Taylor Series Expansion

 

Motivation:

                       For a polynomial   f(x)  of degree n,  we can represent it as

 

                                                      f(x) =

 

                        This can be done by  starting dividing  f(x)  by  (x-c) , and taking the quotient polynomial to be divided by (x-c) again, …  With this process by dividing the quotient polynomial generated from the previous division,  we can find all the  a_i’s .  It can be done via synthetic division. However, that is not the only way to find all the a_i’s .

 

                        At first, we can find  

                                                             f(c) = a₀

 

                        And

                                       f^’(x) =  =

                                  f’(c) = a₁

 

                                      f^”(x) =  =  +2a₂

                                 f”(c) = 2a₂ 

 

                                     …

 

                                     f^(n-1)(x) = a_nn!(x-c) + a_n-1(n-1)!

                                  f^(n-1)(c) = a_n-1(n-1)! 

 

                                    f⁽ⁿ⁾(x) = a_nn!

 

                        Since f  is a polynomial of degree n,  f⁽ⁿ⁾(x)  is a constant. With the equality above,  we still can find  a_n .  In general, we have

 

                                      a_i =   ,   i = 1, 2, 3, …, n

                     And

                                  f(x) = f(c) + f’(c)(x-c)+ … +  (x-c)ⁱ + … +  (x-c)ⁿ

              But this is just for the case that   f(x) is a polynomial.  We would like to find the similar expression when   f(x) is a function other than polynomial.  The result is known as Taylor series expansion for a function f(x) at x=c . Especially, when c=0,  it is known as McLaurin series. 

 

Theorem( Taylor Theorem )

              Let  f   be a function that  it is n times continuously differentiable on a interval I containing x and c , i.e.  f’, f”, … f⁽ⁿ⁾  exist in the interval.  And  f⁽ⁿ⁺¹⁾  is continuous in the interval I.   Then

                               f(x) = f(c) + f’(c)(x-c) + … +  (x-c)ⁿ +  (x-c)ⁿ⁺¹

                                        where b is between x and c .

 

Proof:

                     Let

                                     R(t)= f(x)- f(t) -  

                                    R(x) = 0

 

                     Differentiate with respect to t on both sides:

 

                                   R’(t) = -f’(t)

                                           =   -

 

 

                    Set

                                  Q(t)= R(t) –    R(c)

                              Q(c) = 0,  and Q(x) = R(x) = 0

 

                   From mean value theorem, there exist b between c and x such that

 

                                                             Q’(b) = 0

 

                    And            Q’(t) = R’(t) + (n+1)  R(c)

                             So,    

                                        0 = Q’(b) = R’(b) + (n+1)  R(c)

                              R(c) = - R’(b)  

                                           =

                                           =

 

 

                   Furthermore,      R(t)= f(x)- f(t) -

                     = f(x)- f(c) -

 

                        f(x) =  f(c) +  +

 

                When c=0,  it is known as McLaurin Series.   It is usually used to find the values of many transcendental functions like sin(x), ln(x), ... .

 

 

Example:    Taylor series expansion of  sin(x)  around 0 .

       

                       f(x) = sin(x)  ,  f(0)=0

                       f’(x) = cos(x) , f’(0)=1

                       f’(x) = -sin(x),  f”(0)=0

                       f⁽³⁾(x) = -cos(x) ,  f⁽³⁾(0) = -1

                       f⁽⁴⁾(x) = sin(x),   f⁽⁴⁾(0) = 0

                       f⁽⁵⁾(x) = cos(x),  f⁽⁵⁾(0) = 1

                             ...

 

                    sin(x) = x -  + .... + (-1)^k  + ...

 

 

 

 

Example: Taylor series expansion of  ln(x)  around  1 .

 

                    f(x) = ln(x)

                    f’(x) =   ,  f’(x)=1

                    f”(x) = - ,  f”(1)= -1

                    f⁽³⁾(x) =  ,  f⁽³⁾(1) = 2

                    f⁽⁴⁾(x) = ,  f⁽⁴⁾(1) = - 3!

 

                 ...

                 f⁽ⁿ⁺¹⁾(x) =

 

              So,

                         ln(x) = (x-1) -  (x-1)² ₊  (x-1)³- ... +  (x-1)ⁿ + ...

 

 

Example:   f(x)=e^x  .  Find its Taylor series expansion around 0 .

 

      Sol:

                     f’(x) = e^x  ,  f’(0) = 1

                     f”(x) = e^x ,   f”(0) = 1

                    f⁽³⁾(x) = e^x ,  f⁽³⁾(0) = 1

                    ...

                    f⁽ⁿ⁾(x) = e^x  , f⁽ⁿ⁾(0) = 1

 

              Thus, 

                            e^x = 1 + x +  x² +  x³ + ... +  xⁿ + ...