lacunary functions wander, part one

Consider for a moment:
\prod_{n=1}^{\infty} \sqrt[2^{n}]{1+\frac{1}{2^{n}}}
One fruitful way to generalize this is to consider a function of |q|<1 like so:

(q)=\prod_{n=1}^{\infty} (1+q^{n})^{q^{n}}

\log(q) = \sum_{n=1}^{\infty} q^{n} \log(1+q^{n})

\sum_{n=1}^{\infty} q^{n} \log(1+q^{n}) = \sum_{n=1}^{\infty} q^{n} \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m}q^{mn}

\sum_{n=1}^{\infty} q^{n} \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m}q^{mn} = \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \sum_{n=1}^{\infty} q^{mn} q^{n} = \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \sum_{n=1}^{\infty} q^{n(m+1)}

\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \sum_{n=1}^{\infty} q^{n(m+1)} = \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} \left[\frac{1}{1-q^{m+1}}-1\right]

Which is just
\sum_{m=1}^{\infty} \left[\frac{(-1)^{m+1}}{m(1-q^{m+1})}\right] - \log(2)

Therefore:

(q)=\prod_{n=1}^{\infty} (1+q^{n})^{q^{n}} = \frac{1}{2}\exp\left(\sum_{m=1}^{\infty} \left[\frac{(-1)^{m+1}}{m(1-q^{m+1})}\right]\right)

Here is a plot of अ(q) on the unit disk. There are some artifacts, but nothing major. Enjoy:

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2 responses to “lacunary functions wander, part one

  1. Will you tell me how hard it was to get the LaTeX on your word press? I just made a new blog. I’m at last starting to get generating functions.

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