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ns36
 4 years ago
Find a power series representation for f(x) = ln(1x) and find the radius of convergence. Please help!
ns36
 4 years ago
Find a power series representation for f(x) = ln(1x) and find the radius of convergence. Please help!

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3since f' = 1/(1x) we can do long division to get that power series; then integrate it back up to ln(1x)

ns36
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks but I still don't quite understand how you get the power series from the derivative of the function :(

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3the long division creates a pwer series; an equivalent polynomial. integrateing that takes us back to ln(1x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Remember the geometric series?\[ \frac{1}{1x} = \sum_{n = 0}^{\infty} x^{n}\]for x < 1. So we get (1/(1x)), that is:\[ \sum_{n = 0}^{\infty} x^n = 1 x x^2...\]Integrate this to get ln(1x)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.31xx^2x^3x^4 ....  1x ) 1 (1+x)  x (x+x^2)  x^2 \[D[ln(1x)]=\frac{1}{1x}\] \[D[ln(1x)]=(1+x+x^2+x^3+x^4+...)\] integrate both sides \[ln(1x)=\int (1+x+x^2+x^3+x^4+...) dx\] \[ln(1x)=(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+...)\]

ns36
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks! How does one find the radius of convergence for a power series?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3this gives us the summation of ln(1x) 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
 4 years ago
Best ResponseYou've already chosen the best response.0There is a theorem that states that the derivatives (and integration?) have the same radius of convergence also.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3\[lim\frac{a_{n+1}}{a_{n}}:\ \lim_{n\to inf}\frac{x^{n+1}}{n+1}*\frac{n}{x^{n}}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3\[lim\frac{x*n}{n+1}\to\ x\lim\frac{n}{n+1}\] \[x*1\] set it up such that: \[x<1\]and simplify such that x<R and since theres nothing to simplify: R = 1

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3http://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
 4 years ago
Best ResponseYou've already chosen the best response.0Hahaha thanks so much for all of your help!

ns36
 4 years ago
Best ResponseYou've already chosen the best response.0How would I plug in (x.5) into the power series you found above to express ln2 as a sum of an infinite series?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3when x=1; you get ln(2) right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3http://www.wolframalpha.com/input/?i=sum+%281%29%5E%28n%2B1%29%2F%28n%2B1%29%2C+n+%3D+0+to+inf

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3i think when we go beyond the radius of convergence is when we get iffy in values
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