Prev: math solution, fyi
Next: 2BILLION PLACED PURE PRIME NUMBERS- PROVES THE SUPERIORITY OF NEW -1 TANGENT MATHEMATICS
From: BURT on 11 Jun 2010 20:57
On Jun 11, 4:01 pm, JEMebius <jemeb...(a)xs4all.nl> wrote:
> gudi wrote:
> > On Jun 8, 11:45 pm, JEMebius <jemeb...(a)xs4all.nl> wrote:
> >> gudi wrote:
> >>> If u(x,y) ,v(x,y)  are real/imaginary parts of an analytical function
> >>> f( x + i y), under what conditions does surface Z = u(x,y) ( in Monge
> >>> form ) have a positive  Gauss curvature K? For most of the functions I
> >>> find K is negative. Is it always so? Please give examples when  K >0.
> >>> TIA,
> >>> Narasimham
> >> The real and imaginary parts u and v of an analytic function u + vi = f(x + yi) are
> >> two-dimensional harmonic functions u(x, y) and v(x, y) of x and y, i.e.. they satisfy the
> >> 2D Laplace equation d^2 u/dx^2 + d^2 u/dy^2 = 0; analogously for v(x, y).
>
> >> Harmonic functions have the mean-value property:
> >> the mean value of U(x, y) taken over a circle with centre (X, Y) within the domain of U
> >> equals U(X, Y), the value at the centre.
>
> >> This implies that the 3D graph z = u(x, y) cannot have positive Gaussian curvature at any
> >> point. For instance: consider the neighbourhood of a local maximum or minimum.
>
> >> Seehttp://en.wikipedia.org/wiki/Harmonic_function, in particularhttp://en.wikipedia.org/wiki/Harmonic_function#Mean_value_property
>
> >> Ciao: Johan E. Mebius
>
> > Even after seeing the above references,I cannot find from them how the
> > mean value property implies to a conclusion that every 2 dimensional
> > harmonic function z = u(x, y) should have a negative Gaussian
> > curvature. Can you please help explain?
>
> > Thanking in advance,
>
> > Best Regards,
> > Narasimham
>
> Consider a would-be twice continuously differentiable function U(x, y).
> Consider a circle C with radius R centered at (X, Y) inside the domain of function U.
>
> The Taylor series of U at (X, Y) reads
>
> U(X+h, Y+k) = U(X, Y) + h.Ux + k.Uy + h^2 . Uxx/2 + k^2 . Uyy/2 + h.k.Uxy + o(h^2 , k^2)
>
> The mean-value property remains intact when subtracting the constant term and the terms
> linear in h and k: if U has the mean-value property, then
>
> W(X+h, Y+k) = h^2 . Uxx/2 + k^2 . Uyy/2 + h.k.Uxy + o(h^2 , k^2)
>
> has the mean-value property, and vice versa.
>
> Now assume that the graph of W has a positive curvature at (X, Y), then we have in a
> coordinate system (x', y') based on the principal curvatures Cx' and Cy' the equality
>
> W(X+h, Y+k) = Cx' . x'^2 / 2 + Cy' . y'^2 / 2 + o(x'^2 , y'^2)
>
> with Cx' and Cy' both positive or both negative.
> So the quadratic form is strictly definite positive or negative, and there exists a circle
> around (X, Y) such that W(x, y) is positive or negative on the entire circle.
> And we have W(X, Y) = 0. Therefore the mean-value property fails.
>
> Conclusion: the graph of U(x, y) either has zero curvature, in which case U = A + Bx + Cy
> for some A, B, C; or the graph has negative curvature, in which case the quadratic part of
> the Taylor series is an indefinite quadratic function.
>
> Advice: any textbook on general calculus (univariate and multivariate calculus) treats
> this (IMO rather classical) subject.
>
> Happy studies: Johan E. Mebius- Hide quoted text -
>
> - Show quoted text -

A sphere is an hypercircle. It is round 3D curvature.

