taylor 基数的方程画法

semeng

现在在学习matlab里面的一些命令，
遇到了一些奇怪的问题：
ezplot('4/(5-cos(x))',[-6 6 -1 1]) % no problem


问题在下面的操作：
t=taylor('4/(5-cos(x))',10) % expand 10th Taylor  order
ezplot(t) %  It is no problem also
ezplot(t,[-6 6 -1 1]) % It has so many problem

??? Error using ==> figure
Single input must be an existing figure handle or a scalar integer from 1 to 2033136112
Error in ==> ezplot>determineAxes at 534
    figure(fig);
Error in ==> ezplot>ezplot1 at 428
    cax = determineAxes(fig);
Error in ==> ezplot at 148
    [hp,cax] = ezplot1(cax,f{1},vars,labels,args{:});
Error in ==> sym.ezplot at 57
      h = ezplot(f.s,varargin{:});
不懂了，希望能得到帮助

pcqsl

ezplot('4/(5-cos(x))',[-6 6 -1 1]) % no problem

Since '4/(5-cos(x))' is a string(chars), it  is converted to an inline function as like y=4/(5-cos(x)) .

t=taylor('4/(5-cos(x))',10) % expand 10th Taylor  order
ezplot(t) %  It is no problem also

Note that t is no longer a string, it is a symbolic expression(object) now!  

ezplot(t,[-6 6 -1 1]) % It has so many problem

In this form, t is expected to be a two variable implicit function( or symbolic expression), such as f(x,y).  But you feed in a single variable expression.  It then reports error.   Solution:

1. syms x, y
   
2. t=taylor(4/(5-cos(x)),10)-y
   
3. ezplot(t,[-6 6 -1 1])

复制代码

On the other hand, when handling symbolic expressions, ezplot is not as nice as it handles strings.