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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

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  • $\begingroup$ If $y$ tends to $1$, then $x$ will tend to $12$. $\endgroup$ – Théophile Apr 30 '19 at 16:18
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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);

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

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  • $\begingroup$ That is great, Thanks a lot! $\endgroup$ – amn89 Apr 30 '19 at 18:35

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