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For a polynomial F with dependence on a parameter beta, find the value of beta for which F has 2 positive roots
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Good morning,
I have the following polynomial with degree 4 and its coefficients are quadratic functions of a parameter beta:
FF[z_, β_] := z^4 + z^3 (4.67687 - 0.16 β^2) + z^2 (-0.703991 - 0.652433 β^2) +
z (0.0676982 - 0.049884 β^2) - 0.000608244 β^2 + 0.0000234502;
Consider z > 0 and β > 0. From the code
Manipulate[Plot[FF[z, param], z, 0, 0.2, PlotRange -> All,
PlotLabel -> FF[z, param]],
Column[z /. NSolve[FF[z, param] == 0, z]], CountRoots[FF[z, param], z, 0, ∞],
param, 0.6, 0.18, 0.8]
I know that for some $beta$ (close to 0.6), the quartic polynomial has exactly 2 positive roots. I'd like to find such a value of beta.
I've tried several things. For instance:
FindRoot[NMinimize[FF[z, β], 0.05 < z < 0.15, z], β, 0.6]
and
NSolve[CountRoots[FF[z, β], z, 0, Infinity] == 2, β, 0.6, 0.8]
I appreciate some suggestions.
mathematical-optimization polynomials
edited 48 mins ago
J. M. is computer-less♦
95.1k10295454
asked 13 hours ago
Alex Pereira
161
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add a comment |Â
up vote
3
down vote
favorite
Good morning,
I have the following polynomial with degree 4 and its coefficients are quadratic functions of a parameter beta:
FF[z_, β_] := z^4 + z^3 (4.67687 - 0.16 β^2) + z^2 (-0.703991 - 0.652433 β^2) +
z (0.0676982 - 0.049884 β^2) - 0.000608244 β^2 + 0.0000234502;
Consider z > 0 and β > 0. From the code
Manipulate[Plot[FF[z, param], z, 0, 0.2, PlotRange -> All,
PlotLabel -> FF[z, param]],
Column[z /. NSolve[FF[z, param] == 0, z]], CountRoots[FF[z, param], z, 0, ∞],
param, 0.6, 0.18, 0.8]
I know that for some $beta$ (close to 0.6), the quartic polynomial has exactly 2 positive roots. I'd like to find such a value of beta.
I've tried several things. For instance:
FindRoot[NMinimize[FF[z, β], 0.05 < z < 0.15, z], β, 0.6]
and
NSolve[CountRoots[FF[z, β], z, 0, Infinity] == 2, β, 0.6, 0.8]
I appreciate some suggestions.
mathematical-optimization polynomials
edited 48 mins ago
J. M. is computer-less♦
95.1k10295454
asked 13 hours ago
Alex Pereira
161
New contributor
Alex Pereira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
With brute force search: $beta =0.60705233167945$
– Mariusz Iwaniuk
11 hours ago
@user6014 Well, I'm interested about the positive roots. $beta = 0.6$ and $beta=0.601$ provide only one positive solution ;)
– Alex Pereira
10 hours ago
@mariusz-iwaniuk, can you show us how to do it? I'm asking this because I'll probably have to consider this problem again but with different numbers on the coefficients of F
– Alex Pereira
10 hours ago
1
This article might be useful.
– J. M. is computer-less♦
21 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Good morning,
I have the following polynomial with degree 4 and its coefficients are quadratic functions of a parameter beta:
FF[z_, β_] := z^4 + z^3 (4.67687 - 0.16 β^2) + z^2 (-0.703991 - 0.652433 β^2) +
z (0.0676982 - 0.049884 β^2) - 0.000608244 β^2 + 0.0000234502;
Consider z > 0 and β > 0. From the code
Manipulate[Plot[FF[z, param], z, 0, 0.2, PlotRange -> All,
PlotLabel -> FF[z, param]],
Column[z /. NSolve[FF[z, param] == 0, z]], CountRoots[FF[z, param], z, 0, ∞],
param, 0.6, 0.18, 0.8]
I know that for some $beta$ (close to 0.6), the quartic polynomial has exactly 2 positive roots. I'd like to find such a value of beta.
