How to assign predetermined values for points of discontinuity

Question

Anthony Koning 2021-12-8

1
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1606595-how-to-assign-predetermined-values-for-points-of-discontinuity

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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Answer 1

Matt J 2021-12-9

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https://ww2.mathworks.cn/matlabcentral/answers/1606595-how-to-assign-predetermined-values-for-points-of-discontinuity#answer_851005

编辑：Matt J 2021-12-9

在 MATLAB Online 中打开

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')

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Answer 2

Walter Roberson 2021-12-9

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https://ww2.mathworks.cn/matlabcentral/answers/1606595-how-to-assign-predetermined-values-for-points-of-discontinuity#answer_850910

在 MATLAB Online 中打开

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;

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Answer 3

Steven Lord 2021-12-9

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1606595-how-to-assign-predetermined-values-for-points-of-discontinuity#answer_851025

编辑：Steven Lord 2021-12-9

在 MATLAB Online 中打开

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