Mathematics - 2017-05-08 (page 3 of 4)

Balarka Sen

18:00

@Daminark Something like that. In the $\Bbb C$ world, you can define the winding number as a complex integral.

Daminark

(Now that I think about it we didn't actually prove that it's an integer...)

Ted Shifrin

Sit down and play with it, @Eric. This is actually how I derived first-variation of arclength in my notes, I believe.

Eric

Will do

Ted Shifrin

You just love coming here to get more homework :D

Eric

I just like bashing things out tbh

Ted Shifrin

18:01

bashing?

Eric

like working

Alessandro Codenotti

@Daminark I'm not sure what the angle form is, but we defined the winding number of the point $z$ outside the support of $\gamma$ as $\displaystyle\text{Ind}_\gamma(z)=\frac1{2\pi i}\int\limits_\gamma \frac{\text{d}w}{w-z}$

Daminark

Oh wait that's clever, you can use residue theorem to get that it's an integer

Ted Shifrin

@Alessandro: If you work out $\dfrac{dz}z$, you see that the imaginary part is exactly $\dfrac{-y\,dx+x\,dy}{x^2+y^2} = d\theta$.

That's basically circular reasoning, no pun intended, Demonark.

Balarka Sen

@TedShifrin: Take the Galois closure of $F$ in $K$; that's a subfield of order $|Gal(K/F)|$

Daminark

18:03

Do you use this fact in order to prove the residue theorem?

Balarka Sen

And degree of subextension divides degree of big extension.

Alessandro Codenotti

@Daminark It's a bit more painful to define the winding number for curves which are not piecewise $C^1$

Ted Shifrin

Hmm, I was pondering this, @Balarka. Hold on.

Alessandro Codenotti

you need to prove that every continuous curve can be approximated well enough by a piecewise continuous one

Balarka Sen

@Alessandro Every element of $\pi_1(S^1)$ can be represented by a smooth map from $S^1$, even.

Ted Shifrin

18:04

@Alessandro: It suffices to do piecewise polygonal, with segments parallel to the axes.

Alessandro Codenotti

@TedShifrin yeah, we glossed over some details, but we used piecewise polygonal curves

Ted Shifrin

Right, I agree, @Balarka. Thanks.

Balarka Sen

No problem.

Alessandro Codenotti

@BalarkaSen the obvious one turning around $n$ times in the right direction?

no wait, why a map from $S^1$?

Balarka Sen

@Alessandro Well, that's a much harder theorem.

Elements of $\pi_1(X)$ are maps based homotopy classes of maps $S^1 \to X$

If $M$ is a smooth manifold, every element of $\pi_1(M)$ can be represented by a smooth map from $S^1$. This is smooth approximation stuff.

Alessandro Codenotti

18:11

Hm, I'll trust you on this one :P

Daminark

Oh @Ted now I get it, Cauchy integral formula requires you to evaluate that integral

Liad

18:24

hi , someone here familiar with "the long line" in topology ?

(@TedShifrin hm , i guess you are ? :P )

Daminark

So if I'm not violating any rules, I think you should be able to say that $\frac{1}{z}$ is holomorphic away from $0$ (swap in some $z_0$ in other cases), so that you can homotope your curve away from that point and reparametrize so that it becomes a curve from some $[0,2\pi n]$ such that $t\mapsto re^{it}$, so the integral becomes $\int_0^{2\pi n} \frac{1}{re^{it}}ire^{it} dt = 2\pi i n$

Ted Shifrin

Demonark: Well, the curve of course needn't be a circle, so there's something to prove ...

What's the question, @Liad?

Daminark

Proving that if you have a closed curve, this is homotopic to some $n$-times traversal of the circle?

Liad

im trying to show $L$ is connected ( and path connected which implies connected) and $L \ ^ * $ is also connected but not path connected , @TedShifrin

CausingUnderflowsEverywhere

@TedShifrin I see you have the caring personality. You chose relaxation over laughing at someone

Ted Shifrin

18:28

Now I'm not sure how sarcastic you're being, @Causing.

@Liad: What does the star mean?

Liad

@TedShifrin L with another point , i saw that somewhere it is called 'the extended long line"

Jason Bourne

@CausingUnderflowsEverywhere Hi! Your question isn't really formatted or written properly!

Ted Shifrin

OK, @Liad. I haven't thought about this stuff in forever.

