mathematics-physics-wiki/docs/wiskunde/calculus/taylor-polynomials.md

Taylor polynomials

Linearization

A function f(x) about x = a may be linearized into

P_1(x) = f(a) + f'(a)(x-a),

obtaining a polynomial that matches the value and derivative of f at x = a.

Taylor's theorem

Even better approximations of f(x) can be obtained by using higher degree polynomials if f^{n+1}(t) exists for all t in an interval containing a and x. Thereby matching more derivatives at x = a,

P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2+ ... + \frac{f^{(n)}(a)}{n!}(x-a)^n.

Then the error E_n(x) = f(x) - P_n(x) in the approximation f(x) \approx P_n(x) is given by

E_n(a) = \frac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1},

where s is some number between a and x. The resulting formula

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + ... + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1},

for some s between a and x, is called Taylor's formula with Lagrange remainder; the Lagrange is the error term E_n(x).

Proof:

Observe that the case n=0 of Taylor's formula, namely

f(x) = P_0(x) + E_0(x) = f(a) + \frac{f'(s)}{1!}(x-a),

is just the Mean-value theorem for some s between a and x

\frac{f(x) - f(a)}{x-a} = f'(s).

Using induction to prove for n > 0. Suppose n = k-1 where k \geq 1 is an integer, then

E_{k-1}(x) = \frac{f^{(k)}(s)}{k!}(x-a)^k,

where s is some number between a and x. Consider the next higher case: n=k. Applying the Generalized Mean-value theorem to the functions E_k(t) and (t-a)^{k+1} on [a,x]. Since E_k(a)=0, a number u in (a,x) is obtained such that

\frac{E_k(x) - E_k(a)}{(x-a^{k+1}) - (a-a)^{k+1}}= \frac{E_k(x)}{(x-a)^{k+1}} = \frac{E_k'(u)}{(k+1)(u - a)^k}.

Since

\begin{array}{ll} E_k'(u)&=\frac{d}{dx}(f(x)-f(a)-f'(a)(x-a)-\frac{f''(a)}{2!}(t-a)^2-...-\frac{f^{(k)}(a)}{k!}(t-a)^k)|_{x=u} \ &= f'(u) - f'(a) - f''(a)(u-a)-...-\frac{f^{(k)}(a)}{(k-1)!}(u-a)^{k-1} \end{array}

is just E_{k-1}(u) for the function f' instead of f. By the induction assumption it is equal to

\frac{(f')^{(k)}(s)}{k!}(u-a)^k = \frac{f^{(k+1)}(s)}{k!}(u-a)^k

for some s between a and u. Therefore,

E_k(x) = \frac{f^{(k+1)}(s)}{(k+1)!}(x-a)^{k+1}

Big-O notation

f(x) = O(u(x)) for x \to a if and only if there exists a k > 0 such that

|f(x)| \leq k|u(x)|

For all x in the open interval around x=a.

The following properties follow from the definition:

1. If f(x) = O(u(x)) as x \to a, then Cf(x) = O(u(x)) as x \to a for any value of the constant C.
2. If f(x) = O(u(x)) as x \to a and g(x) = O(u(x)) as x \to a, then f(x) \pm g(x) = O(u(x)) as x \to a.
3. If f(x) = O((x-a)^ku(x)) as x \to a, then \frac{f(x)}{(x-a)^k} = O(u(x)) as x \to a for any constant k.

If f(x) = Q_n(x) + O((x-a)^{n+1}) as x \to a, where Q_n is a polynomial of degree at most n, then Q_n(x) = P_n(x).

Proof: Follows from the properties of the big-O notation

Let P_n be the Taylor polynomial, then properties 1 and 2 of big-O imply that R_n(x) = Q_n(x) - P_n(x) = O((x - a)^{n+1}) as x \to a. It must be shown that R_n(x) is identically zero so that Q_n(x) = P_n(x) for all x. R_n(x) may be written in the form

R_n(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + ... + c_n(x-a)^n

If R_n(x) is not identically zero, then there is a smallest coefficient c_k k \leq n, such that c_k \neq 0, but c_j = 0 for 0 \leq j \leq k -1

R_n(x) = (x-a)^k(c_k + c_{k+1}(x-a) + ... + c_n(x-a)^{n-k}).

