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If, $x+y=1, x^2+y^2=2$ Find $$x^7+y^7=??$$

any help guys please?

Aryabhata
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maya
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6 Answers6

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Hint: $x^{n+1}+y^{n+1}=(x+y)(x^n+y^n)-xy(x^{n-1}+y^{n-1})$

Mark Bennet
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Mark Bennet gave the very good trick.

Since $$x^{n+1}+y^{n+1}=(x+y)(x^n+y^n)-xy(x^{n-1}+y^{n-1})$$ let us define $u_n=x^n+y^n$. So $$u_{n+1}=(x+y)\,u_n-x\,y\,u_{n-1}$$ and you know that $x+y=1$; on the other side $$(x+y)^2=x^2+y^2+2xy$$ means that $1=2+2xy$ so $xy=-\frac12$. This makes the recurrence equation $$u_{n+1}=u_n+\frac12 u_{n-1}$$ The characteristic equation is then $r^2=r+\frac12$ the roots of which being $$r_{1,2}=\frac{1}{2} \left(1\pm\sqrt{3}\right)$$ which gives $$u_n=c_1 r_1^n+c_2 r_2^n$$ Using the conditions $u_1=1$, $u_2=2$, this finally gives $$u_n=\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)^n+\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)^n=\frac{\left(1-\sqrt{3}\right)^n+\left(1+\sqrt{3}\right)^n}{2^n}$$ Now use the binomial expansion (all $\sqrt 3$ will disappear) to get $u_3=\frac{5}{2}$, $u_4=\frac{7}{2}$, $u_5=\frac{19}{4}$, $u_6=\frac{13}{2}$, $u_7=\frac{71}{8}$, $u_8=\frac{97}{8}$, $u_9=\frac{265}{16}$, $\cdots$.

Just to mimic Dietrich Burde (this is a joke) $$u_{125}=\frac{394645107381876211072765516988567419}{4611686018427387904}$$

Claude Leibovici
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    +1 Excellent. $u_{5^3}$ is completely germane, probably even more so than $u_{70}$. – Simon S Jul 04 '15 at 16:15
  • @SimonS Care to explain $u_{5^3}$'s germane-ness? :) – MT_ Jul 04 '15 at 16:44
  • @SimonS. I did not understand what you meant about $u_{53}$. What means germane here ? – Claude Leibovici Jul 04 '15 at 16:55
  • Explaining a joke spoils it, but here it is: Germane in this context means "relevant, pertinent". DB's inclusion of the value of $u_{70}$ was not germane. And I take it you were toying with that idea with your $u_{125}$, which I rewrote as $u_{5^3}$ to create the illusion that $125$ expressed as $5^3$ explained its importance (which it doesn't, because it doesn't have any!). – Simon S Jul 04 '15 at 17:16
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If $x+y=1$ and $x^2+y^2=2$, then $xy=\frac{(x+y)^2-(x^2+y^2)}{2}=-\frac{1}{2}$, so $x$ and $y$ are roots of the polynomial $p(z)=z^2-z-\frac{1}{2}$, or eigenvalues of the (companion) matrix: $$ M=\left(\begin{array}{cc}0 & \frac{1}{2}\\ 1 & 1\end{array}\right)$$ so that: $$ x^7 + y^7 = \text{Tr}(M^7)=\text{Tr}\left( \begin{array}{cc} \frac{15}{8} & \frac{41}{16} \\ \frac{41}{8} & 7 \\ \end{array} \right)=7+\frac{15}{8}=\color{red}{\frac{71}{8}}. $$

In order to compute $M^7$, you may just apply repeated squaring three times, then multiply $M^8$ by $M^{-1}$, or notice that, by setting $s_n=x^n+y^n$, we have:

$$ s_{2n} = s_n^2 - 2(xy)^n = s_n^2-\frac{(-1)^n}{2^{n-1}}, $$ $$ s_{2n+1} = s_n s_{n+1} - (xy)^n(x+y) = s_n s_{n+1}-\frac{(-1)^n}{2^n},$$ so you may compute $(s_{2n},s_{2n+1})$ from $(s_n,s_{n+1})$. We may use this technique for the fast computation of Fibonacci or Lucas numbers.

Jack D'Aurizio
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Use the identity $$ x^7+y^7=(x+y)^7-7\,xy \left( x+y \right) \left( {y}^{2}+xy+{x}^{2} \right) ^{2}. $$ Then, taking into account $x+y=1, x^2+y^2=2, xy=-\dfrac{1}{2}$ we get $$ x^7+y^7=1+7 \cdot \frac{1}{2} \cdot 1 \cdot (2-\frac{1}{2})^2=1+\frac{63}{8}=\frac{71}{8}. $$

Leox
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Substituting $x=1-y$ we obtain $2y^2-2y-1=0$, and then we compute (reducing the polynomials in $y$ by the relation $2y^2=2y+1$) $$x^3+y^3=\frac{5}{2},\; x^4+y^4=\frac{7}{2}, \; x^4+y^4=\frac{19}{4} ,\ldots , x^7+y^7=\frac{71}{8}.$$

By the way, $$x^{70}+y^{70}=\frac{26067787517619401581}{8589934592}. $$

Dietrich Burde
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Note that $xy=\frac12[(x+y)^2-(x^2+y^2)]=-\frac12$. Now $$x^7+y^7=(x^3+y^3)(x^4+y^4)-(xy)^3(x+y)\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$$$=(x+y)(x^2+y^2-xy)[(x^2+y^2)^2-2(xy)^2]-(xy)^3(x+y)$$$\qquad\qquad\qquad\qquad=\frac{71}8.$

John Bentin
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