Jason Bourne

@CausingUnderflowsEverywhere Actually he did answer your question. So what's the problem?

CausingUnderflowsEverywhere

Hi! Welcome to mathematics. Just ask, no need to ask to ask! ....
what? you're not a bot right?

Liad

18:32

@TedShifrin that stuff is not easy for me :/

Ted Shifrin

I didn't think there was a problem, @Jason.

CausingUnderflowsEverywhere

@JasonBourne The problem is ... well you. What's the problem in your eyes?

:/

Ted Shifrin

@Liad: I didn't say it was easy. It requires thinking about transfinite induction or something, and I actually have never taught this stuff.

CausingUnderflowsEverywhere

I wasn't being sarcastic or anything I just looked over your answer and realized the way you wrote it out

Liad

huh, fine im stuck on calculus question that will be easy for you :P @TedShifrin

Ted Shifrin

18:34

OK, no big deal, @Causing :)

Alessandro Codenotti

The way we did this in class was to define $\displaystyle g(h)=\int\limits_a^h\frac{\gamma'(t)}{\gamma(t)-z}\text{d}t$ (which is the same integral as the winding number), then we defined $\varphi(h)=e^{g(h)}$ and we showed that $\varphi(b)=1$, so $g(b)=2\pi i m$ and the winding number is $m$ for some integer $m$ @Dami

Jason Bourne

@CausingUnderflowsEverywhere Well, I just didn't know what you meant by relaxation or laughing at someone, but it seems there is no problem, and so there is no problem.

CausingUnderflowsEverywhere

$ah = ha$ :p

Ted Shifrin

It's weird that you're doing all this totally abstract topology at the same time you're doing basic multivariable calculus, @Liad.

Liad

why it is weird?

Jason Bourne

18:35

@TedShifrin I guess multivariable calculus isn't taught properly in many places.

Ted Shifrin

Well, "properly" depends on the audience.

Daminark

Well you can always homotope your curve to lie on the unit circle so that it becomes $e^{i\alpha(t)}$, say parametrized by $[0,1]$

Ted Shifrin

No you can't, Demonark.

Liad

@TedShifrin i have a curve $\gamma : [0,L] \to R \ ^ 3$ , $||\gamma(t) || = ||\gamma'(t)|| = 1 $ and $L = Len (\gamma)$ , i need to prove that $||\gamma(t)'' || \ge 1$

Ted Shifrin

Demonark: Proving that any path on the circle is homotopic to some $e^{2\pi ikt}$ is nontrivial.

Daminark

18:37

Well if your curve is $c(t)$, and it's lying in the punctured plane, can't you just say $H(s,t) = \frac{c(t)}{1-s+s\|c(t)\|}$?

Ted Shifrin

Something's wrong there, @Liad. Care to proofread?

Liad

i know that $<\gamma' , \gamma '' > = 0$

Jason Bourne

By the way @CausingUnderflowsEverywhere you need to put a \ like \sin to make it look like sin in LaTeX.

Ted Shifrin

Demonark: If your curve never goes through the origin, you can of course homotop it to a curve lying on the unit circle. But it may go forward and backwards and wind a bunch of times.

Liad

and from cauchy we have $<\gamma , \gamma '' > \le ||\gamma'' || $ but i cant go further, @TedShifrin

Jason Bourne

18:39

@CausingUnderflowsEverywhere Compare $sin$ and $\sin$.

Ted Shifrin

@Liad: I agree that $\|\gamma'(t)\|=1$. Do you really know $\|\gamma(t)\|=1$ for all $t$ (so you're on the unit circle)?

Daminark

So what prevents you from kinda saying that this path on the unit circle "lifts" some a parametrizable path in the exponent?

Liad

@TedShifrin yea it is given

Ted Shifrin

OK, sorry, so we're in $\Bbb R^3$ and the curve is on the unit sphere. My apologies.

Right, now the problem makes sense.

Liad

:)

Ted Shifrin

18:41

Have you done Frenet formulas? What have you done?

Liad

no idea what frenet is

Jason Bourne

Frenet and Serret, interesting combo.

Ted Shifrin

OK. So you do know that $\langle\gamma'',\gamma'\rangle = \langle\gamma',\gamma\rangle = 0$.

Liad

i have $<\gamma, \gamma '' > \le ||\gamma''||$ , i cant control the left hand side of this inequality

@TedShifrin i know only that $<\gamma ' , \gamma '' > =0$

Ted Shifrin

You also know the second one, because $\|\gamma(t)\|=1$ for all $t$, you told me.

