1. Sep 22, 2009

### Physics197

If you had something like ln(x+1) is there are way of breaking that term up?

I have a question where I have to find a two term approximation and i'm at a stage similar to this one and I don't know what I can do.

2. Sep 22, 2009

### mathman

What do you mean by breaking it up? There is a power series for ln(1+x)

x-x2/2+x3/3+....

3. Sep 22, 2009

### Physics197

I have y=ln(3u+4) and need a 2 term approximation (we are given a hint 4 = 3+1) so I assume that we must do: y=ln(3u+1+3) and the only way we know to approximate is y=ln(1+u) ~ u so in order to get that I need to somehow get rid of the 3 in y=ln(3u+1+3)

4. Sep 22, 2009

### lurflurf

What kind of approximation.
We know log(x+1)~x
so expansion about x+1=1->x=0 would be nice
let f(x)=log(x+1)
f(x)~f(0)+f'(0)x+f''(0)x^2/2+f''(0)x^3/6+...

5. Sep 22, 2009

### Physics197

The entire question is:

Find a two term approximation for the following function near t=1.

h = 5ln(3t+1)
and after substituting u+1 for t

I get:

h = 5ln(3u+4)
Now it gives us the hint: 4 = 3+1

But im stuck after this.

6. Sep 23, 2009

### mathman

Let v=3u+1, so h=5ln(1+v)~5[v - v2/2 + ...]
Take it from there.