Comments (5)
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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
from esl-cn.
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?
from esl-cn.
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
from esl-cn.
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。
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Related Issues (20)
- Phoneme Data
- Ex. 6.8 HOT 1
- Ex. derive natural spline bases from B-spline bases HOT 2
- Ex. 9.5 HOT 7
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- Ex. 9.6 HOT 6
- broken gravatar for mainland China HOT 1
- disappeared avatar on the n-th (n>=2) level in phones from mainland China
- workflow for Rmd notes
- ok
- Fig. 18.1 HOT 5
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- Ex. 2.2
- Chapter 1.1 - Website URL of ESL is unavailable HOT 1
- Fig. 8.6 and Tab. 8.2: Example for EM HOT 2
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- Ex. 3.7 HOT 1
- Fig. 14.29 HOT 1
- Fig. 14.30 HOT 1
- Can you tell me how the recursive computational spline basis functions of Computations for Splines (B-splines) can be used on natural cubic spline? HOT 4
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