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1 transparent par page - Montefiore

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En substituant :<br />

Puisque<br />

on obtient<br />

x 1<br />

= √<br />

1 − x − x 2<br />

5<br />

<br />

1<br />

1 − α1x −<br />

<br />

1<br />

.<br />

1 − α2x<br />

1<br />

1 − αx = 1 + αx + α2 x 2 + . . . ,<br />

F (x) = 1 <br />

√ (1 + α1x + α<br />

5<br />

2 1x 2 + . . .) − (1 + α2x + α 2 2x 2 + . . .) <br />

Par identification :<br />

[x n ]F (x) = fn = αn 1 − αn 2 √ =<br />

5 1<br />

√<br />

5<br />

<br />

1 + √ n <br />

5 1 −<br />

−<br />

2<br />

√ n 5<br />

.<br />

2<br />

422

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