A difficulty in understanding the solution of a problem.A difficulty in understanding the solution of a problem.2How do you find and classify the critical points of the function?what to do when the multivariable second derivative test is inconclusive?What to do when the second derivative test fails?The Idea behind the Second Partial Derivative TestFind the critical point of the functionFinding the critical points and using the second derivative test to find the local minima or maximaA difficulty in understanding the n-dimensional second order derivative.Prove that if $f$ is a linear transformation then $Df(a)(x) = f(x) $A difficulty in understanding a step in the proof of Thm. 11.5.6 in Petrovic.A difficulty in understanding multivariable Fermat theorem proof.
What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?
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A difficulty in understanding the solution of a problem.
A difficulty in understanding the solution of a problem.2How do you find and classify the critical points of the function?what to do when the multivariable second derivative test is inconclusive?What to do when the second derivative test fails?The Idea behind the Second Partial Derivative TestFind the critical point of the functionFinding the critical points and using the second derivative test to find the local minima or maximaA difficulty in understanding the n-dimensional second order derivative.Prove that if $f$ is a linear transformation then $Df(a)(x) = f(x) $A difficulty in understanding a step in the proof of Thm. 11.5.6 in Petrovic.A difficulty in understanding multivariable Fermat theorem proof.
$begingroup$
The question and its answer is given below:
But I do not understand the following:
1-why all the critical points equal to $ c = frac2n^2 + n +2$?
2-why to test this is the only critical point we compute second order derivative?
3-How is the sequence of determinants $ D_i_i = 1^n$ calculated?
Could anyone explain this for me please?
multivariable-calculus
asked Mar 22 at 8:17
hopefullyhopefully
269215
$endgroup$
add a comment |
$begingroup$
The question and its answer is given below:
But I do not understand the following:
1-why all the critical points equal to $ c = frac2n^2 + n +2$?
2-why to test this is the only critical point we compute second order derivative?
3-How is the sequence of determinants $ D_i_i = 1^n$ calculated?
Could anyone explain this for me please?
multivariable-calculus
asked Mar 22 at 8:17
hopefullyhopefully
269215
$endgroup$
$begingroup$
I don't understand your problem in (3). Are you asking for the $(-1)^m m!(c^cdots)^m$ factor?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:54
$begingroup$
yes I am asking about this @Martín-BlasPérezPinilla and the elements of the matrix also.
$endgroup$
– hopefully
Mar 22 at 8:55
1
$begingroup$
The components of the Hessian are the 2nd-order partial derivatives. For the factor, use a well-know property of determinants.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:58
add a comment |
$begingroup$
The question and its answer is given below:
But I do not understand the following:
1-why all the critical points equal to $ c = frac2n^2 + n +2$?
2-why to test this is the only critical point we compute second order derivative?
3-How is the sequence of determinants $ D_i_i = 1^n$ calculated?
Could anyone explain this for me please?
multivariable-calculus
asked Mar 22 at 8:17
hopefullyhopefully
269215
$endgroup$
The question and its answer is given below:
But I do not understand the following:
1-why all the critical points equal to $ c = frac2n^2 + n +2$?
2-why to test this is the only critical point we compute second order derivative?
3-How is the sequence of determinants $ D_i_i = 1^n$ calculated?
Could anyone explain this for me please?
multivariable-calculus
multivariable-calculus
asked Mar 22 at 8:17
hopefullyhopefully
269215
asked Mar 22 at 8:17
hopefullyhopefully
269215
edited Mar 22 at 8:43
hopefully
asked Mar 22 at 8:17
hopefullyhopefully
269215
asked Mar 22 at 8:17
hopefullyhopefully
269215
asked Mar 22 at 8:17
hopefullyhopefully
269215
269215
$begingroup$
I don't understand your problem in (3). Are you asking for the $(-1)^m m!(c^cdots)^m$ factor?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:54
$begingroup$
yes I am asking about this @Martín-BlasPérezPinilla and the elements of the matrix also.
$endgroup$
– hopefully
Mar 22 at 8:55
1
$begingroup$
The components of the Hessian are the 2nd-order partial derivatives. For the factor, use a well-know property of determinants.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:58
add a comment |
$begingroup$
I don't understand your problem in (3). Are you asking for the $(-1)^m m!(c^cdots)^m$ factor?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:54
$begingroup$
yes I am asking about this @Martín-BlasPérezPinilla and the elements of the matrix also.
