求解思路:
中缀表达式转成后缀表达式,根据后缀表示式生成二叉表达式数,在根据上树生成导函数树。牛顿迭代法求解一元多次方程
牛顿迭代法:
http://www.devacg.com/?post=351
一元非线性方程的求解是高等数学研究的重要课题之一,早在2000多年前,古巴比伦的数学家就能解一元二次方程了,中国的《九章算术》也有对一元二次方程求解的记载。目前人们普遍认为低阶(5阶以下)元非线性方程可以通过求根公式求解,但是高于或等于5阶的一元非线性方程不存在求根公式,要精确求解非常困难。对高阶方程,一般采用迭代法近似求解,牛顿迭代法因为方法简单,迭代收敛速度快而被广泛使用。
牛顿迭代法(Newton’s method)又称牛顿-拉弗森方法(Newton-Raphson method),它是一种在实数域和复数域上近似求解方程的方法。方法使用函数(x)的泰勒级数的前面几项来寻找方程(x)-0的根。
牛顿迭代法逼近示意图:
首先,选择一个接近函数(x)零点的x0作为迭代初始值,计算相应的(x0)和切线斜率(xO)(这里(x)是函数(x)的一阶导函数)。然后我们经过点(xO,f(x))做一条斜率为f(x0)的直线,该直线与x轴有一个交点,可通过以下方程的求解得到这个交点的x坐标:

求解这个方程,可以得到:

我们将新求得的点的x坐标命名为x1,通常x1会比x0更接近方程(X)-0的解。因此我们现在可以利用x1开始下一轮迭代。根据上述方程中x1和x0的关系,可以得到一个求解x的迭代公式:

