# Please check this perturbation solution of polynomial root and truncation order.

I have a quintic polynomial where the coefficients depends on a parameter $c$, i.e. $$a_0(c)+a_1(c)x+a_2(c)x^2+a_3(c)x^3+a_4(c)x^4+x^5$$ I know that the roots of the polynomial are real and non-negative, and that $a_0(c)=\alpha c^2$, so that for $c=0$ one has $a_0(c)=0$ and therefore in this case one root is $x=0$. I want to calculate how the root $x=0$ (the smallest root) is perturbed for $c\neq0$ at the second order in $c$. As far I understand, a typical approach is to consider a solution proportional to $c$ $$x=c z$$ and make the substitution in the polynomial. $$a_0(c)+a_1(c)cz+a_2(c)c^2 z^2+a_3(c)c^3 z^3+a_4(c)c^4 z^4+c^5 z^5,$$ where $a_0(c)=\alpha c^2$. Since I am interested in the second order in $c$, I throw away the last three terms in the polynomial, and I write the first three coefficients as $a_0(c)=\alpha c^2$ (this is exact), $a_1(c)=\beta+\gamma c$, and $a_2(c)=\delta$, that is $$\alpha c^2+(\beta+\gamma c)cz+\delta c^2 z^2+O(c^3)$$

Now my question is: is it possible to neglect the term in $z^2$ and therefore obtain the approximate root $$x=cz=\frac{\alpha c^2}{\beta+\gamma c}?$$

If my reasoning is wrong, can you provide some guidelines?

• No. If your perturbative expansion is of the right form, $z$ should be bounded (away from infinity) but also bounded away from zero. In this case $z^2$ is not small. – Ian Aug 19 '15 at 13:16
• @Ian Ok, so what is the right way to obtain the perturbed root? – sintetico Aug 19 '15 at 16:57
• If the perturbed root is indeed linear in $c$ (which I think it will be, provided $a_1$ is bounded away from zero and away from infinity near $c=0$), then you should solve $a_0+a_1cz+a_2c^2z^2=0$ for $z$ using the quadratic formula, discarding terms as required at the end. – Ian Aug 19 '15 at 16:59
• ...In fact there should be no discarding of terms at all, unless your $\alpha,\beta,\gamma,\delta$ are closely related in certain ways. That's because all three coefficients are $\Theta(c^2)$. – Ian Aug 19 '15 at 17:11