How do you find $h (\frac{n}{2})$ $h(n/2)$ given $h (n) = - 2 n^{3} - 5 n$ $h(n)=-2n^3-5n$ ?

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Answer 1 · 2017-04-28T11:01:34

$h (\frac{n}{2}) = - \frac{n^{3}}{4} - \frac{5 n}{2}$ $h (n/2) = -n^3/4- (5n)/2$

$h (n) = - 2 n^{3} - 5 n$ $h (n) = -2n^3 - 5n$

$h (\frac{n}{2}) = - 2 {(\frac{n}{2})}^{3} - 5 (\frac{n}{2})$ $h (n/2) = -2(n/2)^3 - 5(n/2)$ or

$h (\frac{n}{2}) = - 2 \cdot \frac{n^{3}}{8} - 5 \cdot \frac{n}{2}$ $h (n/2) = -2*n^3/8- 5*n/2$ or

$h (\frac{n}{2}) = - \frac{n^{3}}{4} - \frac{5 n}{2}$ $h (n/2) = -n^3/4- (5n)/2$ [Ans]

How do you find h(n2)h(n/2) given h(n)=−2n3−5nh(n)=-2n^3-5n?