# Show an approximation

1. Jun 9, 2009

### faklif

1. The problem statement, all variables and given/known data
This comes from a book on relativity but it basically comes down to a math problem. The problem is to prove that if
$$T^2 \ll\frac{c^2}{\alpha^2}$$
then
$${t}\approx{T}(1-\frac{\alpha^2{T^2}}{6c^2})$$
given
$$\frac{\alpha{T}}{c}=sinh(\frac{\alpha{t}}{c})$$

2. Relevant equations
See above.

3. The attempt at a solution
I've tried solving for t from the equation
$$\frac{\alpha{T}}{c}=sinh(\frac{\alpha{t}}{c})$$
which gives
$$t=\frac{c}{\alpha}\log(\frac{T\alpha}{c}+\sqrt{1+\frac{T^2\alpha^2}{c^2}})$$
I thought I'd be able to use maclaurin expansion at this point becuase of how the approximation looks but I keep making mistakes and I'm not getting anywhere at the moment so I'd really appreciate some help.

2. Jun 9, 2009

### cepheid

Staff Emeritus
What if you tried writing the hyperbolic sine function in terms of exponential functions first? Their Taylor expansions are easy...

3. Jun 9, 2009

### faklif

Thanks for the reply!

Using the first four terms of each of the expansions I end up with what looks like the some kind of opposite to what I want
$${T}\approx{t}(1+\frac{\alpha^2{t^2}}{6c^2})$$
I can't see that I can get where I want from this and I haven't used all information in the problem to get here, maybe I'm missing something.

4. Jun 9, 2009

### Dick

You need to reverse the roles of t and T. You could also write at/c=arcsinh(aT/c) and use arcsinh(x)=x-x^3/3!+...

5. Jun 9, 2009

### HallsofIvy

Staff Emeritus
No reason in the world to be always writing hyperbolic functions as exponentials!

sihh(0)= 0.
(sinh(x))'= cosh(x) and cosh(0)= 1.
(sinh(x))"= (cosh(x))'= sinh(x) and sinh(0)= 0.
(sinh(x))'''= (sinh(x))'= cosh(x) and cosh(0)= 1, etc.

6. Jun 9, 2009

### Hurkyl

Staff Emeritus
Well, have you at least substituted that approximation into $T(1-\frac{\alpha^2{T^2}}{6c^2})$?

7. Jun 9, 2009

### faklif

Thanks a lot that does it! Also gives me another question and that is how to find that arcsinh(x)=x-x^3/3!+...? My thought of simply solving for t and then looking at what I got didn't really get me there.

How can I do that substitution? I'm feeling very rusty here, I'm sorry.

8. Jun 9, 2009

### Billy Bob

Here's one way. The derivative of arcsinh x is $$(1+x^2)^{-1/2}$$ so use the binomial series to find the series for that, then integrate term by term. You only need a few terms anyway, not the whole series.

9. Jun 10, 2009

### Dick

A physicist would just take
$${T}\approx{t}(1+\frac{\alpha^2{t^2}}{6c^2})$$
and say, hey, that means t=T with correction of higher order. So I can just replace t^2 with T^2 without making any errors I would care about. So I can change that into
$${T}\approx{t}(1+\frac{\alpha^2{T^2}}{6c^2})$$
Now move all the T's to one side and use 1/(1+x) is approximately equal to 1-x for x<<1.

Last edited: Jun 10, 2009
10. Jun 10, 2009

### faklif

Thank you everyone!