Mixing
How to approximate the answer of a polynomial of degree 4?
up vote
0
down vote
favorite
I want to solve this polynomial analytically. I know the useful answer is between 0 and 1. Is there any way I can write the answer based on a, b, and c?
$$
6cdot a cdot x^4 + 2 cdot b cdot x^3-b cdot c=0
$$
Also, an approximate answer is acceptable, for example, an answer with 2% error.
I will appreciate if someone can help me on this subject.
polynomials approximation
asked yesterday
Mostafa
368
add a comment |
up vote
0
down vote
favorite
I want to solve this polynomial analytically. I know the useful answer is between 0 and 1. Is there any way I can write the answer based on a, b, and c?
$$
6cdot a cdot x^4 + 2 cdot b cdot x^3-b cdot c=0
$$
Also, an approximate answer is acceptable, for example, an answer with 2% error.
I will appreciate if someone can help me on this subject.
polynomials approximation
asked yesterday
Mostafa
368
3
There is always Ferrari's formula, one supposes.
– Xander Henderson
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to solve this polynomial analytically. I know the useful answer is between 0 and 1. Is there any way I can write the answer based on a, b, and c?
$$
6cdot a cdot x^4 + 2 cdot b cdot x^3-b cdot c=0
$$
Also, an approximate answer is acceptable, for example, an answer with 2% error.
I will appreciate if someone can help me on this subject.
polynomials approximation
asked yesterday
Mostafa
368
I want to solve this polynomial analytically. I know the useful answer is between 0 and 1. Is there any way I can write the answer based on a, b, and c?
$$
6cdot a cdot x^4 + 2 cdot b cdot x^3-b cdot c=0
$$
Also, an approximate answer is acceptable, for example, an answer with 2% error.
I will appreciate if someone can help me on this subject.
polynomials approximation
polynomials approximation
asked yesterday
Mostafa
368
asked yesterday
Mostafa
368
asked yesterday
Mostafa
368
asked yesterday
Mostafa
368
asked yesterday
Mostafa
368
368
3
There is always Ferrari's formula, one supposes.
– Xander Henderson
yesterday
add a comment |
3
There is always Ferrari's formula, one supposes.
– Xander Henderson
yesterday
3
3
There is always Ferrari's formula, one supposes.
– Xander Henderson
yesterday
There is always Ferrari's formula, one supposes.
– Xander Henderson
yesterday
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
You could use Ferrari's method for solve in general.
https://proofwiki.org/wiki/Ferrari%27s_Method this is an easy algorithmic way to do it.
answered yesterday
Bayesian guy
4210
1
For the third degree equation you could use Cardano's formula proofwiki.org/wiki/Cardano%27s_Formula
– Bayesian guy
yesterday
add a comment |
up vote
0
down vote
Since you say an approximate answer is alright, up to 2% tolerance, it might be a good idea to use a numerical approximation like Newton's method.
The derivative works out to be
$$ f'(x) = 24ax^3 + 6bx^2 = 6x^2 (4ax + b) $$
So you can take an $x_0$ in the range $ (0, 1) $ and try approximating your missing solution, if you know it is there. Just linearize to $ y = f'(x_0) x_1 + f(x_0) $ and solve for $ x_1 $ when $ y = 0 $, and repeat until you have the right confidence.
Newton's method can sometimes instead get you the wrong root. So you would likely have to have some kind of assured interval for the space you are looking at.
If your values are in the wrong range, you won't get a solution in $ (0, 1) $ at all, so there may be some other conditions you are looking at. For instance, suppose for a given triple $ (a, b, c) $, we have a solution to $ f $ in the range $(0, 1)$. Taking the forward image $ f( , (0, 1) , ) = (x , y) $ will give us an upper bound that the function attains on the interval, so if we re-pick $ c $ as
$$ c' = c - frac{y}{b} - 1 $$
then a new function $ g $ with triple $ (a, b, c') $ has no solutions in the range $ (0, 1) $ because of the vertical shift. So not every function with real parameters $a, b, c$ will have the desired property.
answered yesterday
theREALyumdub
890617
1
Thank you for your answer, but as I mentioned, I need to solve it analytically and based on a, b, and c, so the numerical methods like Newton's method doesn't help me. I am looking for a simple analytical method even without an exact answer.
– Mostafa
yesterday
@Mostafa You could use Ferarri's method with the reduced coefficients you have specified. That should reveal the simplicity the solutions have with these coefficients, although it may be tedious and of course completely exact. I edited my answer to explain more that depending $ a, b, c in mathbb{R} $, it may not be possible to have a solution in $ (0, 1)$, so likely any analytic methods you employ could benefit from the implicit constraints on the coefficients.
– theREALyumdub
yesterday
add a comment |
up vote
0
down vote
The exact solution would turn into a "complicated" expression in $a,b,c$.
