Playing with 2018 in the soup

Calculus Level 3

( 2018 + 1 2018 − 2 ) ∑ n = 0 ∞ ( 1 2018 ) n ( cos ⁡ ( n π 4 ) + sin ⁡ ( n π 4 ) ) = ? \displaystyle \left ( 2018 + \frac{1}{2018} - \sqrt{2} \right) \sum_{n = 0}^{\infty} \left(\frac{1}{2018}\right)^n \left (\cos \left(\frac{n \pi}{4}\right) + \sin \left(\frac{n \pi}{4}\right) \right )= ?


The answer is 2018.

Guillermo Templado by Guillermo Templado , 46, Spain

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1 solution

S = ( 2018 + 1 2018 − 2 ) ∑ n = 0 ∞ ( 1 2018 ) n ( cos ⁡ ( n π 4 ) + sin ⁡ ( n π 4 ) ) = ( 2018 + 1 2018 − 2 ) ∑ n = 0 ∞ 2 201 8 n sin ⁡ ( ( n + 1 ) π 4 ) By Euler’s formula: e θ i = cos ⁡ θ + i sin ⁡ θ = ( 2018 + 1 2018 − 2 ) ∑ n = 0 ∞ 2 201 8 n ℑ ( e n + 1 4 π i ) ⟹ the imaginary part ℑ ( e θ i ) = sin ⁡ θ = ( 2018 + 1 2018 − 2 ) 2 ℑ ( e π 4 i ∑ n = 0 ∞ ( e π 4 i 2018 ) n ) = ( 2018 + 1 2018 − 2 ) 2 ℑ ( e π 4 i 1 − e π 4 i 2018 ) = ( 2018 + 1 2018 − 2 ) 2 ℑ ( cos ⁡ π 4 + i sin ⁡ π 4 1 − cos ⁡ π 4 + i sin ⁡ π 4 2018 ) = ( 2018 + 1 2018 − 2 ) 2018 2 ℑ ( 1 + i 2018 2 − 1 − i ) = ( 2018 + 1 2018 − 2 ) 2018 2 ℑ ( ( 1 + i ) ( 2018 2 − 1 + i ) ( 2018 2 − 1 − i ) ( 2018 2 − 1 + i ) ) = ( 2018 + 1 2018 − 2 ) 2018 2 ℑ ( 2018 2 − 2 + 2018 2 i 2 ( 201 8 2 ) − 2 ( 2018 2 ) + 2 ) = ( 2018 + 1 2018 − 2 ) ( 2 ( 201 8 2 ) 2 ( 201 8 2 ) − 2 ( 2018 2 ) + 2 ) = ( 2018 + 1 2018 − 2 ) ( 2018 2018 − 2 + 1 2018 ) = 2018 \begin{aligned} S & = \left(2018+\frac 1{2018}-\sqrt 2\right) \sum_{n=0}^\infty \left(\frac 1{2018}\right)^n \left(\cos \left(\frac {n \pi}4\right)+\sin \left(\frac {n \pi}4\right)\right) \\ & = \left(2018+\frac 1{2018}-\sqrt 2\right) \sum_{n=0}^\infty \frac {\sqrt 2}{2018^n} \color{#3D99F6} \sin \left(\frac {(n+1)\pi}4\right) & \small \color{#3D99F6} \text{By Euler's formula: }e^{\theta i} = \cos \theta + i\sin \theta \\ & = \left(2018+\frac 1{2018}-\sqrt 2\right) \sum_{n=0}^\infty \frac {\sqrt 2}{2018^n} \color{#3D99F6} \Im \left(e^{\frac {n+1}4 \pi i}\right) & \small \color{#3D99F6} \implies \text{the imaginary part }\Im \left(e^{\theta i}\right) = \sin \theta \\ & = \left(2018+\frac 1{2018}-\sqrt 2\right) \sqrt 2 \Im \left(e^{\frac \pi 4 i} \sum_{n=0}^\infty \left(\frac {e^{\frac \pi 4 i}}{2018} \right)^n \right) \\ & = \left(2018+\frac 1{2018}-\sqrt 2\right) \sqrt 2 \Im \left(\frac {e^{\frac \pi 4 i}}{1- \frac {e^{\frac \pi 4i}}{2018}} \right) \\ & = \left(2018+\frac 1{2018}-\sqrt 2\right) \sqrt 2 \Im \left(\frac {\cos \frac \pi 4 + i\sin \frac \pi 4}{1- \frac {\cos \frac \pi 4 + i\sin \frac \pi 4}{2018}} \right) \\ & = \left(2018+\frac 1{2018}-\sqrt 2\right) 2018 \sqrt 2 \Im \left(\frac {1+ i}{2018\sqrt 2 - 1-i} \right) \\ & = \left(2018+\frac 1{2018}-\sqrt 2\right) 2018 \sqrt 2 \Im \left(\frac {(1+ i)(2018\sqrt 2 - 1+i)}{(2018\sqrt 2 - 1-i)(2018\sqrt 2 - 1+i)} \right) \\ & = \left(2018+\frac 1{2018}-\sqrt 2\right) 2018 \sqrt 2 \Im \left(\frac {2018\sqrt 2 - 2+ 2018\sqrt 2 i}{2(2018^2) - 2(2018\sqrt 2)+2} \right) \\ & = \left(2018+\frac 1{2018}-\sqrt 2\right) \left(\frac {2(2018^2)}{2(2018^2) - 2(2018\sqrt 2)+2} \right) \\ & = \left(2018+\frac 1{2018}-\sqrt 2\right) \left(\frac {2018}{2018 - \sqrt 2+\frac 1{2018}} \right) \\ & = \boxed{2018} \end{aligned}

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