摘要: https://www.sciencedirect.com/science/article/pii/S0377042709000909 阅读全文
posted @ 2018-12-02 19:30 lilei9110 阅读(201) 评论(0) 推荐(0) 编辑
摘要: 位置 (2t³ - 3t² + 1) * p0 + (t³ - 2t² + t) * m0 + (-2t³ + 3t²) * p1 + (t³ - t²) * m1 切线 p'(t) = (6t² - 6t)p0 + (3t² - 4t + 1)m0 + (-6t² + 6t)p1 + (3t² - 阅读全文
posted @ 2018-12-02 14:57 lilei9110 阅读(397) 评论(0) 推荐(0) 编辑