I've been given a Polynomial (Cubic)
$$k=\frac16x\cdot(x+1)\cdot(2x+1)$$
If $k$ is given, is there any way to solve for $x$?
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If $k=0$ then $x$ is one of: $0,-1,-\frac12$, for other values it's a bit more complicated – Alice Ryhl Sep 21 '14 at 14:30
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1The shape (sum of squares formula) makes one guess that $x$ is an integer. If so, then $x=\left\lfloor (3k)^{1/3}\right\rfloor$. – André Nicolas Sep 21 '14 at 14:30
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Search for "roots of cubic polynomial". There are formulas, but they are big and ugly. – bubba Sep 21 '14 at 14:32
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@AndréNicolas can you please elaborate – Sep 21 '14 at 14:32
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http://en.wikipedia.org/wiki/Cubic_function – bubba Sep 21 '14 at 14:34
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@AndréNicolas That's a weird result, how woudl you justify it? – Alice Ryhl Sep 21 '14 at 14:35
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I am guessing you are told that the sum of the first $x$ squares is $k$ and you want to find $x$. Then from the formula, we have $2x^3\lt 6k$ and $2(x+1)^3\gt 6k$. – André Nicolas Sep 21 '14 at 14:36
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@AndréNicolas Yes, that does look like sum of squares, But your method doesn't always seem to work. – Sep 21 '14 at 14:39
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It is meant to work only when $x$ is a positive integers, so for $k=1,5,14,30,55$ and so on. – André Nicolas Sep 21 '14 at 14:47
2 Answers
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We solve the problem in a very special case only, where $x$ happens to be a positive integer.
Since $\frac{x(x+1)(2x+1)}{6}\gt \frac{x^3}{3}$, we have $x^3\gt 3k$, and therefore $x\gt (3k)^{1/3}$.
Note that $2(x+1)^3=2x^3+6x^2+6x+2$ while $x(x+1)(2x+1)=2x^3+3x^2+x$. It follows that if $x$ is a positive integer then $\frac{x(x+1)(2x+1)}{6}\lt \frac{(x+1)^3}{3}$, and therefore $x+1\gt (3k)^{1/3}$.
It follows that $x=\lfloor (3k)^{1/3}\rfloor$.
André Nicolas
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In the case where $3888 k^2-1 \gt 0$,that is to say $| k| \gt \frac{1}{36 \sqrt{3}}$ and $k>0$, the only real solution is given by $$x=\frac{1}{6} \left(\sqrt[3]{3 \sqrt{11664 k^2-3}+324 k}+\frac{3^{2/3}}{\sqrt[3]{\sqrt{11664 k^2-3}+108 k}}-3\right)$$ This is Cardano formula applied to your equation.
Now, if $k$ is large, an asymptotic expansion of the solution is given by $$x=\sqrt[3]{3k} -\frac{1}{2}+\frac{\sqrt[3]{\frac{1}{k}}}{12 \sqrt[3]{3}}+O\left(\left(\frac{1}{k}\right)^{4/3}\right)$$ Compare to André Nicolas's answer and comments.
Claude Leibovici
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