给定一组点，什么是适合抛物线给他们以最快的方式？难道这样做的平方计算至少还是有一个迭代的方式？

Given a set of points, what's the fastest way to fit a parabola to them? Is it doing the least squares calculation or is there an iterative way?

感谢

编辑： 我认为，梯度下降是要走的路。平方计算至少会是一点点的税收（不必做QR分解或东西让事情变得稳定）。

I think gradient descent is the way to go. The least squares calculation would have been a little bit more taxing (having to do qr decomposition or something to keep things stable).

推荐答案

如果该点都没有错误有关，您可通过三个点的内插。否则最小二乘或任何同等配方是要走的路。

If the points have no error associated, you may interpolate by three points. Otherwise least squares or any equivalent formulation is the way to go.