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Partial differentiation stationary points watch

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    Hi

    If you found that for a coordinate (x,y), the nature of the stationary point is inconclusive (i.e. D=0 from the test D = f_{xx} \cdot f_{yy}-(f_{xy})^{2}, what alternative methods can I use to determine the nature of the stationary point?

    Any contribution is much appreciated!
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    Is that kinda university maths? I don’t wanna give you an Alevel Maths method because you probably know it!
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    (Original post by Mosa99)
    Is that kinda university maths? I don’t wanna give you an Alevel Maths method because you probably know it!
    Yh, this is partial differentiation lol
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    I can’t help then! 😅😅
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    (Original post by ManLike007)
    Hi

    If you found that for a coordinate (x,y), the nature of the stationary point is inconclusive (i.e. D=0 from the test D = f_{xx} \cdot f_{yy}-(f_{xy})^{2}, what alternative methods can I use to determine the nature of the stationary point?

    Any contribution is much appreciated!
    It depends how full an answer you want. Full information is always contained in the local Taylor expansion. That D=0 (loosely) means the nature of the stationary point is not wholly characterised by the quadratic/degree two part of the expansion.

    But some cases such as

    x^2 + y^4, -x^4-y^4, x^2 - y^4

    can still (around (0,0)) be quickly seen to be a minimum, maximum, saddle, but then you can also get wholly new types of stationary point like

    xy(x-y)

    which is a "monkey saddle" (IIRC its name).
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    (Original post by ManLike007)
    Hi

    If you found that for a coordinate (x,y), the nature of the stationary point is inconclusive (i.e. D=0 from the test D = f_{xx} \cdot f_{yy}-(f_{xy})^{2}, what alternative methods can I use to determine the nature of the stationary point?

    Any contribution is much appreciated!
    Depends on the functions. Sometimes it is quite obvious- like with f(x,y)=x^4+2(xy)^2-y^4-2x^2+3 we have D=0 at (0,0) but notice that

    \begin{aligned} f(x,y)-f(0,0) & = x^4+2(xy)^2-y^4-2x^2 \\ & \leq x^4+2(xy)^2-y^4 - 2x^4 \\ & = -(x^2-y^2)^2 \\ & \leq 0 \end{aligned}

    ....hence f(x,y) \leq f(0,0) and so that point is a max.

    Alternative, you can employ Taylor's expansion in 2-D around your crit point of interest.
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    (Original post by RDKGames)
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    Thanks for your response.

    Just to clarify because I'm slightly baffled with your working out,

    f(0,0)=3 so f(x,y)-f(0,0)=x^{4}+2(xy)^{2}-y^{4}-2x^{2} but

    Second line: Why does +2(xy)^{2} and -2x^{2} become -2(xy)^{2} and -2x^{4} (respectively) ?.
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    (Original post by ManLike007)
    Thanks for your response.

    Just to clarify because I'm slightly baffled with your working out,

    f(0,0)=3 so f(x,y)-f(0,0)=x^{4}+2(xy)^{2}-y^{4}-2x^{2} but

    Second line: Why does +2(xy)^{2} and -2x^{2} become -2(xy)^{2} and -2x^{4} (respectively) ?.
    Typo for the first one.

    Note that in a small neighbourhood around (0,0) (on the xy plane) we have x^4 \leq x^2 therefore -2x^4 \geq -2x^2

    Thus x^4 +2(xy)^2 - y^4 -2x^4 \geq x^4 +2(xy)^2 - y^4 - 2x^2

    EDIT: By small neighbourhood I mean |x|,|y| \leq 1
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    (Original post by RDKGames)
    Note that in a small neighbourhood around (0,0) (on the xy plane) we have x^4 \leq x^2 therefore -2x^4 \geq -2x^2

    Thus x^4 +2(xy)^2 - y^4 -2x^4 \geq x^4 +2(xy)^2 - y^4 - 2x^2
    It still hasn't clicked yet, the "x^4 \leq x^2" and "-2x^4 \geq -2x^2" puts me off, shouldn't the inequality sign be the other way round?

