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).

Example:

Let

f(x) = e2x

Find the best quadratic approximation at  x = 0.

Solution

Note

f '(x) = 2e2x

and

f ''(x) = 4e2x

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 nth degree approximation to f(x) at x = a.  We compare f(x) to

Pn(x) =  a0 + a1(x - a) + a2(x - a)2  + a3(x - a)3 + ... + an(x - a)n

We make the following observations:

f(a) =  Pn(a) =  a0

so that

a0 = f(a)

f '(a) = P'n(a) =  a1 + 2a2(x - a)  + 3a3(x - a)2 + ... + nan(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)an(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 kth derivative of f evaluated at a  then

1
ak =              f(k)(a)
k!

In General

 The Taylor Polynomial The nth degree Taylor polynomial at x = c is      Pn(x) =  f(c) + f '(c)(x - c) + f ''(c)/2!(x - c)2  +                    f (3)(c)/3! (x - c)3 + ... + f (n)(c)/n! (x - c)n               = S f (k)(c)/k! (x - c)k where the sum goes from 0 to n.

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

 The McLaurin Polynomial The McLaurin Polynomial of a differentiable function f(x) is            S f (k)(0)/k! xk   where the sum goes from 0 to n.

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 nth degree Taylor polynomial plus a function that is of order n + 1 near x = c.

 Taylor's Remainder Theorem If f is smooth from a to b, let Pn(x) be the nth degree Taylor polynomial at x = c, then for every x there is a z between x and c with                                   f (n+1)(z)                  f(x) = Pn(x) +                    (x - c)n+1                                       (n + 1)!

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

Solution

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

Plugging in x2 for x, gives

Now integrate to get

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

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