If $x+\frac1x=5$ find $x^5+\frac1{x^5}\,$ [recurrence for power sums]

Question

If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.

Is there any other way find it? $$ \left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110. $$

Thanks

Answer 1

Observe the recurrence relation $$x^{n+1} + x^{-(n+1)} = (x+x^{-1})(x^n+x^{-n}) - (x^{n-1} + x^{-(n-1)}).$$

This immediately gives us the specific recurrence $$f_{n+1} = 5f_n - f_{n-1}, \quad f_0 = 2, \quad f_1 = 5,$$ where $f_n = x^n + x^{-n}$.

Answer 2

\begin{align} x+\frac{1}{x}&=5\\ x^2+1&=5x\\ x^2-5x+1&=0\\ x &=\frac{1}{2} \left( 5 +\sqrt{21}\right)\\ x^5+\frac{1}{x^5}&=\cdots \end{align}

Answer 3

Using the factorization identity $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ we obtain \begin{align} x^5+\frac{1}{x^5}&=\left(x+\frac{1}{x}\right)\left(x^4-x^2+1-\frac{1}{x^2}+\frac{1}{x^4}\right) \\ &=\left(x+\frac{1}{x}\right)\left(\left(x+\frac{1}{x}\right)^4-5x^2-5-5\frac{1}{x^2}\right) \\&=\left(x+\frac{1}{x}\right)\left(\left(x+\frac{1}{x}\right)^4 -5\left(x+\frac{1}{x}\right)^2+5\right)=5(5^4-5\cdot5^2+5)=2525, \end{align} since $$ \left(x+\frac{1}{x}\right)^4=x^4+4x^2+6+\frac{4}{x^2}+\frac{1}{x^4}. $$

Answer 4

Key Idea $ $ The recurrence $f_{n+2} =(a\!+\!b)\,f_{n+1}\!- ab\, f_n\,$ for the power sums $\,a^n+b^n\,$ arises simply by multiplying the first-order recurrences for $\,a^n,\, b^n$ (in operator form). If linear differential or difference operators $L,M\,$ kill $\,f,g,\,$ i.e. $\color{#c00}{Lf} = 0 = \color{#0a0}{Mg}\,\,$ and $\,L,M\,$ have constant coefficients then they commute $\,LM = ML,\,$ so their product (composition) $\,LM\,$ kills $f+g,\,$ viz.

$\qquad\ \begin{align} LM(f\!+\!g) \,&=\, LMf + LMg\ \ \ [\text{the composition $LM$ is linear, by $\,L,M\,$ linear}]\\[.2em] &=\, M\color{#c00}{Lf} + L\color{#0a0}{M g}\ \ \ [\text{by $L,M$ commute}]\\[.2em] &=\,\ \ \ \ \ \ \:\! \color{#c00}0\, +\, \color{#0a0}0 \end{align}$

e.g. with $\, D = \frac{d}{dx}$ the derivative w.r.t. $\,x\,$ and $\,L= D\!-\!i,\ M = D\!+\!i$

$$ (D\!-\!i)e^{ix}\! = 0 = (D\!+\!i)e^{-ix}\Rightarrow\, (D\!-\!i)(D\!+\!i)\,\frac{e^{ix}\!+e^{-ix}}2=0,\ \ {\rm i.e.}\ \ (D^2\!+1)\cos(x) = 0\ \ $$

Similarly, as below, we can discover a second-order recurrence for the power sums $f_n = a^n + \color{#c0f}b^n\,$ by multiplying the first order recurrences for $\,a^n\,$ and $\,b^n.\,$ Specializing $\,\color{#c0f}{b = a^{-1}}\,$ yields a recurrence enabling us to quickly compute OP's sought values from initial values $ f_0 = 1,\ f_1 = 5.$

