How to specify intervals of increasing, decreasing of a function in MATLAB

Hello, I want to specify intervals of increasing & decreasing of a function for a specific range; say -20-20. This is my code here:

clc, close, clear
syms x y real

dy=1-(4/x^2); x=~0; % This is the derivative of the function
% dy=(sin(x)-1)*(2*cos(x)+1); % (0<=x<=2*pi) % Code doesn't work for this.

fplot(dy); hold on; axis([-5, 5, -5, 5]);
j=xline(0); set(j,'color','black','Linewidth', 1.5)
k=yline(0); set(k,'color','black','Linewidth', 1.5)
cp=solve(dy);       % Finding critical points
y_cp=subs(dy,cp);
[~,idx] = unique(cp);
cp = cp(idx);
y_cp = y_cp(idx);
[num, den]=numden(dy);   % Not a conventional way to find other critical points
% I need to account for ties here. The technique I used above for 1st set
% of critical points (idx, unique) isn't working.
cp2=solve(den==0);
disp('The critical points are:')

The critical points are:

for k =1:length(cp)
    fprintf("(%.2f,%.2f)\n",cp(k), y_cp(k))
    plot(cp,y_cp,'*', 'color', 'black');
end

(-2.00,0.00)
(2.00,0.00)

for k =1:length(cp2)
    fprintf("x = %.2f is a critical point, y is undefined\n",cp2(k))
end

x = 0.00 is a critical point, y is undefined
x = 0.00 is a critical point, y is undefined

% y is increasing if dy > 0, decreasing if dy < 0, if dy=0 it's a critical point. Code below doesn't work.
for i = -20:20 
    dy_sub_i=subs(dy,i);
end

Error using sym/subs (line 156)
Division by zero.

if dy_sub_i>0
    fprintf('f is increasing in %s-%s', i(0), i);
elseif dy_sub_i<0
    fprintf('f is decreasing in ');
end

[Image Analyst]

What would "specify" mean to you? Shading plot regions? A vector of where the data is less than the point before, which you can get with diff?

[Walter Roberson]

syms x y real

dy=1-(4/x^2); %x=~0; % This is the derivative of the function
% dy=(sin(x)-1)*(2*cos(x)+1); % (0<=x<=2*pi) % Code doesn't work for this.

fplot(dy); hold on; axis([-5, 5, -5, 5]);
j=xline(0); set(j,'color','black','Linewidth', 1.5)
k=yline(0); set(k,'color','black','Linewidth', 1.5)
cp=solve(dy);       % Finding critical points
y_cp=subs(dy,cp);
[~,idx] = unique(cp);
cp = cp(idx);
y_cp = y_cp(idx);
[num, den]=numden(dy);   % Not a conventional way to find other critical points
cp2=solve(den==0);
disp('The critical points are:')

The critical points are:

for k =1:length(cp)
    fprintf("(%.2f,%.2f)\n",cp(k), y_cp(k))
    plot(cp,y_cp,'*', 'color', 'black');
end

(-2.00,0.00)
(2.00,0.00)

●

for k =1:length(cp2)
    fprintf("x = %.2f is a critical point, y is undefined\n",cp2(k))
end

x = 0.00 is a critical point, y is undefined
x = 0.00 is a critical point, y is undefined

% y is increasing if dy > 0, decreasing if dy < 0, if dy=0 it's a critical point. Code below doesn't work.
for i = -20:20 
    dy_sub_i = limit(dy,x,i);
    if dy_sub_i>0
        fprintf('f is increasing in %g\n', i);
    elseif dy_sub_i<0
        fprintf('f is decreasing in %g\n', i);
    end
end

f is increasing in -20
f is increasing in -19
f is increasing in -18
f is increasing in -17
f is increasing in -16
f is increasing in -15
f is increasing in -14
f is increasing in -13
f is increasing in -12
f is increasing in -11
f is increasing in -10
f is increasing in -9
f is increasing in -8
f is increasing in -7
f is increasing in -6
f is increasing in -5
f is increasing in -4
f is increasing in -3

f is decreasing in -1
f is decreasing in 0
f is decreasing in 1

f is increasing in 3
f is increasing in 4
f is increasing in 5
f is increasing in 6
f is increasing in 7
f is increasing in 8
f is increasing in 9
f is increasing in 10
f is increasing in 11
f is increasing in 12
f is increasing in 13
f is increasing in 14
f is increasing in 15
f is increasing in 16
f is increasing in 17
f is increasing in 18
f is increasing in 19
f is increasing in 20

[rezheen]

@Image Analyst: By specify I mean to display it using disp or fprintf to the user what those intervals are based on the critical points I have found for a specific range say -20:20. Ideally with a for loop. Sorry I vwasn't clear.

