Taylor Polys

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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'(a) = 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

Take derivatives we obtain

P'₂(x) = a₁ + 2a₂x

and

P''₂(x) = 2a₂

We want the derivatives of f and P to match up. We have

   P₂(0)  = a₀ = f(0) = 1

Hence

        a₀ = 1

Next

P₂'(0) = a₁ = f '(0) = 2

Hence

a₁ = 2

Finally

P₂''(0) = 2a₂ = f ''(0) = 4

Hence

a₂ = 2

So

        P₂(x) = 1 + 2x + 2x²

         $Graph of y=e^(2x), y=1, y=1+2x, and y=1+2x+2x^2, all touching at (0,1) and a better approximation as the power increases.$

Notice that the tangent line approximates the curve well for values near x = 0, however the quadratic approximation is a better fit near this point. This leads us to the idea of using higher degree polynomials to get even better fitting curves.

The Taylor Polynomial

Suppose that we want the best n^thdegree 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)

Now we investigate the first derivative.

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)

Taking second derivatives gives

        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

                    1
        a₂ =           f ''(a)
                    2

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^thderivative of f evaluated at a  then

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

We now have the key ingredient for our main result.

The Taylor Polynomial
The n^th degree Taylor polynomial at x = a is

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

$ = \displaystyle\sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x - a)^k $

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

The McLaurin Polynomial

The McLaurin Polynomial of a differentiable function f(x) is

\[ \displaystyle\sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!}x^k \]

Examples:

Find the fifth degree McLaurin Polynomial for sin x.

Solution

We construct the following table to assist in finding the derivatives.

k	f ^(k)(x)	f ^(k)(0)
0	sin x	0
1	cos x	1
2	-sin x	0
3	-cos x	-1
4	sin x	0
5	cos x	1

Now put this into the formula to get

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

                      x³           x⁵
        = x -            +
                      6          120

Notice how well the McLaurin polynomial (in green) approximates y = sin x (in red) for on period.
$undefinedGraph of y=sin(x) and y=x-1/6 x^3 + 1/20 x^5. The graphs are very close near the origin, but move away from each other to the left of x=-2 and to the right of x = 2.$

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.

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

Example

We have

                                   (0.1)³     (0.1)⁵     sin z
        sin(0.1) = 0.1 -            +           -           (0.1)⁶ = .099833416667 + E
                                      6          120         6!

  Where

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

We see that the error quite small.

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