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

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 2

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 3

\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 4

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 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

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 7

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

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