- #1
squenshl
- 479
- 4
Homework Statement
Find two term asymptotic expansions for the roots of the following equations for ε → 0.
$$x^2+x-4ε = 0, \quad x^2+cos(εx)=5 \quad \text{and} \quad εx^3-x+9=0$$
Homework Equations
The 2 term asymptotic expansion is $$x\approx x_0+εx_1.$$
The Attempt at a Solution
Plugging the asymptotic expansion into the first equation gives $$(x_0+εx_1)^2+x_0+εx_1-4ε=0$$
$$x_0^2+2εx_0x_1+ε^2x_1^2+x_0+εx_1-4ε=0.$$
Equation constant and order ε terms one gets $$x_0^2+x_0=0 \quad \text{and} \quad 2x_0x_1+x_1-4=0.$$
$$ x_0 = 0,-1 \Rightarrow x_1=4,-4.$$
Thus our 2 term expansions are
$$x \approx 4ε \quad \text{and} \quad x \approx -1-4ε.$$
First we expand the cos term as a power series
$$\cos(εx) = 1 - \frac{(εx)^2}{2}+\cdots \approx 1.$$
Plugging the asymptotic expansion into the second equation gives $$(x_0+εx_1)^2+1=5$$
$$x_0^2+2εx_0x_1+ε^2x_1^2-4=0.$$
Equation constant and order ε terms one gets $$x_0^2-4=0 \quad \text{and} \quad 2x_0x_1=0.$$
$$ x_0 = \pm 2 \Rightarrow x_1=0.$$
Thus our 2 term expansions are
$$x \approx 2 \quad \text{and} \quad x \approx -2.$$
Plugging the asymptotic expansion into the third equation gives $$ε(x_0+εx_1)^3-x_0-εx_1+9=0$$
$$ε(x_0^3+2εx_0x_1+\cdots+εx_0^2x_1+\cdots)-x_0-εx_1+9=0.$$
Are the first 2 correct and not too sure on how to continue with the final one.
Please help!