Questionc. Justify the statement: $3\left(\frac{n}{4}\right)^{2}+5 n^{2} \in O\left(n^{2}\right)$ .

Studdy Solution

STEP 1

1. We are given the function $f(n) = 3\left(\frac{n}{4}\right)^{2} + 5n^{2}$ .
2. We need to justify that $f(n) \in O(n^2)$ .
3. Big O notation $O(g(n))$ describes an upper bound on the growth rate of a function.

STEP 2

1. Simplify the expression $f(n)$ .
2. Identify the dominant term in $f(n)$ .
3. Use the definition of Big O notation to justify the statement.

STEP 3

Simplify the expression $f(n) = 3\left(\frac{n}{4}\right)^{2} + 5n^{2}$ .
$f(n) = 3\left(\frac{n^2}{16}\right) + 5n^2$ $f(n) = \frac{3n^2}{16} + 5n^2$

STEP 4

Identify the dominant term in $f(n)$ .
The expression $f(n) = \frac{3n^2}{16} + 5n^2$ can be rewritten as:
$f(n) = \left(\frac{3}{16} + 5\right)n^2$ $f(n) = \left(\frac{3}{16} + \frac{80}{16}\right)n^2$ $f(n) = \frac{83}{16}n^2$
The dominant term is $n^2$ .

STEP 5

Use the definition of Big O notation to justify the statement.
By definition, $f(n) \in O(n^2)$ if there exist positive constants $c$ and $n_0$ such that for all $n \geq n_0$ , the inequality $f(n) \leq c \cdot n^2$ holds.
From the previous step, we have:
$f(n) = \frac{83}{16}n^2$
We can choose $c = \frac{83}{16}$ and $n_0 = 1$ (or any positive integer), then for all $n \geq n_0$ :
$f(n) = \frac{83}{16}n^2 \leq \frac{83}{16}n^2$
Thus, $f(n) \in O(n^2)$ .
The statement is justified: $3\left(\frac{n}{4}\right)^{2} + 5n^{2} \in O(n^2)$ .

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