Integral - Sum- Golden ratio

Calculus Level 3

∫ 1 ∞ ( − 1 3 φ n + 2 3 φ 2 n − 3 3 φ 3 n + 4 3 φ 4 n − 5 3 φ 5 n + ⋯ ) d n = a b − ( a + b ) log ⁡ φ \int_ 1^\infty \left(-\frac {1^3}{\varphi^n} + \frac {2^3}{\varphi^{2n}} - \frac {3^3}{\varphi^{3n}} + \frac {4^3}{\varphi^{4n}} - \frac {5^3}{\varphi^{5n}} + \cdots \right) dn = \frac {a\sqrt b-(a+b)}{\log \varphi}

If the equation above holds true for integers a a and b b with b b being square-free, find b − a b-a .

Notation: φ = 1 + 5 2 \varphi = \dfrac {1+\sqrt 5}2 denotes the golden ratio .


The answer is 1.

Srinivasa Raghava by Srinivasa Raghava , 101, India

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1 solution

Chew-Seong Cheong
May 21, 2020

I = ∫ 1 ∞ ∑ k = 1 ∞ ( − 1 ) k k 3 φ k n d n = ∑ k = 1 ∞ ( − 1 ) k k 2 φ k log ⁡ φ Replace 1 φ with x . = x log ⁡ φ ⋅ d d x ( x ⋅ d d x ⋅ 1 1 + x ) = x log ⁡ φ ⋅ d d x ( − x ( x + 1 ) 2 ) = x log ⁡ φ ⋅ x − 1 ( x + 1 ) 3 Replace x with 1 φ . = φ − 1 log ⁡ φ ⋅ φ − 1 − 1 ( φ − 1 + 1 ) 3 Note that 1 φ = φ − 1 = 1 log ⁡ φ ⋅ 1 − φ φ 5 = − 1 φ 6 log ⁡ φ and φ n = F n φ − F n − 1 = − 1 log ⁡ φ ( 8 φ + 5 ) = − 2 log ⁡ φ ( 8 5 + 18 ) = 4 5 − 9 log ⁡ φ \begin{aligned} I & = \int_1^\infty \sum_{k=1}^\infty \frac {(-1)^k k^3}{\varphi^{kn}} dn \\ & = \sum_{k=1}^\infty \frac {(-1)^k k^2}{\varphi^k \log \varphi} & \small \blue{\text{Replace }\frac 1\varphi \text{ with }x.} \\ & = \frac x{\log \varphi} \cdot \frac d{dx} \left(x \cdot \frac d{dx} \cdot \frac 1{1+x} \right) \\ & = \frac x{\log \varphi} \cdot \frac d{dx} \left(-\frac x{(x+1)^2} \right) \\ & = \frac x{\log \varphi} \cdot \frac {x-1}{(x+1)^3} & \small \blue{\text{Replace }x \text{ with } \frac 1\varphi .} \\ & = \frac {\varphi^{-1}}{\log \varphi} \cdot \frac {\varphi^{-1}-1}{(\varphi^{-1}+1)^3} & \small \blue{\text{Note that }\frac 1\varphi = \varphi - 1} \\ & = \frac 1{\log \varphi} \cdot \frac {1-\varphi}{\varphi^5} = - \frac 1{\blue{\varphi^6}\log \varphi} & \small \blue{\text{and }\varphi^n = F_n\varphi - F_{n-1}} \\ & = -\frac 1{\log \varphi(8\varphi +5)} = - \frac 2{\log \varphi(8\sqrt 5+18)} \\ & = \frac {4\sqrt 5 - 9}{\log \varphi} \end{aligned}

Therefore b − a = 5 − 4 = 1 b-a = 5-4 = \boxed 1 .

1 Helpful 0 Interesting 1 Brilliant 0 Confused

well done.

Srinivasa Raghava - 1 year ago

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Glad that you like it. I have solved your previous one involving gamma function

Chew-Seong Cheong - 1 year ago

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