Help solving complicated polynomial equation
4
I am trying to solve the following two equations, but I am having troubles doing so.
Equation 1 :
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) ;
Solve[eq1[n] == 0, n]
If I can find an "n", I would then plug it into the following equation and solve for V.
eq2[n_] := 1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[nFound] == 0, V]
equation-solving
asked Dec 16 at 8:25
james
778519
add a comment |
4
I am trying to solve the following two equations, but I am having troubles doing so.
Equation 1 :
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) ;
Solve[eq1[n] == 0, n]
If I can find an "n", I would then plug it into the following equation and solve for V.
eq2[n_] := 1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[nFound] == 0, V]
equation-solving
asked Dec 16 at 8:25
james
778519
add a comment |
4
4
4
I am trying to solve the following two equations, but I am having troubles doing so.
Equation 1 :
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) ;
Solve[eq1[n] == 0, n]
If I can find an "n", I would then plug it into the following equation and solve for V.
eq2[n_] := 1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[nFound] == 0, V]
equation-solving
asked Dec 16 at 8:25
james
778519
I am trying to solve the following two equations, but I am having troubles doing so.
Equation 1 :
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) ;
Solve[eq1[n] == 0, n]
If I can find an "n", I would then plug it into the following equation and solve for V.
eq2[n_] := 1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[nFound] == 0, V]
equation-solving
equation-solving
asked Dec 16 at 8:25
james
778519
asked Dec 16 at 8:25
james
778519
asked Dec 16 at 8:25
james
778519
asked Dec 16 at 8:25
james
778519
asked Dec 16 at 8:25
james
778519
778519
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
4
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
answered Dec 16 at 9:10
J.W Kang
835
Thanks ! This was exactly what I was looking for.
– james
Dec 16 at 9:13
Never thought about using theValuesfunction in this context. Thanks for the new idea!
– Thies Heidecke
Dec 16 at 14:02
add a comment |
3
Solve can handle simultaneous equations. Solve is an exact solver so use Rationalze to provide exact numbers as input. Solve will work without doing this but will provide a warning that it did it internally. Or use NSolve.
eq1[n_] :=
Evaluate[2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) //
Rationalize[#, 0] & // Simplify];
eq2[n_] := Evaluate[1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2) // Rationalize[#, 0] &];
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}] // N
(* {{n -> 0.00634203 + 0.0986751 I,
V -> -7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 + 0.0986751 I,
V -> 7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> 7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> -7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.587316,
V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real solutions
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}, Reals] // N
(* {{n -> 0.587316, V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real, positive solutions
Solve[{eq1[n] == 0, eq2[n] == 0, V > 0}, {n, V}] // N
(* {{n -> 0.587316, V -> 4.37685*10^9}} *)
answered Dec 16 at 18:26
Bob Hanlon
58.7k23595
This is a great answer ! Thank you very much !
– james
Dec 17 at 14:24
How would you do it with Rationalize ?
– james
Dec 17 at 14:42
Look at the definition ofeq1above. At the end you will see thatRationalizeandSimplifyare used. SinceSetDelayed ( := )is used in the definition rather thanSet ( = ),Evaluateis used to have these evaluated at the time of definition rather than for each use of the functioneq1. Similarly foreq2exceptSimplifywasn't needed. Read the documentation for each of these functions.
– Bob Hanlon
Dec 17 at 15:52
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
answered Dec 16 at 9:10
J.W Kang
835
Thanks ! This was exactly what I was looking for.
– james
Dec 16 at 9:13
Never thought about using theValuesfunction in this context. Thanks for the new idea!
– Thies Heidecke
Dec 16 at 14:02
add a comment |
4
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
answered Dec 16 at 9:10
J.W Kang
835
Thanks ! This was exactly what I was looking for.
– james
Dec 16 at 9:13
Never thought about using theValuesfunction in this context. Thanks for the new idea!
– Thies Heidecke
Dec 16 at 14:02
add a comment |
4
4
4
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
answered Dec 16 at 9:10
J.W Kang
835
Sorry, I am not very sure what you want to ask. Maybe you want to substitute the solution of the first equation into the second equation? Then you can do this.
eq1[n_] :=
2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3);
ansList = Values@Solve[eq1[n] == 0, n]
eq2[n_] :=
1024/5*n + 7133.17 n^3 - 0.06287*V^2*2.21355*10^(-16)/((1 - n)^2);
Solve[eq2[#] == 0, V] & /@ ansList
You also can use the following code to get the value of n.
sol=Solve[eq1[n] == 0, n]
nList=n/.sol
That's all. Please enjoy the fun of Mathematica.
answered Dec 16 at 9:10
J.W Kang
835
answered Dec 16 at 9:10
J.W Kang
835
answered Dec 16 at 9:10
J.W Kang
835
answered Dec 16 at 9:10
J.W Kang
835
835
Thanks ! This was exactly what I was looking for.
