they used to do like talks with invited speakers and stuff! there was evidence of this in the club's office.
also in the fact that the club had an office.
when i was a junior or senior some new students arrived who wanted to make it more focused on contest prep. like putnam or something. there was already a class for that. we were like, OK. i think it just changed which four people went to the meetings.
they had some full-color printouts of posters announcing talks from outside speakers in the office. from the late 80s.
and apparently at one time they made shirts or something as a fundraiser. there was evidence of all of this, it was like the ruins of pompeii in the late 1990s
the late 90s were a weird time to be a math major. dotcom boom and y2k stuff were pushing a lot of people with no interest in computers into the CS major. which was selective, and if you couldn't get into it (e.g. due to bad math grades) the next best thing (in terms of being able to use what you'd taken toward a major) was math.
so, lots of math majors with no real interest in math.
a bit of a culture shock for some of the grad TAs from princeton or yale or wherever, where you only major in math if you actually care about it.
although as a formative experience, maybe more valuable than being surrounded by people who just want to talk about what their favorite book on elliptic curves is, or whatever. more 'realistic' for the world of teaching, or the wider world.
i'm told it's better now, but my information is pretty out of date.
the math club in my senior year was two math ed types who wanted to serve pie on pi day as the club's sole activity, and two people who wanted it to be a putnam thing. the putnam people won. i think the website went offline the year after.
i think a lot of people who go on to be math professors discover an interest in math very young. there is nothing wrong with this. but many of them are only ever socialized, in math, around other people who are like them (i.e., proto-mathematicians).
that might be good for learning math but it's not good for maybe learning how to teach people who aren't miniature versions of yourself.
so, being around people who can do math, in some cases very well, and yet have no interest in specializing it, can be a useful addition to one's perspective.
ideally, encountered before you are teaching for the first time.
Well, we surprisingly had a number of engineers attend the event today in addition to math majors. I hope that ends up being beneficial to the social experience, like you say.
i probably would have preferred that there be slightly more actually-interested-in-math people in my major, but i wouldn't have wanted it to be a thing where absolutely everybody is thinking phd.
yeah. engineering and physics people can be a great addition. i think it's a completely different world, learning or teaching math at a school that doesn't have a big engineering program.
nobody tell them, but they're good to have around.
my closest 'math friends' from undergrad, one does have a phd in math, but teaches biostatistics (her focus) and the other is at a company that designs semiconductors. a good number of my grad school friends are pure math profs, though.
me, i prefer i actively contributing to the downfall of society.
my brother-in-law had a heart attack in his late 40s. it is only a slight exaggeration to say that my daughter's insistence at calling him early in the morning saved him.
Is this (i.imgur.com/ZXfRcOx.jpg) the way to handle inhomogeneous boundary conditions for every linear PDE, despite the temperature-specific terminology, or is it just a heat equation thing?
i last read patrick radden keefe's "empire of pain" about the family behind the opoid epidemic in the US, and elon green's "last call" about a serial killer in NYC during the aids epidemic.
according to my ebook reading history, both were several months ago.
copper the stuff about richardson i found interesting. and there's quite a bit of detail about the german enigma, mathematically, separate from naval derring-do.
if the inner product had some gooftball definition, there might not be vectors we could just immediately grab, without computation, and know they were orthonormal, but here, there are
koro, one approach is first to compute the matrix of T, then from that to compute the matrix of T*, then from that to compute the matrix product T*T by matrix multiplication
Ohh, I make a little loop in y at the bottom but 4 is straight. Now, you'll complain "x" also which should be 'crossed straight lines" but mine has curves. :P
note it would be the same calculation for literally any T on C^n - you have a standard onb, so you don't need a layer of computations to get one before you compute the matrix of T, and the matrix of $T^*$, and the product $T^*T$
we lucked out here in that the matrix of $T^* T$ was diagonal. there's no reason for that to happen in general, but computing its eigenvalues when it's not is "just an eigenvalue problem"
i put just an eigenvalue problem in quotes because this can be hard to do by hand even when n = 3, and under any circumstances when n is super duper large
koro similar definitions work in the infinite dimensional setting. T is isometry iff $T^* T = I$, and unitary iff $T^*T = TT^* = I$, with the two equivalent in finite dimensions because if AB = I then BA = I in finite dimensions
a better way (in finite dimensions, at least) is to think of the svd as an ONB of the range & domain such that the matrix is diagonal with positive entries, ordered in the natural way.
