I just looked in my multivariable text to see how I dealt with this issue. There's no harm in pointing out that $|g(\phi+\Delta\phi)- g(\phi)-g'(\phi)\Delta\phi|\le\epsilon|\Delta\phi|$ for $|\Delta\phi|<\delta$.
Then there's no problem.
I actually don't like Courant's epsilonics. He's thinking of $\epsilon$ as the error, which is a function, so it's good engineering style but bad analysis style.
But for $\Delta \phi = 0$ when $\Delta x \neq 0$, doesn’t $phi^{‘}(x)$ and $\epsilon$ have to equal $0$? Thus, in the final formula $\epsilon \phi^{‘}(x) = 0$ no mattter what the actual value of $\epsilon$ is at zero. Is my logic here incorrect?
Courant’s textbook has been my first dabble in learning calculus properly; I’ve been enjoying it but some of his proofs are very short and go over important details. Hence me trying to fill in the gaps ahahaha
@TedShifrin do you think it’s worth editing my question to include the lemma u mentioned and give a more complete proof? Or should I add that as a comment?
@TedShifrin I’m so sorry for keep asking you questions, but I just edited my question to include my version of the proof. Can you see if that seems ok to you?
Also I would appreciate anyone else’s input as well!
@DavidChoi I also believe, that adding another answer or editing your question is OK until it doesn't undermine other answers. If you think your edit adds a perspective or value, then just do it. It may help someone.
@LucasHenrique Why is it so cool? Sorry, I haven't heard about it before. Wiki tells a bit about it's applications, but it's always better to hear from someone who already dealt with it
The solution with complex analysis is already here. I am looking for solution with other techniques like differentiation under the integral sign, or whatever else.
The differentiation under the integral sign technique only works when one is very, very fortunate. Sometimes you can set it up as an iterated integral and switch the order of integration. But the complex analysis approach is the most robust.
@robjohn The integral for $\phi(x)=x$ diverges even though all conditions are satisfied. Or did I overlook something? Also, can we set $a=0$ and $b=\infty$?
@vitamind In the sense of distributions or fourier transforms $\int_0^\infty e^{i\lambda x}\,\mathrm{d}x=\frac i\lambda$. It is true that $\lim\limits_{N\to\infty}\int_0^N e^{i\lambda x}\,\mathrm{d}x=\lim\limits_{N\to\infty}\frac{e^{i\lambda N}-1}{i\lambda}$ does not exist (it oscillates about the value I gave above). However...
If one uses the polarized version of Plancherel's Theorem, to integrate against a nice function, you get the same thing as integrating the fourier transform of that nice function against $\frac i\lambda$
Since you have sorted out the source of misunderstanding, here I offer an alternative solution, which quite surprisingly* allows you to find the exact roots of the polynomial while avoiding calculus. The method is (much) longer, but I find it interesting and I hope you will too.
(*I say surprisin...
That took a long time to type up but I think it was worth it, it was a lot of fun :)
I have some question that I can't understand. If the real part of $\bar{z}f(z)$ is always positive on the unit circle, then $f$ has only one simple zero on the unit disk.
One thing is that there is no condition on $f$
Hint is to use Argument principle but the hypothesis of that principle is function to be analytic at least on and inside of simple contour
Let $f$ be an analytic function on $\{z:|z|\leq 1\}$ such that $\operatorname{Re}(\overline{z}f(z))>0$ for $|z| =1$. Then $f$ has one simple zero on the unit disk.
My attempt: I tried to show this using argument principle i.e. Claim : $\frac{1}{2\pi i}\int_{|z|=1}\frac{f'(z)}{f(z)}=1$. To see t...
If I have two sets of numbers A & B, and a binary operation, is there a name for the set resulting from doing the binary op on the elements of the Cartesian product of A & B? Eg, let A={10, 20, 30}, B = {4, 5, 6}, and the binary op be addition. Then C = {14, 15, 16, 24, 25, 26, 34, 35, 36}.
@Koro I don't know why you suggest "range". Also, that could be confusing in this context, which is using Python, because range already has a specific meaning in Python. (But of course you weren't to know that I was talking about Python).
even then, i kinda wonder. it isn't clear to me why you'd put complex analysis after real analysis, for example. unless your view is just that you want someone trained in epsilonology before complex analysis.
except for basic familiarity with sets as is common to all mathematics
ofc more advanced branches of number theory such as algebraic or analytic number theory have algebraic or analytic prerequisites specifically
@leslie I would like people to know basic multivariable analysis before taking complex analysis
how are they gonna understand the Cauchy-Riemann equations otherwise
I guess you could make the point that one doesn't need the Cauchy-Riemann equations for most of complex analysis (at least not explicitly), but that would be very artificial
if they ever wanna check whether a function is holomorphic, they should know how to differentiate
you could do it just from power series. i'm not saying this is optimal, just that it could be done. and precisely as you note, because of the cauchy riemann equations, a lot of the pathologies you see fairly early on in real analysis just don't come up.
at my undergrad the complex analysis class tended to be a lot like a calculus class. lots of contour integrals and calculational stuff. at least when the postdocs taught it.
