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12:00 AM
What I really wanted to know there was whether any problems were too tricky for me to figure out. :)
Already 1.14 seems pretty challenging.
 
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.
Yeah, 1.1.14 is challenging.
 
Seems fun. If I have time I'll try it.
 
If he's had basic analysis, 1.1.13 might be fun for him. But none of this is "geometry" yet :P
 
So you want $C^3$, I assume, for 1.1.14.
computing the derivative of curvature
 
Actually, it turns out you don't need $C^2$ even.
 
12:02 AM
Wow.
 
I worked out such a proof, but there's a far more elementary one. Remember this is in the book before the definition of curvature.
 
This is a great format for HW.
I'm going to do this next time I teach upper div.
 
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.
LOL
 
Actually, let's just not discuss this here. Glad to elsewhere.
@TedShifrin Yes, this seems like a good strategy for precisely that reason.
 
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.
 
12:07 AM
Agree. I'm mainly hoping he finds it fun.
That's my third indep study student this semester. I'm at my limit.
 
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.
 
(Two of them only require weekly meetings and ~0 prep, though.)
 
Yikes, well, you're being a good samaritan, especially in these COVID days.
A number of MSE questions have come from my book, it turns out, so a clever student may search and find them.
And, of course, a number are totally standard in every book.
 
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. :)
 
12:11 AM
That's true. I just think that the discipline is up to the student and not the instructor, largely.
I like the problems already. Haven't read any of the text.
 
Well, in general 25-person classes (or even 10) I find that the instructor has to provide a lot of energy to motivate students.
Cool, I'm glad. :)
 
That may be true.
 
If either of you find mistakes or have suggestions to improve, I'd love to have 'em.
I did my major revisions while I was teaching, of course, but there's still room.
 
I'll let you know if so. Might take us a little while to get into the swing.
 
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.
I stated it when I used it.
 
12:17 AM
I'm unhappy with Jack Lee's rigor sometimes, so that's ok.
I learned from his book but I'd never recommend it to a student who isn't excessively careful.
 
Cool, we agree on this point of pedagogy. Diff geo is a good place to "go with the flow" sometimes. (Couldn't resist.)
Anyhow, my compliments to you on your generosity with your students.
 
Thanks. It's the part of my job I most value. And one-on-one communication is fun.
Just... takes time. :)
oh, btw, edwards and penney is an invaluable resource
really appreciate the rec
 
Oh, glad you like it. As I mentioned, the 2nd edition was by far the best.
I have no idea how far it is now. 12th or something. As far as I know, Edwards is still alive, but Penney died some years ago.
 
bleh, I need to brush up on measure theory. If $\int (f-g)\phi = 0$, and $\|\mathbf{1}_E - \phi\|_1 < \epsilon$, I have $|\int (f - g) \mathbf{1}_E| = |\int (f - g) \mathbf{1}_E - \int (f - g) \phi | = |\int (f - g) (\mathbf{1}_E - \phi) |$ ... $\le \|\mathbf{1}_E - \phi\|_1$...?
 
Your first equality is wrong?
You forgot to add back the thing you subtracted.
Oh, it's $0$.
 
12:31 AM
no its $0$
 
Sorrrreeee.
Hmm, $L^1$ bound, not $L^2$? What are you trying to prove?
 
hrmp, $L^1$ for free, but perhaps $L^2$, too, which, then, yes, Hölder's inequality applies
 
I agree you should tell us what you're trying to prove.
 
Let $f \in L^p(\Omega), g \in L^q(\Omega),\ 1 \le p, q \le \infty$. Prove $T_f = T_g \iff f = g$ a.e.
(if $\int_E f = \int_E g$ for all Borel sets $E$, then $f = g$ a.e.)
the parentheses is not part of the question, it's the proof "strategy"
 
$f$ and $g$ are in different $L^j$ spaces?
 
12:35 AM
hrmp, but theyre both $L^1$
 
$\Omega$ is finite measure?
I'm decades too rusty to see anything immediate.
 
doesn't say
 
Well, how do you know they're in $L^1$?
 
$L^p \subset L^1$
by Hölder's inequality
 
I am going to disappear, actually. I have to cook some stuff for dinner before the debate starts.
Isn't that only if $\Omega$ has finite measure?
 
