Help for a Perturbation problem

Can someone please help me. I am studying from what is possibly the worst written book... I do not even know how to start the problem...

Problem:

Let
$\displaystyle f(y, \epsilon) = \frac{1}{(1+\epsilon y)^{\frac{3}{2}}}, y=y_{0}+\epsilon y_{1} +\epsilon^{2} y_{2}+...$

Expand f in powers of $\displaystyle \epsilon$ up to $\displaystyle O(\epsilon^{2})$

I know I let $\displaystyle f(y, \epsilon) =0$

I know I need to find
$\displaystyle O(1), O(\epsilon), O(\epsilon^{2}) $

The problem is that the examples to do this are only for very simple algebraic problems. Any help would be greatly appreciated...

You can use the general binomial theorem

$\displaystyle \frac{1}{(1-x)^s} = \sum_{k=0}^\infty \binom{s+k-1}{s-1}x^k$

Here $\displaystyle s = 3/2$ and $\displaystyle x = -\epsilon y$


$\displaystyle \frac{1}{(1+\epsilon y)^{3/2}} = \sum_{k=0}^\infty \binom{k+1/2}{1/2}(-\epsilon y)^k$

Now it can be shown that

$\displaystyle \binom{k+1/2}{1/2} = \frac{(2k+1)!!}{(2k)!!}$

where !! is the double factorial operator (product of just the odd or even numbers up to and including (2k+1) or (2k)). So

$\displaystyle \frac{1}{(1+\epsilon y)^{3/2}} = \sum_{k=0}^\infty (-1)^k\frac{(2k+1)!!}{(2k)!!}(\epsilon y)^k$

And therefore you can expand in order of epsilon as

$\displaystyle \frac{1}{(1+\epsilon y)^{3/2}} = 1 - \frac{3y}{2}\epsilon + \frac{15y^2}{8}\epsilon^2 + O(\epsilon^3).$

Hope this helps.

Thanks so much. if I understand correctly, this expansion is equivalent to the perturbation expansion. Am I correct? This book is really bad. sorry. :(

Quote:

---

Originally Posted by ginafara

Thanks so much. if I understand correctly, this expansion is equivalent to the perturbation expansion. Am I correct? This book is really bad. sorry. :(

---

My last line is the expansion in terms of y you should still substitute your expansion for y. For example,

$\displaystyle f(y,\epsilon) = 1 - \frac{3y}{2}\epsilon + \frac{15y^2}{8}\epsilon^2 + O(\epsilon^3).$

$\displaystyle y = y_0 + \epsilon y_1 + \epsilon^2 y_2 + O(\epsilon^3).$

Substituting

$\displaystyle f(y,\epsilon) = 1 - \frac{3(y_0 + \epsilon y_1 + \epsilon^2 y_2)}{2}\epsilon + \frac{15(y_0 + \epsilon y_1 + \epsilon^2 y_2)^2}{8}\epsilon^2 + O(\epsilon^3).
$

Simplifying and keeping just terms less than $\displaystyle O(\epsilon^3)$

$\displaystyle f(y,\epsilon) = 1 - \frac{3 y_0}{2}\epsilon + \left(\frac{15y_0^2}{8} - \frac{3 y_1}{2}\right)\epsilon^2 + O(\epsilon^3).
$

So this gives the expansion of $\displaystyle f(y,\epsilon)$ in terms of the expansion for $\displaystyle y$.

That is what I did... Thanks again... I was reading through an example and the author stated that the binomial expansion is equivalent to the perturbation expansion in his example problem... It didn't make sense so I continued on with the rest... Thanks again so much for your help... May I say never use logan's applied mathematics 3rd edition.. Horrible...

Quote:

---

Originally Posted by ginafara

May I say never use logan's applied mathematics 3rd edition.. Horrible...

---

I've never seen it, but I'll bear that in mind for future reference. :)