how can i find convex and concave points
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Cem SARIKAYA
am 15 Mai 2019
Kommentiert: Star Strider
am 16 Mai 2019
how do I find the convex and concave points of the discrete data as in the photo
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Star Strider
am 15 Mai 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
figure
plot(x, f(x))
hold on
plot(xpoint, f(xpoint), 'pg', 'MarkerSize',10, 'MarkerFaceColor','g')
hold off
grid
axis('equal')
xlim([-2.5 2.5])
To illustrate:
Your data may be different, so experiment with different values for the slope to get the result you want.
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Steven Lord
am 15 Mai 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
axis('equal')
xlim([-2.5 2.5])
2 Kommentare
Adam Danz
am 15 Mai 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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