Suppose f :N→N where f(n) ＝ 4n²＋1 find f(s) if s is ｛0,2,5,10｝

We can simply substitute each value in the set {0, 2, 5, 10} into the function f(n) = 4n^2 + 1 to find the corresponding values of f(s):

f(0) = 4(0)^2 + 1 = 1
f(2) = 4(2)^2 + 1 = 33
f(5) = 4(5)^2 + 1 = 101
f(10) = 4(10)^2 + 1 = 401

Therefore, f(s) for s = {0, 2, 5, 10} is {1, 33, 101, 401}.

相关问题

Suppose f :N→N where f(n) ＝ 4n²＋1 is f one-to-one?(explain)

To determine if f is one-to-one, we need to show that for any distinct inputs n₁ and n₂ in the domain N, their corresponding outputs f(n₁) and f(n₂) are also distinct.

Let's assume that there exist two distinct inputs n₁ and n₂ in N such that f(n₁) = f(n₂). Then we have:

4n₁² + 1 = 4n₂² + 1

4n₁² = 4n₂²

n₁² = n₂²

Taking the square root of both sides, we get:

n₁ = ±n₂

Since n₁ and n₂ are distinct, we must have n₁ ≠ n₂. Therefore, the only possibility is that n₁ = -n₂. However, this contradicts the fact that n₁ and n₂ are both in the domain N, which is the set of positive integers. Thus, it is impossible for f(n₁) to be equal to f(n₂) if n₁ and n₂ are distinct, and we can conclude that f is one-to-one.

Therefore, f(n) = 4n² + 1 is one-to-one on the domain N.

Let X, Y, Z be any three nonempty sets and let g : Y → Z be any function. Define the function Lg : Y X → Z X (Lg, as a reminder that we compose with g on the left), by Lg(f) = g ◦f for every function f : X → Y . (i) (2 pts) Show that if Y = Z and g = idY , then LidY (f) = f for all f : X → Y . (ii) (2 pts) Let T be another nonempty set and let h : Z → T be any function. Show that Lh◦g = Lh ◦ Lg. (iii) (2 pts) Show that if g : Y → Z is injective, then Lg : Y X → Z X is also injective. (iv) (2 pts) Show that if g : Y → Z is surjective, then Lg : Y X → Z X is also surjective.

I can answer this question.

(i) If Y = Z and g = idY, then LidY(f) = g ◦ f = idY ◦ f = f for all f : X → Y.

(ii) Let f : X → Y. Then Lh◦g(f) = h ◦ g ◦ f = Lh(Lg(f)) = Lh ◦ Lg(f) for all f : X → Y.

(iii) Let f1, f2 : X → Y. Suppose Lg(f1) = Lg(f2). Then g ◦ f1 = g ◦ f2. Since g is injective, it follows that f1 = f2. Therefore, Lg is injective.

(iv) Let h : Z → Y be a function such that g ◦ h = idZ. Let f : Y → X be any function. Then Lg(h ◦ f) = g ◦ (h ◦ f) = (g ◦ h) ◦ f = idZ ◦ f = f. Therefore, Lg is surjective.