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Let us consider $\beta, k_1,k_2,k_3,k_4,k_5,k_6\in\mathbb R$ and define the equation: $$(\beta k_1+k_2)x^4+(\beta k_3+k_4)x^3+\beta k_5 x+k_6=0$$ I would like to approximate the solution of this equation when $\beta\to+\infty$. I proceeded in this way: $$(\beta k_1+k_2)x^4+(\beta k_3+k_4)x^3+\beta k_5 x+k_6 \approx \beta k_1 x^4+\beta k_3 x^3+\beta k_5 x=0$$ Hence a solution is $x=0$ and the equation $$k_1 x^3+k_3 x^2+ k_5=0$$ remain to be solved.

What do you think of my solution? Would anyone else have done differently?

Thank you!

Mark
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1 Answers1

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More formally, divide by $\,\beta \ne 0\,$:

$$\left(k_1 + \frac{k_2}{\beta}\right)x^4+\left(k_3+\frac{k_4}{\beta}\right)x^3 + k_5 x + \frac{k_6}{\beta}=0 \tag{1}$$

The terms in $\frac{1}{\beta} \to 0$ when $\beta \to \infty$, so in the limit the equation reduces to:

$$x \cdot \left(k_1 x^3+k_3 x^2+ k_5\right)=0 \tag{2}$$

This can be justified because the roots of a polynomial depend continuously on the coefficients, so the roots of $(2)$ will be "close" to the roots of $(1)$ for "large enough" $\beta$.

It should be noted, however, that the "closeness" is measured in the complex plane, so the nature of the roots (real vs. non-real complex) is not guaranteed to be the same between the two equations when $(2)$ has a double real root, because the perturbation term in $(1)$ that was discarded could introduce either two distinct real roots, or two complex conjugate roots.

dxiv
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  • Thank you! With "perturbation term" in (1) do you mean the coefficient $k_6$? – Mark Jul 05 '21 at 07:17
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    @Mark I meant the whole part $,k_2x^4+k_4x^3+k_6,$ which gets dropped in the limit. For a (very) simple example, consider the case $k_1=1, k_2=k_3=k_4=k_5=0, k_6 \ne 0$, then the original equation is $\beta x^4 + k_6=0$ while the equation in the limit is $x^4=0,$. The latter has one quad real root $x=0$, while the former may have two distinct real roots and two complex conjugate roots, or two pairs of complex conjugate roots, depending on the sign of $k_6$. – dxiv Jul 05 '21 at 07:24
  • thank you! It is clear. – Mark Jul 05 '21 at 10:24