Taylor Series for ln(x+1) at x=1

[Thallen]

Find the 4th term of the Taylor series centerd at x=1 for f(x)=ln(x+1)

f(x)=ln(1=x)
f'(x)=(1+x)^-1
f"(x)=(-1)[(1+x)^-2]
f"'(x)=(2)[(1+x)^-3]
f""(x)=-6[(1+x)^-4)]

Plug in 1:
.6931
.5
-.25
.25
-.375

What do I do next? (Also, is the 4th term the 4th term starting with f(x)? or the 4th derivative?)

 

[phoenixthoth]

it would be easier to start with the series for 1/(1+x) which is close to the geometric series. then integrate that to get the series for ln(1+x). then just look at the correct coefficient.

 

[matt grime]

Don't use decimals, please, your teacher should be happier that way. And the 4'th (non-zero) term is exactly that, write down the taylor series, and pick out the 4th term in it (the x^3 term here).