# Homework Solution: Can someone help with number 8?…

Can someone help with number 8?
For each assertion in 7(a)-(c), prove the assertion directly from the definition of the big-O asymptotic notation if it is true by finding values for the constants c and n_0. On the other hand, if the assertion is false, give a counter-example. Then answer the question in 7(d). F denotes the set of all functions from Z^+ to R^+. (a) Let f(n): Z^+ rightarrow R^+. A relation on a set is reflexive if each element is related to itself. The relation "is big-O of" is reflexive over F, In other words, f(n) elementof O(f(n)). (b) Let f(n): Z^+ rightarrow R^+ and g(n): Z^+ rightarrow R^+. A relation on a set is antisymmetric if whenever an element X is related to an element Y and Y is related X, then X = Y. The relation "is big-O of" is antisymmetric over F. In other words, if f(n) elementof O(g(n)) and g(n) elementof O(f(n)), then f(n) = g(n). (c) Let e(n): Z^+ rightarrow R^+, f(n): Z^+ rightarrow R^+ and g(n): Z^+ rightarrow R^+. A relation on a set is transitive if whenever an element X is related to Y and Y is related Z, then X is related to Z. The relation "is big-O of" is transitive over F. In other words, if e(n) elementof O(f(n)) and f(n) elementof O(g(n)), then e(n) elementof O(g(n)). (d) Is "is big-O of" a partial order on F? A relation is a partial order on a set if it is reflexive, antisymmetric and transitive. Given an infinite series s = sigma^infinity_n = 1 f(n), where f(n) is a continuous positive monotonically decreasing function that converges and n elementof Z^+, s can be bounded, using improper integrals, as follows: sigma^k_n = 1 f(n) + integral^infinity_k + 1 f(n) dn lessthanorequalto s lessthanorequalto sigma^k_n = 1 f(n) + integral^infinity_k f(n) dn, k elementof Z^+ Using the inequality in (1) and k = 5, prove that sigma^infinity_n = 1 1/n^3 elementof theta (1).

$\int_{5}^{\infty} \frac{1}{n^3} dn$ converges to a con

Can someone succor with calculate 8?

Coercion each assumption in 7(a)-(c), examine the assumption quickly from the determination of the big-O asymptotic notation if it is gentleman by finding values coercion the fixeds c and n_0. On the other workman, if the assumption is erroneous, impart a counter-example. Then counter-argument the inquiry in 7(d). F denotes the firm of every businesss from Z^+ to R^+. (a) Allow f(n): Z^+ rightarrow R^+. A ratio on a firm is mutual if each part is kindred to itself. The ratio “is big-O of” is mutual balance F, In other language, f(n) partof O(f(n)). (b) Allow f(n): Z^+ rightarrow R^+ and g(n): Z^+ rightarrow R^+. A ratio on a firm is antisymmetric if whenever an part X is kindred to an part Y and Y is kindred X, then X = Y. The ratio “is big-O of” is antisymmetric balance F. In other language, if f(n) partof O(g(n)) and g(n) partof O(f(n)), then f(n) = g(n). (c) Allow e(n): Z^+ rightarrow R^+, f(n): Z^+ rightarrow R^+ and g(n): Z^+ rightarrow R^+. A ratio on a firm is projective if whenever an part X is kindred to Y and Y is kindred Z, then X is kindred to Z. The ratio “is big-O of” is projective balance F. In other language, if e(n) partof O(f(n)) and f(n) partof O(g(n)), then e(n) partof O(g(n)). (d) Is “is big-O of” a particular arrange on F? A ratio is a particular arrange on a firm if it is mutual, antisymmetric and projective. Impartn an infinite course s = sigma^infinity_n = 1 f(n), where f(n) is a consistent fixed monotonically decreasing business that converges and n partof Z^+, s can be terminable, using improper undivideds, as follows: sigma^k_n = 1 f(n) + undivided^infinity_k + 1 f(n) dn lessthanorequalto s lessthanorequalto sigma^k_n = 1 f(n) + undivided^infinity_k f(n) dn, k partof Z^+ Using the dissimilarity in (1) and k = 5, examine that sigma^infinity_n = 1 1/n^3 partof theta (1).

## Expert Counter-argument

$\int_{5}^{\infty} \frac{1}{n^3} dn$ converges to a fixed $\frac{-1}{2*n^2}$ is undivided of this and then we can lawful set-down values and allure acquire a fixed coercion it, allow this ocnstant be k

So

$s =\sum_{n=1}^5 \frac{1}{n^3} + k$

$\sum_{n=1}^5 \frac{1}{n^3} + k$

so

s = fixed = $\theta(1)$