Question

n? - n² - 8 = w(n²) n? = ? 0 ? c.n² < n? - n² - 8 n ? n? cn² < n? - n² - 8 n? - (c+1)n² - 8 > 0 \frac{n?}{2} - (c+1)n² + \frac{n?}{2} - 8 > 0 \frac{n²}{2}(n² - 2(c+1)) + \frac{1}{2}(n? - 16) > 0 + + n ? \sqrt{2(c+1)} n ? 2 n? = max(\sqrt{2(c+1)}, 2) + 1 OR: n? = max(\sqrt{2(c+1)}, 3)

          n? - n² - 8 = w(n²)
n? = ?  0 ? c.n² < n? - n² - 8  n ? n?

cn² < n? - n² - 8
n? - (c+1)n² - 8 > 0
\frac{n?}{2} - (c+1)n² + \frac{n?}{2} - 8 > 0
\frac{n²}{2}(n² - 2(c+1)) + \frac{1}{2}(n? - 16) > 0
+
+
n ? \sqrt{2(c+1)}
n ? 2
n? = max(\sqrt{2(c+1)}, 2) + 1
OR: n? = max(\sqrt{2(c+1)}, 3)

n? - n² - 8 = w(n²)
n? = ? 0 ? c.n² < n? - n² - 8 n ? n?

cn² < n? - n² - 8
n? - (c+1)n² - 8 > 0
(n?)/(2) - (c+1)n² + (n?)/(2) - 8 > 0
(n²)/(2)(n² - 2(c+1)) + (1)/(2)(n? - 16) > 0
+
+
n ? √(2(c+1))
n ? 2
n? = max(√(2(c+1)), 2) + 1
OR: n? = max(√(2(c+1)), 3)

Added by Michael W.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert Varsha Aggarwal

07/02/2022

Step 1

Step 1: Start with the formal definition of ω(n^2): f(n) = ω(g(n)) if for every positive constant c, there exists a positive constant n0 such that f(n) > c * g(n) for all n > n0. Show more…

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Texts: Use the formal definition to show that n^3 + 3n^2 + 1 = ω(n^2). Please solve the question like the example above. n - n - 8 = ω(n^2) nzno n - (c + i)n - 8 > 0 n - (c + i)n^2 - 8 > 0 n(n - 2(c + i) + (n^16)70 2 + + n > 2 n7V2(c + i) no = max (√(c + 1)^2 + 1) OR: no = max(2(c + 1)^3)