2007-11-20 - Geometric Power Series
Today's nonsense
- You must toss the ball to talk. Alyssa is out.
Consider f(x) = 1 / (1-x) . This closely resembles the sum of a geometric series. Series from n=0 to infinity of ar^(n) , for |r| < 1
In other words,
1 / (1-x) = Series from n=0 to infinity of x^(n) in other words, 1 times r^n but that would be 1 times x^n, so this would all equal == 1 + x + x^2 + x^3 + ....
We are going to be able to take these things that are functions that can be written in the form of a / 1 -r and we can write it as a power series, pretty simple.
This is a power series centered at zero. (Thank you, Avery, and Bobby doesn't even know what the question is.)
What about a power series centered at 1 or -1?
Using f(x) = 1 / (1-x), find a power series centered at -1.
1 / (1-x) = something over -(x+1) ......... But now, do you understand that there can't just be an arbitrary +1, so it's like completing the square, in the sense that if you add a number in there you also have to subtract a number. So it's a 2-(x+1) .......... 1 / ( 2 - (x+1) )
= Series from n=0 to infinity of 0.5 * ((x+1)/2)^(n)
= Series from n=0 to infinity of (x+1)^(n) / 2^(n+1)
You want the form a / 1-r
So we are just going to be practicing our manipulation techniques.
Due 2007-11-27: pg. 674 #2-6,8,10,11,13,16,19-21,23,25
ex) Find a power series for f(x) = 4 / (x+2)
4 / (2+x) == 2 / (1 + x/2) == 2 / ( 1 - (-x/2) ) ==
Series from n=0 to infinity of 2(-x/2)^n