Mitch Raemsch
From: gudi on 12 Jun 2010 08:58
On Jun 12, 4:01 am, JEMebius <jemeb...(a)xs4all.nl> wrote:
> gudi wrote:
> > On Jun 8, 11:45 pm, JEMebius <jemeb...(a)xs4all.nl> wrote:
> >> gudi wrote:
> >>> If u(x,y) ,v(x,y)  are real/imaginary parts of an analytical function
> >>> f( x + i y), under what conditions does surface Z = u(x,y) ( in Monge
> >>> form ) have a positive  Gauss curvature K? For most of the functions I
> >>> find K is negative. Is it always so? Please give examples when  K >0.
> >>> TIA,
> >>> Narasimham
> >> The real and imaginary parts u and v of an analytic function u + vi = f(x + yi) are
> >> two-dimensional harmonic functions u(x, y) and v(x, y) of x and y, i.e.. they satisfy the
> >> 2D Laplace equation d^2 u/dx^2 + d^2 u/dy^2 = 0; analogously for v(x, y).
>
> >> Harmonic functions have the mean-value property:
> >> the mean value of U(x, y) taken over a circle with centre (X, Y) within the domain of U
> >> equals U(X, Y), the value at the centre.
>
> >> This implies that the 3D graph z = u(x, y) cannot have positive Gaussian curvature at any
> >> point. For instance: consider the neighbourhood of a local maximum or minimum.
>
> >> Seehttp://en.wikipedia.org/wiki/Harmonic_function, in particularhttp://en.wikipedia.org/wiki/Harmonic_function#Mean_value_property
>
> >> Ciao: Johan E. Mebius
>
> > Even after seeing the above references,I cannot find from them how the
> > mean value property implies to a conclusion that every 2 dimensional
> > harmonic function z = u(x, y) should have a negative Gaussian
> > curvature. Can you please help explain?
>
> > Thanking in advance,
>
> > Best Regards,
> > Narasimham
>
> Consider a would-be twice continuously differentiable function U(x, y).
> Consider a circle C with radius R centered at (X, Y) inside the domain of function U.
>
> The Taylor series of U at (X, Y) reads
>
> U(X+h, Y+k) = U(X, Y) + h.Ux + k.Uy + h^2 . Uxx/2 + k^2 . Uyy/2 + h.k.Uxy + o(h^2 , k^2)
>
> The mean-value property remains intact when subtracting the constant term and the terms
> linear in h and k: if U has the mean-value property, then
>
> W(X+h, Y+k) = h^2 . Uxx/2 + k^2 . Uyy/2 + h.k.Uxy + o(h^2 , k^2)
>
> has the mean-value property, and vice versa.
>
> Now assume that the graph of W has a positive curvature at (X, Y), then we have in a
> coordinate system (x', y') based on the principal curvatures Cx' and Cy' the equality
>
> W(X+h, Y+k) = Cx' . x'^2 / 2 + Cy' . y'^2 / 2 + o(x'^2 , y'^2)
>
> with Cx' and Cy' both positive or both negative.
> So the quadratic form is strictly definite positive or negative, and there exists a circle
> around (X, Y) such that W(x, y) is positive or negative on the entire circle.
> And we have W(X, Y) = 0. Therefore the mean-value property fails.
>
> Conclusion: the graph of U(x, y) either has zero curvature, in which case U = A + Bx + Cy
> for some A, B, C; or the graph has negative curvature, in which case the quadratic part of
> the Taylor series is an indefinite quadratic function.
>
> Advice: any textbook on general calculus (univariate and multivariate calculus) treats
> this (IMO rather classical) subject.
>
> Happy studies: Johan E. Mebius

Just now I tried out this Mma program, without polynomials.3D plot
shows half the area has positive Gauss curvature and the rest
negative,providing what seems to be a counter example.

Expand[(Sin[x] + I Sin[y])^2]
Plot3D[Sin[x]^2 - Sin[y]^2, {y, 0, 2 Pi}, {x, 0, 2 Pi},
Mesh -> {7, 12}, PlotRange -> All]
Plot3D[ Sin[y] Sin[x], {x, 0, 4 Pi}, {y, 0, 4 Pi}, Mesh -> {11, 19},
PlotRange -> All]

What am I missing here?

Narasimham


 | 
Pages: 1
Prev: math solution, fyi
Next: 2BILLION PLACED PURE PRIME NUMBERS- PROVES THE SUPERIORITY OF NEW -1 TANGENT MATHEMATICS