Liad

18:44

why this implies the other one?

Konformist Liberal

I was wondering if all cartesian closed categories are also locally cartesain closed. Besides from this question, is Set locally cartesian closed?

Ted Shifrin

Same proof that you used for yours.

Liad

alright, i believe you. how can i go further :P

Ted Shifrin

Differentiate $\langle\gamma,\gamma'\rangle = 0$?

Liad

hm

Daminark

18:46

@Ted I'm playing here with things I barely understand, but could you use that $\mathbb{R}$ is the universal cover of $S^1$ in order to lift our path such that it can be homotoped there?

Ted Shifrin

I actually assign this in the G&P course, typically, Demonark, with more hint than Guilllemin gives. Yeah, you want to lift the path to a map $[0,1]\to\Bbb R$ and see where it ends up there.

Balarka Sen

Lifting is definitely my favorite proof

Jason Bourne

Guillemin and Pollack is one book I never looked at closely.

Liad

@TedShifrin isn't it $<\gamma' , \gamma' > $ ?

Ted Shifrin

That's only half of it, @Liad.

Liad

18:49

yea multiply by 2

Ted Shifrin

(Also, use \langle and \rangle, not < >.)

Liad

but we have $=0$ so i omitted it

Ted Shifrin

NO, not multiply by 2.

There's a product rule, remember?

Liad

yea im re-writing it now

Jason Bourne

Hello @waiting, how are you? It's J here. You went to the wrong room, I think.

Daminark

18:51

So let's call that map $\alpha$, then naively one would want to directly homotope $\alpha$ to $x(t) = (1-t)\alpha(0) + t\alpha(1)$, and say that this translates back down to a homotopy from our path to an $n$-times traversal of the circle

Does this check out or is there something wrong?

Ted Shifrin

That's correct, Demonark. The hard part is creating $\alpha$ as a smooth map and seeing that the endpoint tells you the degree of the original map :)

Liad

we have $\lim_{h \to 0} \dfrac{ \langle \gamma(t+h ),\gamma'(t+h ) \rangle -\langle \gamma(t ),\gamma'(t ) \rangle } {h}$

Daminark

Alright, I'll table this until tomorrow when we actually define degree, and perhaps wait until we prove the Hopf degree theorem making life even easier :P

Fargle

Hello, chat!

Daminark

Hey @Fargle!

Fargle

18:54

What's going on, @Daminark?

Ted Shifrin

@Liad. What are you doing?!! Just write down the product rule for the derivative of $\langle \gamma(t),\gamma'(t)\rangle$.

Daminark

Everything's doing alright, how about you?

Ted Shifrin

Demonark: You actually need what we're talking about to prove the $n=1$ case, so he might leave that as an exercise.

Liad

@TedShifrin huh, i tried going "by def. "

Ted Shifrin

hi @Fargle\

Fargle

18:55

Alright as well. My goal for today is to demystify Cauchy's integral formulas and residue theorem.

Ted Shifrin

Doing the product rule from the definition is not what I want you to do here, @Liad. Just use it.

Stokes's Theorem, @Fargle :P

Liad

so it is $1 + \gamma(t) \gamma ''(t)$

alright ? @TedShifrin

Ted Shifrin

Right. $1+\langle \gamma(t),\gamma''(t)\rangle$.

And it equals what?

Fargle

@TedShifrin There's still a bit of mystery to that for me too, so I guess what I really need is a multivariate refresher. goes to your book

Daminark

Oh so it's inductive? On the dimension?

Ted Shifrin

18:57

That's how I do it, Demonark.

Liad

wait i have the other term is less then $||\gamma''(t)||$

Ted Shifrin

@Liad. Stop. That sum equals what?

Jason Bourne

Maybe Spivak should write a multivariable calculus book that resembles his calculus book more. =)

Ted Shifrin

That's what my book is.

Spivak ain't writing no more calculus books.

Liad

@TedShifrin what is it ?

Balarka Sen

18:58

@Fargle A noble goal.

Ted Shifrin

What did we differentiate to start with, @Liad?

Balarka Sen

I concur with Ted that Stokes' theorem is the best way to understand Cauchy integral theorem, which is fundamental to all of those.

Liad

huh

Balarka Sen

Or, if you want, Green's theorem.

