Approximate solution for high order polynomial (order 12)

Question

I'm trying to get an approximation value of y by x from the following equation

$\ x = $$\sum_{i=1}^{12} y^i$

The current suggestion is to take y=1+z and z tending to zero or y tending to 1

any ideas

Answer 1

This is mechanized in Maple. Here is an example.

sol := allvalues(solve(x = add(y^i, i = 1 .. 12), y)):series(sol[2], x = 0, 2);

$$ 1/2\sqrt {2+2i\sqrt {3}}+$$ $$512x\times $$ $$ \left( 3\, \left( 2+2\,i\sqrt {3} \right) ^{11/2}-3072\,i\sqrt {3}+10 \, \left( 2+2\,i\sqrt {3} \right) ^{9/2}-3072+32\, \left( 2+2\,i\sqrt {3} \right) ^{7/2}+96\, \left( 2+2\,i\sqrt {3} \right) ^{5/2}+256\, \left( 2+2\,i\sqrt {3} \right) ^{3/2}+512\,\sqrt {2+2\,i\sqrt {3}} \right) ^{-1}+ $$ $$O(x^2) $$