Mathematics - 2022-01-18 (page 2 of 2)

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9:29 PM

3

A: Finding $ \lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{x^2}$ without using series expansion.

The hint: $$\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{x^2}=\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{x}}\left(\frac{x}{1+x}-\ln(1+x)\right)+\frac{ex^2}{2}}{2x^3}=$$ $$=\frac{1}{2}\lim_{x\rightarrow0}\frac{\left((1+x)^{\frac{1}{x}}-e\right)\left(\frac{x}{1+x}-\ln(1+x)\right)+e...

Can anyone please explain why there's 2 in denominator after the first equality?

Why don't you add a comment to the question asking for elaboration?

Doesn't look right to me, but I don't have the energy to disassemble right now.

Have you tried L'Hospital's rule?

9:48 PM

I added the comment.

By L'Hospital's, yes we can do it.

By series expansion too.

But I was wondering how he did it.

the answer is correct so I'm inclined to believe that 2 in denominator is correct.

i presume it is from differentiating $x^2$

but then the power of x^2 should have reduced.

10:06 PM

I assume he's doing a series for $(1+x)^{1/x}$ by exponentiating the series of its log. But it certainly is not obvious. After all, where did the $e$ go?

Well, $(1+x)^{1/x} - e = e^{(1-x/2+x^2/3+\dots)/x} = e(1-x/2+\dots)$ ...

He's definitely not doing L'Hôpital or differentiating.

Typo in what I just did. Forgot to do the $-e$.

Isn't that what OP did in his question?

Just by putting the $1/x^2$ of the numerator in the denominator

you murdered my question @Astyx

sorry

:D

@TedShifrin professor Ted, I thought that he was avoiding LHR and series expansion.

But nope, it's clear now that he used L'Hopital's rule.

@Koro I thought he did this: $\lim f(x)= \lim g(x)$ if $\frac{f(x)}{g(x)}\to 1$

10:24 PM

I hate L’Hôpital, regardless.

l'hopital is a barbaric weapon. it is a crossbow. crude, minimal training needed to apply it.

give me the longbow, every time.

i think sometimes it works even where it shouldn't. i just don't recall such example off the top of my head.

@Astyx how do I show that the denominator of that fraction isn't zero

I've been looking at it for years

You don't

What you can show, however, is that if a solution to the homogeneous DE cancels at a point, then it's zero everywhere

Similarly you can find an expression for a solution of the HDE that cancels nowhere

Yeah, thats what I'm tryna do atm

10:33 PM

the HDE is just the original DE with $g=0$. You have a solution for this one

Oh yeah $\mu(t) - \lambda(t)$

not what I meant

$x(t) = exp(\int_{t_0}^{t}a(t) dt)$

ok

does this function vanish somewhere?

I'd need the integral to diverge no?

to -infinity

10:39 PM

Can that happen?

No

since a is continuous

why not?

yes ok

oh so it can't be zero

11 hours ago, by Koro

Can you show that the IVP $y’=a(t) y, y(t_0)=0$ has a unique solution? Here, a is a continuous function.

I tried to say the same thing.

Yeah i get you @Koro but I required a lot of help beforehand to get to this stage tho

cheers though <3

10:41 PM

Ok. So can you justify division by $\lambda -\mu$ now?

you don't really need that

But that's good for practice.

:)

I'll let Koro take over since I need to go to bed

bye

cheers man @Astyx

Helped me a ton

glad to help

10:43 PM

Astyx, I'll also be going to bed shortly. If you could stay for about 2 mins. I have a question I wanted to ask.

I have difficulty converting region in xy to region in r theta (polar coordiinates). $S=\{(x,y): x^2\le y\le 1, -1\le x\le 1\}$.

$x=r\cos \theta, y=r\sin \theta$. Region in xy is bounded by the parabola $y=x^2$ and the straight line $y=1$.

