series

durham2 · #1 $Old$ October 27th, 2009, 12:42 PM

summation n=0 to infinity r^ncosn(theta)=rcos(theta)-r^2/1-2rcos(theta)+r^2

can neone prove this??

tonio · #2 $Old$ October 27th, 2009, 02:46 PM

Quote:

Originally Posted by durham2 $View Post$

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 $|r|< 1$ , so that then we have the sum of a geometric sequence with quotient less than 1 in absolute value: $r^ne^{n i\theta}=r^n(\cos n\theta + i\sin n\theta), \mbox{using DeMoivre's theorem, so}$

$\sum\limits_{n=0}^\infty(re^{i\theta})^n=\frac{1}{1-re^{i\theta}}=\frac{1-re^{-i\theta}}{r^2+1-2r\cos \theta}$ <--correction done here

Now just separate real and imaginary parts above.

Tonio