MAT187 L06 - Approximating Errors

#MAT187 PCE

Remainder is the difference between the actual value of the function and its approximation.
$R_{n} (x_{}) = f (x) - P_{n} (x)$
Note that the remainder is signed, so it can be either positive or negative at a point.

Error is the absolute value of the remainder.
$E r r o r = | R_{n} (x) | = | f (x) - P_{n} (x) |$

Taylor's Theorem
If you choose a centre at $x = a$ to approximate a function that has $(n + 1)$ continuous derivatives on an interval $I$ close to $x = a$ , then:
$f (x) = p_{n} (x) + R_{n} (x)$ , $\forall x ϵ I$
...where $p_{n} (x)$ is the $n^{t h}$ Taylor polynomial centred at $x = a$ ,
and $R_{n} (x) = \frac{f^{(n + 1)} (c)}{(n + 1)!} (x - a)^{n + 1}$

For the error approximation:

$c$ is some value between $a$ and $x$ that leads to the error approximation equaling the actual error exactly. It's a black box, but we can use it to find upper and lower bounds for the error.
$f^{(n + 1)} (c)$ is the $(n + 1)^{t h}$ derivative of the function at $c$ .

Example: Error of $e^{x}$ .
For the third degree Taylor polynomial of $e^{x}$ , centered at $x = 0$ , which is:
$P_{3} (x) = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3!}$
...we can try to estimate the error $R_{3} (x)$ using:
$R_{3} (x) = \frac{f^{(3 + 1)} (c)}{(3 + 1)!} (x - 0)^{(3 + 1)} = \frac{f^{(4)} (c)}{4!} x^{(4)}$
We don't know what $c$ is, but we know that it is bounded by $0$ and $x$ .
So, let's say that we want to approximate the error at $x = 0.9$ .
Since $0 \leq c \leq x$ , we know that $0 \leq c \leq 0.9$ .
And, since $e^{x}$ is always increasing, $f^{4} (c) = e^{c} = e^{0.9}$ must result in a "worst case scenario" for $R_{3} (x)$ .
i.e. $M a x (R_{3} (0.9)) = \frac{e^{0.9}}{4!} (0.9)^{(4)}$