How may I show $\int_{\mathbb{R}} e^{x^2}D^m(e^{-x^2})D^{n}(e ^{-x^2}) dx = 0$ if $m\neq n$ using IBP? It's supposed to be trivial but I don't see it :>
Let $A$ be a ring, $G$ a finite group, and $A[G]$ the corresponding group ring. Dumb guestion: If $a_1,...,a_k \in A$ and $(a_1 + .... + a_k) \sum_{g \in G} g = 0$, does this imply $a_1 + ... + a_k = 0$?
Maybe I'm not understanding the question, but: that doesn't really make sense, does it? Elements of the group ring are finite $A$-linear combinations of elements of $G$. What do you mean by $\sum_{g \in G} g$?
Is there any simple way to solve for $i_1$ and $i_2$ in $ 20 - 3(i_1+i_2) - 2 \frac{\text{d}}{\text{d}t}(i_1+i_2) - 4i_1 = 0$ and $6i_2 + 3(i_1+i_2) + 2 \frac{\text{d}}{\text{d}t}(i_1+i_2) = 0$? I need it for a physics problem..
In that case, sure: $A[G]$ is a free $A$-module with basis given by elements of $G$, so an element is zero precisely if its coefficients w.r.t. that $G$-basis are all $0$.
Is there a theorem/rule for this: If we have n equations and n variables, then only we can solve for them? (Provided the equations aren't the same like x+y=2 and 2x+2y=4)
But if that's the case, in my equation we had 4 variables right? $i_1$, $i_2$, $\frac{di_1}{dt}$, $\frac{di_2}{dt}$ .. and still we were able to solve for them?
We known that the potential generated by a charge pointwise $q$ is $V(r) = kq/r $ and the equipotential surfaces (in 3D) are spheres centered in the charge with $r\geq 0$ where $r=d(O,P)$, i.e. the distance between the origin and a generic point $P$.
In fact if we are in space, where an orthono...
@Daminark Honestly, for this class, it makes sense to accept late homework (there is a lot of writing, with the opportunity to revise); I just wish that more students would turn in homework on-time---most of them are getting basically no credit, anyway, but I have to look at their work, which takes time. :(
the first question in the book was something like this, let W be the subspace generated by (2,1) and U the subspace generated by (0.1) , show that V= R^2 is a direct sum of U and W
I mean all i need to do is to show that any vector in R ^2 can be written as multple of (2,1) and ( 0,1) in a unique way right?
one thing you can think about: with (0,1) you can "slide" the second coordinate (by adding scalar multiples of it), and you can do that until the first coordinate is twice the second (so then it's a scalar multiple of (2,1)). for instance, (5,5)=(5,2.5)+2.5(0,1)=2.5(2,1)+2.5(0,1)
then the external direct product of U and W is not a subspace of V, but if U and W intersect trivially then there is an internal direct product of U and W within V
if there are only finitely many vector spaces there is no difference between direct product and direct sum so we might as well call them direct sums
For instance, R^(m+n) can be thought of as an internal direct sum of R^m (last n coordinates 0) and R^n (first m coordinates 0). Evidently (m+n)=(m)+(n)
Given two groups A and B, the external direct product is a construction on the Cartesian product AxB that makes it a group. Given a group G with subgroups A and B, we say G is an internal direct product of A and B if every element of A commutes with every element of B and every element of G is uniquely expressible as ab where a is in A and b is in B. Note that AxB is an internal direct product of its subgroups Ax{e} and {e}xB.
Or both are tuples, your mileage may vary. If you want you can view elements of the product as tuples and elements of an internal direct sum as sums. Then $A\times B$ and $A\oplus B$ are canonically isomorphic via $(a,b)\leftrightarrow a+b$
@AlessandroCodenotti So say you have a subset of $\Bbb R^n$ that's convex and centrally symmetric. Furthermore, say that it doesn't hit any lattice points other than zero.
We known that the potential generated by a charge pointwise $q$ is $V(r) = kq/r $ and the equipotential surfaces (in 3D) are spheres centered in the charge with $r\geq 0$ where $r=d(O,P)$, i.e. the distance between the origin and a generic point $P$.
In fact if we are in space, where an orthono...
