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Javulicraft (talk | contribs) m The prior statement was not true. V is positive definite for all r except 0. But r=0 does not imply q=0, qdot=0, but rather edot+alpha*e=0, which is not the same and doesn't tell much. |
m →Example: A function is or isn't *positive definite* everywhere, but *positive* for for specific values. |
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A Control-Lyapunov candidate is then
:<math>
r \mapsto V(r) :=\frac{1}{2}r^2
</math>
which is positive
Now taking the time derivative of <math>V</math>
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