Why is there a significant difference in the assignment results of functions using "subs" and "feval"

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I need a large amount of data for assignment, and I need to use the handle function to accelerate it, but the result calculated using "feval" is unacceptable
clear,clc
syms z
load FF.mat Th3
z3 = 0.0016:0.00022:0.006;
T1 = vpa(subs(Th3,z,z3));
disp(vpa(T1.',14))
Th3f = matlabFunction(Th3);
T2 = feval(Th3f,z3);
disp(T2.')
3.5028 3.9585 4.4094 4.8132 5.1376 5.3597 5.4692 5.4640 5.3344 5.0730 4.6994 3.9508 2.9207 0.9343 -2.8567 -7.1941 -31.3176 -55.4727 -173.9580 -324.4082 -787.7051
  3 个评论
sen
sen 2023-12-6
According to the formula for solving, the last result needs to be close to 0, while the calculated result for “feval” is -787.7051
sen
sen 2023-12-6
Thank you for your reply. Could you please run the code I submitted? I would like to know how to use "feval" to obtain results that are close to those obtained by "subs"

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回答(1 个)

Walter Roberson
Walter Roberson 2023-12-6
编辑:Walter Roberson 2023-12-6
syms z
load FF.mat Th3
z3 = 0.0016:0.00022:0.006;
T1 = vpa(subs(Th3,z,z3));
disp(vpa(T1.',14))
Th3f = matlabFunction(Th3);
T2 = feval(Th3f,z3);
format long g
disp(T2.')
3.50282686149745 3.95854248851538 4.40936088562012 4.8132483959198 5.13755702972412 5.35973119735718 5.46921586990356 5.46400547027588 5.33437728881836 5.07296371459961 4.69937896728516 3.95075988769531 2.92066955566406 0.934326171875 -2.856689453125 -7.194091796875 -31.317626953125 -55.47265625 -173.9580078125 -324.408203125 -787.705078125
T12 = double(T1(:)) - T2(:);
disp(T12)
7.52585900190006e-05 0.000255996608482079 0.000548036419161235 0.00103877265106078 0.00206573817902189 0.00575961188859786 0.0133017952813956 0.0256612421124434 0.0564997901084956 0.119048502399311 0.199439386434167 0.565610566987848 1.12954361764356 2.57470185401177 5.76486565117093 9.46704282663683 32.957595256356 56.5276612838938 174.523944755596 324.618623611357 787.705078125
plot(z3, T12)
  13 个评论

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