Breaking Through the Coulomb Barrier: Solving an Equation

[calum.g]

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 but the trig functions are what I am after

 

[calum.g]

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

 

[Mark44]

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

What do you mean "something like?" If you need to prove some statement, give us exactly the statement that needs to be proved.

What equations are relevant to this problem?

What have you tried? Before anyone can give you any help, you need to show some effort at working the problem.

 

[Cyosis]

Use the identities $\cosh(2x)=\sinh^2(x)+cosh^2(x)$, $\sinh(2x)=2\cosh(x)\sinh(x)$ and $\cosh^2 (2x)-\sinh^2(2x)=1$.

 

[calum.g]

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

 

[calum.g]

am i better off going from the RHS to LHS?

 

[Mark44]

Show us what you've tried.

 

[calum.g]

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 instead gone to

A/2 coth(ax) (2-sech^2(ax))- B/2 cosech(ax)sech(ax)
A coth(ax) -A/2 coth(ax)sech^2(ax)- B/2 cosech(ax)sech(ax)
A coth(ax) -A/2 (cosech(ax) / sech(ax)) sech^2(ax)- B/2 cosech(ax)sech(ax)
A coth(ax) -A/2 cosech(ax) sech(ax)- B/2 cosech(ax)sech(ax)
A coth(ax) -(A+B)/2 cosech(ax) sech(ax)

i don't see any other openings

 

[Mark44]

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)

How did you get to (2) from the right side of (1)? 1 + tanh²(ax) = sech²(ax). If I were doing this, I'd start on the left side and see if I could make it look like the right side.

(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 instead gone to

A/2 coth(ax) (2-sech^2(ax))- B/2 cosech(ax)sech(ax)
A coth(ax) -A/2 coth(ax)sech^2(ax)- B/2 cosech(ax)sech(ax)
A coth(ax) -A/2 (cosech(ax) / sech(ax)) sech^2(ax)- B/2 cosech(ax)sech(ax)
A coth(ax) -A/2 cosech(ax) sech(ax)- B/2 cosech(ax)sech(ax)
A coth(ax) -(A+B)/2 cosech(ax) sech(ax)

i don't see any other openings

I don't know if this would work, but this is what I would try. Starting from (3) I would add (B/2)tanh(ax) - (B/2)coth(ax) and then subtract it off, with the possibility that (-B/2)tanh(ax) + (B/2)coth(ax) -(B/2)cosech(ax)sech(ax) adds up to zero. If so, that what remains is what you want to end up with.

 

[calum.g]

sorry, i meant i started from the LHS

 

[calum.g]

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.

 

[Cyosis]

Start on the left hand side and write the coth and cosech in terms of cosh/sinh. Now apply the three identities from post 4 everywhere you can. And don't forget to show your attempt if you get stuck.

 

[calum.g]

solved it thanks, i was being a fool