If you are looking for an approximated solution, and you know that a real root is near to $1$ (and in fact it is for "normal" positive value of the parameters) , then replace $x$ with $1+y$, retain only the terms of degree $le 2$ and solve for $y$. The shift to $x=1$ is because in $x=0$ the polynomial is quite flat (1st and 2nd derivative null).
Depending on the parameters you might find a better approximation developing instead at $x=1/2$.
answered yesterday
G Cab
16.9k31237
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You could use Ferrari's method for solve in general.
https://proofwiki.org/wiki/Ferrari%27s_Method this is an easy algorithmic way to do it.
answered yesterday
Bayesian guy
4210
1
For the third degree equation you could use Cardano's formula proofwiki.org/wiki/Cardano%27s_Formula
– Bayesian guy
yesterday
add a comment |
up vote
2
down vote
You could use Ferrari's method for solve in general.
https://proofwiki.org/wiki/Ferrari%27s_Method this is an easy algorithmic way to do it.
answered yesterday
Bayesian guy
4210
1
For the third degree equation you could use Cardano's formula proofwiki.org/wiki/Cardano%27s_Formula
– Bayesian guy
yesterday
add a comment |
up vote
2
down vote
up vote
2
down vote
You could use Ferrari's method for solve in general.
https://proofwiki.org/wiki/Ferrari%27s_Method this is an easy algorithmic way to do it.
answered yesterday
Bayesian guy
4210
You could use Ferrari's method for solve in general.
https://proofwiki.org/wiki/Ferrari%27s_Method this is an easy algorithmic way to do it.
answered yesterday
Bayesian guy
4210
edited yesterday
answered yesterday
Bayesian guy
4210
answered yesterday
Bayesian guy
4210
answered yesterday
Bayesian guy
4210
4210
1
For the third degree equation you could use Cardano's formula proofwiki.org/wiki/Cardano%27s_Formula
– Bayesian guy
yesterday
add a comment |
1
For the third degree equation you could use Cardano's formula proofwiki.org/wiki/Cardano%27s_Formula
– Bayesian guy
yesterday
1
1
For the third degree equation you could use Cardano's formula proofwiki.org/wiki/Cardano%27s_Formula
– Bayesian guy
yesterday
For the third degree equation you could use Cardano's formula proofwiki.org/wiki/Cardano%27s_Formula
– Bayesian guy
yesterday
add a comment |
up vote
0
down vote
Since you say an approximate answer is alright, up to 2% tolerance, it might be a good idea to use a numerical approximation like Newton's method.
The derivative works out to be
$$ f'(x) = 24ax^3 + 6bx^2 = 6x^2 (4ax + b) $$
So you can take an $x_0$ in the range $ (0, 1) $ and try approximating your missing solution, if you know it is there. Just linearize to $ y = f'(x_0) x_1 + f(x_0) $ and solve for $ x_1 $ when $ y = 0 $, and repeat until you have the right confidence.
Newton's method can sometimes instead get you the wrong root. So you would likely have to have some kind of assured interval for the space you are looking at.
If your values are in the wrong range, you won't get a solution in $ (0, 1) $ at all, so there may be some other conditions you are looking at. For instance, suppose for a given triple $ (a, b, c) $, we have a solution to $ f $ in the range $(0, 1)$. Taking the forward image $ f( , (0, 1) , ) = (x , y) $ will give us an upper bound that the function attains on the interval, so if we re-pick $ c $ as
$$ c' = c - frac{y}{b} - 1 $$
then a new function $ g $ with triple $ (a, b, c') $ has no solutions in the range $ (0, 1) $ because of the vertical shift. So not every function with real parameters $a, b, c$ will have the desired property.
answered yesterday
theREALyumdub
890617
1
Thank you for your answer, but as I mentioned, I need to solve it analytically and based on a, b, and c, so the numerical methods like Newton's method doesn't help me. I am looking for a simple analytical method even without an exact answer.
– Mostafa
yesterday
@Mostafa You could use Ferarri's method with the reduced coefficients you have specified. That should reveal the simplicity the solutions have with these coefficients, although it may be tedious and of course completely exact. I edited my answer to explain more that depending $ a, b, c in mathbb{R} $, it may not be possible to have a solution in $ (0, 1)$, so likely any analytic methods you employ could benefit from the implicit constraints on the coefficients.
– theREALyumdub
yesterday
add a comment |
up vote
0
down vote
Since you say an approximate answer is alright, up to 2% tolerance, it might be a good idea to use a numerical approximation like Newton's method.
