summation problem.

by hemangsarkar, Sep 9, 2012, 4:57 PM

Q) find the number of real roots of
$\sum_{r=1}^{2014} (x-r)^3 = 0$

my solution :

let $f(x) = \sum_{r=1}^{2014} (x-r)^3$

$f'(x) = 3\sum_{r=1}^{2014} (x-r)^2 > 0$

so, $f(x)$ is an monotonically increasing cubic function with no critical points.
hence only one real root.

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The proof is only complete when you add $f\left(-\infty) <0$ and $f\left(\infty) >0 $

by boywholived, Sep 12, 2012, 2:14 PM

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summation problem.

by hemangsarkar, Sep 9, 2012, 4:57 PM

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