- #76

fresh_42

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You did not prove that this series is the only possibility. You have proven: the Taylor series provides one solution, not all.Sorry, but I don’t quite understand. If ##\mathbf{x}(t_0)=\mathbf{y}(t_0)=\mathbf{x}_0##, then ##\mathbf{x}^{(k)}(t_0)=\mathbf{y}^{(k)}(t_0)=A^k\,\mathbf{x}_0##, which gives rise to the Taylor series, which only depends on ##\mathbf{x}_0, t_0,## and ##t##. I used Taylor’s theorem to prove that any solution to ##\dot{\mathbf{x}}(t)=A\,\mathbf{x}(t)## with ##\mathbf{x}(t_0)=\mathbf{x}_0## converges pointwise to ##e^{(t-t_0)A}\,\mathbf{x}_0##, which includes both ##\mathbf{x}(t)## and ##\mathbf{y}(t)##. Unless I am missing something, this proves uniqueness of the solution I gave, because the difference between any arbitrary solution and my solution converges pointwise to 0.

Edit: corrected a typo; wrote ##\mathbf{x}_0## instead of ##A^k\,\mathbf{x}_0##.