[Walter Roberson]

It is a function. It is sort of increasing for positive x, if you look only at the irrational numbers, but it is everywhere discontinuous, and everywhere decreasing, when you transition from an irrational number to a rational one.

It is not obvious to me that the function is everywhere discontinuous.

If you take any particular irrational input, x, and add to it a rational then the resulting will be irrational, so and are both non-zero (x = 0 is not considered for this purpose because 0 is rational.)

For any rational there exists a smaller rational namely with N->infinity . And would be non-zero for each of those.

The question then becomes, whether for each there exists a such that is rational. Ummm... maybe ? It is not at all obvious to me that such exists.

It is usually said that there are an infinite number of irrational numbers between any two rational numbers. That would surely imply that there are an infinite number of irrational numbers beside each irrational number. So by the "closed compact set" arguments, f(x) is continuous at each irrational x.

[John D'Errico]

@Walter Roberson

Sorry. Your logic fails. All we need is the definition of a continuous function.

https://en.wikipedia.org/wiki/Continuous_function

The definition of a continuous function requires three things.

A function f is continuous at a point c if:

1. f(c) is defined.
2. lim f(x) as x-->c exists
3. lim f(x) as x-->c = f(c)

Certainly f(x) is well defined at all points on the real line.

At all rational points (except zero):

1. For any rational point c, f(c)=0
2. Consider the limit of f(x) as x-->c. There will be irrational points arbitrarily close to c, where f(x)=x. Thus, the limit will approach c, which is non-zero for any rational non-zero c.
3. Therefore, the limit does not equal f(c)=0 unless c=0.

At all irrational points c, we can make a similar argument that the limit will not exist, since there exist rational numbers arbitrarily close to c. This argues that the limit for an irrational c, will approach 0. Of course, we probably don't even need to make this second argument about what happens at an irrational point, since we know what happens at any rational point, and the rationals are dense on the real line.

That means the function is NOT continuous anywhere (except arguably at 0), since the limit is not the same as the value at that given point.

The flippant statement about an infinite number of irrationals between any pair of rational numbers is meaningless in this respect, since the same statement applies in the reverse. You cannot just play around with infinities like that, not in this context.

Answer 1

[Walter Roberson]

Start at the left end of the interval. Record sign() of dy there, along with the current position. Move right one position. Calculate sign() of dy there, along with the current position. If it is the same as the recorded dy, continue moving right.

If you reach the right-hand edge, then emit an appropriate message about the sign of the interval indicating the ends of the interval.

If you reach a place where the sign changes, the emit an appropriate message about the sign of the interval, indicating the locations of the start of the interval and the last location where the interval was the same sign; then record the current location as the start of the new interval.

Note: there is a much more compact way of doing this, using diff() on the sign() of the derivatives

[rezheen]

@ Walter: Are you talking about something like this:

a1=cp(1)-0.01; a2=cp(1)+0.01; b1=cp(2)-0.01; b2=cp(2)+0.01;
dy_a1=subs(dy,a1); dy_a2=subs(dy,a2); 
dy_b1=subs(dy,b1); dy_b2=subs(dy,b2); 
if (dy_a1 > dy_a2) & (dy_b1 < dy_b2)
    fprintf('decreasing in -inf to %.2f\n', cp2(1))
    fprintf('Increasing in %.2f to inf\n', cp2(1))
end

This is like a hardcode or brute force basically. Not a good way to do it at all. There are many more cases to consider.

