Can someone explain to me what a cubic spline is, and how we could use it to interpolate a function?

I have searched on the internet but I would like a simple explanation.


If you have a function $f(x)$ on some interval $[a,b]$, which is divided on $[x_{i-1}, x_i]$ such as $a=x_0< x_1< ... <x_N=b$ then you can interpolate this function by a cubic spline $S(x)$. $S(x)$ is a piecewise function: on each $h_i = x_i - x_{i-1}$ it's a cubic polynomial, which can be written for simplicity as $S_i(x) = a_i + b_i(x - x_i) + {c_i\over2}(x-x_i)^2 + {d_i\over6}(x - x_i)^3 \,\!$. It has to satisfy the next constraints:

1) passing through the knots : $S_i\left(x_{i}\right) = f(x_{i})$

2) be continuous up to the 2nd derivative:

$S_i\left(x_{i-1}\right) = S_{i-1}(x_{i-1}) \\ S'_i\left(x_{i-1}\right) = S'_{i-1}(x_{i-1}) \\ S''_i\left(x_{i-1}\right) = S''_{i-1}(x_{i-1})$

3) for natural splines:

$S''(a) = S''(b) = 0.$

These equations will uniquely define spline coefficients.

A good way to understand this is to take e.g. 3 points and manually solve systems for coefficients of $S_1(x)$ and $S_2(x)$

Finally you should get the next system:

$a_i = f\left(x_{i}\right) \,\!$

$h_ic_{i-1} + 2(h_i + h_{i+1})c_i + h_{i+1}c_{i+1} = 6\left({{f_{i+1} - f_i}\over{h_{i+1}}} - {{f_{i} - f_{i-1}}\over{h_{i}}}\right) \,\!$

$d_i = {{c_i - c_{i-1}}\over{h_i}} \,\!$

$b_i = {1\over2}h_ic_i - {1\over6}h_i^2d_i + {{f_i - f_{i-1}}\over{h_i}}= {{f_i - f_{i-1}}\over{h_i}} + {{h_i(2c_i + c_{i-1})}\over6} \,\!$

  • $\begingroup$ Would it be easy for you to explain it to me through an example?Like if we have a function f(x)=(1/2)|x| and x_0=1,x_1=0,x_2=1 how we interpolate this by a cubic spline with s_1"(x_0)=s_1"(x_2)=0 ?? $\endgroup$ – cube Jan 11 '14 at 14:04
  • $\begingroup$ it has to be s_0''(x_0) = 0 and s_3''(x_2) =0, which gives you c_0 = 0 and c_3=0. then you can use these values in the system, which I've added to the answer. Is it ok? $\endgroup$ – user2575760 Jan 11 '14 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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