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.
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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.
Well wouldn't that imply that one degree equals one radian?!?
That series converges for arbitrary complexQuote:
Originally Posted by jfpeji
, to
where in the
the argument is to be regarded as in radians.
A different series converges to the sin where the argument is in degrees
CB
Ok. You are right. Thank you for your answer.
What is that different series which converges to the sin, where the argument is in degrees, that you mention?
The rather obvious:Quote:
Originally Posted by jfpeji
CB
Ok. You are right. Thank you for your answer.