# How do you find the nth Taylor polynomials for f(x) = sin x centered about a=0?

Nov 25, 2016

 sinx=x/(1!) - x^3/(3!) + x^5/(5!) - x^7/(7!) + ...

or in sigma notation

 sinx= sum_(n=0)^oo (-1)^n/((2n+1)!)x^(2n+1)

#### Explanation:

To be pedantic, a Taylor Series centred about $x = 0$ is a Maclaurin Series. The series is of the form:

 f(x) = f(0) +f'(0)x/(1!) + f''(0)x^2/(2!) + f'''(0)x^3/(3!) + ...

Or in sigma notation:

 f(x) = sum_(n=0)^oo f^((n))x^n/(n!)

So to find the series for $f \left(x\right) = \sin x$, we must find a formula for ${f}^{\left(n\right)}$, and fortunately this is very easy:

 { ( f(x)=sinx, =>f(0),=0 ), ( f'(x)=cosx,=>f'(0),=1 ), ( f''(x)=-sinx,=>f''(0),=0 ), ( f'''(x)=-cosx,=>f'''(0),=-1 ), ( f^((4))(x)=sinx,=>f^((4))(0),=0 ), ( f^((5))(x)=cosx,=>f^((5))(0),=1 ), ( vdots,,vdots ) :}

And so you can see we get an alternating pattern of $+ , - + , -$ terms, so the Maclaurin series for $f \left(x\right) = \sin x$ is:

 sinx=x/(1!) - x^3/(3!) + x^5/(5!) - x^7/(7!) + ...

or in sigma notation

 sinx= sum_(n=0)^oo (-1)^n/((2n+1)!)x^(2n+1)