How would one go about expanding $f(x) = \sqrt{x^2-y^2}$ about $x=y$? You can't do a Taylor series, because $f'(x=y)$ diverges. I tried a few other things, but to no avail. Interestingly enough, Mathematica gives the following expression:
$f(x) = \sqrt{2y(x-y)} + \frac{(x-y)^{\frac{3}{2}}}{2 \sqrt{2y}} + ...$
Can anyone figure out how to produce this result?
Fix y. Write: x^2 - y^2 = (x - y)(x + y) = (x - y)((x- y) + 2y). Let t = x - y, and keep the factor (x - y)^1/2 = t^1/2, and expand the Taylor series for g(t) = (t + 2y)^1/2 at t = 0, then replace t by x - y to get the result.