已知f (1) = 0,f (−1) = −3,f (2) = 4.求函数f (x)过这三点的二次拉格朗日插值 多项式?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 16:56:41
![已知f (1) = 0,f (−1) = −3,f (2) = 4.求函数f (x)过这三点的二次拉格朗日插值 多项式?](/uploads/image/z/6928561-1-1.jpg?t=%E5%B7%B2%E7%9F%A5f+%281%29+%3D+0%2Cf+%28%26%238722%3B1%29+%3D+%26%238722%3B3%2Cf+%282%29+%3D+4.%E6%B1%82%E5%87%BD%E6%95%B0f+%28x%29%E8%BF%87%E8%BF%99%E4%B8%89%E7%82%B9%E7%9A%84%E4%BA%8C%E6%AC%A1%E6%8B%89%E6%A0%BC%E6%9C%97%E6%97%A5%E6%8F%92%E5%80%BC+%E5%A4%9A%E9%A1%B9%E5%BC%8F%3F)
已知f (1) = 0,f (−1) = −3,f (2) = 4.求函数f (x)过这三点的二次拉格朗日插值 多项式?
已知f (1) = 0,f (−1) = −3,f (2) = 4.求函数f (x)过这三点的二次拉格朗日插值 多项式?
已知f (1) = 0,f (−1) = −3,f (2) = 4.求函数f (x)过这三点的二次拉格朗日插值 多项式?
f(x)=5/6x^2+3/2x-7/3
过这三点的拉格朗日插值公式为
F(x)=((1-x)*(-1-x))/((1-2)*(-1-2))*4+((1-x)*(2-x))/((1+1)*(2+1))*(-3)+((2-x)*(-1-x))/((2-1)*(-1-1))*0
计算之后可以得到上面的f(x)