you had to find the right faculty member to get a real class.
i think thats because the sequences and series part of most calculus courses is about getting your hands dirty with using various tests to show convergence/divergence, and this is only a small component of a real analysis class
my analysis instructor wanted the real analysis class to be a point set topology class.
maybe 2/3 of the semester was chapter 2 of rudin PMA. we did technically get to integration and the fundamental theorem of calculus, but it was an afterthought.
yeah that's not ideal, we had a real analysis class which actually covered a surprising amount of baby rudin, pretty much everything up to start of the differential forms bit, and also a class called metric/topological spaces that covered all the point set stuff
For the top answer of this post here: https://math.stackexchange.com/questions/2670/is-the-closedness-of-the-image-of-a-fredholm-operator-implied-by-the-finiteness
This is kind of a slick trick, but I don't really see how the assumption of finite dimensionality of the cokernel was used... any clarification is much appreciated :)
Also, it is not obvious that T' is surjective, if the range isn't already closed, right? I'm just confused...
that's where i'd tell anyone to look. it has the downside of being fairly expensive to purchase. unfortunately, i understand there are ways of obtaining copies of books without purchasing them.
my advisor was writing a paper once and had this really tangled argument. i said, this is done in two clean paragraphs in lax, why not just cite that. he kept it in because he loved it so much.
later i realized the tangled argument also worked for unbounded operators.
sometimes more senior people really do know more than you.
Ah I think I figured it out... The fact that the codim is finite means C is fin dim. This guarantees that the induced map T' (as Jonas calls it in his answer) is bounded. Were C not fin dim, there is no guarantee that this map would actually be bounded, so the OMT would not work
if people notice you retagging to game the system for perms you will probably be suspended (but you'd have to edit a lot of questions and it'd be easier to just answer those questions appropriately)
@Yorch well of course, it's "inappropriate" for a reason
and if you are dupehammering posts that aren't duplicates that is an abuse of power and bannable as well
there are lots of 'meta' mechanics as to math.se that i just don't understand. i leave it in the capable hands of mods, who have the thankless task of keeping the ship afloat.
@copper.hat They bought back swag a few months ago. And it's even retro-active, so if you passed a swag milestone (100k or 250k) during the no-swag period, you're still eligible for swag. There's info about it on the mother meta.
@Yorch Across the network, the general attitude is that such retagging should be organic. That is, do it when you're writing your answer, or when you run into questions that need tag edits. Don't go on a binge of retagging all the old questions you answered and flooding the front page with a bunch of old stuff.
we have a very functional relationship but sometimes we frighten our coworkers. the other day we were asked by someone if we would take someone out for a zoom lunch. there was some toxic back and forth that ended in her emailing me, ccing everybody, "Shut up. YOU DO NOT TALK." i think she forgot that our inside jokes are between us and not generally known.
if i'm being honest, i don't generally care about it. i just sometimes feel silly for weighing in on something only to learn it's only on my radar because someone edited the tex, or whatever, and the OP is long gone.
The Community user bumps old posts that have zero score on the question and its answers, so that we can vote on them. So when you see such bumped posts, vote on them so that Community won't bump them again.
ted cruz is the classic example of someone who isn't working class putting on some carhartt and playing dress-up and for some reason the entire state of texas bought into it. i don't know if they were offered better alternatives.
phony classism really ticks me off but that's another story.
that's a really good map projection.
the new zealand being made out of two tiny south americas is what does it for me.
@leslietownes Americana singer / singwriter / guitarist Gillian Welch was starting to have some success, but some of the country music crowd complained: "How dare you write country music! You grew up in LA!". In retaliation, she wrote this gem: Time (the revelator).
My first time in China proper was on the island of Hainan on Christmas day. I suddenly realised that a map is practically useless in a town unless you can identify the characters.
@leslietownes Excellent. The closing track on that album is the 14m44s epic I Dream A Highway. It's one of the best highway metaphor songs of all time. You have to be in the right mood to appreciate such a long song, but IMHO it's well worth the journey. ;)
as long as it isn't "life is a highway," we can be friends.
i was not familiar with this artist, thank you.
we're at a weird moment with data acquisition. for a while i remember seeing people arguing that we didn't need the US government to be involved in the weather. citing private companies that refined and monetized what was taxpayer funded data.
hello, please is this has a sens : On $\mathbb{R}^n$ Let $E^r$ be the sub vectoriel space from $C$ constituted by functions which have continuous partial derivative until the rang r
the sentence does not identify C. it isn't clear where the functions have their ranges (maybe it is $\mathbb{R}$).
the rest is pretty clear to me although may not be how i would have written it.
with multi-index situations it is sometimes helpful to point out whether 'up to r' means you're summing or maximizing. e.g. is d^2f/dxdy order 2. it probably should be but it's also an iteration of two first order derivatives.
@vitamind The lemma is stated on an interval and, independent of $a$ and $b$ the inequality holds. However, the boundedness of a function does not imply its convergence. So, the lemma implies the bounds on the value one gets from looking at it as a distribution, but it does not imply its convergence.
