Is the origin max or min for $f(x,y)=x^4+frac{1}{4}y^4+4xy^3+4x^2y^2$?
$begingroup$
The gradient is null in (0,0) but the hessian matrix is null.
The function hasn't simmetry.Can I use Taylor expansion?
multivariable-calculus
edited Dec 4 '18 at 15:13
Joshua Mundinger
2,5141026
asked Dec 4 '18 at 14:16
Giulia B.Giulia B.
415211
$endgroup$
add a comment |
$begingroup$
The gradient is null in (0,0) but the hessian matrix is null.
The function hasn't simmetry.Can I use Taylor expansion?
multivariable-calculus
edited Dec 4 '18 at 15:13
Joshua Mundinger
2,5141026
asked Dec 4 '18 at 14:16
Giulia B.Giulia B.
415211
$endgroup$
add a comment |
$begingroup$
The gradient is null in (0,0) but the hessian matrix is null.
The function hasn't simmetry.Can I use Taylor expansion?
multivariable-calculus
edited Dec 4 '18 at 15:13
Joshua Mundinger
2,5141026
asked Dec 4 '18 at 14:16
Giulia B.Giulia B.
415211
$endgroup$
The gradient is null in (0,0) but the hessian matrix is null.
The function hasn't simmetry.Can I use Taylor expansion?
multivariable-calculus
multivariable-calculus
edited Dec 4 '18 at 15:13
Joshua Mundinger
2,5141026
asked Dec 4 '18 at 14:16
Giulia B.Giulia B.
415211
edited Dec 4 '18 at 15:13
Joshua Mundinger
2,5141026
asked Dec 4 '18 at 14:16
Giulia B.Giulia B.
415211
edited Dec 4 '18 at 15:13
Joshua Mundinger
2,5141026
edited Dec 4 '18 at 15:13
Joshua Mundinger
2,5141026
edited Dec 4 '18 at 15:13
Joshua Mundinger
2,5141026
2,5141026
asked Dec 4 '18 at 14:16
Giulia B.Giulia B.
415211
asked Dec 4 '18 at 14:16
Giulia B.Giulia B.
415211
asked Dec 4 '18 at 14:16
Giulia B.Giulia B.
415211
415211
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The function is symmetric in x, and looks like $x^4$ on the $y = 0$ line, so the origin can't be a maximum: it's either a minimum or a saddle point. Along the $x = 0$ line, it looks like $frac{1}{4}y^4$, so again, it looks like a minimum here.
And along any $y = ax$ line, the function looks like $x^4(1 + frac{a^4}{4} + 4a^3 + 4a^2)$. For $a = -3$, that's $x^4(1 + frac{81}{4} - 108 + 36) = frac{-203}{4}x^4$, so along this line, it looks like a maximum.
Thus, it is neither a minimum nor a maximum.
answered Dec 4 '18 at 14:32
user3482749user3482749
4,167919
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The function is symmetric in x, and looks like $x^4$ on the $y = 0$ line, so the origin can't be a maximum: it's either a minimum or a saddle point. Along the $x = 0$ line, it looks like $frac{1}{4}y^4$, so again, it looks like a minimum here.
And along any $y = ax$ line, the function looks like $x^4(1 + frac{a^4}{4} + 4a^3 + 4a^2)$. For $a = -3$, that's $x^4(1 + frac{81}{4} - 108 + 36) = frac{-203}{4}x^4$, so along this line, it looks like a maximum.
Thus, it is neither a minimum nor a maximum.
answered Dec 4 '18 at 14:32
user3482749user3482749
4,167919
$endgroup$
add a comment |
$begingroup$
The function is symmetric in x, and looks like $x^4$ on the $y = 0$ line, so the origin can't be a maximum: it's either a minimum or a saddle point. Along the $x = 0$ line, it looks like $frac{1}{4}y^4$, so again, it looks like a minimum here.
And along any $y = ax$ line, the function looks like $x^4(1 + frac{a^4}{4} + 4a^3 + 4a^2)$. For $a = -3$, that's $x^4(1 + frac{81}{4} - 108 + 36) = frac{-203}{4}x^4$, so along this line, it looks like a maximum.
Thus, it is neither a minimum nor a maximum.
answered Dec 4 '18 at 14:32
user3482749user3482749
4,167919
$endgroup$
add a comment |
$begingroup$
The function is symmetric in x, and looks like $x^4$ on the $y = 0$ line, so the origin can't be a maximum: it's either a minimum or a saddle point. Along the $x = 0$ line, it looks like $frac{1}{4}y^4$, so again, it looks like a minimum here.
And along any $y = ax$ line, the function looks like $x^4(1 + frac{a^4}{4} + 4a^3 + 4a^2)$. For $a = -3$, that's $x^4(1 + frac{81}{4} - 108 + 36) = frac{-203}{4}x^4$, so along this line, it looks like a maximum.
Thus, it is neither a minimum nor a maximum.
answered Dec 4 '18 at 14:32
user3482749user3482749
4,167919
$endgroup$
The function is symmetric in x, and looks like $x^4$ on the $y = 0$ line, so the origin can't be a maximum: it's either a minimum or a saddle point. Along the $x = 0$ line, it looks like $frac{1}{4}y^4$, so again, it looks like a minimum here.
And along any $y = ax$ line, the function looks like $x^4(1 + frac{a^4}{4} + 4a^3 + 4a^2)$. For $a = -3$, that's $x^4(1 + frac{81}{4} - 108 + 36) = frac{-203}{4}x^4$, so along this line, it looks like a maximum.
Thus, it is neither a minimum nor a maximum.
answered Dec 4 '18 at 14:32
user3482749user3482749
4,167919
edited Dec 4 '18 at 14:38
answered Dec 4 '18 at 14:32
user3482749user3482749
4,167919
answered Dec 4 '18 at 14:32
user3482749user3482749
4,167919
answered Dec 4 '18 at 14:32
user3482749user3482749
4,167919
4,167919
add a comment |
add a comment |
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