Therefore,

\lim_{x \to a} \frac{R_n(x)}{(x-a)^k} = c_k \neq 0.

However, by property 3

\frac{R_n(x)}{(x-a)^k} = O((x-a)^{n+1-k}).

Since n+1-k > 0, \frac{R_n(x)}{(x-a)^k} \to 0 as x \to a. This contradiction shows that R_n(x) must be identically zero.

Maclaurin formulas

Some Maclaurin formulas with errors in big-O notation. These may be used in constructing Taylor polynomials from compsite functions. As x \to 0

1. \frac{1}{1-x} = 1 + x + ... + x^n + O(x^{n+1}),

2. \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ... + (-1)^{n-1}\frac{x^n}{n} + O(x^{n+1}),

3. e^x = 1 + x + \frac{x^2}{2!} + ... + \frac{x^n}{n!} + O(x^{n+1}),

4. \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... + (-1)^n\frac{x^{2n+1}}{(2n+1)!} + O(x^{2n+3}),

5. \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... + (-1)^n\frac{x^{2n}}{(2n)!} + O(x^{2n+1}),

6. \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - ... + (-1)^n\frac{x^{2n+1}}{2n+1} + O(x^{2n+3}).

Example

Construct P_4(x) for f(x) = e^{\sin x} around x=0.

e^{\sin x} \approx 1 + (x - \frac{x^3}{3!} + \frac{x^5}{5!}) + \frac{1}{2!}(x - \frac{x^3}{3!} + \frac{x^5}{5!})^2 + \frac{1}{3!}(x - \frac{x^3}{3!} + \frac{x^5}{5!})^3

\begin{array}{ll} P_4(x) &= 1 + x \frac{1}{2}x^2 + (-\frac{1}{6} + \frac{1}{6})x^3 + (-\frac{1}{6} + \frac{1}{4!})x^4 + O(x^5), \ &= 1 + x \frac{1}{2}x^2 - \frac{1}{8}x^4 + O(x^5). \end{array}

Evaluating limits with Taylor polynomials

Taylor and Macluarin polynomials provide a method for evaluating limits of indeterminate forms.

Example

Determine the limit \lim_{x \to 0} \frac{x \arctan x - \ln(1+x^2)}{x \sin x - x^2}.

\begin{array}{ll} x \sin x - x^2 \approx x^2 - \frac{x^4}{6} + O(x^6) - x^2 = - \frac{x^4}{6} + O(x^6) \ x \arctan x - \ln(1+x^2) \approx x^2 - \frac{x^4}{3} + O(x^6) - x^2 + \frac{x^4}{2} + O(x^6) = \frac{x^4}{6} + O(x^6) \end{array}

\lim_{x \to 0} \frac{\frac{x^4}{6} + O(x^6)}{- \frac{x^4}{6} + O(x^6)} = -1

L'Hostpital's rule

Suppose the function f and g are differentiable on the interval (a,b), and g'(x) \neq 0 there. Also suppose that \lim_{x \downarrow a} f(x) = \lim_{x \downarrow a} g(x) = 0 then

\lim_{x \downarrow a} \frac{f(x)}{g(x)} = \lim_{x \downarrow a} \frac{f'(x)}{g'(x)} = L.

The outcome is exactly the same as using Taylor polynomials.

Proof: using Taylor polynomials around x = a.

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x-a)^2 + O((x-a)^3)}{g(a) + g'(a)(x-a) + \frac{g''(a)}{2}(x-a)^2 + O((x-a)^3)}.

If f(a) and g(a) are both zero

\lim_{x \to a} \frac{f'(a)(x - a) + \frac{f''(a)}{2}(x-a)^2 + O((x-a)^3)}{g'(a)(x-a) + \frac{g''(a)}{2}(x-a)^2 + O((x-a)^3)},

enzovoort.