Finding local extrema using taylor series

In summary, the conversation is about finding the local extrema of a function using Taylor Series. The process involves finding the partial derivatives, identifying critical points, and using the Taylor Series expansion to determine the nature of the extrema. The conversation also touches on the concept of constraints and how they affect the extrema of a function. The final part of the conversation discusses a specific example and clarifies the maximum value at a given point.
  • #1
aruwin
208
0
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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  • #2
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:
  • #3
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)?
 
  • #4
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
 
  • #5
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)?
 
  • #6
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
 

FAQ: Finding local extrema using taylor series

What is a local extremum?

A local extremum is a point on a graph where the function reaches its highest or lowest value in a specific region. It is not necessarily the absolute highest or lowest point on the entire graph, but rather within a smaller interval.

How can Taylor series be used to find local extrema?

Taylor series is a mathematical tool that allows us to approximate a function using a polynomial. By finding the critical points of the polynomial, we can determine the local extrema of the original function.

What is the process for finding local extrema using Taylor series?

The process involves taking the derivative of the Taylor series, setting it equal to zero, and solving for the critical points. Then, we can plug these points back into the original function to determine the local extrema.

What is the significance of finding local extrema using Taylor series?

Using Taylor series to find local extrema can help us understand the behavior of a function in a specific region. It can also be useful in optimization problems, where we want to find the maximum or minimum value of a function.

Are there any limitations to using Taylor series to find local extrema?

Yes, Taylor series can only approximate a function within a specific region, so the local extrema found may not necessarily be the true extrema of the original function. Additionally, it may not work for all types of functions, such as those with discontinuities or singularities.

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