Question

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

Answer 1

`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(x)=sinx`, we must find a formula for `f^((n))`, 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(x)=sinx` 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)`