Roots of "almost cyclotomic" polynomials

Question

While looking at the behavior of the probability of getting a run of $k$ heads during $n$ tosses of a fair coin, I ended up needing to know the nature of the roots of polynomials of the form $$ P(x) = x^k - \sum_{m=0}^{k-1} x^m $$ It appears that the largest real root $r(k)$ is always just below $2$, and in fact that it appears that $$ \lim_{k \to \infty} 2^k (2-r(k)) = 1 $$

But I am unsure of my grounds in making these statements.

Does anybody know of any work proving things about the roots of these "almost cyclotomic" polynomials? In particular, is my conjecture a known theorem (or known to be false), and what can be said about the distribution in the complex plane of the other roots?

Answer 1

We have $$P(x)=x^k-\frac{x^k-1}{x-1}=0\iff Q(x)=x^{k+1}-2x^k+1=0$$ (this is indeed a bidirectional implication because the it also holds in the problematic case that $x=1$ is a root, which only happens for $k=1$). Now for $x\ge 2$ we have $Q(x)=(x-2)x^k+1\ge 1$, hence $r(k)<2$ (if there is a real root at all). On the other hand $$Q(2-\epsilon)=1-2^k\epsilon(1-\epsilon/2)^k$$ Hence with $0\le \epsilon \le 2^{-k}$ $$ Q(2-\epsilon)\ge 1-(1-\epsilon/2)^k>0$$ so that $2-r(k)>2^{-k}$. Then continue with the Bernoulli inequality (for $\epsilon<2$) $$Q(2-\epsilon) \le 1-2^k \epsilon(1-k\epsilon/2)=1-2^k\epsilon+k2^{k-1}\epsilon^2$$ Hence if $\epsilon=(1+h)2^{-k}$ with $h=\frac{k}{2^{k-1}}<1$ (for $k>2$) $$Q(2-(1+h)2^{-k})\le-h+\frac{k(1+h)^2}{2^{k+1}}<-h+\frac k{2^{k-1}}=0.$$ This shows that $2^{-k}<2-r(k)<2^{-k}+k2^{-2k+1}$ (and that there is a real root in the first place) for $k>2$ so that indeed $\lim_{k\to\infty}2^k(2-r(k))= 1$.

Complex roots of $Q$: If $|x|>1$ and $|x-2|>1$ then $|Q(x)|>|x-2|-1>0$. Hence all roots are within the union of the closed disks of radius $1$ around $0$ and $2$. If $x$ is a root with $|x|<\sqrt[k]{1/3}$ then both $|x-2|<3$ and $|x-2|=\frac1{|x|^k}>3$, hence the disk with radius $1/\sqrt[k]3$ around $0$ contains no roots.

Answer 2

Let $P_n(x) = x^n - x^{n-1} - \dots - 1$. Note that $P_n(1) = 1 - n \le 0$ when $n \ge 1$. On the other hand, if $r$ is real and $r \ge 2$, then

$$r^n > r^n - 1 \ge \frac{r^n - 1}{r - 1} = r^{n-1} + \dots + 1.$$

Hence any real root $r$ of $P_n(x)$ must be less than or equal to $2$. Since $P_n(2) = 2^n - (2^n - 1) = 1$, by the intermediate value theorem $P_n(x)$ must have a real root between $1$ and $2$, and this is where its largest real root must be.

To estimate this largest real root $r$ (dependence on $n$ suppressed because I hate writing $r_n^n$), we can rearrange the identity $r^n = \frac{r^n - 1}{r - 1}$ to

$$2 - r = \frac{1}{r^n}.$$

Write $\varepsilon = 2 - r$. Once we know, for example, that $r \ge 1.5$ for $n \ge 2$ (which is true), we immediately conclude that $\varepsilon \le \frac{1}{1.5^n}$, and hence this identity lets us bootstrap bad lower bounds on $r$ into much better upper bounds on $\varepsilon$. In fact, once we know that $\varepsilon$ decays exponentially in $n$ at all it follows immediately that

$$\lim_{n \to \infty} \frac{r^n}{2^n} = \lim_{n \to \infty} \left( 1 - \frac{\varepsilon}{2} \right)^n = 1$$

(in fact to get this we just need to know that $\varepsilon$ decays faster than $\frac{1}{n}$) and the desired result follows. I used this argument explicitly for the purpose of analyzing the behavior of getting runs during coin tosses here.

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Here is another argument which uses fewer of the special features of the problem, although it is not very careful. Below $\approx$ means two things are equal up to an error of $O(\varepsilon^2)$, and (mea culpa!) I am not tracking the dependence of the error on $n$, which I really should be. We have

$$r^n = (2 - \varepsilon)^n \approx 2^n - n \cdot 2^{n-1} \varepsilon$$

and

$$\frac{r^n - 1}{r - 1} \approx ((2^n - 1) - n \cdot 2^{n-1} \varepsilon)(1 + \varepsilon) \approx (2^n - 1) - (n \cdot 2^{n-1} - (2^n - 1)) \varepsilon$$

and hence equating the two we get

$$1 \approx (2^n - 1) \varepsilon.$$

More generally, suppose $f(x)$ is a smooth function and $r_0$ is a real number such that $f(r_0)$ is small. You want to estimate the size of a root $r$ close to $r_0$. Writing $r = r_0 - \varepsilon$, we have

$$f(r_0 - \varepsilon) \approx f(r_0) - f'(r_0) \varepsilon$$

and hence

$$\varepsilon \approx \frac{f(r_0)}{f'(r_0)}.$$

This approximation is the basis of Newton's method.

Answer 3

Two excellent answers show that the most positive real root is less than 2, which in fact I had surmized from the following probability argument: The number of sequences among $N$ coin flips containing a run of at least $k$ heads is $h_k(N) = 2^N - s_k(n)$ where $s_k(n)$ obeys a recursion $$ s_k(n) = \sum_{m=0}^{k-1}s_k(n-m-1) $$ The solution is $$s_k(n) = \sum_i a_i [r_{(i)}(k)]^n$$ where $r_{(i)}(k)$ is the $i$-th solution of that almost-cyclotomic equation. $2^{-n}s_k(n)$ is the probability that there will not be a run of $k$ heads in a sequence of $n$ tosses.

Now consider the probability that there will be a run of at least $\frac{n}{2}$ heads in a sequence of $n$ tosses. For large $n$, this will be very nearly zero. Than implies that none of the roots of the equation is greater than or equal to 2.

The distribution of roots is also interesting. Apparently, the roots other than the one real root just below $x=2$ is that they lie on an ellipse (perhaps a ciercle) centered at $z=0$, of radius that varies with $k$ but approaches $1$ from below. The spacing seems to be almost uniform, except that for even $k$ the point on the positive real axis is missing (replaced by the point near $z=2$), and for odd $k$ the gap near the positive real axis is twice the other gaps.

The uniformity of angular spacing is not quite perfect, the center of the "circle" is slightly positive (about $0.025$ for $k = 19$) and the radiuss is not quite perfectly even. But all three of these effects get smaller as $k$ grows.