# Thread: Convergence of the Taylor series for the sine function

1. ## 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.

2. ## Re: Convergence of the Taylor series for the sine function

Well wouldn't that imply that one degree equals one radian?!?

3. ## Re: Convergence of the Taylor series for the sine function

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

5. ## 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?

6. ## Re: Convergence of the Taylor series for the sine function

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