Perhaps. Some get challenging, but many of those didn't even make it to my pyramid list. Some did. I am willing to provide a hint if you ask, but I won't force it.
This is the only course in which I could really pull that off. It made grading homework for 25 students a pain, though.
The required problems, of course, were things that I tended to put in some form on exams.
I always had a variety of students. Some were weak, some were very strong. I thought this was the best way for them to optimize their enjoyment/learning.
I have an innocent comment. In all my upper level undergraduate courses, there were occasional beginning graduate students and Honors students taking them for grad credit. So I expected them to do some harder problems. That's a way of handling this in an easier fashion.
Anyhow, I hope your student finds it interesting and learns a lot. Whether you do is less important :P
Balarka worked a number of my exercises some years ago, I think.
My good students loved the course, and even some of the weaker ones got a lot out of it. Some of the proofs with Codazzi and Gauss equations are less than inspiring, so don't belabor that too much.
Yes, I'm sure one can. At this level of education I've never really cared. It's up to the student to decide if they're actually learning when they do that. If they are, great. If not, well, they're gonna have a bad time.
Well, especially with independent study, it takes a disciplined student. But I think my text is relatively readable, with a number of pictures and examples.
Anyhow, my feelings won't be hurt if you guys don't like it. :)
Caveat: I purposely did not write presuming the sophistication of someone who's studied real analysis. So I'm a bit sloppier, on purpose. Jack Lee was unhappy places with my lack of rigor. I didn't revise to please him :)
And I certainly assumed things like maximum value thm on compact (=closed/bounded) once or twice.
basically he's stuck on the same issue I have. He has convergence of $\phi_n \rightarrow f$ in $L^1$, but that's not enough (in my case $f = \mathbf{1}_E$)
This is interesting. One can prove Darboux's theorem on symplectic manifolds using Weinstein's splitting theorem. And Weinstein's splitting theorem's proof does not use any deformation type of argument (i.e Moser's Trick)
Let $M_n(F)$ be the ring of matrices over a field $F$. Then $M_n(F)$ is a topological ring, and $\det : M_n(F) \to F$ is continuous. Is $\operatorname{GL}_n(F)$ open in $M_n(F)$ ? If I knew whether or not $F^\times$ was open in $F$ (as a topological field) then I could argue that $\operatorname{GL}_n(F)$ is the preimage of $F^\times$.. but I don't know that
I showed yesterday or the day before or smth that for a non-archimedean local field $F$, the additive group $F$ and the multiplicative group $F^\times$ are both locally profinite groups. For an $n \in \Bbb N$ the vector space $F^n$ carries the product topology and that makes it into a locally profinite group too, so $M_n(F)$ is a locally profinite group under addition, in which multiplication of matrices is continuous.
The group $\operatorname{GL}_n(F)$ is then claimed to be an open subset of $M_n(F)$
I think my problem is that I'm struggling to track which structures these mfs are talking about
Let $S$ be some ring, and let $F$ be a free $S$-module with basis $e_1,e_2,...$. I am trying to show that the following functions form a basis for $Hom_{S}(F,F)$ as a $Hom_{S}(F,F)$-module. Let $x_1(e_i) = e_{i/2}$ if $i$ is even and $x_1(e_i) = e_i$, and let $x_2(e_i) = 0$ if $i$ is even and $x_2(e_{(i+1)/2}$ if $i$ is odd.
I've already shown that $x_1$ and $x_2$ are linearly independent, but I am having trouble showing they span $Hom_{S}(F,F)$. I'm pretty certain it spans it if and only if they span the identity map, so I just tried showing that they span the identity map...but to no avail.
So, I am trying to find $f, g \in Hom_{S}(F,F)$ such that $id_{F} = f \circ x_1 + g \circ x_2$. I could use some help.
@Thorgott Yeah, I think you're right. But this was the basis my professor suggested.
