Recent content by Inquisitive_Mind

I suspect that the a formal proof will be quite technical. However, to understand it intuitively, one may try to discretise the integral in the definition of y(t). The decretised integral is a linear combination of jointly Gaussian variables, and is therefore Gaussian. Each of the y(ti) can be...

This semester I took a course in Quantum Field Theory. It is difficult (the professor assumes you know everything) and I have so many questions... Starting with a lagrangian density, I was told that canonical quantisation is a procedure where we impose the usual commutation relation between...

Are there any other suggestions, pls?

Thanks. I somehow can convince myself that the mass matrix has to be positive definite. But specifically I need to show that the principle of least action *requires* the mass matrix to be positive definite. So is there a way? N.B. By principle of least action I mean, of all the possible paths...

Lagrangian Mass Matrix Question, Pls Help! "Assume that the lagrangian is given in the form L(q_{i}, q_{i}', t) = \sum_{m, n}T_{mn}(q_{i})q_{m}'q_{n}' - V(q_{i}) Show that in order for the principle of least action to hold, T_{mn}(q_{i}) has to be positive definite." My first...

For Q2, just consider the open interval of (-1, 1) in R. Consider an open cover like the collection of intervals, { (x-w, x+w), w=(1-|x|)/2: x from (-1, 1) }. If this collection has a finite subcover, would there be any points not covered by the subcovering (and thus a contradiction)?

Use t method? (i.e. set t=tan(x/2), such that cos(x)=(1-t^2)/(1+t^2))

I would suggest that you consider the function g(x)=x-\frac{f(x)}{f^{'}(x)}, using Taylor's inequality. Note that R_n(x) can be expressed as |x_{n+1}-r|. Also, after differentiating g(x) once, you may need an estimate for |f(x)|. You may obtain such by noting that f(x)=\int^{x}_{r}f^{'}(y)dy

Hint for Q2: what's f(1) and f(-1)?

Hint for Q2: Investigate the absolute difference |b_{n}-b_{n-1}|, noting that b_{n}>0 for all n. You should get something like |b_{n+1}-b_{n}|\leq 6|b_{n}-b_{n-1}|. Then use Cauchy's criterion for convergence.

Yes, that is correct. You may consider the vectors normal to the curve u(x,y)=c1 and v(x,y)=c2. To construct normal vectors you just take the gradients of the functions u(x,y) and v(x,y). The condition (du/dx)(du/dy) + (dv/dx)(dv/dy) = 0 means that the two sets of normal vectors are orthogonal...

Thanks for your replies! To selfAdjoint, Maybe you mistakably thought that I was talking about the ordinary path integral? Or did I miss something? To Haelfix, Path integrals are new to me. May you explain why the non-differentiability of paths does not matter? I can imagine that there...

In path integrals, how does one deal with non-differentiable paths? Obviously non-differentiable paths are allowed, but with Feymann's formulation, one has to calculate the action for a path, and then sum over all possible paths. How is the action defined (if it is defined at all) for a...