# Taylor series (very easy but have a problem)

1. Jun 24, 2011

### FermatPell

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

series expansion at $$c=2$$ of $$ln(x^2+x-6)$$

2. Relevant equations

3. The attempt at a solution

After substituting $$y= x -2$$ we get ln(y^2+5y) = ln(y) + ln(y+5) but im not kinda sure how to use the taylor series of ln(1+x)...

2. Jun 24, 2011

### gb7nash

Hi Fermatpall,

ln(x^2+x−6) = ln((x+3)(x-2)) = ln(x+3) + ln(x-2)

The problem tells us to expand the taylor series around c = 2. The formula we need is:

$$\sum_{n = 0} ^ {\infty} \frac{f^{(n)}(c)}{n!} \, (x - c)^{n}$$

Let's crank out some derivatives then:

$$f'(x) = \frac{1}{x+3} + \frac{1}{x-2}$$
.
.
.

Calculate a few more derivatives and you'll find a pattern. After this is done, throw your result into the formula.

3. Jun 24, 2011

### I like Serena

Hmm, I don't think there is a Taylor expansion around x=2, since $\ln(2^2 + 2 -6)=-\infty$.

4. Jun 24, 2011

### gb7nash

Good call. I didn't check to see if the point was even valid.