An occult sum

Calculus Level 5

∑ n = 1 ∞ n n ! cos ⁡ ( π n 5 ) \sum _{ n=1 }^{ \infty }{ \frac { n }{ n! } } \cos { \left( \frac { \pi n }{ 5 } \right) }

If the value of the above sum can be expressed in the form e φ cos ⁡ ( β π + γ α ( γ − γ ) δ ) \sqrt { { e }^{ \varphi } } \cos { \left( \frac { \beta \pi +\gamma \sqrt { \alpha \left( \gamma -\sqrt { \gamma } \right) } }{ \delta } \right) } with α \alpha , β \beta , and γ \gamma positive divisors of δ \delta ; α \alpha a divisor of β \beta ; and α \alpha and γ \gamma prime numbers, find β × δ α × γ \dfrac { \beta \times \delta }{ \alpha \times \gamma } .

Notations :

  • e e denotes Euler's number , the base of the natural logarithm.

  • φ \varphi represents the golden ratio , or 1 + 5 2 \dfrac { 1+\sqrt { 5 } }{ 2 } .

This problem is original.
Image Credit: Pentacle 1 by Nyo, Wikipedia .


The answer is 8.

Jonas Katona by Jonas Katona , 22, 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

∑ n = 1 ∞ n n ! cos ⁡ ( n π 5 ) = ∑ n = 1 ∞ e i n π 5 + e − i n π 5 2 ( n − 1 ) ! = 1 2 ( e i π 5 ∑ n = 0 ∞ e i n π 5 n ! + e − i π 5 ∑ n = 0 ∞ e − i n π 5 n ! ) = 1 2 ( e i π 5 e e i n π 5 + e − i π 5 e e − i n π 5 ) = 1 2 ( [ cos ⁡ π 5 + i sin ⁡ π 5 ] e cos ⁡ π 5 + i sin ⁡ π 5 + [ cos ⁡ π 5 − i sin ⁡ π 5 ] e cos ⁡ π 5 − i sin ⁡ π 5 ) = 1 2 e cos ⁡ π 5 ( cos ⁡ π 5 [ e i sin ⁡ π 5 + e − i sin ⁡ π 5 ] + i sin ⁡ π 5 [ e i sin ⁡ π 5 − e − i sin ⁡ π 5 ] ) = e cos ⁡ π 5 ( cos ⁡ π 5 cos ⁡ ( sin ⁡ π 5 ) − sin ⁡ π 5 sin ⁡ ( sin ⁡ π 5 ) ) = e cos ⁡ π 5 cos ⁡ ( π 5 + sin ⁡ π 5 ) = e 1 + 5 4 cos ⁡ ( π 5 + 5 − 5 8 ) = e φ cos ⁡ ( 4 π + 5 2 ( 5 − 5 ) 20 ) \begin{aligned} \sum_{n=1}^\infty \frac{n}{n!} \cos \left(\frac{n \pi}{5} \right) & = \sum_{n=1}^\infty \frac{e^{i\frac{n \pi}{5}} + e^{-i\frac{n \pi}{5}}}{2(n-1)!} \\ & = \frac{1}{2}\left( e^{i\frac{\pi}{5}} \sum_{n=0}^\infty \frac{e^{i\frac{n \pi}{5}}}{n!} + e^{-i\frac{\pi}{5}} \sum_{n=0}^\infty \frac{e^{-i\frac{n \pi}{5}}}{n!} \right) \\ & = \frac{1}{2}\left(e^{i\frac{\pi}{5}} e^{{e^{i\frac{n \pi}{5}}}} + e^{-i\frac{\pi}{5}} e^{{e^{-i\frac{n \pi}{5}}}} \right) \\ & = \frac{1}{2}\left(\left[\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} \right] e^{\cos \frac{\pi}{5} + i \sin \frac{\pi}{5}} + \left[\cos \frac{\pi}{5} - i \sin \frac{\pi}{5} \right] e^{\cos \frac{\pi}{5} - i \sin \frac{\pi}{5}} \right) \\ & = \frac{1}{2} e^{\cos \frac{\pi}{5}} \left(\cos \frac{\pi}{5} \left[ e^{i \sin \frac{\pi}{5}} + e^{-i \sin \frac{\pi}{5}} \right] + i\sin \frac{\pi}{5} \left[ e^{i \sin \frac{\pi}{5}} - e^{-i \sin \frac{\pi}{5}} \right] \right) \\ & = e^{\cos \frac{\pi}{5}} \left(\cos \frac{\pi}{5} \cos \left(\sin \frac{\pi}{5} \right) - \sin \frac{\pi}{5} \sin \left(\sin \frac{\pi}{5} \right) \right) \\ & = e^{\cos \frac{\pi}{5}} \cos \left( \frac{\pi}{5} + \sin \frac{\pi}{5} \right) \\ & = e^{\frac{1+\sqrt{5}}{4}} \cos \left(\frac{\pi}{5} + \sqrt{\frac{5-\sqrt{5}}{8}} \right) \\ & = \sqrt{e^\varphi} \cos \left(\frac{4 \pi + 5\sqrt{2(5-\sqrt{5})}}{20} \right) \end{aligned}

⇒ β ˙ δ α ˙ γ = 4 ˙ 20 2 ˙ 5 = 8 \Rightarrow \dfrac{\beta \dot{} \delta}{\alpha \dot{} \gamma} = \dfrac{4 \dot{} 20}{2 \dot{} 5} = \boxed{8}

5 Helpful 0 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...