Why would it feel any more weird than a $u = xt \rightarrow du = x dt \rightarrow \frac{1}{x} du = dt$ substitution? In this case $t = u \rightarrow dt = du$
In solving problems of mathematics, I am met with a seemingly impassable wall extending indefinitely in all directions. I find windows too small to fit through, and all that appears to be an entrance is a one-way exit. I honestly weary of trying to break through. On the other side, however, is paradise, and it is the paradise of complete mathematical freedom which I seek; and I have over 3000 years of hindsight nearly untapped to get there.
That being said, if I wanted to learn math based on elementary intuition of elementary mathematics alone, where would I go?
We're talking being unpracticed in elementary algebra but nevertheless grasping only as much as the elementary arithmetic operations up to and including the exponential and logarithm. I would say I know more, but everything else I learned in Algebra 2 and Geometry I forgot.
I was prevented from being able to go beyond that anyways, not that that would have mattered seeing how much I've forgotten.
we don't build up to the same things anymore. hardy published his stuff about a century ago and he introduced real analysis thinking to the english world
so if you grab stuff from a century ago chances are you won't be preparing to deal with up to date stuff
i mean the above specifically for pre-real analysis maths
what you want is a good precalc book with lots of exercises and then calculus with lots of exercises
I had a wonderful Dover book by Henry Burchard Fine called College Algebra when I was in high school. I don't know if it's still available. It's old-fashioned but very thorough.
One mistake you made is when you said $\lfloor\frac m2\rfloor=\lfloor\frac{m-1}{2}\rfloor +1$. This is only true when $m$ is even.
I find it is much easier to split into cases based on whether $m$ is even or odd.
If $m$ is even, so that $m=2k$ for some $k\in \mathbb N$, then
$$
\begin{align}
F_{m...
For reference, please refer to the above post, I made, where I used this identity.
There, some users are of the opinion that this is valid, if m is odd while some are of the opinion that the identity is valid, when m is even.
Ok, now, I am sure, that this identity is valid for m even only. But the answers are a bit confusing. If I find something not clear, I will come back again.
in general, neither implies the other (or the negation of the other).
i dunno really what to say about it, except, the relevant definitions don't correspond very well with "plain English" understanding of either term. which is fine because maybe there isn't a plain english understanding of point sets.
the thing that is generally true is, if a set is closed, its complement is guaranteed to be open, and vice versa.
which is already maybe tricky because 'complement in what?' and the openness or closedness can very much depend upon how you answer that question.
doors can be open or closed, but i don't think there's a plain english understanding of the complement of a door, let alone the complement of a door in some other, perhaps larger door.
@Koro The one line description is that rep theory is a way to turn questions about groups into questions in linear algebra, and the latter is much more well understood that the former, so it's a good translation
But this doesn't really make rep theory justice, it's a nice subject (that I know too little about)
Koro. Not all institutes are worst and not all professors. You can take a look at IISC's official website.The course structure there is as good as any foreign universities ( honestly I am not comparing with few std US universities).
@onepotatotwopotato the answerer is a very nice professor as per what I have heard.
I've not taken any class from him as he doesn't teach at my college.
So our functional analysis teacher was 'teaching' something and except some 2 or 3, the whole class was silent. He asked 'how many of you have understood?' Only two or 3 people raised hands. He still continued.
What was the point of asking 'how many of you have understood?'?
So if $G = GL_2(\Bbb Z/9)$ then $|G| = 2^4\cdot 3^5$. By the proof I linked, $g\in G$ has $3$-power order if and only if its image in $GL_2(\Bbb Z/3)$ does. Now the problem is asking: Show that a Sylow $2$-subgroup of $G$ is isomorphic to $\Bbb F_9^\times\rtimes(\Bbb Z/2)$ where the nontrivial element $1\in\Bbb Z/2$ acts on $\Bbb F_9^\times$ via $x\mapsto x^3$.
Is there a topological embedding $f:(X\times {0,1})\times [0,1)\rightarrow X\times \mathbb{R}$ such that $f$ has open image and contains $X\times {0,1}$?
@Koro Actually the situation is even worse here (in IIT Kgp), at your institute if you're leaned towards pure mathematics then at least you get to choose from the options, here you don't get that privilege.
Among the electives, there are only two courses on pure mathematics and you have no other options than to take those courses.
@shintuku Well books I'm really only inclined to resort to because I don't know where I can find a good teacher instead outside of a university to just learn everything optimally.
