I want to plot region for z numbers that satisfy D<10^-4.

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How do I plot the region showing z=x+iy values that make the difference between a polynomial and an exponential function less than 10^-4?
For example;
syms z
[x,y]=meshgrid(-6:.1:2,-6:.1:6);
z=x+1i*y;
R = - 0.0000000000000042786173933199698996786792332228*z.^16 + 0.000000000000093679201874795131075154866430805*z.^15 ...
- 0.0000000000018102462775238750488561630646573*z.^14 + 0.000000000070607888835529581094992975687635*z.^13 ...
- 0.00000000047430380749604348847250198255397*z.^12 + 0.0000000060302178555555310479780219749732*z.^11 ...
- 0.00000014850754796722791834393401507891*z.^10 - 0.0000011975088237244406931082370289037*z.^9 ...
- 0.00000486020688657407415163842024264*z.^8 + 0.000027465820312500018146735337769556*z.^7 ...
+ 0.0010023328993055556912781216101668*z.^6 + 0.0080891927083333337897961540664676*z.^5 ...
+ 0.041666666666666667576168711195926*z.^4 + 0.16666666666666666756898545265065*z.^3 + 0.5*z.^2 + 1.0*z + 1.0;
T = exp(z);
D = R-T;
I want to plot region for z numbers that satisfy D<10^-4.
How can I do that? I would be appreciate if you help.

采纳的回答

Torsten
Torsten 2022-6-8
编辑:Torsten 2022-6-8
R1 = @(x,y)- 0.0000000000000042786173933199698996786792332228*(x+1i*y).^16 + 0.000000000000093679201874795131075154866430805*(x+1i*y).^15 ...
- 0.0000000000018102462775238750488561630646573*(x+1i*y).^14 + 0.000000000070607888835529581094992975687635*(x+1i*y).^13 ...
- 0.00000000047430380749604348847250198255397*(x+1i*y).^12 + 0.0000000060302178555555310479780219749732*(x+1i*y).^11 ...
- 0.00000014850754796722791834393401507891*(x+1i*y).^10 - 0.0000011975088237244406931082370289037*(x+1i*y).^9 ...
- 0.00000486020688657407415163842024264*(x+1i*y).^8 + 0.000027465820312500018146735337769556*(x+1i*y).^7 ...
+ 0.0010023328993055556912781216101668*(x+1i*y).^6 + 0.0080891927083333337897961540664676*(x+1i*y).^5 ...
+ 0.041666666666666667576168711195926*(x+1i*y).^4 + 0.16666666666666666756898545265065*(x+1i*y).^3 + 0.5*(x+1i*y).^2 + 1.0*(x+1i*y) + 1.0;
R2 = @(x,y) sum((x+1i*y).^(0:16)./factorial(0:16));
T = @(x,y) exp(x+1i*y);
fun1 = @(x,y)norm(R1(x,y)-T(x,y))-1e-4;
fun2 = @(x,y)norm(R2(x,y)-T(x,y))-1e-4;
fimplicit(fun1)
Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.
hold on
fimplicit(fun2)
Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.
  4 个评论
Leia
Leia 2022-6-10
This polynomial have found as an approximate solution to a differential equation by a spectral numerical method.

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