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给定函数值为: f(0.25)=0.5000, f(0.3)=0.5477, f(0.39)=0.6245, f(0.45)=0.6708, f(0.53)=0.7280 其二阶导数值若表示为: f''(0.25)=M0, f''(0.3)=M1, f''(0.39)=M2, f''(0.45)=M3, f''(0.53)=M4 请利用计算三次样条插值时的三弯矩方程,求出该函数满足自然边界条件(即M0=M4=0)时,M1=______, M2=______, M3=______。(结果用小数表示,保留小数点后4位)
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参考解析:
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