- #1
andert
- 12
- 0
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?
[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?