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

##### 1 Answer

# 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

# 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

And so you can see we get an alternating pattern of

# 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) #