The quadratic spline
I'd like to fit the data in table as blow
x f(x)
3.0 2.5
4.5 1.0
7.0 2.5
9.0 0.5when $x=5$, I want to find value of $f(x)$ by using Quadratic Spline.
this is my text book example. but there are some confused point which is I can't understand equation.question
How to set the equation?
- $20.25a_1 + 4.5b_1 + c_1 = 1.0$
- $20.25a_2 + 4.5b_2 + c_2 = 1.0$
- $49a_2 + 7b_2 + c_2 = 2.5$
- $49a_3 + 7b_3 + c_3 = 2.5$
I meant that my question is that why does equation to be
- $a_1 + b_1 + c_1$
- $a_2 + b_2 + c_2$
- $a_2 + b_2 + c_2$
- $a_3 + b_3 + c_3$
instead of
- $a_1 + b_1 + c_1$
- $a_2 + b_2 + c_2$
- $a_3 + b_3 + c_3$
- $a_4 + b_4 + c_4$
Also
- $9a_1 + 3b_1 + c_1 =2.5$
- $81a_3+9b_3+c3 = 0.5 $
?
update2
as you can see from the picture, there is one equation 18.33 But I'm not understand it. how to that condition can be represented to 18.33 equation ? would you please give any hint?
update3.
finally, I understand almost but except the matrix. In the text, they are use the matrix for find to each coefficients. So My question is how to set this matrix. I can't find quite a bit relationships.
Would you please give me any hint?
$\endgroup$ 12 Answers
$\begingroup$Here is a cubic-spline interpolation for the $4$ points given in your question:
$ f(x)= \begin{cases} +0.1 \overline3(x-3.0)^3 -1.3 (x-3.0)+2.5 & \text{$3.0 \leq x \leq 4.5$}\\ -0.15\overline1(x-4.5)^3+0.6 (x-4.5)^2+0.0\overline4(x-4.5)+1.0 & \text{$4.5 \leq x \leq 7.0$}\\ +0.0 \overline8(x-7.0)^3-0.5\overline3(x-7.0)^2-0.2\overline8(x-7.0)+2.5 & \text{$7.0 \leq x \leq 9.0$}\\ \end{cases} $
Here is a piece of C code for any given number of points $(x_0,y_0),(x_1,y_1),\ldots,(x_N,y_N)$:
void Spline(double x[N+1],double y[N+1], // input double A[N],double B[N], // output double C[N],double D[N]) // output
{ double w[N]; double h[N]; double ftt[N+1]; for (int i=0; i<N; i++) { w[i] = (x[i+1]-x[i]); h[i] = (y[i+1]-y[i])/w[i]; } ftt[0] = 0; for (int i=0; i<N-1; i++) ftt[i+1] = 3*(h[i+1]-h[i])/(w[i+1]+w[i]); ftt[N] = 0; for (int i=0; i<N; i++) { A[i] = (ftt[i+1]-ftt[i])/(6*w[i]); B[i] = ftt[i]/2; C[i] = h[i]-w[i]*(ftt[i+1]+2*ftt[i])/6; D[i] = y[i]; }
}
void PrintSpline(double x[N+1], // input double A[N],double B[N], // input double C[N],double D[N]) // input
{ for (int i=0; i<N; i++) { printf("%f <= x <= %f : f(x) = ",x[i],x[i+1]); printf("%f(x-%f)^3 + ",A[i],x[i]); printf("%f(x-%f)^2 + ",B[i],x[i]); printf("%f(x-%f) + " ,C[i],x[i]); printf("%f\n" ,D[i]); }
}Here is a piece of Python code for any given number of points $(x_0,y_0),(x_1,y_1),\ldots,(x_N,y_N)$:
class Point: def __init__(self,x,y): self.x = 1.0*x self.y = 1.0*y
def Spline(points): N = len(points)-1 w = [(points[i+1].x-points[i].x) for i in range(0,N)] h = [(points[i+1].y-points[i].y)/w[i] for i in range(0,N)] ftt = [0]+[3*(h[i+1]-h[i])/(w[i+1]+w[i]) for i in range(0,N-1)]+[0] A = [(ftt[i+1]-ftt[i])/(6*w[i]) for i in range(0,N)] B = [ftt[i]/2 for i in range(0,N)] C = [h[i]-w[i]*(ftt[i+1]+2*ftt[i])/6 for i in range(0,N)] D = [points[i].y for i in range(0,N)] return A,B,C,D
def PrintSpline(points,A,B,C,D): for i in range(0,len(points)-1): func = str(points[i].x)+' <= x <= '+str(points[i+1].x)+' : f(x) = ' components = [] if A[i]: components.append(str(A[i])+'(x-'+str(points[i].x)+')^3') if B[i]: components.append(str(B[i])+'(x-'+str(points[i].x)+')^2') if C[i]: components.append(str(C[i])+'(x-'+str(points[i].x)+')') if D[i]: components.append(str(D[i])) if components: func += components[0] for i in range (1,len(components)): if components[i][0] == '-': func += ' - '+components[i][1:] else: func += ' + '+components[i] print func else: print func+'0'
def Example(): points = [Point(3.0,2.5),Point(4.5,1.0),Point(7.0,2.5),Point(9.0,0.5)] A,B,C,D = Spline(points) PrintSpline(points,A,B,C,D)Please note that the two pieces of code above assume $x_0 < x_1 < \ldots < x_N$.
$\endgroup$ 1 $\begingroup$Your textbook basically is creating one quadratic spline in between each two points. So, let's define the quadratic splines as
$f_1(x)=a_1x^2+b_1x+c_1$ for $3 \le x \le 4.5$,
$f_2(x)=a_2x^2+b_2x+c_2$ for $4.5 \le x \le 7$ and
$f_3(x)=a_3x^2+b_3x+c_3$ for $7 \le x \le 9$,
then we can obtain 4 equations from the position continuity at $x=4.5$ and $x=7$:
$f_1(4.5)=1.0$,
$f_2(4.5)=1.0$,
$f_2(7)=2.5$,
$f_3(7)=2.5$.
This will result in the 4 equations in your original post. Since there are only 3 quadratic splines involved, there are no $a_4$, $b_4$ and $c_4$.
The other two equations $9a_1+3b_1+c1=2.5$ and $81a_3+9b_3+c_3=0.5$ come from $f_1(3)=2.5$ and $f_3(9)=0.5$.
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