The derivative works out to be
$$ f'(x) = 24ax^3 + 6bx^2 = 6x^2 (4ax + b) $$
So you can take an $x_0$ in the range $ (0, 1) $ and try approximating your missing solution, if you know it is there. Just linearize to $ y = f'(x_0) x_1 + f(x_0) $ and solve for $ x_1 $ when $ y = 0 $, and repeat until you have the right confidence.
Newton's method can sometimes instead get you the wrong root. So you would likely have to have some kind of assured interval for the space you are looking at.
If your values are in the wrong range, you won't get a solution in $ (0, 1) $ at all, so there may be some other conditions you are looking at. For instance, suppose for a given triple $ (a, b, c) $, we have a solution to $ f $ in the range $(0, 1)$. Taking the forward image $ f( , (0, 1) , ) = (x , y) $ will give us an upper bound that the function attains on the interval, so if we re-pick $ c $ as
$$ c' = c - frac{y}{b} - 1 $$
then a new function $ g $ with triple $ (a, b, c') $ has no solutions in the range $ (0, 1) $ because of the vertical shift. So not every function with real parameters $a, b, c$ will have the desired property.
answered yesterday
theREALyumdub
890617
1
Thank you for your answer, but as I mentioned, I need to solve it analytically and based on a, b, and c, so the numerical methods like Newton's method doesn't help me. I am looking for a simple analytical method even without an exact answer.
– Mostafa
yesterday
@Mostafa You could use Ferarri's method with the reduced coefficients you have specified. That should reveal the simplicity the solutions have with these coefficients, although it may be tedious and of course completely exact. I edited my answer to explain more that depending $ a, b, c in mathbb{R} $, it may not be possible to have a solution in $ (0, 1)$, so likely any analytic methods you employ could benefit from the implicit constraints on the coefficients.
– theREALyumdub
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
Since you say an approximate answer is alright, up to 2% tolerance, it might be a good idea to use a numerical approximation like Newton's method.
The derivative works out to be
$$ f'(x) = 24ax^3 + 6bx^2 = 6x^2 (4ax + b) $$
So you can take an $x_0$ in the range $ (0, 1) $ and try approximating your missing solution, if you know it is there. Just linearize to $ y = f'(x_0) x_1 + f(x_0) $ and solve for $ x_1 $ when $ y = 0 $, and repeat until you have the right confidence.
Newton's method can sometimes instead get you the wrong root. So you would likely have to have some kind of assured interval for the space you are looking at.
If your values are in the wrong range, you won't get a solution in $ (0, 1) $ at all, so there may be some other conditions you are looking at. For instance, suppose for a given triple $ (a, b, c) $, we have a solution to $ f $ in the range $(0, 1)$. Taking the forward image $ f( , (0, 1) , ) = (x , y) $ will give us an upper bound that the function attains on the interval, so if we re-pick $ c $ as
$$ c' = c - frac{y}{b} - 1 $$
then a new function $ g $ with triple $ (a, b, c') $ has no solutions in the range $ (0, 1) $ because of the vertical shift. So not every function with real parameters $a, b, c$ will have the desired property.
answered yesterday
theREALyumdub
890617
Since you say an approximate answer is alright, up to 2% tolerance, it might be a good idea to use a numerical approximation like Newton's method.
The derivative works out to be
$$ f'(x) = 24ax^3 + 6bx^2 = 6x^2 (4ax + b) $$
So you can take an $x_0$ in the range $ (0, 1) $ and try approximating your missing solution, if you know it is there. Just linearize to $ y = f'(x_0) x_1 + f(x_0) $ and solve for $ x_1 $ when $ y = 0 $, and repeat until you have the right confidence.
Newton's method can sometimes instead get you the wrong root. So you would likely have to have some kind of assured interval for the space you are looking at.
If your values are in the wrong range, you won't get a solution in $ (0, 1) $ at all, so there may be some other conditions you are looking at. For instance, suppose for a given triple $ (a, b, c) $, we have a solution to $ f $ in the range $(0, 1)$. Taking the forward image $ f( , (0, 1) , ) = (x , y) $ will give us an upper bound that the function attains on the interval, so if we re-pick $ c $ as
$$ c' = c - frac{y}{b} - 1 $$
then a new function $ g $ with triple $ (a, b, c') $ has no solutions in the range $ (0, 1) $ because of the vertical shift. So not every function with real parameters $a, b, c$ will have the desired property.
answered yesterday
theREALyumdub
890617
edited yesterday
answered yesterday
theREALyumdub
890617
answered yesterday
theREALyumdub
890617
answered yesterday
theREALyumdub
890617
890617
1
Thank you for your answer, but as I mentioned, I need to solve it analytically and based on a, b, and c, so the numerical methods like Newton's method doesn't help me. I am looking for a simple analytical method even without an exact answer.
