Help!!

Question

Help!!

dramaqueen1201 · Answer

I have a question that says: when the function
$$f(x) = \frac{ 1 }{ x(1-x) }$$ is expanded in the powers of $$x - \frac{ 2 }{ 3 }$$
what will be the radius of convergence of the resulting power series?

Can someone just explain to me what they are asking exactly? I would like to try and solve the problem by myself first though. Just guide me please? Am I supposed to expand
$$[\frac{ 1 }{ x(1-x) }] ^{x-\frac{ 2 }{ 3 }}$$?

irishboy123 · Answer

dramaqueen

post the original question or link it

if you can :-)

if not, mmmm

dramaqueen1201 · Answer

@IrishBoy123

irishboy123 · Answer

ok

i think they want you to expand this after a linear sub $u = x - \frac{2}{3}$

how long have you got on this?!?!  because i can have a go in a while.  

confession: i have forgotten how to do this, but i can work it out :-)  no problema!

dramaqueen1201 · Answer

it's nothing hurried, i'm practicing for my exam. what are you supposed to do, i could trying doing it too

dramaqueen1201 · Answer

@IrishBoy123 ???

irishboy123 · Answer

Hi @dramaqueen1201 

i had a chance to look at this on the train today.  i can come back in a few hours and latex some stuff

right now, quickly, the awful way to do it is to churn through the Taylor expansion:

$f(x-a) = f(a) + (x-a) f'(a) + ...$ etc but the derivatives are horrible and spotting the pattern just as bad

so i think the way to go is either break the expression up using partial fractions so the deriv's are really easy....

...or to just start with $f(x-3)$ and maybe do a  binomial expansion instead. cos that always easier imho.

dramaqueen1201 · Answer

@IrishBoy123 so if i'm doing this using partial fractions, i get $$\frac{ 1 }{ x(1-x) } = \frac{ 1 }{ x } + \frac{ 1 }{ 1-x }$$ and then i expand 1/x and 1/(1-x) and add them, and then? Put them in series form? I mean with this $$\sum_{?}^{?}$$?

dumbcow · Answer

$$\frac{1}{1-x} = \frac{1}{1-(x-\frac{2}{3})} = \sum_{}^{}(x-\frac{2}{3})^n$$
$$\frac{1}{x} = \frac{1}{x-\frac{2}{3}} = \frac{1}{1-(\frac{5}{3} -x)} = \sum_{}^{}(\frac{5}{3}-x)^n$$
Adding together
$$\sum_{}^{} (x-\frac{2}{3})^n + (\frac{5}{3} - x)^n$$
$$|x- \frac{2}{3}| < 1 AND |\frac{5}{3} -x| < 1$$
$$\frac{2}{3} < x< \frac{5}{3}$$
The radius of convergence is 1

irishboy123 · Answer

FWIW

irishboy123 · Answer

hi @dramaqueen1201 

i think dumbcow has provided the method.  it is a mix of the 2 things i tried but failed to mix.  split out the function via partial fractions - BUT then do a binomial on the sum of those.  

i have deleted my long post but the pdf will hang there as i would like to learn how to get the ratio test going.  but this is definitely an easier way to do it.

Roll with 

$\dfrac{ 1 }{ x(1-x) } = \dfrac{ 1 }{ x } + \dfrac{ 1 }{ 1-x }$

but then try and turn each term on the RHS into a binomial expansion of $\dfrac{1}{1 - u} = \Sigma u^n$ converging for $|u| < 1$ 


it's slightly easier for the expression $\dfrac{1}{1-x}$

because we can say that 

$\dfrac{1}{1-x} $

$= \dfrac{1}{1-x - \frac{2}{3} + \frac{2}{3}}$

$= \dfrac{1}{1-(x - \frac{2}{3}) - \frac{2}{3}}$

$= \dfrac{1}{\frac{1}{3}-(x - \frac{2}{3})}$

$= \dfrac{3}{1-3(x - \frac{2}{3})}$

$=3 \left(  \dfrac{1}{1-3(x - \frac{2}{3})} \right)$

so $u = 3(x - \frac{2}{3})$

this one converges for

$-1 < 3(x - \frac{2}{3}) < 1$

$-\frac{1}{3} < x - \frac{2}{3}< \frac{1}{3}$

$\frac{1}{3} < x  < 1$

hope this is somehow helpful :-)

dramaqueen1201 · Answer

okay. thanks a lot! i'll try this out in a few. thank you so much @IrishBoy123

irishboy123 · Answer

for the second one

$\dfrac{1}{x} = \dfrac{1}{x - \frac{2}{3} + \frac{2}{3}}$



$ = \dfrac{1}{\frac{2}{3} - (\frac{2}{3} - x)}$

$ = \dfrac{\frac{3}{2}}{1 - \frac{3}{2}(\frac{2}{3} - x)}$

thank you so much, learned loads today!  hope this helps in some way!