| 查看: 186 | 回复: 2 | |||
| 当前主题已经存档。 | |||
| 【有奖交流】积极回复本帖子，参与交流，就有机会分得作者 dxyhn1979 的 2 个金币 | |||
[交流]
【求助】非线性方程的求多根的数值方法？
|
|||
|
请问各位大虾： 牛顿法，牛顿弦截法能求非线性方程的多根么？求非线性方程的多根应用什么数值方法？ |
回复此楼» 猜你喜欢
导师想让我从独立一作变成了共一第一
已经有9人回复
-
博士读完未来一定会好吗
已经有23人回复
-
到新单位后，换了新的研究方向，没有团队，持续积累2区以上论文，能申请到面上吗
已经有11人回复
-
读博
已经有4人回复
-
JMPT 期刊投稿流程
已经有4人回复
-
心脉受损
已经有5人回复
-
Springer期刊投稿求助
已经有4人回复
-
小论文投稿
已经有3人回复
-
申请2026年博士
已经有6人回复
hitzhang
木虫 (正式写手)
- 仿真EPI: 1
- 应助: 0 (幼儿园)
- 贵宾: 2.15
- 金币: 1376.7
- 散金: 969
- 红花: 8
- 帖子: 862
- 在线: 226.4小时
- 虫号: 390575
- 注册: 2007-06-02
- 性别: GG
- 专业: 无机非金属类电介质与电解
★
dxyhn1979(金币+4):先谢了哥们，先给几个币币，我试试 2010-03-17 14:01
adu886886(金币+1):谢谢提供意见 2010-03-17 14:27
dxyhn1979(金币+4):先谢了哥们，先给几个币币，我试试 2010-03-17 14:01
adu886886(金币+1):谢谢提供意见 2010-03-17 14:27
|
给你个函数 function P = InterX(L1,varargin) %INTERX Intersection of curves % P = INTERX(L1,L2) returns the intersection points of two curves L1 % and L2. The curves L1,L2 can be either closed or open and are described % by two-row-matrices, where each row contains its x- and y- coordinates. % The intersection of groups of curves (e.g. contour lines, multiply % connected regions etc) can also be computed by separating them with a % column of NaNs as for example % % L = [x11 x12 x13 ... NaN x21 x22 x23 ...; % y11 y12 y13 ... NaN y21 y22 y23 ...] % % P has the same structure as L1 and L2, and its rows correspond to the % x- and y- coordinates of the intersection points of L1 and L2. If no % intersections are found, the returned P is empty. % % P = INTERX(L1) returns the self-intersection points of L1. To keep % the code simple, the points at which the curve is tangent to itself are % not included. P = INTERX(L1,L1) returns all the points of the curve % together with any self-intersection points. % % Example: % t = linspace(0,2*pi); % r1 = sin(4*t)+2; x1 = r1.*cos(t); y1 = r1.*sin(t); % r2 = sin(8*t)+2; x2 = r2.*cos(t); y2 = r2.*sin(t); % P = InterX([x1;y1],[x2;y2]); % plot(x1,y1,x2,y2,P(1,  ,P(2,,'ro')% Author : NS % Version: 2.0, 26/11/09 % Two words about the algorithm: Most of the code is self-explanatory. % The only trick lies in the calculation of C1 and C2. To be brief, this % is essentially the two-dimensional analog of the condition that needs % to be satisfied by a function F(x) that has a zero in the interval % [a,b], namely % F(a)*F(b) <= 0 % C1 and C2 exactly do this for each segment of curves 1 and 2 % respectively. If this condition is satisfied simultaneously for two % segments then we know that they will cross at some point. % Each factor of the 'C' arrays is essentially a matrix containing % the numerators of the signed distances between points of one curve % and line segments of the other. %...Argument checks and assignment of L2 error(nargchk(1,2,nargin)); if nargin == 1, L2 = L1; hF = @lt; %...Avoid the inclusion of common points else L2 = varargin{1}; hF = @le; end %...Preliminary stuff x1 = L1(1, '; x2 = L2(1,;y1 = L1(2, '; y2 = L2(2,;dx1 = diff(x1); dy1 = diff(y1); dx2 = diff(x2); dy2 = diff(y2); %...Determine 'signed distances' S1 = dx1.*y1(1:end-1) - dy1.*x1(1:end-1); S2 = dx2.*y2(1:end-1) - dy2.*x2(1:end-1); C1 = feval(hF,D(bsxfun(@times,dx1,y2)-bsxfun(@times,dy1,x2),S1),0); C2 = feval(hF,D((bsxfun(@times,y1,dx2)-bsxfun(@times,x1,dy2))',S2'),0)'; %...Obtain the points where an intersection is expected [i,j] = find(C1 & C2); dx2=dx2'; dy2=dy2'; L = dy2(j).*dx1(i) - dy1(i).*dx2(j); i = i(L~=0); j=j(L~=0); L=L(L~=0); %...Avoid divisions by 0 %...Solve system of eqs to get the common points P = unique([dx2(j).*S1(i) - dx1(i).*S2(j)', ... dy2(j).*S1(i) - dy1(i).*S2(j)']./[L L],'rows')'; function u = D(x,y) u = bsxfun(@minus,x(:,1:end-1),y).*bsxfun(@minus,x(:,2:end),y); end end |
2楼2010-03-17 11:41:18
3楼2010-03-17 14:31:04