another version that avoids the notational issue of 'n' in an interesting way. i don't think it would make sense on an in-class quiz. math.stackexchange.com/questions/2775599/…
one from 2012. math.stackexchange.com/questions/226347/…. i like distinguishing the index in the affine combinations from the dimension of the space. i don't like only implicitly defining 'affine combination' in the problem statement.
what a weirdly rich tapestry of different ways of posing essentially the same problem.
yes. it's hard to envision teaching a class like that without "some" linear algebra, if only at the level of "matrices multiply each other and vectors"
@leslietownes does this require a proof? I thought that since every finitely generated subgroup of Q is cyclic and total subgroups of Q are uncountable so set of non finitely generated subgroups is uncountable - countable which is uncountable.
Unless linear alg is a prerequisite, which means they have to go through several calc. Our quantum computing course, joint math/Cs/phys,did have serious prereq.
Also, showing $Q$ and $Q\times Q$ are not isomorphic turns out to be bit tricky. For showing isomorphism, I look for cardinality (for example, R and Q can't be isomorphic), or that domain group has elements of certain order but the codomain group has no element of that order etc.
@leslietownes does this require a proof? I thought that since every finitely generated subgroup of Q is cyclic and total subgroups of Q are uncountable so set of non finitely generated subgroups is uncountable - countable which is uncountable.
I say, suppose on the contrary that $Q\sim Q\times Q$, there exists a bijection $f: Q\to Q\times Q$ so let $f(p/q)=(1,0)$ and $f(r/s)=(0,1)$. It follows that $f(p/q)=f(ps. \frac 1{qs})=ps f(\frac 1{qs})=(1,0)$ and similarly $f(r/s)=rq f(\frac 1{qs})=(0,1)$. These two relations force $f(\frac 1{qs})=(0,0) $ whereby we get $f(p/q)=f(r/s)=(0,0)$ which is a contradiction.
Also, showing $Q$ and $Q\times Q$ are not isomorphic turns out to be bit tricky. For showing isomorphism, I look for cardinality (for example, R and Q can't be isomorphic), or that domain group has elements of certain order but the codomain group has no element of that order etc.
right. so (switching to the D notation b/c i'm tired of writing d/dx) we see that $x^2 D^2$ maps monomials back to themselves, up to the factor $k(k-1)$
@Koro the 'up to isomorphism' question is a little subtler and not just a subgroup count. it is generally possible for large numbers of distinct subgroups of an abelian group to be isomorphic to one another.
consider e.g. the group G of all functions [0,1] to R, componentwise addition as the operation, and for x in [0,1] let G_x = {f in G: f(t) = 0 for all t \neq x}. the subgroups {G_x: x in [0,1]} are mutually distinct, there are uncountably many of them, and yet they're all isomorphic to one another.
@leslietownes does this require a proof? I thought that since every finitely generated subgroup of Q is cyclic and total subgroups of Q are uncountable so set of non finitely generated subgroups is uncountable - countable which is uncountable.
namely, such an operator always sends monomials to monomials. so if you've got something like $(ax^3 D^3+bx^2D^2+cxD+1)y=x^n$, then the answer will still be some multiple of $x^n$.
another cute operator observation one can make here: $x^2D^2+xD=(xD)^2$
hello, is there some resource where one can learn to use generating function to find sum of series.. series like partial series, for example $\sum_{i=1}^{m}\binom{n}{i}i^p$ where n may be greater or lesser than n
hi, could someone help me out with this ? Let $\Omega \subset \mathbb{C}$ be a region( open + connected), and suppose $p dx + q dy$ is a locally exact differential in $\Omega$, suppose $f(z) = \int_{\gamma_{z_0 \rightarrow z}} p dx + q dy$, where $\gamma_{z_0 \rightarrow z}$ is any path from $z_0$ to $z$, and assume $f$ is well-defined in the sense that for any two such paths, the difference of that integral is $2 \pi i k$ for some integer $k$ depending on the paths.
