Bourbaki

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) of x5+xx^5 + x (for xx tending to +∞+\infty); putting x5=tx^5 = t one is led to seek an expansion of the inverse function u(t)u(t) of t+t1/5t + t^{1/5} (for tt tending to +∞+\infty), that is, to apply prop. 5 of V, p. 256, to the case where g(t)=−t1/5g(t) = -t^{1/5}. Let us, for example, calculate, u2(t)u_2(t); one has

u2(t)=t−(t−t1/5)1/5=t−t1/5+15t−3/5+225t−7/5+o1(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).
Moreover, by (11) (V, p. 256)
u(t)−u2(t)∼−125t−7/5 u(t) - u_2(t) \sim -\frac{1}{25} t^{-7/5}
whence
u(t)=t−t1/5+15t−3/5+125t−7/5+o2(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)
and one deduces the expansion
v(x)=u(x)1/5=x1/5−15x−3/5−125x−7/5+o3(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)
sought.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

Transcribed from the printed volume. Not translated.