2
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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 $p_i(x) \in \mathbb{Q}[x]$ ,in other word, $p_i(x)$ is polynomial with coefficients of rational numbers.

$$p_0(x) + p_1(x) \cdot y + p_2(x)\cdot y^2 + \cdots + p_n(x)\cdot y^n $$ 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 $b_i\in \mathbb{N}\bigcup 0$, or can we find the condition under which the Taylor expansion of the function is series with coefficients of natural number?

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  • $\begingroup$ It is from math.stackexchange.com/posts/2596567, since some body there say it is fit in this site. $\endgroup$ – XL _At_Here_There Jan 9 '18 at 1:19
  • $\begingroup$ That link doesn't work for me. Maybe math.stackexchange.com/questions/2596567/… is better. $\endgroup$ – Gerry Myerson Jan 9 '18 at 2:24
  • $\begingroup$ I don't understand the big equation you wrote. You don't need elliptic functions to solve quartics. In your deleted question I was mentioning the fact $x \mapsto x_2 = \wp_\tau(\frac{m}{n}\wp_\tau^{-1}(x))$ is $\in \overline{\mathbb{C}(x)}$ such that $[m](x,f(x)^{1/2})=\pm [n](x_2,f(x_2)^{1/2})$ with $[n]$ the multiplication by $n$ in the elliptic curve $y^2 = f(x),f(x)=4x^3-g_2(\tau) x-g_3(\tau)$ and $\wp_\tau^{-1}(x) = \int_\infty^x \frac{dt}{f(t)^{1/2}}dt$. $\endgroup$ – reuns Jan 10 '18 at 22:15
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    $\begingroup$ @reuns why do you think I ask about quartic? $\endgroup$ – XL _At_Here_There Jan 10 '18 at 23:13
  • $\begingroup$ @reuns I almost know the answer, but I am sure the exact formal answer, lets wait for someone to answer, possibly we get a beautiful formula or answer. $\endgroup$ – XL _At_Here_There Jan 11 '18 at 0:44

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