Power Series Problem?
Find a power series representation for the function and determine the interval of convergence.
a) f(x) = (1+x)/(1-x)
b) f(x) = ln(5-x)
OK so I understand the gist of how to do these problems however I just can't seem to figure out the algebra that goes with a and b. I know a) comes out as the (1+2) * (sum from n=1 to infinity of x^n) with the interval of convergence as (-1,1). Could someone please help me out?
Comments
Summary of what appears below the first line of asterisks:
(a) For this one, I present an argument not dependent on the classic Taylor series method, but valid all the same.
(b) For this one, I attack through the classic Taylor series method. It turns out the answer I came up with was lacking a factor of (-1)^(n) that it needed -- this becomes obvious when I try plugging in an actual x value to see if the series converges to the "right" value of the function. So I get the right series in the end, but I leave it to you to find out where I messed up the minus signs in the derivation.
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For (a), I'm going to start with a well-known result for the sum of a geometric series. E.g., in the Wikipedia article entitled "geometric series," it is shown that the sum from n=0 to infinity of x^n is 1/(1-x). Evidently, then, the sum from n=1 to infinity of x^n must be 1/(1-x) - 1 = 1/(1-x) - (1-x)/(1-x) = x/(1-x).
Also note that (1+x)/(1-x) = (1-x)/(1-x) + (2x)/(1-x). Hence,
(1+x)/(1-x) = 1 + 2[ x/(1-x) ],
but we showed above that x/(1-x) = sum from 1 to inf of x^n.
As for the radius of convergence, it should be obvious that the series can't converge if |x| is 1 or more, as the terms won't shrink
(b) This one I'll attempt in the standard way, which is to find the first several derivatives of the given function.
f(x) = ln(5-x)
f'(x) = -1/(5-x) or -(5-x)^(-1)
f"(x) = -(5-x)^(-2)
f'''(x) = -2(5-x)^(-3)
f""(x) = -6(5-x)^(-4)
The nth derivative will be
-(5-x)^(-n) (n-1)!
The general formula for a Taylor series in
the neighborhood of x=a is
f(x) = f(a) + sum from n = 1 to infinity of
f^(n)(a) * (x-a)^n * n!
Let's try the neighborhood of x=0:
f(x) = ln(5-x) =
= ln(5) + the summation from n=1 to infinity of
-(5-x)^(-n)*(n-1)!*x^n/n!
= ln(5) + the summation from n=1 to infinity of
- [x/(5-x)]^n / n
To see if this answer makes any sense, let's see what happens at x=1. Here we should have
ln(4) = ln(5) + summation from n=1 to infinity of
-(1/4)^n / n
ln(4) = ln(5) - 1/4 - 1/32 - 1/192 - 1/1024 - 1/5120 - 1/24576 -...
The decimal value of ln(4) is about 1.386, and that of ln(5) is about 1.609. Apparently 1.609 - 0.25 is already less than 1.386; but I have an idea that the ALTERNATING series might converge properly:
ln(4) = ln(5) - 1/4 + 1/32 - 1/192 + 1/1024 - 1/5120 + 1/24576 -...
Is it true that 1.3863 is about equal to
1.6094 - 0.25 + 0.03125 - 1/192 + 1/1024 - 1/5120 etc?
YES!!!!
OK, so there is some mistake about minus signs, in my work, but the series you want is:
ln(5) + summation from 1 to infinity of
{ (-1)^(n) * [x/(5-x)]^n * (1/n) }
This will converge in some neighborhood of x=0. It's obvious the radius of convergence is no more than 5/2, because that would make x and 5-x equal, but you may have to do some fiddling around to find out whether it's that large.