Try to expand your series expansion at (3,5) and see if you get the original f(x,y)!
i) Derive the two variable second order Taylor series approximation,
below, to f(x,y) = x3 + y3 – 5xy centred at (a,b) = (3,5).
(ii) Calculate and state this approximate value at (x,y) = (5,7).
(iii) Calculated and state the actual value of f(x,y) at (5,7).
(iv) Calculated and state the error, Q(x,y) - f(x,y) at (5,7).
This is a simple one,
i've managed to work out part (i) and (ii) , provided my working isn't wrong.
anyway, i'll show you what i've done.
for (a,b) = (3,5) , f(x,y) = 77+2(x-3) + 60(y-5) + 1/2!(18)(x-3)^2 + 1/2!(22)(x-3)(y-5) + 1/2!(70)(x-3)(y-5) + 1/2!(30)(y-5)^2 .....
so before I move on, I need to check if part(ii) has the same procedure as part(i)?? they changed the variable from (a,b) to (x,y) so do I still just sub in x,y = 5,7 like how i subbed in a,b = 3,5 for part i?
part (iii) whats the difference between an actual and approximate value and how do u go about calculating it? does it have anything to do with truncation errors?
(iv) i guess you just subtract part (iii) and (ii)
I meant have you expanded, factored, and simplified this:
f(x,y) = 77+2(x-3) + 60(y-5) + 1/2!(18)(x-3)^2 + 1/2!(22)(x-3)(y-5) + 1/2!(70)(x-3)(y-5) + 1/2!(30)(y-5)^2 .....