So first, I would like to map $y=1$. For that $r\sin\theta=1$ so can I say that $y=1$ got mapped to $g(r,\theta)=0$, where $g(r,\theta)= r\sin\theta-1$?. Then I would like to map $y=x^2$ that is $r\sin \theta=r^2\cos^2\theta\implies r=0$ or $r\cos\theta =\\sin \theta$

I don't know how to find the region in $r\theta$.

Simply put, complex analysis comes to the rescue here. I'm just going to mention two words and the solution should become obvious: complex argument.

@Koro

Polar sucks, but you can do it. You need three separate integrals.

@AMDG Huh?

10:59 PM

He said he has difficulty mapping square coordinates to circular coordinates.

You didn’t read, huh?

I guess not enough.

If we're just talking about mapping coordinates from cartesian to polar, though, that's pretty straightforward in principle, regardless of context. $\arctan(y, x)$ maps a point $(x, y)$ to an angle.

$r$ is just the distance to the origin

@Koro you have $r=\tan\theta$ and $r=1/\sin\theta$, but for what $\theta$ intervals?

Why not compute the regions in xy and convert back to polar? Wouldn't that be easier to integrate anyways?

Ted, that's the difficulty I am facing. I don't know where to stop when we convert to polar coordinates. I mean how do you know that there will be $\theta$ intervals.

@AMDG I almost understand you. In complex, we have straight lines get mapped to .... etc.

But I have difficulty in polar.

So in complex, you have a recipe to know what region is going to become what.

11:14 PM

Draw the region in the $xy$-plane. What values of $\theta$ in the whole region? Draw a ray of fixed $\theta$. For which values do we exit at the parabola? At the horizontal line?

There are examples in my videos, of course.

in later video lectures where I have not yet reached.

Is complex analysis supposed to be much harder than real analysis?

The region looks like side view of a cup (without a holding handle) with a lid on its top. I'm not sure how theta varies overall because behavior of theta becomes strange around origin.

I'm taking it now as a summer course, and I have taken and did well in the per-requisites (analysis, multivariable calculus and linear algebra) and the professor is using some things outside that

this week he defined differential forms

Oh wait. $y=x^2$ so $y'=2x$ and so $y'(0)=0$ hence $\theta_0=0$.

11:23 PM

derivative: it really depends on the instructor. at my undergrad complex analysis had the reputation of being the 'easiest' of all of the classes required for the major, but it certainly wasn't easy when some people were teaching it.

@leslietownes it's a summer course, so things are a little different than I'm used to. The professor said it was an introductory level class, but introductory for a master's. But he also said we just needed to have taken those 3 classes

Anyone here savvy with derivatives and combinatorics?

Not a homework question btw

user image

I'm looking at ways in which this could be obvious

But I fail to find any

@Koro: Draw pictures. $\theta$ goes from $0$ to ?

:')

@1010011010 obvious is a strong word

but you can probably prove that with induction

11:27 PM

Without a doubt

The problem is that I cannot defend posing it as a conjecture (because I cheated with WolframAlpha :p)

you can conjecture anything you like, doesn't mean that it will be correct

in fact in math you often have to conjecture something first to be able to prove it

derivative: in my experience, instructors are the least reliable authorities as to how difficult their classes are :)

that's...funny

I mean, I get that

But let's just say that I refuse to believe this was just somebody's hunch one day

@leslie I was always a reliable authority.

11:30 PM

I checked the individual terms in this summation for a few specific cases because I wanted to see what was going on under the hood

It didn't get me anywhere, other than confirming that everything adds up to the answer in some mysterious way

@1010011010 I don't think it would be quite so hard to guess if you did the first 10 or so cases by hand

@TedShifrin from $0$ to $arc tan 2$ (if we're talking parabola). But if we consider $y=1$ then $\theta =\pi$ or $2\pi$. :(

(people like to complain about computation, but I think it's kind of important in cases like this)

@Derivative You'd be surprised

It's completely non obvious

No. You aren’t looking at the picture at all.

Where are the corners of the region?