Moreover , suppose there is a harmonic function $u$ in $\Omega$, such that locally, $du + i (pdx+qdy)$ is of the form $g dz$ for some holomorphic $g$, (i.e. $pdx + q dy$ is the conjugate harmonic differential of $u$), then why is $exp(f)$ holomorphic in $\Omega$?
I can see that $exp(f)$ is now unambiguously defined, but I fail to see why it is holomorphic
sorry i mistyped the question
Let $\Omega \subset \mathbb{C}$ be a region( open + connected), and suppose $p dx + q dy$ is a locally exact differential in $\Omega$ ($p,q$ real-valued and continuous, say), suppose $f(z) = \int_{\gamma_{z_0 \rightarrow z}} du + i( p dx + q dy)$, where $\gamma_{z_0 \rightarrow z}$ is any path from $z_0$ to $z$, and assume $f$ is multi-valued in the sense that for any two such paths, the difference of that integral is $2 \pi i k$ for some integer $k$ depending on the paths. Moreover , suppose $u$ is a harmonic function in $\Omega$, such that locally, $du + i (pdx+qdy)$ is of the form $g dz$…
@Ishwaran trig functions are based on geometric investigation/experience, but can also be derived totally by theoretical means. So they are and they are not.
it depends on how you approach them, but I'd say they are mostly theoretical with a bit of empirical background to them.
@Ishwaran almost all of the formulas for trigonometric formulas are theoretical. That sine is increasing from $0$ to $1$ in the first quadrant is an empirical observation, but to actually derive that $\sin(x)=\sum\limits_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}$ needs more than simple observation.
@TedShifrin I am doing a bit better, the pup ate quite a bit yesterday, so we are hopeful. The thing is that she will like something one day and not want to look at it the next.
real quick, if $(A_n)$ is a sequence of events such that $P(A_n) \rightarrow 0$, and $X$ is a random variable, then for $E(X 1_{A_n}) \rightarrow 0$, we need $X < \infty$, don't we?
that condition is missing from the problem statement, and I seem to have a counter example without it
When we try to find dy/dx then we take delta fx/deltax and make it smaller and smaller. We get better rate. But that rate never stops at a fixed value. So we take limit and and we assign dy/dx a value which it can never have ie the limit of delta fx/deltax.
@TedShifrin I think so, why not. An rv is just a measurable function
and for the second thing you said, let $P$ be the uniform probability measure on $[0,1]$, and consider the sequence $A_n = (0, \frac{1}{n})$, with $X(\omega) = \frac{1}{\omega}$
@TedShifrin Yes, they can, but they won't, they just want to trick us in all possible ways. I was asked to give 2 possible and unique proof to the solution. And yes, I got two proofs
@TedShifrin I derived myself by proving similarity btw triangle pto and qso where o is the point of intersection of line segments tq and ps, another one is given by Paramanand in Basic Maths room by taking pq as diameter
The worst part is I have to take class to my juniors on this problem and explain the 2 proofs I got . The more points they give for me, the more internal marks I get
Let me see if I can finish. I just want to prove angle congruence. So call the right bottom angles $\alpha,\beta$, left ones $\gamma,\delta$, and the two angles in our little triangle $x,y$. Relate them all because of right angles and sum of angles, etc.
Hmm, no, I must have a mistake. Let me go back.
Found the mistake.
What is correct is $\angle SQT = \angle TPS$. That should do it, I think.
that's the nice version, although i was also making a somewhat tasteless joke about charles schulz's inability to draw non-wobbly curves in his last years.
peanuts had sort of come and gone by the time i was reading comics as a kid. i knew him as the guy who owned the only ice skating rink i had ever seen in northern california.
Lol. I'm taking on a data scientist role, going the food science route. Won't be revealing the company name, but anyone who knows where I live should be able to guess which one.
