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how can i find convex and concave points

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how do I find the convex and concave points of the discrete data as in the photoWhatsApp Image 2019-05-15 at 5.41.47 PM.jpeg


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Accepted Answer

Star Strider
Star Strider on 15 May 2019
It depends on how you want to define them.
Here, I define them as points where the slope is -0.5:
f = @(x) 1-(x./sqrt(1+x.^2)); % Create Function
x = linspace(-10, 10);
h = x(2)-x(1); % Step Interval
dfdx = gradient(f(x),h); % Derivative
[~,infpt] = min(dfdx);
xpoint(1) = interp1(dfdx(1:infpt-1),x(1:infpt-1),-0.5); % Slope = -0.5
xpoint(2) = interp1(dfdx(infpt+1:end),x(infpt+1:end),-0.5); % Slope = -0.5
plot(x, f(x))
hold on
plot(xpoint, f(xpoint), 'pg', 'MarkerSize',10, 'MarkerFaceColor','g')
hold off
xlim([-2.5 2.5])
To illustrate:
how can i find convex and concave points - 2019 05 15.png
Your data may be different, so experiment with different values for the slope to get the result you want.


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Star Strider
Star Strider on 16 May 2019
If you say 'x' to the vertical section and 'y' to the horizontal section ...’
In my code, ‘x’ is the independent variable and ‘y’ is the dependent variable.
I want to find slopes which constantly changes
You can set the slope value (I chose -0.5 here) to be whatever you like, within limits. (The slope has to have that value somewhere in the region of interest.) My code (specifically the ‘xpoint’ interpolations) should be reasonably robust to your choices.
can give a range?
I am not certain what you intend. See the documentation for the interp1 (link) function (that I use to calculate the ‘xpoint’ values) to understand what the function can do, and how to use it.
Cem SARIKAYA on 16 May 2019
Thank you very much for your time.

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More Answers (1)

Steven Lord
Steven Lord on 15 May 2019
Depending on what you want to do with this information (which is not clear from the question) you may find the ischange function useful.
f = @(x) 1-(x./sqrt(1+x.^2)); % Create Function
x = linspace(-10, 10);
y = f(x);
changes = ischange(y, 'linear', 'SamplePoints', x);
plot(x, y, '-', x(changes), y(changes), 'gp')
grid on
xlim([-2.5 2.5])


Cem SARIKAYA on 15 May 2019
my data does not have a function, all of my data in the matrix and manually entered values. I need to derive from here or do I need a different code? i actually don't understand. mybe this picture tells you betterWhatsApp Image 2019-05-16 at 1.21.44 AM.jpeg
Adam Danz
Adam Danz on 15 May 2019
@Cem SARIKAYA, Steven Lord's proposal is similar to Star Strider's. In the function ischange(), when the method is set to 'linear', the slope of the line is considered and it searches for abrupt changes in the slope.
Again, take a moment to grasp these concepts conceptually before you worry about implementing the code.

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