Find global minima of vector valued quadratic equation [closed]

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I have the following equation, with 5 $mathbbR^3$ vectors A, B, C, D, and P, and a scalar, $t$;

$-At^3 + 3Bt^3 + Ct^3 - 3Dt^3 + 3At^2 - 6Bt^2 + 3Dt^2 - 3At + 3Bt + A - P$

I'm trying to find the minimum of this equation's absolute magnitude, but I'm neither sure how to apply the pythagorean formula to a function, nor am I sure how to find the minima and maxima of a vector valued function. If anyone could provide some advice, it would be much appreciated.

edited Jul 15 at 17:22

asked Jul 15 at 16:51

Maurdekye

1376

closed as unclear what you're asking by amWhy, Cesareo, JosÃ© Carlos Santos, Xander Henderson, max_zorn Jul 16 at 0:27

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, itÃ¢Â€Â™s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

Well, it isn't a quadratic, and I'm not sure what the Pythagorean formula would have to do with it, even if it were a quadratic.
â€“Â Cameron Buie
Jul 15 at 17:52

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up vote
1
down vote

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I have the following equation, with 5 $mathbbR^3$ vectors A, B, C, D, and P, and a scalar, $t$;

$-At^3 + 3Bt^3 + Ct^3 - 3Dt^3 + 3At^2 - 6Bt^2 + 3Dt^2 - 3At + 3Bt + A - P$

edited Jul 15 at 17:22

asked Jul 15 at 16:51

Maurdekye

1376

closed as unclear what you're asking by amWhy, Cesareo, JosÃ© Carlos Santos, Xander Henderson, max_zorn Jul 16 at 0:27

Well, it isn't a quadratic, and I'm not sure what the Pythagorean formula would have to do with it, even if it were a quadratic.
â€“Â Cameron Buie
Jul 15 at 17:52

add a commentÂ |Â

up vote
1
down vote

favorite

I have the following equation, with 5 $mathbbR^3$ vectors A, B, C, D, and P, and a scalar, $t$;

$-At^3 + 3Bt^3 + Ct^3 - 3Dt^3 + 3At^2 - 6Bt^2 + 3Dt^2 - 3At + 3Bt + A - P$

edited Jul 15 at 17:22

asked Jul 15 at 16:51

Maurdekye

1376

I have the following equation, with 5 $mathbbR^3$ vectors A, B, C, D, and P, and a scalar, $t$;

$-At^3 + 3Bt^3 + Ct^3 - 3Dt^3 + 3At^2 - 6Bt^2 + 3Dt^2 - 3At + 3Bt + A - P$

edited Jul 15 at 17:22

asked Jul 15 at 16:51

Maurdekye

1376

edited Jul 15 at 17:22

asked Jul 15 at 16:51

Maurdekye

1376

asked Jul 15 at 16:51

Maurdekye

1376

asked Jul 15 at 16:51

Maurdekye

1376

closed as unclear what you're asking by amWhy, Cesareo, JosÃ© Carlos Santos, Xander Henderson, max_zorn Jul 16 at 0:27

Well, it isn't a quadratic, and I'm not sure what the Pythagorean formula would have to do with it, even if it were a quadratic.
â€“Â Cameron Buie
Jul 15 at 17:52

add a commentÂ |Â

Well, it isn't a quadratic, and I'm not sure what the Pythagorean formula would have to do with it, even if it were a quadratic.
â€“Â Cameron Buie
Jul 15 at 17:52

Well, it isn't a quadratic, and I'm not sure what the Pythagorean formula would have to do with it, even if it were a quadratic.
â€“Â Cameron Buie
Jul 15 at 17:52

add a commentÂ |Â

1 Answer
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Essentially you have four vectors $vecv_k,, k=0,3$ and the vector-valued function

$$ vecf(t)=sum_k=0^3vecv_kt^k $$

and you would like to know how close the path of this function gets to $vec0$.

So you should minimize $vertvecf(t)vert^2$.

If we let $vecv_k=(x_k,y_k,z_k)$ then

$$ vertvecf(t)vert^2 =left(sum_k=0^3x_kt^kright)^2+left(sum_k=0^3y_kt^kright)^2+left(sum_k=0^3z_kt^kright)^2=g(t)$$
This is a sixth degree non-negative polynomial so we know it has a minimum value. But to find it we must solve the fifth degree polynomial equation $g^prime(t)=0$.

begineqnarrayfrac12g^prime(t)&=&left(sum_k=1^3kx_kt^k-1right)left(sum_k=0^3x_kt^kright)\
&+&left(sum_k=1^3ky_kt^k-1right)left(sum_k=0^3y_kt^kright)\
&+&left(sum_k=1^3kz_kt^k-1right)left(sum_k=0^3z_kt^kright)\
&=&0
endeqnarray

As you know, this fifth-degree polynomial equation cannot be solved analytically so you will have to rely on numerical methods.

edited Jul 17 at 15:49

answered Jul 15 at 21:25

John Wayland Bales

12.8k21135

I'm not sure what half those words mean, but i'm going to assume it means there isn't a mathematical formula for what I want to do.
â€“Â Maurdekye
Jul 17 at 4:02

That is correct. However, it resolves into a polynomial equation, so there are standard computer algorithms to find the extreme points. Note that in the last equation I had a couple of equal signs where I intended plus signs. I am correcting that.
â€“Â John Wayland Bales
Jul 17 at 15:47

add a commentÂ |Â

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

Essentially you have four vectors $vecv_k,, k=0,3$ and the vector-valued function

$$ vecf(t)=sum_k=0^3vecv_kt^k $$

and you would like to know how close the path of this function gets to $vec0$.

