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Revision 1123 by francois, Mon Aug 15 21:28:10 2022 UTC vs.
Revision 1124 by francois, Fri Sep 1 15:21:57 2023 UTC

# Line 219 | Line 219 | Q_{exa}&=&Q_{app}\left(\frac{h}{2}\right
219   \end{eqnarray*}
220   En soustrayant les deux dernières équations il vient
221   \begin{eqnarray*}
222 < (2^n-1)Q_{exa}&=&2^nQ_{app}\left(\frac{h}{2}\right)-Q_{app}(h) -\frac{1}{2}C_{n+1}\frac{h^{n+1}}{2} -\frac{3}{4}C_{n+2}\frac{h^{n+2}}{2^2} +...\\
223 < Q_{exa}&=&\frac{2^nQ_{app}\left(\frac{h}{2}\right)-Q_{app}(h)}{2^n-1}+\frac{ -\frac{1}{2}C_{n+1}\frac{h^{n+1}}{2} -\frac{3}{4}C_{n+2}\frac{h^{n+2}}{2^2} +...}{2^n-1}\\
222 > (2^n-1)Q_{exa}&=&2^nQ_{app}\left(\frac{h}{2}\right)-Q_{app}(h) -\frac{1}{2}C_{n+1}h^{n+1} -\frac{3}{4}C_{n+2}h^{n+2} +...\\
223 > Q_{exa}&=&\frac{2^nQ_{app}\left(\frac{h}{2}\right)-Q_{app}(h)}{2^n-1}+\frac{ -\frac{1}{2}C_{n+1}h^{n+1} -\frac{3}{4}C_{n+2}h^{n+2} +...}{2^n-1}\\
224   &=&\frac{2^nQ_{app}\left(\frac{h}{2}\right)-Q_{app}(h)}{2^n-1}+O(h^{n+1})\\
225   \end{eqnarray*}
226   \begin{encadre2}

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