12:37 AM
ohhh yes
true true
man I was really hoping to get this done before the debate :\ bon apetite
 
Oh, so you can choose $\phi$ to approximate $\mathbb 1_E$ in any norm you want.
I'll let my subconscious ponder it while I am cleaning green beans.
Bye.
 
bye
$\Rightarrow$ is the only hard part, obviously..
distributions in $\mathcal{D}'(\Omega, \mathbb{R})$
 
Yeah I figured as much.
@JoeShmo Ah and this makes sense more or less because these functions are locally $L^1$
 
hrmp, how do we know?
I found a cuter proof of this fact online, but there's a gap in that proof I'm trying to bridge, and I want to get to the bottom of this proof anyway
 
Like you said, $L^p \subset L^1$ by Holder's on a space of finite measure, and certainly $\Bbb R^n$ is locally of finite measure
 
12:43 AM
oh, sure
can I approximate any $L^p$ function by a continuous, compactly supported function from below?
 
iunno, does that matter?
 
well, there's an alternative (cuter) proof of this fact here -- math.stackexchange.com/questions/593701/…
but in order to invoke the dominated convergence theorem, I need $\phi_n \nearrow f$, or something
 
i think you need much less than that, but i'm not gonna think too hard about measure theory
 
12:59 AM
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 $f = \mathbf{1}_E$)
 
 
4 hours later…
5:15 AM
@Everstudent i try to just write the essential definitions theorems and ideas of proofs, so it doesn't take much mental energy when writing
@Everstudent by the way if u can solve that problem please reach out to me i'm also in need for such help
 
 
1 hour later…
6:23 AM
Mornin'
 
Good morning does anyone want to hear a song I made?
 
 
3 hours later…
9:46 AM
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)
 
 
3 hours later…
1:10 PM
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
idk if that makes any sense
 
1:25 PM
$F$ is a topological field?
$F$ being a field means $F^{\times}=F\setminus\{0\}$
 
Yeah so
inversion being continuous should mean $F^\times$ is open right?
 
So the question is whether $\{0\}$ is closed
 
or am I stupid
ahhh
 
Since translation is a homeomorphism, $\{0\}$ is closed iff all singletons are closed
and this is equivalent to $F$ being $T_1$
@Edward what's your definition of topological field, actually
 
A field $F$ together with a topology for which addition, multiplication, and inversion of non-zero elements are continuous
 
1:29 PM
ok, that's actually stronger than "a topological ring which is a field"
interestingly
 
But really what you mean is a topological ring and restricting the inversion operation to $F^\times$ is continuous wrt. subspace topology
 
any field $F$ is a topological field with the indiscrete topology, which isn't $T_1$
so what topology does $M_n(F)$ carry if $F$ is indiscrete
 
What's the closure of $0$?
 
if $F$ is indiscrete? the entire space
 
err well the topology on $M_n(F)$ is the product topology after identifying with $F^{n^2}$
I'm a topology noob btw
so be patient
 
1:37 PM
ok, so let's take $F=\mathbb{F}_2$ with the indiscrete topology
 
I'm pretty sure some separation is assumed for topological fields in any sane definition
 
oh wait
the product of indiscrete topologies is of course indiscrete
so if $F$ is indiscrete, $GL_n(F)$ will never be open in $M_n(F)$
however, according to this, if $F$ is a non-Hausdorff topological field, it is necessarily indiscrete
so that's the only bad thing that can happen
 
Maybe I'll give you the context
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
 
There is only one topology on $M_n(F)$ in what you wrote
 
just the product topology right
 
1:44 PM
Yep
 
so it's a union of products of open subguys of $F$
 
$GL_n$ is open because it is the preimage of $F\setminus\{0\}$ through $\det$
 
i said that above
but why is $F^\times$ open in $F$
 
Because $F$ better be $T_1$ in your definition of a topological field
(already for topological groups $T_1$ implies completely regular by the way)
 
$T_1$ is equivalent to all singletons being closed?
 
1:48 PM
yes
 
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.
 
aight
 
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.
 
you'd have an easier time with a better choice of $x_1,x_2$, I believe
but in any case, you can just write down what $f,g$ ought to do on the basis
 
2:10 PM
@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.
 
oh, I see the issue now
these aren't linearly independent anymore
so the original choices are probably the simplest
turns out that, unsurprisingly, your professor put more thought into this than I did
 
Lol...So how do I find $g$? I think I was able to find $f$ (see my above definition), but I don't know how to find $g$>
 
2:25 PM
essentially the same way as to how you found f
just reverse the roles of odd and even
 
Oh, so $g(e_i) = 0$ if $i$ is even and $g(e_i) = e_{2i}$ if $i$ is odd?
 
plug in the basis and see what works out
 
2:43 PM
I think I figured it out: $g(e_i) = 0$ if $i$ is even and $g(e_i) = e_{2i - 1}$ if $i$ is odd.
 