11:33 PM

there's also this thing en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula

The first Pochhammer factor is obvious from the general form without massaging anything

I'm looking at the picture. It's right in front of me. At x=1, the slope from parabola is $y'(1)=2x|_{x=1}=2\implies \theta_1=arc tan 2$ and $\theta_{-1}=arc tan (-2)$.

The corners are $(-1,1)$ and $(1,1)$.

You are not understanding what $\theta$ is, clearly.

Slope is irrelevant.

So what is $\theta$ at the point $(1,1)$?

By what I said above (wrongly), it should be arc tan 2. But using $arc tan (1/1)$, it should be $\frac {\pi}4$.

Right.

11:37 PM

Ah, so you mean to say that theta doesn't depend upon graph.

As I said, you are misunderstanding $\theta$, or were.

Only on point.

Rays from the origin!

Ok. So theta intervals are done. Now, how do we know that r will or will not have intervals?
I mean how do we know whether 1) $r=f(\theta)$ to $r=g(\theta)$ where $f$ and $g$ are non constant or 2) $r=c_1$ to $r=c_2$. I ask because sometimes it may be possible that one is not able to solve the equation for $r=$value 1 and $r=$ value 2.

Never mind, I'll think about that professor Ted.

You need to work out lots of examples.

11:42 PM

@TedShifrin wu had a similar status at berkeley. it was exactly as promised.

one of my wife's friends was a math major who took complex analysis maybe four times and just could not pass it. she ended up switching majors. i couldn't figure that out.

thanks a lot Ted. At least theta makes some sense to me now. :)

I certainly had majors who failed undergrad complex the one time I taught it. I taught the grad course numerous times.

Angle is a linear measure of a difference in direction relative to some other direction.

Theta represents this angle.

the undergrad class at berkeley was a common 'at least one postdoc will be teaching it like a calculus class' offering.

@Koro The first integration lecture on polar coords will help you for sure.

11:45 PM

so it was weird to see my wife's friend fail repeatedly. for some of them it was just 50 contour integrals and done.

I taught it without expecting understanding of uniform convergence, leslie. Those who did, fine.

i could never figure it out. she did fine in her other classes, but complex was required for the major.

Computation is not trivial for most people, anyhow,

i didn't intervene because i didn't want to mess their friendship up.

that's true, it is a stumbling block in ODE and beginning linear algebra. i struggled a lot with matrix computation by hand.

I find matrices pain compared to using circular functions.

Can never get a grip on it in spite of the fact that it's used all the time for shaders.

11:48 PM

I think I asked for the integral of $\bar z dz$ around the circle of radius 2 on the final. Some people told me 0 by Cauchy.

oof, you can't fix that.

Uh huh.

Funny how I remember this from 25+ years ago.

Not a killer final exam question.

"ted shifrin is so crazy he asks these problems that nobody knows the answer to on the final"

that's supposed to be 0 when integrand is analytic?

Well I don't know, I saw Faa di Bruno elsewhere in relation to the trace expansion of a determinant and indeed didn't think to apply it here

11:50 PM

Yes.

I read complex analysis a long time back.

That's when I started using this site :)

yes and ted had the devious idea of choosing the one function that is example 1 of not being analytic.

because he's crazy.

Meanie!

i heard he's on probation and if he gives one more F they will fire him or something.

can you fire an emeritus professor?

11:52 PM

LOL

yes. it's a special procedure.

koro has seen through my not-too-well-thought-out rumor about prof. ted.

Leslie specializes in firing unemployed people.

user image

I think the way forward is to use just the part containing sum (function of x) as a first guess and see where it goes from there, that seems reasonable with the Faa di Bruno beast in the back of my head

ted, it's a niche practice but we like what we do.

@TedShifrin Thanks, I'll take a look at that for sure. I'm in home quarantine for some days anyways. :)

11:56 PM

Ugh. Be well!

You kick people when they’re down, leslie. This is where munchkin learns.

@TedShifrin thanks :).

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