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10

If I understand the question correctly, you wish to obtain a parameterized solution {U2[U1], W2[W1]} from the equation in the question, so that you can vary that parameter to obtain a "nice" solution. One approach is as follows. Define exp = 32 + 8 a^2 (2 + b) + 4 a (2 + b) (6 + b) - b (-4 (8 + U1 - W1 + U2[U1]) + b (-8 + (-2 + U1) U1 + W1^2 - 2 U2[...


6

Introduce a new variable $z=x^py^q$. eqs1 = {(y + 1) (a - z) == 0, (y + 1) z - c == 0} res1 = Solve[eqs1, {y, z}] eq2 = (z == x^p y^q /. First[res1]) Solve[eq2, x] The answer is $y=\frac{c-a}{a}$ and $x=\left(a \left(\frac{c-a}{a}\right)^{-q}\right)^{1/p}$.


5

sp1 = Sphere[{0, 0, 0}, 1]; sp2 = Sphere[{1, 1, 1}, 1.5]; ri1 = RegionIntersection[sp1, sp2]; l = MeshCoordinates[DiscretizeRegion[ri1]]; Show[Graphics3D[{Opacity[0.5], sp1, sp2, Thick, Red, Line@l[[Last[FindShortestTour[l]]]]}]] Note that you can find the center and radius of the circle with: o = Mean[l]; r = Mean[Table[Norm[l[[i]] - Mean[l]], {i, ...


3

As Mr. Wizard supposed in a comment, one can indeed use ContinuedFractionK[] here: With[{A = 3., B = 2., x = 0.1}, 1/(1 + I A x + ContinuedFractionK[-n^2/(4 n^2 - 1) x^2 A (1 - I 2 B x), 1 + I A x, {n, 2, 2000}])] 0.9197103744410972 - 0.28251974414934944 I However, if what you want is to approximate the ...


3

A common way, for a long time, of denoting a (generalized) continued fraction is to list the partial numerators and denominators, sometimes with $+$ and fraction bars like this: $$ F = b_0+ \frac{a_1}{b_1+}\, \frac{a_2}{b_2+}\, \frac{a_3}{b_3+}\cdots $$ It is also an efficient way to store a continued fraction (cf. ContinuedFraction). What is needed is a way ...


3

This answer uses the circle3D function written by Taiki, found here. But it specifically does not use any of the Region functionality to find the intersection. Instead, I just grabbed some formulas from this MathWorld page circle3D[centre_: {0, 0, 0}, radius_: 1, normal_: {0, 0, 1}, angle_: {0, 2 Pi}] := Composition[ Line, Map[RotationTransform[{...


2

If you want to simply sample the intersection, you can do this: RandomPoint[ ImplicitRegion[ RegionMember[ RegionIntersection[Sphere[{0, 0, 0}, 1], Sphere[{1, 1, 1}, 1.5]], {x, y, z}], {x, y, z}], 100] // Point // Graphics3D This should work, but maybe it doesn't because of the single-dimensional nature of the sampling region: ...


2

This is a solution to the problem with exactly the boundary conditions as stated: Dc = 1460 Kc = 9.41/(10^2) LaplacEquation = Dc*D[u[x, y], {x, 2}] + Dc*D[u[x, y], {y, 2}] - Kc*u[x, y] == 0 v[x] /. First[DSolve[{LaplacEquation /. u -> (v[#] &), v[0] == 1}, v[x], x]] (* ==> E^(-0.0080282 x) (1. - 1. C[1] + E^(0.0160564 x) C[1]) *) All I did ...


2

DSolve has limited abilities with PDEs. I can get part way there, maybe, but something is fishy about the result Mathematica returns. First, since your equation is linear and homogeneous and has a simple relationship among the coefficients, we can rewrite it as a BVP over a unit square: $$\nabla^2u + a\, u=0, u(x,0)=u(x,1),\ u(0,y)=1$$ It is apparently ...


2

Using the results from this MathWorld page: BlockRandom[SeedRandom["spherical"]; c1 = RandomReal[1, 3]; c2 = RandomReal[1, 3]; r2 = RandomReal[3/2]; r1 = RandomReal[3/2]; d = EuclideanDistance[c1, c2]; u = (d^2 + r1^2 - r2^2)/(2 d^2); cc = {1 - u, u}.{c1, c2}; rc = Sqrt[r1^2 - d^2 u^2]; ...


1

You can use Floor and Solve for part a. Solve[Floor[5 x] == 2 x + 1, x, Reals] x -> 1/2 FindInstance for part b (or a). FindInstance[Floor[1/5 x] > 1 + 2 x, {x}, Reals] x -> -(109/5)


1

Why not just generate the parametric equations for the intersection of the spheres? You know that the intersection points form a circle in a plane perpendicular to the vector between the centers of the spheres. sphereIntersection[{x1_, r1_}, {x2_, r2_}, th_] := Module[{dvec, d, d1, r, rr, ss, nx, ny, nz}, dvec = x2 - x1; d = Norm@dvec; dvec = {...


1

Or use my favorite command: FindInstance[ {(x - 1)^2 + (y - 1)^2 + (z - 1)^2 == 49, x != 1, y != 1,z != 1}, {x, y, z},Integers, 100]



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