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I want to know how to find analytic functions $f(z)$ that are asymptotic and analytic on and near the real line of functions of the type $\ln(C +\exp(P(z^2)))$ where $C$ is a complex constant and $P$ is polynomial.

I want the approximation to satisfy $\frac{abs(f(z))}{abs(\ln(C +\exp(P(z^2))))} < 1+\frac{1}{1+abs(z)^2}$ in the strip $-1 < Im(z) < 1$ and I want it to be analytic there.

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Could you write what you want with quantificators? –  Davide Giraudo Sep 13 '12 at 13:12
    
@DavideGiraudo What did you edit ? And why do you want quantificators ? Isnt the question clear ? Do you want bounds on how good the asymptotic needs to be ? Or are you doubting the existance and want existence quantifiers ? –  mick Sep 13 '12 at 13:23
    
I just fixed tex (I should have written that when I edited but I forgot). Yes, I'm asking what would look like the bounds you are looking for. –  Davide Giraudo Sep 13 '12 at 13:25
    
I did not think about that yet , lets say 1+1/abs(z) as the ratio between the abs(asymptotic) and the abs(function) valid in the strip -1 < Im(z) < 1. –  mick Sep 13 '12 at 13:53
    
Ok replace $1 + \frac{1}{abs(z)}$ by $1 + \frac{1}{1+abs(z)^2}$ in the previous comment. ( sorry ) –  mick Sep 13 '12 at 15:16
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1 Answer 1

up vote -2 down vote accepted

The basic problem is too find an analytic function that approximates $ln(E(z))$ near the real line ( strip described in OP ) where $E(z)$ is a given entire function. From the Weierstrass Product Theorem we know that $E(z) = exp(F(z))(z-z_1)(z-z_2)...$

Hence if $E(z)$ has no zero's within the strip and $E(z)$ is of the form $E(z) = exp(F(z)) (z^4 + a_1)(z^4 + a_2)...$

Then we can write $ln(E(z))$ ~ $F(z) + log(z^4 + a_1) + log(z^4 + a_2) + ...$ And then we only need to find an entire function that approximates $ln(z)$ near real $x > 1$. Such an example is $-(ln(2 sinh(z)/z) - z)$. Hence this way we construct our solution if we know enough about the location of the zero's of $E(z)$.

For those not familiar with entire functions : Weierstrass Product Theorem. Notice that we can also use what we did here to give an entire function that approximates sqrt($A(z)$) for some entire $A(z)$. ( use sqrt($x$) = $exp(\frac{ln(x)}{2})$ )

It is possible to get the error term as small as we want by using better asymptotics for $ln(z)$. Im unaware of better conceptual solutions to this problem but they might exist. In case our $E(z)$ has a finite numbers of zero's then the problem is trivial ( we can compute $F(z)$ easily ). I have not seen $ln(2 sinh(z)/z) - z$ often in the literature but imho its an interesting function (or I did not read enough).

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Did you answer your own question incorrectly and then write that your answer is incorrect and then accept your own incorrect answer? Or am I just not understanding? –  Todd Wilcox Nov 21 '12 at 14:27
    
No. I answered my question on sep 17 as you can read. And today I discovered a tiny mistake. It is still a correct answer in the strip , but it is not entire. Imho this is a good answer but perhaps not the best. Since it is the only answer and It answers the OP completely apart the entire condition I think it deserves the correct flag. If anyone gives a better answer I might reconsider. –  mick Nov 21 '12 at 14:39
    
Things should be clear now. –  mick Feb 10 '13 at 19:00
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