maybe it’s just me imagining, but I’ve been seeing lots of questions on how to evaluate pullbacks, and it almost saddens/annoys me that everyone tries to do it directly from first principles and definitions rather than learning the few simple rules
Well, this is because of books and bad teaching. Mathematicians who aren't proficient with calculations with forms do it and teach it wrong :)
In fact, in my book, the formula with pushing forward tangent vectors is only in an exercise (which I never assigned unless someone asked me about it).
Ted you stated that $f(x) = x - g(x)/g'(x)$ will be a contraction mapping when $|gg''/(g')^2| \leq c < 1$. How did you determine that to be the lower bound for $c$? Is that what we are trying to accomplish in question 8?
Speaking of forms, here's a cool fact: For a Riemannian manifold $TM$, under the "Legendre transform" $STM \to ST^*M$ between the unit sphere and cosphere bundles, the geodesic flow on $STM$ turns into the Reeb flow on $ST^*M$ with respect to the contact form $\sum p_i dq_i$, $q_1, \cdots, q_n, p_1, \cdots, p_n$ being the generalized position-momentum coordinates.
This is why you see cusp-like things in your coffee cup when light is reflected on it.
@TedShifrin ChatGPT actually managed to solve my first year calculus problem sets pretty decently, and with some prodding, it refined its answers and I would have given it a 90-95%. sadly, some students blindly copied its first iteration, and ended up with almost crap
I'm confused, @D.C. Yes, we want to use the MV Theorem/inequality to argue that a such a bound on the derivative implies contraction. That's in the book.
I will be Balarka, even has its own footnote. Yes it is in the book in the readings, but I am just trying to see how you got it. Because you state the condition, but don't derive it
@Balarka I only learned it by trying to teach out of Hubbard and Hubbard before I wrote my book. I think it's a wonderful result, but I still wouldn't teach IFT that way to freshmen.
Well, I do not like H&H for its exposition or its exercises, but the world sure likes it more than my book. I think they're on their 6th edition or something.
@TedShifrin Out of curious im just playing around with the problem, and as a sanity check, what if I say for all small $r>0$, $B_{f(x)}(cr) \subseteq f(B_{x}(r))$, where $c>0$ is some fixed constant, then $f|_{K}^{-1}$ is Lipschitz right?
@D.C.theIII well, the question seemed too long, and I saw the picture, and it seemed plausible. That’s why I skipped it. if the picture wasn’t there I probably would have done it
Maybe it's more natural to put the $c$ on the other side, but yes, monoidal.
That exercise appeared because I gave Spivak a very hard time for not having Newton's Method in his book. (I also added lots on "typical" applications problems, not to mention using Taylor polynomials and other things.)
With regard to D.C.'s earlier discussion, numerical analysts care deeply about the rate at which iterated processes converge :) Not that I ever took a numerical analysis course.
Speaking about confusion in posts, poor Nate is demanding that I draw him a picture with MS Paint. I really am handicapped by no longer having a graphics program on my computer.
I had purchased Microsoft Office and Adobe Illustrator many years ago with an academic discount. However, Apple's improvements (64-bit technology) made them bricks. And there is no way to replace them, since Microsoft and Adobe have adopted a "monthly rental fee" model. The hell with them.
I used Illustrator heavily for all my books. Now I'm screwed if I have to revise/update or add a diagram.
My thesis is on $h$-principles for Legendrian submanifolds, @TedShifrin, don't worry. The last chapter will use a little bit of microlocal sheaf theory.
No. Certainly Adobe does not, and I think I might have checked on updating Excel. For that I'm happy using Apple's Numbers, although it wouldn't open some very old Excel files.
@Balarka Well, maybe you'll motivate me to learn some of that. Nah, I'm too old to learn.
Yea when I said hard copy I should've said they provided you with a code to download the software to your system, but they are slick and reduced the amount of computers you can install the software to down to one computer......unless you pay this handsome "monthly fee"
It requires more effort and patience to write a coherent story for a research paper. That is why most people don't bother to put the effort, and write badly.
In my MSc thesis I'm taking a (quasi-)geodesic path to tell the story I want to tell.
Before the days of MSE, students would have gone to the professor's office hours and asked politely, "What the hell were you talking about when you said ... ?"
Hi @Ted, I am sslightly stuck Assume $X$ and $Y$ are metric spaces such that $f:X\rightarrow Y$ is continuous, and there exists a $c>0$ such that for all $x\in X$, there exists an $r'>0$ such that for $r<r'$, where $r>0$, $B(f(x),cr)\subseteq f(B(x,r))$, then for $A\subseteq X$, suppose $f|_{A}$ is injective.
