Let f(x)=ln(1-3x). Find a quadratic polynomial p(x) so that p(0)=f(0), p'(0)=f'(0) and p''(0)=f''(0).

I don't know how to do this :/

2. Originally Posted by Velvet Love
Let f(x)=ln(1-3x). Find a quadratic polynomial p(x) so that p(0)=f(0), p'(0)=f'(0) and p''(0)=f''(0).

I don't know how to do this :/

First a quick note if you know how to find a Taylor series centered at x=0 you are done. Just take the first 3 terms.

So lets suppose you don't so we want a quadratic function

$p(x)=ax^2+bx+c$ and we want

$f(0)=p(0) \iff ln(1)=c \iff c=0$

So now we have $p(x)=ax^2+bx \implies p'(x)=2x+b$

$f'(0)=\frac{-3}{1-3(0)}=2(0)+b=p'(0) \implies b=-3$

so now we have $p''(x)=2a$

$f''(0)=\frac{-9}{(1-3(0))^2}=2a \implies a=-\frac{9}{2}$

So we get

$p(x)=-\frac{9}{2}x^2-3x$

3. Thank you a lot. I've never learned how to do that type of problem. What is that Taylor series?