2007-11-16
Today's nonsense
- Homework due Monday and Tuesday.
Power Series
Series from n=0 to infinity of a_(n) * x^(n) = a_(0) + a_(1) * x + a_(2) * x^(2) + ... a_(n) * x^(n) ...
is called a power series. More generally,
Series from n=0 to infinity of a_(n) * (x-c)^(n) = a_(0) + a_(1)(x-c) + a_(2)(x-c)^(2)+ ... + a_n(x-c)^n + ...
Which is called a power series centered at c. So now do we have a Taylor power series and the Maclaurin power series? Yes.
ex. a) Series from n=0 to infinity of (x^(n)) / n! = 1 + x + (x^(2))/2! + (x^(3))/3! + (x^(n))/n! + ....
is a power series centered at 0.
b) Series from n=0 to infinity of (-1)^(n)(x+1)^(n) = 1 - (x+1) + (x+1)^(2) - (x+1)^(3) + .....
is a power series centered at -1. It's what makes the stuff inside the parenthesis (x+1) to be zero. So it's -1.
c) Series from n=1 to infinity of (1/n) (x-1)^(n) = (x-1) + (1/2)(x-1)^(2) + (1/3)(x-1)^(3) + ....
is a power series centered at 1.
Radius and Interval of Convergence
For what values of x does this series converge? X over 2 to the n. Clearly, x/2 would be my r, and the absolute value of r has to be less than one, so the absolute value of x/2 is less than one, therefore the absolute value of x is less than two. Anyway, you did something like that, whether you know what I am talking about or not. Geometric series, indeed, the absolute value of r has to be less than one. You either didn't do this or got these problems from somebody or you didn't do evens, or you know what I am talking about, etc. What values of x work for series? You are given a series, and you want to know for what values of x does the series converge? Well, that's our interval of convergence - where it's centered, and then a certain distance from the center. If it's centered at 3 and the radius is 2, then it's 1 to 5 for the interval. You have to plug in the end points to figure out if the end points are continuous or not, etc. You have to know where it is centered before you go far from the left or far from the right, first.
The three cases of solutions for Radius and Interval of Convergence:
- 1. Where does this thing converge? It could be that it converges at a single point and what is that single point? The center, c. Well, it if only converges at a single point, it must be at the center. The radius is zero. Zero to the left, zero to the right -> that's your solution set.
- 2. An interval, so on an interval, you're at your center, and you are going to go a certain distance to your left and a certain distance to your right, how far to the left? So it's either (c-r) or (c+r). Your radius is r.
- 3. The real line Your radius is = infinite.
Finding the Radius of Convergence
ex 1) Series from n=0 to infinity of n!x^(n)
Use the ratio test. Lim as n->infinity of |(n+1)!x^(n+1) / (n! x^(n)) | = lim as n->infinity of |(n+1)x| = infinity > 1 ...
We're looking for all the values of x that will make this thing converge. Now, in order to converge, the limit has to be less than one. The ratio test says that the ratio has to be less than one. So, nothing will make this, x will not change this limit, unless x was zero. Then, of course, zero is less than one. So, this series will never converge unless x=0, then the series will be nothing but a summation of zeroes. So this is a single-pointer and therefore R=0. c=0. Your solution set is just the number zero.
ex 2) Series from n=0 to infinity of 3(x-2)^(n)
Use the ratio test again. You get x-2. So. So. This thing, what are the conditions in order for this thing to converge, it has to be less than one. So |x-2| < 1. So c=2, and R=1. Think about the numbers that you can plug into this, okay, alright, keep in mind the absolute value is distance from so you're saying it's centered at two, and this is saying that your distance from has to be less than one, so I can go anywhere up to three and anywhere back to one, and that number there is going to be my radius, so then you're saying that your radius is always going to be one or infinity, and the answer is no, but if you said I have a fraction, suppose I did the, suppose I did the ratio test and got this for my answer | (x-3)/5 | if I got this for my answer, where is it centered? It's centered at three, now what's the radius? Well, this still has to be less than one (the whole thing), for the ratio test this is given, but I want to manipulate this thing so that I just have my |x-c| here .. so I want |x-c| < so then I would multiply by five, and my raidus would be five in this case (|x-3|) and so what would my interval be? It would be centered at three, and five in each direction, so the interval is -2 to 8. Any number in that interval, well, anything in there would make it converge. Everything else will diverge. Then we have to plug in the endstops of the interval to figure out the convergence/divergence at those points.
3) Series from n=0 to infinity of (-1)^(n) * x^(2n+1) / (2n+1)! ....
Use the ratio test. R=infinity.
Endpoint Convergence
ex) Find the interval of convergence
Series from n=1 to infinity of x^(n) / n .......
This is where we are going to have to discuss the endpoints, whether or not the interval is opened or closed. You get "x" when you do this.
And what happens when you have the radius of convergence in two dimensions, instead of just one 'c' and 'R' value on the horizontal number line?