Series Expansion: Proving √(y)√(1+y) - ln[√(y)+√(1+y)]=2y^(3/2)/3

[t_n_p]

Homework Statement

I want to show/prove that

√(y)√(1+y) - ln[√(y)+√(1+y)] = 2y^(3/2)/3 when y<<1 by series expansion.

Homework Equations

√(1+y) = 1+y/2 - (y^2)/8 + ...
and
ln[√(y)+√(1+y)] = ln[1 + √(y) + y/2 -(y^2)/8 + ...]

The Attempt at a Solution

I'm thinking I sub in the expansions above into the initial equation. My question then, is what do I do next?

 

[Gib Z]

Use [tex]\log_e (1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} [/tex] and let [tex] x = \sqrt{y} + \sqrt{1+y} -1 [/tex]. The first two terms of that expansion should clean up very nicely with the [itex]\sqrt{y}\sqrt{1+y}[/itex] from the original question, only slightly hard part is expanding that cubic term. Good luck

 

[t_n_p]

Cool, I understand this method, just wondering how I would go about it if I were to use the relevant equations though.

 

[Gib Z]

Well the problem changes to showing [tex]\log_e \left( 1 + \sqrt{y} + \frac{y}{2} - \frac{y^2}{8} ... \right) \approx \sqrt{y} - \frac{y^{3/2}}{6} - \frac{y^{5/2}}{8} ...[/tex].

Using the log (1+x) expansion to only a linear term doesn't work, so I might try using two terms, but that might take a while, Sorry I can't see anything better at the moment.

 

[t_n_p]

that's ok, its just that those specific results are provided, so I'm eager to utilize them!

 

[t_n_p]

After referring to a textbook, it seems a more appropriate way would be to let Q = √(y) + y/2 - y^2/8 with ln(1+Q) = Q - (Q^2)/2 + (Q^3)/3 - ... [similiar to what was initially mentioned].

I expand (without worrying about terms higher than 3/2) and am left with

http://img145.imageshack.us/img145/8342/69018112db1.jpg
does this appear correct?
pardon my forgetfulness, but can fractional numbers raised to powers be added?

 

[t_n_p]

*to the top*

help please!