We read every piece of feedback, and take your input very seriously.
To see all available qualifiers, see our documentation.
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
The text was updated successfully, but these errors were encountered:
Sorry, something went wrong.
d23653a
第三问这个构造有问题吧,这样构造出来的\ell 二阶导存在吗, 如果是连续函数的话,显然在knots处一阶导都不存在
线性函数为啥不存在二阶导?
第三问这个构造有问题吧,这样构造出来的\ell 二阶导存在吗, 如果是连续函数的话,显然在knots处一阶导都不存在 线性函数为啥不存在二阶导?
\ell只是在[x1,x2]上是线性的吧,在其他knots处取值应该是0,不然对于i=3,...,n 怎么保证 (y_i-f(x_i))^2=(y_i-f_1(x_i))^2
第三问这个构造有问题吧,这样构造出来的\ell 二阶导存在吗, 如果是连续函数的话,显然在knots处一阶导都不存在 线性函数为啥不存在二阶导? \ell只是在[x1,x2]上是线性的吧,在其他knots处取值应该是0,不然对于i=3,...,n 怎么保证 (y_i-f(x_i))^2=(y_i-f_1(x_i))^2
谢谢,我意识到错误了!
当初我可能是误解了 “with knots at each of the x_i”,以为要对 (x_i, y_i) 进行 interpolate,但其实并不一定。应该可以直接由 (a)(b) 结论得出,对于任意 f,总有过 (x_i, f(x_i)) 的 natural cubic spline 比 f 要好,则 minimizer 一定是 natural cubic spline。
No branches or pull requests
The text was updated successfully, but these errors were encountered: