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The task is: if $x+ \frac{1}{x}= 1$ find $x^5 + \frac{1}{x^5} $.

I used the binomial formula and proved that $x^5 + \frac{1}{x^5} = 1$, but I have a question about following method, I am not sure if it's correct. If I take square of the the first equality, I get: $x^2 +2 + \frac{1}{x^2} = 1$ so $x^2 +\frac{1}{x^2} = -1$. Now, the sum of 2 squares is nonnegative and the right side is negative, so when I come to this part does it mean that this method is wrong?

In general, when proving such equalities, when are we allowed to take square (and we don't know if one side of equality is positive as in this task)? Thanks in advance.

Michael Rozenberg
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User154
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    Perhaps $x$ isn’t real – J. W. Tanner Oct 23 '20 at 21:08
  • You are right, I was thinking about real numbers all the time...Thanks! – User154 Oct 23 '20 at 21:09
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    You can already see from $x+{1\over x}=1$ that $x$ can't be real: It certainly can't be negative (or $0$), and if $x$ is positive then either it or $1/x$ is already at least $1$. (In fact, for positve values of $x$ we have $x+{1\over x}\ge2$, with the minimum value occuring at $x=1$.) – Barry Cipra Oct 23 '20 at 21:24
  • See https://math.stackexchange.com/a/3444908/589 – lhf Oct 23 '20 at 21:55

6 Answers6

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The fact is that

$$x+ \frac{1}{x}= 1 \implies x^2-x+1=0 \implies x=\frac{1\pm\sqrt{-3}}{2}=\frac{1\pm i\sqrt{3}}{2} \in \mathbb{C}$$

In general, when proving equalities, we are always allowed to take square. We need to pay attention when we are solving an equation for $x$, in these case infact squaring both sides can produce some extra solutions which must be checked with respect to the original equation.

user
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If you think on $x + \frac{1}{x} = 1$ as $x^2-x+1=0$, a nicer fact is:

For any cuadratic equation $ax^2+bx+c=0$ with roots $x_1,x_2$. If you define $S_t = x_1^t + x_2^t$ for $t\in\mathbb{R}$, this relation always hold:

$$aS_{t+1}+bS_{t}+cS_{t-1} = 0$$

(You can think like changing $x^t$ for $S_t$).

Since always $S_0 = 2$, and $S_1=x_1+x_2 = -b/a$, you can easy find $S_{-1}$. Later with some recursion you cand find $S_{-2},S_{-3},S_{-4}$ and finally $S_{-5} = x_1^5+x_2^{5 }$.

This relation holds for all polynomials (even complexs one).

L F
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$x+\frac{1}{x} =1$

$\Rightarrow $ $ (x+\frac{1}{x}) ^5=1$

$\Rightarrow $ $x^5+(\frac{1}{x})^5 +5(x^3+(\frac{1}{x})^3)+10 (x+\frac{1}{x})=1$

$\Rightarrow $ $x^5+(\frac{1}{x})^5=-9-5(x^3+(\frac{1}{x})^3)(*)$

Let's find :$ x^3+(\frac{1}{x})^3$

$ (x+(\frac{1}{x}))^3 =1$

$\Rightarrow $ $x^3+(\frac{1}{x})^3+3(x+\frac{1}{x})=1$

$\Rightarrow $ $x^3+(\frac{1}{x})^3=-2(**)$

After$(*,**)$ we can see that :

$x^5+(\frac{1}{x})^5=-9+10=1$

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$x+\frac{1}{x}=1$ gives $x^2-x+1=0,$ which gives $x^3+1=0$ or $x^3=-1$.

Thus, by your work $$x^5+\frac{1}{x^5}=-x^2-\frac{1}{x^2}=1.$$

Now, about your last question.

You used the following statement.

If $x+\frac{1}{x}=1$ so $\left(x+\frac{1}{x}\right)^2=1$.

It's true because if $a=b$ so $a-b=0$ and from here $$a^2-b^2=(a-b)(a+b)=0,$$ which gives $$a^2=b^2.$$

Id est, we proved the following claim: $$a=b\Rightarrow a^2=b^2$$ is true.

Michael Rozenberg
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If $\,x+\frac1x=1,$ then $\,x^2-x+1=0$, which has two complex solutions:

$$ \alpha = \frac{1+i\sqrt3}2 = \cos(\pi/3)+i\sin(\pi/3) = e^{i\pi/3} $$

$$ \beta = \frac{1-i\sqrt3}2 = \cos(-\pi/3)+i\sin(-\pi/3) = e^{-i\pi/3} = \overline{\alpha}. $$

Consider the solution $\alpha$. For any $\,n\in\Bbb N\,$ we have:

$$ \alpha^n+\alpha^{-n} = e^{in\pi/3}+e^{-in\pi/3} = 2\cos(n\pi/3) \tag{*}\label{soluzione}. $$

If we consider the solution $\,\beta$, we obtain the same result. In fact

$$ \beta^n+\beta^{-n} = {\overline\alpha}^{\,n} + {\overline\alpha}^{\,-n} = e^{-in\pi/3}+e^{in\pi/3} = \alpha^n+\alpha^{-n}. $$

Therefore equation \eqref{soluzione} provides a closed solution to your problem, without resorting to any recursion:

$$ x^n + \frac1{x^n} = 2\cos(n\pi/3). \tag{**}\label{soluzione finale}$$

In particular,

$$ x^5 + \frac1{x^5} = 2\cos(5\pi/3) = 1. $$

$\,$

Note. From equation \eqref{soluzione finale} it follows that the sequence $\,(x^n+1/x^n)_{n=0,1,2,\ldots}\,$ is periodic with period 6:

$$ x^{6n+k} + \frac1{x^{6n+k}} = x^k + \frac1{x^k} \qquad\quad (k=0,1,2,\ldots,5), $$

as it easy to verify. The results consist in repeating the block $\;2,1,-1,-2,-1,1\;$ corresponding to $\;k=0,1,2,3,4,5\;$ and $\;n=0$.

gpassante
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Let $x$ be solution of $x+\frac 1x=1\iff x^2-x+1=0$.

Then $u_n=x^n+\dfrac 1{x^n}$ is solution of $\begin{cases}u_0=2\\u_1=1\\u_n=u_{n-1}-u_{n-2}\end{cases} $ from the characteristic equation above.

From there we can calculate:

  • $u_2=u_1-u_0=1-2=-1$
  • $u_3=u_2-u_1=-1-1=-2$
  • $u_4=u_3-u_2=-2+1=-1$
  • $u_5=u_4-u_3=-1+2=1$
zwim
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