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Y1 eval - PowerPoint PPT Presentation

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????. ??? Seidel ?????????. f(3)= -0.9107. f (3)= 0.2440. ??????????. function z=newton(f,x0,y0,e) x = sym('x'); y=sym('y') Jx=det([diff(f,y,1),f]); Jy=det([f,diff(f,x,1)]); J=det([diff(f,x,1),diff(f,y,1)]); x=x0; y=y0; n=1 x1=x0+ eval (Jx)/ eval (J); y1=y0+ eval (Jy)/ eval (J);

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Y1 eval - PowerPoint PPT Presentation

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