Skip to main content

Am I missing anything? Solving Equations


I want to know what $ a^4+b^4+c^4 $ would be when


$ a+b+c=1, a^2+b^2+c^2=2, a^3+b^3+c^3=3. $


I tried


Solve[{a + b + c == 1, a^2 + b^2 + c^2 == 2, 
a^3 + b^3 + c^3 == 3}, {a, b, c}]


a^4 + b^4 + c^4 /. {a -> ... }

I copied and pasted one instance from the result of Solve[] into {a -> ...}


The solution and the result of Simplify[] of it was not quite what I expected.


So as a test, I tried a+b+c /. {a -> ...}


Still it was strange, it didn't come up with 1 which is obvious correct answer.


So am I missing anything here?


Is there any better way?



Answer



Solve your equations with respect to another variable d == a^4 + b^4 + c^4 eliminating given ones



Solve[{ a + b + c == 1,       a^2 + b^2 + c^2 == 2, 
a^3 + b^3 + c^3 == 3, a^4 + b^4 + c^4 == d}, {d}, {a, b, c}]


{{d -> 25/6}}

This tutorial will be helpful Eliminating Variables.


Edit


For the sake of completeness we add another methods of dealing with the problem. Let's point out even simpler way than the first one, just add assumptions to Simplify:


Simplify[ a^4 + b^4 + c^4, {a + b + c == 1, a^2 + b^2 + c^2 == 2, a^3 + b^3 + c^3 == 3}]



25/6

Of course we can use FullSimplify as well. In this case, Simplify works nicely, but the first method is stronger and should work in more general cases.


SymmetricReduction is more appropriate when we would like to get symbolic results, i.e. here we would like to express a^4 + b^4 + c^4 in terms of these polynomials: {a + b + c, a^2 + b^2 + c^2, a^3 + b^3 + c^3}


First we need to get rid of unknown symmetric polynomials (we know only a + b + c):


 SymmetricPolynomial[#, {a, b, c}] & /@ Range[3]



{a + b + c, a b + a c + b c, a b c}

form the last reduction using the previous ones:


First @ SymmetricReduction[ a^# + b^# + c^#, {a, b, c}] & /@ Range[4] // Column


a + b + c

(a + b + c)^2 - 2 (a b + a c + b c)


3 a b c + (a + b + c)^3 - 3 (a + b + c) (a b + a c + b c)

4 a b c (a + b + c) + (a + b + c)^4 - 4 (a + b + c)^2 (a b + a c + b c)
+ 2 (a b + a c + b c)^2

Now we have a^4 + b^4 + c^4 expressed in terms of given polynomials:


str = 
First[ SymmetricReduction[ a^4 + b^4 + c^4, {a, b, c}]] //. {
(a b + a c + b c) -> 1/2 (a + b + c)^2 - 1/2 (a^2 + b^2 + c^2),
(a b c) -> 1/3 (a^3 + b^3 + c^3) - 1/3 (a + b + c)^3 + (a + b + c) (a b + a c + b c)}



 (a + b + c)^4 - 4 (a + b + c)^2 (1/2 (a + b + c)^2 + 1/2 (-a^2 - b^2 - c^2))
+ 2 (1/2(a + b + c)^2 + 1/2(-a^2 - b^2 - c^2))^2 + 4(a + b + c) (-(1/3)(a + b + c)^3
+ 1/3 (a^3 + b^3 + c^3) + (a + b + c) (1/2 (a + b + c)^2 + 1/2 (-a^2 - b^2 - c^2)))

of course


Simplify @ %



a^4 + b^4 + c^4

and finally:


str /. {a + b + c -> 1, (-a^2 - b^2 - c^2) -> -2, (a^3 + b^3 + c^3) -> 3}


25/6

Comments

Popular posts from this blog

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 - 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 - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],