How to approximate the answer of a polynomial of degree 4?











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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







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    There is always Ferrari's formula, one supposes.
    – Xander Henderson
    yesterday















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







share|cite|improve this question


















  • 3




    There is always Ferrari's formula, one supposes.
    – Xander Henderson
    yesterday













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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









Mostafa

368




368








  • 3




    There is always Ferrari's formula, one supposes.
    – Xander Henderson
    yesterday














  • 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










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.






share|cite|improve this answer



















  • 1




    For the third degree equation you could use Cardano's formula proofwiki.org/wiki/Cardano%27s_Formula
    – Bayesian guy
    yesterday




















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.






share|cite|improve this answer



















  • 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




















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 $x=1/2$.






share|cite|improve this answer





















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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    active

    oldest

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    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.






    share|cite|improve this answer



















    • 1




      For the third degree equation you could use Cardano's formula proofwiki.org/wiki/Cardano%27s_Formula
      – Bayesian guy
      yesterday

















    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.






    share|cite|improve this answer



















    • 1




      For the third degree equation you could use Cardano's formula proofwiki.org/wiki/Cardano%27s_Formula
      – Bayesian guy
      yesterday















    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.






    share|cite|improve this answer














    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    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
















    • 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












    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.






    share|cite|improve this answer



















    • 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

















    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.






    share|cite|improve this answer



















    • 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















    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.






    share|cite|improve this answer














    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    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
















    • 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












    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 $x=1/2$.






    share|cite|improve this answer

























      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 $x=1/2$.






      share|cite|improve this answer























        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 $x=1/2$.






        share|cite|improve this answer












        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 $x=1/2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        G Cab

        16.9k31237




        16.9k31237






























             

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