$endgroup$
– hopefully
Mar 22 at 8:55
1
$begingroup$
The components of the Hessian are the 2nd-order partial derivatives. For the factor, use a well-know property of determinants.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:58
$begingroup$
I don't understand your problem in (3). Are you asking for the $(-1)^m m!(c^cdots)^m$ factor?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:54
$begingroup$
I don't understand your problem in (3). Are you asking for the $(-1)^m m!(c^cdots)^m$ factor?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:54
$begingroup$
yes I am asking about this @Martín-BlasPérezPinilla and the elements of the matrix also.
$endgroup$
– hopefully
Mar 22 at 8:55
$begingroup$
yes I am asking about this @Martín-BlasPérezPinilla and the elements of the matrix also.
$endgroup$
– hopefully
Mar 22 at 8:55
1
1
$begingroup$
The components of the Hessian are the 2nd-order partial derivatives. For the factor, use a well-know property of determinants.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:58
$begingroup$
The components of the Hessian are the 2nd-order partial derivatives. For the factor, use a well-know property of determinants.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hints:
(1) Use that $x_1 = x_2 = cdots = A$:
$$A = (1 - x_1 - 2 x_2 - cdots - n x_n) = (1 - A - 2 A - cdots - n A) = cdots$$
(2) Because we want know if is max/min?
(3) The second-order partial derivatives are already calculated...
answered Mar 22 at 8:51
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
$begingroup$
for (2) why this is the way to say that we do not have critical points more than this?
$endgroup$
– hopefully
Mar 22 at 8:57
2
$begingroup$
In (1) it was already established that there is a single critical point (for positive $x_i$). Now it is just a matter of deciding if it is a minimum, maximum or saddle point.
$endgroup$
– PierreCarre
Mar 22 at 9:00
$begingroup$
@hopefully, see the comment by PierreCarre.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:01
1
$begingroup$
@PierreCarre, start your comment like me (with @...) and the user will be pinged.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:03
$begingroup$
Got it @Martín-BlasPérezPinilla !
$endgroup$
– PierreCarre
Mar 22 at 9:10
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hints:
(1) Use that $x_1 = x_2 = cdots = A$:
$$A = (1 - x_1 - 2 x_2 - cdots - n x_n) = (1 - A - 2 A - cdots - n A) = cdots$$
(2) Because we want know if is max/min?
(3) The second-order partial derivatives are already calculated...
answered Mar 22 at 8:51
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
$begingroup$
for (2) why this is the way to say that we do not have critical points more than this?
$endgroup$
– hopefully
Mar 22 at 8:57
2
$begingroup$
In (1) it was already established that there is a single critical point (for positive $x_i$). Now it is just a matter of deciding if it is a minimum, maximum or saddle point.
$endgroup$
– PierreCarre
Mar 22 at 9:00
$begingroup$
@hopefully, see the comment by PierreCarre.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:01
1
$begingroup$
@PierreCarre, start your comment like me (with @...) and the user will be pinged.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:03
$begingroup$
Got it @Martín-BlasPérezPinilla !
$endgroup$
– PierreCarre
Mar 22 at 9:10
add a comment |
$begingroup$
Hints:
(1) Use that $x_1 = x_2 = cdots = A$:
$$A = (1 - x_1 - 2 x_2 - cdots - n x_n) = (1 - A - 2 A - cdots - n A) = cdots$$
(2) Because we want know if is max/min?
(3) The second-order partial derivatives are already calculated...
answered Mar 22 at 8:51
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
$begingroup$
for (2) why this is the way to say that we do not have critical points more than this?
$endgroup$
– hopefully
Mar 22 at 8:57
2
$begingroup$
In (1) it was already established that there is a single critical point (for positive $x_i$). Now it is just a matter of deciding if it is a minimum, maximum or saddle point.
$endgroup$
– PierreCarre
Mar 22 at 9:00
$begingroup$
@hopefully, see the comment by PierreCarre.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:01
1
$begingroup$
@PierreCarre, start your comment like me (with @...) and the user will be pinged.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:03
$begingroup$
Got it @Martín-BlasPérezPinilla !
