Solution of a fourth degree equation












0












$begingroup$


Is there a viable strategy to solve the following equation in an analytic way, without using numeric methods?



$(1+frac{1}{8}x^2)^2=frac{p^2}{2}(sqrt{1+x^2}+1)$



Edit: When I substitute $t=sqrt{1+x^2}$, I get:



$frac{1}{64}(49+14t^2+t^4)=frac{p^2}{2}(t+1)$






quartic-equations







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$endgroup$












  • $begingroup$
    @KarnWatcharasupat But didn't you forget the square root in there? Or am I missing something?
    $endgroup$
    – Wild Feather
    Dec 23 '18 at 13:42








  • 1




    $begingroup$
    Try to substitute $$t=sqrt{1+x^2}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 23 '18 at 13:43










  • $begingroup$
    @Dr.SonnhardGraubner I did that and the result seems to be a quartic equation where the term with $t^3$ is missing. But is there a way to solve that type of equation? I'll edit my question to show what I got when I did the substitution you suggested.
    $endgroup$
    – Wild Feather
    Dec 23 '18 at 13:48








  • 1




    $begingroup$
    The quartic equation is the best what you can get.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 23 '18 at 13:49
















0












$begingroup$


Is there a viable strategy to solve the following equation in an analytic way, without using numeric methods?



$(1+frac{1}{8}x^2)^2=frac{p^2}{2}(sqrt{1+x^2}+1)$



Edit: When I substitute $t=sqrt{1+x^2}$, I get:



$frac{1}{64}(49+14t^2+t^4)=frac{p^2}{2}(t+1)$






quartic-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    @KarnWatcharasupat But didn't you forget the square root in there? Or am I missing something?
    $endgroup$
    – Wild Feather
    Dec 23 '18 at 13:42








  • 1




    $begingroup$
    Try to substitute $$t=sqrt{1+x^2}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 23 '18 at 13:43










  • $begingroup$
    @Dr.SonnhardGraubner I did that and the result seems to be a quartic equation where the term with $t^3$ is missing. But is there a way to solve that type of equation? I'll edit my question to show what I got when I did the substitution you suggested.
    $endgroup$
    – Wild Feather
    Dec 23 '18 at 13:48








  • 1




    $begingroup$
    The quartic equation is the best what you can get.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 23 '18 at 13:49














0












0








0





$begingroup$


Is there a viable strategy to solve the following equation in an analytic way, without using numeric methods?



$(1+frac{1}{8}x^2)^2=frac{p^2}{2}(sqrt{1+x^2}+1)$



Edit: When I substitute $t=sqrt{1+x^2}$, I get:



$frac{1}{64}(49+14t^2+t^4)=frac{p^2}{2}(t+1)$






quartic-equations







share|cite|improve this question











$endgroup$




Is there a viable strategy to solve the following equation in an analytic way, without using numeric methods?



$(1+frac{1}{8}x^2)^2=frac{p^2}{2}(sqrt{1+x^2}+1)$



Edit: When I substitute $t=sqrt{1+x^2}$, I get:



$frac{1}{64}(49+14t^2+t^4)=frac{p^2}{2}(t+1)$






quartic-equations




quartic-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 13:50







Wild Feather

















asked Dec 23 '18 at 13:36









Wild FeatherWild Feather

16611




16611












  • $begingroup$
    @KarnWatcharasupat But didn't you forget the square root in there? Or am I missing something?
    $endgroup$
    – Wild Feather
    Dec 23 '18 at 13:42








  • 1




    $begingroup$
    Try to substitute $$t=sqrt{1+x^2}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 23 '18 at 13:43










  • $begingroup$
    @Dr.SonnhardGraubner I did that and the result seems to be a quartic equation where the term with $t^3$ is missing. But is there a way to solve that type of equation? I'll edit my question to show what I got when I did the substitution you suggested.
    $endgroup$
    – Wild Feather
    Dec 23 '18 at 13:48








  • 1




    $begingroup$
    The quartic equation is the best what you can get.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 23 '18 at 13:49


















  • $begingroup$
    @KarnWatcharasupat But didn't you forget the square root in there? Or am I missing something?
    $endgroup$
    – Wild Feather
    Dec 23 '18 at 13:42








  • 1




    $begingroup$
    Try to substitute $$t=sqrt{1+x^2}$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 23 '18 at 13:43










  • $begingroup$
    @Dr.SonnhardGraubner I did that and the result seems to be a quartic equation where the term with $t^3$ is missing. But is there a way to solve that type of equation? I'll edit my question to show what I got when I did the substitution you suggested.
    $endgroup$
    – Wild Feather
    Dec 23 '18 at 13:48








