Taylor Polynomials

Taylor Polynomials

Review of the Tangent Line

Recall that if f(x) is a function, then f '(a) is the slope of the tangent line at x = a.  Hence

        y - f(a) = f '(a)(x - a)  

or

        P₁(x) = y = f(a) + f '(a)(x - a) 

is the equation of the tangent line.  We can say that this is the best linear approximation to f(x) near a.

Note:      P₁'(x) = f '(a).
 

Quadratic Approximations

Example:  

Let 

        f(x) = e^2x

Find the best quadratic approximation at  x = 0.

Solution  

Note 

        f '(x) = 2e^2x  

and 

        f ''(x) = 4e^2x

Let 

        P₂(x) = a₀ + a₁x + a₂x²  

Note 

         P'₂(x) = a₁ + 2a₂x  

and 

        P''₂(x) = 2a
Hence 

        a₀ = 1
        

        P₂'(0)  =  a₁  =  f '(0)  =  2 

Hence

        a₁ = 2

        P₂''(0) = 2a₂ = f ''(0) = 4 

Hence  

        a₂ = 2

So

        P₂(x) = 1 + 2x + 2x²


        

The Taylor Polynomial

Suppose that we want the best n^th degree approximation to f(x) at x = a.  We compare f(x) to

        P_n(x) =  a₀ + a₁(x - a) + a₂(x - a)²  + a₃(x - a)³ + ... + a_n(x - a)ⁿ

We make the following observations:

        f(a) =  P_n(a) =  a₀  

so that  

        a₀ = f(a)

        f '(a) = P'_n(a) =  a₁ + 2a₂(x - a)  + 3a₃(x - a)² + ... + na_n(x - a)^n-1 at x = a 

so that 

        a₁ = f '(a)

        f ''(a) = P''_n(a) =   2a₂  + (3)(2)a₃(x - a)+ ... + n(n - 1)a_n(x - a)^n-2 at x = a 

so that 

        a₂ = 1/2 f ''(a)

Note Each time we take a derivative we pick up the next integer in other words
        
                        1
        a₃ =                  f '''(a) 
                    (2)(3)

If we define f^(k)(a) to mean the k^th derivative of f evaluated at a  then

                     1
        a_k =              f^(k)(a)
                    k!

In General

The special case when a = 0 is called the McLaurin Series

Examples:

Find the fifth degree McLaurin Polynomial for sin x

        f(0) = sin(0) = 0

        f '(0) = cos(0) = 1

        f ''(0) = -sin(0) = 0

        f ⁽³⁾(0) = -cos(0) = -1

        f ⁽⁴⁾(0) = sin(0) = 0

        f ⁽⁵⁾(0) = cos(0) = 1

So that 

                                1              0               -1              0               5
        P₅(x) = 0 +            x +          x² +           x³ +          x⁴ +          x⁵ 
                             1!             2!               3!              4!             5!

                     x³             x⁵
        =  x -             +                       
                    6              120
 

Taylor's Remainder

Taylor's Remainder Theorem says that any smooth function can be written as an n^th degree Taylor polynomial plus a function that is of order n + 1 near x = c.

Example

We have

                                 (.1)³           (.1)⁵        -sin z
        sin(.1) = .1  -              +               +                = .099833416667 + E
                                  6             120            6!

Where 

                   1
        E <           (.1)⁶  = .0000000014
                  6!

Example

Use an 11th degree Taylor polynomial to approximate 

        \( \int_1^2 e^{x^2} dx \)

Solution

First notice that there is no elementary antiderivative.  Hence, we find the Taylor polynomial and then integrate.  We have

        \( e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + ... \)

Plugging in x² for x, gives

        \( e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \frac{x^8}{24} + \frac{x^{10}}{120} + ... \)

Now integrate to get

        \( \int e^{x^2} dx = x + \frac{x^3}{3} + \frac{x^5}{10} + \frac{x^7}{42} + \frac{x^9}{216} + \frac{x^{11}}{1320} + ... \)        

We ignore all terms after the 11th power term to get

        1.46253

The actual integral up to five decimal places of accuracy is 

        1.46265

Back to the Sequence and Series and Newton's Method Page