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

1 Answer
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(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) #