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 photo
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Star Strider
2019-5-15
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
2019-5-15
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 个评论
Adam Danz
2019-5-15
@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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