Taylor Approximation (I think) on Transmission Coefficient

[erok81]

Homework Statement

I have this equation:

T=(1+\frac{U_{0}^{2}}{4E(U_{0}-E)}sinh^{2}(2 \alpha L))^{-1}

Where α is given by:

\alpha = \sqrt{ \frac{2m(U_{0}-E)}{\hbar^{2}}}

I have to show that in the limit αL>>1 my equation is approximately given by:

T=\frac{16E(U_{0}-E)}{U_{0}^{2}}e^{(-4 \alpha L)}

Homework Equations

n/a

The Attempt at a Solution

I am horrible with taylor approximations and sometimes am not even sure when to use them (hence the "I think" in the subject).

I would assume that we are approximating due to the question. I would also assume that because of that, we want to taylor expand. I didn't try this by hand, but I did put it into Maple and it said that there wasn't a taylor expansion for this particular item.

Am I approaching it correctly? Taylor expansions and when to use them are my goal between spring and summer semester, that's for sure.

 

[erok81]

Okay I lied. I tried the expansion again from 1 to 1 instead of 0 to 1. This got me closer, assuming I am doing it right. Except my α term is gone now.

T=\frac{16(e^{2L})^{2}E(U_{0}-E)}{(1+U_{0}^{2})((e^{2L})^{2}-1)^{2}}

With a term involving what looks like the letter O. I've never used Maple to do these, so I'm not familiar with that term.

It looks like if I multiply that all out, I might have something. With the exception of my missing α terms.EDIT: I mistyped this into maple so the solution isn't right. I corrected it, but now I'm not even close. So I have to be doing something drastically long.

 

[erok81]

Oh...I also expanded over α.

I tried simplifying it and got close but can't quite get there. Not to mention I am missing the α term I need.

 

[ideasrule]

You don't need to Taylor expand, nor do you need to substitute in the expression for alpha. Think about it this way: sinh(aL)=1/2(e^aL - e^-aL). What happens when aL approaches infinity? No calculation required!

Now, what happens to sinh(aL)^2 when aL approaches infinity?

 

[erok81]

Well...when e^-aL has -aL goes to infinity it goes to zero. When e^aL goes to inifinity, it diverges to infinity...right?

Then trying the next part, the infinity throws me off again. So I think I am missing something.

Plus if I try it that way, I can't get my 16 out front. At least I don't think I can.

 

[erok81]

Ok. I've tried and tried approximate this thing, but I am still not getting anywhere. I've attached my latest attempt in Maple. It's close, but not quite there.

What do you guys think?

 

[SammyS]

Start with sinh(x) = (e^x - e^-x)/2

so sinh²(x) = (e^x - e^-x)²/4 = ?

 

[erok81]

That gives me (e^2x+e^-2x-2)/4

 

[SammyS]

Do the same for sinh²(2αL).

Presumably, e^2(2αL) >> 2 >> e^-2(2αL),

Now, justify getting rid of the 1

 

[erok81]

Hmm...can I just get rid of the one because it really has no impact on the final values since it's such a small number. Especially when dealing with infinities on exponents.

So now I have a 4 in the numerator and two exponentials. I have no idea how I can get what I need from that.

 

[SammyS]

Do the same for sinh²(2αL).

Presumably, e^2(2αL) >> 2 >> e^-2(2αL),

What the above implies is that for αL >> 1,
sinh²(2αL) → (e^2(2αL))/4

So, if you drop the "1 +", you should have your result.

 

[erok81]

Ok. I'm getting close to understanding this. :shy:

You get sinh²(2αL) → (e^2(2αL))/4 because as αL >> 1 the negative exponent drops off to zero?

Assuming that is correct. Do you have any suggestions on what to look up, topics/subjects, in order to get more practice identifying and approximating stuff? Doesn't have to be this type of physics, just anything.

 

[SammyS]

Very often the Taylor series approach works fine. This was just an unusual situation. It helped to have a result to match. Sometimes you must simply "play around" with the functions. Graphing can also help.

 

[SammyS]

Here's a little more straight forward way to show this.

Starting with sinh(x) = (e^x - e^-x)/2 = (e^x/2)(1-e^-2x) → (e^x/2) for x >> 1 .

So, sinh²(x) → e^2x/4 for x >> 1 .

Put that result into T.

As x → ∞, 1/(1+x) → 1/x-(1/x)^2+(1/x)^3-(1/x)^4+…

Keeping the first term should give you the result.