2

Consider $z^3 − z^2 = \lambda$ where $z ∈ \mathbb{C}$, $\lambda \geq 0$. Find the first 2 leading terms in the asymptotic expansion of all the roots of the equation near $\lambda = 0$.

My approach

For $\lambda =0$, the equation has a repeated root, $z=0$.

Let $z(\lambda) \sim z_0 + \sqrt\lambda z_1 + \lambda z_2 + \lambda^{3/2} z_2 + \cdot \cdot$

Can someone comment on my approach?

Atul Anurag Sharma
  • 682
  • How do you know the expansion is in this form? By the way, if $\lambda > 0$, the roots can be expressed in radical form, and one can get the asymptotic expansion. – River Li Sep 29 '20 at 03:18
  • I think that you want an approach not via closed form (radical form)? For example, if one need the asymptotic expansion of real solution of $x-\sin(x) = \lambda$ where $\lambda > 0$. – River Li Sep 29 '20 at 03:28
  • Yes, I was looking in that way. @RiverLi – Atul Anurag Sharma Sep 29 '20 at 16:49

3 Answers3

2

Let us first solve $z^2-z^3=a^2$ for small complex $a$ and small complex $z$. It is equivalent to $$z(1-z)^{1/2}=\pm a.$$ It is well known how to solve $z(1-z)^{1/2}=b.$ Observe that for $|z|<1$, the function $f(z)=z(1-z)^{1/2}$ is holomorphic and satisfies $f(0)=0,f'(0)=1$ Therefore, is has a holomorphic inverse function $f^{-1}$ in a certain neighborhood of $0$. Then $z(1-z)^{1/2}=b$ is equivalent to $z=f^{-1}(b).$ The solutions of $z^2-z^3=a^2$ are therefore $z=f^{-1}(\pm a)$.
This implies that the solutions of $z^3-z^2=\lambda$ or $z^2-z^3=-\lambda$ are $z=f^{-1}(\pm i\sqrt{\lambda})$, where $\sqrt{\lambda}$ denotes the principal value of the square root in the complex domain. Given that the holomorphic function $f^{-1}(y)$ in the neighborhood of $y=0$ has a convergent power series expansion, this justifies the approach in the question for small $\lambda$ and small $z$, where we have actually $$f^{-1}(y)=c_1 y+c_2 y^2+...$$ Altogether the expansion for the two roots of $z^3-z^2=\lambda$ close to $0$ is $$z=\pm c_1i\sqrt\lambda-c_2\lambda+...,$$ where the coefficients can be found by inserting into $z^3-z^2=\lambda$. Of course $c_1$ is not uniquely determined since there are two roots, but once $c_1$ has been chosen, the other coefficients are determined. Except for this sign ambiguity, we find $c_1=1$ and $c_2=1/2$.
The equation $z^3-z^2=\lambda$, $\lambda$ small has a third root near $z=1$. Here, the implicit function theorem can be applied directly because $g(z)=z^3-z^2$ satisfies $g(1)=0$ and $g'(1)=1$. Hence there exist a holomorphic function $h$ in the neighborhood of $0$ with $h(0)=1$ such that $z=h(\lambda)$. Here we have an expansion $$z=h(\lambda)=1+h_1\lambda+h_2\lambda^2+...$$ for small $\lambda$. The coefficients can again be found by inserting into the equation, for example $h_1=1$.

Helmut
  • 5,109
2

For the only real root

Clearly, for $\lambda > 0$, the equation $z^3 - z^2 = \lambda$ always has one real root $z_1$ and two non-real conjugate complex roots $z_2, z_3$, since the discriminant is $-\lambda(27\lambda+4) < 0$.

Let us analyze the only real root $z_1$. By using the Lagrange inversion theorem [1], it is easy to get \begin{align} z_1 &= 1 + \sum_{n=1}^\infty \frac{(3n-2)!}{n!(2n-1)!}(-1)^{n-1} \lambda^n\\ &= 1+\lambda-2\lambda^2+7\lambda^3-30\lambda^4+143\lambda^5-728\lambda^6 + \cdots \end{align} (Note: If $0 < \lambda < \frac{4}{27}$, the series is convergent.)

Remark: Also see Interval of convergence of Lagrange's infinite series

Reference

[1] https://en.wikipedia.org/wiki/Lagrange_inversion_theorem

River Li
  • 49,125
1

I think that for this problem, brute force could be a simple solution.

