2
$\begingroup$

I am learning about natural splines and basis functions and am struggling with it a lot.

I understand the concepts of knots being the part where first and second derivatives are equal on either side. But when it comes to calculating basis functions and eventually a basis matrix I get confused.

I understand that for a cubic spline $d_i(x) = \frac{(x-x_i)^{3}_+ - (x-x_n)^{3}_+}{x_n - x_i}, i=1,\dots,n-1.$

for the example n = 4 ,

$d_i(x) = \frac{(x-x_i)^{3}_+ - (x-x_4)^{3}_+}{x_4 - x_i}$

From here onwards I get lost. How do I find $d_1(x_1)$? I would assume you just replace i with 1 and x with $x_1$ however this would result in $$d_1(x_1) = \frac{(x_1-x_i)^{3}_+ - (x_1-x_4)^{3}_+}{x_4 - x_1} = \frac{(0)^{3}_+ - (x_1-x_4)^{3}_+}{x_4 - x_1} = \frac{- (x_1-x_4)^{3}_+}{x_4 - x_1}$$

However the text book says the answer is 0. For $d_1(x_2)$ it gives an answer of $\frac{(x_2-x_1)^3}{x_4-x_1}$.

I do not understand why it appears that for every answer it doesn't include the second term of the numerator.

Any clarification as to how $d_1(x_1)$ or $d_3(x_2)$ is 0 would be very appreciated.

$\endgroup$
1

1 Answer 1

0
$\begingroup$

Not sure what book or reference you are using, but it is standard to assume you ordered the data so that $x_1 \leq x_2 \leq \dots \leq x_n$. So for $d_1(x_1)$, recalling that $(w)_+ = \max(w,0)$, we have

$$d_1(x_1) = \frac{(x_1 - x_4)_+^3}{x_4-x_1} = \frac{\big(\max(0,x_1-x_4)\big)^3}{x_4-x_1} = 0$$

where the last equality followed because $x_1-x_4 \leq 0$ so that the numerator is $0$.

Also, just in case, this question on the site may be of interest to you: Why are the basis functions for natural cubic splines expressed as they are? (ESL)

$\endgroup$
2
  • $\begingroup$ Thankyou, I must have missed that if the answer would result in less than 0 it becomes 0. $\endgroup$ Mar 26, 2021 at 1:16
  • $\begingroup$ yep, that's the purpose of the little plus sign below the parenthesis $(x)_+$ $\endgroup$
    – doubled
    Mar 26, 2021 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.