  • 1




    $begingroup$
    The quartic equation is the best what you can get.
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 23 '18 at 13:49
















$begingroup$
@KarnWatcharasupat But didn't you forget the square root in there? Or am I missing something?
$endgroup$
– Wild Feather
Dec 23 '18 at 13:42






$begingroup$
@KarnWatcharasupat But didn't you forget the square root in there? Or am I missing something?
$endgroup$
– Wild Feather
Dec 23 '18 at 13:42






1




1




$begingroup$
Try to substitute $$t=sqrt{1+x^2}$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 23 '18 at 13:43




$begingroup$
Try to substitute $$t=sqrt{1+x^2}$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 23 '18 at 13:43












$begingroup$
@Dr.SonnhardGraubner I did that and the result seems to be a quartic equation where the term with $t^3$ is missing. But is there a way to solve that type of equation? I'll edit my question to show what I got when I did the substitution you suggested.
$endgroup$
– Wild Feather
Dec 23 '18 at 13:48






$begingroup$
@Dr.SonnhardGraubner I did that and the result seems to be a quartic equation where the term with $t^3$ is missing. But is there a way to solve that type of equation? I'll edit my question to show what I got when I did the substitution you suggested.
$endgroup$
– Wild Feather
Dec 23 '18 at 13:48






1




1




$begingroup$
The quartic equation is the best what you can get.
$endgroup$
– Dr. Sonnhard Graubner
Dec 23 '18 at 13:49




$begingroup$
The quartic equation is the best what you can get.
$endgroup$
– Dr. Sonnhard Graubner
Dec 23 '18 at 13:49










1 Answer
1






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oldest

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1












$begingroup$

As you wrote, using $t=sqrt{1+x^2}$, you end with
$$t^4+14 t^2-32 p^2 t-(32 p^2-49)=0$$ which can be solved using Ferrari's method.



If you use the steps given here, you should find that
$$Delta=-1048576, p^4 left(p^2-1right) left(27 p^2+98right)$$ which is negative if $p^2 lt 1$.



We also get $P=112$, $Q=-256 p^2$, $Delta_0=784-384 p^2$, $D=-2048 p^2$ and you can analyze the number and nature of roots depending on $p$. Now, play with the radicals (have fun) and come back to $x$ (more fun).



Personally, I would prefer to consider the function



$$f(x)=frac{left(1+frac{x^2}{8}right)^2}{2 left(sqrt{x^2+1}+1right)}$$ which is symmetric and shows a minimum value equal to $frac 14$ at $x=0$. This means that if $p^2 lt frac 14$ no root at all; if $p^2 = frac 14$, a quadruple root equal to $0$ and if $p gt frac 14$, two real roots (good to know that they are symmetric) and two complex conjugate roots.



If you do not want to play with nasty radicals, just use a numerical method considering that you look for the zero of function
$$g(y)=frac{left(1+frac{y}{8}right)^2}{2 left(sqrt{y+1}+1right)}-p^2$$



If you consider Newton method, series expansion built at $x=0$ or $xto infty$ followed by series reversion will give quite good estimates.






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    $begingroup$

    As you wrote, using $t=sqrt{1+x^2}$, you end with
    $$t^4+14 t^2-32 p^2 t-(32 p^2-49)=0$$ which can be solved using Ferrari's method.



    If you use the steps given here, you should find that
    $$Delta=-1048576, p^4 left(p^2-1right) left(27 p^2+98right)$$ which is negative if $p^2 lt 1$.



    We also get $P=112$, $Q=-256 p^2$, $Delta_0=784-384 p^2$, $D=-2048 p^2$ and you can analyze the number and nature of roots depending on $p$. Now, play with the radicals (have fun) and come back to $x$ (more fun).



    Personally, I would prefer to consider the function



    $$f(x)=frac{left(1+frac{x^2}{8}right)^2}{2 left(sqrt{x^2+1}+1right)}$$ which is symmetric and shows a minimum value equal to $frac 14$ at $x=0$. This means that if $p^2 lt frac 14$ no root at all; if $p^2 = frac 14$, a quadruple root equal to $0$ and if $p gt frac 14$, two real roots (good to know that they are symmetric) and two complex conjugate roots.



    If you do not want to play with nasty radicals, just use a numerical method considering that you look for the zero of function
    $$g(y)=frac{left(1+frac{y}{8}right)^2}{2 left(sqrt{y+1}+1right)}-p^2$$



    If you consider Newton method, series expansion built at $x=0$ or $xto infty$ followed by series reversion will give quite good estimates.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      As you wrote, using $t=sqrt{1+x^2}$, you end with
      $$t^4+14 t^2-32 p^2 t-(32 p^2-49)=0$$ which can be solved using Ferrari's method.