To do so it is convenient to express recurrences in polynomial operator form using the linear shift operator $ S $ such that $\ S g_n = g_{n+1}.\,$ Notice that $\,g_n = a^n\,$ satisfies $\, S a^n = a^{n+1} = a a^n\,$ therefore $\, \color{#c00}0 = Sa^n - a a^n = \color{#c00}{(S-a)a^n}.\,$ Similarly $\,\color{#0a0}{(S-b)b^n = 0}.\,$ $\,S\!-\!a\,$ and $\,S\!-\!b\,$ commute because their coefficients $\,a,b\,$ are constants w.r.t. $\,n\,$ i.e. $\,Sag = a Sg,\,$ and $\, Sbg = bSg,\,$ hence

$\qquad (S\!-\!a)(S\!-\!b)(a^n\!+b^n)\, =\, (S\!-\!b)\color{#c00}{(S\!-\!a)a^n} + (S\!-\!a)\color{#0a0}{(S\!-\!b)b^n}\, =\, \color{#c00}0 + \color{#0a0}0\, =\, 0$

Thus $\ \ 0 = (S\!-\!a)(S\!-\!b) f_n = (S^2\!-(a\!+\!b)S+ab)f_n = \underbrace{f_{n+2}\!-(a\!+\!b)\,f_{n+1}\!+ab\, f_n}_{\large{\rm recurrence\ \ for}\ \ f_n}.$

Many further examples are in the "Linked" questions list in the sidebar $\Longrightarrow$

Answer 5

Hint: Expand (binomial formula) $\left(x+\frac{1}{x}\right)^3$ and then $\left(x+\frac{1}{x}\right)^5$. (No need for even powers.)

Answer 6

Hint: $$\left( x + \frac{1}{x}\right)^5 = x^5 + \frac{1}{x^5} + 5\left(x^3 + \frac{1}{x^3}\right) + 10\left(x + \frac{1}{x} \right)$$ $$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x} \right)\left(x^2 + \frac{1}{x^2} - 1 \right)$$

$$\left( x + \frac{1}{x}\right)^2 - 2 = x^2 + \frac{1}{x^2}$$

Answer 7

Since the other answers have worked directly with powers of $x$, an alternative perspective may be interesting. Let $\tau =\ln x$. Then $x^n+x^{-n}=e^{n\tau}+e^{-n\tau}=2\cosh n\tau$. For $n=1$ in particular we have $2\cosh\tau=x+x^{-1}=5$ by assumption, and so we may write $$x^n+x^{-n}=2\cosh(n\tau)=2\cosh(n\cosh^{-1}(5/2)).$$ So this problem can be interpreted as in terms of particular inversion of the hyperbolic cosine. This in fact serves as one definition of the Chebyshev polynomials of the first kind as $T_n(x)=\cosh(n\cosh^{-1}x))$. So what we're after is $2T_5(5/2)$.

One can develop recursive methods of computing $T_n(x)$, but I'll settle with quoting the Wikipedia page to get $$T_5(x)=16x^5-20x^3+5x\implies 2T_5(5/2)=5^5-5^4+5^2=\boxed{2525}.$$

Answer 8

OP seems to have found a fairly efficient means of calculating (or there is one implicit in the question) as follows:

Let $a_n=x^n+x^{-n}$

Then we have $a_n a_{n-1}=a_{2n-1}+a_1$ or $$a_{2n-1}=a_na_{n-1}-a_1$$

And $a_n^2=a_{2n}+2$ or $$a_{2n}=a_n^2-2$$

So we have, with $a_1=5$ $$a_2=a_1^2-2=23$$

And $$a_3=a_2a_1-a_1=110$$

So that $$a_5=a_3a_2-a_1=2525$$

I put this up as an answer, because the method looks to work pretty efficiently for higher powers.

Answer 9

First find out $x^2 + x^{-2}$ and then $x^3 + x^{-3}$ by expanding and then at last find the required answer by taking power $5$.

Answer 10

There is the famous explicit formula

$$ \frac{x^n+\dfrac{1}{x^n}}{\left(x+\dfrac{1}{x}\right)^{n}}=\sum_{k=0}^{\frac{n}{2}} (-1)^k \frac{n}{n-k} \binom{n-k}{k}\left(x+\frac{1}{x}\right)^{-2k}. $$