# Thread: Simplify expression, Is it correct?

1. ## Simplify expression, Is it correct?

Attached is the expression and my solution.

I get a solution of zero, is it correct?

2. Sorry the attachment is not clear. Can't really tell u from a in the attachment. Can attach a clearer version.

3. There is no a? All of the exponents are u (positive and negative).

4. the correct answer is 4/((e^u+e^(-u))^2)

5. This is the approach i took.

(e^u+e^-u)(e^u+e^-u)-(e^u-e^-u)(e^u-e^-u) / (e^u+e^-u)^2
(e^2u+2+e^-2u)-(e^2u-2+e^-2u)/ (e^u+e^-u)^2
(e^2u+2+e^-2u-e^2u+2-e^-2u)/ (e^u+e^-u)^2
4 / (e^u+e^-u)^2

Originally Posted by nrslmz

6. Attached is a better version.

7. Ok noted. Your approach is not quite right. Please look at my previous post for further clarification.

8. Ahh, I see now. Thanks!

9. Actually, I noticed that you didn't simplify your denominator.

The denominator simplified is 3 because (e^2u + e^0 + e^0 + e^-2u).

10. You are right. I forgot to simplify the denominator

In that case it should be

(e^u+e^-u)(e^u+e^-u)-(e^u-e^-u)(e^u-e^-u) / (e^u+e^-u)^2
(e^2u+2+e^-2u)-(e^2u-2+e^-2u)/ (e^u+e^-u)^2
(e^2u+2+e^-2u-e^2u+2-e^-2u)/ (e^u+e^-u)^2
4 / (e^2u+2+e^-2u)
4 / 3

11. Thanks!

12. Originally Posted by mwok
Attached is the expression and my solution.

I get a solution of zero, is it correct?
Hi,

If I am not mistaken, this can be solved using hyperbolic functions.
I'll break this problem into many sections so you can see what I am doing

If you consider that the hyperbolic functions are

$\frac{e^{x} - e^{-x}}{2} = sinh(x)$

and that

$\frac{e^{x} + e^{-x}}{2} = cosh(x)$

then we can conclude that

$(e^{x} - e^{-x}) = 2sinh(x)$

$e^{x} + e^{-x} = 2cosh(x)$

------------------------------------------------------
Which means that if you use them in your problem you will get

$(e^{u} + e^{-u})*(e^{u}+e^{-u}) = 4 cosh^{2}(u)$

$(e^{u} - e^{-u})*(e^{u}-e^{-u}) = 4 sinh^{2}(u)$

$(e^{u} - e^{-u})^{2} = 4 cosh^{2}(u)$

$4 cosh^{2}(u) - 4 sinh^{2}(u) = 4$

$\frac{4}{4 cosh^{2}(u)} = \frac{1}{cosh^{2}(u)} = sech^{2}(u)
$