    Also are you basically saying that for a stationary point (a,b), if f(x,y) \leq f(x,y) - f(a,b) then it's maximum point and vice verse for the minimum point?
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    (Original post by ManLike007)
    It still hasn't clicked yet, the "x^4 \leq x^2" and "-2x^4 \geq -2x^2" puts me off, shouldn't the inequality sign be the other way round?

    Also are you basically saying that for a stationary point (a,b), if f(x,y) \leq f(a,b) then it's maximum point and vice verse for the minimum point?
    You should know from A-Level that if x>y then multiplying through by -1 changes the inequality to -x<-y (or if STILL unconvinced, just subtract (x+y) from both sides to get -y>-x which is the same thing, different orientation).
    So, x^4 \leq x^2 \Rightarrow -x^4 \geq -x^2 \Rightarrow -2x^4 \geq -2x^2

    Obviously, if f(x,y) \leq f(a,b) in the vicinity of (a,b) then it is a local maximum. If the inequality is true for ALL (x,y) \in D (D is your domain) then f(a,b) is your global max.
    Similarly, for min points with f(x,y) \geq f(a,b)
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    (Original post by RDKGames)
    Obviously, if f(x,y) \leq f(a,b) in the vicinity of (a,b) then it is a local maximum. If the inequality is true for ALL (x,y) \in D (D is your domain) then f(a,b) is your global max.
    Similarly, for min points with f(x,y) \geq f(a,b)
    I should've asked this in my previous post but why does the value of f(x,y)-f(a,b) determines the nature of the stationary point (like you showed in your first post)?
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    (Original post by ManLike007)
    I should've asked this in my previous post but why does the value of f(x,y)-f(a,b) determines the nature of the stationary point (like you showed in your first post)?
    The value of that can be shown to be always either +ve or -ve in the vicinity of (a,b) so then then after getting \leq 0 (or \geq 0) you can just say f(x,y)-f(a,b) \leq 0 then add f(a,b) to both sides which tells you that f(a,b) is a max.
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    (Original post by RDKGames)
    The value of that can be shown to be always either +ve or -ve in the vicinity of (a,b) so then then after getting \leq 0 (or \geq 0) you can just say f(x,y)-f(a,b) \leq 0 then add f(a,b) to both sides which tells you that f(a,b) is a max.
    I see

    What if f(a,b)=0, is this an inflexion point or a saddle or what?
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    (Original post by ManLike007)
    I see

    What if f(a,b)=0, is this an inflexion point or a saddle or what?
    The value of f(a,b) doesn't dictate what the nature of the point is.

    As said, you need to use Taylor's expansion to determine whether a point is max/min/saddle/etc for a point that fails the second derivative test.
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    (Original post by RDKGames)
    The value of f(a,b) doesn't dictate what the nature of the point is.
    I'm very sorry, I can't tell if you've answered my question or not

    As in going back to what you said,

    (Original post by RDKGames)
    The value of that can be shown to be always either +ve or -ve in the vicinity of (a,b) so then then after getting \leq 0 (or \geq 0) you can just say f(x,y)-f(a,b) \leq 0 then add f(a,b) to both sides which tells you that f(a,b) is a max.
    What if f(a,b) =0? In this case it's neither maximum or minimum so what now?

    I'm awfully sorry for these string of questions!
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    (Original post by ManLike007)
    What if f(a,b) =0? In this case it's neither maximum or minimum so what now?

    I'm awfully sorry for these string of questions!
    I'm saying that you cannot draw a conclusion like that from merely saying that f(a,b) = 0. Don't assume the method I posted in post 6 applies to all functions!

    Going back to that example, I can redefine f(x,y) = x^4+2(xy)^2-y^4-2x^2 which still has a crit point at (0,0,0) (so f(0,0)=0) and D=0 and going through the same procedure shows that it is a max point - so f(a,b)=0 doesn't imply anything about the nature of the point (a,b,0) if it's a crit point!
 
 
 
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