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This is a test. Anyone should understand why the nonlinear system $\dot{x} = -x^5$ is asymptotically stable around the equilibrium point $x = 0$ even though its linearized tangent system $\dot{x} = 0$ at $x=0$ is Lyapunov-stable but not asymptotically stable. $$ E =\gamma m c^2$$

Posted 8 months ago
Edited 8 months ago
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Using the button file, I would like to preview $E = \gamma m c^2$ where $\gamma = \frac{1}{\sqrt{1-\beta^2}}$ and $\beta = \frac{v}{c}$.

Posted 8 months ago
Edited 8 months ago

Trying paste of an image again like this image: file

Posted 8 months ago

Your answer

2 Answers
8 months ago
8 months ago