能否用牛顿法求四次方程最小正实根？收敛性及初始值咨询

Great question—let's break this down step by step since Newton-Raphson (NR) can be finicky with higher-degree polynomials like quartics, especially when targeting a specific root.

Convergence Guarantees (When Setup Correctly) #

Newton-Raphson has local quadratic convergence for simple roots (where the first derivative at the root isn't zero) if your initial guess is sufficiently close to the root. For a quartic (Ax^4 + Bx^3 + Cx^2 + Dx + E = 0):

• First, confirm the smallest positive real root (r) is simple (most cases unless the polynomial has a repeated factor like ((x-r)^2)). If it's a simple root, (f'(r) \neq 0), which satisfies NR's key convergence condition.
• You also need to ensure that between your initial guess and (r), the function doesn't have any critical points (where (f'(x)=0)) that would cause NR to diverge or jump to another root. If you can pick an initial guess in a region where (f(x)) is monotonic (either strictly increasing or decreasing) and convex/concave consistently, NR will reliably converge to (r).

Is Starting at (x=0) Feasible? #

It depends entirely on the coefficients of your quartic:

• Let's compute (f(0) = E) and (f'(0) = D).
  • If (E > 0) (since (f(r)=0) and (r>0), (f(x)) must cross from positive to negative at (r)) and (D < 0) (so (f(x)) is decreasing at (x=0)), the first NR step will give (x_1 = 0 - \frac{f(0)}{f'(0)} = -\frac{E}{D}), which is positive. If this (x_1) is close enough to (r) and the function stays monotonic between 0 and (r), NR will converge nicely.
  • But if (D > 0), the first step will land you at a negative (x_1), which is moving away from the positive root. NR will likely diverge or converge to a negative root instead.
  • If (E = 0), then (x=0) is already a root—but you're looking for the smallest positive root, so this only applies if 0 is a root and there's a larger positive one (but then 0 isn't positive, so you'd need to adjust your initial guess slightly above 0).

In short: (x=0) works in some cases, but it's not a universal solution.

Should You Use a Different Method? #

If you're struggling with initial guesses for NR, or need more reliability regardless of coefficients, consider these alternatives:

• Bisection Method: If you can find an interval ([a, b]) where (f(a)) and (f(b)) have opposite signs (e.g., (a=0) and (b) large enough that (f(b)) has the opposite sign of (f(0))), bisection will always converge to a root in that interval. It's slower than NR but 100% reliable.
• Jenkins-Traub Algorithm: This is a specialized method for real-coefficient polynomials (perfect for quartics). It's designed to find all roots (real and complex) with excellent stability, and it doesn't require careful initial guesses. It's often the go-to for polynomial root-finding in practice.
• Secant Method: Similar to NR but uses finite differences instead of exact derivatives. It has superlinear convergence (faster than bisection, slower than NR) and doesn't require computing (f'(x)), which might be handy if deriving the derivative is a hassle.

内容的提问来源于stack exchange，提问作者spraff