zoukankan      html  css  js  c++  java
  • 转载:Cubic interpolation

    https://www.paulinternet.nl/?page=bicubic

    Cubic interpolation

    If the values of a function f(x) and its derivative are known at x=0 and x=1, then the function can be interpolated on the interval [0,1] using a third degree polynomial. This is called cubic interpolation. The formula of this polynomial can be easily derived.

    A third degree polynomial and its derivative:

    f(x) = ax^3 + bx^2 + cx + d

    f'(x) = 3ax^2 + 2bx + c

    The values of the polynomial and its derivative at x=0 and x=1:

    f(0) = d

    f(1) = a + b + c + d

    f'(0) = c

    f'(1) = 3a + 2b + c

    The four equations above can be rewritten to this:

    a = 2f(0) - 2f(1) + f'(0) + f'(1)

    b = -3f(0) + 3f(1) - 2f'(0) - f'(1)

    c = f'(0)

    d = f(0)

    And there we have our cubic interpolation formula.

    Interpolation is often used to interpolate between a list of values. In that case we don't know the derivative of the function. We could simply use derivative 0 at every point, but we obtain smoother curves when we use the slope of a line between the previous and the next point as the derivative at a point. In that case the resulting polynomial is called a Catmull-Rom spline. Suppose you have the values p0, p1, p2and p3 at respectively x=-1, x=0, x=1, and x=2. Then we can assign the values of f(0), f(1), f'(0) and f'(1) using the formulas below to interpolate between p1 and p2.

    f(0) = p_1

    f(1) = p_2

    f'(0) = dfrac{p_2 - p_0}{2}

    f'(1) = dfrac{p_3 - p_1}{2}

    Combining the last four formulas and the preceding four, we get:

    a = -	frac{1}{2}p_0 + 	frac{3}{2}p_1 - 	frac{3}{2}p_2 + 	frac{1}{2}p_3

    b = p_0 - 	frac{5}{2}p_1 + 2p_2 - 	frac{1}{2}p_3

    c = -	frac{1}{2}p_0 + 	frac{1}{2}p_2

    d = p_1

    So our cubic interpolation formula becomes:

    f(p_0,p_1,p_2,p_3,x) = (-	frac{1}{2}p_0 + 	frac{3}{2}p_1 - 	frac{3}{2}p_2 + 	frac{1}{2}p_3)x^3 + (p_0 - 	frac{5}{2}p_1 + 2p_2 - 	frac{1}{2}p_3)x^2 + (-	frac{1}{2}p_0 + 	frac{1}{2}p_2)x + p_1

    For example:

    plot


    For the green curve:

    a = -	frac{1}{2}cdot2 + 	frac{3}{2}cdot4 - 	frac{3}{2}cdot2 + 	frac{1}{2}cdot3 = 	frac{7}{2}

    b = 2 - 	frac{5}{2}cdot4 + 2cdot2 - 	frac{1}{2}cdot3 = -	frac{11}{2}

    c = -	frac{1}{2}cdot2 + 	frac{1}{2}cdot2 = 0

    d = 4

    f(x) = 	frac{7}{2}(x-2)^3 - 	frac{11}{2}(x-2)^2 + 4

    The first and the last interval

    We used the two points left of the interval and the two points right of the inverval as inputs for the interpolation function. But what if we want to interpolate between the first two or last two elements of a list? Then we have no p0 or no p3. The solution is to imagine an extra point at each end of the list. In other words, we have to make up a value for p0 and p3 when interpolating the leftmost and rightmost interval respectively. Two ways to do this are:

      • Repeat the first and the last point.
        Left: p0 = p1
        Right: p3 = p2
      • Let the end point be in the middle of a line between the imaginary point and the point next to the end point.
        Left: p0 = 2p1 - p2
        Right: p3 = 2p2 - p1
  • 相关阅读:
    gulp serve 报错 gulp.ps1
    执行git命令时出现fatal: 'origin' does not appear to be a git repository错误
    利用 SASS 简化 `nth-child` 样式的生成
    git的一些常用命令
    回调函数
    匿名函数
    css消除行内元素的间隙
    @click.native的使用
    Element-ui 下拉列表 选项过多时如何解决卡顿问题
    vue组件通信(父子之间,兄弟之间)
  • 原文地址:https://www.cnblogs.com/fellow1988/p/9992026.html
Copyright © 2011-2022 走看看