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Fixed arithmetic error in CLRS 7.4-3 #39

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Fixed arithmetic error in CLRS 7.4-3 #39

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3 changes: 2 additions & 1 deletion other/clrs/07/04/03.markdown
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Original file line number Diff line number Diff line change
@@ -1,13 +1,14 @@
> Show that the expression $q^2 + (n - q - 1)^2$ achieves a maximum over $q =
> 0, 1, \ldots, n-1$ when $q = 0$ and $q = n - 1$.


$$ \begin{align}
f(q) &= q^2 + (n - q - 1)^2 \\\\
f'(q) &= 2q - 2(n - q - 1) = 4q - 2n + 2 \\\\
f''(q) &= 4 \\\\
\end{align} $$

$f'(q) = 0$ when $q = \frac{1}{2}n - \frac{1}{4}$. $f'(q)$ is also continious.
$f'(q) = 0$ when $q = \frac{1}{2}n - \frac{1}{2}$. $f'(q)$ is also continuous.
$\forall q : f''(q) > 0$, which means that $f'(q)$ is negative left of $f'(q) =
0$ and positive right of it, which means that this is a local minima. In this
case, $f(q)$ is decreasing in the beginning of the interval and increasing in
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