Euler function

The coefficient $p(k)$ in the formal power series expansion for $1/\backslash phi(q)$ gives the number of partitions of k. That is,

:$\backslash frac\{1\}\{\backslash phi(q)\}=\backslash sum\_\{k=0\}^\backslash infty\; p(k)\; q^k$

where $p$ is the partition function.

The Euler identity, also known as the Pentagonal number theorem, is

:$\backslash phi(q)=\backslash sum\_\{n=-\backslash infty\}^\backslash infty\; (-1)^n\; q^\{(3n^2-n)/2\}.$

$(3n^2-n)/2$ is a pentagonal number.

The Euler function is related to the Dedekind eta function as

:$\backslash phi\; (e^\{2\backslash pi\; i\backslash tau\})=\; e^\{-\backslash pi\; i\backslash tau/12\}\; \backslash eta(\backslash tau).$

The Euler function may be expressed as a q-Pochhammer symbol:

:$\backslash phi(q)\; =\; (q;q)\_\{\backslash infty\}.$

The logarithm of the Euler function is the sum of the logarithms in the product expression, each of which may be expanded about q = 0, yielding

:$\backslash ln(\backslash phi(q))\; =\; -\backslash sum\_\{n=1\}^\backslash infty\backslash frac\{1\}\{n\}\backslash ,\backslash frac\{q^n\}\{1-q^n\},$

which is a Lambert series with coefficients -1/n. The logarithm of the Euler function may therefore be expressed as

:$\backslash ln(\backslash phi(q))\; =\; \backslash sum\_\{n=1\}^\backslash infty\; b\_n\; q^n$

where $b\_n=-\backslash sum\_\{d|n\}\backslash frac\{1\}\{d\}=$ -[1/1, 3/2, 4/3, 7/4, 6/5, 12/6, 8/7, 15/8, 13/9, 18/10, ...] (see OEIS [http://oeis.org/A000203/table A000203])

On account of the identity $\backslash sigma(n)\; =\; \backslash sum\_\{d|n\}\; d\; =\; \backslash sum\_\{d|n\}\; \backslash frac\{n\}\{d\}$ , where $\backslash sigma(n)$ is the sum-of-divisors function, this may also be written as

:$\backslash ln(\backslash phi(q))\; =\; -\backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\{\backslash sigma(n)\}\{n\}\backslash \; q^n$.

Also if $a,b\backslash in\backslash mathbb\{R\}^+$ and $ab=\backslash pi\; ^2$, thenBerndt, B. et al. "The Rogers–Ramanujan Continued Fraction"

:$a^\{1/4\}e^\{-a/12\}\backslash phi\; (e^\{-2a\})=b^\{1/4\}e^\{-b/12\}\backslash phi\; (e^\{-2b\}).$

The next identities come from Ramanujan's Notebooks:{{Cite book |last1=Berndt |first1=Bruce C. |title=Ramanujan's Notebooks Part V |publisher=Springer |year=1998 |isbn=978-1-4612-7221-2}} p. 326

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

: $\backslash phi(e^\{-2\backslash pi\})=\backslash frac\{e^\{\backslash pi/12\}\backslash Gamma\backslash left(\backslash frac14\backslash right)\}\{2\backslash pi^\{3/4\}\}$

: $\backslash phi(e^\{-4\backslash pi\})=\backslash frac\{e^\{\backslash pi/6\}\backslash Gamma\backslash left(\backslash frac14\backslash right)\}\{2^\{\{11\}/8\}\backslash pi^\{3/4\}\}$

: $\backslash phi(e^\{-8\backslash pi\})=\backslash frac\{e^\{\backslash pi/3\}\backslash Gamma\backslash left(\backslash frac14\backslash right)\}\{2^\{29/16\}\backslash pi^\{3/4\}\}(\backslash sqrt\{2\}-1)^\{1/4\}$

Using the Pentagonal number theorem, exchanging sum and integral, and then invoking complex-analytic methods, one derives{{Cite OEIS|A258232}}

: $\backslash int\_0^1\backslash phi(q)\backslash ,\backslash mathrm\{d\}q\; =\; \backslash frac\{8\; \backslash sqrt\{\backslash frac\{3\}\{23\}\}\; \backslash pi\; \backslash sinh\; \backslash left(\backslash frac\{\backslash sqrt\{23\}\; \backslash pi\; \}\{6\}\backslash right)\}\{2\; \backslash cosh\; \backslash left(\backslash frac\{\backslash sqrt\{23\}\; \backslash pi\; \}\{3\}\backslash right)-1\}.$

{{reflist}}

• {{Apostol IANT}}

{{Leonhard Euler}}

Category:Number theory

Category:Q-analogs

Category:Leonhard Euler