2007-11-14

Today's nonsense
- Laptop's green screen of death.
- "You need to be like a Jerry Springer show because you're just great at pissing people off." - Phoebe

Taylor's theorem

If a function f is differentiable through order n+1 in an interval I containing c, then for each x in I there exists Z between x and c such that: f(x) = f(c) + f'(c)(x-c) + f''(c)(x-c)^2 / 2! + ... + f^n of (c) (x-c)^n / n! + R(c)

where R_n (x) = f^(n+1) of(z) (x-c)^(n+1) / (n+1)! z falls somewhere between x and c

The remainder in this theorem is the Lagrange term of the remainder.

Remainder of a Taylor Polynomial

f(x) = P_{_(n)}(x) also known as the approximate value + R_{_n} (x) also known as the remainder = The exact value.
Error = |R_{_n} (x)| = |f(x) - P_{_n}|

How can we estimate the remainder associated with a Taylor polynomial?

Use Taylor's thorem to determine the accuracy of the approximation.
1) e is approximately .. well e^x is 1 + x + x^2 / 2! + x^3 / 3! ...
1 + 1 + 1^2 / 2! + 1^3 / 3! + 1^3 / 3! + 1^4 / 4! + 1^5 / 5! and if all of these are added, how accurate is this approximation? You have to consider the sixth. So what is the sixth derivative?
f(x) = e^x
f''''''(x) = e^x
R₅ < = (the worst possible scenario) is going to be e^z * (1)^6 / 6!.
So then the question comes in: what is e^z? What are my possible values for z?
0 < = z ,< = 1
Worst case scenario of e^z is going to be e ...