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Rather than asking the most general question possible, I will frame it in terms of what I believe is an illustrative example.

Let $\epsilon>0$ be a small parameter, let $a,b>0$ and $x\in [-\epsilon,\epsilon]$. If $y\geq 0$ is sufficiently small in terms of $\epsilon$, then we can solve the equation $$a x^2+b x^4=y$$ explicitly for $x$. One solution is $$x_1=\sqrt{-\frac{a}{2 b} + \frac{\sqrt{a^2 + 4 b y}}{2 b}}.$$ Suppose that instead we want to solve the perturbed equation $$a x^2+b x^4+\phi(x)=y$$ for $x$, where $\phi$ is some smooth function satisfying $\phi(x)=O(x^5)$ as $|x|\rightarrow 0$. In general, an explicit solution like before is not available. How to proceed? In particular, if $\widetilde{x}_1$ denotes the nonnegative solution of the perturbed equation, what can be said about $|x_1-\widetilde{x}_1|$? Is it $O(\epsilon^5)$?

Thank you.

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In a small neighborhood of $x_1$, write

$$ \tilde x_1 = x_1 + r(x_1),\quad r(x) = O(x^N)$$ and since $\phi$ is smooth, write also $$ \phi(x)=\alpha x^5+O(x^6). $$

We want to find $N$ such that the following identity holds for all small $r=\tilde x_1 - x_1$:

$$ a \tilde x_1^2 + b \tilde x_1^4 + \phi(\tilde x_1) = y = ax_1^2+bx_1^4. $$

Now we open parenthesis and collect powers of $r$. If my calculations are correct, this amounts to something like this: $$ r(ax_1+2bx_1^3) + \alpha(x_1^5+3x_1^4 r + O(r^2)) + O(x_1^6) = 0 $$ Since this should hold for all small $r$, the coefficients of the expansion must be identically zero. In particular, since $a>0$, and assuming $\alpha\neq 0$, the term corresponding to the smallest degree of $r$ is $$ rax_1+\alpha x_1^5 $$ from where $$ N = 4.$$

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Why do you assume $\phi(x) \sim \alpha x^5$? – Antonio Vargas May 24 '12 at 18:02
@Antonio: the assumption $\phi(x)=O(x^5)$ plus smoothness seems to imply that. – dima_b May 25 '12 at 1:27
But what if $\phi(x) = x^6$? In that case $\phi(x) = O(x^5)$ as $|x| \to 0$ and $\phi$ is smooth, but $\phi(x) \not\sim \alpha x^5$. Your assumption is unwarranted. – Antonio Vargas May 25 '12 at 2:33
@Antonio: you are right, technically it can also be that $\phi$ is identically zero. I would say that stronger assumption on $\phi$ is required for one. On the other hand, even if $\phi=x^6$ the final answer is still correct, i.e. $r=O(x^4)$. The big-O bounds are not tight to begin with. – dima_b May 25 '12 at 6:33

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