Taylor Polynomials

I.  Quiz

II.  Homework

III. 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 that P₁'(x) = f'(a).

IV.  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) = 2a₂

P₂(0)  =  a₀ = f(0) = 1 Hence a₀ = 1

P₂'(0) = a₁ = f'(0) = 2 Hence  a₁ = 2

P₂''(0) = 2a₂ = f''(0) = 4 Hence  a₂ = 2

We will demonstrate this on the screen.

V.  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 = 1/2f''(a)

Note each time we take a derivative we pick up the next integer in other words

a₃ = 1/(2)(3) f'''(a) 

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

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

In General

P_n(x) =  f(a) + f'(a)(x - a) + f''(a)/2!(x - a)²  + f⁽³⁾(a)/3! (x - a)³ + ... + f⁽ⁿ⁾(a)/n! (x - a)ⁿ

= sum from k = 0 to n of f^(k)(a)/k! (x - a)^k

This is called the n^th degree taylor polynomial at x = a.

When a = 0,

= sum from k = 0 to n of f^(k)(0)/k! x^k is called the n^th McLaurin Series.

Examples:

Find the fifth degree McLaurin Polynomial for sinx

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

x - x³/6 + x⁵/120

VI .  Taylor's Remainder

Theorem:  If f is smooth from a to b, let P_n(x)  be the n^th degree Taylor polynomial at x = c, then for every x there is a z between x and c with

f(x) = P_n(x) + f⁽ⁿ⁺¹⁾(z)/(n + 1)! (x - c)ⁿ⁺¹   

Example

We have

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

Where E < 1/6! (.1)⁶  = .0000000014