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Solution to Quartic Equation with small first coefficient

  1. May 22, 2013 #1
    I want to solve a depressed quartic:

    [tex]ax^4 + bx^2 + cx + d = 0[/tex]

    Assume [tex]|a|\ll |b|,|c|,|d|[/tex].

    I would like to find the solutions by expanding around the solution to the quadratic. If you try to solve in, say, Maple, and expand around a in a series you get something that blows up. That seems silly to me, why would the solution blow up as [tex]a\rightarrow 0[/tex]. Only because the solution method assumed a was not zero. Clearly, if you plotted it on the complex plane, the quartic solutions must approach the quadratic. Anyway, so my idea was to solve the quadratic equation

    [tex]bx^2+cx+d=0[/tex] to get

    [tex]x_0 = \frac{-c\pm\sqrt{c^2-4bd}}{2b}[/tex]

    Then insert [tex]x_0[/tex] into the quartic part like so,

    [tex]ax_0^4 + bx^2+cx+d = 0[/tex]

    Then solve this quadratic to get,

    [tex]x_1 = \frac{-c\pm\sqrt{c^2-4b(d+ax_0^4)}}{2b}[/tex].

    Does anyone see a problem with this approach (assuming I just want some sort of approximation)?

    Secondly, this question just occurred to me: I see this generates 4 solutions, but will all solutions be closer to the exact solutions than the initial quadratic solutions are?
     
  2. jcsd
  3. May 22, 2013 #2

    mfb

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    Small a will lead to two large solutions, which you do not catch with that approach. The solutions you can get are close to the solutions for a=0. If a is small enough and the other coefficients are not special in some way, I would expect that the last recursion relation converges.
     
  4. May 22, 2013 #3
    Yes, I just thought of that too. The quartic solution x would be proportional to 1/a^(1/4) unfortunately, so that term would not be "small" after all.
     
  5. May 22, 2013 #4
    Take this special case though: I'm only interested in the cases where a>0, b>0, c real, and d< 0. There should be (by Descartes rule of signs) two real solutions. When I plot the graph, the solutions of the quartic appear to converge to the quadratic solutions. So it appears that my recursion would be valid in that case. [No proof yet though.]
     
  6. May 22, 2013 #5

    mfb

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    With a>0, b>0, d<0 and c not too large, the two "large" solutions are imaginary, as they lie in the region where bx^2+cx+d is positive.
     
  7. May 24, 2013 #6
    Why not just use Newton's method with x0 as an initial guess? This is easy to set up, and should converge rapidly.
     
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