Ted Shifrin

@Fargle: If you look at my very last 3510 lecture, I actually did some of this complex variables stuff in there.

Liad

18:59

i thought you asked about $\gamma (t) \gamma''(t)$

Fargle

Yeah, I'm pretty sure the proof in my book (Zill) used Stokes'.

Liad

it is equals zero

Ted Shifrin

Right. We differentiated the constant function $0$, so the answer is $0$. (By the way, this technique is used thousands of times in differential geometry.)

Jason Bourne

@TedShifrin I should really take a look at it some day. I could never find a copy on Russian servers for evaluation. =)

Ted Shifrin

So you have $1+\langle \gamma(t),\gamma''(t)\rangle = 0$. Now play with it and you're done.

Liad

19:00

@TedShifrin yea i wasn' sure you talked about this , sorry

Jason Bourne

@BalarkaSen Concur is such a formal word that I only use agree.

Alessandro Codenotti

@BalarkaSen it looks like half of the stuff I learnt in the analysis and geometry courses is just a special case of Stokes' theorem

Secret

[Weird topology idea] (elaboration available shortly)

Ted Shifrin

Mr. Bond, you're getting downright obnoxious.

Secret

Balarka Sen

19:02

@Fargle Do you remember the Green's theorem? I can tell you a quick proof based on that and point out which bits you should understand, if you want.

Liad

@TedShifrin we still have $\gamma(t) \gamma''(t) \le ||\gamma''(t)|| $ , and this completes the proof doesn't it?

Ted Shifrin

Not quite, @Liad, but close. How are you completing?

Fargle

@BalarkaSen Not well enough where I'd find it very enlightening. We only glossed over it in my multivariable class, which I'm finding out was a cardinal sin.

Balarka Sen

@Alessandro Hah

Ted Shifrin

Most multivariable calc classes suck.

(And I'm not saying that because of lack of rigor. I'm saying it because key concepts are not well taught. And the most important stuff at the end hardly ever gets covered.)

Jason Bourne

19:04

@Fargle I wonder if an ordinal sin is more severe than a cardinal sin.

@TedShifrin Exactly. Leaving one not well prepared for complex analysis etc.

Fargle

@TedShifrin Right, this is exactly what happened to us. Stuff like Green's, Stokes', etc. were presented as Useful Tools rather than as generalizations of FTC.

Secret

Consider 3 copies of a topological space "superimposing on each other" i.e. if given a coordinate system, they will share the same (x,y)coordinates

Now those circles are clearly boundaries for two of the topological spaces.

The rules of this topological space is the usual eucledian metric for anywhere except the closure of those red and blue circles

Now:

Liad

@TedShifrin i think i got it - $| \gamma(t) \gamma''(t) | = |-1| \le ||\gamma''(t)|| $

Ted Shifrin

There you go, @Liad. Well done.

Liad

thanks , you are great teacher :P

Ted Shifrin

19:06

I'm gone for now. Bye, all.

Liad

see ya

Daminark

See you @Ted!

Jason Bourne

You all behave without Ted around, lol.

Balarka Sen

@Fargle Ah. Well, I can tell you how I think about Green's.

Jason Bourne

I always think whether I should write Gauss' theorem or Gauss's theorem and Stokes' theorem or Stokes's theorem.

Daminark

19:09

I usually suggest Stokes' theorem

Balarka Sen

"Stoke's"

Secret

We define the following rules for sets in this set of 3 topological spaces as follows:

1. Each colored discs (which is a closed set) is surrounded by some open set concentric to the discs of some fixed radius. We call this open set the open set carried by that disc
2. If the discs are moved around, then the open set it carries follows it provided rule 3 does not apply
3. If any discs entered the interior of a different colored open set, the open set it carries will stay put even if the disc continues to move around (in which case, the disc is said to be moving within the topological space …

Jason Bourne

Well, I think all four forms are correct.

Daminark

@Balarka angreacts only

Balarka Sen

@Daminark Stoke's theorem is a great theorem.

You should try it sometime.

Daminark

19:12

I'm so stoked to learn it

lel

Secret

Example trajectories:

Blue is moving, red is stationary

Anonymous

@JasonBourne I've seen all of them being used in books. Spellings are one of the least important things in Physics/Maths and that too when they are proper nouns.

