Taylor Series Expansion as x→∞?

Question

I know that the Taylor Series formula is:
f(x) = Σ[n=0,∞] fⁿ(x₀) * (x - x₀)ⁿ / n!

However, what if I want to take the Taylor Series as x tends towards infinity? ∴ x₀ = ∞, and this formula becomes nonsensical. Is there a specific formula to perform this operation?

Example:
I have the function:
f(x) = 1/√[1 - 1/(4x²)]
When I ask Wolfram for the Taylor Series, it gives me:
1 + 1/(8x²) + 3/(128x^4) + ...

[Source: http://www.wolframalpha.com/input/?i=taylor+series+%281+-+1%2F%284x%5E2%29%29%5E%28-1%2F2%29 ]

How did they get this answer?

Alfredo Kraus · Accepted Answer

Simple.  Just put y=1/x^2 and you have
f(y) = (1 - (y/4)) ^(-1/2) = 1 - (-1/2)(y/4) + (1/2!) (-1/2)( (-1/2) -1) (y/4)^2 + ...
= 1 + (1/(8x^2)) + (1/(128x^4)) + ...

Taylor Series Expansion as x→∞?

1 Answer