[Walter Roberson]

syms x y real

dy=1-(4/x^2); %x=~0; % This is the derivative of the function

cp=solve(dy);       % Finding critical points
[~,idx] = unique(cp);
cp = cp(idx);

a1 = round(min(cp)-1);
a2 = round(max(cp)+1);
a3 = a1:0.1:a2;
a4 = arrayfun(@(V)limit(dy, x, V), a3);

plot(a3, max(a4,-50));
yline(0);

●

sa4 = sign(a4);
starts = strfind([0 sa4], [0 1]);
stops = strfind([sa4 0], [1 0]);
for K = 1 : length(starts)
    fprintf('f is decreasing on %g (%g) : %g (%g)\n', a3(starts(K)), a4(starts(K)), a3(stops(K)), a4(stops(K)));
end

f is decreasing on -3 (0.555556) : -2.1 (0.0929705)
f is decreasing on 2.1 (0.0929705) : 3 (0.555556)

starts = strfind([0 sa4], [0 -1]);
stops = strfind([sa4 0], [-1 0]);
for K = 1 : length(starts)
    fprintf('f is increasing on %g (%g) : %g (%g)\n', a3(starts(K)), a4(starts(K)), a3(stops(K)), a4(stops(K)));
end

f is increasing on -1.9 (-0.108033) : 1.9 (-0.108033)

[rezheen]

Hi Walter. I appreciate your help and code. Shouldn't f be increasing in -3 to -2.1 (dy >0), 2.1 to 3 (dy >0), and decreasing in -1.9 to 1.9 (dy < 0)? When I swap the 1 to -1 in your starts and stops vectors, I get what I stated above:

syms x y real
dy=1-(4/x^2); cp=solve(dy); [~,idx] = unique(cp); cp = cp(idx);
a1 = round(min(cp)-1); a2 = round(max(cp)+1); a3 = a1:0.1:a2; 
a4 = arrayfun(@(V)limit(dy, x, V), a3); sa4 = sign(a4);
starts = strfind([0 sa4], [0 -1]); stops = strfind([sa4 0], [-1 0]);
for K = 1 : length(starts)
    fprintf('f is decreasing on %g (%g) : %g (%g)\n', a3(starts(K)), a4(starts(K)), a3(stops(K)), a4(stops(K)));
end

f is decreasing on -1.9 (-0.108033) : 1.9 (-0.108033)

starts = strfind([0 sa4], [0 1]); stops = strfind([sa4 0], [1 0]);
for K = 1 : length(starts)
    fprintf('f is increasing on %g (%g) : %g (%g)\n', a3(starts(K)), a4(starts(K)), a3(stops(K)), a4(stops(K)));
end

f is increasing on -3 (0.555556) : -2.1 (0.0929705)
f is increasing on 2.1 (0.0929705) : 3 (0.555556)

[Walter Roberson]

@rezheen

I think you are right about that.

[rezheen]

@Walter Roberson

When I change my dy function to new one such as:

syms x y real
dy=-3*x^2+4*x; % new function
fplot(dy); hold on; axis([-5, 5, -5, 5]);
j=xline(0); set(j,'color','black','Linewidth', 1.5)
k=yline(0); set(k,'color','black','Linewidth', 1.5)
cp=solve(dy); y_cp=subs(dy,cp); [~,idx] = unique(cp);
cp = cp(idx); y_cp = y_cp(idx);
disp('The critical points are:')

The critical points are:

for k =1:length(cp)
    fprintf("(%.2f,%.2f)\n",cp(k), y_cp(k))
    plot(cp,y_cp,'*', 'color', 'black');
end

(0.00,0.00)
(1.33,0.00)

a1 = round(min(cp)-1);
a2 = round(max(cp)+1);
a3 = a1:0.1:a2;
a4 = arrayfun(@(V)limit(dy, x, V), a3);
sa4 = sign(a4);
starts = strfind([0 sa4], [0 -1]);
stops = strfind([sa4 0], [-1 0]);
for K = 1 : length(starts)
    fprintf('f is decreasing on %g (%g) : %g (%g)\n', a3(starts(K)), a4(starts(K)), a3(stops(K)), a4(stops(K)));
end

f is decreasing on -1 (-7) : -0.1 (-0.43)

starts = strfind([0 sa4], [0 1]);
stops = strfind([sa4 0], [1 0]);
for K = 1 : length(starts)
    fprintf('f is increasing on %g (%g) : %g (%g)\n', a3(starts(K)), a4(starts(K)), a3(stops(K)), a4(stops(K)));
end

Index exceeds array bounds.

I get that error above. The decreasing works (partially) but the rest of it doesn't (increasing in 0-1.33, decreasing in 1.33 to inf) How can I correct it?

