I have a real number $I_h$ depending on a small parameter $h>0$. I want to show that it has an asymptotic expansion in integer powers $h$, i.e. there exists a sequence $(J_k)_{k}$ such that

$$ I_h \sim \sum_{k=0}^\infty \ h^{k} \ J_{k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$$

Assume I am not able to show this directly, but that I can construct for every
$\alpha $ say in $[\frac 12,1) $ a double sequence $(J_{k,m}^{(\alpha)})_{k,m}$ such that asymptotically

$$I_h \sim \sum_{k,m=0}^\infty \ h^{k + \alpha m} \ J^{(\alpha)}_{k,m}$$

This should imply what I want (I consider for example $\alpha_1 =\frac 12$ and $\alpha_2=\frac{\sqrt 2}{ 2}$, so only the coefficients of integer powers can be nonzero, otherwise I have a contradiciton. )

My question is (given that what I wrote is correct): is this somehow a standard trick in asymptotic analysis? Can you give me examples of situations where this argument is used?

EDIT 1: I changed $\alpha\in[\frac 12,1]$ to $\alpha\in [\frac 12,1)$. If I had also $\alpha=1$ I would be done of course.

EDIT 2: The coefficients $(J_{k,m}^{(\alpha)})_{k,m}$ are very complicated and I would say it is hopeless to write them all down in a closed formula (they arise from a combiantion of Laplacae asymtptotics and several other expansions). And even if one manages to write down a formula, there appear quantities derived from a WKB expansion, for which it seems hard to me to get directly much more information then just existence (to show directly $(*)$ I would need to know that some complicated combinations of arbitrary high derivatives vanish at some point... ).

In brief: even if there is a direct argument to prove $(*)$, the undirect argument is much shorter and painless.

  • 2
    $\begingroup$ I find it hard to believe that this can be useful as stated. My feeling is that the analysis of your proof of such double expansion would uncover the reason for analyticity of $I_h$ in a more direct way. $\endgroup$ – user31373 Aug 18 '12 at 5:39
  • $\begingroup$ Something escapes me: if $I_h=J^{(\alpha)}_{0,0}+J^{(\alpha)}_{0,1}h^{\alpha}+o(h^{\alpha})$ and $J^{(\alpha)}_{0,1}\ne0$, then $I_h=J_0+J_1h+o(h)$ is impossible, for example. In short, the expansion of $I_h$ in powers of $h$ is unique, no? $\endgroup$ – Did Aug 18 '12 at 9:23
  • $\begingroup$ @did. Exactly. It is impossible, except when it happens that $J_{0,1}^{(\alpha)} = 0$. I cannot prove the latter property directly, but it is a consequence of the undirect argument. $\endgroup$ – Hans Aug 18 '12 at 9:36
  • $\begingroup$ @Leonid Kovalev. Yes, I admit it seems strange and I'm not very happy with it since of course it would be nicer to give a direct proof (but...see EDIT 2). This is why I look for consolation given by other similar cases. $\endgroup$ – Hans Aug 18 '12 at 9:40

Here is a fact which might help:

Let $h\mapsto I_h$ denote a function defined at least in an interval $[0,h_0)$ and such that there exists two increasing nonnegative sequences $(i_n)_{n\geqslant0}$ and $(j_n)_{n\geqslant0}$ and some nonzero coefficients $(a_n)_{n\geqslant0}$ and $(b_n)_{n\geqslant0}$ such that $$ I_h=\sum_{n=0}^{+\infty}a_nh^{i_n}=\sum_{n=0}^{+\infty}b_nh^{j_n}, $$ in the sense that, for every $k\geqslant0$, when $h\to0^+$, $$ I_h-\sum_{n=0}^{k}a_nh^{i_n}=o(h^{i_k}),\qquad I_h-\sum_{n=0}^{k}b_nh^{j_n}=o(h^{j_k}). $$ Then, $i_n=j_n$ and $a_n=b_n$ for every $n\geqslant0$.

To prove this fact, assume the result is not true, consider $k=\min\{n\mid (i_n,a_n)\ne(j_n,b_n)\}$ and treat separately the cases when $i_k\ne j_k$ and when $i_k=j_k$ but $a_k\ne b_k$.

To apply this result to your case, write $$ I_h=\sum_{k=0}^{+\infty}J_kh^k=\sum_{n=0}^{+\infty}a_nh^{i_n},\qquad I_h=\sum_{k,m=0}^{+\infty}J^{(\alpha)}_{k,m}h^{k + \alpha m}=\sum_{n=0}^{+\infty}b_nh^{j_n}, $$ where $(i_n)_n$ enumerates the set of $k$ such that $J_k\ne0$ and $(j_n)_n$ enumerates the set of $j$ such that $$ \sum_{k+\alpha m=j}J^{(\alpha)}_{k,m}\ne0. $$ If $\alpha$ is irrational, for example $\alpha=\frac{\sqrt2}2$, this shows that $J^{(\alpha)}_{k,m}=0$ for every $m\ne0$. If $\alpha=\frac12$, this shows that, for every $i\geqslant0$, $$ \sum_{k=0}^iJ^{(\alpha)}_{k,2i-2k+1}=0. $$


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