713 reputation
413
bio website
location
age 21
visits member for 2 years, 2 months
seen Dec 12 at 22:34

Dec
11
asked Singularities of complex exponential and asymptotic expansion
Dec
10
comment Asymptotic expansion of integral (Laguerre)
ah nice, what is the surface of $\text{Re}f(t)$ here? It is proportional to $r^2 \cos 2\phi$?
Dec
10
comment Asymptotic expansion of integral (Laguerre)
ok thanks, I guess I am just trying to see how this example in question is not a contour along a constant altitude
Dec
10
comment Asymptotic expansion of integral (Laguerre)
Oh right, is that because in this case, we have a saddle point on the real axis?
Dec
10
comment Asymptotic expansion of integral (Laguerre)
Thank you, sorry can I ask why?
Dec
10
comment Asymptotic expansion of integral (Laguerre)
sorry, I meant $\phi = \pi/2$.
Dec
10
comment Asymptotic expansion of integral (Laguerre)
ah so this is the stationary phase method? I expand $t - t_o = re^{i \phi}$ and so $(t-t_o)^2 = r^2e^{i2\phi} \Rightarrow \sin 2 \phi = 0$ for stationary phase, giving $\phi = 0, \pi/2, \pi, 3\pi/2$, I choose $\phi = \pi$ since this gives $\text{Re}(f(t)) < 0$.Seem okay?
Dec
10
comment Asymptotic expansion of integral (Laguerre)
Ok thanks, can I just check though, shouldn't I have obtained a negative for the second derivative? Otherwise in the end I will have a diverging integral?
Dec
10
comment Asymptotic expansion of integral (Laguerre)
Yes I got the same upon doing another recheck, thanks. Perhaps more importantly, if I can ask, why do we then approximate the original integral over a contour by a linear integral extending from -infinity to infinity. The infinite bounds make sense if I can understand why we can do a linear integral in the first place. Is it because we deform the contour to pass through $t_o$ and then subsequently put a tangent at this point that goes from $-\infty$ to $+\infty$. Then the dominant contribution is near the saddle point, hence the allowed limit extension?
Dec
10
comment Asymptotic expansion of integral (Laguerre)
@AntonioVargas: Hi, $t_o$ is the saddle point and I got it to be $1/2$. Were you getting the same? Thank you for your interest.
Dec
10
asked Asymptotic expansion of integral (Laguerre)
Dec
10
accepted Generating function of the Laguerre Polynomials
Dec
10
comment Generating function of the Laguerre Polynomials
sure, no problem and thanks for your help! +1,accepted
Dec
10
comment Generating function of the Laguerre Polynomials
Thank you! It makes sense to me. From this we can determine that $$L_n = \frac{1}{2\pi i } \oint_{C'} \frac{1}{(1-t)^{\alpha+1} t^{n+1}} e^{-\frac{xt}{1-x}} dt$$ Now set $\alpha = n$ and I want to find the leading asymyptotic behaviour for large $n$. For this the saddle point method could be applied. Are you familiar with this? I thought about reexpressing the terms dependent on $n$ in the integrand there by exponentials so I could use the method but my choice results in a vanishing second derivative of the function multiplying $n$. Should I update the main post for this? Thanks again!
Dec
10
asked Generating function of the Laguerre Polynomials
Nov
22
comment Expansion of integrand before integration?
I am modelling the atom-atom van der waals interaction as $U = -C/r^6$. where $r$ is the distance between the atoms. Really I am wondering if expanding before integration is allowed?
Nov
22
comment Expansion of integrand before integration?
I should also point out that the setting up of the problem and obtaining the answer via whatever means (I.e by expansion before or after integration) is only worth 1 mark! My prof seems happy for me to expand but if I do so in another configuration of wires I obtain the same answer as the above configuration which I know is not true. So expansion is introducing errors I think.
Nov
22
comment Expansion of integrand before integration?
Thanks, but I need to use the van der walls interaction model to solve the problem. I have been advised elsewhere that an expansion before integration introduces large errors so the advice was to integrate exactly and then expand. The problem I am having is how to integrate exactly my expression in my last post. Polar coordinates looks good but there is a cross term which introduces further difficulties and I am not sure how to write the limits in polar coordinates for a square region in $x_1 x_2$ plane?
Nov
22
comment Expansion of integrand before integration?
@TonyPiccolo Any thoughts?
Nov
21
comment Expansion of integrand before integration?
Hi, I am computing the van der waals interaction between two parallel wires separated by a distance $d$ and length L and the pairwise interaction between atoms is $U = -C/r^6$ so my integral should read (I know the form is right, and I think the limits are also good)$$-\int_{-L/2}^{L/2} \int_{-L/2}^{L/2}\frac{C}{((x_2-x_1)^2 + d^2)^3} dx_1 dx_2$$ where the $x_i$ are coordinates for the wires.