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$t^n=a$, we get one solution to the equation: $$t=e^{\frac{1}{n}\int^a_1 \frac{1}{x}}$$, generalizing this result by replacing the exponential with an elliptic modular function and the integral with hyperelliptic integrals, we can get a solution to an algebraic equation $a_0+a_1x+a_2x^2+\cdots+a_nx^n=0$ with degree above 5 by formulation of modular function and hyperelliptic integral(both formulated by Siegel Theta functions): $$x=\frac{\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4}{2\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4}+\frac{\theta\left( \begin{array}{cccccc} 0 & 0 & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} 0 & \frac{1}{2} & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4}{2\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4}-\frac{\theta\left( \begin{array}{cccccc} 0 & 0 & \cdots & 0 \\ \frac{1}{2} & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} 0 & \frac{1}{2} & \cdots & 0 \\ \frac{1}{2} & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4}{2\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4}$$

or

$$x=\frac{1}{2}+\frac{\theta\left( \begin{array}{cccccc} 0 & 0 & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} 0 & \frac{1}{2} & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4}{2\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4}-\frac{\theta\left( \begin{array}{cccccc} 0 & 0 & \cdots & 0 \\ \frac{1}{2} & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} 0 & \frac{1}{2} & \cdots & 0 \\ \frac{1}{2} & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4}{2\theta\left( \begin{array}{cccccc} \frac{1}{2} & 0 & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4\theta\left( \begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & \cdots & 0 \\ 0 & 0 &\cdots & 0 \\ \end{array} \right)(\Omega)^4}$$ where $\Omega$ is the period matrix of the hyperelliptic curve $\mathbb{C}$, please see David Mumford Tate lecture on Theta $\textrm{II}$ Jacobian page 266 for more detail.

Now let us extend this result to algebraic equation with coefficients of $p_i(x) \in Q[x]$ where $Q[x]$ is a ring.

$$p_0(x) + p_1(x) \cdot y + p_2(x)\cdot y^2 + \cdots + p_n(x)\cdot y^n $$ is an algebraic polynomial where $p_i(x)$ are the polynomials with rational coefficients. When $$p_0(x) + p_1(x) \cdot y + p_2(x)\cdot y^2 + \cdots + p_n(x)\cdot y^n =0$$, we have solution to the equation in which $y$ is formulated by modular function and hyperelliptic integrals with $x$ as variable, like $y= \phi(x)$, in another word, $$p_0(x) + p_1(x) \cdot \phi(x) + p_2(x)\cdot \phi(x)^2 + \cdots + p_n(x)\cdot \phi(x)^n =0$$

My question is when $y$ is expanded as power series (Taylor expansion) in $x$,as $$y=\sum_0^{\infty }b_i x^i$$, or $$\phi(x) = \sum_0^{\infty }b_i x^i$$, under what condition ( formulated by modular function and hyperelliptic integrals ) can we have $b_i\in \mathbb{N}\bigcup 0$ ?

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    $\begingroup$ Do you understand that in order to even begin to understand what the Tata Lectures on Theta talk about, leave alone what's going on on page 266 of the volume II of that book, has must already know a whole lot of very, very advanced stuff before that? Meaning: your question is a Research Level one, more fit for Overflow than for this site. It also looks like a question that requires lots of background for anyone to get interested in it, so you'll probably need to write a lot more than what you did to explain yourself. $\endgroup$
    – DonAntonio
    Jan 8, 2018 at 8:54
  • $\begingroup$ @DonAntonio. I don't know, I just think it is a hard question, thank you for your comment. $\endgroup$ Jan 8, 2018 at 9:03
  • $\begingroup$ Now posted to MO, mathoverflow.net/questions/290251/… $\endgroup$ Jan 9, 2018 at 2:24

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