9
$\begingroup$

Let be the polynomial:

$$P_n (x)=x^{n+1} - (x^{n-1}+x^{n-2}+\cdots+x+1)$$

I want to prove that it has a single positive real root we'll denote by $x_n$, and then to compute: $$\lim_{n\to\infty} x_{n}$$

  • 0
    Numerically the root converges to $\phi = \dfrac{1+\sqrt{5}}2$.2012-06-24
  • 0
    @Marvis: very interesting. This is the golden number.2012-06-24
  • 0
    Really? if you evaluate the expression at $\phi$, you get $2/(\sqrt 5-1)$. Kind of funny.2012-06-24
  • 2
    By summing the geometric series, we see that $P_n(x) = 0$ iff $x^n (x^2-x-1) = -1$, so if $x>1$ and $n$ is large, then $x^2-x-1$ has to be very small, which explains why $\phi$ appears.2012-06-24
  • 0
    @HansLundmark There's an additional root at $x = 1$ that you get that way : $x^n(x^2 - x - 1) + 1 = (x-1)P_n(x)$2012-06-24
  • 0
    @Cocopuffs: Ah yes, that's true, so "iff" was a bit sloppy there. But the point wasn't the root $x=1$, but the root $x \approx \phi$. :-)2012-06-24
  • 3
    To prove it has a single positive root, you can use Descartes' Rule of Signs.2012-06-24
  • 0
    Thanks for all your solutions / suggestions.2012-06-25

3 Answers 3

3

Since it's not much more work, let's study the roots in $\mathbb{C}$.

Note that $x=1$ is not a solution unless $n=1$, since $P_n(1) = 1-n$.

Since we are interested in the limit $n\to\infty$, we can assume $x\ne 1$. Sum the geometric series, $$\begin{eqnarray*} P_n (x) &=& x^{n+1} - (x^{n-1}+x^{n-2}+\cdots+x+1) \\ &=& x^{n+1} - \frac{x^n-1}{x-1}. \end{eqnarray*}$$ The roots will satisfy $$x_n^{n}(x_n^2-x_n-1) = -1.$$

(Addendum: If there are concerns about convergence of the sum, think of summing the series as a shorthand that reminds us that $(x-1)P_n(x) = x^{n}(x^2-x-1) + 1$ for all $x$.)

If $0\le |x_n|<1$, $\lim_{n\to\infty} x^n = 0$, thus, in the limit, there are no complex roots in the interior of the unit circle.

If $|x_n|>1$, $\lim_{n\to\infty} 1/x^n = 0$, thus, in the limit, the roots must satisfy $$x_n^2 - x_n - 1 = 0.$$ There is one solution to this quadratic equation with $|x_n|>1$, it is real and positive, $$x_n = \frac{1}{2}(1+\sqrt{5}).$$ This is the golden ratio. It is the only root exterior to the unit circle.

The rest of the roots must lie on the boundary of the unit circle.

Contour plot of <span class=$|P_{15}(x+i y)|$.">

Figure 1. Contour plot of $|P_{15}(x+i y)|$.

  • 0
    interesting approach and nice graph. :-)2012-06-25
  • 0
    Another interesting question!2012-06-25
7

$$P_n(x)=x^{n+1}-(x^{n-1}+x^{n-2}+...x+1)=x^{n+1}-\sum_{k=0}^{n-1}x^{k}$$ By the formula for the sum of a finite geometric series, we have, if x $\ne$ 1: $$P_n(x)=x^{n+1}-\frac{x^n-1}{x-1}$$ Equating to zero, $$x^{n+1}=\frac{x^n-1}{x-1} \implies x^{n+1}(x-1)=x^{n+2}-x^{n+1}=x^n-1 \implies x^{n+2}-x^{n+1}-x^{n}=-1$$ So, provided x $\ne$ 0, we also have $x^2-x-1=\frac{-1}{x^n}$
Now we can limit both sides as n approaches infinity, to get (assuming x>1) $$x^2-x-1=0 \implies x=\frac{1+\sqrt{5}}{2}$$