3:16 PM
Hey... what is this apparent contradiction? $i\ln(e) = i$? wolframalpha.com/input/?i=iln%28e%29
$\ln(e)$ just "magically" disappears?
 
ln(e) =1
$e=e^1$
 
Checks the amount of coffee had this morning and sees I haven't had enough
Hi, yes, where do I delete my account?
Oh man, that's the best one yet for me.
I'm trying to figure out if there isn't a hack for $e^{ix}$ using the identity $a^{b} = e^{b\ln(a)}$
 
$e^{ix}=\cos x+i\sin x$
 
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.
 
sounds like you're either looking for nonsense or tautologies
 
3:29 PM
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.
 
What do you mean by "computing the circular functions" ?
 
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)
 
there are plenty of known methods to compute sine and cosine numerically
 
And the exponential gives you a pretty good approximation
Complex exponentiation is pretty straightforward if you want a numerical approximation
 
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).
 
3:36 PM
$e^x = \sum_{n=0}^\infty x^n/n!$
Truncate at arbitrarily large n
 
I think this is what's usually used: en.wikipedia.org/wiki/CORDIC
 
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'm aware of CORDIC. I want better.
 
I don't understand what you want, I'm out
 
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?
 
I have other stuff to do
 
3:43 PM
part of what makes a problem solvable is actually having a proper formulation of the problem
perhaps you could get better answer if there was a mathematically precise formulation of what you want
 
That doesn't help me, then, considering I'm terrible at communication, let alone formally.
But I guess in all the earth, there isn't a man who has the patience to help me.
Or the time for that matter.
 
I think the problem is that there isn't a man who knows what you actually want
4
 
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".
 
You can't have an algorithm of constant depth generating an irrational number with arbitrary precision
Simply because the amount of results you can get in constant depth is finite
 
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.
 
3:56 PM
An arbitrarily close rational approximation will tell you all the digits of a number
As soon as $|q-r|<10^{-n}$, q and r share n (maybe n-1 ?) digits
 
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 $e^{x}$.
 
Then the sum I gave above, truncated to a certain step, gives what you're asking for
 
$(2.7)^{x}$ is a poor approximation of $e$, but if it's all you need in terms of accuracy, then that is sufficient.
"I need more accuracy". Then add another digit to the rational approximation.
Do you get my point now?
I want to be able to do that with the circular functions.
 
Do you get mine ?
 
Yes, I do, and it doesn't meet my requirement of "constant number of terms".
 
4:00 PM
define "number of terms"
 
the number of terms cannot determine the accuracy of the results.
 
On a certain subset of complex numbers it does
 
Term as in term of a polynomial, but also in general, um... I don't know how to describe it generally for what I mean.
 
If you take the unitary circle (which is all you need WLOG to be fair) you can bound the error very easily
 
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.
(For my requirements)
 
4:03 PM
define h(x)=f(x)+g(x) and then suddenly there's one term on the left side and two on the right side, so this doesn't really work in general
 
Ok, but thing is, if you can implement h(x) as-is, then h(x) is now a single, constant number of terms and we no longer care about f(x) or g(x).
so therefore, we must say instead h(x) := f(x) + g(x)
 
if you can implement f and g, you can implement h
so in the end every possible expression is just a single term and this becomes superfluous
 
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.
 
that still doesn't define "number of terms"
 
I'll give a simple example, then.
 
4:09 PM
examples aren't definitions
 
It's the best I can do if I can't give a formal definition...
2 = sin^2(x) + sin^2(x) + cos^2(x) + cos^2(x)
2 = 1/2 + 1/2
err lol 2 = 1 + 1
 
you haven't answered my message
 
it's clear what you want intuitively, but what I'm trying to tell you that it isn't all that easy to reasonably abstract that
 
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.
@Astyx what message was that?
 
@AMDG My answer to this
 
4:16 PM
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
 
@Astyx well then I have to say I don't quite understand your answer. I was expressing a constraint, not making an assertion.
 
