swift算法手记-7

    @IBAction func compute(sender: AnyObject) {
        //  19*x^7-31*x^5+16*x^2+7*x-90=0
        //  newton迭代法求一元方程的解,最大求解范围[-100000,100000]
       
     mytitle.stringValue="19*x^7-31*x^5+16*x^2+7*x-90=0"
        let trycount = 120
        var accuracy: Double = 1e-15
        var answer: Double?=nil
        
        // 估计解范围
        var leftbound:Double?=nil
        var rightbound:Double?=nil
        
        for var bound:Double=1;bound<10000000;bound*=10{
            let leftres=comresult(-bound)
            let rightres=comresult(bound)
            if  (leftres*rightres) < 0 {
                leftbound = (-bound)
                rightbound = bound
                break
            }
            else if leftres==0{
                answer=leftbound
                break
            }
            else if rightres==0{
                answer=rightbound
                break
            }
        }
        if (leftbound==nil || rightbound==nil){
            return
        }
        var center=leftbound!+(rightbound!-leftbound!)/2
        let centres:Double=comresult(center)
        if  centres==0 {
            answer=center
        }
        
        if centres*comresult(rightbound!)<0{
            leftbound=center
        }
        else if centres*comresult(leftbound!)<0{
            rightbound=center
        }
    

        if answer==nil{
            //计算方程的解
            var p0=leftbound!+(rightbound!-leftbound!)/2            
            var p:Double
            for i in 1...trycount{
                
                p = newtoncompresult(p0)
                if abs(p-p0) < accuracy {
                    answer=p0
                    break
                }
                p0=p
            }
        }
        if let ans=answer{
            //方程有解
            result.stringValue="解："+String(stringInterpolationSegment: ans)+"   "
            result.stringValue += "解代入方程的值："+String(stringInterpolationSegment:comresult(ans))
        }     
    }

用牛顿迭代法解非线性方程

本博客所有内容是原创，如果转载请注明来源

http://blog.csdn.net/myhaspl/