Show there exists an $C$ such that for $x,y\in A$, $d(f(x),f(y))\geq Cd(x,y)$.
$f|_{A}$ injective means $f|_{A}^{-1}$ exists, where $f|_{A}$ is function from $A$ onto $f(A)$.
Let $x,y\in A$. It is enough to show $d(f^{-1}(x),f^{-1}(y))\leq \frac{d(f(x),f(y))}{C}$ for some $C$.
Well, you were so busy pedantic that you wrote it wrong :) I think I would forget about $f^{-1}$ and try to do this directly. The thing we actually want to prove is the statement about $C$ and the inequality?
If so, I'm tempted to argue by contradiction. That will localize it at a particular $x$.
Well, I was thinking locally Lipschitz, not globally. I don't know.
If $A$ is compact, I get an easy proof by contradiction. Suppose that you have $x_n$ and $y_n$ with $d(f(x_n),f(y_n))<\frac1n d(x_n,y_n)$. Pass to convergent subsequences and apply injectivity.
Okay your argument is all I need: Assume WLOG $x_n\rightarrow x$ and $y_n\rightarrow y$. Then, $d(f(x),f(y))=0$ so $x=y$ and why this gives contradiction?
A week or so ago I found a nice family of approximations for $e$. They can also be used for $e^x$. They aren't the most efficient way to compute $e$, but they're ok.
$\lim_{x\to\infty}(1+1/x)^x<e<(1+1/x)^{x+1}$ so $e\approx(1+1/(x-1/2))^x$. If we make $x$ an integer power of 2 we can do the exponentiation by repeated squaring.
Even better, $e\approx(1+1/(x-1/2+1/12x))^x$. We can generalise this using the continued fraction of $e^{1/x}$, which has a simple form, A110185. I assume Euler knew this stuff, since he was a master with continued fractions.
Writing $e\approx(1+1/f_k(x))^x$, here are the first few values of $f_k(x)$.
No issue, I just wanted to verify that the solution I worked out is the right way of going about it.
It feels like what I did was right, but there may be some notation issues perhaps.
If I'm thinking about this properly, I could look at $A^{-1}$ as being my function $g$ such that $g: \Omega \subset \mathbb{R}^n \to \mathbb{R}^n$ and then $f:g(\Omega) \to \mathbb{R}$.
This allows me to use the change of variables formula $\int_{\Omega} f \circ g(x) |\det Dg| dV$. Now $A^{-1}$ is a linear map. And the derivative of the linear map is the linear map itself. So $\det(Dg) = \det(A^{-1})$
I went about calculating the determinant for $A$. So after some row reduction and all that I get $\det(A) = (n-1)!$. Which means that $\det(A^{-1}) = 1/(n-1)!$.
Bringing this all together then:
$$\int_{\Omega}f(A^{-1}x)|\det(A^{-1})|dV = \int_{\Omega}f(A^{-1}x)| \frac{1}{(n-1)!} dV = \frac{1}{(n-1)!} \int_{\Omega}f(A^{-1}x)$$
and since $\int{\mathbb{R}^n}f dV = 1$ I am left with $\frac{1}{(n-1)!}$, as the solution. My only concern is with how I treated $f$ on the last step. Was that the way to go about it?
well no. So I would have to manipulate the function we're discussing to get it into the $f(x)$ form. Oh...this goes back to the idea of shifting the intergral. I'm trying to find the exercise in SPivak that I'm talking about
Oh, wait. If $d(f(x_n),f(y_n)) < \frac 1n d(x_n,y_n)$, and $x_n,y_n$ are in an $r$-ball centered at $x$, isn't $d(x_n,y_n) > nd(f(x_n),f(y_n))$ contradicting the stuff about the $r$-ball at $x$ and $f(x)$?
Yes, but to prove by contradiction, you asserted $x_n,y_n$ are in $K$. No reason for $x_n,y_n$ to be in r-ball centered at $x$. You don't have control on where x_n,y_n live, no? Unless im being silly.
I got caught in what you told me to be aware of Ted, fixated on the mechanics. Like you said I should've just let $g = A$. Then consider $g$ mapping from the image of $A^{-1}x$. then I can properly apply the hypothesis because $f \circ g = f(A(A^{-1}x))$. So what I'll have instead is $(n-1)! \int{\Omega}f(A(A^{-1}x)) = (n-1)! \int_{\Omega}f(x)$.
