calculus – Partial derivatives of $v(x,y) = x f(x+y) + y g(x,y) $

For $v(x,y) = x f(x+y) + y g(x,y)$ where $f$ and $g$ are twice differentiable, I need to prove that
$$ v_{xx} – 2v_{xy} + v_{yy} = 0 $$
(it is understood that $v_{xx} = frac{partial^2 v}{partial x^2}$ and $v_{xy} = frac{partial^2 v}{partial x partial y}$.)

My attempt:
$$v_{xx} = 2f_x(x+y) + xf_{xx}(x+y) + yg_{xx}(x+y),$$
$$v_{yy} =xf_{yy}(x+y) + 2g_y(x+y) + yg_{yy}(x+y),$$
$$-2v_{xy} = -2(f_y(x+y)+ xf_{xy}(x+y)+g_x(x+y)+yg_{xy}(x+y) ).$$

Only the problem is, these terms don’t add up to zero! I can’t see why. Your help will be appreciated!

Is $f_{XY}(x,y)$ the same as $f(x,y)$?

I have a slight confusion with the notation, sometimes I find the joint density function like this:
and on other occasions like this:
Is the same or what is the difference?

elementary set theory – Is it possible to define the standard function notation $f(x)=y$ in terms of an arbitrary relation?

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In elementary set theory (as far as I know) $f$ is a function if, and only if $f$ is a relation: $forall x(x in f rightarrow exists y exists z(x = langle y,z rangle))$, and
$f$ is many-one: $forall x forall y forall z(langle x,y rangle in f land langle x,z rangle in f rightarrow y=z)$

Now, in maths, it is very unusual to see the notation $langle x,y rangle in f$ being used, with the notation $f(x)=y$ being preferred. However, this type of notation doesn’t work for any relation. Specifically, it only works if the $y$ is unique for any given $x$. As an example, taking an relation $A={ langle 1,2 rangle, langle 1,1 rangle }$, using the notation $A(1) = 2$ and $A(1) = 1$, we can conclude that $1=2$, which is an absurd. This notation only works if the relation is a function. So we can define the notation as:

$$f(x)=y leftrightarrow f text{ is a function } land langle x,y rangle in f$$

Which is consistent.

However it would be very interesting if we could define a notation $f(x)$ for any relation, and then prove that it’s unique only for functions, and then be able to use it accordingly. A definition like:
$$f(x) leftrightarrow langle x,y rangle in f$$

is not really valid, since y is free but only appears on one side of the definition. We could define it like:
$$f(x)_y leftrightarrow langle x,y rangle in f$$

Which seems valid. From there, it is clear that for any relation $f$, $forall x forall y forall z (f(x)_y land f(x)_z rightarrow y=z) leftrightarrow f text{ is a function }$, which shows that the value of the subscript is unique. However, it doesn’t tell us how to “remove” it to use the unique value $f(x)$, or even that $f(x)$ is a variable. We could try:

$$f(x)=y leftrightarrow langle x,y rangle in f$$

However, that leads us right back to $1=2$.

So my question is, is it possible for us to define this type of notation for an arbitrary relation, and then prove that $f(x)=y leftrightarrow langle x,y rangle in f$ if, and only if, $f$ is a function?
Or is it necessary to embed the function status of $f$ in the definition?

The rings $ F[x,y]/ (y^{2} – x) $ and $F[x,y]/( y^{2} – x^{2}) $ for any field F

The rings $ F(x,y)/ (y^{2} – x) $ and $F(x,y)/( y^{2} – x^{2}) $ are not isomorphic for any field F.

finite element method – Mesh Refinement for interpolation of f[x,y] fails for all but the simplest f

I would like to interpolate a function, f(x,y,z), on a mesh of the unit cube, starting with a coarse tetrahedral mesh and refining tetrahedra as needed to reduce the interpolation error below a tolerance. The problem: DiscretizeRegion(Cuboid(),MeshRefinementFunction->mrf) ignores my mrf. In an attempt to understand, I have simplified to a toy version of the problem (2D, on a square instead of a cube, with a simple form for f). I discovered that of the following 5 mathematically equivalent forms for f, the last works and the others do not:

fDirect({x_, y_}) = Cos(2*Pi*x)*Sin(2*Pi*y)
fDelayed({x_, y_}) := Cos(2*Pi*x)*Sin(2*Pi*y)
fEmbedded1 = Function({p}, fDirect(p))
fEmbedded2 = fDirect(#1) & 
fPureFun = Function({p}, Cos(2*Pi*p((1)))*Sin(2*Pi*p((2))))

Here is what the unrefined mesh looks like:

sq = DiscretizeRegion(bm, MaxCellMeasure -> Infinity)

Coarse mesh of unit square without refinement

The resulting mesh has 4 nodes and 2 triangular elements.

Now we define a refinement function. For the toy problem I have adopted a simplified form of the error estimate. I check the error only at the center of each tetrahedron. The center is at Mean(vertices). The true value there is f(Mean(vertices)). My “interpolated value” there is just the mean of f at the individual vertices, Mean(f/@vertices). When Abs(error)>tolerance, the function returns True. A diagnostic print statement lets us see when the mesh refinement function is called.

