Is this a correct taylor series representation centered at 1

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Homework Help Overview

The discussion revolves around the Taylor series representation of the function f(x) = 1/(1-x^2)^(1/2). Participants are examining whether the series is correctly centered at 1 and exploring the implications of this choice given the function's domain.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants question the validity of centering the Taylor series at 1, noting that 1 may not be within the function's domain. There are discussions about the possibility of centering at 0 instead and considerations of using the binomial series or Maclaurin series for expansion.

Discussion Status

The discussion is active, with participants raising questions about the assumptions made regarding the center of the series. Some have suggested alternative centers and approaches, indicating a productive exploration of the problem.

Contextual Notes

There is a focus on the domain of the function, particularly the implications of evaluating at x=1, which is outside the interval of convergence for the series. Participants are considering the constraints imposed by the function's definition.

nameVoid
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f(x)=1/(1-x^2)^(1/2)
1/x^(1/2)=1+ sum(( (-1)^n 1*3*5*7...(2n-1)(x-1)^n )/(2^n n! ) , n=1, infty )
thus 1/(1-x^2)^(1/2) = 1+ sum(( 1*3*5*7...(2n-1)(x^2)^n )/(2^n n! ) , n=1, infty )
is this a correct taylor series representation centered at 1
 
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nameVoid said:
f(x)=1/(1-x^2)^(1/2)
1/x^(1/2)=1+ sum(( (-1)^n 1*3*5*7...(2n-1)(x-1)^n )/(2^n n! ) , n, infty )
thus 1/(1-x^2)^(1/2) = 1+ sum(( 1*3*5*7...(2n-1)(x^2)^n )/(2^n n! ) , n, infty )
is this a correct taylor series representation centered at 1

Are you sure its supposed to be centered at 1? And you are sure its 1/sqrt(1-x^2)? The answer you seek is in the imaginary domain.
 
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well you for a=1 on D -1<x<1
 


nameVoid said:
well you for a=1 on D -1<x<1

So if f(x) = 1/sqrt(1-x^2)

What is f(1) ?
 


|x|<1
 


So it can't be centered at 1, a number not in the domain.
 


Mark44 said:
So it can't be centered at 1, a number not in the domain.

Would make sense that it is centered at 0 then :biggrin:
 


is the binomial series only way to go here
 


No, but it is probably easier. Another approach is using the definition of the Maclaurin series (since you are expanding in powers of x): f(x) = f(0) + f'(0)x + f''(0)x^2/2! + ...
 

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