Inspired by Fibonacci et al

Calculus Level 5

1 2 + 1 4 − 3 8 − 7 16 + 5 32 + 33 64 + 13 128 + ⋯ + g ( n ) 2 n + ⋯ \dfrac{1}{2}+\dfrac{1}{4}-\dfrac{3}{8}-\dfrac{7}{16}+\dfrac{5}{32}+\dfrac{33}{64}+\dfrac{13}{128}+\cdots+\dfrac{g(n)}{2^n}+\cdots

The numerators g ( n ) g(n) are given by g ( n ) = g ( n − 1 ) − 4 g ( n − 2 ) g(n)=g(n-1)-4g(n-2) for n > 2 n>2 , with g ( 1 ) = g ( 2 ) = 1 g(1)=g(2)=1 .

If M M and m m are the supremum and the infimum , respectively, of the partial sums of the series above, find M + m M+m . As your answer, enter ⌊ 1000 ( M + m ) ⌋ \lfloor 1000(M+m) \rfloor .

Enter 2016 if you come to the conclusion that the supremum or the infimum fail to exist.

Notation : ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .

This problem is a follow-up to this question .


The answer is 666.

Otto Bretscher by Otto Bretscher , 65, USA

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1 solution

Otto Bretscher
May 10, 2016

The terms of the series are given by 2 sin ⁡ ( n arccos ⁡ ( 1 / 4 ) ) 15 \frac{2\sin(n \arccos(1/4))}{\sqrt{15}} . Using Lagrange's identity for ∑ n = 1 N sin ⁡ ( k θ ) \sum_{n=1}^{N}\sin(k\theta) , we find that M M and m m are 1 15 cot ⁡ ( θ / 2 ) ± 1 15 sin ⁡ ( θ / 2 ) = 1 3 ± 2 2 3 5 \frac{1}{\sqrt{15}}\cot(\theta/2)\pm\frac{1}{\sqrt{15}\sin(\theta/2)}=\frac{1}{3}\pm \frac{2\sqrt{2}}{3\sqrt{5}} where θ = arccos ⁡ ( 1 / 4 ) \theta=\arccos(1/4) , with M + m = 2 3 M+m=\frac{2}{3} . The answer is 666 \boxed{666} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you explain how you got your very first sentence?

Pi Han Goh - 5 years, 1 month ago

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As I mention here , I'm using two basic properties of the Chebyshev polynomials of the second kind, U n ( x ) U_n(x) ... you know that I love those ;)

We have the recursive equation

U n + 1 ( x ) = 2 x U n ( x ) − U n − 1 ( x ) U_{n+1}(x)=2xU_n(x)-U_{n-1}(x)

and we observe that the terms of our series satisfy this equation for x = 1 4 x=\frac{1}{4} . We need to fine tune the formula to get the given initial values: the n n th term of our series is 1 2 U n − 1 ( 1 4 ) \frac{1}{2}U_{n-1}(\frac{1}{4}) . Now we can use

U n ( cos ⁡ θ ) = sin ⁡ ( ( n + 1 ) θ ) sin ⁡ θ U_n(\cos\theta)=\frac{\sin((n+1)\theta)}{\sin\theta}

with θ = arccos ⁡ ( 1 4 ) \theta=\arccos(\frac{1}{4}) and sin ⁡ θ = 15 4 \sin\theta=\frac{\sqrt{15}}{4} .

Did you look at this one , Comrade? I think this is one of my better and more challenging problems, and I'm curious to see whether I will get a solution ;)

Otto Bretscher - 5 years, 1 month ago

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Thanks for your explanation.

Yep, I noticed your other question already and I have no idea how to solve it.

Pi Han Goh - 5 years, 1 month ago

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@Pi Han Goh – A one-word hint: Legendre

Otto Bretscher - 5 years, 1 month ago

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