1. ## Approximation of ln

My question is about e) $\displaystyle x*ln(x)$. So to find the approximation. It is said:

$\displaystyle x*ln(x) = (1+h)ln(1+h) \approx (1+h)(h-\frac{h^2}{2}) \approx h + \frac{h^2}{2}$

My question is about $\displaystyle (h-\frac{h^2}{2})$. I thought I studied my approximations manual fairly in depth however I don't recall if anything was said about ln or anything about $\displaystyle (h-\frac{h^2}{2})$? I am wondering why ln(1+h) seems to be approximately equal to $\displaystyle (h-\frac{h^2}{2})$ or should I just accept that it is?

Thanks for any responses...

2. ## Re: Approximation of ln

Hey sepoto.

What do you have for your Taylor series approximation (up to quadratic term) for ln(1+x)?

[deleted]

4. ## Re: Approximation of ln

I am trying to solve the problem without using the Taylor series because I just have not gotten there yet. Is (h-h^2/2) the quadratic approximation of ln(1+h) perhaps? The use of ln is throwing me off I think.

P.S. I dug this out of the lecture notes. I think this is the identity that I was looking for:

I am trying to get to the Taylor Series but I have so much to do still.