Finding local extrema using taylor series

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Discussion Overview

The discussion revolves around finding local extrema of functions using Taylor series expansion, specifically for the function f(x,y) = x^4 + y^4 - 2(x - y)^2 under the constraint f(x,y) ≤ 0. Participants explore the application of Taylor series in identifying local maxima and minima, comparing it to traditional derivative methods.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about using Taylor series for finding extrema, contrasting it with their usual method involving derivatives.
  • Participants identify critical points (0,0), (√2,-√2), and (-√2,√2) based on partial derivatives.
  • There is an attempt to expand the function using Taylor series at the origin, leading to a discussion about local minima and maxima based on the sign of the function.
  • Another participant questions how to determine if f(x,y) is greater than or equal to f(0,0) based on various substitutions.
  • A later reply discusses a different function, f(x,y) = e^(-x^2 - y^2) * (2x^2 - 3y^2), and its approximation, leading to a query about the maximum value at a stationary point.
  • There is mention of the quadratic term's signs indicating a local maximum at (1,0) for the second function, but uncertainty remains about the maximum value.
  • Participants seek clarification on the stationary points and their identification process, indicating a need for further exploration of the topic.

Areas of Agreement / Disagreement

Participants express differing views on the application of Taylor series for finding extrema, with some uncertainty about the implications of their findings. The discussion remains unresolved regarding the maximum values and the identification of all stationary points.

Contextual Notes

Limitations include potential misunderstandings of the Taylor series application, dependence on the definitions of local extrema, and unresolved steps in the mathematical reasoning presented.

Who May Find This Useful

This discussion may be useful for students and practitioners interested in advanced calculus, particularly those exploring methods for finding extrema of multivariable functions using Taylor series.

aruwin
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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 let's 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 going to be negative and that means they're always going to 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).
 
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aruwin said:
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 let's 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 going to be negative and that means they're always going to 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:
CaptainBlack said:
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)?
 
aruwin said:
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
 
CaptainBlack said:
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)?
 
aruwin said:
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
 

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