Should you find theta? (II)

Geometry Level 4

x n = 2 n cos ⁡ n θ \large x_n=2^n \cos n\theta

For the given sequence we know that x 2 = − 28 9 \large x_2=-\frac{28}{9} and x 3 = − 184 27 . \large x_3=- \frac{184}{27}.

If x 5 = A B , x_5=\dfrac{A}{B}, where A A and B B are coprime positive integers, find A + B . A+B.

See Part 1 .


The answer is 7955.

View wiki
Arturo Presa by Arturo Presa , 62, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Arturo Presa
Dec 27, 2015

First we can find a recurrence relation for x n x_n . Indeed, since x n = ( 2 e i θ ) n + ( 2 e − i θ ) n 2 , x_n=\frac {(2e^{i\theta})^n+(2e^{-i\theta})^n}{2}, then x n x_n satisfies a linear recurrence relation with characteristic polynomial ( r − 2 e i θ ) ( r − 2 e − i θ ) = (r-2e^{i\theta})(r-2e^{-i\theta})= = r 2 − 4 r cos ⁡ θ + 4. =r^2- 4r\cos \theta +4. Then this implies that the sequence satisfies the recurrence: x n = 4 ( cos ⁡ θ ) x n − 1 − 4 x n − 2 for all n ≥ 2. x_n=4(\cos \theta)x_{n-1}-4x_{n-2}\quad \text{for all}\quad n\geq 2. For details you can see the last example of the wiki Linear Recurrence Relations with repeated roots .

Making n = 3 n=3 in the recurrence, substituting x 3 x_3 and x 2 x_2 by the given values, and x 1 x_1 by 2 cos ⁡ θ , 2\cos\theta, we obtain the equation − 184 27 = 4 ( cos ⁡ θ ) ( − 28 9 ) − 8 cos ⁡ θ . -\frac{184}{27}=4(\cos\theta) (-\frac{28}{9})-8\cos\theta. Solving the equation we obtain that cos ⁡ θ = 1 3 . \cos\theta=\frac{1}{3}. Now, making n = 4 n=4 in the recurrence and substituting the cos ⁡ θ \cos \theta we obtain that x 4 = 4 ( cos ⁡ θ ) x 3 − 4 x 2 = 4 ( 1 3 ) ( − 184 27 ) − 4 ( − 28 9 ) = 272 81 . x_4=4(\cos \theta)x_3-4x_2=4(\frac{1}{3})(-\frac{184}{27})-4(-\frac{28}{9})=\frac{272}{81}.
and x 5 = 4 ( cos ⁡ θ ) x 4 − 4 x 3 = 4 ( 1 3 ) ( 272 81 ) − 4 ( − 184 27 ) = 7712 243 . x_5=4(\cos \theta)x_4-4x_3=4(\frac{1}{3})(\frac{272}{81})-4(-\frac{184}{27})=\frac{7712}{243}. Then the solution to the problem is 7955 . \boxed{7955}.

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Very nice ! (+1) . Not as elegant as yours but it is fairly easy to first find cos ⁡ 5 θ \cos { 5\theta } by noticing that if:

θ ∈ [ 0 , 2 π ] \theta \quad \in \quad \left[ 0,2\pi \right]

cos ⁡ 3 θ < 0 , cos ⁡ 2 θ < 0 ⇔ θ ∈ ( π 4 , π 2 ) ⇒ sin ⁡ 2 θ > 0 ⇒ s i n 2 θ = 1 − cos ⁡ 2 2 θ = 4 2 9 ∵ cos ⁡ 2 θ = − 7 9 < − 1 2 , θ > π 3 ⇒ 3 θ ∈ ( π , 3 π 2 ) ⇒ sin ⁡ 3 θ < 0 ⇒ sin ⁡ 3 θ = − 1 − cos ⁡ 2 3 θ = − 10 2 27 ∴ cos ⁡ 5 θ = cos ⁡ 3 θ cos ⁡ 2 θ − sin ⁡ 2 θ sin ⁡ 3 θ = 241 243 ⇒ x 5 = 7712 243 \quad \quad \cos { 3\theta } <0\quad ,\quad \cos { 2\theta } <0\quad \Leftrightarrow \quad \theta \quad \in \quad \left( \frac { \pi }{ 4 } ,\frac { \pi }{ 2 } \right) \Rightarrow \quad \sin { 2\theta \quad >\quad 0 } \Rightarrow \quad sin2\theta \quad =\quad \sqrt { 1-\cos ^{ 2 }{ 2\theta } } =\frac { 4\sqrt { 2 } }{ 9 } \\ \because \quad \cos { 2\theta } =\quad \frac { -7 }{ 9 } <\frac { -1 }{ 2 } ,\quad \theta \quad >\frac { \pi }{ 3 } \quad \Rightarrow \quad 3\theta \quad \in \quad \left( \pi ,\frac { 3\pi }{ 2 } \right) \Rightarrow \sin { 3\theta } <0\Rightarrow \sin { 3\theta = } -\sqrt { 1-\cos ^{ 2 }{ 3\theta } } =\frac { -10\sqrt { 2 } }{ 27 } \\ \therefore \quad \cos { 5\theta } =\cos { 3\theta } \cos { 2\theta } -\sin { 2\theta } \sin { 3\theta } =\quad \frac { 241 }{ 243 } \quad \Rightarrow \quad { x }_{ 5 }\quad =\quad \boxed { \frac { 7712 }{ 243 } } \\

Aditya Dhawan - 3 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...