Asked by CHRIS
what is the limit n to infinite of
cos1*cos(1/2)*cos(1/4)*cos(1/8)*cos(1/16)*...*cos(1/2^n)
cos1*cos(1/2)*cos(1/4)*cos(1/8)*cos(1/16)*...*cos(1/2^n)
Answers
Answered by
Steve
Since sinc(1) = cos(1/2)*cos(1/4)*cos(1/8)*cos(1/16)*...*cos(1/2^n)
the limit here is cos(1)sinc(1) = cos(1)sin(1) = 0.4546...
You can read about sinc(x) in various places. This function doesn't usually pop up in introductory calculus courses.
the limit here is cos(1)sinc(1) = cos(1)sin(1) = 0.4546...
You can read about sinc(x) in various places. This function doesn't usually pop up in introductory calculus courses.