Trevor rides Jake piggyback III

Calculus Level 5

It's a beautiful Monday morning. The wind is violent but relaxing, the sun warm and fellow, and the Trevor is riding Jake piggyback upon the viridian turf of the golf course.

Starting at ( 0 , 0 ) (0,0) , Trevor tells his serf to advance by L n 5 n \dfrac{L_{n}}{\sqrt{5}^{n}} meters and turn 9 0 ∘ 90^{\circ} clockwise at the nth step, where L 1 = 1 L_{1} = 1 , L 2 = 3 L_{2} = 3 , and L n + 2 = L n + 1 + L n L_{n+2} = L_{n+1}+L_{n} .

What is the magnitude of the displacement in meters Trevor and Jake have travelled from the origin? If it is of the form a c b \dfrac{a}{\sqrt[b]{c}} where a a and b ≥ c > 1 b \geq c > 1 are positive integers, find a + b + c a+b+c .


The answer is 46.

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Jake Lai by Jake Lai , 21, Singapore

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1 solution

Gopinath No
Jul 9, 2016

Here's a sketch of the solution:

Find the generating function of the series: 0 , 1 , 3 , 4 , 7 , ⋯ , 0, 1, 3, 4, 7, \cdots, which is G ( x ) = x + 2 x 2 1 − x − x 2 G(x) = \frac{x+2\, x^2}{1-x-x^2}

Now, we need to find the sums ∑ n = 0 ∞ L n 5 n × cos ⁡ ( π 2 n ) \sum_{n=0}^{\infty} \frac{L_n}{\sqrt{5}^n} \times \cos{\left(\frac{\pi}{2}n\right)} ∑ n = 0 ∞ L n 5 n × sin ⁡ ( π 2 n ) \sum_{n=0}^{\infty} \frac{L_n}{\sqrt{5}^n} \times \sin{\left(\frac{\pi}{2}n\right)} which are the real and imaginary parts of G ( e π 2 I / 5 ) G\left(\mathbf{e}^{\frac{\pi}{2} \mathbf{I}} \bigg/ \sqrt{5}\right) and answer is the magnitude of that, 3 41 \frac{3}{\sqrt{41}}

and a + b + c = 3 + 2 + 41 = 46 a+b+c = 3+2+41=46

0 Helpful 0 Interesting 0 Brilliant 0 Confused

@Calvin Lin I made a silly mistake in computing the final answer. Please convert this to a solution,

We see the coordinates of the point are ( ∑ n = 1 ∞ ( − 1 ) n + 1 L 2 n x 2 n , ∑ n = 1 ∞ ( − 1 ) n + 1 L 2 n − 1 x 2 n − 1 ) , x = 5 ∴ D i s p l a c e m e n t = ζ = { ∑ n = 1 ∞ ( − 1 ) n + 1 L 2 n x 2 n } 2 + { ∑ n = 1 ∞ ( − 1 ) n + 1 L 2 n − 1 x 2 n − 1 } 2 L e t f ( x ) = ∑ n = 1 ∞ L n x n L n + L n + 1 = L n + 2 ⟹ ∑ n = 1 ∞ L n x n + ∑ n = 1 ∞ L n + 1 x n = ∑ n = 1 ∞ L n + 2 x n ⇒ f ( x ) + x f ( x ) − 1 = x 2 f ( x ) − x − 3 ⇒ f ( x ) = x + 2 x 2 − x − 1 − f ( i x ) = ∑ n = 1 ∞ ( − 1 ) n + 1 L 2 n x 2 n + i ∑ n = 1 ∞ ( − 1 ) n + 1 L 2 n − 1 x 2 n − 1 = − ( i x + 2 ) − x 2 − i x − 1 ∴ ζ = ∣ − f ( i x ) ∣ = ∣ i x + 2 { x 2 + 1 } + i x ∣ = x 2 + 4 x 4 + 3 x 2 + 1 = 3 41 \left( \sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\frac { { L }_{ 2n } }{ { x }^{ 2n } } } ,\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 } } \frac { { L }_{ 2n-1 } }{ { x }^{ 2n-1 } } \right) ,\quad x=\sqrt { 5 } \\ \therefore \quad Displacement=\quad \zeta \quad =\quad \sqrt { { \left\{ \sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\frac { { L }_{ 2n } }{ { x }^{ 2n } } } \right\} }^{ 2 }+{ \left\{ \sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 } } \frac { { L }_{ 2n-1 } }{ { x }^{ 2n-1 } } \right\} }^{ 2 } } \\ \qquad Let\quad f\left( x \right) =\quad \sum _{ n=1 }^{ \infty }{ \frac { { L }_{ n } }{ { x }^{ n } } } \\ { L }_{ n }{ +L }_{ n+1 }={ L }_{ n+2 }\Longrightarrow \sum _{ n=1 }^{ \infty }{ \frac { { L }_{ n } }{ { x }^{ n } } } +\sum _{ n=1 }^{ \infty }{ \frac { { L }_{ n+1 } }{ { x }^{ n } } } =\sum _{ n=1 }^{ \infty }{ \frac { { L }_{ n+2 } }{ { x }^{ n } } } \\ \Rightarrow f\left( x \right) +xf\left( x \right) -1={ x }^{ 2 }f\left( x \right) -x-3\\ \Rightarrow f\left( x \right) =\frac { x+2 }{ { x }^{ 2 }-x-1 } \\ -f\left( ix \right) =\quad \sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\frac { { L }_{ 2n } }{ { x }^{ 2n } } } +i\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 } } \frac { { L }_{ 2n-1 } }{ { x }^{ 2n-1 } } =\quad \frac { -(ix+2) }{ { -x }^{ 2 }-ix-1 } \\ \therefore \quad \zeta =\quad \left| -f\left( ix \right) \right| =\left| \frac { ix+2 }{ { \{ x }^{ 2 }+1\} +ix } \right| =\sqrt { \frac { { x }^{ 2 }+4 }{ { x }^{ 4 }+3{ x }^{ 2 }+1 } } =\quad \boxed { \frac { 3 }{ \sqrt { 41 } } } \\

Aditya Dhawan - 4 years, 7 months ago

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