I've tried several things. For instance:
FindRoot[NMinimize[FF[z, β], 0.05 < z < 0.15, z], β, 0.6]
and
NSolve[CountRoots[FF[z, β], z, 0, Infinity] == 2, β, 0.6, 0.8]
I appreciate some suggestions.
mathematical-optimization polynomials
edited 48 mins ago
J. M. is computer-less♦
95.1k10295454
asked 13 hours ago
Alex Pereira
161
New contributor
Alex Pereira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Good morning,
I have the following polynomial with degree 4 and its coefficients are quadratic functions of a parameter beta:
FF[z_, β_] := z^4 + z^3 (4.67687 - 0.16 β^2) + z^2 (-0.703991 - 0.652433 β^2) +
z (0.0676982 - 0.049884 β^2) - 0.000608244 β^2 + 0.0000234502;
Consider z > 0 and β > 0. From the code
Manipulate[Plot[FF[z, param], z, 0, 0.2, PlotRange -> All,
PlotLabel -> FF[z, param]],
Column[z /. NSolve[FF[z, param] == 0, z]], CountRoots[FF[z, param], z, 0, ∞],
param, 0.6, 0.18, 0.8]
I know that for some $beta$ (close to 0.6), the quartic polynomial has exactly 2 positive roots. I'd like to find such a value of beta.
I've tried several things. For instance:
FindRoot[NMinimize[FF[z, β], 0.05 < z < 0.15, z], β, 0.6]
and
NSolve[CountRoots[FF[z, β], z, 0, Infinity] == 2, β, 0.6, 0.8]
I appreciate some suggestions.
mathematical-optimization polynomials
mathematical-optimization polynomials
edited 48 mins ago
J. M. is computer-less♦
95.1k10295454
asked 13 hours ago
Alex Pereira
161
New contributor
Alex Pereira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 48 mins ago
J. M. is computer-less♦
95.1k10295454
asked 13 hours ago
Alex Pereira
161
New contributor
Alex Pereira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 48 mins ago
J. M. is computer-less♦
95.1k10295454
edited 48 mins ago
J. M. is computer-less♦
95.1k10295454
edited 48 mins ago
J. M. is computer-less♦
95.1k10295454
95.1k10295454
asked 13 hours ago
Alex Pereira
161
New contributor
Alex Pereira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 13 hours ago
Alex Pereira
161
asked 13 hours ago
Alex Pereira
161
161
New contributor
Alex Pereira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alex Pereira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alex Pereira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
With brute force search: $beta =0.60705233167945$
– Mariusz Iwaniuk
11 hours ago
@user6014 Well, I'm interested about the positive roots. $beta = 0.6$ and $beta=0.601$ provide only one positive solution ;)
– Alex Pereira
10 hours ago
@mariusz-iwaniuk, can you show us how to do it? I'm asking this because I'll probably have to consider this problem again but with different numbers on the coefficients of F
– Alex Pereira
10 hours ago
1
This article might be useful.
– J. M. is computer-less♦
21 mins ago
add a comment |Â
With brute force search: $beta =0.60705233167945$
– Mariusz Iwaniuk
11 hours ago
@user6014 Well, I'm interested about the positive roots. $beta = 0.6$ and $beta=0.601$ provide only one positive solution ;)
– Alex Pereira
10 hours ago
@mariusz-iwaniuk, can you show us how to do it? I'm asking this because I'll probably have to consider this problem again but with different numbers on the coefficients of F
– Alex Pereira
10 hours ago
1
This article might be useful.
– J. M. is computer-less♦
21 mins ago
With brute force search: $beta =0.60705233167945$
– Mariusz Iwaniuk
11 hours ago
With brute force search: $beta =0.60705233167945$
– Mariusz Iwaniuk
11 hours ago
@user6014 Well, I'm interested about the positive roots. $beta = 0.6$ and $beta=0.601$ provide only one positive solution ;)
– Alex Pereira
10 hours ago
@user6014 Well, I'm interested about the positive roots. $beta = 0.6$ and $beta=0.601$ provide only one positive solution ;)
– Alex Pereira
10 hours ago
@mariusz-iwaniuk, can you show us how to do it? I'm asking this because I'll probably have to consider this problem again but with different numbers on the coefficients of F
– Alex Pereira
10 hours ago
@mariusz-iwaniuk, can you show us how to do it? I'm asking this because I'll probably have to consider this problem again but with different numbers on the coefficients of F
– Alex Pereira
10 hours ago
1
1
This article might be useful.