      If you use the steps given here, you should find that
      $$Delta=-1048576, p^4 left(p^2-1right) left(27 p^2+98right)$$ which is negative if $p^2 lt 1$.



      We also get $P=112$, $Q=-256 p^2$, $Delta_0=784-384 p^2$, $D=-2048 p^2$ and you can analyze the number and nature of roots depending on $p$. Now, play with the radicals (have fun) and come back to $x$ (more fun).



      Personally, I would prefer to consider the function



      $$f(x)=frac{left(1+frac{x^2}{8}right)^2}{2 left(sqrt{x^2+1}+1right)}$$ which is symmetric and shows a minimum value equal to $frac 14$ at $x=0$. This means that if $p^2 lt frac 14$ no root at all; if $p^2 = frac 14$, a quadruple root equal to $0$ and if $p gt frac 14$, two real roots (good to know that they are symmetric) and two complex conjugate roots.



      If you do not want to play with nasty radicals, just use a numerical method considering that you look for the zero of function
      $$g(y)=frac{left(1+frac{y}{8}right)^2}{2 left(sqrt{y+1}+1right)}-p^2$$



      If you consider Newton method, series expansion built at $x=0$ or $xto infty$ followed by series reversion will give quite good estimates.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        As you wrote, using $t=sqrt{1+x^2}$, you end with
        $$t^4+14 t^2-32 p^2 t-(32 p^2-49)=0$$ which can be solved using Ferrari's method.



        If you use the steps given here, you should find that
        $$Delta=-1048576, p^4 left(p^2-1right) left(27 p^2+98right)$$ which is negative if $p^2 lt 1$.



        We also get $P=112$, $Q=-256 p^2$, $Delta_0=784-384 p^2$, $D=-2048 p^2$ and you can analyze the number and nature of roots depending on $p$. Now, play with the radicals (have fun) and come back to $x$ (more fun).



        Personally, I would prefer to consider the function



        $$f(x)=frac{left(1+frac{x^2}{8}right)^2}{2 left(sqrt{x^2+1}+1right)}$$ which is symmetric and shows a minimum value equal to $frac 14$ at $x=0$. This means that if $p^2 lt frac 14$ no root at all; if $p^2 = frac 14$, a quadruple root equal to $0$ and if $p gt frac 14$, two real roots (good to know that they are symmetric) and two complex conjugate roots.



        If you do not want to play with nasty radicals, just use a numerical method considering that you look for the zero of function
        $$g(y)=frac{left(1+frac{y}{8}right)^2}{2 left(sqrt{y+1}+1right)}-p^2$$



        If you consider Newton method, series expansion built at $x=0$ or $xto infty$ followed by series reversion will give quite good estimates.






        share|cite|improve this answer











        $endgroup$



        As you wrote, using $t=sqrt{1+x^2}$, you end with
        $$t^4+14 t^2-32 p^2 t-(32 p^2-49)=0$$ which can be solved using Ferrari's method.



        If you use the steps given here, you should find that
        $$Delta=-1048576, p^4 left(p^2-1right) left(27 p^2+98right)$$ which is negative if $p^2 lt 1$.



        We also get $P=112$, $Q=-256 p^2$, $Delta_0=784-384 p^2$, $D=-2048 p^2$ and you can analyze the number and nature of roots depending on $p$. Now, play with the radicals (have fun) and come back to $x$ (more fun).



        Personally, I would prefer to consider the function



        $$f(x)=frac{left(1+frac{x^2}{8}right)^2}{2 left(sqrt{x^2+1}+1right)}$$ which is symmetric and shows a minimum value equal to $frac 14$ at $x=0$. This means that if $p^2 lt frac 14$ no root at all; if $p^2 = frac 14$, a quadruple root equal to $0$ and if $p gt frac 14$, two real roots (good to know that they are symmetric) and two complex conjugate roots.



        If you do not want to play with nasty radicals, just use a numerical method considering that you look for the zero of function
        $$g(y)=frac{left(1+frac{y}{8}right)^2}{2 left(sqrt{y+1}+1right)}-p^2$$



        If you consider Newton method, series expansion built at $x=0$ or $xto infty$ followed by series reversion will give quite good estimates.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 23 '18 at 15:33

























        answered Dec 23 '18 at 14:32









        Claude LeiboviciClaude Leibovici

        126k1158134




        126k1158134






























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