taylor series

Question

Find the Taylor series for f(x)=sin x expanded at about x=(pi/2) and prove that the series converges to sin x for all x

Answer 1

Hi Kee,

 

First, the Taylor series for a function f(x) centered at a point a is given by

 

T = ∑^∞_n=0 (f⁽ⁿ⁾(a)/n!)*(x-a)ⁿ

 

where f⁽ⁿ⁾(x) is the n^th derivative of f, and f⁽⁰⁾(x)=f(x). So, to begin, let's examine f⁽ⁿ⁾(π/2) for n = 0 to n = 4:

 

f(π/2) = sin(π/2) = 1

 

f⁽¹⁾(π/2) = cos(π/2) = 0

 

f⁽²⁾(π/2) = -sin(π/2) = -1

 

f⁽³⁾(π/2) = -cos(π/2) = 0

 

f⁽⁴⁾(π/2) = sin(π/2) = 1

 

For the 5^th and subsequent derivatives, we will continue this pattern of 1, 0, -1, 0, 1, 0, -1, 0, ...

 

Let us write out the first few terms of the Taylor Series

 

T = ∑^∞_n=0 (f⁽ⁿ⁾(a)/n!)*(x-a)ⁿ

 

   = 1/0! * (x-π/2)⁰ + 0/1! * (x-π/2)¹ + -1/2! * (x-π/2)² + 0/3! * (x-π/2)³ + ...

 

   = 1/0! * (x-π/2)⁰ + 0 - 1/2! * (x-π/2)² + 0 + ...

 

We have enough above to notice a pattern; every other term is zero, the numerator alternates between 1 and -1 (this indicates a (-1)ⁿ term in the Taylor Series), the denominator of the non zero coefficients increases by 2 at every iteration, and the powers of the (x-π/2) tern increase by 2. So, we can write out our Taylor Series centered at x = π/2 as:

 

sin(x) = ∑^∞_n=0 ( (-1)ⁿ/(2n)! * (x-π/2)²ⁿ )

 

The above is our Taylor Series for sin(x) centered at x = π/2.

 

To find the radius of convergence, which is what I think the second part of your question is asking, we must do a ratio test:

 

lim_n→∞ | a_n+1/a_n | (where a_n is the n^th term in our series)

 

lim_n→∞ | ( (-1)ⁿ⁺¹ * (x-π/2)²⁽ⁿ⁺¹⁾/(2(n+1))! ) / ( (-1)ⁿ * (x-π/2)²ⁿ/(2n)! ) |

 

= lim_n→∞ | ( (-1)ⁿ⁺¹ * (x-π/2)²ⁿ⁺²/(2n+2)! ) / ( (-1)ⁿ * (x-π/2)²ⁿ/(2n)! ) |

 

Because of the absolute value, we can disregard the (-1)ⁿ terms above. Flipping, multiplying, and simplifying the above yields

 

= lim_n→∞ | (x-π/2)²ⁿ⁺²/(2n+2)! * (2n)!/(x-π/2)²ⁿ |  

 

= lim_n→∞ | (x-π/2)²/( (2n+2)*(2n+1) ) |

 

When examining the above limit, the x term is remaining "fixed", while n tends to infinity. Hence, the above limit is zero. Since the limit of |a_n+1/a_n| = 0, we have that the radius of convergence is (-∞,∞), i.e. for all real numbers x, our Taylor Series

 

sin(x) = ∑^∞_n=0 ( (-1)ⁿ/(2n)! * (x-π/2)²ⁿ )

 

holds.