2007-11-12
Today's nonsense
- 32 hours of labor.
- "Find the one with the best review, the best review, cheapest best reviews. Alright. Shhh. Elven? Thirteen, fourteen, counting- ? I'm saying, counting, but I'm counting everyone that is on the roll. Yeah, we said that about the. Alright, are you ready?"
- Dakota's rear thermometer
Homework due Wednesday: pg. 656 #1-4,13,16,17,19-21,25,28-30
Polynomial Approximations
You did this last year, but only in the form of a linear approximation.
ex 1) For f(x) = e^(x), find a first degree polynomial, in other words you are finding the equation of the tangent line. Sincei t's first degree, we're going to say P_(1) = a_(1)x + a_(0) whose value and slope agree with the value and slope of f at x=0. So check f(0) and now d[f(0)]. So P_(1)(x) = x+1. The slope is 1, so it's 1x + something, but when you plug in zero, you should get one, and so what do you have to do to get the constant, so ... the constant is one.
First degree polynomial = linear. There are no fractional exponents in any polynomial.
Second degree polynomial = quadratic.
2) Find a second degree approximation for f(x)=e^(x), P_(2) = a_(2)x^(2) + a_(1)x + a_(0).
So we want all of these conditions to be the same. So we need f(x), f'(x), and f''(x). We want f(0), f'(0) and f''(0) to all match for e^(x) and our approximation. In the linear approximation they have the same point and the same slope. The second derivatives must equal each other, so 2a_(2) = f''(0). When I plug 0 into P' what does it equal = a_(1), so a_(1) must be 1. If I plug 0 into P, what do you get? a_0 = 1. So P_(2) = (1/2)x^2 + x + 1. So look at the pattern ... the cubic approximation would be (1/6)x^(3) + (1/2)x^2 ... etc. It's a factorial for that first coefficient. 24 = 4!. The exponent matches the 1/the_factorial.
Taylor and Maclaurin Polynomials
Definition of n^th Taylor polynomial and Maclaurin polynomial.
These are polynomials that will approximate a function. There is in fact a pattern. You were totally oblivious to what you were doing, and then at least you will believe that there is a pattern.
If f has n derivatives at c, then the polynomial:
P_(n)(x) = f(c) + f'(c)(x-c) + f''(c) /2! * (x-c)^2 + ... + fn(c)(x-c)^(n) / n!
And this is known as the nth Taylor polynomial for f at c.
If c=0, then it's a Maclaurin polynomial and it reduces down to
P_(n) = f(0) + f'(0)x + f''(0) x^(2) / 2! + f'''(0)x^(3) / 3! + .... + f(n)(0)x_(n) / n!
is called the nth Maclaurin polynomial for f.