This does not work for the first definition of f. Here is what that looks like:

 mrf = Function({vertices, area}, 
   Module({fest, ftrue, error, tolerance, test}, 
    fest = Mean(fDirect /@ vertices); 
    ftrue = fDirect(Mean(vertices)); 
    error = fest - ftrue; tolerance = 0.43; 
    test = Abs(error) > tolerance; 
    Print({vertices, area, fest, ftrue, error, tolerance, test}); 
sq = DiscretizeRegion(bm, MeshRefinementFunction -> mrf, 
  MaxCellMeasure -> Infinity)
Print("The mesh has ", Length(mc = MeshCoordinates(sq)), " nodes.")
Print("The mesh has ", 
 Length(mcells = MeshCells(sq, 2)), " triangular cells.")

Failed mesh refinement using a direct definition of f(x,y)

Notice that the mrf is called only once (there is output for only one diagnostic Print()) despite that there are two triangles. In that one call the error exceeds tolerance and the mrf returns true (i.e., do a refinement), yet there is no subsequent refinement. Definitions 2 – 4 produce the same result.

It does work for definition 5. Here is what it’s supposed to look like when it works:

mrf = Function({vertices, area}, 
   Module({fest, ftrue, error, tolerance, test},
    fest = Mean(fPureFun /@ vertices);
    ftrue = fPureFun(Mean(vertices));
    error = fest - ftrue;
    tolerance = 0.43;
    test = Abs(error) > tolerance;
    Print({vertices, area, fest, ftrue, error, tolerance, test});
sq = DiscretizeRegion(bm, MeshRefinementFunction -> mrf, 
  MaxCellMeasure -> (Infinity))
Print("The mesh has ", Length(mc = MeshCoordinates(sq)), " nodes.")
Print("The mesh has ", 
 Length(mcells = MeshCells(sq, 2)), " triangular cells.")

Working mesh refinement with f defined as a pure function

In the working case, we get a diagnostic print output for each cell of the mesh with repeats for several iterations, until all triangular elements have interpolation error estimates below tolerance.

It seems odd that the change that makes a difference here is a difference in the definition of f, which is internal to a module within a function definition. This seems a failure of encapsulation.

Definitions 3 and 4 were attempts to hide one of the apparently unacceptable forms of definition inside of the acceptable pure function definition form, but this did not work.

Unfortunately, my non-toy version of f is not easily written as a pure function (the one mode that worked above). That function is the end result of tens of pages of development within the notebook. It calls a number of functions, uses a numerical differential equation solver, an iterative root finder,… It’s not a simple function, which is why I want to tabulate and interpolate it.

Is this a bug? Is there a workaround–another way to access mesh refinement within Mathematica? Or would this work if only I did it correctly? (How does one do it correctly?)

multivariable calculus – Find all points (a,b,c) where the graph z = f(x,y) has a horizontal tangent plane.

Given f(x,y)=x^2 + 4xy + y^2 – 2x + 2y + 1

I tried setting F(x,y,z) = f(x,y) – z and found the partial derivatives Fx, Fy, and Fz.

Fx = 2x + 4y -2

Fy = 4x + 2y + 2

Fz = -1

Apparently the gradient is perpendicular to the tangent plane, so it must be parallel to <0,0,1> which the normal vector of the xy plane. So I set Fx = 0 and Fy = 0 and solved for x and y, which turns out to be -1 and 1 respectively. But I don’t know where to go from here, or whether what I have done so far is correct at all. Please help.

limits – Show that $lvert lvert (x,y) rvert rvert rightarrow infty$ implies $f(x,y) = x^2-4xy+4y^2+y^4-2y^3+y^2 rightarrow infty$

I am working on the following exercise:

Consider the function
$$f(x,y) = x^2-4xy+4y^2+y^4-2y^3+y^2.$$

Show that $$lvert lvert (x,y) rvert rvert rightarrow infty quad Longrightarrow quad f(x,y) rightarrow+infty.$$

I do not see how I could prove this in a “nice way”. Is there any way to avoid a lot of case distinctions, so to avoid considering each case like

$$x rightarrow +infty, y rightarrow -infty $$
$$x rightarrow -infty, y rightarrow -infty $$

and so on separately?

abstract algebra – Showing $ F[x,y]/ (ax ^ 2 + by ^ 2-1) $ is a Dedekind domain

Try to follow this track …

$ (i) $ $ R $ is Noetherian since it is quotient of a Noetherian domain, that is, the polynomial ring $ F (x, y) $.
Let us remember that this last fact follows from Hilbert's Base Theorem.

$ (ii) $ The fraction field of $ R $, which we denote by $ k $, is given by $ k = F (x) (y) / (ax ^ 2 + by ^ 2-1) $, therefore you can verify that $ R $ is fully closed from $ F (x) $ is.

$ (iii) $ Canonical inclusion. $ F (x) hookrightarrow R $ it is integral, therefore $ dim R = dim F (x) = 1 $, where $ dim $ is Krull's dimension

Observation Yes $ f colon A longrightarrow B $ it is injective and integral so $ dim A = dim B $, this continues Ascent theorem.