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anonymous
 4 years ago
find the radius of convergence of the taylor series around x=0 for ln(1/(1+2x))
anonymous
 4 years ago
find the radius of convergence of the taylor series around x=0 for ln(1/(1+2x))

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i spose we would nee to generate the power series for it first

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[[ln\frac{1}{1+2x}]'=\frac{1/(1+2x)'}{1/(1+2x)}\] \[[ln\frac{1}{1+2x}]'=\frac{2/(1+2x)^2}{1/(1+2x)}\] \[[ln\frac{1}{1+2x}]'=\frac{2/(1+2x)}{1/1}\to\ \frac{2}{1+2x}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1lol ln(1+2x) = 2/(1+2x)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.12+4x8x^2+16x^332x^4+64x^5 ...  1+2x ) 2 (24x) 4x (4x+8x^2) 8x^2 (8x^216x^3)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1somewhere i had seen something like this integration dln(1+2x)/dx and expand it as power series.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1soo, a relevant power series would be:\[ln(\frac{1}{1+2x})=\int\sum (1)^{n+1}\ 2^{2n}x^{n}dx\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[ln(\frac{1}{1+2x})=\sum_{0}^{inf}\frac{(1)^{n+1}}{n+1}x^{n+1}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[ln(\frac{1}{1+2x})=\sum_{0}^{inf}\frac{(1)^{n+1}2^{n+1}}{n+1}x^{n+1}\] \[lim\frac{(1)^{n+1}2^{n+1}x^{n+1}}{n+1}\frac{n}{(1)^{n}2^{n}x^{n}}\] \[lim\frac{2x}{n+1}\frac{n}{1}\] \[xlim\frac{2n}{n+1} = 2\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1prolly shoulda taken the  with the x :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1all non  ns should be vacated \[2x\ lim\frac{n}{n+1}=2x\] \[2x<1;\ x<\frac{1}{2}\]
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