割线法未收敛问题咨询：Matlab求f(x)=exp(2x)+3x-4根程序排查

Hey Maarten, let's break down why your program isn't converging—first, let's start with some key facts about your function to set the stage:

• Your function is strictly monotonically increasing (its derivative f’(x) = 2exp(2x) + 3 is always positive, since exp(2x) is never negative)
• As x → -∞, f(x) → -∞; as x → +∞, f(x) → +∞
• This means there is exactly one real root—so in theory, a correctly implemented method should converge. The issue is almost certainly either a bug in your code, or a poor choice of iteration method/initial value.

1. Most Likely: Program Defects #

Here are the common code mistakes that cause non-convergence:

• Incorrect derivative calculation: If you're using Newton-Raphson (the most common root-finding method), the formula requires dividing by f’(x). A typo like omitting the +3 in 2exp(2x) + 3 will completely break the iteration. Double-check your derivative code—make sure it's df = 2*exp(2*x) + 3; in MATLAB.
• Bad termination conditions: If you set too few maximum iterations, or a convergence threshold that's unreasonably strict (like 1e-12 when your iteration is only getting to 1e-6), the program will stop before converging. Also, make sure you're checking either |x_new - x_old| < tolerance or |f(x_new)| < tolerance (both is even better).
• Extreme initial value: While your function is monotonic, an initial value way outside the root's range (like x=10) will cause exp(2x) to overflow to Inf, breaking the iteration. The root is between 0 (where f(0)=-3) and 0.5 (where f(0.5)≈0.218), so start with a value in this interval (e.g., 0.3).
• Unchecked numerical errors: If your iteration produces NaN or Inf at any step, the rest of the loop will fail. Add a check to catch these and terminate early with a message.

2. Could the Method Itself Fail? #

It depends on which method you used:

• Newton-Raphson: For your strictly convex function (f''(x)=4exp(2x) > 0 everywhere), Newton-Raphson will converge to the root from any initial value that doesn't cause numerical overflow. If this method isn't converging, you definitely have a code bug.
• Fixed-point iteration: If you rearranged the equation to x = (4 - exp(2x))/3, this will diverge! The derivative of the iteration function g(x) = (4 - exp(2x))/3 is g’(x) = -2exp(2x)/3. Near the root (~0.449), |g’(x)| ≈ 1.5 > 1, which violates the fixed-point convergence condition (needs |g’(x)| < 1 in the root's neighborhood). This is a case where the method itself won't work—you'll need to pick a different rearrangement or method.
• Bisection method: This method is guaranteed to converge as long as your initial interval contains the root. If bisection isn't working, your code has a logic error (like updating the interval incorrectly).

3. Quick Tests to Diagnose the Issue #

• Verify the root first: Use MATLAB's built-in fzero function to get the correct root:

  fzero(@(x) exp(2*x) + 3*x - 4, 0.3)
  

  You should get a result around 0.449—this confirms the root exists and is computable.
• Print iteration steps: Add disp([x_n, f(x_n)]) inside your loop to see how x and f(x) are changing. If x is oscillating or blowing up, you'll spot the issue immediately.
• Test with a known-good initial value: Start with x=0.3 (right in the root's interval) and see if the iteration converges. If it does, your original initial value was the problem; if not, your iteration formula is wrong.

内容的提问来源于stack exchange，提问作者p.late