I'd like to show $P(z)=1-2z^2-2z^3-2z^4-2z^5$ has a unique root inside the disk $|z|<0.6$.

I tried using Rouche's theorem, which worked for polynomials of this form $1-2z^2-2z^3-2z^4...-2z^n$ but of higher degrees, but the same method did not work on $n=5$ or smaller $n$.

For higher degrees, I used it as follows, multiplying by $z-1$ we obtain the polynomial $-2z^{n+1}+2z^2+z-1$.

By defining $f(z)=2z^2+z-1=2(z+1)(z-\frac{1}{2})$ and $g(z)=-2z^{n+1}$ one can show (using the regular and reverse triangle inequality) that on $|z|=0.6$ we have $|f(z)|>|g(z)|$ for $n\geq6$. However for $n=3,4,5$ this method failed.

I'm wondering if there's another way besides Rouche's theorem, or maybe a different use of Rouche here, or even some idea on why Rouche doesn't work on those values.

Just note that $0.6$ is not especially important, it's what I thought to use myself to find a proof of a certain claim. Similar radii (not far from $0.6$) that guarantee a unique root inside it would also be helpful.

  • $\begingroup$ Have you tried using winding numbers? $\endgroup$
    – Toby Mak
    Apr 4 '20 at 5:33
  • $\begingroup$ A short explanation on winding numbers : math.stackexchange.com/q/3139148 $\endgroup$
    – Jean Marie
    Apr 4 '20 at 6:14
  • $\begingroup$ 3Blue1Brown's video on winding numbers is also a great introduction to the subject. $\endgroup$
    – Toby Mak
    Apr 4 '20 at 6:37
  • 1
    $\begingroup$ Maybe you could try to actually minimize $\big|f(z)\big|$ on $|z|=0.6$, instead of just using a weak lower bound $2\cdot (1-0.6)\cdot(0.6-0.5)$. It turns out that the minimum value is $0.32$ (see here). This means: Roche's thm works for $n=3,4,5$ as well. An application of Lagrange multipliers doesn't seem to be too difficult. $\endgroup$ Apr 4 '20 at 7:13

I will expand on my comment. Following the OP's attempt, I will minimize $$F(x,y)=\big|f(x+yi)\big|=2\sqrt{\big((x+1)^2+y^2\big)\big((x-1/2)^2+y^2\big)}$$ subject to $x^2+y^2=r^2$ ($r$ is a non-negative constant). Let $$\mathcal{L}(x,y,\lambda)=\frac14\big(F(x,y)\big)^2+\lambda(x^2+y^2-r^2).$$ We set $$0=\frac{\partial \mathcal{L}}{\partial x}=2(x+1)\big((x-1/2)^2+y^2\big)+2(x-1/2)\big((x+1)^2+y^2)+2\lambda x,\tag{1}$$ $$0=\frac{\partial \mathcal{L}}{\partial y}=2y\big((x-1/2)^2+y^2\big)+2y\big((x+1)^2+y^2)+2\lambda y.$$ For the second equation, we have either that $y=0$ or $$(x-1/2)^2+(x+1)^2+2y^2+\lambda=0.\tag{2}$$ We also have the constraint condition $$y^2=r^2-x^2.\tag{3}$$ Thus $y=0$ yields solutions $$(x,y)=(\pm r,0).$$ We have $$a(r)=F(r,0)=|2r^2+r-1|$$ and $$b(r)=F(-r,0)=|2r^2-r-1|$$

