如何实现分段函数，然后在MATLAB中以一定的间隔绘制它

Question

我实际上是在尝试编写三次样条插值的代码。三次样条曲线归结为一系列n-1段，其中n是最初给出的原始坐标的数量，并且每个段都由一些三次函数表示。

我已经弄清楚如何获取每个段的所有系数和值，存储在向量a,b,c,d中，但我不知道如何在不同的时间间隔上将该函数绘制为分段函数。到目前为止，这是我的代码。最后一个for循环是我试图绘制每个段的地方。

%initializations
x = [1 1.3 1.9 2.1 2.6 3.0 3.9 4.4 4.7 5.0 6 7 8 9.2 10.5 11.3 11.6 12 12.6 13 13.3].';
y = [1.3 1.5 1.85 2.1 2.6 2.7 2.4 2.15 2.05 2.1 2.25 2.3 2.25 1.95 1.4 0.9 0.7 0.6 0.5 0.4 0.25].';

%n is the amount of coordinates
n = length(x);
%solving for a-d for all n-1 segments
a = zeros(n,1);
b = zeros(n,1);
d = zeros(n,1);

%%%%%%%%%%%%%% SOLVE FOR a's %%%%%%%%%%%%%
%Condition (b) in Definition 3.10 on pg 146
%Sj(xj) = f(xj) aka yj
for j = 1: n
    a(j) = y(j);
end

%initialize hj
h = zeros(n-1,1);

for j = 1: n-1
    h(j) = x(j+1) - x(j);
end

A = zeros(n,n);
bv = zeros(n,1); %bv = b vector

%initialize corners to 1
A(1,1) = 1;
A(n,n) = 1;

%set main diagonal
for k = 2: n-1
    A(k,k) = 2*(h(k-1) + h(k));
end

%set upper and then lower diagonals
for k = 2 : n-1
    A(k,k+1) = h(k); %h2, h3, h4...hn-1
    A(k,k-1) = h(k-1); %h1, h2, h3...hn
end

%fill up the b vector using equation in notes
%first and last spots are 0
for j = 2 : n-1
    bv(j) = 3*(((a(j+1)-a(j)) / h(j)) - ((a(j) - a(j-1)) / h(j-1)));
end

%augmented matrix
A = [A bv];



%%%%%%%%%%%% BEGIN GAUSSIAN ELIMINATION %%%%%%%%%%%%%%%
offset = 1;
%will only need n-1 iterations since "first" pivot row is unchanged
for k = 1: n-1
  %Searching from row p to row n for non-zero pivot
  for p = k : n
      if A(p,k) ~= 0;
          break;
      end
  end

  %row swapping using temp variable
  if p ~= k
      temp = A(p,:);
      A(p,:) = A(k,:);
      A(k,:) = temp;
  end


  %Eliminations to create Upper Triangular Form
  for j = k+1:n
     A(j,offset:n+1) = A(j,offset:n+1) - ((A(k, offset:n+1) * A(j,k)) / A(k,k));
  end
  offset = offset + 1;
end

c = zeros(n,1); %initializes vector of data of n rows, 1 column

%Backward Subsitution
%First, solve the nth equation
c(n) = A(n,n+1) / A(n,n);

%%%%%%%%%%%%%%%%% SOLVE FOR C's %%%%%%%%%%%%%%%%%%
%now solve the n-1 : 1 equations (the rest of them going backwards
for j = n-1:-1:1 %-1 means decrement
    c(j) = A(j,n+1);
    for k = j+1:n
        c(j) = c(j) - A(j,k)*c(k); 
    end
    c(j) = c(j)/A(j,j);
end

%%%%%%%%%%%%% SOLVE FOR B's and D's %%%%%%%%%%%%%%%%%%%%
for j = n-1 : -1 : 1
    b(j) = ((a(j+1)-a(j)) / h(j)) - (h(j)*(2*c(j) + c(j+1)) / 3);
    d(j) = (c(j+1) - c(j)) / 3*h(j);
end

%series of equation segments

for j = 1 : n-1
    f = @(x) a(j) + b(j)*(x-x(j)) + c(j)*(x-x(j))^2 + d(j)*(x-x(j))^3;
end
plot(x,y,'o');

假设我已为每个段正确计算了向量a,b,c,d。如何绘制每个立方体片段，使它们都显示在单个图上？

我很感激帮助。

Answer 1

这很简单。您已经通过定义一个匿名函数完成了一半的工作，该函数用于每个间隔之间的三次样条函数。但是，您需要确保函数中的操作是元素。您目前使用标量操作，或者假设我们正在使用矩阵运算。不要这样做。使用.*代替*和.^代替^。你需要这样做的原因是为了更容易地生成样条上的点，接下来我的下一点。

接下来要做的就是在相邻x关键点定义的区间内定义一堆x点，并将它们替换为你的函数，然后绘制结果......所以像这样的东西：

figure;
hold on;
for j = 1 : n-1
    f = @(x) a(j) + b(j).*(x-x(j)) + c(j).*(x-x(j)).^2 + d(j)*(x-x(j)).^3; %// Change function to element-wise operations - be careful
    x0 = linspace(x(j), x(j+1)); %// Define set of points
    y0 = f(x0); %// Find output points
    plot(x0, y0, 'r'); %// Plot the line in between the key points
end
plot(x, y, 'bo');

我们生成一个新的数字，然后使用hold on，这样当我们多次调用plot时，我们会将结果附加到同一个数字。接下来，对于我们拥有的每组三次样条系数，定义一个样条函数，然后生成一堆x值x，它们位于当前linspace关键点和旁边的关键点之间它。默认情况下，x(j)在起点（即x(j+1)）和终点（即linspace(x(j), x(j+1), 25);）之间生成100个点。您可以通过指定第三个参数来控制要生成的点数（因此类似x生成25个点）。我们使用这些y值并将其替换为我们的样条方程以获取{{1}}值。然后，我们使用红线在图上绘制此结果。完成后，我们将关键点绘制为曲线顶部的蓝色圆圈。

作为奖励，我使用上面的绘图机制运行你的代码，这就是我得到的：