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Simplifying an expression using limits

  • #1

Homework Statement


Given that

##T = \frac{4E(V-E)}{4E(V-E)+V^2\sinh^2 (ka)}##

Simplify the expression for T when a is large but not infinite, and again for the case when a tends to zero and the potential tends to infinity, such that ##d = Va## is a real constant. k is a constant, as is E (which is assumed to be less than V).


Homework Equations




The Attempt at a Solution


To try and simplify it for the first case, I used the exponential definition of sinh, squared it and got that
##\sinh^2(x) = \frac{1}{4}(e^{2x}+e^{-2x}-2)##

For large x the ##e^{-2x}## term would get small very quickly, especially compared to the other exponential term which would increase rapidly. So this becomes

##\sinh^2(x) = \frac{1}{4}(e^{2x}-2)##

I don't know if that's what the question meant. It doesn't seem massively simpler. Is there something else that can be done to simplify the expression?

And the second part, I don't know where to start! There should be some sort of constant ##d = Va##, but I thought ##sinh^2(0)=0##, which would mean that ##a## disappears. As far as I can see that expression simplifies to 1 for the second set of limits! Why is that wrong?
 

Answers and Replies

  • #2
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I don't know if that's what the question meant. It doesn't seem massively simpler. Is there something else that can be done to simplify the expression?
What is the dominating term in the denominator?
And the second part, I don't know where to start! There should be some sort of constant ##d = Va##, but I thought ##sinh^2(0)=0##, which would mean that ##a## disappears. As far as I can see that expression simplifies to 1 for the second set of limits! Why is that wrong?
The limit is such that a tends to zero while V tends to infinity at the same time, so the limit of the term [itex]V^2 \sinh^{2}(ka)[/itex] cannot be done simply by setting a = 0 as you've done. A good starting point is to Taylor expand [itex]\sinh^{2}(ka)[/itex], and see where that gets you to, before taking the limits.
 
  • #3
Ray Vickson
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Homework Statement


Given that

##T = \frac{4E(V-E)}{4E(V-E)+V^2\sinh^2 (ka)}##

Simplify the expression for T when a is large but not infinite, and again for the case when a tends to zero and the potential tends to infinity, such that ##d = Va## is a real constant. k is a constant, as is E (which is assumed to be less than V).


Homework Equations




The Attempt at a Solution


To try and simplify it for the first case, I used the exponential definition of sinh, squared it and got that
##\sinh^2(x) = \frac{1}{4}(e^{2x}+e^{-2x}-2)##

For large x the ##e^{-2x}## term would get small very quickly, especially compared to the other exponential term which would increase rapidly. So this becomes

##\sinh^2(x) = \frac{1}{4}(e^{2x}-2)##

I don't know if that's what the question meant. It doesn't seem massively simpler. Is there something else that can be done to simplify the expression?

And the second part, I don't know where to start! There should be some sort of constant ##d = Va##, but I thought ##sinh^2(0)=0##, which would mean that ##a## disappears. As far as I can see that expression simplifies to 1 for the second set of limits! Why is that wrong?
Even better in the large-##a## case would be to write ##\sinh^2(x) \sim e^{2x}/4##, because ##e^{2x}## is overwhelmingly large compared to the 2. Also, if ##E(E-V)## remains constant, it is overwhelmingly small compared with the second term in the denominator, and that leads to even more simplification.
 
  • #4
vela
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It would help to note that
$$k^2 = \frac{2m(V-E)}{\hbar^2}.$$ Since ##ka## depends on both ##a## and ##V##, you need to first figure out how it behaves in the limit as ##a \to 0## and ##V \to \infty## before writing down an approximation for ##\sinh ka##. Start by writing ##V = d/a## and getting everything in terms of ##a##.
 
  • #5
It would help to note that
$$k^2 = \frac{2m(V-E)}{\hbar^2}.$$ Since ##ka## depends on both ##a## and ##V##, you need to first figure out how it behaves in the limit as ##a \to 0## and ##V \to \infty## before writing down an approximation for ##\sinh ka##. Start by writing ##V = d/a## and getting everything in terms of ##a##.
Oh, I completely forgot about that! Thanks, I'll rewrite in terms of ##a## now.
 
  • #6
What is the dominating term in the denominator?

The limit is such that a tends to zero while V tends to infinity at the same time, so the limit of the term [itex]V^2 \sinh^{2}(ka)[/itex] cannot be done simply by setting a = 0 as you've done. A good starting point is to Taylor expand [itex]\sinh^{2}(ka)[/itex], and see where that gets you to, before taking the limits.
In terms of a, ##ka = \frac{\sqrt{2mad - 2ma^2E}}{\hbar}##.

That still tends to zero, given d is a real constant. So ##sinh^2(ka)## tends to zero, since its taylor expansion is ##x^2 + \frac{x^4}{3} + \frac{2x^6}{45}##.

Taking the ##V^2## term into account, how do I know what happens? I've never dealt with composite limits before, it kind of looks like V tends to infinity just as fast as the ##sinh^2## term tends to zero!
 
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  • #7
Sorry, misplaced # in the post above.

In terms of a, ##ka = \frac{\sqrt{2mad - 2ma^2E}}{\hbar}##.

That still tends to zero, given d is a real constant. So ##sinh^2(ka)## tends to zero, since its taylor expansion is ##x^2 + \frac{x^4}{3} + \frac{2x^6}{45}##.

Taking the ##V^2## term into account, how do I know what happens? I've never dealt with composite limits before, it kind of looks like V tends to infinity just as fast as the ##sinh^2## term tends to zero!
 
  • #8
It would help to note that
$$k^2 = \frac{2m(V-E)}{\hbar^2}.$$ Since ##ka## depends on both ##a## and ##V##, you need to first figure out how it behaves in the limit as ##a \to 0## and ##V \to \infty## before writing down an approximation for ##\sinh ka##. Start by writing ##V = d/a## and getting everything in terms of ##a##.
Is getting everything in terms of a equivalent to getting it in terms of V, or do both cases have to be considered?
 
  • #9
954
117
That still tends to zero, given d is a real constant. So ##sinh^2(ka)## tends to zero, since its taylor expansion is ##x^2 + \frac{x^4}{3} + \frac{2x^6}{45}##.
Taking the ##V^2## term into account, how do I know what happens? I've never dealt with composite limits before, it kind of looks like V tends to infinity just as fast as the ##sinh^2## term tends to zero!
Why not try multiplying [itex]V^{2}[/itex] to your Taylor series expansion and see what you get? There'll be some simplifications that allow you to take the limit properly.
[Hint: use Va = d]
 

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