Taylor series radius of convergence

In summary: Expand \sqrt{1+t} about t=0 to get the result. Substituting t=-x^2 did not work for me because it changed the differentiation of each term and effects of that additional 2x would compound as we went higher order. Thanks for explaining it.
  • #1
saminny
9
0
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
 
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  • #2
One can't find the radius of convergence if one can't estimate the nth term.
 
  • #3
[tex]\sqrt{1-x^2} = \sum_{n=0}^{\infty} \frac{ (2n)! x^{2n}}{4^n(1-2n)(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
Gib Z said:
[tex]\sqrt{1-x^2} = \sum_{n=0}^{\infty} \frac{ (2n)! x^{2n}}{4^n(1-2n)(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.

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 [itex]\sqrt{1+t}[/itex] 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
[tex]\log{1-x^2}[/tex]
[tex]\exp(1-x^2)[/tex]
 
  • #5
saminny said:
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 [itex]\sqrt{1+t}[/itex] 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
[tex]\log{1-x^2}[/tex]
[tex]\exp(1-x^2)[/tex]


I understood it..
 

What is a Taylor series radius of convergence?

A Taylor series radius of convergence is a mathematical concept that determines the interval of values for which the Taylor series of a function will converge, or approach the correct value of the function. It is represented by a radius, which can be positive or zero, and indicates the distance from the center of the series where the function will converge.

How is the radius of convergence calculated?

The radius of convergence is typically calculated using the Ratio Test, which involves taking the limit of the absolute value of the ratio of consecutive terms in the Taylor series. If this limit is less than one, the series will converge, and the radius of convergence can be calculated by taking the reciprocal of the limit.

What does it mean when the radius of convergence is infinite?

When the radius of convergence is infinite, it means that the Taylor series will converge for all values of the variable. This usually occurs when the function being approximated by the series is an entire function, meaning it is analytic everywhere.

Can the radius of convergence be negative?

No, the radius of convergence cannot be negative. The radius is always represented by a positive value, as it measures the distance from the center of the series where the function will converge. If the radius is negative, it is simply interpreted as zero.

Why is the radius of convergence important?

The radius of convergence is important because it determines the validity of the Taylor series approximation. If the value of the variable falls within the radius of convergence, the series will converge and provide an accurate approximation of the function. However, if the value falls outside the radius, the series will diverge and the approximation will not be valid.

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