For $f$, I think it should be $f(e_i) = e_{2i}$ if $i$ is even and $f(e_i) = 0$ if $i$ is odd; but I couldn't figure out $g$.
@Thorgott What if I just define $x(e_i) = e_i$ if $i$ is even and $x(e_i) = 0$ if $i$ is odd, and $y(e_i) = 0$ if $i$ is even and $y(e_i) = e_i$ if $i$ is odd. Wouldn't $id_{F} = x + y$?
and then I'd be done? Really I am just trying to show that $Hom_{S}(F,F)$ is rank $2$, so you're right that the problem would be easier if I found a simpler basis.
Yes, well I'm looking for $\cos x$ and $\sin x$ without using something so expensive as complex exponentiation. Would be rather easy if I had real-valued forms of $x^{i}$ and i'th root of x.
Why do you say that? Do you claim there isn't a more trivial means of computing the circular functions? They're functions that describe a very simple form of motion which is trivial to perform; there is no indication that the circular functions shouldn't be simple and trivial to compute.
Sine and cosine are circular functions. To compute them, you take an approximation of one of their definitions to get approximate values of sine and cosine--or you get a right triangle and compute the ratios directly if you have all the side lengths (which is simple and trivial).
I'm looking for a definition of the circular functions that allows approximations to be of arbitrary accuracy such that the accuracy increases with the number of digits provided for the input and the transcendental numbers used rather than the number of terms.
(I had previously said arbitrary precision, but that was an inaccuracy)
And which of them is most trivial and gives the desired accuracy as I have just described? With something like $e^{x}$, if I want to improve the accuracy, then I just increase the number of digits that I use for $e$ as a rational approximation of $e$. I want the same for the circular functions (in principle).
That infinite sum doesn't matter in terms of implementation. If I have a function $a^{b}$ for all real values of $a$ and $b$, then I can provide a rational approximation of $e$ as the value for $a$ without an infinite series assuming $a^{b}$ is not an infinite series.
I don't get it. Why do people just give up when they don't understand something, eh? There's this thing called interrogation... or how else do you think all of these problems were solved?
Yes, and when I attempt to explain, no one seems to understand, and after just a little bit more of my feeble attempts to explain, they just give up.
What I want is a real-valued, arbitrary accuracy, closed form set of circular functions whose accuracy does not depend on the number of terms.
I don't care if the stupid thing has a million terms in it, or if it isn't even a polynomial at all (since I can make just about anything periodic using modulus). It just has to remain constant in the... I don't know how you would call it... the length and height and depth of the algebraic expression that defines the function, I guess?
(x + iy)^n... that expands into many terms. On the contrary, if the number of terms it expands to is constant in number, then that is what I mean by "constant algebraic expression".
Yes, but... you can't have irrational numbers be represented by a finite number of digits, so that is a moot point. It has to be a rational approximation.
We're dealing with computers of a fixed number of bits.
Yes, up to a certain number of digits, sure, and that is what I'm aiming for. If I want more accuracy, then I improve the accuracy of the function itself like I said with $e^{x}$.
If I have a + b, a and b are both terms. if I have f(x) + g(x), then the functions f and g are both terms. In fact, f and g don't even need to be functions.
Yes, ok, so then I think it would be better to say that both f(x) and g(x) expand to a finite number of terms in their definition, and there cannot be an indefinite number of nested terms in either f or g.
i.e. f(x) can be defined by an infinite series, but you can't use an infinite series definition to define f(x) strictly in this context. You have to use a different definition.
I want to use 1 + 1 = 2, clearly, because even though I can represent the constant 1 using an infinite series, I have it right there directly as a finite number of terms, but sine and cosine, as far we know at this time, since I must use a finite number of terms, and sine and cosine would require an infinite series, then I must use 1 + 1 = 2.
you'd likely have to do something like settle on a set of primitive functions and then only consider those functions which can be formed from those primitive functions by, say, composition, multiplication, addition, inversion, division or some other set of operations and then define the number of terms as the shortest number of primitive functions appearing in one way of writing such a function down in terms of these operations applied to primitive functions
Well I would say your definition there is more than sufficient. So that then begs the question of (since we're dealing with waves) what set of primitive wave functions can we use to compute the circular functions?
