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Finding local extrema using taylor series

aruwin

Member
Jul 4, 2012
121
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.

My attempt:
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)

Second derivatives,
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).
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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.

My attempt:
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)

Second derivatives,
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).
See >>here<< or better >>here<<. Also check which of your stationary points satisfy the constraint.



CB
 
Last edited:

aruwin

Member
Jul 4, 2012
121
See >>here<< or better >>here<<. Also check which of your stationary points satisfy the constraint.



CB
Thanks for the links,they are helpful. But I need you to check this for me,please.

Look at this approxiamtion:
f = 2e^(-1), f_x = 0, f_y = 0, f_xx = -8e^(-1), f_xy = 0, f_yy = -10e^(-1)

Hence, f(x, y) = 2e^(-1) + 0(x - 1) + 0y + (1/2!) [-8e^(-1) (x - 1)^2 + 0(x - 1)y - 10e^(-1)y^2] + ...
Since the quadratic term is [-4e^(-1) (x - 1)^2 - 5e^(-1)y^2 (with negative signs and no cross term), we have a local maximum at (1, 0). My question is, what is the maximum value? Is it 2e^(-1)?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Thanks for the links,they are helpful. But I need you to check this for me,please.

Look at this approxiamtion:
f = 2e^(-1), f_x = 0, f_y = 0, f_xx = -8e^(-1), f_xy = 0, f_yy = -10e^(-1)

Hence, f(x, y) = 2e^(-1) + 0(x - 1) + 0y + (1/2!) [-8e^(-1) (x - 1)^2 + 0(x - 1)y - 10e^(-1)y^2] + ...
Since the quadratic term is [-4e^(-1) (x - 1)^2 - 5e^(-1)y^2 (with negative signs and no cross term), we have a local maximum at (1, 0). My question is, what is the maximum value? Is it 2e^(-1)?
Sorry but you now seem to have a different question in mind. Your maximum value is \(f(x_s,y_s)\) where \((x_s,y_s)\) is the stationary point, which we may guess from your post is \((1,0)\) . But as you have not said what \(f(x,y)\) is we can take this no further.

CB
 

aruwin

Member
Jul 4, 2012
121
Sorry but you now seem to have a different question in mind. Your maximum value is \(f(x_s,y_s)\) where \((x_s,y_s)\) is the stationary point, which we may guess from your post is \((1,0)\) . But as you have not said what \(f(x,y)\) is we can take this no further.

CB
Ops, sorry. Yes, this is another question but they're the same thing. I'm going to write the complete question,then.
f(x, y) = e^(-x^2 - y^2) * (2x^2 - 3y^2)

approximtion:
f = 2e^(-1), f_x = 0, f_y = 0, f_xx = -8e^(-1), f_xy = 0, f_yy = -10e^(-1)

Hence, f(x, y) = 2e^(-1) + 0(x - 1) + 0y + (1/2!) [-8e^(-1) (x - 1)^2 + 0(x - 1)y - 10e^(-1)y^2] + ...
Since the quadratic term is [-4e^(-1) (x - 1)^2 - 5e^(-1)y^2 (with negative signs and no cross term), we have a local maximum at (1, 0). Is the max value 2e^(-1)?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Ops, sorry. Yes, this is another question but they're the same thing. I'm going to write the complete question,then.
f(x, y) = e^(-x^2 - y^2) * (2x^2 - 3y^2)

approximtion:
f = 2e^(-1), f_x = 0, f_y = 0, f_xx = -8e^(-1), f_xy = 0, f_yy = -10e^(-1)

Hence, f(x, y) = 2e^(-1) + 0(x - 1) + 0y + (1/2!) [-8e^(-1) (x - 1)^2 + 0(x - 1)y - 10e^(-1)y^2] + ...
Since the quadratic term is [-4e^(-1) (x - 1)^2 - 5e^(-1)y^2 (with negative signs and no cross term), we have a local maximum at (1, 0). Is the max value 2e^(-1)?
What are the stationary points? How did you find them? What I am asking is where are the others?

CB