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

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.

Answer 1

[Torsten]

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.

[Leia]

thank you!

[Torsten]

What kind of expansion of exp(z) is R ? Obviously, it's not the usual Taylor expansion.

[Leia]

This polynomial have found as an approximate solution to a differential equation by a spectral numerical method.

[Torsten]

And the differential equation has exp(z) as solution ?