Either I find a teacher for one-on-one, or I resort to books.
Either way, I need a learning plan for getting from elementary arithmetic to the depths of discrete mathematics, real analysis, complex analysis, geometry, and I'll definitely want topology.
Having a teacher is safer; getting the wrong book will probably leave me in misery. I say "from a century ago" as a guess because the world was a bit more sane back then, and I know first hand what modern "pedagogy" so-called is, and I want none of it. I cannot fathom a multiple choice question existing a thousand years ago in a university.
Khan Academy is full of that stuff. I wasted a year completing half the algebra 1 course on there when I had already taken the course 80% of the way through; and I had already taken prealgebra in like 4th grade, but the papers were mixed up, so I was put in the wrong class; then in the right class; but then, again, said 80% through the course, I was taken back into prealgebra because the teacher was afraid I would fail the final exam.
I can't fathom what her reasoning was because that probably only happened since I didn't bother with the homework. I struggled to find the motivation to do so, especially when I already grasped the concepts.
idk, man. School is just a painful recollection and reminder that no one knows how to teach today, and having become Catholic and learned me some Filosofie, I finally know why too. When I was in some sort of precalc or geometry class, the calculus class was down the hall. I went over after class and asked for the homework and he denied it to me. Bruh. The same teacher, when I had him for geometry, questioned my ability to learn calculus by probing my knowledge of terminology instead of ideas.
*taken prealgebra in 7th grade (I remember I was going into 8th grade).
@shintuku I ended our conversation in this regard for a reason. For the sake of your reputation, I will not be stating what that reason is. Suffice it to say that the thing you have rejected infallibly refutes whatever "evidence" you might present to me on the subject.
Just for the record, such "great" philosophers as Kant said that 12 is a "synthesis a priori" concerning the sum 5 + 7 = 12. He is literally claiming that 5 + 7 is distinct in essence from the essence of 12--and the whole world seems to trust this man LOL.
^ just yet one more reason for me to distrust modern pedagogy and discipline in general, and is most probably also why my own profession is in ruins.
Maybe as before, $g\in GL_2(\Bbb Z/p^2)$ has order $p^5$ (order of Sylow $p$ subgroup) if and only if $\overline{g}\in GL_2(\Bbb Z/p)$ has order $p$ (order of Sylow $p$ subgroup) under natural sujection.
@SoumikMukherjee I'm not so surprised. If the courses which they are offering here at my college have nothing to do with one's interest (say, they are floating only algebra related courses), then having 'options' does not matter. At kgp, if you saw the brochure of your programme, and the brochure said courses x,y,z will be taught, then they keep those words. Now, if one goes to the college they know what subjects will be there. But here, as I said: they are not floating all courses
which were mentioned in the brochure.
So if someone reads the brochure of the programme and says ohh, I'll go here as they offer courses x and y too. But upon reaching there, they come to know that neither x nor y are being offered. Instead, they have no other choice but to choose only from algebra courses.
But you could go to your students affairs section and ask them to. I'm sure they'll listen because that's IIT. You could inform that there are n number of students who want to take this course, please let it be offered.
we don't have any SA section here.
IITs are very student friendly. I never faced such issues during my bachelors.
@Franklin I think that the identity is true for all $m$. What makes you think it's not true for $m$ odd? Perhaps your comment above was made before you worked stuff out in the question.
@SoumikMukherjee During my bachelors, during festivals (Holi, Diwali etc.), we used to go to some professors' homes also, it was so much fun :-). That's how I know professors there are really nice :-). I could even go to some professors' office to ask my doubts.
But I'm not talking about that. I expected an environment completely different from IIT here before coming here. But not floating all the electives that were mentioned in the brochure is what disappointed me.
I saw a few theorems where they assume that the curve $\gamma$ is null homotopic (for example argument principle ). Why do we need that? I'm following Rudin, and he doesn't assume null homotopic.
@robjohn Yeah, you are right. I was actually talking about this idenity:$\lfloor\frac{m}{2}\rfloor=\lfloor\frac{m-1}{2}\rfloor + 1$. It is invalid for all odd $m$, say $m=3$ as pointed out in the comments of OP.
Can anyone please help me, understand the Taylor Series expansion for f(x)=cos(a+h)x ?
They are expanding f(x) about which point ?
I think there's a lack of clarity in here...
