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ns36
Find a power series representation for f(x) = ln(1-x) and find the radius of convergence. Please help!
since f' = -1/(1-x) we can do long division to get that power series; then integrate it back up to ln(1-x)
Thanks but I still don't quite understand how you get the power series from the derivative of the function :(
the long division creates a pwer series; an equivalent polynomial. integrateing that takes us back to ln(1-x)
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)
-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 \[ln(1-x)=\int -(1+x+x^2+x^3+x^4+...) dx\] \[ln(1-x)=-(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+...)\]
Thanks! How does one find the radius of convergence for a power series?
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
There is a theorem that states that the derivatives (and integration?) have the same radius of convergence also.
\[lim\frac{a_{n+1}}{a_{n}}:\ \lim_{n\to inf}\frac{-x^{n+1}}{n+1}*\frac{n}{-x^{n}}\]
\[lim\frac{-x*n}{n+1}\to\ |-x|\lim\frac{n}{n+1}\] \[|-x|*1\] set it up such that: \[|-x|<1\]and simplify such that |x|<R and since theres nothing to simplify: R = 1
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
Hahaha thanks so much for all of your help!
How would I plug in (x-.5) into the power series you found above to express ln2 as a sum of an infinite series?
when x=-1; you get ln(2) right?
http://www.wolframalpha.com/input/?i=sum+-%28-1%29%5E%28n%2B1%29%2F%28n%2B1%29%2C+n+%3D+0+to+inf
i think when we go beyond the radius of convergence is when we get iffy in values