Using Cardano method, the three roots are given by $$z_1=\frac{1}{3} \left(\frac{\sqrt[3]{3 \sqrt{3} \sqrt{27 \lambda ^2+4 \lambda }+27 \lambda +2}}{\sqrt[3]{2}}+\frac{\sqrt[3]{2}}{\sqrt[3]{3 \sqrt{3} \sqrt{27 \lambda ^2+4 \lambda }+27 \lambda +2}}+1\right)$$ $$z_2=-\frac{\left(1-i \sqrt{3}\right) \sqrt[3]{3 \sqrt{3} \sqrt{27 \lambda ^2+4 \lambda }+27 \lambda +2}}{6 \sqrt[3]{2}}-\frac{1+i \sqrt{3}}{3\ 2^{2/3} \sqrt[3]{3 \sqrt{3} \sqrt{27 \lambda ^2+4 \lambda }+27 \lambda +2}}+\frac{1}{3}$$ $$z_3=-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{3 \sqrt{3} \sqrt{27 \lambda ^2+4 \lambda }+27 \lambda +2}}{6 \sqrt[3]{2}}-\frac{1-i \sqrt{3}}{3\ 2^{2/3} \sqrt[3]{3 \sqrt{3} \sqrt{27 \lambda ^2+4 \lambda }+27 \lambda +2}}+\frac{1}{3}$$

Now, let us work the common term $$A=\sqrt[3]{3 \sqrt{3} \sqrt{27 \lambda ^2+4 \lambda }+27 \lambda +2}$$ $$A^3=3 \sqrt{3} \sqrt{27 \lambda ^2+4 \lambda }+27 \lambda +2$$ and use Taylor series $$A^3=2+6 \sqrt{3} \sqrt{\lambda }+27 \lambda +\frac{81}{4} \sqrt{3} \lambda ^{3/2}-\frac{2187}{64} \sqrt{3} \lambda ^{5/2}+O\left(\lambda ^{7/2}\right)$$ Then, the binomial theorem $$A=\sqrt[3]{A^3}=\sqrt[3]{2}+\sqrt[3]{2} \sqrt{3} \sqrt{\lambda }+\frac{3 \lambda }{2^{2/3}}-\frac{5 \sqrt{3} \lambda ^{3/2}}{4\ 2^{2/3}}-3 \sqrt[3]{2} \lambda ^2+\frac{231 \sqrt{3} \lambda ^{5/2}}{64\ 2^{2/3}}+\frac{21 \lambda ^3}{2^{2/3}}+O\left(\lambda ^{7/2}\right)$$ Now, long division $$\frac 1 A=\frac{1}{\sqrt[3]{2}}-\frac{\sqrt{3} \sqrt{\lambda }}{\sqrt[3]{2}}+\frac{3 \lambda }{2 \sqrt[3]{2}}+\frac{5 \sqrt{3} \lambda ^{3/2}}{8 \sqrt[3]{2}}-\frac{3 \lambda ^2}{\sqrt[3]{2}}-\frac{231 \sqrt{3} \lambda ^{5/2}}{128 \sqrt[3]{2}}+\frac{21 \lambda ^3}{2 \sqrt[3]{2}}+O\left(\lambda ^{7/2}\right)$$ AT this point, we have all the elements required for the expansion of the roots. This should lead to $$z_1=1+\lambda -2 \lambda ^2+7 \lambda ^3+O\left(\lambda ^{7/2}\right)$$ $$z_2=i \sqrt{\lambda }-\frac{\lambda }{2}-\frac{5}{8} i \lambda ^{3/2}+\lambda ^2+\frac{231}{128} i \lambda ^{5/2}-\frac{7 \lambda ^3}{2}+O\left(\lambda ^{7/2}\right)$$ $$z_3=-i \sqrt{\lambda }-\frac{\lambda }{2}+\frac{5}{8} i \lambda ^{3/2}+\lambda ^2-\frac{231}{128} i \lambda ^{5/2}-\frac{7 \lambda ^3}{2}+O\left(\lambda ^{7/2}\right)$$

Checking $$z_1+z_2+z_3=1+O\left(\lambda ^{7/2}\right)$$ $$z_1z_2+z_1z_3+z_2z_3=O\left(\lambda ^{7/2}\right)$$ $$z_1z_2z_3=\lambda +O\left(\lambda ^4\right)$$

Claude Leibovici
  • 289,558