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Euler function is defined, for $|x|\le 1$, as follows: $$\phi(x)=\prod_{i=1}^\infty(1-x^i)$$

Upper bounds for $\phi$ can be simply derived from ending the product early, e.g. $$\phi(x)<\prod_{i=1}^2(1-x^i)=1-x-x^2+x^3$$

What lower bounds are known for $\phi(x)$?


I'm especially interested in bounds which can be computed efficiently as this is needed for a computer application.

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    $\begingroup$ mpmath can compute it exactly, does this help? $\endgroup$ – joro Nov 16 '15 at 16:34
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    $\begingroup$ @joro - Seems good, if it's fast enough. Thanks ! $\endgroup$ – R B Nov 16 '15 at 16:36
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    $\begingroup$ It is very efficient with 100 decimal digits of precision. Answered with details. $\endgroup$ – joro Nov 16 '15 at 17:07
  • $\begingroup$ Also, mpmath is open source, written in Python. $\endgroup$ – joro Nov 16 '15 at 17:21
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mpmath can compute it very efficiently.

The function is called mpmath.qp. Here is a sage session for $\phi(e^{-\pi})$ with precision 100 decimal digits which takes about 360 microseconds on and old machine:

sage: time mpmath.qp(mpmath.exp(-mpmath.pi))
CPU times: user 0 ns, sys: 0 ns, total: 0 ns
Wall time: 361 µs
0.9549187899876741037512339781102910776327153738078052831487991916760940356867145395349815186744610988

This agrees with wikipedia's closed form for $$\phi(e^{-\pi})=\frac{e^{\pi/24}\Gamma\left(\frac14\right)}{2^{7/8}\pi^{3/4}}$$

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You get concrete bounds from above or from below just stopping the well-known expansion after two positive terms, resp. two consecutive negative terms; see this answer.

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