Taylor series (very easy but have a problem)

[FermatPell]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

 

[gb7nash]

Hi Fermatpall,

Try factoring x^2+x−6 instead. So:

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.

 

[I like Serena]

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

 

[gb7nash]

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

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