A Trigonometric Sum | Problem 291
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- čas přidán 29. 08. 2024
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The Schaum Outline Series are fabulous books to own and use. I currently own over two hundred of them.
Wow!
cos(nθ) = (eᶥⁿᶿ + e⁻ᶥⁿᶿ)/2
eᶥᶿ = a => eᶥⁿᶿ = aⁿ
cos(nθ) = (aⁿ + 1/aⁿ)/2
E = 1 + cos(θ)/2 + cos(2θ)/4 + cos(3θ)/8 + .....
E = (a⁰ + 1/a⁰)/2 + (a¹ + 1/a¹)/4 + (a² + 1/a²)/8 + (a³ + 1/a³)/16 + .....
E = (a⁰/2 + a¹/4 + a²/8 + ..) + [1/(2a⁰) + 1/(4a¹) + 1/(8a²) + 1/(16a³) + ..]
E₁= (1/2)(a⁰/2⁰ + a¹/2¹ + a²/2² + a³/2³ + ..)
E₂= (1/2)[1/(2⁰a⁰) + 1/(2¹a¹) + 1/(2²a²) + 1/(2³a³) + ..]
E₁= (1/2)[1/(1 - a/2)]
E₂= (1/2)[1/(1 - 1/2a)]
E₁= 1/(2 - a)
E₂= a/(2a - 1)
E = 1/(2 - a) + a/(2a - 1)
E = (2a - 1 + 2a - a²)/[(2 - a)(2a - 1)]
E = (-a² + 4a - 1)/(4a - 2 - 2a² + a)
E = (a² - 4a + 1)/(2a² - 5a + 2)
*E = (e²ᶥᶿ - 4eᶥᶿ + 1)/(2e²ᶥᶿ - 5eᶥᶿ + 2)*
E = (eᶥᶿ - 4 + e⁻ᶥᶿ)/(2eᶥᶿ - 5 + 2e⁻ᶥᶿ)
E = (2cosθ - 4)/(4cosθ - 5)
*E = (4 - 2cosθ)/(5 - 4cosθ)*
Wow!
The first method calls for a substitution. Let x = cos(theta), factor out 1/2, resolve the series.
The leftover terms are the issue with the first method, not the powers of cosine over 2^n.
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Nice!
Thanks!