@TedShifrin Actually I think I have a nice way to explain what a fibration is. I'm basically gonna give the definition of a fibration really quickly, and since there's a theorem that says for any continuous map $p : E \to B$ we can turn this in a fibration, I'll say "You don't need to worry too much about the definition, just remember fibrations exist in abundance since we can turn any continuous map into one"
False theorem: Inverses of bijective smooth functions are smooth. Counterexample: $\Bbb{R} \to \Bbb{R}$ given by $x \mapsto x^3$. What is the smooth structure on $\Bbb{R}$? I.e., what are charts/maps?
I'm actually almost 30. Five years older than graduate students going for their doctorate. Spent too much time on other vocations where I wasn't suited.
It's OK, @Rithaniel. You're more motivated, for sure.
I got a Facebook message from an undergraduate at UGA who's taking "my" course and thanked me for my videos. Turns out he's your age. I have no idea what he was up to before.
You would have probably done better than me as a chef. I'm all about the simple recipes. Like, I made a burrito mix this week for my lunches. It's delicious (perhaps a bit too salty) but has no elegance to it. It's all brute force.
I tried going to culinary arts classes, but they kind of expected a higher level of entry proficiency than I had.
The conclusion is that $G$ is free abelian, which is equivalent to saying that it has a nonempty basis, but my approach was not going to work because we don't know that the hypothetical generating subset will necessarily be a basis.
Well, it always hurts a little bit when a person claims a domain for a function is $\{(x,y)\mid x^2>y\}$ but then sketches the regions for $\{(x,y)\mid x^2<y\}$
my personal hell today was getting people to realize that the prediction "$v_f = \frac{m_A}{m_A+m_B}v_i$" meant that, if you plotted $v_f$ on the vertical axis and $v_i$ on the horizontal axis, then the graph should be a line through the origin with slope $m_A/(m_A+m_B)$
looking at the exam draft the prof gave me, I find myself frustrated.
in intro physics kinematics, there's three main equations you use: $x=x_0+v_0 t+\frac12 at^2$, $v=v_0+at$, and $v^2=v_0^2+2a (x-x_0)$
the first two I'm fine with. I sorta detest the last one, as useful as it is, because it's really just energy conservation in disguise
So having a quiz problem for which the most efficient solution is to use that formula repeatedly, rubs me the wrong way. they'll be doing plenty of energy conservation arguments in the future, so it seems counter-productive to teach them to use the emaciated special case of it
Semiclassical, yea i saw the pattern but didnt wana go through the hassle on my homework because i didnt wana risk losing marks for something I didn't "thoroughly explain"
(4/9tw^−2)^−3 (4/9tw^−2)^3 I looked up the answer in the back of my book and it says that it's 1, but I don't understand. I know that this is a pathetically easy question, but I have no math background, and I'm learning on my own.
The parentheses mean that I'm supposed to multiply, but when I multiply the fractions, I get 16/81. and w^-6 subtracted by w^-6 is negative 12. I think I'm supposed to flip one of them to the denominator, but which one? Because then it would give me -6 minus -6 which would make it addition, thus 0, right? and that would leave me with just w.
Sorry if I'm not writing it clearly, I'm just confused.
Think about if you just looked at (4/9tw^-2) as a single chunk. It appears twice, each time with a particular power, so you can imagine everything in the parentheses as just a "thing" being raised to a particular power.
