AngelofMusic
Apr12-04, 08:03 PM
Hi,
I've got a question about the use of dummy variables in Taylor Series.
We are asked to expand:
g(x) = xlnx
In terms of (x-2). So originally, I used a dummy variable approach to try and find an answer.
Let t = x-2, so x = t+2.
g(x) = (t+2)ln(2+t)
We're given:
ln(1+x) = \sum_{k=1}^{\infty}\frac{(-1)^k}{k}x^k
So, I re-arrange g(x) = (t+2)ln[2(1+t/2)] = (t+2)(ln2 + ln(1+t/2)).
Expanding, I get: g(x) = tln2 + tln(1+t/2) + 2ln2 + 2ln(1+t/2).
So g(x) = 2ln2 + (x-2)ln2 + (x-2)\sum_{k=1}^{\infty}\frac{(-1)^k}{k}(x/2-1)^k + 2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}(x/2-1)^k.
This is where I get stuck because the book's answer says it should be:
g(x) = 2ln2 + (1+ln2)(x-2) + \sum_{k=2}^{\infty}\frac{(-1)^k}{k(k-1)2^{k-1}}(x-2)^k
I'm sort of close to the answer, but can't quite make the final manipulations to make it work. The book used the method of taking successive derivatives to arrive at that formula, so I know how to do it that way - but is my original method correct as well? If not, why not?
I've got a question about the use of dummy variables in Taylor Series.
We are asked to expand:
g(x) = xlnx
In terms of (x-2). So originally, I used a dummy variable approach to try and find an answer.
Let t = x-2, so x = t+2.
g(x) = (t+2)ln(2+t)
We're given:
ln(1+x) = \sum_{k=1}^{\infty}\frac{(-1)^k}{k}x^k
So, I re-arrange g(x) = (t+2)ln[2(1+t/2)] = (t+2)(ln2 + ln(1+t/2)).
Expanding, I get: g(x) = tln2 + tln(1+t/2) + 2ln2 + 2ln(1+t/2).
So g(x) = 2ln2 + (x-2)ln2 + (x-2)\sum_{k=1}^{\infty}\frac{(-1)^k}{k}(x/2-1)^k + 2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}(x/2-1)^k.
This is where I get stuck because the book's answer says it should be:
g(x) = 2ln2 + (1+ln2)(x-2) + \sum_{k=2}^{\infty}\frac{(-1)^k}{k(k-1)2^{k-1}}(x-2)^k
I'm sort of close to the answer, but can't quite make the final manipulations to make it work. The book used the method of taking successive derivatives to arrive at that formula, so I know how to do it that way - but is my original method correct as well? If not, why not?