Recent content by calum.g

solved it thanks, i was being a fool

I should also point out i don't actually know if it possible to solve this, the paper I am reading claims it is but I am not convinced.

sorry, i meant i started from the LHS

need to show (1) A coth(2ax) - B cosech(2ax)= (A+B)/2 tanh(ax)+(A-B)/2 coth(ax) starting from the RHS (2) A/2 (1+tanh^2(ax))/tanh(ax) - B/2 cosech(ax)sech(ax) (3) A/2 coth(ax) + A/2 tanh(ax) - B/2 cosech(ax)sech(ax) from here i just starting going in circles from line 2 i could have...

am i better off going from the RHS to LHS?

the exact statement is A coth(2ax) + B cosech(2ax)= (A+B)/2 tanh(ax)+(A-B)/2 coth(ax) i have tried using the double angle formulas but seem to be going in circles

Homework Statement something like A coth(2x) + B cosech(2x)= (A+B)/2 tanh(x)+(A_B)/2 coth(x) the a's and b's may not be right but the trig is what's important Homework Equations anything The Attempt at a Solution

im generating potentials, however the coulomb is proving tricky. stumbling block is finding a way to write A coth(2x) + B cosech(2x)= (A+B)/2 tanh(x)+(A_B)/2 coth(x) i am at the RHS and need to get the tanh to come out which I am told can be done. the a's and b's might not be quite right...

E in terms of n

Homework Statement im trying to solve this integral, though it is solvable using mathematica the solution is horrible. wondered if anyone has any cunning tricks or ideasHomework Equations integral between \int_{-\sqrt{E}}^\sqrt{E} \frac{\sqrt{E-W^2}}{BW^2+a}\,dW = n \pi