Convergence of the Taylor series for the sine function

**jfpeji** · November 20th 2011, 06:22 AM

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.

**TheChaz** · November 20th 2011, 07:08 AM

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

**CaptainBlack** · November 20th 2011, 07:58 AM

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

**jfpeji** · November 20th 2011, 08:28 AM

Ok. You are right. Thank you for your answer.

**jfpeji** · November 20th 2011, 08:42 AM

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

**CaptainBlack** · November 20th 2011, 11:26 AM

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

**jfpeji** · November 20th 2011, 11:36 AM

Ok. You are right. Thank you for your answer.

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