Compare to (4/9tw^−2)^−2 (4/9tw^−2)^3=(4/9tw^-2) Or (4/9tw^−2) (4/9tw^−2)=(4/9tw^−2)^2
I have 10 different balls I want to distribute among 3 different boxes, problem is that there are too much different combination of balls and boxes, I have no idea to approach it
then do all possible box cases (using only one of the tree boxes,exactly 2 and using all the 3 boxes ) and multiply by the possible permutations for every case
doubt is that star and bar is applicable for identical things, it just example, actually I want to understand how to distribute distinct things among distinct people so that each get at least something
yeah but don't forget you got to many solution caus you should remove the cases where box are empty and the case when the sames balls are in the same box with different order
I think for you case it is easier to just do 3^10 since you have 3 choice for every balls
@rapasite a small question advice rolled 4 time, the number are listed, the number of different throws, such that largest number appearing in list is not 4
Total possible pairs are 6.6.6.6=1296
- where the highest number list may be 4
It mean rest of three position there are only 3 option that is 3,2,1
$$
\text { Find the volume of the solid bounded by the surfaces } x=0 \text { and } y^{2}+z^{2}=4 \text { and } x+z=4
$$
The part which I am confused is what are limits of integration over here $x$ goes from $ 0 $ to $4-z$ ; $ y $ goes from $0$ to $ 4 - z^2$ and $ z $ from $0$ to $z$. Is it cor...
Hmm... 1. Nuking reflexivity will not do much here as you are not interested in equalities 2. You already considered what happened when you nuke connexivity, the resulting example you gave (1 < 2 > 3) is an example of a tree. Trees are important in computational sciences as they represent data structures and filing systems 3. You can also nuke transitivity. One of the ways gives you (2), the other way (anti transitivity), gives you functions such that the resulting partial order or preorder is cyclic (i.e. 1 < 2 < 3 < 1). This kind of function will fit the usual criteria to be monotone in t…
anti transitivity is when a < b and b < c, a > c
obviously that cannot hold for all elements otherwise you have b > c and c < b but (b,c) can be anything
Thus, the near counterexamples I can came up with are trees with cycles conjoined in some elements such that e.g. 1 < 2 > 3 > 4 > 5 > 3 etc.
What is the smooth manifold structure on $\Bbb{R}$? I want to show that $x \mapsto x^3$ is a smooth function but it's inverse isn't. What are the chart/atlases on $\Bbb{R}$? The identity function an all open sets?
@MatheinBoulomenos That's the catch, I cannot use diagonizable. This was given to me in an exercise where only basic ring theory has been assumed including Lagrange's theorem
@MatheinBoulomenos What exactly is a smooth manifold? I'm having trouble finding a concise definition. I know you start with a topological manifold (and I know what a topological manifold is). And you then you obtain a smooth manifold by putting a certain structure on it, which involves choosing certain open sets and maps into some Euclidean space.
well, the diagonizable case is easy, you can just check it for a diagonal matrix. In the non-diagonizable non-invertible case, you're going to have $B^2=0$
Okay. So the open sets have to cover the topological manifold, call them $M_{\alpha}$. Then there are maps $\varphi_{\alpha}$ (homeomorphisms?) from $M_{\alpha}$ to some open set $U_{\alpha} \subseteq \Bbb{R}^n$ (assuming the $M$ is an $n$-dimensional topological manifold?).
Rank 1 matrices can be written as a product [[a], [b]] * [1, s], so you can take this to the whatever power and refactor it as mostly products [1, s]*[[a], [b]], which are just numbers
And then it reduces to the original fact that x^p = x in your field; can actually do this over any finite field I think, not just Z/p
A less labor-intensive approach is to recognize that that matrix is of the form $D+u u^\top$ where $D$ is a diagonal matrix and $u$ is a column vector of ones
At which point one can do stuff like appeal to the matrix determinant lemma
that's got a cute description: $A$ is the matrix exponential of $\begin{pmatrix} s & 0 \\ 0 & -s\end{pmatrix}$
(that's less interesting than it may sound: the matrix exponential of a diagonal matrix is just the matrix you get by replacing each matrix element with its exponentiation)
where things get more interesting is when you deal with non-diagonal matrices
