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\begin{document}


{\bf Question}

Show that the roots of
$x^4+(4-\varepsilon)x^3+(6-2\varepsilon)x^2+(4+\varepsilon)x^2
+(4+\varepsilon)x+1-\varepsilon^2=0$ have the form

$$x=-1+(2\varepsilon)^{\frac{1}{4}}e^{i\pi\frac{(2n+1)}{4}}+O\left(\varepsilon^{\frac{1}{2}}\right),\
n=1,2,3,4,\ \varepsilon\to 0.$$



\vspace{.5in}

{\bf Answer}

$x^4+(4-\varepsilon)x^3+*(6+2\varepsilon)x^2+(4+\varepsilon)x+1-\varepsilon^2=0$

Put $\varepsilon=0$: $x^4+4x^3+6x^2+4x+1=0$

Pascal's Triangle should come to mind (look at 1,4,6,4,1 pattern).

Thus we have $(x+1)^4=0$ i.e., $x=-1$ four times degeneracy.

So try ansatz

$$x=x_0+x_1\varepsilon^{\frac{1}{4}}+x_2\varepsilon^{\frac{1}{2}}
+O(\varepsilon^{\frac{3}{4}})$$

and substitute.

$\
[x_0+x_1\varepsilon^{\frac{1}{4}}+x_2\varepsilon^{\frac{1}{2}}+O(\varepsilon^{\frac{3}{4}})]^4
+(4-\varepsilon)(x_0+\varepsilon^{\frac{1}{4}}x_1+\varepsilon^{\frac{1}{2}}x_2
+O(\varepsilon^{\frac{3}{4}})^3$

$+(6+2\varepsilon)(x_0+\varepsilon^{\frac{1}{4}}x_1+\varepsilon^{\frac{1}{2}}x_2
+O(\varepsilon^{\frac{3}{4}}))^2+(4+\varepsilon)(x_0+\varepsilon^{\frac{1}{4}}
+\varepsilon^{\frac{1}{2}}x_2+O(\varepsilon^{\frac{3}{4}})+1-\varepsilon^2=0$

$\
x_0^4+4x_0^3x_1\varepsilon^{\frac{1}{4}}+6x_0^2x_1^2\varepsilon^{\frac{1}{2}}
+4x_0^3x_2\varepsilon^{\frac{1}{2}}+O(\varepsilon^{\frac{3}{4}})$

$+(4-\varepsilon)(x_0^3+3x_0^2\varepsilon^{\frac{1}{4}}x_1+3x_0\varepsilon^{\frac{1}{2}}x_1^2
+3x_0^2\varepsilon^{\frac{1}{2}}x_2+O(\varepsilon^{\frac{3}{4}})$

$+(6+2\varepsilon)(x_0^2+2x_0x_1\varepsilon^{\frac{1}{4}}+2x_0x_2\varepsilon^{\frac{1}{2}}
+\varepsilon^{\frac{1}{2}}x_2^2+O(\varepsilon^{\frac{3}{4}}))$

$+(4+\varepsilon)(x_0+\varepsilon^{\frac{1}{4}}x_1+\varepsilon^{\frac{1}{2}}x_2
+O(\varepsilon^{\frac{3}{4}}))$

$+1-\varepsilon^2$

Balance at

$O(\varepsilon^0): x_0^4+4x_0^3+6x_0^2+4x_0+1=0 \Rightarrow$ roots
$-1 (\times 4)$ as above

$O(\varepsilon^{\frac{1}{4}}): 4x_0^3x_1+12x_0^2x_1+12x_0x_1+4x_1
\Rightarrow$ put $\alpha x_0=-1$ and see

$0=-4x_1+12x_1-12x_1+4x_1$. Therefore no information on $x_1$.
Must go to $O(\varepsilon^{\frac{1}{2}})$.

$O(\varepsilon^{\frac{1}{2}}):
6x_0^2x_1^2+4x_0^3x_2+12x_0x_1^2+12x_0^2x_2+12x_0x_2+4x_2+6x_1^2=0$

$x_0=-1 \Rightarrow 6x_1^2-4x_2-12x_1^2+12x_2-12x_2+4x_2+6x_1^2=0$

\un{$0=0$}!!

What's happening? It's just an extreme version of what we have met
before: degenerate roots lead to extra work in the sense that you
have to go to higher orders. In fact
$O(\varepsilon^{\frac{3}{4}})$ also leads to 0=0. Only when you
get to $O(\varepsilon)$ can you get: $2+x_1^4=0 \Rightarrow
x_1=2^{\frac{1}{4}}e^{-\frac{\pi}{4}(2n+1)}$.

Thus
$x=-1+2^{\frac{1}{4}}e^{\frac{i\pi(2n+1)}{4}}\varepsilon^{\frac{1}{4}}
+O(\varepsilon^{\frac{1}{2}}),\ n=0,1,2,3$

To check this just substitute back into equation and observe the
cancellations up to $O(\varepsilon^{\frac{1}{4}})$.




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