Now, when all excitement has gone, I’ll repeat several questions.
Have you ever encountered a problem of inattention? I think my death will come while I try to diagnosticate mistakes in my solutions. How could an infected one be cured from this?
Congratulations on the tie breaker @MikeMiller * The leaderboard counts hats earned across the network. While all four earned the maximum network-wide, the tiebreaker goes to the person who scored the most hats on their home site. Mike Miller’s 36 hats on Mathematics was the maximum he could earn. The two he missed on Math were HairBoat, since Abby hadn’t posted on his site and Kofia, which is awarded to brand-new posters only.
Well, I dealt with a drugged out flunking student on probation and a new student who surely doesn't know enough to pass my course, @robjohn ... And then I went to start retirement :)
@skullpatrol wow... I only got 10 hats. There were several hats that I couldn't get because they were for first time occurrences that had already happened, and some that required downvoting that I wouldn't get.
@evinda Then after swapping A[n] and A[i], you have destroyed more than you should have, A[i+1,...,n]has no root node anymore, your parent-child relations are lost. If you have the root at A[1], you should build the sorted array from back to front, first swap the largest element into A[n], heapify A[1,...,n-1], swap the largest remaining element into A[n-1] etc.
@evinda You make a max-heap first. Then the largest element is in A[1]. Swap A[1] and A[n], heapify (max) A[1,...,n-1] so you have the largest remaining element in A[1].
@evinda If your Initialize_Heap and Heapify are for max-heaps, all you have to do is change indices, and let the loop go down from n to 2. Otherwise you need to change more. But you should test your code, then you can check whether it does what it is intended to do in some example cases, a good testing strategy would catch most errors.
@evinda Quite a bit. If you build a min-heap, you have the smallest element in A[1] at the start. So you'd leave that in place. But then you have two unrelated min-heaps with roots at A[2] and A[3]. You would need to join these two heaps to become one min-heap, which is not trivial.
Of course, you just need to flip a few comparisons to change the code to produce max-heaps.
@DanielFischer The root should contain the smallest key and so the key of each child should be greater and this has to hold for each subtree.
@DanielFischer So at the beginning we have one element that will be the smallest, then we get an other and we have to compare its value with this from the root and if it is smaller we have to swap them, then we do the same for the right child of the root.And then we have to follow the same procedure for the left and right child being the root, right? But we have to compare the new entries again with the root..Or am I wrong ?
@DanielFischer We want an increasing order of the keys, right? So don't we have to apply the heap-sort for a max-heap and the other two functions for the min-heap? Or am I wrong?
@evinda I'm not sure what you mean. But you can answer these questions yourself by reading a good description of a heap-sort (e.g. wikipedia) and playing around with the code, seeing what results you get. To understand the algorithms, you need to play around with them.
After defining what it means for a variety to be irreducible at a point, they show that if f is irreducible in the local ring O_n then its zero locus is irreducible at the point 0. With me so far?
OK. To prove this, they suppose V = {f(z) = 0} is the union of two varieties V_1 and V_2. Now here's where I get stuck. They claim there is f_1 in O_n which vanishes on V_1 and is non-zero on V_2, and f_2 in O_n which is non-zero on V_1 and vanishes on V_2. How do we know that V_1 must be the zero locus of a single function rather than the common zero locus of a collection of functions? Likewise for V_2.
Oh, because a hypersurface is equidimensional ... It can't have pieces of different dimensions. This follows from the stuff G&H say after this proof, if you don't buy it.
complex analysis question, if anyone's interested: how could I solve for z in this equation? $e^{z^2}=4-4i$. Do I literally just approach this as an algebra problem? I'd think there's something else to do (probably an identity of some sort) as the answer, I believe, is periodic
The trouble is, one must memorize them inside out, and be able to use them analytically, like algebra.
Having not been introduced to this sort of thing before, I find myself constantly relying upon actually peeking at the identities while doing problems.
Brute Memorization unfortunately is not for everyone. I need some sort of advice to help think of them better.
We don't really use textbooks in class. Our teacher types up all the information on sheets and hands them out to us... It seems that they don't focus upon proving them and explaining why they are the way they are, but instead they focus upon solving problems analytically with them. My textbook might have what you are saying, let me check.
Hmm, I don't think my textbook actually derives the identities, but merely states them.
I'd recommend something that isn't math/comp. sci. I took a lot of math and comp sci in college, but I wish I had taken more liberal arts - history and music in particular.
I'm not really good in math so. Mind helping me with this one. I can't hold on how to start. And I'm really curious how to solve this one.
A Pythagorean triplet is a set of three natural numbers, $a < b < c$, for which,
$a^2 + b^2 = c^2$
For example, $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.
I need a ...
More precisely, let $S^3 \to \Bbb C \cup \{\infty\}$ be the map sending $(z_0, z_1)$ to $z_0/z_1$. But if you take the inverse image of $\infty$, you get $(0, z_1)$ where $|z_1| = 1$, which is precisely a circle!
In fact, preimage of every point is a circle.
So there is your fiber bundle $S^1 \hookrightarrow S^2 \to S^3$
I am still having a bit trouble to visualize this. I was given the hint that $S^1 * S^1 = S^3$
Let $\phi: E^n \to E^n$ be an isometry and set $v_0 = \phi(0)$. If $v_0 \ne 0$, then let $\rho_0$ be the reflection about the hyperplane $P(\frac{v_0}{|v_0|},\frac{|v_0|}{2})$. Then it is claimed by a textbook that $\rho(v_0) = 0$, however isn't it $v_0$?
