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Sep 27, 2022 at 14:17 comment added Dani Roca Gonzalez Sure, does make sense. I definitely can fill the gaps from this, thank you.
Sep 26, 2022 at 22:08 history edited Anthony Quas CC BY-SA 4.0
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Sep 26, 2022 at 15:22 comment added Anthony Quas I didn't mean $\lambda$ to be the Lyapunov exponent; just some number. But if I did, I guess I would write $e^{n(\lambda-\epsilon)}\|e\|\le \|A^n e\|\le e^{n(\lambda+\epsilon)}\|e\|$.
Sep 26, 2022 at 14:50 vote accept Dani Roca Gonzalez
Sep 26, 2022 at 13:33 comment added Dani Roca Gonzalez I am essentially convinced, but do you mean $e^{-\epsilon n} \lambda^n \|e\|\le \|A^ne\|\le e^{\epsilon n} \lambda^n \|e\|$ and $\|A^nf\|\le \mu^n\|f\|$
Sep 25, 2022 at 0:36 history edited Anthony Quas CC BY-SA 4.0
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Sep 24, 2022 at 19:45 history answered Anthony Quas CC BY-SA 4.0