15

Let $x+x^{-1}=\dfrac{1+\sqrt{5}}{2}$. Find $x^{2000}+x^{-2000}$.

How many nice methods do you know for solving this problem? Thank you everyone.

My method: because $x+\dfrac{1}{x}=2\cos{\dfrac{2\pi}{5}}$, so $$x^{2000}+\dfrac{1}{x^{2000}}=2\cos{\dfrac{2000\pi}{5}}=2.$$

Can you think of other nice methods? Or this problem has not used Euler's theorem: $(\cos{x}+i\sin{x})^n=\cos{nx}+i\sin{nx}$

Jyrki Lahtonen
  • 133,153
math110
  • 93,304
  • 13
    I don't follow your method. Why does it hold that $x^{2000} + \frac{1}{x^{2000}}$ is what you say it is? – Hammerite Jun 10 '13 at 15:32

5 Answers5

24

Here is an algebraic way avoiding trig functions: note that your number $x$ satisfies $$x^2-(\frac{1+\sqrt{5}}{2})x+1=0 \quad \implies \quad \text{by multiplying by the conjugate} \quad x^4-x^3+x^2-x+1=0$$ and then use the factorization $$x^{10}-1=(x^6+x^5-x-1)(x^4-x^3+x^2-x+1)$$ to see that $x^{10}=1$.

TMM
  • 9,976
Stephen
  • 14,811
  • @Steve, instead of $x^6+x^5-x-1$ , why did not you multiply $x^4-x^3+x^2-x+1=0$ by $x+1$ which is much more natural multiplier? – lab bhattacharjee Jun 10 '13 at 15:23
  • @labbhattacharjee, Only because I divided $x^{10}-1$ by $x^4-x^3+x^2-x+1$. You are right that it would have been easier to get $x^5+1$. Oh well! – Stephen Jun 10 '13 at 18:50
4

Just one note.

If $x \in \mathbb{R}$, then $x+x^{-1} \geqslant 2$. So, there are no $x$, such that $x+x^{-1}=\varphi=\dfrac{1+\sqrt{5}}{2}$.

If $x \in \mathbb{C}$, then denote $x = m (\cos \alpha + i\sin \alpha)$, where $m,\alpha \in \mathbb{R}$.
So, $x^{-1} = m^{-1}(\cos\alpha - i\sin \alpha)$, and $$ x+x^{-1} = (m+m^{-1})\cos \alpha + i (m - m^{-1})\sin\alpha \implies $$ $ m = m^{-1}=1, \cos \alpha = \varphi/2$.

Yes, if $x\in \mathbb{C}$, then there is sense.

Oleg567
  • 17,295
4

$$x+x^{-1}=\frac{\sqrt5+1}2$$

$$\implies 2x^2-x+2=\sqrt5x$$

$$\text{On squaring,} (2x^2-x+2)^2=5x^2$$

$$\implies 1-x+x^2-x^3+x^4=0$$

$$\text{ or, }x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2\cdot x\cdot\frac1x=\left(\frac{\sqrt5+1}2\right)^2-2=\frac{\sqrt5-1}2$$ $$\implies x+\frac1x-\left(x^2+\frac1{x^2}\right)=1\implies 1-x+x^2-x^3+x^4=0$$

which is a Geometric Series with common ratio $=-x$ and the first term being $=1$ $$\implies 1-x+x^2-x^3+x^4=\frac{1+x^5}{1+x}\implies 1+x^5=0$$

$$\text{ or } 1+x^5=(1+x)(1-x+x^2-x^3+x^4)=0$$

$\implies x^5=-1\implies x^{10n}=(x^5)^{2n}=(-1)^{2n}=1$

Put $n=200,-200$

lab bhattacharjee
  • 274,582
3

Just adding a different way of looking at it (fully endorsing Steve's excellent answer based on factoring the tenth cyclotomic polynomial over $\mathbb{Q}[\sqrt5]$).

Let $t=x+\dfrac1x$. Let us denote $t_n=x^n+\dfrac1{x^n}$. We calculate $$ t^3=x^3+3x+\frac3x+\frac1{x^3}=t_3+3t, $$ $$ t^5=x^5+5x^3+10x+10\frac1x+5\frac1{x^3}+\frac1{x^5}=t_5+5t_3+10t. $$ From these we can easily solve $$ t_3=t^3-3t,\qquad \text{and}\qquad t_5=t^5-5t_3-10t=t^5-5t^3+5t. $$ This time $t=(1+\sqrt5)/2$, so $t^3=2+\sqrt5$ and $t^5=(11+5\sqrt5)/2$, and thus from the above formula we get $$ t_5=\cdots=-2. $$ From the equation (a quadratic in the unknown $x^5$) $$ x^5+\frac{1}{x^5}=t_5=-2 $$ we get $x^5=-1$ as the only solution. The OP's result follows easily.

Anyway, what I wanted to add is that the so called Dickson polynomials give us (among other things) formulas for $t_n$ as a degree $n$ polynomial of $t$. They occasionally come in handy with problems like this.

Jyrki Lahtonen
  • 133,153
  • 1
    We can utilize $$\left(x^n+\frac1{x^n}\right)\left(x^m+\frac1{x^m}\right)=\left(x^{n+m}+\frac1{x^{n+m}}\right)+\left(x^{n-m}+\frac{1}{x^{n-m}}\right)$$ $$\iff \left(x^{n+m}+\frac1{x^{n+m}}\right)=\left(x^n+\frac1{x^n}\right)\left(x^m+\frac1{x^m}\right)-\left(x^{n-m}+\frac1{x^{n-m}}\right)$$ to find $x^5+\frac1{x^5}$ – lab bhattacharjee Jun 10 '13 at 15:48
  • 1
    @lab: A good idea - in particularly if one needs to calculate those $t_n$:s. This allows something like a "double-and-add" -algorithm (w.r.t. $n$). – Jyrki Lahtonen Jun 13 '13 at 14:22
1

Depending on what one wants to do with this, one can treat the process as an isoseries, based on $a(n+1) = k \cdot a(n) - a(n-1)$. There are algorithms that can generate these numbers for any form of multiplication, including the modular form (ie $(a\cdot b ) mod c$.

The following iterations show the value for finding $a(37)$. The first two columns are actual integers as shown. 37 is odd, which produces a '1'. This means that we keep the values in O0 and O1. The value of 37 divided by 2 gives 18, an even number. We keep the valuse in E0 and E1.

The values in brackets are the values of $a(n)$, shown as (n). The third column is powers of 2, by $a(2n) = a(n)^2 - 2$. The values in columns 4 onwards show the powers, in steps of the powers. We have, eg $a(5) = $a(2) \cdot a(3) - a(1)$.

$a(0)=2 = x^0 + x^{-0}$, and $a(1) = a$.

                  P2   E0     O0     E1    O1
       37   1    (1)   (0)   (1)    (2)    (3)
       18   0    (2)   (1)   (3)    (5)
        9   1    (4)   (1)   (5)    (9)   (13)
        4   0    (8)   (5)   (13)   (21)
        2   0    (16)  (5)   (21)   (37)
        1   1    (32)  (5)   (37)
wendy.krieger
  • 6,856