How do I find the extrema using Taylor Series?? I am so used to find extrema just by finding the first derivative (make it =0) and then finding the second derivative and then just use the formula f_xx.f_yy - f_xy and just look at the sign but this time I need to use taylor expansion. I hope you guys can guide me.I have done partial solution but I am not so sure if it's right so help me.
Find the local extrema of the function f(x,y) = x^4 + y^4 - 2(x - y)^2 where f(x,y)≤ 0.
from the partial derivatives,
f_x = 4x^3 - 4(x-y) = 0
f_y = 4y^3 + 4(x-y) = 0
we get (x,y) = (0,0),(√2,-√2),(-√2,√2) (lets name these critical points O,A,B)
f_xx= -4, f_xy= 4, f_yy= -4
Expanding the function using taylor series at the origin O = (0,0),
1/2.[fxx(0,0)x2 + 2.fxy(0,0)xy + fyy(0,0)y2 ] = -2(x-y)^2
From the definition:
For all values of (x,y) near (0,0),
f(x,y)≥f(0,0) → At (0,0) f(x,y) has a local minimum
f(x,y)≤f(0,0) → At (0,0) f(x,y) has a local maximum
What I understand is that from the taylor expansion, we know that f(0,0) is always negative (or 0).In other words, f(x,y) is always negative(or 0) at (0,0). Right?And since at the beginnging of the problem, the constraint f(x,y) ≤ 0 is given, it is also negative or 0 but how to know that f(x,y) bigger or than f(0,0)? So lets say there are points
P=(-1,-2) and Q = (1,2). Subtituting these I get -2 and no matter what values I put,they're always gonna be negative and that means they're always gonna be maximum because it can never go beyond 0,right? I am going to apply what I think and please correct me if I am wrong.
The function at O is -2(x-y)^2.
Substituting values of x and y with 0,
so f(0,0) = 0 and
substituting values of x and y with 1 and 2,
so f(1,2) = -2 and
thus, f(1,2) is lower than f(0,0). Generalizing it, we have f(x,y) lower than or equals to(when we substitute with 0) f(0,0) and so, f(x,y) has a local maximum value of 0 at (0,0).