I am wondering, if I can find a good approximant for this function

$$f(z)=\log \left[ \frac{1-z^2}{z \left(3-z^2\right)}\sinh \left\{\frac{z \left(3-z^2\right)}{1-z^2}\right\}\right]$$

assuming $z \in [0,1)$. A first naive approach would be to take the Taylor expension. But as this function has a pole at one I would have to take more and more terms into account as z takes values closer to one. Is there a suitable method to approximate this function over the whole interval?

EDIT: I add this plot to show the proposed assymptotics down below

enter image description here

But the basic question remains. Can I find a single function, that is a good approximant of $f$?

EDIT: Changed the plot to account for the small mistake.

enter image description here

  • 1
    $\begingroup$ What do you mean by "good" approximant -i.e. How many terms do you need? Have you tried splitting the expression up using laws of logs? $\endgroup$ – David Quinn Jul 28 '15 at 12:50
  • $\begingroup$ I have something like a Pade approximant in the back of my mind. A "good" approximant would be something that for no value of $z \in [0,1)$ returns a relative error of more than 5%and that most impotantly shows all the limiting features of the original function (for small z defined by the first Taylor expansion term and for z close to 1 defined by a diverging rational function ). To answer your question, I have not yet tried to split the logarithm. $\endgroup$ – Asking Questions Jul 28 '15 at 12:56

For $z \rightarrow 0$ a taylor approximation is fine. I leave this part to you.

For $z\rightarrow 1$ we may observe that the relevant contributions arise from the terms containing a $1-z^2$. The other contributions will only give finite corrections in this limit and can be approximated by their values at $z=1$ . Furthermore we observe that $\sinh(x)\sim \frac{e^x}{\bf2} $ as $x\rightarrow \infty$. Taken all this into account we may write

$$ \lim_{z\rightarrow 1_-}f(z)\sim \log\left(\frac{1-z^2}{2}\frac{e^{\frac{2}{1-z^2}}}{{\bf2}}\right)=\frac{2}{1-z^2}+\log(1-z^2)-\log(4) $$

Edit: A forgotten $\frac{1}{\bf2}$ was inserted. Now the graph looks even better!

  • $\begingroup$ Thank you very much, this is a great explanation of the assymptotics. Can I derive from it a simple closed form of an approximant that is valid over the whole interval $[0,1)$? Let me add the first taylor expansion terms: $$f(z)=\frac{3}{2}z^2+\frac{31}{20}z^4+\mathcal{O}(z^6)$$ $\endgroup$ – Asking Questions Jul 28 '15 at 13:15
  • $\begingroup$ I added a plot to the original post showing the two limiting behavior, but also stated that my acceptance of the answer may have been premature. $\endgroup$ – Asking Questions Jul 28 '15 at 14:04
  • $\begingroup$ The last comment I don't understand: $$\frac{3-z^2}{1-z^2}+\log \left(1-z^2\right)-\log (4)$$ does not give a good approximation when I plot it against $f(z)$. Am I missing something? $\endgroup$ – Asking Questions Jul 28 '15 at 15:27
  • $\begingroup$ Unfortunatly plotting this it does not match the original functions graph very well. $\endgroup$ – Asking Questions Jul 28 '15 at 15:39
  • $\begingroup$ Sorry i'm a little bit unconcentrated today. I will work out this part correctly when i feel better $\endgroup$ – tired Jul 28 '15 at 15:42

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