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    Apologies it's not in latex. Working out stationary points of:

    z=4xy - x^4 - y^4

    dz/dx = 4y-4x^3 = 0

    dz/dy = 4x-4y^3 = 0

    I solve dz/dx and get x = y^1/3. I substitute this into dz/dy and get y = 0, y = 1. I'm missing a result.The stationary points should be (0,0), (-1,-1), ), (1,1)Where does the (-1,-1) come from?!

    I'm being really stupid..
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    (Original post by hezzlington)
    Apologies it's not in latex. Working out stationary points of:

    z=4xy - x^4 - y^4

    dz/dx = 4y-4x^3 = 0

    dz/dy = 4x-4y^3 = 0

    I solve dz/dx and get x = y^1/3. I substitute this into dz/dy and get y = 0, y = 1. I'm missing a result.The stationary points should be (0,0), (-1,-1), ), (1,1)Where does the (-1,-1) come from?!

    I'm being really stupid..
    Initially I assumed you would have accidentally lost a solution by having incorrectly divided through by a variable somewhere in your simplification, but doing so would ommit the (0,0) solution and not the (-1,-1) solution so I'm not quite sure what you've done?

    In any case from dz/dx = 4y-4x^3 = 0 we get x=y^1/3 or x^3=y which then when subbed into dz/dy = 0 gives x=x^9 (or y=y^9).

    y=y^9 is satisfied by y=0,-1,1 which correspondingly gives solutions (x,y) = (0,0), (-1,-1) and (1,1) so I'm not sure about your issue?

    A mistake might have been to write y=y^9 => 1=y^8 => y=+-1 which loses the (0,0) solution because you have artificially lowered the order of the equation but since you are missing (-1,-1) I'm not sure whats gone wrong without seeing your workings...

    Hope this helps.

    Edit: Unless you've just made the elementary oversight of forgetting -1^(2n-1) = -1 for all positive integers n.
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    (Original post by In One Ear)
    A mistake might have been to write y=y^9 => 1=y^8 => y=+-1 which loses the (0,0) solution because you have artificially lowered the order of the equation but since you are missing (-1,-1) I'm not sure whats gone wrong without seeing your workings...
    I'm pretty sure the OP got to the point y^{1/3} = y^3 and then simply failed to realise (-1,-1) was a solution; looks fairly clear they were just spotting roots by inspection and it seems fairly easy to fail to recognize that y being -ve is an option here even though we're taking the cube root.

    (Original post by hezzlington)
    ..
    Without wanting to "nag", I'll point out that even if you didn't find (-1, -1) yourself, you really should have been able to verify that (-1, -1) was a solution to your equation. Since verification is usually easier than finding solutions, it's a good habit to get into, both with your own solutions (to make sure you've not made a silly mistake), and also solutions provided to you. (For a provided solution, you are not so much checking the solution is correct, but that your equations do actually "work" for the solution).
 
 
 
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