– J. M. is computer-less♦
21 mins ago
This article might be useful.
– J. M. is computer-less♦
21 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
You can brute force this using a method inspired by binary search.
Your function:
FF[z_, b_] :=
z^4 + z^3 (4.67687 - 0.16 b^2) +
z^2 (-0.703991 - 0.652433 b^2) +
z (0.0676982 - 0.049884 b^2) - 0.000608244 b^2 +
0.0000234502;
Function which cuts the min/max range in half each time, outputting the new min/max to search:
f[x1_, x2_] :=
Module[v1, v2, v3, min = N[x1, 30], max = N[x2, 30], vals,
v1 = NSolve[FF[z, min] == 0, z, Reals];
v2 = NSolve[FF[z, max] == 0, z, Reals];
v3 = NSolve[FF[z, (min + max)/2] == 0, z, Reals];
If[Length@v3 <= 3,
vals = (min + max)/2, max,
vals = min, (min + max)/2
];
vals
]
And using our custom function above with FixedPoint we get a range that is hopefully specific enough for your application (with the floating point numbers in your original equations we're restricted to $MachinePrecision):
In[9]:= FixedPoint[f, .6, .61]
Out[9]= 0.6070523316794493, 0.6070523316794494
So as @Mariusz Iwaniuk said in the comments, you're looking for a number in the ballpark of $0.607052331679449$.
Note that my implementation isn't very general and relies on your specific FF being the format it is. No doubt it could be generalized to fit a broader scope of function definitions, though.
There may be a more elegant math-y way to solve it, but I just resort to computational/numerical methods above.
answered 5 hours ago
user6014
2,9261123
add a comment |Â
up vote
1
down vote
First an extended comment:.
You could use ContourPlot to get an overview.
eq=z^4 + z^3 (4.67687 - 0.16 [Beta]^2) +z^2 (-0.703991 - 0.652433 [Beta]^2) +z (0.0676982 - 0.049884 [Beta]^2) -0.000608244 [Beta]^2 +0.0000234502
ContourPlot[0 ==eq, [Beta], 0, 1, z, -6, 1, MaxRecursion -> 5 ,FrameLabel -> [Beta],z]
The solutionplot shows two ranges [Beta]<0.6 and [Beta]>0.78 with only two solutions!
solution
The intersting branch of solution is in the range z0<z<z1
[Beta]0, z0 = 0, z /. Solve[eq == 0 /. [Beta] -> 0, z][[2]]
(*0, -0.000345152*)
[Beta]1, z1 = 1, z /. Solve[eq == 0 /. [Beta] -> 1, z][[4]]
(*1, 0.271234*)
Now it's easy to calculate
beta = [Beta] /. Solve[eq == 0, [Beta]][[2]] (* beta[z]*)
ParametricPlot[z, beta, z, z0, z1, AxesLabel -> "z", "[Beta]"]
The two extrema define the beta-values we are looking for
beta /. NSolve[D[beta, z] == 0, 0 < z < z1, z]
(*0.779012, 0.607052*)
That's it. Hope it helps!
answered 1 hour ago
Ulrich Neumann
5,230413
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You can brute force this using a method inspired by binary search.
Your function:
FF[z_, b_] :=
z^4 + z^3 (4.67687 - 0.16 b^2) +
z^2 (-0.703991 - 0.652433 b^2) +
z (0.0676982 - 0.049884 b^2) - 0.000608244 b^2 +
0.0000234502;
Function which cuts the min/max range in half each time, outputting the new min/max to search:
f[x1_, x2_] :=
Module[v1, v2, v3, min = N[x1, 30], max = N[x2, 30], vals,
v1 = NSolve[FF[z, min] == 0, z, Reals];
v2 = NSolve[FF[z, max] == 0, z, Reals];
v3 = NSolve[FF[z, (min + max)/2] == 0, z, Reals];
If[Length@v3 <= 3,
vals = (min + max)/2, max,
vals = min, (min + max)/2
];
vals
]
And using our custom function above with FixedPoint we get a range that is hopefully specific enough for your application (with the floating point numbers in your original equations we're restricted to $MachinePrecision):
In[9]:= FixedPoint[f, .6, .61]
Out[9]= 0.6070523316794493, 0.6070523316794494
So as @Mariusz Iwaniuk said in the comments, you're looking for a number in the ballpark of $0.607052331679449$.