From now on suppose that $y\ne 0$. Therefore $(2)$ holds. Plug $(3)$ into $(1)$ and $(2)$ to get $$(x+1)\big((x-1/2)^2-x^2+r^2\big)+(x-1/2)\big((x+1)^2-x^2+r^2\big)+\lambda x=0\tag{4}$$ and $$(x-1/2)^2+(x+1)^2-2x^2+2r^2+\lambda=0.$$ The previous equation gives $$x=-2r^2-\lambda-5/4.\tag{5}$$ Plug $(5)$ into $(4)$ to get $$-\left(2r^2+\lambda+\frac14\right)\left(3r^2+\lambda+\frac32\right)+\left(2r^2+\lambda+\frac74\right)\left(3r^2+2\lambda+\frac32\right)-\lambda\left(2r^2+\lambda+\frac54\right)=0.$$ That is, $$\lambda=-\frac{9(2r^2+1)}{8}.$$ This means $$x=-2r^2+\frac{9(2r^2+1)}{8}-\frac{5}{4}=\frac{2r^2-1}{8}.$$ Therefore $$y=\pm\frac{\sqrt{-4r^4+68r^2-1}}{8},$$ which is real only if $$0.12132\approx \frac{3\sqrt{2}-4}{2}\le r \le \frac{3\sqrt{2}+4}{2}\approx 4.12132.$$ Observe that \begin{align}c(r)&=F\left(\frac{2r^2-1}{8},\pm\frac{\sqrt{-4r^2+68r-1}}{8}\right)\\&=2\sqrt{\left(r^2+2\cdot\frac{2r^2-1}{8}+1\right)\left(r^2-\frac{2r^2-1}{8}+\frac14\right)}\\&=\frac{3(2r^2+1)}{2\sqrt2}.\end{align} We have $$\big(c(r)\big)^2-\big(a(r)\big)^2=\frac{(2r^2-8r-1)^2}{8}\geq 0$$ and $$\big(c(r)\big)^2-\big(b(r)\big)^2=\frac{(2r^2+8r-1)^2}{8}\geq 0.$$ Therefore, $a(r)\leq c(r)$ and $b(r)\leq c(r)$ always.

Therefore, the minimum of $F(x,y)$ with $x^2+y^2=r^2$ is $$m(r)=\min\{a(r),b(r)\}=\min\big\{|2r^2+r-1|,|2r^2-r-1|\big\}.$$ Since $m(0.6)=0.32$, we see that $$\big|g(z)\big|=\big|-2z^{n+1}\big|=2\cdot 0.6^{n+1}\leq 2\cdot 0.6^4=0.2592<0.32\leq \big|f(z)\big|$$ for $n\geq 3$ and $|z|=0.6$. Because $f(z)=2z^2+z-1$ has exactly one root $z=1/2$ inside the disk $|z|<0.6$, by Rouche's theorem, $$1-2z^2-2z^3-\ldots-2z^n=\frac{f(z)+g(z)}{z-1}$$ has exactly one root inside $|z|<0.6$. (If you replace $0.6$ by $0.7$, the statement is still true.)

Question: This left me wonder whether this is true. Let $f(z)$ be a polynomial function with only real roots. Is it true that the minimum value of $\big|f(z)\big|$ on any circle $|z|=r$ is attained at $z=r$ or $z=-r$ (clearly $z=\pm r$ are critical points)? Can someone prove or disprove this?

Edit: Fixed miscalculations.

  • $\begingroup$ Thank you for the elaborate answer, though I did not understand how $\lambda=-\frac{9(2r^2+1)}{8}$ from the equation that preceeded it, can you elaborate? $\endgroup$
    – user7610
    Apr 5 '20 at 4:14
  • $\begingroup$ Just expand the whole thing. The $\lambda^2$ terms cancel, leaving only the linear terms and the constant terms. $\endgroup$ Apr 5 '20 at 4:33
  • 1
    $\begingroup$ How did you get $-\lambda^2$? The first term would give $-\lambda^2$, the second term would give $2\lambda^2$, and the third $-\lambda^2$. Are you forgetting that one of the $\lambda$s in the second term has $2$ in front of it? $\endgroup$ Apr 5 '20 at 4:36
  • $\begingroup$ I see my mistake. Thanks. Yes, you are right, I forgot the 2 infront. $\endgroup$
    – user7610
    Apr 5 '20 at 4:38
  • $\begingroup$ Thank you, Lagrange multipliers worked well. $\endgroup$
    – user7610
    Apr 6 '20 at 3:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.