– james
Dec 16 at 9:13
Never thought about using theValuesfunction in this context. Thanks for the new idea!
– Thies Heidecke
Dec 16 at 14:02
add a comment |
Thanks ! This was exactly what I was looking for.
– james
Dec 16 at 9:13
Never thought about using theValuesfunction in this context. Thanks for the new idea!
– Thies Heidecke
Dec 16 at 14:02
Thanks ! This was exactly what I was looking for.
– james
Dec 16 at 9:13
Thanks ! This was exactly what I was looking for.
– james
Dec 16 at 9:13
Never thought about using the
Values function in this context. Thanks for the new idea!– Thies Heidecke
Dec 16 at 14:02
Never thought about using the
Values function in this context. Thanks for the new idea!– Thies Heidecke
Dec 16 at 14:02
add a comment |
3
Solve can handle simultaneous equations. Solve is an exact solver so use Rationalze to provide exact numbers as input. Solve will work without doing this but will provide a warning that it did it internally. Or use NSolve.
eq1[n_] :=
Evaluate[2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) //
Rationalize[#, 0] & // Simplify];
eq2[n_] := Evaluate[1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2) // Rationalize[#, 0] &];
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}] // N
(* {{n -> 0.00634203 + 0.0986751 I,
V -> -7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 + 0.0986751 I,
V -> 7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> 7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> -7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.587316,
V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real solutions
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}, Reals] // N
(* {{n -> 0.587316, V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real, positive solutions
Solve[{eq1[n] == 0, eq2[n] == 0, V > 0}, {n, V}] // N
(* {{n -> 0.587316, V -> 4.37685*10^9}} *)
answered Dec 16 at 18:26
Bob Hanlon
58.7k23595
This is a great answer ! Thank you very much !
– james
Dec 17 at 14:24
How would you do it with Rationalize ?
– james
Dec 17 at 14:42
Look at the definition ofeq1above. At the end you will see thatRationalizeandSimplifyare used. SinceSetDelayed ( := )is used in the definition rather thanSet ( = ),Evaluateis used to have these evaluated at the time of definition rather than for each use of the functioneq1. Similarly foreq2exceptSimplifywasn't needed. Read the documentation for each of these functions.
– Bob Hanlon
Dec 17 at 15:52
add a comment |
3
Solve can handle simultaneous equations. Solve is an exact solver so use Rationalze to provide exact numbers as input. Solve will work without doing this but will provide a warning that it did it internally. Or use NSolve.
eq1[n_] :=
Evaluate[2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) //
Rationalize[#, 0] & // Simplify];
eq2[n_] := Evaluate[1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2) // Rationalize[#, 0] &];
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}] // N
(* {{n -> 0.00634203 + 0.0986751 I,
V -> -7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 + 0.0986751 I,
V -> 7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> 7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> -7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.587316,
V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real solutions
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}, Reals] // N
(* {{n -> 0.587316, V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real, positive solutions
Solve[{eq1[n] == 0, eq2[n] == 0, V > 0}, {n, V}] // N
(* {{n -> 0.587316, V -> 4.37685*10^9}} *)
answered Dec 16 at 18:26
Bob Hanlon
58.7k23595
This is a great answer ! Thank you very much !
– james
Dec 17 at 14:24
How would you do it with Rationalize ?
– james
Dec 17 at 14:42
Look at the definition ofeq1above. At the end you will see thatRationalizeandSimplifyare used. SinceSetDelayed ( := )is used in the definition rather thanSet ( = ),Evaluateis used to have these evaluated at the time of definition rather than for each use of the functioneq1. Similarly foreq2exceptSimplifywasn't needed. Read the documentation for each of these functions.