for instance: the matrix exponential of $\begin{pmatrix} 0 & s \\ s & 0\end{pmatrix}$ is $$\exp\begin{pmatrix} 0 & s \\ s & 0\end{pmatrix} = \begin{pmatrix} \cosh s & \sinh s \\ \sinh s & \cosh s\end{pmatrix}$$
this is more algebraic than analytic/differential-geometric, but the group $\left\{ \begin{pmatrix} t & 0 \\ 0 & t^{-1} \end{pmatrix}, t \in \Bbb R^\times \right\}$ is a maximal abelian Zariski-connected subgroup of $\mathrm{SL}_2(\Bbb R)$, i.e. an algebraic torus
though the real fun one is doing $M=\begin{pmatrix} 0 & s \\ -s & 0\end{pmatrix}$ and $A=\exp M = \begin{pmatrix} \cos s & \sin s \\ -\sin s & \cos s\end{pmatrix}$
the matrix I wrote earlier with cosh/sinh doesn't have an interpretation as a rotation
but it does show up in special relativity, where it plays the role of a boost (i.e. changing your frame of reference to one that's moving faster/slower)
(for motion along one spatial direction in SR, anyways. in general you have 3 dimensions of space and 1 of time, so the matrix should be 4-by-4. but that's tedious to write out so w/e)
there's also the whole Lorentz group vs. restricted Lorentz group business
which I think is basically the equivalent of O(3) vs. SO(3)
you might dig this bit: "The restricted Lorentz group SO+(1, 3) is isomorphic to the projective special linear group PSL(2,C) which is, in turn, isomorphic to the Möbius group, the symmetry group of conformal geometry on the Riemann sphere."
had a bit of SR in high school, but without 4-vectors or anything, we just derived time dilatation and length contraction and stuff like that from a few axioms
"In other words, [a Lorentz transformation] transforms the celestial sphere (changes the appearance of the night sky) in exactly the same way that [the corresponding Mobius transformation] transforms the Riemann sphere."
Prove that these linear programming problems are bounded by $O(k^{1/2})$
The expanded partial sums of the Möbius inverse of the Harmonic numbers have two out of three properties in common with this set of linear programming problems:
$$\begin{array}{ll} \text{minimize} & \displaystyle\sum_{n=1}...
(I prefer the proof shown on 3B1B's channel because it's much more readily generalized - I bet you could solve $\sum1/n^4$ with the same techniques - but this^ is pretty cool)
(Though would you be able to do it under exam conditions? :P)
@Akiva DogAteMy: There are various tricks for this one (another, which I learned from Simmons's calculus book, is a double integral with change of variables — it's an exercise in my multivariable book). But I prefer Fourier analysis and/or residues, as they generalize immediately.
@schn The zeroes don't do anything special, although if the size of the matrix is arbitrary, perhaps this is a good set-up for mathematical induction.
@Simone Schaums Outlines Advanced Calculus is nice, at least for convergence tests for sums. I don't know/remember if it has much about differential equations.
@Simone: If you like Spivak for rigor, then most advanced calculus books will be disappointing. The ones with "differential equations and stuff" are oriented toward engineering, rather than rigorous mathematics.
No, don't do that. It's way too terse and assumes you already know multivariable calculus, truly. You might look at my book or my YouTube lectures on multivariable calculus/analysis and linear algebra. Do NOT look at Rudin for this, either.
You can find info in my profile. For differential equations, most texts are very engineering-style. I like Simmons's book on differential equations. He explains beautifully and has interesting applications.
@MikeMiller there's one more way you can prove it without considering rank or diagonalizibility. Just by using the characteristic polynomial and splitting it into two cases of 0 and non zero determinant
I have a question: imagine your light a Rope it burn in a time X (seconds), now if make a circle the same rope gonna burn in X/2 , is there In any "geometry or topology" a way to make you rope burn in X/3 making any shape or twist or whatever you want? If not why and how do you prove it?
I post another question about integral dependence...much interesting than the previous one I think. The integral closure behaves as a “weaker” closure operator. But I haven’t figure out what this observation can do though...
$$
\text { Find the volume of the solid bounded by the surfaces } x=0 \text { and } y^{2}+z^{2}=4 \text { and } x+z=4
$$
The part which I am confused is what are limits of integration over here $x$ goes from $ 0 $ to $4-z$ ; $ y $ goes from $0$ to $ 4 - z^2$ and $ z $ from $0$ to $z$. Is it cor...
