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  • one year ago

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  1. anonymous
    • one year ago
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    \[\frac{\partial^2f(x,y)}{\partial x \partial y}=\frac{\partial^2f(x,y)}{\partial y \partial x}\] Is this always true???

  2. anonymous
    • one year ago
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    @Loser66

  3. Loser66
    • one year ago
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    NOPPPPPe. It's true IF AND ONLY IF f(x,y) is a symmetric one.

  4. anonymous
    • one year ago
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    What's that mean ??

  5. Loser66
    • one year ago
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    like \(f(x,y)= x^2 + y^2 + x +y\) what happens to x, it happens to y. if \(f(x,y) = x^2 + y \) then partial x, y is different from partial y, x

  6. anonymous
    • one year ago
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    wait symmetric as in do u mean f(x,y)=f(y,x)??

  7. Loser66
    • one year ago
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    kind of

  8. Loser66
    • one year ago
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    Other example: \(f(x,y) = 2x^2 + 5xy + 2y^2\) partial w.r.t x = \(f'_x= 4x +5y\), then w.r.t y \(f"_{x,y}= 5\) partial w.r.t.y = \(f'_y =4y +5x\) then, w.r.t x \(f"_{y,x}= 5\)

  9. Loser66
    • one year ago
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    to higher degree of a function, the letters switch to each other, but the answer are the same like \(f(x,y) = x^3+ 20x^2y + y^3\), it is a symmetric one, to find \(f"_{x,y}~~f"_{y,x}\), you can save time by doing just one, then switch the letters.

  10. IrishBoy123
    • one year ago
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    .

  11. anonymous
    • one year ago
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    hmm kk

  12. IrishBoy123
    • one year ago
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    according to the Gospel of Mary (Boas : Mathematical Methods in Physical Sciences [p190, 3rd Edition, if you have access] ), so long as \(f_x, f_y, f_{xx} and f_{yy}\) are continuous, then \(f_{xy} = f_{yx}\) here's an example of an exception: http://www.math.tamu.edu/~tvogel/gallery/node18.html and if you google something like "mixed partials not equal" i think you will find more stuff. my sense FWIW is that if you are doing physical/applied stuff you will be more than aware of the discontinuity and can probably even find a way around it.

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