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

2 Answers
Apr 1, 2017

Answer:

#(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

Answer:

#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)#