Define a function f: Z* to Z* by the rule: for each integer n, f(n) = the sum of the positive divisors of n. This function is O None of the choices onto only one-to-one only neither one-to-one nor onto one-to-one and onto
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- Define a function S : Z+ → Z+ as follows. For each positive integer n, S(n) =the sum of the positive divisors of n. S (7) ?The question describes a function S(k) which is defined as the sum of the positive divisors of a positive integer k, minus k itself. The function S(1) is defined as 1, and for any positive integer k greater than 1, S(k) is calculated as S(k) = σ(k) - k, where σ(k) is the sum of all positive divisors of k. Some examples of S(k) are given: S(1) = 1 S(2) = 1 S(3) = 1 S(4) = 3 S(5) = 1 S(6) = 6 S(7) = 1 S(8) = 7 S(9) = 4 The question then introduces a recursive sequence a_n with the following rules: a_1 = 12 For n ≥ 2, a_n = S(a_(n-1)) Part (a) of the question asks to calculate the values of a_2, a_3, a_4, a_5, a_6, a_7, and a_8 for the sequence. Part (b) modifies the sequence to start with a_1 = k, where k is any positive integer, and the same recursion formula applies: for n ≥ 2, a_n = S(a_(n-1)). The question notes that for many choices of k, the sequence a_n will eventually reach and remain at 1, but this is not always the case. It asks to find, with an explanation, two specific…Define upper bound. If f(n) = 3n * 5 for what values of C and n this function is said to be upper bounded.
- Given g = {(1,c),(2,a),(3,d)}, a function from X = {1,2,3} to Y = {a,b,c,d}, and f = {(a,r),(b,p),(c,δ),(d,r)}, a function from Y to Z = {p, β, r, δ}, write f o g as a set of ordered pairs.Give an example of a function from Z to N that is one-to-one, but not onto.Consider the function f : N × N → N given byf(m, n) = 2m-1(2n − 1), (m, n) ∈ N × NShow that f is bijective
- Determine whether the proposed definition isa valid recursive definition of a function f from the setof nonnegative integers to the set of integers. If f is welldefined, find a formula for f(n) when n is a nonnegativeinteger and prove that your formula is valid. f(0) = 1, f(n) = −f(n − 1) for n ≥ 1How do we define that a function f(n) has an upper bound g(n), i.e., f(n) ∈ O(g(n))?The sequence of values: n, c(n), c(c(n)), c(c(c(n))), c(c(c(c(n)))),... is known as the hailstone sequence starting at n. Implement c(n) as a function. Use predicates to test the parity of each input n.
- Simplify the complement of the following function: F(A,B,C,D)=(0,2,4,5,8,9,10,11) Your answer: F=((A'B'D)' (BC)'(AB)')' F=((A'BD)'(BC)'(AB)')' F=((A'B'D)'(B'C)'(AB)') F=((A'B'D')' (BC)'(AB)')A common way of implementing a max function is to look at the sign of a - b. In this case, we can't use a comparison operator on this sign, but we can use multiplication. Let k equal the sign of a - b such that if a - b >= 0, then k is 1. Else, k 0. Let q be the inverse of k.n implement the code:Determine whether f is a function from Z to R if a)f(n)=±n. b)f(n)=square root n^2+1. c)f(n)=1/(n^2-4)