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Dec
8
comment Proof of a theorem about oscillation
You are right. I see it now. Thanks.
Dec
8
comment Proof of a theorem about oscillation
Can you include this calculation in your answer ? I am not sure if the method of variation of parameters applies to nonlinear equations
Dec
8
comment Proof of a theorem about oscillation
Hmm.. but I can't see how the derivative of second equation is same as the first?
Dec
8
comment Proof of a theorem about oscillation
I think the difference of x and y satisfies $\displaystyle \frac {d (x-y)}{dt}=A(x-y)+R$?
Dec
8
answered Proof of a theorem about oscillation
Dec
8
comment Proof of a theorem about oscillation
Since the first derivatives of $x$ and $y$ are given so we can write a Taylor series expansion of $x(t)$ and $y(t)$. It gives a result of the form $x(t)-y(t)= x_0^2 h(t,x_0)$ for some function $h$. If one can show that $h$ is bounded in a rectangular region $[0,T]\times[-a,a]$ with some upper bound $M$, then the theorem will follow. Since then $|x(t)-y(t)|<\delta \epsilon$ if we take $|x_0|<\delta$ and $\delta<\epsilon/M$
Sep
1
awarded  Scholar
Sep
1
accepted Probability for the sum of two random numbers being a prime number?
Aug
31
comment Probability for the sum of two random numbers being a prime number?
If the choice of numbers is in the range $N_1$ to $N_2$ (where both $N_1$ and $N_2$ are 'large') then the probability will be roughly $N_2/((N_2-N_1)log(N_2))−N_1/((N_2-N_1)log(N_1))$. Right ?
Aug
30
awarded  Commentator
Aug
30
asked Probability for the sum of two random numbers being a prime number?
Jun
28
awarded  Yearling
Jun
28
answered why symplectic form should be closed when we work on a manifold
Jun
28
comment why symplectic form should be closed when we work on a manifold
Now when you are asking for an example I think my statement is not correct;) It should be other way around - every symplectic vector space is also a symplect manifold. Doesn't the condition $d\omega=0$ follow from bilinearity? Since matrix elements of $\omega$ will have no dependence on coordinates.
Jun
28
comment why symplectic form should be closed when we work on a manifold
I guess that "symplectic vector space" and "symplectic manifold" are two different notions. In particular not every symplectic vector space is a symplectic manifold.
May
13
awarded  Caucus
Jul
31
awarded  Enthusiast
Jul
24
awarded  Editor
Jul
23
comment Projective representations of loop groups
But from Lie algebra point of view study of projective representations will also give you ordinary ones. Right ?
Jul
23
comment Projective representations of loop groups
I am not sure. If there is any mathematical reason I too would like to know :)