1. amistre64

since f' = -1/(1-x) we can do long division to get that power series; then integrate it back up to ln(1-x)

2. ns36

Thanks but I still don't quite understand how you get the power series from the derivative of the function :(

3. amistre64

the long division creates a pwer series; an equivalent polynomial. integrateing that takes us back to ln(1-x)

4. bmp

Remember the geometric series?$\frac{1}{1-x} = \sum_{n = 0}^{\infty} x^{n}$for |x| < 1. So we get -(1/(1-x)), that is:$- \sum_{n = 0}^{\infty} x^n = -1 -x -x^2...$Integrate this to get ln(1-x)

5. amistre64

-1-x-x^2-x^3-x^4- .... -------------------- 1-x ) -1 (-1+x) ------ -x (-x+x^2) --------- -x^2 $D[ln(1-x)]=\frac{-1}{1-x}$ $D[ln(1-x)]=-(1+x+x^2+x^3+x^4+...)$ integrate both sides $ln(1-x)=\int -(1+x+x^2+x^3+x^4+...) dx$ $ln(1-x)=-(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+...)$

6. ns36

Thanks! How does one find the radius of convergence for a power series?

7. amistre64

this gives us the summation of ln(1-x) as:$\sum_{n=0}^{inf}\frac{-x^{n+1}}{n+1}$ the radius of convergence is the limit as n goes to inf of the ratio of an+1/an

8. bmp

There is a theorem that states that the derivatives (and integration?) have the same radius of convergence also.

9. amistre64

$lim\frac{a_{n+1}}{a_{n}}:\ \lim_{n\to inf}\frac{-x^{n+1}}{n+1}*\frac{n}{-x^{n}}$

10. amistre64

$lim\frac{-x*n}{n+1}\to\ |-x|\lim\frac{n}{n+1}$ $|-x|*1$ set it up such that: $|-x|<1$and simplify such that |x|<R and since theres nothing to simplify: R = 1

11. amistre64
12. ns36

Hahaha thanks so much for all of your help!

13. ns36

How would I plug in (x-.5) into the power series you found above to express ln2 as a sum of an infinite series?

14. amistre64

when x=-1; you get ln(2) right?

15. amistre64
16. amistre64

i think when we go beyond the radius of convergence is when we get iffy in values