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Algebra Level 5

In Fibonacci sequence, terms are defined ∀ k ∈ N , k ≥ 2 \forall k \in \mathbb{N}, k \geq {2} , F 1 : = 1 F_{1}:=1 , F 2 : = 1 F_{2}:=1 and F k + 1 : = F k + F k − 1 F_{k+1}:=F_{k}+F_{k-1} . The number ϕ \phi is given as the limit of the ratio between F k + 1 F k \frac{F_{k+1}}{F_{k}} , as k k increases indefinitely. Find the value of f ( 100 ) f(100) , where:

f ( n ) = c o s h [ ( n + 1 ) ⋅ a ] − c o s h [ ( n − 1 ) ⋅ a ] s i n h ( n ⋅ a ) f(n)=\frac{cosh[(n+1)\cdot a]-cosh[(n-1)\cdot a]}{sinh(n\cdot a)} , a = l n ϕ a= ln \ \phi and n ∈ R − { 0 } n \in \mathbb{R}-\{0\} .


The answer is 1.

Mikael Marcondes by Mikael Marcondes , 24, Brazil

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2 solutions

The terms of Fibonacci sequence ∀ k ∈ N \forall k \in \mathbb{N} are defined as:

F 1 = 1 F_1 = 1 , F 2 = 1 F_2=1 and F k + 1 = F k + F k − 1 F_{k+1} = F_k + F_{k-1} for k ≥ 2 k\ge 2 .

If lim ⁡ k → ∞ F k + 1 F k = ϕ \displaystyle \lim_{k \to \infty} {\dfrac {F_{k+1}}{F_k}} = \phi , find f ( 100 , ln ⁡ ϕ ) f(100, \ln{\phi}) , where:

f ( n , a ) = cosh ⁡ [ ( n + 1 ) a ] − cosh ⁡ [ ( n − 1 ) a ] sinh ⁡ ( n a ) f(n, a) = \dfrac {\cosh{[(n+1)a]}-\cosh{[(n-1)a]}}{\sinh{(na)}} for n , a ∈ R − { 0 } n, a \in \mathbb{R} - \{0\} .

ϕ = lim ⁡ k → ∞ F k + 1 F k = lim ⁡ k → ∞ F k + F k − 1 F k = lim ⁡ k → ∞ ( 1 + F k − 1 F k ) = 1 + lim ⁡ k → ∞ F k − 1 F k = 1 + 1 ϕ \begin{aligned} \phi & = \displaystyle \lim_{k \to \infty} {\dfrac {F_{k+1}}{F_k}} = \lim_{k \to \infty} {\dfrac {F_{k}+F_{k-1}}{F_k}} = \lim_{k \to \infty} {\left( 1 + \dfrac {F_{k-1}}{F_k} \right)} \\ & = 1 + \displaystyle \lim_{k \to \infty} {\dfrac {F_{k-1}}{F_k}} = 1 + \dfrac {1}{\phi} \end{aligned}

⇒ ϕ = 1 + 1 ϕ ⇒ ϕ 2 − ϕ − 1 = 0 ⇒ ϕ = 1 + 5 2 = φ , golden ratio \Rightarrow \phi = 1 + \dfrac {1}{\phi} \quad \Rightarrow \phi^2 - \phi - 1 = 0 \quad \Rightarrow \phi = \dfrac {1+\sqrt{5}}{2} = \varphi \text{, golden ratio}

f ( 100 , ln ⁡ φ ) = cosh ⁡ ( 101 ln ⁡ φ ) − cosh ⁡ ( 99 ln ⁡ φ ) sinh ⁡ ( 100 ln ⁡ φ ) = 1 2 ( e 101 ln ⁡ φ + e − 101 ln ⁡ φ − e 99 ln ⁡ φ − e − 99 ln ⁡ φ ) 1 2 ( e 101 ln ⁡ φ − e − 100 ln ⁡ φ ) = φ 101 + φ − 101 − φ 99 − φ − 99 φ 100 − φ − 100 = φ 100 ( φ − φ − 1 ) − φ − 100 ( φ − φ − 1 ) φ 100 − φ − 100 = φ − 1 φ = 1 \begin{aligned} f(100, \ln{\varphi}) & = \dfrac {\cosh{(101\ln{\varphi})}-\cosh{(99\ln{\varphi})}}{\sinh{(100\ln{\varphi})}} \\ & = \dfrac {\frac {1}{2} \left( e^{101\ln{\varphi}}+e^{-101\ln{\varphi}}-e^{99\ln{\varphi}} -e^{-99\ln{\varphi}}\right)}{\frac {1}{2} \left( e^{101\ln{\varphi}}-e^{-100\ln{\varphi}} \right)} \\ & = \dfrac {\varphi^{101}+\varphi^{-101}-\varphi^{99}-\varphi^{-99}}{\varphi^{100}-\varphi^{-100}} \\ & = \dfrac {\varphi^{100} \left( \varphi-\varphi^{-1} \right) -\varphi^{-100} \left( \varphi-\varphi^{-1} \right)}{\varphi^{100}-\varphi^{-100}} \\ & = \varphi - \dfrac {1}{\varphi} = \boxed{1} \end{aligned}

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Mikael Marcondes , denke für den Satz: "Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk." Ich habe gelernt, ein bisschen Deutsch.

I only learned now that it is originally German. I have suggested changes to the phrasing this problem. Hope it is useful.

Chew-Seong Cheong - 6 years, 2 months ago

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Dankeschön für der Vorschlag, @Chew-Seong Cheong c:

I was a little lazy to write it out as a limit and I typed it instead, hahaha.

I don't believe in the existence of any God, but I think this quote is beautiful from the viewpoint that it makes us a reminder of how mathematics have been developed and observed by us throughout the centuries.

I used the Binet's formula to arrive in the initial statement that this fraction would be valued as one, for n different than zero. Plugging into exponential form of hyperbolic trigonometric functions is a good and easier way to find the value of this function.

Mikael Marcondes - 6 years, 2 months ago
Khushal Sethi
Aug 31, 2015

in the definition of f(n) it is written senh(na) in denominator. i thought it to be sech(na). please correct it.

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No, it's sinh(na), indeed!

Mikael Marcondes - 5 years, 9 months ago

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