sometimes you also want just the singular values and not the SVD. in finite dimensions this means solving a polynomial equation only and not grabbing eigenvectors
copper: For now, I understand that using SVD we can diagonalize any operator (on finite dimensional vector space) if we are allowed to consider two bases to write matrix of T.
koro i forget if axler introduces the operator norm in an inner product space but as copper mentioned earlier this is one way you would often actually compute it in concrete cases
@Koro well, i am an engineer, so my focus different from maths, but to me the essence of the svd its use as a decomposition that has meaning in terms of distances to certain classes of matrices (such as singular, etc).
i am not sure what you mean. you can write the series $\sum_n a_n x^n$ so it exists in that sense, but the radius of convergence is zero so there is not much you can do with it.
people just throw the tag convex out there purely like shouting squirrel to a dog.
alex, for any sequence (a_n) whatsoever, there is a smooth function f with f^{(n)}(0) = a_n, for all n
most of these are not represented by their taylor series, but that's a separate question
it may help to phrase the problem at this level of generality
there's also no uniqueness (again because these things tend not to be represented by their power series)
so your constructions will involve arbitrary choices, and a lot of them
try some recipe like sum a_n c(x/c_n) x^n where c( ) is a smooth bump function with compact support in an interval around 0 and identically 1 on some subinterval around 0 and choose the coefficients a_n and c_n inductively to get the thing to work out
or some variation on that theme
lots and lots of arbitrary choices
some version of this is done in vol. 1 of hormander's linear PDE book, very near the beginning
i think i mean sum a_n/n! c(x/c_n) x^n above, if (a_n) is the given input sequence
same difference
for more on this and other math tips and tricks, buy lesliecoin now
Let $T\in L(V)$ where V is finite dimensional vector space. It is given that there exists an isometry S in L(V) and a positive operator R in L(V) such that $T=SR$ then it is to be proven that $R=\sqrt {T^*T}$
My solution: Since T=SR, it follows that $T^*T=R^*R$. Since R is positive, it is self adjoint by definition and hence $R^*=R$. It follows that $R^2=T^*T$.
So R is a square root of $T^*T$, which is a positive operator and hence has a unique positive square root. It follows that $R=\sqrt{T^*T}$.
can anyone please review this and let me know if it is correct? Thanks.
I use this definition for positive operator: If T in L(V) , where V is finite dimensional vector space, is self adjoint and is such that $\langle Tv, v\rangle \ge 0$ for all v in V, then T is called a positive operator.
Some people use the word "semi positive definite" for positive operator.
The answer is clear, but I guess I'm.not sure how what you said relates to the quote I put into my question
> This allowed Gödel to show a correspondence between statements about natural numbers and statements about the provability of theorems about natural numbers, the key observation of the proof
I have 2 basis in R2. The first is A = {(1,0),(0,1)}, the second one is B = {(0,1), (-1,0)}. So B is obtained by rotating A by 90 degrees counterclockwise (that’s what I see if I draw the vectors in R2). The change of basis matrix from A to B is First row: 0 1 Second row: -1 0 But the matrix which represents the 90 degrees rotation counterclockwise is First row: 0 -1 Second row: 1 0 Why are they different? Why are they not the same matrix?
In that context, I guess my 'tweaking' suggestion is 'can you define a Prov(n) that says P=NP (or Fermat's last theorem etc.) Is true?' Can you then work backwards to see what that implies about the logical steps required?
So in a way, what I'm asking is if the true Prov(n) is statements 1 through 12, is there an (incorrect) Prov(n) that says 'statement 2 is not compatible with this Prov(n)'
@LeakyNun but reframing a question can often lead to a novel way of solving?
