By considering 1+ cistheta + cis2theta + cis3theta + .... + cisntheta as a geometric series, can you find: sum_(r=0)^n cosrtheta?

2 Answers
Apr 1, 2017

(sin((n+1)theta/2)cos(ntheta/2))/(sin(theta/2))

Explanation:

Rewording the question, we are asked to find an equation for 1+costheta+cos2theta+...+cosntheta

To solve this you need De Moirve's theorem: (cistheta)^n=cisntheta=cosntheta+isinntheta

Notice that the sum 1+cistheta+cis2theta+...+cisntheta is the same as 1+cistheta+(cistheta)^2+...+(cistheta)^n; a geometric series.
Seeing that the first term is 1, the ratio cistheta and the number of terms is n+1, we can represent this neatly:
1+cistheta+(cistheta)^2+...+(cistheta)^n=1((cistheta)^(n+1)-1)/(cistheta-1)
1+cistheta+(cistheta)^2+...+(cistheta)^n=((cis(n+1)theta)-1)/(cistheta-1)

I am going to use a trick when simplifying cistheta-1 here:
cisntheta-1=c
=1-2sin^2(ntheta/2)+i2sin(ntheta/2)cos(ntheta/2)-1
=-2sin(ntheta/2)(sin(ntheta/2)+icos(ntheta/2))
=-2sin(ntheta/2)(cos(90-ntheta/2)+isin(90-ntheta/2))
=-2sin(ntheta/2)cis(90-ntheta/2)

Notice that the left hand side can be expanded to:
1+costheta+isintheta+cos2theta+isin2theta+...+cosntheta+isinntheta, and that the real part of this is what we want. Thus, we need to find the real component of ((cis(n+1)theta)-1)/(cistheta-1)

((cis(n+1)theta)-1)/(cistheta-1)
(-2sin((n+1)theta/2)cis(90-(n+1)theta/2))/(-2sin(theta/2)cis(90-theta/2))
(sin((n+1)theta/2)cis(90-(n+1)theta/2-(90-theta/2)))/(sin(theta/2))
(sin((n+1)theta/2)cis(-ntheta/2))/(sin(theta/2))
Getting the real part, we see get
(sin((n+1)theta/2)cos(-ntheta/2))/(sin(theta/2))
(sin((n+1)theta/2)cos(ntheta/2))/(sin(theta/2))
and we're done

Apr 1, 2017

sum_(k=0)^n cos(kx) = 1/2(cos(n x) + sin(n x)(cos(x/2)/sin(x/2))+1)

Explanation:

sum_(k=0)^n e^(ik x) = sum_(k=0)^n cos(kx)+i sum_(k=0)^nsin(kx)
sum_(k=0)^n e^(-ik x) = sum_(k=0)^n cos(kx)-i sum_(k=0)^nsin(kx)

so

sum_(k=0)^n cos(kx) = 1/2(sum_(k=0)^n e^(ik x)+sum_(k=0)^n e^(-ik x) )

but

sum_(k=0)^n e^(ik x)=(e^(i(n+1)x)-1)/(e^(ix)-1) = (cos((n+1)x)-1+isin((n+1)x))/(cosx-1+isinx)

and

sum_(k=0)^n e^(-ik x)=(e^(-i(n+1)x)-1)/(e^(-ix)-1) = (cos((n+1)x)-1-isin((n+1)x))/(cosx-1-isinx)

so after simplifications

sum_(k=0)^n e^(ik x)+sum_(k=0)^n e^(-ik x) =cos(n x) + sin(n x)(cos(x/2)/sin(x/2))+1

and finally

sum_(k=0)^n cos(kx) = 1/2(cos(n x) + sin(n x)(cos(x/2)/sin(x/2))+1)