讲解使用牛顿迭代法求一元三次方程的根《C程序设计第五版》谭浩强著第五章第138页第14道题14.用牛顿迭代法求下面方程在1.5附近的根2x^3 - 4x^2 3x - 6 0向AI提问得到的答案高三《数学》第三册第62页的内容求导数的方法回顾上一道题的牛顿迭代法的解题思路《讲解使用牛顿迭代法求x的平方根》https://blog.csdn.net/weixin_42145772/article/details/163803077?spm1001.2014.3001.5502[suozhanglocalhost test5_14]$ cat test5_14.c#includestdio.hintmain(){floatf0,f1,x0,x1,d;intn;x01.0;f02*x0*x0*x0-4*x0*x03*x0-6;f16*x0*x0-8*x03;x1x0-f0/f1;n2;printf(第1次迭代\n);printf(x0%f\n,x0);printf(x1%f\n,x1);printf(d%f - %f %f\n,x0,x1,x0-x1);printf(\n);do{x0x1;f02*x0*x0*x0-4*x0*x03*x0-6;f16*x0*x0-8*x03;x1x0-f0/f1;dx0-x1;printf(第%d次迭代\n,n);printf(x0%f\n,x0);printf(x1%f\n,x1);printf(d%f - %f %f\n,x0,x1,d);printf(\n);nn1;}while(d1e-5);printf(x1%f\n,x1);return(0);}[suozhanglocalhost test5_14]$ gcc -g -Wall test5_14.c -o test.bin[suozhanglocalhost test5_14]$ ./test.bin第1次迭代x01.000000x16.000000d1.000000 - 6.000000 -5.000000第2次迭代x06.000000x14.245614d6.000000 - 4.245614 1.754386第3次迭代x04.245614x13.109505d4.245614 - 3.109505 1.136109第4次迭代x03.109505x12.423687d3.109505 - 2.423687 0.685819第5次迭代x02.423687x12.092294d2.423687 - 2.092294 0.331392第6次迭代x02.092294x12.005691d2.092294 - 2.005691 0.086604第7次迭代x02.005691x12.000023d2.005691 - 2.000023 0.005667第8次迭代x02.000023x12.000000d2.000023 - 2.000000 0.000023第9次迭代x02.000000x12.000000d2.000000 - 2.000000 0.000000x12.000000