Transform given problem and eliminate x and y

[leothorn]

Homework Statement

The problem is tough to type out correctly. Pasting problem statement image

http://postimg.org/image/a0r92a0wl/

http://postimg.org/image/a0r92a0wl/

The Attempt at a Solution

I just need to know how to proceed with the problem. Not the answer. This is the scan of my attempt.
http://postimg.org/image/uun8gt2sz/

http://postimg.org/image/uun8gt2sz/

I stuck about how i eliminate the x and y terms. I thought replacing it with GP series sum would help but it unnecessarily introduces a restriction on x and y.

Is the initial process in my worked out solution correct?
if not what direction do you suggest ?

 

[TSny]

Hello.
In your final equation, you can sub for x in terms of y. Then collect all terms of order y², all terms of order y³, etc.

 

[leothorn]

hmm i have x^n in the last line. How exactly do i sub and with what ? Wont it just explode the number of terms ?
Also Bnxn term is not in terms of yn to replace it ...I don't see how it can be done

 

[TSny]

You "only" want to go up to y⁶. So you only need to keep the appropriate finite number of terms in the sums.

If you want, you can bring your first summation on the right over to the left and combine it with the summation already on the left. [EDIT: Sorry, I see you already brought the sum on the right to the left. You can combine those two sums.]

The coefficient of a certain power of y on the left must match the coefficient of the same power of y on the right.

Start by matching lowest power of y on left with same power of y on right.

It will get tedious as you work your way up to y⁶.

 

[leothorn]

Are you referring to the last step ?. Well i separated y^n terms and x^n terms on either side of the equation.
I started with n=2. I notice that x and y terms don't go away. ..How do i get rid of them ?

 

[haruspex]

Are you referring to the last step ?.

Going from the last line of your attempt, combine the two sums on the left into a single power series in y, as TSny suggested.
You can use equation (2) to substitute for x on the right. This gives a nasty double sum, but it does get rid of x. Expanding, only one term will contribute to y², three will contribute to y³, eight will contribute to y⁴...

 

[TSny]

Yes, I see now that you got all the y's over to the left. You can combine the two sums on the left.

For $x$ on the right, you will need to substitute $y +\displaystyle \sum_{k=2} A_k y^k$.
You will have a sum over k inside a sum over n.

You will have to consider how far to carry out the sums over n and the sum over k to get all terms up to order 6.

It does get tedious. I worked out through A₄ by hand and then turned to Mathematica to help with A₅ and A₆.

 

[leothorn]

I am curious about introduction of k.
Once say we have summed over k how does one double sum over n ..now that n's gone. I see the light though thanks.

 

[TSny]

The right hand side is

$\displaystyle \sum_{n=2}^{\infty} b_nx^n = \displaystyle \sum_{n=2}^{\infty} b_n ( y +\displaystyle \sum_{k=2}^{\infty} A_k y^k)^n$

or

$\displaystyle \sum_{n=2}^{\infty} b_nx^n = b_2 (y +\displaystyle \sum_{k=2}^{\infty} A_k y^k)^2 + b_3 (y +\displaystyle \sum_{k=2}^{\infty} A_k y^k)^3+b_4 (y +\displaystyle \sum_{k=2}^{\infty} A_k y^k)^4+...$

So, suppose you are trying to collect together all the $y^2$ terms. Inspection shows that only the first term on the right (the $b_2$ term) will produce any $y^2$ terms.

For $y^3$ you can see that both the $b_2$ term and the $b_3$ term will produce $y^3$ terms.

 

[leothorn]

Thank you for the detailed help. I got it !