Let us establish the uniqueness of this root for x>1: $$P_n'(x)=(n+1)x^{n+1}-((n-1)x^{n-1}+(n-2)x^{n-2}+...2x+1)=0$$ Has exactly one positive root, by Descartes' rule of Signs. Note that $P_n'(0)=1$ and $p_n'(1)=n+1-(n-1+n-2+n-3...)<0$ so the root must lie between 0 and 1. As a result, the curve intersects the positive x axis in exactly one place.

$$P_n(\phi)=\frac{1}{\phi^n(\phi-1)}(\phi^{2}-\phi-1+\frac{1}{\phi^n})$$ Now, note that $\phi^2-\phi-1=0$ by definition, and $\frac{1}{\phi^n}=(\phi-1)^n$, so $$P_n(\phi)=\frac{(\phi-1)^{n-1}}{\phi^n}=(\phi-1)^{2n-1}$$ Since, $|\phi-1|<1$, as n tends to infinity, $P_n(\phi)$ tends to zero.
I've never done any proper real analysis (I'm 17) so forgive me for not being rigorous. Is this sufficient?

  • 1
    That doesn't show the uniqueness, though. Not sure about existence, but that's not hard to show anyhow.2012-06-24
  • 1
    For uniqueness, look at the roots of $(x-1)x^{n+1}-x^n+1$, and use Rolle's theorem. Also, since $P_n(1)=-(n-1)<0$ and $\lim_{x\to +\infty}P_n(x)=+\infty$, we get that $x_n>1$.2012-06-24
  • 2
    For the limit, you should add more details (in the equation it's $x_n$, not $x$).2012-06-24
  • 0
    Some simple key facts to show uniqueness: (1) $\lim_{n\to\infty} x^n = 0$ for $|x|<1$, (2) $x=1$ is not a solution of the original equation, (3) $\lim_{n\to\infty} 1/x^n = 0$ for $|x|>1$.2012-06-24
  • 0
    @Davide Giraudo: if you're willing then you may post your own solution with the way you described above.2012-06-25
  • 0
    @DanLitt: thanks for your solution.2012-06-25
  • 1
    As Davide pointed out, this solution is very incomplete in that it just assumes there is a limit $>1$. First, there might not be a limit at all, second, even if there is a limit, we might have $x_n\to 1$ and therefore $-1/x_n^n$ could converge to a non-zero limit, so the equation on $x$ would become different.2012-06-25
  • 0
    I've had a go at making the answer more complete - let me know if it's sufficient.2012-06-25
5

Descartes' rule of signs shows that $P_n$ has at most one positive root, and since $P_n(1)=1-n<0$ and $\lim_{+\infty} P_n=+\infty$, $P_n$ has exactly one positive root $x_n>1$. Let $$R_n(x)=x^{-n}(x-1)P_n(x)=x^2-x-1+x^{-n}$$ Then $R_n$ has exactly two positive roots 1 and $x_n$, so that it is negative on the interval $(1,x_n)$ and positive on $(x_n,+\infty)$: for all $x>1$, $$\operatorname{sgn} (x-x_n)=\operatorname{sgn} R_n(x)$$

For all $x>1$, $R_n(x)$ converges pointwise to $$Q(x)=x^2-x-1$$

But $$\operatorname{sgn} Q(x)=\operatorname{sgn} (x-\phi)$$ where $\phi=(1+\sqrt 5)/2$, so that whenever $x\ne\phi$, $\operatorname{sgn} Q(x)\ne 0$ and we have: $$\lim_{n\to\infty} \operatorname{sgn} (x-x_n)=\lim_{n\to\infty} \operatorname{sgn} R_n(x)=\operatorname{sgn} Q(x)=\operatorname{sgn} (x-\phi)$$ and therefore $x_n$ converges to $\phi$.

  • 0
    this is another interesting approch.2012-06-25