@XanderHenderson You need to construct a "form letter-auto reply to all email from any student in your class." "Sorry, Currently en route to the Bahamas. Enjoy your semester break!"
To get an idea why context is relevant: If you were tensoring complex numbers over $\mathbb C$, then $a\otimes bi = abi(1\otimes 1)$ (in fact $\mathbb C\otimes_{\mathbb C}\mathbb C\cong \mathbb C$). Over $\mathbb R$, you could do $a\otimes bi = ab(1\otimes i)$ (and $\mathbb C\otimes_{\mathbb R}\mathbb C\cong \mathbb R^4$ as a vector space).
there was a really good free command line version of pdftk. i can't find it on their site now, everything refers to some windows installer and a lot of it appears to require money
@leslietownes it's like a law of nature, the most-upvoted answers will be bizarre and nonsubstantive. Yes, just like the most abstract definition will be used by Bourbaki.
@copper.hat .......Marlin 4........I got up sold. They told me the Trek 820 which I originally wanted was back ordered until Winter 2022....So I went up a level....Don't regret it one bit and actually enjoying the disc brakes
@amWhy the ireland i grew up in was very conservative in some ways. after 30 years in the last family house we were still referred to as the new people.
Yea I'm not mad at it. Gradually getting all the other needed amenitities such as patches, pump, etc. Made sure to get my winter gear for riding though.
@copper.hat Yes, indeed! My great grandpa was a boot-legger during prohibition, here in WI. He died from gangrene after leaping from a third story, to escape a husband who caught him sleeping with the husband's wife!
on another note, i have never got email from doordash, but seconds after a recent convo with a brother in ireland in which i mentioned doordash, i got an email. i guess mark meta is listening to every convo.
@TedShifrin I think the final went better than expected. There were 3 questions that I'm kind of kicking myself for, but out of 12, I think that's pretty good.
Are Fourier Series really needed to solve Laplace's Equation by separation of variables, as my book always does? In the two-dimensional case, for example, I seem to be able to use the two homogeneous boundary values to solve for $\lambda$ and one arbitrary constant, then lump away another arbitrary constant, then use the remaining two boundary conditions to solve for the remaining two arbitrary constants. I'm not sure if such a solution is correct, but it seems algebraically valid.
At a high level, it seems to me that the reason Fourier Series are needed for the heat or wave equation is because there's one fewer boundary conditions, because time goes on to infinity.
Okay, so I was solving $u_{xx} + u_{yy} = 0, u(x, 0) = 1, u(x, 1) = 2, u(0, y) = 0, u_x(1, y) = 0$. Via separating variables I got $u = (Ae^{-\sqrt\lambda x} + Be^{\sqrt\lambda x})(Ce^{-\sqrt{-\lambda} y} + De^{\sqrt{-\lambda} y})$. I used the latter two boundary conditions to solve for $A$ and $\lambda$, then I could lump $B$ into $C$ and $D$, then I used the first two boundary conditions to solve for the new $C$ and $D$ to arrive at a solution.
I don't really understand the point of the weight function/weighted inner product/generalized Fourier Series as they relate to these types of PDEs. The book talks about it for a while, says the weight is $w(x) = 1$ in the first example, and never mentions it again after that. The problems I've worked seem to be solvable via the familiar separation of variables approach I'd been using all along.
Does this theory require modifying one's solution technique for certain types of problems, or is it just a different way to describe the same solution technique?
I have a question please: Let C[0,1] denote the metric space of all real continuous functions on [0,1] under the metric d(f,g)=$\sup \{|f(x)-g(x)|: x\in [0,1]\}$. If $S\subset C[0,1]$ is the set of all polynomials in which coefficient of $x^2$ is zero then S is dense in $C[0,1]$. How do we prove this?
Is the following correct?
S is a real algebra, [0,1] is a compact set. Since S separates points on [0,1] and does not vanish on [0,1], it follows that uniform closure of S is C[0,1].
which in other words means that S is dense in [0,1]. Hence proved.
But this seems so straightforward that I think this is not correct. I really don't understand the role played by coefficient of x^2 being zero.
The same argument works even if S is an algebra of polynomials with coefficient of x^3 equal to zero. That's why I doubt the correctness of my "alleged" proof.
koro: the same types of argument can work because the result is still true. those generalizations of the stone weierstrass theorem are just that powerful. you could give more constructive proofs than the general theorem in this case but why be bothered.
but, please check the hypotheses of whatever theorem you are using.
you just say S has these properties. does it? does your version of S-W allow you to get there?
it might. there are some very super powered versions of it out there.
I'm trying to complete what Ted suggested. I thought earlier that: For every f in C[0,1] , there exists sequence of polynomials {P_n(x)} such that $\lim_n P_n(x)= f(x)$ so I was trying to manipulate LHS somehow to get elements of S. I think that's what Ted wants me to do for $f(x)=x^2$.
@leslietownes I have only 3 axioms to see if the given structure is an algebra. For any h,g in S: 1) f+g is in S. (true), 2) fg is in S (hmm, needs verification), 3) cf is in S (true)
If I enumerate the rationals in $[0,1]$, I can construct a polynomial $P_n$ which has roots at the first $n$ rationals and a coefficient $-1$ for $x^2$ by scaling. Could this work?
One has $P_{n+1}(x) = c_{n+1}(x-a_{n+1})P_n(x)$. We have $|x-a|\le 1$ on $[0,1]$. The coefficient of $x^2$ in $(x-a_1)(x-a_2)\cdots(x-a_n)$ only increases, so $|c_{n+1}|<1$ I think.
