How to assign predetermined values for points of discontinuity

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Anthony Koning
Anthony Koning 2021-12-8
编辑： Steven Lord 2021-12-9
Does anyone know how I would be able to assign a pretermined value for a point of discontinuity? I am currently working on writing a script to determine values at cetrain points of v, and with the exception of v=10, the script is working just fine. at v=10, we are left with the indeterminate of 0/0. Using basic calculus and applying L'Hopital's rule, we can determine that at v=10, the output is .1. However, I don't know how to add this to my script, and am still getting the output of NaN. Any help on how to fix this would be appreciated.
v = 0:5:50
a_n=(0.01.*(10-v))./(exp((10-v)./10)-1)
b_n= exp(-v./80)./8
if v == 10
then a_n = 0.1
end

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采纳的回答

Matt J
Matt J 2021-12-9
编辑：Matt J 2021-12-9
Ran in:
v0 = 10+(-3:0.01:3)*1e-8;
v=v0;
a_n=(0.01.*(10-v))./expm1((10-v)./10);
idx=abs(v-10)<1e-8;
v=v(idx);
a_n(idx)=(0.01)./(0.1+(10-v)./200); %use Taylor approx
plot(v0,a_n,'x')

更多回答（2 个）

Walter Roberson
Walter Roberson 2021-12-9
v = 0:5:50
a_n=(0.01.*(10-v))./(exp((10-v)./10)-1)
b_n= exp(-v./80)./8
a_n(v == 10) = 0.1;
Steven Lord
Steven Lord 2021-12-9
编辑：Steven Lord 2021-12-9
Ran in:
v = 0:5:50
v = 1×11
0 5 10 15 20 25 30 35 40 45 50
a_n=(0.01.*(10-v))./(exp((10-v)./10)-1)
a_n = 1×11
0.0582 0.0771 NaN 0.1271 0.1582 0.1931 0.2313 0.2724 0.3157 0.3609 0.4075
a_n = fillmissing(a_n, 'constant', 0.1) % Fill missing values (NaN) in a_n with a constant 0.1
a_n = 1×11
0.0582 0.0771 0.1000 0.1271 0.1582 0.1931 0.2313 0.2724 0.3157 0.3609 0.4075
b_n= exp(-v./80)./8
b_n = 1×11
0.1250 0.1174 0.1103 0.1036 0.0974 0.0915 0.0859 0.0807 0.0758 0.0712 0.0669

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