I need a confirmation on: "Radius of sphere will be equal to half the diagonal of parallelepiped if all vertices of parallelepiped are on the surface of that sphere"
@TedShifrin: I'm going through some previous definitions and I'm a bit wondering what's happening here: Let $\varphi: M \to \bar{M}$ be a differentiable map between two manifolds. Let $\bar{\omega}$ be a 1-form (covector field) on $\bar{M}$. Then, we define the 1-form $\varphi^* \bar{\omega}$ on $M$ by $$\varphi^*: d\bar{f} \mapsto d(\bar{f} \circ \varphi)$$ for $\bar{f} \in \mathcal{F}(\bar{M})$. However, iirc, only covectors always can be written das $\omega = df$, but not all 1-forms, no?
@TedShifrin: Or are we implicitly looking at the covector $\omega_p \in T_p^*$ which is defined through $$\omega(X)(p) = \langle \omega_p, X_p \rangle$$ and then write it in the suggested form to locally define $\varphi^*$?
Maybe a bit of a trivial question, but I'm working through some lecture notes where they wrote that some square matrix A was smaller than or equal to the identity matrix. What determines the 'norm' of such a matrix? The matrix was diagonalizable, so maybe the trace in the diagonal form?
Well, I should have thought about it a bit more. From the context it seems likely that they were using the trace norm, which in this case just corresponds to the sum of the absolute eigenvalues
No, I have to admit I'm not. I was looking at en.wikipedia.org/wiki/Matrix_norm where I saw the Schatten norm section, which mentions the trace norm. That one seemed relevant to the context (it's a proof on distinguishability for quantum states, which uses trace all the time) and is particularly nice as the matrix is hermitian, but I'm actually quite sure now that this is not what they mean
Well its about POVM's in finite dimensions. They are written as $\Lambda_x$ and they obey two conditions: $\Lambda_x \geq zeromatrix$ and $\sum_x{\Lambda_x} = identitymatrix$
so any single Lambda_x must be smaller than the identity matrix, but I don't understand what
I want to see the canonical action of $S^1$ on $S^1 * S^1$. The latter is just $[0, 1] * [0, 1]$ (i.e., a tetrahedron) with appropriate identifications. I believe if you make a small square inside this tetrahedron, then it breaks up into two bits homeomorphic to the solid torus. So I guess the action should be on the little square.
@Huy No, no indeed. Your answer is the correct one, which reduces to the one I found in this specific case, so I got lucky (or unlucky as I would have done it wrong in the future)
@BalarkaSen So you basically have $S^1\times [0,1]\times S^1$ with identifications for $t = 0$ and $t = 1$. The canonical action would be $(\lambda, \langle \xi, t, \eta\rangle) \mapsto \langle \lambda\cdot \xi, t, \lambda\cdot\eta\rangle$ I'd think. How to sketch that is beyond me, however.
@BalarkaSen We know that there is no non-zero polynomial in $f$ such that $f(X)=0$ and we want to show that there is a non-zero polynomial in $E$ such that $g(X)=0$, right?? WE have the extension $F < E \leq F(X)$, right?? But how could we conlcude using these information that such a polynomial $g$ exists?? I don't really have a idea... :/
D'Alambert defines his operator as $$\square^{2}=\frac{\partial^{2}}{\partial t^{2}}-\nabla^{2}$$
My question is what is the intuition behind this definition? When I asked my professor this question he took his shotgun and shot 3 times while I was escaping the room, I was $\epsilon$ near a death....
^ quoting the question: "When I asked my professor this question he took his shotgun and shot 3 times while I was escaping the room, I was $\epsilon$ near a death...." lol
@Huy: To respond quickly to your question. Covector field is a synonym for $1$-form. Even locally one cannot necessarily write a $1$-form (or covector field) as $df$. I think that was meant to be only an example. If you want to write pullback without any local coordinates, you do it by $(\varphi^*\omega)(p)(X_p) = \omega(\varphi(p))(d\varphi_p(X_p))$.
Hi everyone! First, my best wishes! I would like to ask a question about terminology. I would like to formally say that "we take the eigenanalysis of this covariance matrix". So, how should I express that? Should I write "By performing eigenanalysis on this ...", or what? Thanks a lot for your help!
@Chris'ssis I saw that, but I don't see how you computed the integral. I used that same identity, but had to do some tricky integration by parts to get the integral
@TedShifrin I imagined the day I'll publish my book ... I'll be so happy that day ... I'm still working on a lot of problems, trying to add to my book the best ones.
After that moment I'll take a break, maybe a longer one.
@TedShifrin The hardest part in my book will definitely be the way of explaining to the reader all I do, I mean explaining things in a very nice, lovely way such that all can be easily understood.
I would think that the organization of your book will take lots of trial and error. Do you organize by themes of problems, methods of solutions (probably more instructive), etc.
In general my proofs look completely different from what I use to post here. They are very nicely arranged and all English used is carefully considered. Yeah, sometimes I ask for help to be sure everything is fine.
Probably the publisher will need an editor to work on your English. And then you'll have to correct things that the editor messes up for not knowing enough math.
I'm not really good in math so. Mind helping me with this one. I can't hold on how to start. And I'm really curious how to solve this one.
A Pythagorean triplet is a set of three natural numbers, $a < b < c$, for which,
$a^2 + b^2 = c^2$
For example, $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.
I need a ...
