i'd like to ask is there any function to find asymptotes of an equation y = f(x) that satisfies x = x(t) and y = y(t). I'm new to matlab.

Question

Nhat Huy 2023-11-25

1
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2051882-i-d-like-to-ask-is-there-any-function-to-find-asymptotes-of-an-equation-y-f-x-that-satisfies-x

评论： John D'Errico 2023-12-1

在 MATLAB Online 中打开

syms t

x = (t+1)./(t-1);

y = (t^2+2)./(t^2-t);

% Find horizontal asymptotes

syms x0

tx = solve(x == x0, t);

num_ha = 0;

sh = [];

for id = 1:length(tx)

yx = subs(y, t, tx(id));

for s = 0:1

infi = (-1)^s*inf;

ya = limit(yx, x0, infi);

if isfinite(ya)

fprintf('Tiem can ngang la: y = %f\n', ya)

num_ha = num_ha + 1;

end

fprintf('Do thi ham f(x) co %d tiem can ngang\n------------------\n', num_ha)

Do thi ham f(x) co 0 tiem can ngang ------------------

% Find slant asymptotes

num_sa = 0;

ss = [];

for id = 1:length(tx)

yx = subs(y, t, tx(id));

for s = 0:1

infi = (-1)^s*inf;

a = limit(yx/x0, x0, infi);

b = limit(yx - a*x0, x0, infi);

c = [a b];

if isfinite(a) && isfinite(b)

new = 1;

for k = 1:size(ss, 1)

if all(c == ss(k, :))

new = 0;

break

end

if new

ss = [ss; c];

fprintf('Tiem can xien la: y = x * %f + %f\n', a, b)

num_sa = num_sa + 1;

end

Tiem can xien la: y = x * 1.500000 + -2.500000

fprintf('Do thi ham f(x) co %d tiem can xien\n-------------------\n', num_sa)

Do thi ham f(x) co 1 tiem can xien -------------------

% Find vertical asymptotes

syms y0

ty = solve(y == y0, t);

num_va = 0;

sv = [];

for s = 0:1

infi = (-1)^s * inf;

ti = subs(ty, y0, infi);

for id = 1:length(ti)

t0 = ti(id);

if ~isreal(t0)

continue

end

xa = limit(x, t, t0);

if isfinite(xa)

new = 1;

for k = 1:size(sv, 1)

if xa == sv(k)

new = 0;

break

end

if new

sv = [sv xa];

fprintf('Tiem can dung la: x = %f\n', xa)

num_va = num_va + 1;

end

fprintf('Do thi ham f(x) co %d tiem can dung\n-------------------\n', num_va)

Do thi ham f(x) co 0 tiem can dung -------------------

figure

hold on

fplot(x, y)

xlabel('Ox')

ylabel('Oy')

grid on

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John D'Errico 2023-12-1

编辑：John D'Errico 2023-12-1

Please don't keep on reposting your question. I closed it as a pure duplicate.

As I said in my answer, your question is not even a question about MATLAB. Until you know how to approach a problem, you cannot write code to solve it. There is no tool in MATLAB to solve your problem.

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Answer 1

John D'Errico 2023-11-25

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2051882-i-d-like-to-ask-is-there-any-function-to-find-asymptotes-of-an-equation-y-f-x-that-satisfies-x#answer_1359562

编辑：John D'Errico 2023-11-25

Its an interesting question of mathematics, maybe not really a question about MATLAB in my eyes, because until you know how to solve a problem using mathematics, you cannot hope to write code.

Is there any function? No. Sorry, but I don't see any way to do so, since x and y can be literally anything. And since they could be arbitraily complicated, messy functional forms, it might even be possible to find some pair of functions where we cannot prove that an asymptote exists or not.

You have made an interesting attempt at a solution, where you look for points where x or y approach infinity, but I could counter with examples where x or y go to infinity at infinitely many points.

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Answer 2

Walter Roberson 2023-12-1

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2051882-i-d-like-to-ask-is-there-any-function-to-find-asymptotes-of-an-equation-y-f-x-that-satisfies-x#answer_1363464

在 MATLAB Online 中打开

syms t;

x_t = (t+1)./(t-1);

y_t = (t^2 + 2)./(t^2 - t);

limit(y_t, t, -inf)

ans =

1

limit(y_t, t, 0, 'left')

ans =

∞

limit(y_t, t, 0, 'right')

ans =

limit(y_t, t, 1, 'left')

ans =

limit(y_t, t, 1, 'right')

ans =

∞

limit(y_t, t, inf)

ans =

1

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John D'Errico 2023-12-1

But this does not answer the question of how to identify a possibly linear asymptote. Or what to do if you cannot easily identify poles of the functions for a completely general problem.

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