# Lower bound for Euler's function

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.

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

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 $$\phi(e^{-\pi})=\frac{e^{\pi/24}\Gamma\left(\frac14\right)}{2^{7/8}\pi^{3/4}}$$

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.