@AchillesRamNakirekanti To give you a simple answer as to the "practical" use: speed (in the physical sense) is the derivative of position, and acceleration is the derivative of sped
Go look up more about factoring online, get familiar with it, then come back. I'm more than happy to help, but I'm not a teacher and teaching such things on a chat on the internet isn't easy @Lagranian
I think I've had a dumb idea but I feel like there's a chance it's not. For rank nullity, given a linear map $A$, such that $A:V \to W$, if I define an equivalence relation E, such that for $q, p \in V$, $qEp$ iff $A(q) = A(p)$, then take the quotient space of V with E, $V/E$ and then just use the fact that for a quotient space between R^n / R^m, the dimension of resulting space is $n-m$, or that it's bijective to $R^{n-m}$. So then it follows that $dim(V) = dim(V/E) + dim(E)$?
could anyone please give me feedback on whether I'm barking up the wrong tree or not? :^)
Hi everyone! What about my problem: the numbers $b_1, b_2, b_3, b_4$ represent a geometric sequence, while $(b_1)+1, (b_2)+1, (b_3)+4, (b_4)+13$ an arithmetic one find the common ratio. I used some formulae $S_n=\frac{2a_1+d(n-1)}{2}, S_n-S_(n-1)=S_n$ but I couldn't figure it out, any comments?
@Phase That sounds right. This is related to the first isomorphism theorem for vector spaces. You probably also want to say something about why it is that $E$ ends up being the nullspace of $A$ and similarly $V/E$ the image.
@Tug'Tegin My hint is that since the $b$ are a geometric sequence, we can rewrite all of them using just $b_1$ and some unknown ratio. Similarly we can take the difference between terms in your arithmetic sequence and they will all be equal to some unknown difference.
@Tug'Tegin That gives 3 unknowns, $b_1$, the ratio, and the difference. And take the differences of adjacent terms in your arithmetic sequence gives 3 equations. So sounds like you should be able to solve that system of 3 equations with 3 unknowns.
By inspection, this surface contains the points (1,0,0), (0,1,0), (0,0,1) and (1,1,1). Moreover, it in fact contains the six line segments between these four points.
Now, one thing that one can show by algebra is that the above equation is invariant under $(x,y,z)\to (1-x,1-y,z)$.
is there a way to express this equation so as to make that obvious?
In case it comes up somewhere else, let me tell you how I got it. First I separated out all the terms that depended only on $z$ or constants. Then I found that there were 2 separate terms that were each invariant under the reflection, $-2x+x^2-2y+y^2+2xy$ and $2xz + 2 yz - 4xyz$. I then factored those 2 terms by making sure to treat $x$ and $y$ completely symmetrically (like breaking up $2xy = xy+xy$ and putting 1 $xy$ with the $x$ terms and the other with the $y$ terms).
if the Sylow $p$-subgroup is not cyclic then there are infinite dimensional indecomposables, and this implies that there are infinitely many indecomposables
For example denoting the restriction to a subspace by $|N$ rater than $|_N$
In general I tend to avoid too old books, as the notation will often have solidified to be something slightly different, and everyone will use that different notation now
Also, many concepts will have been more thoroughly studied in the mean time, allowing for better exposition, as it becomes clear why certain otherwise random-seeming things are as they are
@MatheinBoulomenos Frobenius algebras are essentially algebras with a "trace" map to the group field
plus some other stuff.
I remember the higher version plays a big role in the proof of Soergels conjecture
Ahh, now I remember the idea: If you have a ring $A$ with a subring $B$ then you get induction and restriction of modules, which are adjoint on one side. $A$ is a Frobenius algebra over $B$ (or a Frobenius extension) if these also have adjoints on the other side, and one of these is the "trace"
Or rather, the trace should be a map from $B$ to $A$ giving rise to the other two adjoints
Is it really true that there are no infinite irreducible representations of a finite group?
I was thinking a lot about this question but I haven't found any such representations.
@MatheinBoulomenos Actually, I am fairly certain all finite dimensional algebras have the property you ask about, as they have finitely many simples, and the Ext-spaces are all finite-dimensional (I think)
Probably group algebras should still be fine, as those are Hops algebras, so they have internal Ext-groups, and I think this should imply that Ext between finite dimensionals is itself finite-dimensional
@MatheinBoulomenos Hops algebras are basically the ones that allow you to take duals and tensor products with respect to the ground field and still get representations
Cocommutative Hops algebra tend to come as group algebras or enveloping algebras of Lie algebras
whereas commutative ones are (the same as) the coordinate algebra of algebraic groups
and those that are neither usually come in the form of quantum groups
That doesn't require you to take the group algebra over a field, right? We're doing that with $\Bbb Z[G]$-modules as well in our group cohomology course
I would need to think some more about why it should work. The idea is that the dimension of $Ext^1$ is the number if iso classes of extensions between those two reps
so if these are always finite-dimensional, then there are only finitely many ways to make things of a given length
The thing is, I am fairly certain there should only be finitely many ways to construct extensions between two simple $G$-modules for an algebraic group $G$, but when restricting to algebraic reps
So probably the restriction to algebraic reps messes everything up, since I do agree that elements of Ext correspond to exensions
Any idea how to compute $\operatorname{Ext}^1_{R}(R/m,R/m)$ for $R=k[x,y]/(x,y)^2$ and $m=(x,y)$? If what you say about Ext groups is true, this needs to be at least two-dimensional
maybe i will start and someone can continue :P for every $(a,b) \subset \Bbb R$ $f \ ^ {-1}( (a,b) ) = A \cup B $ where $A\in \Sigma$ and $B$ is $\mu' $ null set. how should i deinfe $f' $ ? i thought $f(x)$ where $x \in A$ but then $f'$ wouldn't be well-defined
What's $\pi_1$ of the Klein bottle in terms of direct or semidirect product or other operations on known groups? The presentation is $\langle a,b|abab^{-1}\rangle$
It's the semidirect product of two copies of $\Bbb Z$, where the morphism $\Bbb Z \to \operatorname{Aut}(\Bbb Z)$ sends $1$ to inversion
The presentations says $bab^{-1}=a^{-1}$, so conjugating with one generator inverts the other one
This equation implies that $\langle a \rangle$ is normal
@Liad here's how I would proceed: first split $f$ into positive and negative part, so that you can assume wlog that $f$ is non-negative. If $f$ is non-negative, it can be approximated by a monotone sequence of simple functions. For any simple function, it's easy to give a $\mu$-measurable function which agrees $\mu'$-a.e. using the decomposition of $\mu'$-measurable sets into $\mu$-measurable sets and $\mu'$ null-sets.
Thus, we only need to change the values on a countable union of null-sets, which is again a null-set. The resulting sequence will only converge a.e., but that doesn't matter, you can just set $f'$ to $0$ where it doesn't converge
The joint probability of two random variables A and B is P(A and B) = P(A | B) * P(B) = P(B | A) * P(A).
I am curious to know how you, intuitively and mathematically (but especially intuitively), interpret that the joint probability of A and B is the PRODUCT of two other probabilities? Why not the sum or any other operation?
Is it really true that there are no infinite irreducible representations of a finite group?
I was thinking a lot about this question but I haven't found any such representations.
