When plotting functions like 1/sin(x), how can I remove the vertical lines at the points of discontinuity?

Hi there! When I plot functions such as 1/sin(x), there are vertical lines that appear at the points of discontinuity. How can I best remove these vertical lines, if I wanted to see the graph for a bigger domain, say, from -2pi to 2pi? Thanks in advance.

Answer 1

[Sam Chak]

Use fplot().

fplot(@(x) 1./sind(x), [-359 359]), grid on, xlabel('degree')

% d   = -345:30:345;
% x   = deg2rad(d);
% y   = 1./sin(x);
% plot(x, y)

[Walter Roberson]

Or use symbolic fplot with the option ShowPoles off

[Sam Chak]

Hi @Walter Roberson

I believe that the display of asymptotes at poles is acceptable because these gray dashed lines merely indicate that the curve approaches infinity and are not part of the mathematical function itself. The original poster likely intended to remove the continuity of the graph (vertical lines) at the poles when using the plot() function; however, this does not make sense at the singularities.

Anyhow, thanks for the option to hide the asymptotes.

fplot(@(x) 1./sind(x), [-359 359], 'ShowPoles', 'off'), grid on, xlabel('degree')

[Noob]

Hi Sam!

What if I weren't doing symbolic math, and am instead just using plot( )?

I am currently reading Mathworks documentation to understand fplot( ).

Thanks!

[Walter Roberson]

@Noob

There are two versions of fplot() . One of them https://www.mathworks.com/help/matlab/ref/fplot.html works with function handles; the other https://www.mathworks.com/help/symbolic/fplot.html works with symbolic expressions or symbolic functions.

[Noob]

Hi @Walter Roberson! Do you feel that one should truncate the intervals of interest and only plot the function where it is known to be continous (putting aside the issue of smoothness and differentibability, for the moment)? In my work, I am sort of aiming to demonstrate a function's behavior for an extended domain (say, for larger angles than previously considered). Perhaps removing the discontinuitity is unnecessary, and maybe even wrong to do, and that one should aim to remove the vertical lines that do not actually represent function values: They are merely extraneous lines from Matlab connecting function values, I think. What do you think? Thanks!

Answer 2

[Manish]

Hi,

I understand that you want to remove the vertical lines that appear at the points of discontinutity for the function y=1/sin(x).

This can be achieved setting 'y' values to NAN where '1/sin(x)' is close to zero.Use a small threshold to identify the values that are very close to zero and adjust the threshold accordingly.

This approach results in skipping the vertical lines.

Refer the code sample below for the better understanding:

x = linspace(-2*pi, 2*pi, 1000);
y = 1 ./ sin(x);
threshold=1e-6;% small threshold
% Identify points where sin(x) is zero and set corresponding y values to NaN
y(abs(sin(x)) < threshold) = NaN;  
figure;
plot(x, y, 'b-', 'LineWidth', 1.5);
ylim([-5, 5]);
xlabel('x');
ylabel('1/sin(x)');
title('Plot of 1/sin(x) without Vertical Lines');
grid on;

Hope this helps!

[Sam Chak]

But this approach does not "remove" the vertical lines.

[Manish]

Adjust the threshold to 1e-2.

x = linspace(-2*pi, 2*pi, 1000);
y = 1 ./ sin(x);
threshold=1e-2;
% Identify points where sin(x) is zero and set corresponding y values to NaN
y(abs(sin(x)) < threshold) = NaN;  % Use a small threshold to account for numerical precision
figure;
plot(x, y, 'b-', 'LineWidth', 1.5);
ylim([-5, 5]);
xlabel('x');
ylabel('1/sin(x)');
title('Plot of 1/sin(x) without Vertical Lines');
grid on;

●

[Noob]

Hi Manish!

Interesting solution, but I'm afraid the threshold is not good enough; I would imagine a much smaller number needs to be used, such as 1e-15, but then the spikes to infinity and -infinity would still appear.

[Torsten]

I would imagine a much smaller number needs to be used, such as 1e-15

so you want the value of the vertical axis to run up to

1/sin(1e-15)

ans = 1.0000e+15

?

Answer 3

[Voss]

x = linspace(-2*pi,2*pi,1000);
y = 1./sin(x);

figure
% put NaNs where y changes sign:
y_plot = y;
y_plot([false diff(sign(y))~=0]) = NaN;
plot(x,y_plot)
ylim([-5 5])
grid on

[Noob]

Hi Voss!

What does this line of code mean:

y_plot([false diff(sign(y))~=0]) = NaN

I understand diff represents symoblic differentiation.

I undertand sign( ) checks the sign of a number.

I understand NaN = Not a number.

I understand y_plot( ... ) = NaN means setting function values for y_plot, if certain conditions are met.

But, what does false and ~= mean?

And, do you think I really need to use symbolic math?

Can I do it without symbolic math?

Thanks!

[Walter Roberson]

I understand diff represents symoblic differentiation.

Not in this case. There are two major diff() functions. When diff() is applied to a sym or symfun or symmatrix then diff() means symbolic differentiation.

However, when diff() is applied to numeric values, diff(A) means roughly (A(2:end)-A(1:end-1)) -- the consecutive differences function. (There is some nuance to this having to do with row vectors compared to column vectors compared to arrays, and there are options having to do with multiple differences at the same time.)

In this particular context, diff() is "consecutive differences". sign(y) returns -1 where y is negative, 0 where y is 0, and +1 where y is positive. diff()~=0 applied to sign() is looking for places where the sign() is changing

false is a function that called by itself returns a scalar logical false. It is equivalent to logical(0) in this context.

~= is "not equal" .