Well for that you need to define what you mean by accurate. If a set of values approach a fixed a value, then you get closer and closer to the value as you increase the number of values you consider in that set. Think "sequences" and limit of sequences
Sequences could be one way to define accuracy, what do you want to define as accuracy?
Maybe there is a probabilistic definition of accuracy, like hitting darts on a board, but I do not know how one would define something like that for things like functions and their values
The thing with numerical mathematics (or any form of mathematical modelling) is that there is not a single god-given metric that tells which algorithm/which model is the "best" one to do anything, so an integral part of doing this type of mathematics is to start by deciding on which metric you want to use to determine how good your algorithm/model is and then actually go and construct it.
Well, by closed form, I mean something that does not strictly require an infinite series as the only kinds of definitions for which the function can be composed of. For example, the infinite sum x - x^3 + x^5 + O(x^6) has the following closed form: x/(1+x^2)
I don't want the infinite sum, I just want x/(1+x^2). Now what is that for the circular functions? That's my question.
@SayanChattopadhyay well I could use $\f(x)=\frac{x}{1+x^{2}}$ as an example. substitute 1 for some variable, I'll call it $a$, such that we now have $\g(x)=\frac{x}{a+x^{2}}$. Then as $a$ approaches 1, $\g(x)$ approaches $\f(x)$. $\g(x)$ = $\f(x)$ iff $a$ = 1. Let the error of $\g(x)$ with respect to $\f(x)$ be $|\f(x) -\g(x)|$. Then the accuracy of $\g(x)$ increases as the error of $\g(x)$ with respect to $\f(x)$ approaches 0.
I don't know how to generalize that :)
I can then say that $\g(x)$ is formally an approximation of $\f(x)$. Actually I could probably generalize this to all functions by saying $\f(x) = \delta(x)$ for some behavior $\delta(x)$ and $\delta(x) \approx \g(x)$.
Then I could restate my question as "What is $\g(x)$ for any circular function, $\f(x)$?"
@AMDG What I can understand is that you seem to be talking about something like uniform convergence for a sequence of functions. Anyway, the term thing that you were going about doesn't make sense in this constant, because to approach closer you are going closer to it by increasing the no. of terms you consider in your function.
Also not that such a $g$ is not unique. So I don't know what you mean by what such a $g$ is for a $f$. You can approximate a function by a lot of different kinds of sequence of functions (depending upon wether you want pointwise convergence or uniform)
Let $P$ be a left-cancellative monoid, $q \in P$, and define $qP := \{qp \mid p \in P\}$. I conjecture that $qP = P$ implies $q$ is invertible in $P$ (the converse is trivial). Note that $1 \in P = qP$, so there exists $p \in P$ such that $1 = qp$, so $q$ has a right inverse. However, I don't see how to prove that it has a left inverse.
@jeea These are the columns of a matrix? Column operations are invertible. If you ended up with a column of zeroes, you could never go back to the original vectors.
Ok, yes, that does give a good approximation for the exponential function, however, for that finite sum, you still need N terms for which, like you said, $a$ digits of accuracy corresponding to some $N$.
Instead, what I would like is: $$\forall \alpha \in \mathbb{N},\forall \omega \in \mathbb{N}, \exp(x)=\sum_{\alpha}^{N}{\delta_{\alpha}(x)},(\exists g(x)=\sum_{\omega}^{1}\delta_{\omega}(x) : \exp(x) = g(x))$$.
@TedShifrin do you know offhand why people tend to discuss $n$-colourings of knots where $n$ is specifically a prime, rather than letting it be any natural number >2?