I'm saying $\sum_{n=0}^N x^n/n!$ gives a numerical approximation of $e^x$ for $|x|=1$
 
Oh, I see what you mean now.
 
If you want it up to $k$ digits, you can find a $N$ that works
 
I still can't have the accuracy be improved by increasing the number of terms.
 
4:20 PM
The number of terms of what?
 
@Thorgott well I'm glad you understand what I want now.
 
You have the accuracy improved by increasing the number of terms
Not sure what you mean
 
no, I do not understand what you want
 
@Thorgott Then what did you mean by this?
 
I mean what I said
Then I outlined one possible way of abstracting that notion
But ultimately you are the one who has to provide the definition you want
 
4:23 PM
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?
 
Do you want something like a fourier series thing to approximate functions using sine and cosine?
 
If it requires adding more terms to make it more accurate, then no.
 
It's your decision which functions you want to allow
You should also define what you actually mean by "closed form"
 
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.
 
4:29 PM
You still haven't answered my message
 
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.
 
So unless you give a precise definition of what you want your numerical approximation method to do, this is very pointless
$x-x^3+x^5+O(x^6)$ isn't an infinite sum
your definition of closed form isn't precise at all
 
@AMDG x - x^3 + x^5 is also a closed form of x - x^3 + x^5 + O(x^6)
 
oh, well, then just look at this for the series expansion at x=0: wolframalpha.com/input/?i=infinite+series+x%2F%281%2Bx%5E2%29
 
Is this still happening
 
4:35 PM
You make that sound like a bad thing.
 
it is
 
What specifically is bad about this?
 
Idk it’s all a bit flimsy and ill-defined and idk what’s going on
You seem
to know what you want
 
@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, $\f(x)$?"
Does that... make sense?
 
5:11 PM
@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)
 
Uniform convergence does seem to fit the description, though would need a bit of tweaking to make it exactly what I want.
I'm going to eat and do a few things and then I'll be back.
Well, in the meantime, I can at least say for now that $f$ is the limit of some unique function $g$ where $f$ is not rational and $g$ is rational.
Is there a general notation for "any series" whether product, sum, or otherwise?
I can at least use a specific definition in the meantime
I my question is better (though probably not perfectly) described as a conjecture: \exists \sum^{N}{f(x)} | \sum^{\infty}{f(x)} = \sum^{N}{f(x)}
I don't even know if conjecture is the correct word.
There exists a sum of N terms of f(x) such that the sum of N terms of f(x) is equal to the sum of infinite terms of f(x) for any integer N.
The sum can converge or diverge.
And f(x) is not equal to zero.
 
 
1 hour later…
6:56 PM
@SayanChattopadhyay how's this? $$\forall \alpha \in \mathbb{N},\forall \omega \in \mathbb{N},\exists f(x)=\sum_{\alpha}^{\infty}{\delta_{\alpha}(x)},\exists g(x)=\sum_{\omega}^{n}\delta_{\omega}(x) : f(x) = g(x)$$
 
is there simple way to see that any if we perform any column operation on n independent vectors they always remain independent
 
@SayanChattopadhyay Then for $f(x) = \cos(x)$ what is $g(x)$?
Actually, no, let's go simpler: $f(x) = \arctan(x)$.
 
7:31 PM
@Thorgott Is it compulsory for international students to know German in order to get admission in a University in Germany?
 
@Knight I had to take a German exam
 
@EdwardEvans Hmm... I mean was it of A1 level or what?
 
I took C2 but I think you need B2 minimum or smth
 
And I think (from google searches) that German Universities take international students for almost masters only.
They seem to be taking very less international students for bachelors, what could be the reason?
 
8:02 PM
from what I understand a radon transform attempts to reconstruct a 3d figure from a 2d figure. Is that basically the idea?
 
8:14 PM
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.
 
8:34 PM
Is there a function of a parabola whose focus lies on the line x=y?
 
@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.
 
hi chat
 
o/
@Astyx so what was your message again? Can you restate it for me so I can answer it?
I think I can do better this time given the latex I demonstrated above.
 
From what I understand, you're asking for a function of bounded "depth" that gives approximation of exponential up to a certain decimal
I'm saying that $\sum_{n=0}^N x^n/n!$ works
For $|x|$ bounded
Which is all you need if you're solely interested in cos and sin
(as a matter of fact sin and cos have their own series which you can truncate)
 
8:50 PM
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?
The >2 part makes sense.
The prime part---not sure.
 
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