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Taylor series radius of convergence 
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#1
Jul2710, 12:05 AM

P: 9

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 


#2
Jul2710, 07:09 AM

P: 336

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



#3
Jul2710, 11:02 AM

HW Helper
P: 3,352

[tex]\sqrt{1x^2} = \sum_{n=0}^{\infty} \frac{ (2n)! x^{2n}}{4^n(12n)(n!)^2}[/tex]
You may have seen this more easily if you tried to expand [tex]\sqrt{1+t}[/tex] about t=0 then substituted [itex]t=x^2[/itex]. Or, still easier, used the generalized binomial theorem. 


#4
Jul2710, 12:02 PM

P: 9

Taylor series radius of convergence
Some more examples [tex]\log{1x^2}[/tex] [tex]\exp(1x^2)[/tex] 


#5
Jul2710, 04:15 PM

P: 9

I understood it.. 


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