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Mathoverflow.net news digest

  • 0 days

    Reciprocal of q-binomial theorem

    The $q$-binomial theorem states$$\frac{(az;q)_\infty}{(z;q)_\infty}=\sum_{n\geq 0}\frac{(a;q)_\infty}{(q;q)_\infty}z^n,$$where $(a;q)_n=\prod_{r=0}^{n-1}(1-aq^r)$ is the $q$-Pochhammer symbol.
    Is there a closed-form equation for its reciprocal? That is, a formula for the coefficients $c_n$ in$$\frac{(z;q)_\infty}{(az;q)_\infty}=\sum_{n\geq 0}c_nz^n?$$...

  • 0 days

    Does off-diagonal trace randomness force a $t$-design?

    Let $d\ge 2$ and let $\mathcal U=\{U_1,\dots,U_K\}\subset U(d)$ be a finite set of unitaries. For an integer $t\ge 1$ consider the $t$-th moment operator$$M_t := \frac{1}{K}\sum_{a=1}^K U_a^{\otimes t}\otimes \overline{U_a}^{\otimes t}\in \mathrm{End...

  • 0 days

    How does the Galois groups of Taylor polynomials of entire functions relate to t...

    How does the Galois groups of Taylor polynomials of entire functions relate to the Weierstrass factorization?
    Setup: Let $f$ be a non-polynomial entire function with Weierstrass factorization:$$f(z)=z^m e^{g(z)}\prod_{n=1}^\infty E_{p_n}\!\left(\frac{z}{a_n}\right)$$where $$E_k(u)=(1-u)\exp\!\left(\sum_{j=1}^k u^j/j\right).$$ Define $\hat f(z)=\frac{f(z)}{e...

  • 6 months

    Do all commutative semirings satisfy the strong rank condition?

    It is well known that all nonzero commutative rings satisfy the strong rank condition (and in fact also the potentially stronger Orzech property; whether they are equivalent for rings is the subject of another question). Recently, Thomas Browning verified...

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