How should I show that this root is given approximately by this?

[Math100]

Homework Statement

By sketching the graphs of $y=a^2-x^2$ and $y=\epsilon\sinh x$ for $x\geq 0$, show that the equation $a^2-x^2=\epsilon\sinh x, 0<\epsilon<<1$, has a root near $x=a$. Use perturbation theory to show that this root is given approximately by $x=a-\frac{\epsilon}{2a}\sinh a+\frac{\epsilon^{2}}{16a^3}(2a\sinh 2a-\cosh 2a+1)+O(\epsilon^{3})$.

Relevant Equations

None.

Proof:

Consider the equation of $a^2-x^2=\epsilon\sinh x$ for $0<\epsilon<<1$.
Let $\epsilon=0$.
Then the unperturbed equation is $a^2-x^2=0$.
This gives $a^2-x^2=0\implies (a+x)(a-x)=0\implies x=\pm a$ with the root $x=x_{0}$
such that $x_{0}(a)=a$ because $x\geq 0$ implies that $x\neq -a$.
Applying the perturbation theory produces $x=a+\epsilon x_{1}+\epsilon^{2}x_{2}+O(\epsilon^{3})$
for $0<\epsilon<<1$.
By direct substitution of this expansion $x=a+\epsilon x_{1}+\epsilon^{2}x_{2}+O(\epsilon^{3})$
into the equation $a^2-x^2=\epsilon\sinh x, 0<\epsilon<<1$ for $x\geq 0$,
we get $a^2-(a+\epsilon x_{1}+\epsilon^{2}x_{2}+O(\epsilon^{3}))^{2}= \epsilon\sinh(a+\epsilon x_{1}+\epsilon^{2}x_{2}+O(\epsilon^{3}))$ for $0<\epsilon<<1$
where $x\geq 0$.
Observe that $a^2-(a+\epsilon x_{1}+\epsilon^{2}x_{2}+O(\epsilon^{3}))^{2}= a^2-(a^2+a\epsilon x_{1}+a\epsilon^{2}x_{2}+aO(\epsilon^{3})+a\epsilon x_{1}+\epsilon^{2}x_{1}^{2} +\epsilon^{3}x_{1}x_{2}+\epsilon x_{1}O(\epsilon^{3})+a\epsilon^{2}x_{2} +\epsilon^{3}x_{1}x_{2}+ \epsilon^{4}x_{2}^{2}+\epsilon^{2}x_{2}O(\epsilon^{3})+aO(\epsilon^{3})+\epsilon x_{1}O(\epsilon^{3}) +\epsilon^{2}x_{2}O(\epsilon^{3})+(O(\epsilon^{3}))^{2}$.
Thus, $-2a\epsilon x_{1}-2a\epsilon^{2}x_{2} -\epsilon^{2}x_{1}^{2}-2\epsilon^{3}x_{1}x_{2}-\epsilon^{4}x_{2}^{2}+O(\epsilon^{3})$.
Similarly, we will use the Taylor's series to expand $\sinh(a+\epsilon x_{1}+\epsilon^{2}x_{2}+O(\epsilon^{3}))$
where the Taylor's series expansion for the hyperbolic function is given by
$\sinh(x)=x+\frac{x^3}{3!}+\frac{x^5}{5!}+...$.

For the first part of the question/problem, I've already obtained the sketch of the graphs of those two given
equations from my graphing calculator but I have doubt about the second part of the question/problem
where the proof is involved because I need to find both $x_{1}, x_{2}$ in order to show the given
approximation but I want to know, how to use this direct substitution by using the Taylor's series expansion
of $\sinh(x)=x+\frac{x^3}{3!}+\frac{x^5}{5!}+...$? Or is this method even correct? May anyone please
check/verify/confirm to see if the work/proof shown above is correct?

 

[anuttarasammyak]

Expansion of x by $\epsilon$
x=a+\epsilon x^{(1)}+\epsilon^2x^{(2)}+...
\sinh x= \sinh a + \epsilon x^{(1)} \cosh a + ...
by Taylor series. The equation
x^2-a^2=-\epsilon \sinh x
(\epsilon x^{(1)}+\epsilon^2 x^{(2)}+...)(2a+\epsilon x^{(1)}+\epsilon^2 x^{(2)}+...)
=-\epsilon\sinh a - \epsilon^2 x^{(1)} \cosh a + ...
Comparing the both sides in order $\epsilon$ terms
2ax^{(1)}=-\sinh a
in order $\epsilon^2$ terms
2ax^{(2)}+(x^{(1)})^2=- x^{(1)} \cosh a

 

[Math100]

Expansion of x by $\epsilon$
x=a+\epsilon x^{(1)}+\epsilon^2x^{(2)}+...
\sinh x= \sinh a + \epsilon x^{(1)} \cosh a + ...
by Taylor series. The equation
x^2-a^2=-\epsilon \sinh x
(\epsilon x^{(1)}+\epsilon^2 x^{(2)}+...)(2a+\epsilon x^{(1)}+\epsilon^2 x^{(2)}+...)
=-\epsilon\sinh a - \epsilon^2 x^{(1)} \cosh a + ...
Comparing the both sides in order $\epsilon$ terms
2ax^{(1)}=-\sinh a
in order $\epsilon^2$ terms
2ax^{(2)}+(x^{(1)})^2=- x^{(1)} \cosh a

I finally solved this problem!