– Mostafa
yesterday
@Mostafa You could use Ferarri's method with the reduced coefficients you have specified. That should reveal the simplicity the solutions have with these coefficients, although it may be tedious and of course completely exact. I edited my answer to explain more that depending $ a, b, c in mathbb{R} $, it may not be possible to have a solution in $ (0, 1)$, so likely any analytic methods you employ could benefit from the implicit constraints on the coefficients.
– theREALyumdub
yesterday
add a comment |
1
Thank you for your answer, but as I mentioned, I need to solve it analytically and based on a, b, and c, so the numerical methods like Newton's method doesn't help me. I am looking for a simple analytical method even without an exact answer.
– Mostafa
yesterday
@Mostafa You could use Ferarri's method with the reduced coefficients you have specified. That should reveal the simplicity the solutions have with these coefficients, although it may be tedious and of course completely exact. I edited my answer to explain more that depending $ a, b, c in mathbb{R} $, it may not be possible to have a solution in $ (0, 1)$, so likely any analytic methods you employ could benefit from the implicit constraints on the coefficients.
– theREALyumdub
yesterday
1
1
Thank you for your answer, but as I mentioned, I need to solve it analytically and based on a, b, and c, so the numerical methods like Newton's method doesn't help me. I am looking for a simple analytical method even without an exact answer.
– Mostafa
yesterday
Thank you for your answer, but as I mentioned, I need to solve it analytically and based on a, b, and c, so the numerical methods like Newton's method doesn't help me. I am looking for a simple analytical method even without an exact answer.
– Mostafa
yesterday
@Mostafa You could use Ferarri's method with the reduced coefficients you have specified. That should reveal the simplicity the solutions have with these coefficients, although it may be tedious and of course completely exact. I edited my answer to explain more that depending $ a, b, c in mathbb{R} $, it may not be possible to have a solution in $ (0, 1)$, so likely any analytic methods you employ could benefit from the implicit constraints on the coefficients.
– theREALyumdub
yesterday
@Mostafa You could use Ferarri's method with the reduced coefficients you have specified. That should reveal the simplicity the solutions have with these coefficients, although it may be tedious and of course completely exact. I edited my answer to explain more that depending $ a, b, c in mathbb{R} $, it may not be possible to have a solution in $ (0, 1)$, so likely any analytic methods you employ could benefit from the implicit constraints on the coefficients.
– theREALyumdub
yesterday
add a comment |
up vote
0
down vote
The exact solution would turn into a "complicated" expression in $a,b,c$.
If you are looking for an approximated solution, and you know that a real root is near to $1$ (and in fact it is for "normal" positive value of the parameters) , then replace $x$ with $1+y$, retain only the terms of degree $le 2$ and solve for $y$. The shift to $x=1$ is because in $x=0$ the polynomial is quite flat (1st and 2nd derivative null).
Depending on the parameters you might find a better approximation developing instead at $x=1/2$.
answered yesterday
G Cab
16.9k31237
add a comment |
up vote
0
down vote
The exact solution would turn into a "complicated" expression in $a,b,c$.
If you are looking for an approximated solution, and you know that a real root is near to $1$ (and in fact it is for "normal" positive value of the parameters) , then replace $x$ with $1+y$, retain only the terms of degree $le 2$ and solve for $y$. The shift to $x=1$ is because in $x=0$ the polynomial is quite flat (1st and 2nd derivative null).
Depending on the parameters you might find a better approximation developing instead at $x=1/2$.
answered yesterday
G Cab
16.9k31237
add a comment |
up vote
0
down vote
up vote
0
down vote
The exact solution would turn into a "complicated" expression in $a,b,c$.
If you are looking for an approximated solution, and you know that a real root is near to $1$ (and in fact it is for "normal" positive value of the parameters) , then replace $x$ with $1+y$, retain only the terms of degree $le 2$ and solve for $y$. The shift to $x=1$ is because in $x=0$ the polynomial is quite flat (1st and 2nd derivative null).
Depending on the parameters you might find a better approximation developing instead at $x=1/2$.
answered yesterday
G Cab
16.9k31237
The exact solution would turn into a "complicated" expression in $a,b,c$.
If you are looking for an approximated solution, and you know that a real root is near to $1$ (and in fact it is for "normal" positive value of the parameters) , then replace $x$ with $1+y$, retain only the terms of degree $le 2$ and solve for $y$. The shift to $x=1$ is because in $x=0$ the polynomial is quite flat (1st and 2nd derivative null).
Depending on the parameters you might find a better approximation developing instead at $x=1/2$.
answered yesterday
G Cab
16.9k31237
answered yesterday
G Cab
16.9k31237
answered yesterday
G Cab
16.9k31237
answered yesterday
G Cab
16.9k31237
16.9k31237
add a comment |
add a comment |
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3
There is always Ferrari's formula, one supposes.
– Xander Henderson
yesterday