So you should minimize $vertvecf(t)vert^2$.

If we let $vecv_k=(x_k,y_k,z_k)$ then

As you know, this fifth-degree polynomial equation cannot be solved analytically so you will have to rely on numerical methods.

edited Jul 17 at 15:49

answered Jul 15 at 21:25

John Wayland Bales

12.8k21135

I'm not sure what half those words mean, but i'm going to assume it means there isn't a mathematical formula for what I want to do.
â€“Â Maurdekye
Jul 17 at 4:02

That is correct. However, it resolves into a polynomial equation, so there are standard computer algorithms to find the extreme points. Note that in the last equation I had a couple of equal signs where I intended plus signs. I am correcting that.
â€“Â John Wayland Bales
Jul 17 at 15:47

add a commentÂ |Â

up vote
1
down vote

accepted

Essentially you have four vectors $vecv_k,, k=0,3$ and the vector-valued function

$$ vecf(t)=sum_k=0^3vecv_kt^k $$

and you would like to know how close the path of this function gets to $vec0$.

So you should minimize $vertvecf(t)vert^2$.

If we let $vecv_k=(x_k,y_k,z_k)$ then

As you know, this fifth-degree polynomial equation cannot be solved analytically so you will have to rely on numerical methods.

edited Jul 17 at 15:49

answered Jul 15 at 21:25

John Wayland Bales

12.8k21135

I'm not sure what half those words mean, but i'm going to assume it means there isn't a mathematical formula for what I want to do.
â€“Â Maurdekye
Jul 17 at 4:02

That is correct. However, it resolves into a polynomial equation, so there are standard computer algorithms to find the extreme points. Note that in the last equation I had a couple of equal signs where I intended plus signs. I am correcting that.
â€“Â John Wayland Bales
Jul 17 at 15:47

add a commentÂ |Â

up vote
1
down vote

accepted

Essentially you have four vectors $vecv_k,, k=0,3$ and the vector-valued function

$$ vecf(t)=sum_k=0^3vecv_kt^k $$

and you would like to know how close the path of this function gets to $vec0$.

So you should minimize $vertvecf(t)vert^2$.

If we let $vecv_k=(x_k,y_k,z_k)$ then

As you know, this fifth-degree polynomial equation cannot be solved analytically so you will have to rely on numerical methods.

edited Jul 17 at 15:49

answered Jul 15 at 21:25

John Wayland Bales

12.8k21135

Essentially you have four vectors $vecv_k,, k=0,3$ and the vector-valued function

$$ vecf(t)=sum_k=0^3vecv_kt^k $$

and you would like to know how close the path of this function gets to $vec0$.

So you should minimize $vertvecf(t)vert^2$.

If we let $vecv_k=(x_k,y_k,z_k)$ then

As you know, this fifth-degree polynomial equation cannot be solved analytically so you will have to rely on numerical methods.

edited Jul 17 at 15:49

answered Jul 15 at 21:25

John Wayland Bales

12.8k21135

edited Jul 17 at 15:49

answered Jul 15 at 21:25

John Wayland Bales

12.8k21135

answered Jul 15 at 21:25

John Wayland Bales

12.8k21135

answered Jul 15 at 21:25

John Wayland Bales

12.8k21135

I'm not sure what half those words mean, but i'm going to assume it means there isn't a mathematical formula for what I want to do.
â€“Â Maurdekye
Jul 17 at 4:02

That is correct. However, it resolves into a polynomial equation, so there are standard computer algorithms to find the extreme points. Note that in the last equation I had a couple of equal signs where I intended plus signs. I am correcting that.
â€“Â John Wayland Bales
Jul 17 at 15:47

add a commentÂ |Â

I'm not sure what half those words mean, but i'm going to assume it means there isn't a mathematical formula for what I want to do.
â€“Â Maurdekye
Jul 17 at 4:02

That is correct. However, it resolves into a polynomial equation, so there are standard computer algorithms to find the extreme points. Note that in the last equation I had a couple of equal signs where I intended plus signs. I am correcting that.
â€“Â John Wayland Bales
Jul 17 at 15:47

I'm not sure what half those words mean, but i'm going to assume it means there isn't a mathematical formula for what I want to do.
â€“Â Maurdekye
Jul 17 at 4:02

That is correct. However, it resolves into a polynomial equation, so there are standard computer algorithms to find the extreme points. Note that in the last equation I had a couple of equal signs where I intended plus signs. I am correcting that.
â€“Â John Wayland Bales
Jul 17 at 15:47

add a commentÂ |Â

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