Growth estimates on x ^ y
for complex x
, y
#
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Let l
be a filter on ℂ
such that complex.re
tends to infinity along l
and complex.im z
grows at a subexponential rate compared to complex.re z
. Then
-
complex.is_o_log_abs_re_of_subexponential_im_re
:real.log ∘ complex.abs
iso
-small ofcomplex.re
alongl
; -
complex.is_o_cpow_mul_exp
: $z^{a_1}e^{b_1 * z} = o\left(z^{a_1}e^{b_1 * z}\right)$ alongl
for any complexa₁
,a₂
and realb₁ < b₂
.
We use these assumptions on l
for two reasons. First, these are the assumptions that naturally
appear in the proof. Second, in some applications (e.g., in Ilyashenko's proof of the individual
finiteness theorem for limit cycles of polynomial ODEs with hyperbolic singularities only) natural
stronger assumptions (e.g., im z
is bounded from below and from above) are not available.
- tendsto_re : filter.tendsto complex.re l filter.at_top
- is_O_im_pow_re : ∀ (n : ℕ), (λ (z : ℂ), z.im ^ n) =O[l] λ (z : ℂ), rexp z.re
We say that l : filter ℂ
is an exponential comparison filter if the real part tends to
infinity along l
and the imaginary part grows subexponentially compared to the real part. These
properties guarantee that (λ z, z ^ a₁ * exp (b₁ * z)) =o[l] (λ z, z ^ a₂ * exp (b₂ * z))
for any
complex a₁
, a₂
and real b₁ < b₂
.
In particular, the second property is automatically satisfied if the imaginary part is bounded along
l
.
Alternative constructors #
Preliminary lemmas #
If l : filter ℂ
is an "exponential comparison filter", then $\log |z| =o(ℜ z)$ along l
.
This is the main lemma in the proof of complex.is_exp_cmp_filter.is_o_cpow_exp
below.
Main results #
If l : filter ℂ
is an "exponential comparison filter", then for any complex a
and any
positive real b
, we have (λ z, z ^ a) =o[l] (λ z, exp (b * z))
.
If l : filter ℂ
is an "exponential comparison filter", then for any complex a₁
, a₂
and any
real b₁ < b₂
, we have (λ z, z ^ a₁ * exp (b₁ * z)) =o[l] (λ z, z ^ a₂ * exp (b₂ * z))
.
If l : filter ℂ
is an "exponential comparison filter", then for any complex a
and any
negative real b
, we have (λ z, exp (b * z)) =o[l] (λ z, z ^ a)
.
If l : filter ℂ
is an "exponential comparison filter", then for any complex a₁
, a₂
and any
natural b₁ < b₂
, we have (λ z, z ^ a₁ * exp (b₁ * z)) =o[l] (λ z, z ^ a₂ * exp (b₂ * z))
.
If l : filter ℂ
is an "exponential comparison filter", then for any complex a₁
, a₂
and any
integer b₁ < b₂
, we have (λ z, z ^ a₁ * exp (b₁ * z)) =o[l] (λ z, z ^ a₂ * exp (b₂ * z))
.