Recent content by Andy_ToK

your solution was right, all u have to do then is to plug it into your calculator and calculate arctan4/3

Thanks, Avodyne, but I'm not sure what u mean by "the Hilbert space consists of functions on the interval where the potential is zero". Could you elaborate a little bit?

Hi, I have a question about the momentum eigenstates in a 1D infinite square well example. First of all, are there any eigenstates at all in this example? By explicitly applying the wavefunction(stationary states) which can be easily obtained from the boundary conditions, it can shown that the...

Thank you. HallsofIvy

Thanks. well, if let h=r \cos \theta, k = r \sin \theta, lim_{(h,k)->(0,0)} \sqrt{\frac{hk}{h^2+k^2}}} can be simplified to lim_{r->0}\sqrt{\cos \theta\sin \theta} which is dependent on theta however, how should I proceed then?

Hi, I'm sort of new to the calculation of limits of functions defined in R^2( or R^n, n>=2) exp. the limit of sqrt{hk/(h^2+k^2)} as (h,k) --> (0,0) I think it should be 1/sqrt(2) if we let h=k. but I'm not sure if this is the right approach. what about other limits in the form "0/0" as...

Thanks quasar987 and ZioX I forgot to use T(x)-T(y)=T(x-y) for the linear transformation... let d=e/C then, |x-y|<d --> |T(x)-T(y)|=|T(x-y)<cd=e

oops, it should be <=, thanks. AM=arithmetic mean RMS=root mean square AM<=RMS (the equality holds when |x1|=|x2|...=|xm|)

1.Thanks for pointing that out, it should be the absolute value.:smile: 2.sorry i made a mistake there, it should be sum_i|xi|<sqrt(m)*sqrt(sum_i(xi^2)) by AM<RMS

x is an element of R^m |T(x)|=|sum_i(xi*T(ei))|<=sum_i{xi(|T(ei)|)} let A=max{|T(e1)|,|T(e2)|...|T(em)|) then |T(x)|<=A*sum_i(xi)<A/sqrt(m)*sqrt(sum_i(xi^2))=A/sqrt(m)*|x| (AM<RMS) sorry, i don't know how to use latex here:frown:

T is a linear transformation from R^m->R^n, prove that T is continuous. I have proved that there's always a positive real number C that |T(x)|<=C|x|. How shall I proceed then? Thanks~

Thank you all.

[0,1] is but A isn't, i think. because A doesn't contain those irrational numbers between 0 and 1.

Hi, here is the question, if A is a closed set that contains every rational number r: [0,1], show that [0,1] is a subset of A. But, how could A be closed? If A is closed, R^n-A is open, so any point in R^n-A would have a open sphere around it and this open sphere wouldn't intersect A...

hi, thanks. I have the answer but wonder how to solve it without using calculator. Sorry if I wasn't clear.