@PedroTamaroff Imagine a matrix with just two diagonals, say 1 on the main diagonal, and -2 in the diagonal below. Solving a system with that matrix involves successive multiplications with 2 (and additions). Numerical errors thus get exponentially amplified, the computed solution is unusable. Same can happen in an IIR filter due to its feedback loop.
Hi I'd like to check something. I think I found a mistake in one exam. This says: Check that all the elements in $Q/Z(Q)$ has order 4. But this doesn't make sense (Q is the quaternion grouo) because the quotient is isomorphic to the klein 4 group doesn't? and all the elements different to the identity has order 2
No, sorry the problem says that the order of each of the elements. I'm sure is a typo. Of course the order of the quotient is 4. I think someone makes a mistake.
Here is other I really like show that any subgroup of the S_5 of order 10 is in the alternating group. I don't sure but to avoid the terrible computation I think that the easiest way is proving that all the groups 2p where p is a prime are either cyclic or dihedral. In this case is trivial to show that the group is the dihedral and is generated by any 5 cycle which of course is in the alternating group and any other group of order 2 since the group is dihedral
the element of order two should satisfy the relatation t p t = p^-1
WLOG suppose the 5 cycle is (12345) it invers is (54321) so the element t must be of the form (15)(24) which indeed is in the alternating grouo
I think is the easiest way which I know so far.
@anon I'm completely sure is a mistake and the wanted to say the order of the quotient and not the order of the elements. Anyway the exercise is very simple in any case.
@anon Glad you asked! Academy-wise, I'm taking one course -- complex analysis -- and looking forward to sit for at least two more finals. Also slowly chugging on Pete's algebra notes.
@JoseAntonio once you classify the groups of order 2p, you simply need to disprove there is an elt of order 10 in S5. (also you mean dihedral not alternating in first sentence.)
then sure, the subgroup of order 10 must be generated by an elt of order 2 and of order 5; the latter is necessarily a cycle, disprove the first is a 2-cycle (so it must be a product of two 2-cycles)
By the way, some days ago, talking with friends someone thought in the following what functions can live in C_b(Q) continuous and bounded functions from Q to R . Seems to be very pathological ones. Does someone knows a good place to check or study this types or problems?
@TedShifrin in intermediate/college algebra we had Test 1 last week (we have four tests + final in those classes over the semester). in the abstract algebra class I T/A the takehome portion is due this Thursday I believe, and so I'll be grading with the teacher on Friday afternoon.
Really dumb question. Whenever we talk about the functional limit of a function between metric spaces (X,d) and (Y,w) , does the limit of the function as its dependent variable approaches some point in the domain, have to be in Y ?
In the same spirit as Robert Israel's answer and continuing Raymond Manzoni's answer (both of them deserve the credit because of inspiring my answer) we have
$$
\sum_{n=1}^\infty \frac{H_nx^n}{n^2}=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operat...
Just asking ... nothing more ...
I mean $$\sum_{n=1}^\infty \frac{H_nx^n}{n^3}$$ formula doesn't work ...
@robjohn Then I like the idea in $(2)$, but how do we really recover the remaining part with that variable change? Working with indefinite integrals can be a a very powerful way.
@Committingtoaname: I have a good suggestion for you. I would have used it for myself if there is no thing like 1 name per month. Anyway, how's "where is my water?"
Interesting, I will take a look at them. Also, it feels very strange to refer to you by your current name, and as a result I haven't been using it at all
@DanielFischer Your chat profile isn't linking to anything but Stack Overflow, but I was certain I had seen posts from you, and I have seldom used Stack Overflow
Well, I don't know if there's an "official" way. Fudging it is possible, but I don't think the result will be pretty. However, I don't see a need for two-column posts here, so meh.
@robjohn Note now that $$\sum_{i=1}^{\infty}x^{i^2 } =(1/2)*(-1 + EllipticTheta[3, 0, x])$$ Ramanujan might have some work on these sums, I need to check.
In the same spirit as Robert Israel's answer and continuing Raymond Manzoni's answer (both of them deserve the credit because of inspiring my answer) we have
$$
\sum_{n=1}^\infty \frac{H_nx^n}{n^2}=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operat...
really works.
It simply works because of the specific value $x$ that is $x=1/2$
@BalarkaSen usually $(2)$ shouldn't make sense if you know what I mean. Actually, what is the meaning of that variable change in that context, and how do we make sure we recover then what is added or lost?
@BalarkaSen: I have a doubt on the integration constant. I have a by-part formula which I roughly state as the following : $$ \int{uv} = u\int v - \int(u'\int v)$$
When I do $\int v$, my instructor has told me not to use the C
because it messes up the results. i can't explain it, he can't explain it.
The above can easily be derived from the product rule $$(ab)' = ab' + ba' \implies \int{ab'} = \int{(ab)}' - \int{ba'} = ab - \int{a'b}$$ Let $a = u$, $b = \int v$
You get the formula I first stated
I've been told to put the C at the final result after all the calculations.
But I get this feeling that sometimes somewhere the constant of integration for that $\int v$ gets multiplied with an $x$ from $u$ and stops being a constant.
This would mean many of my answers using the by-parts formula is wrong.
I also know that the constant can at many instances affect the result. $$\text{Example: }\int \sin x \cos x \, dx = -\frac{1}{2}\cos^2 x + C_1 = \frac{1}{2} \sin^2 x + C_2 $$