– Bob Hanlon
Dec 17 at 15:52
add a comment |
3
3
3
Solve can handle simultaneous equations. Solve is an exact solver so use Rationalze to provide exact numbers as input. Solve will work without doing this but will provide a warning that it did it internally. Or use NSolve.
eq1[n_] :=
Evaluate[2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) //
Rationalize[#, 0] & // Simplify];
eq2[n_] := Evaluate[1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2) // Rationalize[#, 0] &];
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}] // N
(* {{n -> 0.00634203 + 0.0986751 I,
V -> -7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 + 0.0986751 I,
V -> 7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> 7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> -7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.587316,
V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real solutions
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}, Reals] // N
(* {{n -> 0.587316, V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real, positive solutions
Solve[{eq1[n] == 0, eq2[n] == 0, V > 0}, {n, V}] // N
(* {{n -> 0.587316, V -> 4.37685*10^9}} *)
answered Dec 16 at 18:26
Bob Hanlon
58.7k23595
Solve can handle simultaneous equations. Solve is an exact solver so use Rationalze to provide exact numbers as input. Solve will work without doing this but will provide a warning that it did it internally. Or use NSolve.
eq1[n_] :=
Evaluate[2.21355*10^(-16)/((1 - n)^2)*(1024/5 + 3*7133.17 n^2) -
4.42709*10^(-16)/((1 - n)^3)*(1024/5*n + 7133.17 n^3) //
Rationalize[#, 0] & // Simplify];
eq2[n_] := Evaluate[1024/5*n + 7133.17 n^3 -
0.06287*V^2*2.21355*10^(-16)/((1 - n)^2) // Rationalize[#, 0] &];
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}] // N
(* {{n -> 0.00634203 + 0.0986751 I,
V -> -7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 + 0.0986751 I,
V -> 7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> 7.58581*10^8 - 6.22509*10^8 I}, {n -> 0.00634203 - 0.0986751 I,
V -> -7.58581*10^8 + 6.22509*10^8 I}, {n -> 0.587316,
V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real solutions
Solve[{eq1[n] == 0, eq2[n] == 0}, {n, V}, Reals] // N
(* {{n -> 0.587316, V -> -4.37685*10^9}, {n -> 0.587316, V -> 4.37685*10^9}} *)
For real, positive solutions
Solve[{eq1[n] == 0, eq2[n] == 0, V > 0}, {n, V}] // N
(* {{n -> 0.587316, V -> 4.37685*10^9}} *)
answered Dec 16 at 18:26
Bob Hanlon
58.7k23595
answered Dec 16 at 18:26
Bob Hanlon
58.7k23595
answered Dec 16 at 18:26
Bob Hanlon
58.7k23595
answered Dec 16 at 18:26
Bob Hanlon
58.7k23595
58.7k23595
This is a great answer ! Thank you very much !
– james
Dec 17 at 14:24
How would you do it with Rationalize ?
– james
Dec 17 at 14:42
Look at the definition ofeq1above. At the end you will see thatRationalizeandSimplifyare used. SinceSetDelayed ( := )is used in the definition rather thanSet ( = ),Evaluateis used to have these evaluated at the time of definition rather than for each use of the functioneq1. Similarly foreq2exceptSimplifywasn't needed. Read the documentation for each of these functions.
– Bob Hanlon
Dec 17 at 15:52
add a comment |
This is a great answer ! Thank you very much !
– james
Dec 17 at 14:24
How would you do it with Rationalize ?
– james
Dec 17 at 14:42
Look at the definition ofeq1above. At the end you will see thatRationalizeandSimplifyare used. SinceSetDelayed ( := )is used in the definition rather thanSet ( = ),Evaluateis used to have these evaluated at the time of definition rather than for each use of the functioneq1. Similarly foreq2exceptSimplifywasn't needed. Read the documentation for each of these functions.
– Bob Hanlon
Dec 17 at 15:52
This is a great answer ! Thank you very much !
– james
Dec 17 at 14:24
This is a great answer ! Thank you very much !
– james
Dec 17 at 14:24
How would you do it with Rationalize ?
– james
Dec 17 at 14:42
How would you do it with Rationalize ?
– james
Dec 17 at 14:42
Look at the definition of
eq1 above. At the end you will see that Rationalize and Simplify are used. Since SetDelayed ( := ) is used in the definition rather than Set ( = ), Evaluate is used to have these evaluated at the time of definition rather than for each use of the function eq1. Similarly for eq2 except Simplify wasn't needed. Read the documentation for each of these functions.– Bob Hanlon
Dec 17 at 15:52
Look at the definition of
eq1 above. At the end you will see that Rationalize and Simplify are used. Since SetDelayed ( := ) is used in the definition rather than Set ( = ), Evaluate is used to have these evaluated at the time of definition rather than for each use of the function eq1. Similarly for eq2 except Simplify wasn't needed. Read the documentation for each of these functions.– Bob Hanlon
Dec 17 at 15:52
add a comment |
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