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'Hôpital'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.