Secret

Note as soon blue entered red, it is moving within the red copy of the topological space (imagine the whole rectangle is red when you start tracing the red trajectories). It cannot escape from it unless it moves back out from where it entered red

Daminark

@Eric Classes are up, though none of the professors are there

Balarka Sen

@blue Interestingly, Gauss and Stokes both have theorem/lemmas/laws named after them in physics also.

Semiclassical

19:18

hi chat.

Balarka Sen

Hi @SemiC

Anonymous

@BalarkaSen Hehe, yeah :P They were multitalented it seems :D

Semiclassical

Well, Gauss was also a physicist in his day.

Balarka Sen

Well in those times math and physics weren't completely distinct disciplines

Semiclassical

That too.

I think Gauss did do some experiments, in fact.

Anonymous

19:19

In those days maths+physics+some other stuff=philosophy :D

Balarka Sen

natural sciences, not philosophy, I think

Semiclassical

So calling div E = - rho as Gauss's law seems appropriate

I think back then you would've called it "natural philosophy".

Balarka Sen

rho/epsilon yeah?

Semiclassical

Right, careless.

You don't see Stokes referenced in electromagnetism, by contrast, but you do see him mentioned in fluid mechanics i.e. Stokes flow.

Balarka Sen

Yup.

Semiclassical

19:21

And, actually, you do see him referenced when it comes to certain optical phenomena (Stokes parameters) though I don't remember what those actually mean. Something something polarization of EM waves.

Balarka Sen

Stokes' law predicts the force exerted by viscous fluids on freely falling, small, spherical particles

Semiclassical

(Per Wikipedia: The Stokes parameters are a set of values that describe the polarization state of electromagnetic radiation. Very precisely stated /s)

To be fair, Stokes parameters are a real pain in the arse.

As evidence, try to look at this page without getting a headache: en.wikipedia.org/wiki/Stokes_parameters

@BalarkaSen This is a rather nice line about Stokes parameters, actually: "From a geometric and algebraic point of view, the Stokes parameters stand in one-to-one correspondence with the closed, convex, 4-real-dimensional cone of nonnegative Hermitian operators on the Hilbert space C^2."

Sha Vuklia

Guys, I don’t understand why $A^*=B$ makes $x(t)$ real. I’m guessing we want $C$ to be real? But the complex conjugate of $C$ is
$$
C^*=\frac{A^*+B^*}{\cos\phi},
$$
so how does $A^*=B$ help?

Semiclassical

@ShaVuklia It's simpler than that. What's the complex conjugate of $e^{-\gamma t}(Ae^{i \omega t}+Be^{-i \omega t})$?

(I dropped the tilde overhead because it's annoying.)

Sha Vuklia

give me 1 sec, I'm not good with complex numbers:P

Balarka Sen

19:30

@Semiclassical Interesting. I don't know anything about Stokes parameters though.

Semiclassical

Keep in mind that $t,\omega,\gamma$ are all real quantities.

@BalarkaSen Yeah, I wouldn't advise looking at them if you don't have to.

But that one statement is cute.

Sha Vuklia

$e^{-\gamma t}(A^*e^{-i\omega t}+B^*e^{i\omega t})$

Semiclassical

Right.

Sha Vuklia

oh right

ahh

I see now :P

Semiclassical

Yep.

Sha Vuklia

19:32

thx!

Semiclassical

np

Fargle

Maybe I'm just lazy, but handwriting vec symbols over everything is something I find very tedious, though necessary.

Semiclassical

Yeah.

Sha Vuklia

I don't do it always

Fargle

I also just have bad handwriting, so it turns into a veritable deluge of horizontal ticks.

That's why I prefer to do all my math homework in $\LaTeX$.

Semiclassical

19:35

One convention I wish was used more was underlines under vectors instead of the little arrows, e.g. $\underline{r}$

Especially since then you can indicate matrices as two underlines.

But you typically only would have that as a convention in physics where people don't like distinguishing row/column vectors. (Which is actually pretty annoying.)

Sha Vuklia

physics<3

Anonymous

@Fargle Doesn't that take way more time?

Fargle

I like that, except for how it intersects with letters with a tail. It might not be immediately obvious that $\underline{g}$ is underlined in handwritten sources, for example.

Sha Vuklia

Hey @Waiting :)

Semiclassical

Fair.

Fargle

19:37

@blue Sure, it takes more time, but I'm fine sacrificing time for readability to counteract my laziness.

Semiclassical

I prefer to just make clear at some point what stuff is in R^1 and what stuff is in R^n.

