Dummy Variables in Taylor Series

[AngelofMusic]

Taylor Series in x-a

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:

[tex]ln(1+x) = \sum_{k=1}^{\infty}\frac{(-1)^k}{k}x^k[/tex]

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 [tex]g(x) = 2ln2 + (x-2)ln2 + (x-2)\sum_{k=1}^{\infty}\frac{(-1)^k}{k}(x/2-1)^k + [/tex][tex]2\sum_{k=1}^{\infty}\frac{(-1)^k}{k}(x/2-1)^k.[/tex]

This is where I get stuck because the book's answer says it should be:

[tex]g(x) = 2ln2 + (1+ln2)(x-2) + \sum_{k=2}^{\infty}\frac{(-1)^k}{k(k-1)2^{k-1}}(x-2)^k[/tex]

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?

 

[cookiemonster]

Your method should work out.

Before we work too much on it, why don't you try putting the two sums you have into one and see if it doesn't match up then?

cookiemonster

 

[M98Ranger]

Hi there,

Thank you for sharing your approach to expanding g(x) = xlnx in terms of (x-2) using dummy variables. Your approach is definitely correct and it is a valid way of solving this problem. However, the book's answer uses a different approach which involves taking successive derivatives of g(x) and using the general formula for the Taylor Series expansion of a function.

The reason why your answer might not match exactly with the book's answer is because of the use of dummy variables. When we use dummy variables, we are essentially substituting x with another variable (in this case, t). This can sometimes lead to slight differences in the final answer compared to the traditional approach of taking derivatives and using the general formula.

But don't worry, both approaches are correct and it's always good to have multiple ways of solving a problem. Keep up the good work!