Skip to main content

equation solving - How do I get all possible solutions in an underdetermined system?



I have two problems which I'd like to solve with Mathematica.


If I have a system of two equations with three unknowns, how can I get to list all possible solutions for the unknowns?


Here is what I have tried:


Solve[{ a + b + c == 5, 1/a + 1/b + 1/c == 1/5}, { a, b, c}]



Solve::svars: Equations may not give solutions for all "solve" variables. >> 

{{a -> 5, c -> -b}, {b -> 5, c -> -a}, {b -> -a, c -> 5}}

What would I change in this specific instance?


Here are the problems:


I


Suppose that $a, b, c$ are real numbers satisfying $a+b+c=5$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=+\frac{1}{5}$.

Find the greatest possible value of $a^3+b^3+c^3$


If I list all solutions I'll be able to choose all solutions maximizing $a^3+b^3+c^3$.


II


Finding integers $x, y$ and $z$ that satisfy this system:
$$\quad x^2 y + y^2 z + z^2 x = 2186 $$
$$\quad x y^2 + y z^2 + z x^2 = 2188$$.
evaluate $x^2+y^2+z^2$


The both problems can be found here (see exercises $27$ and $30$ ).



Answer



I



Let's write down an appropriate system we would like to solve,
i.e. we are to maximize a^3 + b^3 + c^3 knowing that a + b + c == 5 and 1/a + 1/b + 1/c == 1/5, thus the most direct approach uses Maximize with adequate conditions:


Maximize[{a^3 + b^3 + c^3, a + b + c == 5, 1/a + 1/b + 1/c == 1/5}, {a, b, c}]


{125, {a -> 1, b -> 5, c -> -1}}

With Maximize we can get only a specific solution, an example can be found here : How do I determine the maximum value for a polynomial, given a range of x values?, nevertheless we can remedy this problem using Lagrange multipliers, see e.g. How can I implement the method of Lagrange multipliers to find constrained extrema?.
However since there is a symmetry between a, b and c we can conclude that any permutation of this triple {a -> 1, b -> 5, c -> -1} is also a solution.


There are another ways to solve the problem which can be examined with the answers to these questions: Am I missing anything? Solving Equations

Efficient code for solve this equation


Let's provide the simplest:


Simplify[ a^3 + b^3 + c^3, {{a + b + c == 5, 1/a + 1/b + 1/c == 1/5}}]


125

II


Another question provides a nice example where a simple usage of Solve and Reduce with an appropriate domain specification will not be sufficient.


E.g. this yields a complicated system returning the solution but it doesn't clarify if another solutions really exist.



Reduce[ x^2 y + y^2 z + z^2 x == 2186 && x y^2 + y z^2 + z x^2 == 2188 && 
(x | y | z) ∈ Integers, {x, y, z}]

Thus we should approach the problem in a different way.
Let's notice that:


Simplify[ x y^2 + y z^2 + z x^2 - (x^2 y + y^2 z + z^2 x)]


-(x - y) (x - z) (y - z)


Now we can conclude that using slightly different system we can find an appropriate solution:


 x^2 + y^2 + z^2 /. Normal @ 
Solve[ x - y == a && x - z == b && y - z == c &&
x^2 y + y^2 z + z^2 x == 2186 && -a b c == 2, {x, y, z}, Integers]//Union//First


245    

Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....