Eric

oh cool @Daminark

Semiclassical

(and I always take vectors to be column vectors unless there's a T)

Of course, there's a difference between an experienced user debating whether or not to use vec, versus a novice user.

Balarka Sen

I have written smallish notes in latex. It doesn't take time if you don't give a damn about formatting

but most of the time that's hard to do

Semiclassical

For the novice I do think it's better to include vec's if only to avoid silly things like solving $m = x\cdot y\implies x=m/y$ when $x,y$ are vectors.

Once you have the distinction in your head it's fine. But until then it'd better be enforced on the page.

Eric

19:41

Ah @Daminark looks like algebraic geometry is indeed there, so probably will be taking that. I wonder when they put up the graduate classes though.

Sha Vuklia

what is meant by "envelope of the motion" ?

(at the bottom of the picture)

if it's the dashed curve

Semiclassical

What's the largest value that the cosine part of the solution can take?

Sha Vuklia

then that seems like the amplitude for me

Semiclassical

Hard to call it an amplitude if it's changing.

Sha Vuklia

$e^{-\gamma t}C$

yea okay, instantaneous amplitude?

Semiclassical

19:43

Right. That's the largest positive value, at any rate. The largest negative is $-e^{-\gamma t} C$

So the dashed curves are just those two lines.

But yeah, it's basically the instantaneous amplitude.

Sha Vuklia

yea but why are those dashes curves the curves that pass through the extremes of the motions?

and why don't they write $C$ in figure 4.4?

Semiclassical

Because those curves are $y=\pm e^{-\gamma t}C$.

As for why they're not in the figure...hrm.

Sha Vuklia

sorry I meant:

why are they not the curves that pass through the extremes

because that what they text seems to say at the bottom

i'll add the other part too

Semiclassical

Oh, right.

Sha Vuklia

Semiclassical

19:45

Suppose you want to find the max value. For that, you'd differentiate and set that to zero.

Sha Vuklia

yes

Semiclassical

So that'd be $y'(t)=-\gamma e^{-\gamma t}C \cos(\omega t+\phi)-\omega e^{-\gamma t}C \sin(\omega t+\phi)=0$

So $\tan(\omega t +\phi) = -\gamma/\omega$.

Sha Vuklia

ohhhh.. wow I wouldn't have seen that myself

Semiclassical

In which case, you can write cos/sin using trig stuff.

Ted Shifrin

@Semiclassic: I presume the professor remained incompetent and didn't correct the exam?

Semiclassical

19:48

Changed one of the two problems.

Second one is still geometric optics but now is just a refraction problem not a mirror problem.

Ted Shifrin

@Balarka: NOOOOOO ... The man's name was STOKES. You must write either Stokes's or Stokes'.

You and your damned screwed-up apostrophes.

Sha Vuklia

lol

Ted Shifrin

Oh good, @Semiclassic. At least he responded.

Semiclassical

First one I still think is really dumb, though, upon looking at it again. It's not just a lens problem but one which deals with radius of curvature of a lens

And that's really specialized.

So I'm still annoyed but I'll have to live with it.

1 out of 20 of the problems being bad is better than 2 out of 20.

Also, as far as I know I'm the only TA who raised a complaint. (But if they replied to him only then I wouldn't know, so that's not without ambiguity.)

Daminark

@Eric Gleaned some info about who's teaching what next year, didn't ask about algebraic geometry though

Semiclassical

19:52

@ShaVuklia Continuing on, from that value of tan we have $\cos(\omega t+\phi) = \pm \omega/\sqrt{\gamma^2+\omega^2}$ depending on whether it's a maximum or a minimum.

Hence the max/min values are $y(t)=e^{-\gamma t}C\cdot \frac{\pm \omega}{\sqrt{\gamma^2+\omega^2}}$

Daminark

Basically set to die third quarter tho...

Semiclassical

If $\omega\gg \gamma$, then this last factor is approximately just $\pm 1$ and you have $e^{-\gamma t}C$ as the maximum.

That's the formal answer. For an intuitive answer, I'm pretty sure that $e^{-\gamma t}C$ intersects $y(t)$ tangentially.

Since each of those tangents is sloped down, $e^{-\gamma t}C$ would therefore not pass through the peak but would rather meet the curve a bit to the right of that.

If you look at that, the yellow line doesn't pass through the peak; rather, it'll have to intersect just below and to the right. That becomes more obvious if I zoom in:

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$Mathematics$ Mathematics