Taylor series radius of convergence

[saminny]

Hi,

We need a generic expression of a taylor series nth term to find out the radius of convergence of the series. However, there are series where I don't think it is even possible to find a generic term. How do we find the radius of convergence in such cases?

e.g. sqrt (1 - x^2)

There is no generic nth term..

Please help.

thanks,

Sam

 

[Eynstone]

One can't find the radius of convergence if one can't estimate the nth term.

 

[Gib Z]

\sqrt{1-x^2} = \sum_{n=0}^{\infty} \frac{ (2n)! x^{2n}}{4^n(1-2n)(n!)^2}

You may have seen this more easily if you tried to expand \sqrt{1+t} about t=0 then substituted t=-x^2. Or, still easier, used the generalized binomial theorem.

 

[saminny]

\sqrt{1-x^2} = \sum_{n=0}^{\infty} \frac{ (2n)! x^{2n}}{4^n(1-2n)(n!)^2}

You may have seen this more easily if you tried to expand \sqrt{1+t} about t=0 then substituted t=-x^2. Or, still easier, used the generalized binomial theorem.

eehhh.. how did you get that result? I've been trying to find the nth term for so many hours with no success. If you expand \sqrt{1+t} about t=0, how can you just substitute t for x^2 since that would change the differentiation of each term and effects of that additional 2x will compound as we go higher order. if you could provide some explanation, that would be really helpful. thanks.

Some more examples
\log{1-x^2}
\exp(1-x^2)

 

[saminny]

eehhh.. how did you get that result? I've been trying to find the nth term for so many hours with no success. If you expand \sqrt{1+t} about t=0, how can you just substitute t for x^2 since that would change the differentiation of each term and effects of that additional 2x will compound as we go higher order. if you could provide some explanation, that would be really helpful. thanks.

Some more examples
\log{1-x^2}
\exp(1-x^2)

I understood it..