Why matlab gives graph of exp(1/x) and exp(-1/x) wrong?

17 ビュー (過去 30 日間)
ali yaman
ali yaman 2021 年 5 月 17 日
コメント済み: ali yaman 2021 年 5 月 17 日
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

採用された回答

Stephen23
Stephen23 2021 年 5 月 17 日
編集済み: Stephen23 2021 年 5 月 17 日
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 件のコメント
Stephen23
Stephen23 2021 年 5 月 17 日
"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.
ali yaman
ali yaman 2021 年 5 月 17 日
Dear @Stephen Cobeldick thanx for all your answer, i got it.
i can't thank you enough!

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeAnnotations についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by