summation n=0 to infinity r^ncosn(theta)=rcos(theta)-r^2/1-2rcos(theta)+r^2 can neone prove this??
Follow Math Help Forum on Facebook and Google+
Originally Posted by durham2 summation n=0 to infinity r^ncosn(theta)=rcos(theta)-r^2/1-2rcos(theta)+r^2 can neone prove this?? This is true iff , so that then we have the sum of a geometric sequence with quotient less than 1 in absolute value: <--correction done here Now just separate real and imaginary parts above. Tonio
Last edited by tonio; Oct 27th 2009 at 02:58 PM.
View Tag Cloud