Note that my implementation isn't very general and relies on your specific FF being the format it is. No doubt it could be generalized to fit a broader scope of function definitions, though.
There may be a more elegant math-y way to solve it, but I just resort to computational/numerical methods above.
answered 5 hours ago
user6014
2,9261123
add a comment |Â
up vote
3
down vote
You can brute force this using a method inspired by binary search.
Your function:
FF[z_, b_] :=
z^4 + z^3 (4.67687 - 0.16 b^2) +
z^2 (-0.703991 - 0.652433 b^2) +
z (0.0676982 - 0.049884 b^2) - 0.000608244 b^2 +
0.0000234502;
Function which cuts the min/max range in half each time, outputting the new min/max to search:
f[x1_, x2_] :=
Module[v1, v2, v3, min = N[x1, 30], max = N[x2, 30], vals,
v1 = NSolve[FF[z, min] == 0, z, Reals];
v2 = NSolve[FF[z, max] == 0, z, Reals];
v3 = NSolve[FF[z, (min + max)/2] == 0, z, Reals];
If[Length@v3 <= 3,
vals = (min + max)/2, max,
vals = min, (min + max)/2
];
vals
]
And using our custom function above with FixedPoint we get a range that is hopefully specific enough for your application (with the floating point numbers in your original equations we're restricted to $MachinePrecision):
In[9]:= FixedPoint[f, .6, .61]
Out[9]= 0.6070523316794493, 0.6070523316794494
So as @Mariusz Iwaniuk said in the comments, you're looking for a number in the ballpark of $0.607052331679449$.
Note that my implementation isn't very general and relies on your specific FF being the format it is. No doubt it could be generalized to fit a broader scope of function definitions, though.
There may be a more elegant math-y way to solve it, but I just resort to computational/numerical methods above.
answered 5 hours ago
user6014
2,9261123
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You can brute force this using a method inspired by binary search.
Your function:
FF[z_, b_] :=
z^4 + z^3 (4.67687 - 0.16 b^2) +
z^2 (-0.703991 - 0.652433 b^2) +
z (0.0676982 - 0.049884 b^2) - 0.000608244 b^2 +
0.0000234502;
Function which cuts the min/max range in half each time, outputting the new min/max to search:
f[x1_, x2_] :=
Module[v1, v2, v3, min = N[x1, 30], max = N[x2, 30], vals,
v1 = NSolve[FF[z, min] == 0, z, Reals];
v2 = NSolve[FF[z, max] == 0, z, Reals];
v3 = NSolve[FF[z, (min + max)/2] == 0, z, Reals];
If[Length@v3 <= 3,
vals = (min + max)/2, max,
vals = min, (min + max)/2
];
vals
]
And using our custom function above with FixedPoint we get a range that is hopefully specific enough for your application (with the floating point numbers in your original equations we're restricted to $MachinePrecision):
In[9]:= FixedPoint[f, .6, .61]
Out[9]= 0.6070523316794493, 0.6070523316794494
So as @Mariusz Iwaniuk said in the comments, you're looking for a number in the ballpark of $0.607052331679449$.
Note that my implementation isn't very general and relies on your specific FF being the format it is. No doubt it could be generalized to fit a broader scope of function definitions, though.
There may be a more elegant math-y way to solve it, but I just resort to computational/numerical methods above.
answered 5 hours ago
user6014
2,9261123
You can brute force this using a method inspired by binary search.
