2}) + 2xu = 12x^{3}$.

Semiclassical

21:12

yep. with that in mind, what can you deduce about the LHS?

Owatch

Ah, let me fix that.

I deduce that it is the derivative of $x^{2}u$. Given that that would be what is on the LHS.

Semiclassical

right. so now you can integrate

Owatch

$u = x^{2}, u' = 2x, v = u, v' = u'$ = $x^{2}u' + 2xu$.

Yes, I shall.

$u = 3x^{2} + \frac{1}{x^{2}}c$.

Semiclassical

yep. and while you're not done quite yet, all that's left is easy stuff

Owatch

Which I can substitute $y' $ back in for of course, but I do not have the correct 'form' to again, repeat.

Must I not get $y$ ?

Semiclassical

21:21

if you've got $y'=P(x)$, what's $y$?

Owatch

Ah.

What about C?

Can just ignore it I guess.

Move it in front, being a constant.

Mike Miller

@Balarka: There is no aense in which that converges to the delta function. It converges to zero or doesn't converge.

Owatch

$y = x^{3} - \frac{C}{x} + (D)emiclassica$

Semiclassical

you'll want to pick a different label for the other C

Mike Miller

In the sup norm it does not converge. In the $L^2$ norm it does.

Semiclassical

21:23

say, as $D$

i'm the label, eh? :)

Evinda

Hi @Semiclassical

Semiclassical

hello

Owatch

Immortalized as a constant, yes.

Evinda

@Semiclassical Could I ask you something?

Balarka Sen

@MikeMiller Eh, I guess I am thinking of pointwise convergence. $x^n$ is $1$ when $x = 1$ for all $n$.

Semiclassical

21:25

that is the point of this room

which is to say, go ahead..

Mike Miller

@BalarkaSen: So, let's rephrase again. I asked you to show that continuous functions, with the $L^2$ norm, are not complete.

Evinda

@Semiclassical I want to draw the following relations

$$6x_1+3x_2\leq180 \\ 4x_1+5x_2\leq200 \\ 5x_1+2x_2 \geq 100 \\ 2x_1+4x_2 \geq 100$$

Semiclassical

okay?

Evinda

@Semiclassical What points could I pick so that I can do it by hand?

Mike Miller

That is, I wanted you to find a Cauchy sequence $f_n$ in the $L^2$ norm that does not converge (in the $L^2$ norm). You just gave me a sequence with $\|f_n\| = 1/(n+1)$. Would you say that does not have a limit in the $L^2$ norm?

Semiclassical

21:27

usual way i'd do it is draw the equalities first

and then pick points based on the regions formed by that

Evinda

@Semiclassical Yes, I know..

Balarka Sen

yep, agree with you it converges to the $0$ function.

Evinda

I got high values, i.e. 50 and 40 @Semiclassical

Semiclassical

not sure there's really anything more sophisticated than that

Evinda

So I should consider greater numbers at the graph, right?

Mike Miller

21:29

Great. So you're not done yet, in my eyes.

Semiclassical

i guess? i'm not really seeing the question

Balarka Sen

Right.

I think I should be able to do a modification to this.

Semiclassical

just to check for myself: the partial sums of the fourier series expansion for the sign function on $[-\pi,\pi]$ do converge in the L^2 norm, right?

Mike Miller

yes

Semiclassical

mmkay.

Balarka Sen

21:39

I am not sure what Fourier series of sign function means though, because it's not continuous.

Mike Miller

That's fine.

Balarka Sen

Hmm $f_n(x) = x^{1/(3n)}$ works, I'd think.

Mike Miller

Why the 3?

That appears to converge to one of my favorite functions. I call it "One".

Semiclassical

me myself and identity

Balarka Sen

I mean, in $[-1, 1]$.

Admittedly I didn't try to prove it. I am mostly thinking of pointwise convergence, and trying to see if it works in $L^2$ too. Let me check.

Must be a huge difference between the two if this doesn't work.

@MikeMiller The $3$ is there to make it converge to the sign function.

Mike Miller

21:46

OK. Fine, that will work. You should prove it.

Well, no.

You want $(2n+1)$

not 3n

And in practice it seems less painful to use $x^{2n+1}$. :P

Balarka Sen

Yes, sorry.

Ted Shifrin

@MikeM: You got wet. I just walked 3.5 miles in 20° weather. We're even?

Balarka Sen

three and half a mike.

Mike Miller

Sure.

Balarka Sen

that's a lot of them

Ted Shifrin

21:47

smacks impudent Balarka

Mike Miller

Writing Thurs talk. I forgot how fun Nevanlinna theory is.

Balarka Sen

I wasn't showing impudence. Just trying to make mostly bad puns out of your typos.

Ted Shifrin

I headed in that direction (Griffiths style) for one year of grad school, Mike.

Mike Miller

I'm not really sure what direction that is, @Ted. My reference is Rubel's book.

Semiclassical

What's Nevanlinna theory? I remember it's in complex analysis but not much else

Mike Miller

21:51

@Semiclassical: study a meromorphic function by studying the way its values are distributed

Semiclassical

hmm!

is there a simple example?

Mike Miller

not sure there's a simple example, but it has a so-called main theorem

which is one big inequality

an immediate corollary of that inequality is the picard theorem

Semiclassical

little or great (or big, i forget the terminology) ?

Mike Miller

little; I made some minor efforts to get the great one but couldn't see how

it doesn't work for functions with essential singularities

Semiclassical

mmkay.

Mike Miller

21:53

here's the theorem that really blows my mind that I'll try to cover thurs

let $f$, $g$ be meromorphic functions

let $z_1, z_2, z_3, z_4, z_5$ be distinct elements of $\Bbb C \cup \infty$

if $f^{-1}(z_i) = g^{-1}(z_i)$ for all $i$, $f=g$

that's wild. five values agree and they agree everywhere

Semiclassical

yeah.

Mike Miller

this is tight: $e^z$ and $e^{-z}$ share $0, \infty, 1, -1$

Huy

cool

Semiclassical

what threw me at first was the inverse

Ted Shifrin

@MikeM: this Very geometric.

Balarka Sen

21:57

@MikeMiller OK, yeah, it does not converge, but is Cauchy.

Mike Miller

@TedShifrin: Still have a copy? :)

@BalarkaSen: Why does it not converge to any continuous function?

Ted Shifrin

Nope ... Someone took it.

Mike Miller

Darn.

Semiclassical

the sense in which those two functions share those points is where i'm having to sit down and think through it

Ted Shifrin

I don't know if it's in his complete works volumes, which I do still have at home.

Mike Miller

21:59

@Semiclassical: neither ever take on $0$ or $\infty$. each take on the value $1$ precisely at the points $n \cdot 2\pi i$. each take on the value $-1$ at the points $n \cdot 2\pi i + i$

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$Mathematics$ Mathematics