Why matlab gives graph of exp(1/x) and exp(-1/x) wrong?
3 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
when trying to find graph of exp(1/x) and exp(-1/x) i am getting graphs that definitely not belong to exp(1/x) and exp(-1/x).
my codes are:
syms x
y=exp(-1/x) % or y=exp(1/x)
fplot(y)
when i run this, i get the graphs that as in below, figure 1 is belong to exp(-1/x) and figure 2 is belong to exp(1/x)
figure 1=exp(-1/x)
figure 2=exp(1/x)
But, this is absolutely ridiculous. The graphs of exp(1/x) and exp(-1/x) must be like :
Is there anybody that will help me? I do not know where i am wrong, please if you know help me
thanks
0 commentaires
Réponse acceptée
Stephen23
le 17 Mai 2021
Modifié(e) : Stephen23
le 17 Mai 2021
fun = @(x) exp(-1./x);
fplot(fun)
ylim([0,10])
fun = @(x) exp(1./x);
fplot(fun)
ylim([0,10])
I doubt there is much that can be done to change the automatic discontinuity detection algorithm, but feel free to read the documentation and see.
5 commentaires
Stephen23
le 17 Mai 2021
"Now, how can we decide the limitations, to get a correct shape of curve?"
The shape is always "correct", as explained above. The only difference is how humans like to focus on particular parts or features of the curve: the computer does not know which parts you are interested in, only you know that. Possibly someone has published a heuristic algorithm to predict Y-limits for arbitrary functions.
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Line Plots dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!