$endgroup$
– PierreCarre
Mar 22 at 9:10
add a comment |
$begingroup$
Hints:
(1) Use that $x_1 = x_2 = cdots = A$:
$$A = (1 - x_1 - 2 x_2 - cdots - n x_n) = (1 - A - 2 A - cdots - n A) = cdots$$
(2) Because we want know if is max/min?
(3) The second-order partial derivatives are already calculated...
answered Mar 22 at 8:51
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
$endgroup$
Hints:
(1) Use that $x_1 = x_2 = cdots = A$:
$$A = (1 - x_1 - 2 x_2 - cdots - n x_n) = (1 - A - 2 A - cdots - n A) = cdots$$
(2) Because we want know if is max/min?
(3) The second-order partial derivatives are already calculated...
answered Mar 22 at 8:51
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
answered Mar 22 at 8:51
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
answered Mar 22 at 8:51
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
answered Mar 22 at 8:51
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
35.4k42972
$begingroup$
for (2) why this is the way to say that we do not have critical points more than this?
$endgroup$
– hopefully
Mar 22 at 8:57
2
$begingroup$
In (1) it was already established that there is a single critical point (for positive $x_i$). Now it is just a matter of deciding if it is a minimum, maximum or saddle point.
$endgroup$
– PierreCarre
Mar 22 at 9:00
$begingroup$
@hopefully, see the comment by PierreCarre.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:01
1
$begingroup$
@PierreCarre, start your comment like me (with @...) and the user will be pinged.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:03
$begingroup$
Got it @Martín-BlasPérezPinilla !
$endgroup$
– PierreCarre
Mar 22 at 9:10
add a comment |
$begingroup$
for (2) why this is the way to say that we do not have critical points more than this?
$endgroup$
– hopefully
Mar 22 at 8:57
2
$begingroup$
In (1) it was already established that there is a single critical point (for positive $x_i$). Now it is just a matter of deciding if it is a minimum, maximum or saddle point.
$endgroup$
– PierreCarre
Mar 22 at 9:00
$begingroup$
@hopefully, see the comment by PierreCarre.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:01
1
$begingroup$
@PierreCarre, start your comment like me (with @...) and the user will be pinged.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:03
$begingroup$
Got it @Martín-BlasPérezPinilla !
$endgroup$
– PierreCarre
Mar 22 at 9:10
$begingroup$
for (2) why this is the way to say that we do not have critical points more than this?
$endgroup$
– hopefully
Mar 22 at 8:57
$begingroup$
for (2) why this is the way to say that we do not have critical points more than this?
$endgroup$
– hopefully
Mar 22 at 8:57
2
2
$begingroup$
In (1) it was already established that there is a single critical point (for positive $x_i$). Now it is just a matter of deciding if it is a minimum, maximum or saddle point.
$endgroup$
– PierreCarre
Mar 22 at 9:00
$begingroup$
In (1) it was already established that there is a single critical point (for positive $x_i$). Now it is just a matter of deciding if it is a minimum, maximum or saddle point.
$endgroup$
– PierreCarre
Mar 22 at 9:00
$begingroup$
@hopefully, see the comment by PierreCarre.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:01
$begingroup$
@hopefully, see the comment by PierreCarre.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:01
1
1
$begingroup$
@PierreCarre, start your comment like me (with @...) and the user will be pinged.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:03
$begingroup$
@PierreCarre, start your comment like me (with @...) and the user will be pinged.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 9:03
$begingroup$
Got it @Martín-BlasPérezPinilla !
$endgroup$
– PierreCarre
Mar 22 at 9:10
$begingroup$
Got it @Martín-BlasPérezPinilla !
$endgroup$
– PierreCarre
Mar 22 at 9:10
add a comment |
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$begingroup$
I don't understand your problem in (3). Are you asking for the $(-1)^m m!(c^cdots)^m$ factor?
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:54
$begingroup$
yes I am asking about this @Martín-BlasPérezPinilla and the elements of the matrix also.
$endgroup$
– hopefully
Mar 22 at 8:55
1
$begingroup$
The components of the Hessian are the 2nd-order partial derivatives. For the factor, use a well-know property of determinants.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 22 at 8:58