Your function:
FF[z_, b_] :=
z^4 + z^3 (4.67687 - 0.16 b^2) +
z^2 (-0.703991 - 0.652433 b^2) +
z (0.0676982 - 0.049884 b^2) - 0.000608244 b^2 +
0.0000234502;
Function which cuts the min/max range in half each time, outputting the new min/max to search:
f[x1_, x2_] :=
Module[v1, v2, v3, min = N[x1, 30], max = N[x2, 30], vals,
v1 = NSolve[FF[z, min] == 0, z, Reals];
v2 = NSolve[FF[z, max] == 0, z, Reals];
v3 = NSolve[FF[z, (min + max)/2] == 0, z, Reals];
If[Length@v3 <= 3,
vals = (min + max)/2, max,
vals = min, (min + max)/2
];
vals
]
And using our custom function above with FixedPoint we get a range that is hopefully specific enough for your application (with the floating point numbers in your original equations we're restricted to $MachinePrecision):
In[9]:= FixedPoint[f, .6, .61]
Out[9]= 0.6070523316794493, 0.6070523316794494
So as @Mariusz Iwaniuk said in the comments, you're looking for a number in the ballpark of $0.607052331679449$.
Note that my implementation isn't very general and relies on your specific FF being the format it is. No doubt it could be generalized to fit a broader scope of function definitions, though.
There may be a more elegant math-y way to solve it, but I just resort to computational/numerical methods above.
answered 5 hours ago
user6014
2,9261123
edited 5 hours ago
answered 5 hours ago
user6014
2,9261123
answered 5 hours ago
user6014
2,9261123
answered 5 hours ago
user6014
2,9261123
2,9261123
add a comment |Â
add a comment |Â
up vote
1
down vote
First an extended comment:.
You could use ContourPlot to get an overview.
eq=z^4 + z^3 (4.67687 - 0.16 [Beta]^2) +z^2 (-0.703991 - 0.652433 [Beta]^2) +z (0.0676982 - 0.049884 [Beta]^2) -0.000608244 [Beta]^2 +0.0000234502
ContourPlot[0 ==eq, [Beta], 0, 1, z, -6, 1, MaxRecursion -> 5 ,FrameLabel -> [Beta],z]
The solutionplot shows two ranges [Beta]<0.6 and [Beta]>0.78 with only two solutions!
solution
The intersting branch of solution is in the range z0<z<z1
[Beta]0, z0 = 0, z /. Solve[eq == 0 /. [Beta] -> 0, z][[2]]
(*0, -0.000345152*)
[Beta]1, z1 = 1, z /. Solve[eq == 0 /. [Beta] -> 1, z][[4]]
(*1, 0.271234*)
Now it's easy to calculate
beta = [Beta] /. Solve[eq == 0, [Beta]][[2]] (* beta[z]*)
ParametricPlot[z, beta, z, z0, z1, AxesLabel -> "z", "[Beta]"]
The two extrema define the beta-values we are looking for
beta /. NSolve[D[beta, z] == 0, 0 < z < z1, z]
(*0.779012, 0.607052*)
That's it. Hope it helps!
answered 1 hour ago
Ulrich Neumann
5,230413
add a comment |Â
up vote
1
down vote
First an extended comment:.
You could use ContourPlot to get an overview.
eq=z^4 + z^3 (4.67687 - 0.16 [Beta]^2) +z^2 (-0.703991 - 0.652433 [Beta]^2) +z (0.0676982 - 0.049884 [Beta]^2) -0.000608244 [Beta]^2 +0.0000234502
ContourPlot[0 ==eq, [Beta], 0, 1, z, -6, 1, MaxRecursion -> 5 ,FrameLabel -> [Beta],z]
The solutionplot shows two ranges [Beta]<0.6 and [Beta]>0.78 with only two solutions!
solution
The intersting branch of solution is in the range z0<z<z1
[Beta]0, z0 = 0, z /. Solve[eq == 0 /. [Beta] -> 0, z][[2]]
(*0, -0.000345152*)
[Beta]1, z1 = 1, z /. Solve[eq == 0 /. [Beta] -> 1, z][[4]]
(*1, 0.271234*)
Now it's easy to calculate
beta = [Beta] /. Solve[eq == 0, [Beta]][[2]] (* beta[z]*)
ParametricPlot[z, beta, z, z0, z1, AxesLabel -> "z", "[Beta]"]
The two extrema define the beta-values we are looking for
beta /. NSolve[D[beta, z] == 0, 0 < z < z1, z]
(*0.779012, 0.607052*)
That's it. Hope it helps!
answered 1 hour ago
Ulrich Neumann
5,230413
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First an extended comment:.
