I attempted to solve the equation $x^5-x+41=0$ using the rational zero theorem. I soon found out that the equation has no rational zeros. Many sources say when presented with a polynomial with no rational zeros, use the Newton approximation method. I didn't want an approximation of the polynomial. Is there a way to solve $x^5-x+41=0$ without approximating? I am aware of the Abel-Ruffini theorem which states that there is no solution in radicals to general polynomial equations of degree five or higher with arbitrary coefficients. Is there possibly a polar or trigonometric solution to this equation?
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root is negative if a real root exists. – Oct 06 '19 at 16:08
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1Solution with integration from post – Dmitry Ezhov Oct 06 '19 at 17:58
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Watch in awe: https://math.stackexchange.com/q/540964/96384 (Notice that your polynomial is already in Bring-Jerrard form.) Also compare https://math.stackexchange.com/q/291909/96384. – Torsten Schoeneberg Oct 07 '19 at 05:49
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@RoddyMacPhee: Every odd-degree real polynomial has a real root, as per the Intermediate Value Theorem. – Torsten Schoeneberg Oct 07 '19 at 06:00
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There is a solution using generalized hypergeometric series.
Writing $x = -41^{1/5} t$, the equation becomes $$ t^5 - s t - 1 = 0 \tag 1$$ where $s = 41^{-4/5}$. There is a series for the root of (1) near $t=1$ in powers of $s$:
$$ \eqalign{t &= \sum _{k=0}^{\infty }{\frac {{3125}^{-k}{16}^{k}\Gamma \left( 2\,k-1/ 10 \right) \Gamma \left( 2\,k+2/5 \right) \Gamma \left( 4/5 \right) \Gamma \left( 3/5 \right) {s}^{5\,k}}{\Gamma \left( -1/10 \right) \Gamma \left( 4/5+k \right) \Gamma \left( 3/5+k \right) \Gamma \left( 2/5+k \right) k!}}\cr &+1/5\,{\frac {{3125}^{-k}{16}^{k}\Gamma \left( 2\,k+ 4/5 \right) \Gamma \left( 2\,k+3/10 \right) \Gamma \left( 6/5 \right) \Gamma \left( 3/5 \right) {s}^{5\,k+1}}{\Gamma \left( 3/10 \right) \Gamma \left( k+6/5 \right) \Gamma \left( 4/5+k \right) \Gamma \left( 3/5+k \right) k!}}\cr &-1/25\,{\frac {{3125}^{-k}{16}^{k}\Gamma \left( 2\,k +6/5 \right) \Gamma \left( 7/5 \right) \Gamma \left( 4/5 \right) {s}^{ 5\,k+2}}{\Gamma \left( {\frac{7}{10}} \right) \Gamma \left( k+7/5 \right) \Gamma \left( k+6/5 \right) \Gamma \left( 4/5+k \right) k!} \Gamma \left( 2\,k+{\frac{7}{10}} \right) }\cr &+{\frac {{3125}^{-k}{16}^{k }\Gamma \left( 2\,k+8/5 \right) \Gamma \left( 7/5 \right) \Gamma \left( 6/5 \right) {s}^{5\,k+3}}{125\,\Gamma \left( {\frac{11}{10}} \right) \Gamma \left( k+8/5 \right) \Gamma \left( k+7/5 \right) \Gamma \left( k+6/5 \right) k!}\Gamma \left( 2\,k+{\frac{11}{10}} \right) }} $$
Robert Israel
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Can you elaborate more on how to utilize the hypergeometric series? I am new to this. – VINCENT ZHANG Oct 07 '19 at 01:10