这就是牛顿迭代公式。目前已经证明,如果(x)的一阶导函数(x)是连续函数,并且待求的零点x是孤立的,则在零点x周围存在一个区间,只要初始值x0位于这个区间,牛顿迭代必定收敛。并且,只要(x)0,牛顿送代法具有平方收敛的性能。这意味着每迭代一次,结果的有效数字将增加一倍,这比二分逼近法的线性收敛速度快了一个数量级。
导函数的求解与近似公式
using System;namespace NewtonTest{class Program{public delegate double Function(double x);const double PRECISION = 0.1;//计算精度static void Main(string[] args){double x = NewtonRaphson(TestFun, -8);Console.WriteLine("f({0})={1}", x, TestFun(x));Console.Read();}//示例方程:一元二次方程//求解y=0时,x的值.public static double TestFun(double x){//注意:要确保这个方程是有解的。double y = 5 * Math.Pow(x, 2) + 8 * x + 2;return y;}/// <summary>/// 计算函数f在x附近的一阶导数值/// </summary>/// <param name="f">原函数</param>/// <param name="x">x值</param>/// <returns></returns>public static double CalcDerivative(Function f, double x){return (f(x + 0.000005) - f(x - 0.000005))/0.00001;}/// <summary>/// 牛顿迭代法/// 用于求解一元非线性方程/// </summary>/// <param name="f">原函数</param>/// <param name="x0">迭代初始值</param>/// <returns></returns>public static double NewtonRaphson(Function f, double x0){double x1 = x0 - f(x0) / CalcDerivative(f, x0);while(Math.Abs(x1 - x0) > PRECISION){x0 = x1;x1 = x0 - f(x0) / CalcDerivative(f, x0);}return x1;}}}
中缀表达式转后缀表达式—C#代码实现
https://blog.csdn.net/Czhenya/article/details/78067542
版主没用递归来实现,有些繁琐
namespace 玩嗨的练习{class Program{static void Main(string[] args){//运行时用建议输入空格Console.WriteLine("请输入您要转换的表达式:");string inputstr = Console.ReadLine();//测试用//string inputstr = "9 * (3 - 1) + (10 - 1) / 2";//string inputstr = "1 - 2 - 3 * 4 + 10 / 5";Console.WriteLine("您转换后的结果为:");Change(inputstr);Console.ReadKey();}private static void Change(string inputstr){Stack<char> arrStark = new Stack<char>();char[] arrChar = inputstr.ToCharArray();int a = 4; //默认状态是什么都没有,这里是4状态,,for (int i = 0; i < arrChar.Length; i++){//数字空格就直接输出if (arrChar[i] <= '9' && arrChar[i] >= '0'){Console.Write(arrChar[i]);}if (arrChar[i] == ' '){Console.Write(" ");}//运算符走状态机switch (a){//=============================状态1 * / ====================================case 1: //表示当前是+- 状态if (arrChar[i] == '+' || arrChar[i] == '-'){a = 1;Console.Write(arrStark.Pop());arrStark.Push(arrChar[i]);}elseif (arrChar[i] == '*' || arrChar[i] == '/'){a = 2;arrStark.Push(arrChar[i]);}elseif (arrChar[i] == '('){a = 3;arrStark.Push(arrChar[i]);}elseif (arrChar[i] == ')'){a = 4;for (int j = 0; j < arrStark.Count; j++){//输出括号中间的符号,,,if (arrStark.Peek() == '('){//把左括号,,弹出栈外arrStark.Pop();break;}else{Console.Write(arrStark.Pop());}}}break;//=============================状态2 * / ====================================case 2: //表示当前是* / 状态if (arrChar[i] == '+' || arrChar[i] == '-'){a = 1;for (int j = 0; j <= arrStark.Count; j++){//Console.Write(arrStark.Count);Console.Write(arrStark.Pop());}arrStark.Push(arrChar[i]);}elseif (arrChar[i] == '*' || arrChar[i] == '/'){a = 2;Console.Write(arrStark.Pop());arrStark.Push(arrChar[i]);}elseif (arrChar[i] == '('){a = 3;arrStark.Push(arrChar[i]);}elseif (arrChar[i] == ')'){a = 4;for (int j = 0; j < arrStark.Count; j++){if (arrStark.Peek() == '('){//把左括号,,弹出栈外arrStark.Pop();break;}else{//输出括号中间的符号,,,Console.Write(arrStark.Pop());}}}break;//=============================状态3 ) ====================================case 3: //表示当前是( 状态if (arrChar[i] == '+' || arrChar[i] == '-'){arrStark.Push(arrChar[i]);}elseif (arrChar[i] == '*' || arrChar[i] == '/'){arrStark.Push(arrChar[i]);}elseif (arrChar[i] == '('){a = 3;arrStark.Push(arrChar[i]);}elseif (arrChar[i] == ')'){a = 4;for (int j = 0; j < arrStark.Count; j++){if (arrStark.Peek() == '('){//把左括号,,弹出栈外arrStark.Pop();break;}else{//输出括号中间的符号,,,Console.Write(arrStark.Pop());}}}break;//=============================状态4 ( ====================================case 4: //表示当前是 ) 状态if (arrChar[i] == '+' || arrChar[i] == '-'){if (arrStark.Count == 0){a = 1;arrStark.Push(arrChar[i]);}else{if (arrStark.Peek() == '+' || arrStark.Peek() == '-'){a = 1;arrStark.Push(arrChar[i]);}elseif (arrStark.Peek() == '*' || arrStark.Peek() == '/'){a = 1;for (int j = 0; j < arrStark.Count; j++){Console.Write(arrStark.Pop());}arrStark.Push(arrChar[i]);}}}elseif (arrChar[i] == '*' || arrChar[i] == '/'){if (arrStark.Count == 0){a = 2;arrStark.Push(arrChar[i]);}else{if (arrStark.Peek() == '+' || arrStark.Peek() == '-'){a = 2;arrStark.Push(arrChar[i]);}else if (arrStark.Peek() == '*' || arrStark.Peek() == '/'){a = 2;arrStark.Push(arrChar[i]);}}}elseif (arrChar[i] == '('){a = 3;arrStark.Push(arrChar[i]);}if (arrChar[i] == ')'){a = 4;for (int j = 0; j < arrStark.Count; j++){if (arrStark.Peek() == '('){//把左括号,,弹出栈外arrStark.Pop();break;}else{//输出括号中间的符号,,,Console.Write(arrStark.Pop());}}}break;default:Console.WriteLine("Wrong");break;}}//遍历栈中剩余的符号输出,并清空栈,foreach (char item in arrStark){Console.Write(" " + item);}arrStark.Clear();}}}————————————————版权声明:本文为CSDN博主「Czhenya」的原创文章,遵循CC 4.0 BY-SA版权协议,转载请附上原文出处链接及本声明。原文链接:https://blog.csdn.net/Czhenya/article/details/78067542
