如何提升牛顿均差插值精度？针对指数函数的改进方案问询

Hey there, let's tackle your questions about Newton's divided difference interpolation—super relevant when you're chasing better accuracy, especially for specific functions like $y=e^{-x}-x$. Here's a breakdown:

There are several practical, battle-tested ways to crank up the accuracy of Newton's divided differences:

• Ditch equidistant nodes for Chebyshev nodes: Equidistant nodes are convenient, but they suffer from Runge's phenomenon—huge errors near the endpoints of your interval. Chebyshev nodes (spaced as $\cos\left(\frac{(2k-1)\pi}{2n}\right)$ for an interval $[-1,1]$, scaled to your target range) spread out more evenly in terms of function curvature, drastically reducing endpoint errors.
• Boost numerical stability in calculations: When computing divided differences, avoid subtracting nearly equal numbers (a classic source of floating-point error). Use the recursive formula carefully, and consider using higher-precision floating-point types—like Python's decimal.Decimal with increased precision or C++'s long double—instead of standard 64-bit floats.
• Add adaptive node refinement: Don't stick to a fixed set of nodes. First interpolate with a base set, then estimate the interpolation error (using the next divided difference as an error bound, or comparing to a higher-degree interpolant). Add extra nodes in regions where the error is above your threshold.
• Handle ill-conditioned systems: If your nodes are too close together, the divided difference table can become numerically unstable. Regularize the problem slightly (e.g., add a tiny epsilon to node positions) or reselect nodes to ensure they're sufficiently spaced.

Absolutely! The key here is to leverage the specific behavior of $y=e^{-x}-x$ to tailor your interpolation approach:

• Split the function and interpolate only the non-linear part: Notice that $y=e^{-x} - x$ is a combination of an exponential term and a linear term. The linear term $-x$ can be computed exactly—no need to interpolate it. Focus your Newton's divided difference interpolation solely on $e^{-x}$, then subtract $x$ from the result. This eliminates any unnecessary error from interpolating a perfectly linear component.
• Piecewise adaptive interpolation based on function curvature: The function $y=e^{-x}-x$ has a second derivative $y''=e^{-x}$, which decays exponentially to 0 as $x$ increases. That means:
  • For small $x$ (e.g., $x < 4$), the function curves more sharply—use Chebyshev nodes with a higher-degree interpolant (or adaptive refinement) here.
  • For large $x$ (e.g., $x \geq 4$), $e^{-x}$ is negligible, so $y \approx -x$. A simple linear interpolant (or even just using the exact linear approximation) will be more than accurate enough, and avoids overfitting with high-degree polynomials.
• Use weighted divided differences: Since $e^{-x}$ decays rapidly, standard interpolation can struggle to capture its behavior for large $x$ without adding tons of nodes. Instead, use a weighted Newton's divided difference approach with a weight function like $w(x)=e^x$. This transforms the function to $w(x)y(x)=1 - x e^x$, which is a smoother, better-behaved function to interpolate—you can then divide by $w(x)$ to get back the original $y(x)$.
• Precompute exponential values with high precision: For nodes where you need $e^{-x}$, use a high-precision exponential function (like Python's math.exp is already reliable, but libraries like mpmath can go further) instead of relying on interpolated values for the exponential term. This reduces error at the source.

内容的提问来源于stack exchange，提问作者Surely Advert