You could use ContourPlot to get an overview.
eq=z^4 + z^3 (4.67687 - 0.16 [Beta]^2) +z^2 (-0.703991 - 0.652433 [Beta]^2) +z (0.0676982 - 0.049884 [Beta]^2) -0.000608244 [Beta]^2 +0.0000234502
ContourPlot[0 ==eq, [Beta], 0, 1, z, -6, 1, MaxRecursion -> 5 ,FrameLabel -> [Beta],z]
The solutionplot shows two ranges [Beta]<0.6 and [Beta]>0.78 with only two solutions!
solution
The intersting branch of solution is in the range z0<z<z1
[Beta]0, z0 = 0, z /. Solve[eq == 0 /. [Beta] -> 0, z][[2]]
(*0, -0.000345152*)
[Beta]1, z1 = 1, z /. Solve[eq == 0 /. [Beta] -> 1, z][[4]]
(*1, 0.271234*)
Now it's easy to calculate
beta = [Beta] /. Solve[eq == 0, [Beta]][[2]] (* beta[z]*)
ParametricPlot[z, beta, z, z0, z1, AxesLabel -> "z", "[Beta]"]
The two extrema define the beta-values we are looking for
beta /. NSolve[D[beta, z] == 0, 0 < z < z1, z]
(*0.779012, 0.607052*)
That's it. Hope it helps!
answered 1 hour ago
Ulrich Neumann
5,230413
First an extended comment:.
You could use ContourPlot to get an overview.
eq=z^4 + z^3 (4.67687 - 0.16 [Beta]^2) +z^2 (-0.703991 - 0.652433 [Beta]^2) +z (0.0676982 - 0.049884 [Beta]^2) -0.000608244 [Beta]^2 +0.0000234502
ContourPlot[0 ==eq, [Beta], 0, 1, z, -6, 1, MaxRecursion -> 5 ,FrameLabel -> [Beta],z]
The solutionplot shows two ranges [Beta]<0.6 and [Beta]>0.78 with only two solutions!
solution
The intersting branch of solution is in the range z0<z<z1
[Beta]0, z0 = 0, z /. Solve[eq == 0 /. [Beta] -> 0, z][[2]]
(*0, -0.000345152*)
[Beta]1, z1 = 1, z /. Solve[eq == 0 /. [Beta] -> 1, z][[4]]
(*1, 0.271234*)
Now it's easy to calculate
beta = [Beta] /. Solve[eq == 0, [Beta]][[2]] (* beta[z]*)
ParametricPlot[z, beta, z, z0, z1, AxesLabel -> "z", "[Beta]"]
The two extrema define the beta-values we are looking for
beta /. NSolve[D[beta, z] == 0, 0 < z < z1, z]
(*0.779012, 0.607052*)
That's it. Hope it helps!
answered 1 hour ago
Ulrich Neumann
5,230413
edited 7 mins ago
answered 1 hour ago
Ulrich Neumann
5,230413
answered 1 hour ago
Ulrich Neumann
5,230413
answered 1 hour ago
Ulrich Neumann
5,230413
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Alex Pereira is a new contributor. Be nice, and check out our Code of Conduct.
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Alex Pereira is a new contributor. Be nice, and check out our Code of Conduct.
Alex Pereira is a new contributor. Be nice, and check out our Code of Conduct.
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With brute force search: $beta =0.60705233167945$
– Mariusz Iwaniuk
11 hours ago
@user6014 Well, I'm interested about the positive roots. $beta = 0.6$ and $beta=0.601$ provide only one positive solution ;)
– Alex Pereira
10 hours ago
@mariusz-iwaniuk, can you show us how to do it? I'm asking this because I'll probably have to consider this problem again but with different numbers on the coefficients of F
– Alex Pereira
10 hours ago
1
This article might be useful.
– J. M. is computer-less♦
21 mins ago