2007-11-26
Today's nonsense
- Ketteman's dough "has to thaw"
- Calculus and physics homework due tomorrow.
- Math club pictures on Thursday during second period.
- Spiller is against croutons in salads because you can't fork them. If you're going to throw bread into a salad you might as well eat a sandwhich.
- The next candy-prize Ketteman gets goes to charity (a.k.a. other members of the class) because he has (today) commited a form of candy-abuse, the worst of its kind, a hideous, horrible crime.
ex 2) Find the (he spoke "the" but didn't write anything there) power series for f(x) = 1/x centered at 1. We're trying to get to the form of a/1-r.
Centered at 1 means that there is the need for the term of (x-1). So, we're starting with 1/x, but we want that to be an x-1 in the denominator. You can't just do that in the denominator, without putting a +1, so it looks like 1/x = 1 / (1 + (x-1)) ... and now we just need that to be a minus sign and so : = 1 / ( 1 - (-x(-1) ) ) ... and so then a=1 and r = -(x-1)
The answer is basically going to look like:
Series from n=0 to infinity of ar^n
Series from n=0 to infinity of 1[-(x-1)]^n = < ---------- so that it is in the form that we want
Series from n=0 to infinity of (-1)^n * (x-1)^n
ex 3) Find a power series centered at 0 for f(x) = (3x-1) / (x^2 - 1)
Partial fractions.
Perhaps maybe possibly you thought thery had gone away, they haven't gone anywhere - partial fractions, sum of the power series, break them up, you see, people love these, except Bill who doesn't know what's going on,
3x - 1 = A(x-1) + B(x+1)
A=2 and B=1
Now we have: 2/(x+1) + 1/(x-1) and each one of these are going to be in the form of a/(1-r) in the end ...
2 / (1+x) + 1 / (-1+x)
Now remember, that has to be a one in the denominator, so all you have to do here is right this as 1 - a negative.
2 / (1-(-x)) + -1/(1-x) (you do this by distributing a negative)
What are the two power series?
Series from n=0 to infinity of 2(-x)^n + Series from n=0 to infinity of (-1)x^n
= Series from n=0 to infinity of [2(-1)^n - 1]x^n
If you wrote out some terms ...
Interval of convergence, could we do it? Perhaps. Somewhat chance of. Interval of convergence is much easier than you might think. Do the ratio test, first. So here your r is x, r has to be less than one in order to converge, therefore your interval convergence is going to be -1 to 1 and it's not going to include the end points because in a geometric series the end points wouldn't matter because of the definition for the values for which r makes the series converge. So this is always going to be parenthesis, parenthesis, a closed interval (not an inclusive interval). The end points are not going to be included.
ex 4) Find power series for f(x) = lnx centered at 1.
Take the derivative.
f'(x) = 1/x = Series from n=0 to infinity of (-1)^n * (x-1)^n because we just did this today, a couple of examples ago. So what do we have to do to f'(x) to get back to f(x)? Take the integral.
ln x = Integral of (1/x)dx = Integral of Series from n=0 to infinity of (-1)^n * (x-1)^n dx
ln x = c + Series from n=0 to infinity of (-1)^n * (x-1)^(n+1) all over (n+1)
"Ceeriously." - Bill
Let's find c. When x=1, ln 1 = c + Series of 0
Therefore ln x = Series from n=0 to infinity of (-1)^n * (x-1)^(n+1) all over (n+1)
So in these cases, c=0
(0,2) <------- interval of convergence. It's centered at 1.
ex 5) Find power series f(x) = 1 / (2x-5), c= -3
1 / (2x-5) = 1 / (2(x+3)-5 - 6 (because we have to keep it the same as before)) =
1 / (2(x+3)-11)
-1 / (1-2(x+3))
= (-1/11) / (1-(2/11)(x+3))
a = -1/11
r = (2/11)(x+3)
Series from n=0 to infinity of -1/11 * ( (2/11) ( x+3) )^n = Series from n=0 to infinity of - (2^n) / 11^(n+1) * (x+3)^n
Radius of convergence = R = 11/2 because remember: | (2/11)(x+3) | < 1
|x+3| < 11/2
c = -3
Interval of convergence? (-17/2, 5/2) and it's closed because it's geometric
ex 6) Given 1 / (1+x) = Series from n=0 to infinity of (-1)^n x^n
Find (-1) / (x+1)^2 ...
Well. That's the derivative of your given. So the derivative of the series of the original will be the series that you need to find.
Series from n=1 to infinity of (-1)^n n x^(n-1) ... but the first term of the original series was 1. So the derivative is going to mean that you're going to have one less term, so you'll start at n=1 for your derivative. Every time you take a derivative ... all you do is plug in the first term, if you plug it in and you get a constant, you have to go the next term on the derivative, but if you have a variable then you're fine.