Creating a polynomial fit expression using just the order number

Question

Jason 2025-11-18，21:26

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2181354-creating-a-polynomial-fit-expression-using-just-the-order-number

编辑： dpb 2025-11-20，16:37

Hello. Im performing a fit to data using e.g a 3rd order polynomial and the expresion below. For cases when i want e.g a 4th or 5th order fit, rather than use a switch / case approach is there a way to construct the expression below simply by passing in n the polynomial order?

a123 = [x.^3, x.^2, x]\y;

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

dpb 2025-11-18，21:38

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2181354-creating-a-polynomial-fit-expression-using-just-the-order-number#answer_1572087

编辑：dpb 2025-11-18，21:57

在 MATLAB Online 中打开

c=x.^[n:-1:1]\y;

I presume leaving off the intercept is intentional? Otherwise, there's polyfit

5 个评论
显示 3更早的评论隐藏 3更早的评论

Jason 2025-11-19，19:31

编辑：Jason 2025-11-19，19:59

在 MATLAB Online 中打开

I need to be able to "force fit" so it passes through a certain point

   ax=app.UIAxes3;
   [x,y] = getDataFromGraph(app,ax,1);   % My function to extract x,y data
   hold(ax,"on");
   D=[x,y];
   D=rmmissing(D);        % Remove rows with any NaNs or missing data
   x=D(:,1);   y=D(:,2); 
   %  Force polyfit thru first point
   x0=x(1,1);
   y0=y(1,1);
   ReportMessage(app,['Forced Fit Thru Point: (',num2str(x0),', ',num2str(y0),')'])
   % assume model of the form y=a1*x^3+a2*x^2+a3*x
   % we can fit simply as P=[a123;0]  
   % but now if we want to force the model it so passes thru x0,y0
   % essentially have to fit the model  (just transfor like this)
   % y-y0=a1*(x-xo)^3+a2(x-x0)^2+a3(x-x0)
   % so when x==x0, y==y0
   xtr=x-x0;
   % acoeffs=[xtr.^5,xtr.^4,xtr.^3,xtr.^2,xtr]\(y-y0)     %acoeffs=[xtr.^7,xtr.^6,xtr.^5,xtr.^4,xtr.^3,xtr.^2,xtr]\(y-y0)
   acoeffs=[xtr.^7,xtr.^6,xtr.^5,xtr.^4,xtr.^3,xtr.^2,xtr]\(y-y0);
   %Evaluation of the model is easy, you can still use polyval
   % just rememeber to subtract x0 from x first
   P=[acoeffs;y0];
   ypred=polyval(P,x-x0);
   plot(ax,x,ypred,'y--','LineWidth',2);

dpb 2025-11-19，21:22

With a 7th order polynomial, are you forcing it through a set of points, maybe? Would a spline be an alternative?

Torsten 2025-11-19，23:53

编辑：dpb 2025-11-20，16:37

在 MATLAB Online 中打开

@Jason

xtr=x-x0;
% acoeffs=[xtr.^5,xtr.^4,xtr.^3,xtr.^2,xtr]\(y-y0)     %acoeffs=[xtr.^7,xtr.^6,xtr.^5,xtr.^4,xtr.^3,xtr.^2,xtr]\(y-y0)
acoeffs=[xtr.^7,xtr.^6,xtr.^5,xtr.^4,xtr.^3,xtr.^2,xtr]\(y-y0);

will give you a polynomial that passes through (x0,y0), but will have a constant term - thus will no longer be of the form you used earlier.

Thus the property of passing through (x0,y0) is payed by losing the property of passing through (0,0).

请先登录，再进行评论。

Creating a polynomial fit expression using just the order number

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

5 个评论
显示 3更早的评论隐藏 3更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

Creating a polynomial fit expression using just the order number

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

5 个评论 显示 3更早的评论隐藏 3更早的评论

更多回答（0 个）

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

5 个评论
显示 3更早的评论隐藏 3更早的评论