Editing Talk:1047: Approximations
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::::* Let z = exp(i*pi/9) = cos(pi/9) + i sin(pi/9). We have z^18-1 = 0, and z^9-1 and z+1 are not 0, so using the same factorisation, Phi_18(z) = z^8-z^7+z^6-z^5+z^4-z^3+z^2-z+1 = 0. | ::::* Let z = exp(i*pi/9) = cos(pi/9) + i sin(pi/9). We have z^18-1 = 0, and z^9-1 and z+1 are not 0, so using the same factorisation, Phi_18(z) = z^8-z^7+z^6-z^5+z^4-z^3+z^2-z+1 = 0. | ||
::::* Hence, the conclusion follow from: 2(cos(pi/9) + cos(3pi/9) + cos(5pi/9) + cos(7pi/9)) = exp(i*pi/9) + exp(-i*pi/9) + exp(i*3pi/9) + exp(-i*3pi/9) + exp(i*5pi/9) + exp(-i*5pi/9) + exp(i*7pi/9) + exp(-i*7pi/9) = -Phi_18(z)+1 = 1. | ::::* Hence, the conclusion follow from: 2(cos(pi/9) + cos(3pi/9) + cos(5pi/9) + cos(7pi/9)) = exp(i*pi/9) + exp(-i*pi/9) + exp(i*3pi/9) + exp(-i*3pi/9) + exp(i*5pi/9) + exp(-i*5pi/9) + exp(i*7pi/9) + exp(-i*7pi/9) = -Phi_18(z)+1 = 1. | ||
− | ::: Well, well. I hope you kinda see the pattern. Dgbrt, I know you hate typos, and I'm pretty sure that in this long text lay many of them. So I apologize, and I will correct them later. The following paragraph was posted after I started my text but before I finished mine. It wasn't signed so I will just leave it down there. It's another valid straightforward proof | + | ::: Well, well. I hope you kinda see the pattern. Dgbrt, I know you hate typos, and I'm pretty sure that in this long text lay many of them. So I apologize, and I will correct them later. The following paragraph was posted after I started my text but before I finished mine. It wasn't signed so I will just leave it down there. It's another valid straightforward proof.[[User:Varal7|Varal7]] ([[User talk:Varal7|talk]]) 21:50, 17 May 2014 (UTC) |