Editing Talk:1047: Approximations
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:::: *Note that exp(i*pi/7)*exp(i*6pi/7)= exp(i*pi)=-1. So the inverse of z is -exp(i*6pi/7). But we also know that it is exp(-i*pi/7). Well. That was just a fancy way to prove that exp(-i*pi/7) = - exp(i*6pi/7). Good enough. The same holds for exp(-i*3pi/7) = exp(i*14pi/7)*exp(-i*3pi/7)=exp(i*11pi/7)=exp(i*7pi/7)*exp(i*4pi/7)=-exp(i*4pi/7). And the exact same calculation shows that exp(-i*5pi/7)=-exp(i*2pi/7). Alright. | :::: *Note that exp(i*pi/7)*exp(i*6pi/7)= exp(i*pi)=-1. So the inverse of z is -exp(i*6pi/7). But we also know that it is exp(-i*pi/7). Well. That was just a fancy way to prove that exp(-i*pi/7) = - exp(i*6pi/7). Good enough. The same holds for exp(-i*3pi/7) = exp(i*14pi/7)*exp(-i*3pi/7)=exp(i*11pi/7)=exp(i*7pi/7)*exp(i*4pi/7)=-exp(i*4pi/7). And the exact same calculation shows that exp(-i*5pi/7)=-exp(i*2pi/7). Alright. | ||
:::: *Now, use that for any x, we have cos(x) = (exp(ix)+exp(-ix))/2. See [http://en.wikipedia.org/wiki/Euler%27s_formula#Relationship_to_trigonometry here]. Let's calculate twice the sum of the left hand side. 2(cos(pi/7)+cos(3pi/7)+cos(5pi/7))= exp(i*pi/7) + expi(-i*pi/7) + exp(3pi/7) + exp(-3pi/7) + exp(5pi/7) +exp(-5pi/7) = exp(i*pi/7)-exp(i*2pi/7)+exp(i*3pi/7)-exp(i*4pi/7)+exp(i*5pi/7)-exp(i*6pi/7) = -Phi_14(z) +1 = 1. | :::: *Now, use that for any x, we have cos(x) = (exp(ix)+exp(-ix))/2. See [http://en.wikipedia.org/wiki/Euler%27s_formula#Relationship_to_trigonometry here]. Let's calculate twice the sum of the left hand side. 2(cos(pi/7)+cos(3pi/7)+cos(5pi/7))= exp(i*pi/7) + expi(-i*pi/7) + exp(3pi/7) + exp(-3pi/7) + exp(5pi/7) +exp(-5pi/7) = exp(i*pi/7)-exp(i*2pi/7)+exp(i*3pi/7)-exp(i*4pi/7)+exp(i*5pi/7)-exp(i*6pi/7) = -Phi_14(z) +1 = 1. | ||
− | :::: * So dividing both sides by 2, we get what we want. | + | :::: * So dividing both sides by 2, we get what we want. Pfiou. |
::: '''Why is 7 so special? Well it isn't.''' Let's prove it for 9. | ::: '''Why is 7 so special? Well it isn't.''' Let's prove it for 9. | ||
::::* 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. |