Neither do I understand, the rest part after that expansion ...🥲
@Shinrin-Yoku I assume you're referring to the non-vanishing determinant of the appropriate part of the derivative matrix? Yes, look at the case of $x^2+y^2=1$. Why can you not write $y=f(x)$ locally near $(\pm 1,0)$? If the tangent line (plane) is vertical, you cannot write the level set locally as a graph.
@TedShifrin Am restricting to the 2D case to have better intuition. I have seen that example, but there are others like the infinity symbol where one can have a function, but still the center is not a regular point...
I guess I am trying to say that regular point does not always correspond with vertical tangent, so its not clear how one would come up with the regulairty condition
(I also don't have good intuition for when a point is not regular when the function is from R^2 to R)
@TedShifrin Umm...Now, that you mention it, but it was not so clear, though. Also, you mean "expanding about the point $a$" instead of "expanding about the point $ax$", right ?
@TedShifrin The thing that is bothering me, is that it's said, $f(x)=(a+h)x$. Now, if we look at the first few terms of the Taylor series expansion about ax: f(ax)=\cos(a+h)ax$ , $\frac{f'(ax)(x-ax)}{1!}=\frac{-((a+h)a\sin((a+h)ax))(x-ax)}{1!}$ and so on. These were only the 1st two terms, but they aren't alike to thing written there ? Am I missing something?
We’re going around in circles. Maybe you should watch some of my lectures. And, instead of being lazy, do the exercise I gave you instead of repeating yourself endlessly.
@Shinrin-Yoku here are some implicit functions of $x$ on the infinity loop: https://imgur.com/a/XEYcKMc here are some implicit functions of $y$ on the infinity loop: https://imgur.com/a/mmgapVc
Solve the following differential equation: $\frac{d^2y}{dx^2}+a^2y=\cos ax.$
The solution given is as follows:
The complementary function is $c_1\cos ax + c_2\sin ax$; the particular integral is $\frac{1}{D^2+a^2}(\cos ax)=\frac{1}{-a^2+a^2}\cos ax$; and thus the method fails. In this case, chan...
@TedShifrin Good point. That makes more sense to me. But it looks like Franklin's book expands $\cos(ax+hx)$, so he probably got a bit confused by your suggestion.
We're talking about intuition, I thought. Yai0Phah is giving a famous example which shows that $g=0$ is the same as $g^2=0$ which is the same as $g^3=0$, etc.
Suppose that $k$ is a field extension of field $f$.
I don't understand one step in the proof of $[k:f]<\infty\implies |Gal(k/f)|<\infty.$
Proof: If $[k:f] < \infty$, then $k$ is finitely generated extension of $f$, i.e., $k= f(a_1, ..., a_k)$. I know that $k$ automorphisms are determined by actio...
koro: each a_i is the root of some polynomial. and by homomorphism-ness, any automorphism has to send a_i to a root of that polynomial. so, at most as many homomorphisms as there are maps from {a_1, ..., a_n} to a finite set. the comment shows how you can get a better bound if you otherwise know that the extension is simple.
for what it's worth, i don't see something that implies the primitive element theorem in your hypotheses (and we don't need the PET for finiteness, although you might need it for that bound on the degree in the other answer).
and we're not using 'automorphism,' same argument shows finitely many homomorphisms. :)
Or should I say that: If $\omega\in Q(2^{1/3})$, then $Q\subset Q(\omega)\subset Q(2^{1/3})\implies [Q(2^{1/3}):Q]=[Q(2^{1/3}): Q(\omega)][Q(\omega): Q]$
LHS= 3, the second term on RHS is 2 so the first term on RHS becomes non integer, contradiction.
@leslietownes ok, thanks :-). I meant 'a' correct proof.
also fine as a line of argument if you explain why that's a problem (e.g., uh, why is the second term on the RHS 2?)
if someone pickier were here, they'd also maybe just start out by noting that it would help to pin down more explicitly what you mean by "the cube root of unity." something other than 1 that cubes to 1?
well, if you're regarding all of this as a subfield of C, i don't know why you're uncomfortable using the fact that Q(2^(1/3)) is a subfield of R.
you can certainly do all of this without referring to i or complex exponentials. i was wondering if that's what you wanted. but there's no reason you'd have to avoid that.
you can do all of this without reference to R, also, if you want. it's just where you are comfy working.
probably the differences between various books/treatments in this area will come down to how they introduce the subject, for some people's exposition, everything is a subfield of C unless otherwise indicated, in others, you don't assume the existence of more than Q or try to do as much as possible as closely as possible to the field axioms. and maybe some authors shift focus depending on what they are working on.