Skip to main content

Solving system of equations with Root


I have this system of equations. I want to express all variables in terms of x as you can see. x is a parameter.


Solve[{y (z - x) == x^6, x zg - z zg == x z^2 za zb - x z za zb zg, 
za zg - zb zg == z za zb - z za^2 zb, zb - zg == z za zb - z zb^2,
za zb (z - zg) == zg (-zb + zg), za > 0, zb > 0, zg > 0, z > 0}, {y, za, zb, zg, z}, Reals]


However Mathematica doesn't know how to solve it. Solutions show me something called Root and I can't make sense out of any of the expressions I get. What I get is:


y -> ConditionalExpression[((
x^6)/(-x +
Root[-x +
2 x^2 + (1 - 3 x - x^2) #1 + (1 + 2 x - 4 x^2) #1^2 + (-1 +
5 x - x^2) #1^3 + (-1 + x - x^2) #1^4 + x^2 #1^5 &, 3]))

I also tried using reduce which gives me a set of solutions for y:


y == Root[

x^26 + (-x^18 + x^19 - x^20 + 5 x^21) #1 + (-x^12 + x^13 + 3 x^14 -
4 x^15 + 10 x^16) #1^2 + (x^6 - x^7 + 5 x^8 + 3 x^9 - 6 x^10 +
10 x^11) #1^3 + (1 - x + 3 x^3 + x^4 - 4 x^5 +
5 x^6) #1^4 + (-1 + x) #1^5 &, 1]

I've read this thread: How do I work with Root objects? but it doesn't solve my problem since I have this free parameter x that ruins everything. For example if I try ToRadicals or numerical value I don't get anything. How do I make sense of it? How can I proceed to get the full expression for y in terms of x without Root? I tried FindInstance but it doesn't work in my case since I have this free parameter x. If this helps I'm only interested in y, the rest of the parameters don't really matter. Any help is greatly appreciated.



Answer



From a comment, it seems maybe the OP would like to rationalize the expression:


Product[y - 
Root[x^26 + (-x^18 + x^19 - x^20 + 5 x^21) #1 + (-x^12 + x^13 +

3 x^14 - 4 x^15 + 10 x^16) #1^2 + (x^6 - x^7 + 5 x^8 +
3 x^9 - 6 x^10 + 10 x^11) #1^3 + (1 - x + 3 x^3 + x^4 -
4 x^5 + 5 x^6) #1^4 + (-1 + x) #1^5 &, i], {i,
5}] // FullSimplify
(*
(1/(-1 + x))(x^26 + x^18 (-1 + x - x^2 + 5 x^3) y +
x^12 (-1 + x (1 + x (3 + 2 x (-2 + 5 x)))) y^2 +
x^6 (1 + x (-1 + x (5 + x (3 + 2 x (-3 + 5 x))))) y^3 + (1 - x +
3 x^3 + x^4 - 4 x^5 + 5 x^6) y^4 + (-1 + x) y^5)
*)


Set this equal to zero to represent the locus. Note that the product may or may not have introduced extraneous solutions.




Addendum


Silly me, this is equivalent to, with less work, the following:


First[
Root[x^26 + (-x^18 + x^19 - x^20 + 5 x^21) #1 + (-x^12 + x^13 +
3 x^14 - 4 x^15 + 10 x^16) #1^2 + (x^6 - x^7 + 5 x^8 +
3 x^9 - 6 x^10 + 10 x^11) #1^3 + (1 - x + 3 x^3 + x^4 -
4 x^5 + 5 x^6) #1^4 + (-1 + x) #1^5 &, 1]

]@y
(*
x^26 + (-x^18 + x^19 - x^20 + 5 x^21) y + (-x^12 + x^13 + 3 x^14 -
4 x^15 + 10 x^16) y^2 + (x^6 - x^7 + 5 x^8 + 3 x^9 - 6 x^10 +
10 x^11) y^3 + (1 - x + 3 x^3 + x^4 - 4 x^5 + 5 x^6) y^4 + (-1 +
x) y^5
*)

There is still the general caveat about rationalizing leads potentially to extraneously solutions.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...