[Walter Roberson]

syms x y real
dy=-3*x^2+4*x; % new function
fplot(dy); hold on; axis([-5, 5, -5, 5]);
j=xline(0); set(j,'color','black','Linewidth', 1.5)
k=yline(0); set(k,'color','black','Linewidth', 1.5)
cp=solve(dy); y_cp=subs(dy,cp); [~,idx] = unique(cp);
cp = cp(idx); y_cp = y_cp(idx);
disp('The critical points are:')

The critical points are:

for k =1:length(cp)
    fprintf("(%.2f,%.2f)\n",cp(k), y_cp(k))
    plot(cp,y_cp,'*', 'color', 'black');
end

(0.00,0.00)
(1.33,0.00)

a1 = round(min(cp)-1);
a2 = round(max(cp)+1);
a3 = a1:0.1:a2;
a4 = arrayfun(@(V)limit(dy, x, V), a3);
sa4 = sign(a4);
starts = strfind([0, sa4<0], [0 1]);
stops = strfind([sa4<0, 0], [1 0]);
for K = 1 : length(starts)
    fprintf('f is decreasing on %g (%g) : %g (%g)\n', a3(starts(K)), a4(starts(K)), a3(stops(K)), a4(stops(K)));
end

f is decreasing on -1 (-7) : -0.1 (-0.43)
f is decreasing on 1.4 (-0.28) : 2 (-4)

starts = strfind([0, sa4>0], [0 1]);
stops = strfind([sa4>0, 0], [1 0]);
for K = 1 : length(starts)
    fprintf('f is increasing on %g (%g) : %g (%g)\n', a3(starts(K)), a4(starts(K)), a3(stops(K)), a4(stops(K)));
end

f is increasing on 0.1 (0.37) : 1.3 (0.13)

Answer 2

[John D'Errico]

This can get fairly messy, if your function is carefully designed to cause you problems. For example, you may need to find a singularity. And singularities can come in many different forms, not just as a root of the denominator of a rational polynomial. As such, numden is worthless to you. A derivative singularity can cause problems, for example:

fplot(@(x) sqrt(abs(x - sqrt(2))),[-20,20])
grid on

The singularity resides here at x==sqrt(2), an irrational number. So just testing at isolated points will not help you a lot, nor will the function numden.

Or your function may just be the simple

fplot(@tan,[-20,20])
grid on

which is everywhere increasing, except at some isolated (transcendental) locations, where it transitions from +inf to -inf.

And of course, discontinuities can cause problems all of their own. Next is a cute one, which will cause all hell to break loose if you are not careful, as it is everyehere constant, except at some isolated locations, and those locations do not occur at any set of convenient points on the real line. (I could easily make that worse too.)

fplot(@(x) round(2*sin(x.^2/10)),[-20,20])
grid on

Do you see the issues? And I've not even scratched the surface of the nasty behavior such a tool could encounter. How about the simple function f(x), defined as:

f(x) = 0 if x is rational

f(x) = x if x is irrational

It is a function. It is sort of increasing for positive x, if you look only at the irrational numbers, but it is everywhere discontinuous, and everywhere decreasing, when you transition from an irrational number to a rational one. (As if the word "transition" makes any sense in this context.) And if you consider only the rational numbers, then it is everywhere constant at zero. And since all numbers that can be represented in double precision are inherantly rational, then f(x) will be arguably identically zero, when implemented in MATLAB. And worse, you can't even locate the uncountably many transition points. As I said, nasty.

In order to do anything with your question, you will need to severely restrict the behavior of your allowed functions. For example, you may want to require continuity, even some degree of differentiability, as otherwise, one can always create some function that will play hell with any code you write.

Remember that not all functions are polynomials, even though those are the ones you tend to see in a class.

All of this gets much easier if you are willing to require some degree of continuity. If you get rid of any breaks in continuity, then it gets simpler. If you exclude derivative singularities and breaks at that level, then too, you start to exclude most of the problematic functions I showed. But that also might exclude things like rational polynomials. Even something as simple as 1/x is an issue, as if that is allowed, then you need to properly handle functions like tan(x). Again, writing code to handle general problems like this becomes tricky. It requires you to carefully set boundaries on what class of problems you would agree to handle.