Find a power series representation for f(x) = ln(1-x) and find the radius of convergence. Please help!

Question

amistre64 · Answer

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

ns36 · Answer

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

amistre64 · Answer

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

anonymous · Answer

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)

amistre64 · Answer

-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
$$ln(1-x)=\int -(1+x+x^2+x^3+x^4+...) dx$$
$$ln(1-x)=-(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+...)$$

ns36 · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

http://www.wolframalpha.com/input/?i=sum+-x%5E%28n%2B1%29%2F%28n%2B1%29%2C+n+%3D+0+to+inf the wolf agree

ns36 · Answer

Hahaha thanks so much for all of your help!

ns36 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

http://www.wolframalpha.com/input/?i=sum+-%28-1%29%5E%28n%2B1%29%2F%28n%2B1%29%2C+n+%3D+0+to+inf

amistre64 · Answer

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