Power series for ln(x) centered at x=1

[Icedfire01]

I need to know how to get the first 6 terms of the power series for:
f(x)=ln(x) centered at x=1. Thanks

 

[Corneo]

Well

$$f'(x)=\frac {1}{x}$$
$$f''(x)= - \frac {1}{x^2}$$
$$f'''(x)= \frac {2}{x^3}$$
$$f^4(x)=- \frac {6}{x^4}$$
$$f^5(x)=\frac {24}{x^5}$$
$$f^6(x)=- \frac {120}{x^6}$$

so $$f(1)=0, \ f'(1)=1, \ f''(1)=-1 , \ f'''(1)=2, \ f^4(1)=-6 , ...$$

so $$\ln x \approx 0 + (x-1) + (-1) \frac {(x-1)^2}{2} + (2) \frac {(x-1)^3}{3!} + (-6) \frac {(x-1)^4}{4!}+ (24)\frac {(x-1)^5}{5!}$$

 

[Icedfire01]

thanks, and that 2nd term answer isn't supposed to be 2! on the bottom instead of 2 is it?

 

[arildno]

That's correct (even though, of course 2!=2); the full series expansion of ln(x) about
x=1 is
$$ln(x)=\sum_{i=1}^{\infty}(-1)^{i+1}\frac{(x-1)^{i}}{i}$$

 

[mathwonk]

the usual way to do this is to note that 1/(1+x) = 1 - x + x^2 - x^3 + x^4 -+ ...

Then integrate both sides. On the left you get ln(1+x) and on the right you get

well you do it... This will be valid wherever the series on the right converges, i.e. for |x| < 1. but also for x = 1, which yields a nice series for ln(2).