Library / Functions of a Real Variable / Chapter V / Appendix 0
Example 1
Tag 08H6, permanent. Cite it as [Bourbaki, Tag 08H6].
Functions of a Real Variable, chapter V, Appendix 0. Hardy Fields. (H) Functions
Let us seek an asymptotic expansion for the inverse function v ( x ) v(x) v ( x ) of x 5 + x x^5 + x x 5 + x (for x x x tending to + ∞ +\infty + ∞ ); putting x 5 = t x^5 = t x 5 = t one is led to seek an expansion of the inverse function u ( t ) u(t) u ( t ) of t + t 1 / 5 t + t^{1/5} t + t 1/5 (for t t t tending to + ∞ +\infty + ∞ ), that is, to apply prop. 5 of V, p. 256, to the case where g ( t ) = − t 1 / 5 g(t) = -t^{1/5} g ( t ) = − t 1/5 . Let us, for example, calculate, u 2 ( t ) u_2(t) u 2 ( t ) ; one has
u 2 ( t ) = t − ( t − t 1 / 5 ) 1 / 5 = t − t 1 / 5 + 1 5 t − 3 / 5 + 2 25 t − 7 / 5 + o 1 ( t − 7 / 5 ) .
u_2(t) = t - \left( t - t^{1/5} \right)^{1/5} = t - t^{1/5} + \frac{1}{5} t^{-3/5} + \frac{2}{25} t^{-7/5} + o_1\left( t^{-7/5} \right).
u 2 ( t ) = t − ( t − t 1/5 ) 1/5 = t − t 1/5 + 5 1 t − 3/5 + 25 2 t − 7/5 + o 1 ( t − 7/5 ) . Moreover, by (11) (V, p. 256)
u ( t ) − u 2 ( t ) ∼ − 1 25 t − 7 / 5
u(t) - u_2(t) \sim -\frac{1}{25} t^{-7/5}
u ( t ) − u 2 ( t ) ∼ − 25 1 t − 7/5 whence
u ( t ) = t − t 1 / 5 + 1 5 t − 3 / 5 + 1 25 t − 7 / 5 + o 2 ( t − 7 / 5 )
u(t) = t - t^{1/5} + \frac{1}{5} t^{-3/5} + \frac{1}{25} t^{-7/5} + o_2\left( t^{-7/5} \right)
u ( t ) = t − t 1/5 + 5 1 t − 3/5 + 25 1 t − 7/5 + o 2 ( t − 7/5 ) and one deduces the expansion
v ( x ) = u ( x ) 1 / 5 = x 1 / 5 − 1 5 x − 3 / 5 − 1 25 x − 7 / 5 + o 3 ( x − 7 / 5 )
v(x) = u(x)^{1/5} = x^{1/5} - \frac{1}{5} x^{-3/5} - \frac{1}{25} x^{-7/5} + o_3\left( x^{-7/5} \right)
v ( x ) = u ( x ) 1/5 = x 1/5 − 5 1 x − 3/5 − 25 1 x − 7/5 + o 3 ( x − 7/5 ) sought.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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