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i have the following System of Differential Equations $$ \begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix} = \begin{pmatrix} 0 && 1\\ -1 && 0 \end{pmatrix} \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} \ and \ \begin{pmatrix} x(0) \\ y(0) \end{pmatrix}= \begin{pmatrix} 2 \\ 0 \end{pmatrix} $$ When i use the Picard-Iteration, i get ($ s:= \begin{pmatrix} x(t) \\ y(t)\end{pmatrix} $ ) $$ s_1 = \begin{pmatrix} 2 \\ -2t\end{pmatrix} $$ $$ s_2 = \begin{pmatrix} 2-t^2 \\ -2t\end{pmatrix} $$ $$ s_3 = \begin{pmatrix} 2-t^2 \\ \frac{1}{3} t^3-2t\end{pmatrix} $$ I assume that $ s_{\infty} = \begin{pmatrix} 2cos t \\ -2sint \end{pmatrix} $ But when i consider the series for sin and cos, for example for the index 1, $ s_1$ will be $ s_1 = \begin{pmatrix} 2-t^2 \\ \frac{1}{3} t^3-2t\end{pmatrix} $ How do i get this right?

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    $\begingroup$ Writing $x(x)$ is confusing (even mildly infuriating), so please use $x(t)$ or something similar. $\endgroup$ – MisterRiemann Nov 24 '18 at 15:26
  • $\begingroup$ I changed it. I am sorry. $\endgroup$ – Steven33 Nov 24 '18 at 16:53
  • $\begingroup$ You get the linear terms correct in the first integration, the quadratic terms in the second, and the cubic terms in the third integration. Why do you think you could get (correct or at all) cubic terms in the first integration? $\endgroup$ – LutzL Nov 24 '18 at 17:21
  • $\begingroup$ I mean it differently. When i consider, $ 2 cos t = \sum_{k=0}^{\infty} \frac{(-1)^k t^{2k} }{(2k)!} $, then i get for k=1: $ 2- t^2 \ne s_{11}=2$ Dont i have to get the indices synchronised? $\endgroup$ – Steven33 Nov 24 '18 at 17:34
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There is no guarantee that taking finitely many steps of Picard iteration will result in successively matched Taylor polynomials between the approximant and the solution. This does happen to occur for $y'=y,y(0)=1$ but this is in some sense a coincidence.

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  • $\begingroup$ In this case, my solutions work. Its easily to proof. But do you understand my index problem? $\endgroup$ – Steven33 Nov 25 '18 at 10:38

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