Question

When taking the quadratic approximation of the product of 2 functions(as in (e^(-3x))(1+x)^(-1/2) ) the third power or higher terms can be canceled out because when you take the the quadratic approximation of the product of the functions you take the quadratic approximation of the product of the of the quadratic approximations of the 2 functions(Q(f(x)g(x)) = Q(Q(f(x))Q(g(x))) ). But this is only true when you take the approximation near x=0. If you were to take the quadratic approximation near, say, x=1 all terms would stay, correct?

Answer 1

I may not be following you, but if you create a linear approximation (for example), you toss all terms higher than order 1. And if you create a quadratic approximation, you toss all terms order 3 and higher. It does not matter where your starting point is (i.e. x=0 vice x=1)

the difference between using 0 or 1 (in a linear approximation) is this:

f(x) = f(0) + f'(0) x <-- near 0
f(x) = f(1) + f'(1) (x-1) <-- near 1

we use (x-1) rather than x to measure how far we are from the "center point"

Answer 2

a quadratic approximation around x=a would be
f(x) = f(a) + f'(a)(x-a) +1/2 f''(a)(x-a)^2