Taylor series of ln(x+2)

1. Oct 28, 2013

zzmanzz

1. The problem statement, all variables and given/known data

Using power series, expand ln(x + 2) about a = 0 (Taylor series)

2. Relevant equations

3. The attempt at a solution

Is this appropriate?

$$ln(x+2) = ln((x+1)+1)$$

$$x' = x+1$$

$$ln(x'+1) = \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x')^{n+1}$$

or

$$ln(x+2) = ln(\frac{x}{2}+1)$$

$$x' = \frac{1}{2}x$$

$$ln(x'+1) = \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x')^{n+1}$$

Last edited: Oct 28, 2013
2. Oct 28, 2013

vanhees71

Well, you rather have
$$\ln(2+x)=\ln[2(1+x/2)].$$

3. Oct 28, 2013

zzmanzz

So would your formula then be:

$$ln2 + ln(1+(x/2)) = ln2 + \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x/2)^{n+1}$$

or

$$ln(2(x/2+1)) = 2*\sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x/2)^{n+1}$$

Last edited: Oct 28, 2013
4. Oct 28, 2013

vanhees71

This you should be able to answer yourself by giving the reasoning behind each of the two claims you made. Of course only one of them can be right!

5. Oct 28, 2013

zzmanzz

I believe that the first one is right because

$$ln(AB) = lnA + lnB$$

$$A = 2, B = (1+x/2)$$

6. Oct 28, 2013

Staff: Mentor

I think you want to do the expansion around x = -1, so that a = 0.

So, the first few terms are: $$ln(x+2) = 0 + (x-1)f'(-1)+...=(x-1)+...$$

7. Oct 28, 2013

vanhees71

It was asked to do the expansion around $0$, and of course the first equation is correct, i.e.,
$$\ln(x+2)=\ln 2 + \ln(1+(x/2)) = \ln 2 + \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}(x/2)^{n+1}.$$
Finally you should think about the domain, where the series is convergent!

8. Oct 28, 2013

Ray Vickson

Why not just use the Maclaurin series formula
$$f(x) = f(0) + f'(0)\,x + f''(0)\, \frac{x^2}{2!} + \cdots \:\:?$$
It is really very simple to use in this case.

9. Oct 28, 2013

zzmanzz

Yes, that's what I had in mind. For the x terms in your expansion. Can I do:

$$f(x) = f(0) + f'(0)(\frac{x}{2}) + f''(0)\frac{1}{2!}(\frac{x}{2})^2 + ...$$

10. Oct 28, 2013

Ray Vickson

Your could write that, but you would be WRONG. The Correct form of the expansion is exactly as I wrote it. Your series above is for f(x/2), not for f(x).

Last edited: Oct 28, 2013
11. Oct 28, 2013

zzmanzz

ok I think I get it now.

I use your mclaurin series to expand the formula..

Then based on the series outcome, we build the sum. So the following sum is still correct:

$$ln2 + \sum_{n=0}^{\inf} \frac{(-1)^n}{n+1}(x/2)^{n+1}$$

12. Oct 28, 2013

Staff: Mentor

Is that what it means when it says "expand ln(x + 2) about a = 0?" I'm not familiar with this type of terminology. What the heck is "a"?

Chet

13. Oct 28, 2013

Ray Vickson

My guess would be that it was a typo, and should have said "...about x = 0".