short question

M.A · **Joined:** Thu, 26 Jul 2007 19:08:57 UTC **Posts:** 48

Hi
Given a system of two equations:
$\begin{array}{rl}x-z+y^2&=0 \\ -2x +2y +z^2 -y^2 &= 0 \end{array}$
It is required to show that we can parametrize x,z in terms of y up to third order as:
$\begin{array}{rl}x&=y+y^3+O(y^4) \\ z&=y-y^2-y^3+O(y^4) \end{array}$

Thanks

outermeasure · **Posted:** Thu, 10 May 2012 04:04:20 UTC

M.A wrote:

Hi
Given a system of two equations:
$\begin{array}{rl}x-z+y^2&=0 \\ -2x +2y +z^2 -y^2 &= 0 \end{array}$
It is required to show that we can parametrize x,z in terms of y up to third order as:
$\begin{array}{rl}x&=y+y^3+O(y^4) \\ z&=y-y^2-y^3+O(y^4) \end{array}$

Thanks

Note that elimininating

gives

which gives you other possibilities.

First check that y is a local coordinate near y=0

(exercise). So, expanding near y=0, $x=a_0+a_1y+a_2y^2+a_3y^3+O(y^4),\quad z=b_0+b_1y+b_2y^2+b_3y^3+O(y^4)$ gives
$\begin{aligned} (a_0+a_1y+a_2y^2+a_3y^3)-(b_0+b_1y+b_2y^2+b_3y^3)+y^2&=O(y^4)\\ -2(a_0+a_1y+a_2y^2+a_3y^3)+2y+(b_0+b_1y+b_2y^2+b_3y^3)^2-y^2&=O(y^4)\\ \end{aligned}$
So equating coefficients, ....

M.A · **Joined:** Thu, 26 Jul 2007 19:08:57 UTC **Posts:** 48

thanks alot

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