Taylor series for 1/(1-x²)^(1/2) centered at 1

[nameVoid]

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

 

[cronxeh]

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.

 

[nameVoid]

well you for a=1 on D -1<x<1

 

[cronxeh]

well you for a=1 on D -1<x<1

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

What is f(1) ?

 

[nameVoid]

|x|<1

 

[Mark44]

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

 

[cronxeh]

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

Would make sense that it is centered at 0 then

 

[nameVoid]

is the binomial series only way to go here

 

[Mark44]

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! + ...