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The problem is as follows:

Let $f_{n}(z)=n + (n-1)z +(n-2)^2z+ ... z^{n-1}$.

Show that $f_{n}(z)$ has no roots in the unit disk $|z| \leq 1$.

My proof is as follows: first I show that $f_{n}(z)$ has no roots on $|z|=1$ as $z^{n-1}f_{n}(1/z)\neq f_{n}(z)$.

Then I proceed by induction, the base case when $n=1$ is $f(z)=1$ which clearly has no roots on the unit disk. Then I assume for some natural number k, $f_{k}(z)\neq 0$ on $|z|<1$, and I show that $f_{k+1}(z)$ also has no roots there.

To Show this I use Rouché's Theorem, which states that if $|f(z)-g(z)|<|f(z)|+|g(z)|$ on the boundary of some simply connected set $K$, then $f$ and $g$ have the same number of roots inside $K$.

I set $f = f_{k+1}(z)$ and $g = f_{k}(z)$, and I show that the inequality is always true by the Triangle Inequality. The strict inequality is proven by showing that if the RHS and LHS were equal, then $f_{k+1}(z) = \alpha f_{k}(z)$ for some real $\alpha$, which cannot be true as $f_{k+1}(z)$ is of a higher order, so it cannot be a scalar multiple of the other.

Therefore $f_{k+1}(z)$ and $f_{k}(z)$ have the same number of roots in $|z|<1$, but by our assumption $f_{k}(z)$ had no roots and therefore so does $f_{k+1}(z)$, proving our claim by induction.

Does this proof hold, or have I made a mistake somewhere?

Riccardo Caiulo
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    To get roots on the unit circle you do not need that the polynomial is symmetric. The idea to use the reverse polynomial to get a general expression might be useful. – Lutz Lehmann Jul 08 '21 at 14:28
  • show that $(z-1)f_n(z) = z^{n-1} + z^{n-2} + ... + z - n$ and from here you can conclude that there are no zeros in unit disk given that $f_n(1) > 0$ – sku Jul 09 '21 at 06:11

2 Answers2

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The claim is wrong. $f_5$ has two roots inside the unit circle, as has $f_6$. Next $f_7$ has four roots inside the unit circle.

Test script in python:

def fn(n): p=[(n-k)**k for k in range(n)]; p[0]=n; return p[::-1]

for n in range(3,8): p = fn(n); r = np.roots(p); print(r); print(abs(r))

resulting in

[-1.+1.41421356j -1.-1.41421356j]
[ 1.73205081      1.73205081    ]

[-3.46750386+0.j         -0.26624807+1.04051888j -0.26624807-1.04051888j]
[ 3.46750386              1.07404263              1.07404263            ]


[-6.73562588+0.j         -1.32293275+0.j          0.02927932+0.74850573j   0.02927932-0.74850573j]
[ 6.73562588              1.32293275              0.74907817               0.74907817            ]


[-14.17301564+0.j          -1.04276743+0.48860844j   -1.04276743-0.48860844j

   0.12927526+0.55002184j   0.12927526-0.55002184j                          ]
[ 14.17301564               1.15156508                1.15156508

   0.56500983               0.56500983                                      ]


[-29.30994691+0.j          -1.7066998 +0.j          -0.65398717+0.52327898j

  -0.65398717-0.52327898j   0.16231052+0.41608778j   0.16231052-0.41608778j]
[ 29.30994691               1.7066998                0.83756797
   0.83756797               0.44662484               0.44662484            ]

However, if your polynomial were $f_n(z)=n+(n-1)z+(n-2)z^2+...+(n-k)z^k+..+z^{n-1}$, then Let $a_n$ be a decreasing sequence. Prove that the power series $\sum a_n x^n$ has no roots in $A=\{z\in C:|z|<1\}$ and linked applies.

Or simply that the reverse polynomial $\tilde f_n(z)=1+2z+...+nz^{n-1}$ is the derivative of $g(z)=1+z+z^2+...+z^n$ with all roots on the unit circle. It is a common exercise to show that the roots of the derivative $g'=\tilde f_n$ lie inside the convex hull of the roots of $g$.

Lutz Lehmann
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In proving that the strict inequality holds, you claim that $f_{k+1}(z) \neq \alpha f_k(z)$ due to the difference in degrees. However, I don't see how does that helps you, since we need to show strict inequality for all $z$ with $|z|=1$.

I agree that there isn't an $\alpha$ such that $f_{k+1}(z) = \alpha f_k(z)$ for all $|z|=1$, but for a given one, there may be an $a_z$ such that $f_{k+1}(z) = \alpha_zf_k(z)$.

Dosidis
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