# Convergence of the Taylor series for the sine function

• Nov 20th 2011, 07:22 AM
jfpeji
Convergence of the Taylor series for the sine function
I would like to know if the Taylor series for the sine function,

sin(x) = x - x^3/3! + x^5/5! -...

is convergent if the argument of the function, x , is expressed in degrees instead of radians.
• Nov 20th 2011, 08:08 AM
TheChaz
Re: Convergence of the Taylor series for the sine function
Well wouldn't that imply that one degree equals one radian?!?
• Nov 20th 2011, 08:58 AM
CaptainBlack
Re: Convergence of the Taylor series for the sine function
Quote:

Originally Posted by jfpeji
I would like to know if the Taylor series for the sine function,

sin(x) = x - x^3/3! + x^5/5! -...

is convergent if the argument of the function, $x$, is expressed in degrees instead of radians.

That series converges for arbitrary complex $x$, to $\sin(x)$ where in the $\sin$ the argument is to be regarded as in radians.

A different series converges to the sin where the argument is in degrees

CB
• Nov 20th 2011, 09:28 AM
jfpeji
Re: Convergence of the Taylor series for the sine function
• Nov 20th 2011, 09:42 AM
jfpeji
Re: Convergence of the Taylor series for the sine function
What is that different series which converges to the sin, where the argument is in degrees, that you mention?
• Nov 20th 2011, 12:26 PM
CaptainBlack
Re: Convergence of the Taylor series for the sine function
Quote:

Originally Posted by jfpeji
What is that different series which converges to the sin, where the argument is in degrees, that you mention?

The rather obvious:

$\sin_d(x)=\left(\frac{\pi}{180}\right)x- \left(\frac{\pi}{180}\right)^3\frac{x^3}{3!}+\left (\frac{\pi}{180}\right)^5\frac{x^5}{5!}- ...$

CB
• Nov 20th 2011, 12:36 PM
jfpeji
Re: Convergence of the Taylor series for the sine function