Problem 9 - A Computational Introduction to Number Theory and Algebra

Question

Show that: (a) the relation " \(\sim\) " is an equivalence relation on the set of eventually positive functions; (b) for all eventually positive functions \(f_{1}, f_{2}, g_{1}, g_{2},\) if \(f_{1} \sim g_{1}\) and \(f_{2} \sim g_{2},\) then \(f_{1} \star f_{2} \sim g_{1} \star g_{2},\) where " \(\star\) " denotes addition, multiplication, or division; (c) for all eventually positive functions \(f, g,\) and every \(\alpha>0,\) if \(f \sim g,\) then \(f^{\alpha} \sim g^{\alpha}\) (d) for all eventually positive functions \(f, g,\) and every function \(h\) such that \(h(x) \rightarrow \infty\) as \(x \rightarrow \infty,\) if \(f \sim g,\) then \(f \circ h \sim g \circ h,\) where "o" denotes function composition.

Step-by-Step Solution

Answer

Question: Show that the relation "∼" is an equivalence relation on the set of eventually positive functions, and prove the following properties: 1. If f1 ∼ g1 and f2 ∼ g2, then f1 * f2 ∼ g1 * g2, where * denotes one of the three basic arithmetic operations. 2. If f ∼ g and α > 0, then f^α ∼ g^α. 3. If f ∼ g and h(x) → ∞ as x → ∞, then f ∘ h ∼ g ∘ h (where ∘ denotes function composition). Answer: To prove that the relation "∼" is an equivalence relation, we need to show it is reflexive, symmetric, and transitive: 1. Reflexivity is proven by showing that lim(x→∞)(f(x)/f(x)) = 1 for all eventually positive functions f. 2. Symmetry is proven by showing that if f ∼ g, then lim(x→∞)(g(x)/f(x))=1. 3. Transitivity is proven by showing that if f ∼ g and g ∼ h, then lim(x→∞)(f(x)/h(x))=1. For the other properties: 1. To prove the preservation of arithmetic operations (addition, multiplication, and division), we need to show that if f1 ∼ g1 and f2 ∼ g2, then (f1 ⋆ f2) ∼ (g1 ⋆ g2). 2. To prove the preservation of exponentiation, show that if f ∼ g and α > 0, then lim(x→∞)((f(x)/g(x))^α) = 1. 3. To prove the preservation of function composition, show that if f ∼ g and h(x) → ∞ as x → ∞, then lim(x→∞)(f(h(x))/g(h(x))) = 1.

Step 1: Show that f ∼ f for all eventually positive functions f

By the definition of the relation, f ∼ g if lim(x→∞)(f(x)/g(x)) = 1. Let's work with f(x)/f(x): lim(x→∞)(f(x)/f(x)) = 1, since anything divided by itself is equal to 1. Hence, f ∼ f, which proves reflexivity. #a. Symmetry#

Step 2: Show that if f ∼ g, then g ∼ f

Suppose that f ∼ g. Then, by definition: lim(x→∞)(f(x)/g(x)) = 1 Reciprocal rule leads to: lim(x→∞)(g(x)/f(x)) = 1 Which means that g ∼ f. This proves symmetry. #a. Transitivity#

Step 3: Show that if f ∼ g and g ∼ h, then f ∼ h

Suppose that f ∼ g and g ∼ h. Then, by definition, we have the following limits: lim(x→∞)(f(x)/g(x)) = 1 and lim(x→∞)(g(x)/h(x)) = 1 Now, we need to show that lim(x→∞)(f(x)/h(x)) = 1. Let's multiply the two given limits: lim(x→∞)(f(x)/g(x)) * lim(x→∞)(g(x)/h(x)) = 1 * 1 By limit laws, we get: lim(x→∞)(f(x)/h(x)) = 1 Thus, f ∼ h, proving transitivity. #b. Preservation of Arithmetic Operations#

Step 4: Prove that addition, multiplication, and division are preserved under the relation

We need to show that if f1 ∼ g1 and f2 ∼ g2, then (f1 ⋆ f2) ∼ (g1 ⋆ g2) for addition, multiplication, and division. To be concise, we demonstrate that addition is preserved: lim(x→∞)((f1(x) + f2(x))/(g1(x) + g2(x))) = lim(x→∞)(1 + f2(x)/g2(x)) * lim(x→∞)(1 + g1(x)/f1(x)) = 1 * 1 = 1 This shows that addition is preserved. Similar steps can be followed for multiplication and division. #c. Preservation of Exponentiation#

Step 5: Prove that if f ∼ g and α > 0, then f^α ∼ g^α

Let's show that if f ∼ g, then f^α ∼ g^α, where α > 0. Since f ∼ g: lim(x→∞)(f(x)/g(x))=1 Now, let's raise both sides to the power of α. Using limit laws: lim(x→∞)((f(x)/g(x))^α) = 1^α = 1 Now, using the fact that a^m/n=(a^m)^(1/n) lim(x→∞)((f(x))^α/(g(x))^α)=1 Hence, f^α ∼ g^α. #d. Preservation of Function Composition#

Step 6: Prove that if f ∼ g and h(x) → ∞ as x → ∞, then f ∘ h ∼ g ∘ h

Let's prove this by showing that: lim(x→∞)((f(h(x))/g(h(x)))) = 1 Since f ∼ g, we have: lim(h(x)→∞)(f(h(x))/g(h(x))) = 1 Given that h(x) → ∞ as x → ∞, we can replace h(x) with x in the limit: lim(x→∞)(f(x)/g(x))=1 Thus, f ∘ h ∼ g ∘ h.

Key Concepts

Eventually Positive Functions

An eventually positive function is a function that becomes strictly positive beyond a certain point. Imagine a function that may not be positive initially, but as you move to the right along the x-axis, it eventually stays positive from that point forward. This property is important when considering limits because it ensures that the function behaves consistently over time.

• Eventually positive doesn't mean positive everywhere, just after a certain point.
• Examples of such functions might include certain logarithmic or polynomial expressions that are negative or zero only in a limited initial range.

Understanding this concept is crucial for establishing relationships between functions and performing operations on them, as they ensure the stability and positivity required for certain types of mathematical analysis.

Arithmetic Operations in Limits

When dealing with limits, particularly as functions approach infinity, arithmetic operations such as addition, multiplication, and division must be handled carefully. The main idea is that if two functions are equivalent in their limits as they approach infinity, certain arithmetic operations on these functions will maintain this equivalence.

• For addition and multiplication, if two pairs of functions are equivalent, their sums and products will also be equivalent.
• Division requires more caution; even though the equivalence holds, it's crucial that the denominator does not become zero, as this would invalidate the operation.

For example, if we have two functions, say, \(f_1 \sim g_1\) and \(f_2 \sim g_2\), and we know their limits are equal, then \(f_1 + f_2 \sim g_1 + g_2\) holds true as the limits of these sums will equal 1. This is a valuable tool in calculus, facilitating the simplification of complex expressions.

Function Composition

Function composition involves applying one function to the results of another function. In mathematical language, if you have two functions, \(f\) and \(g\), function composition is expressed as \(f \circ g\), which means \(f(g(x))\). In the context of limits and equivalence, function composition needs special consideration.

• For equivalence relations, even if \(f \sim g\), to ensure that \(f \circ h \sim g \circ h\), it's necessary that \(h(x)\) approaches infinity as \(x\) does.
• This means that as you apply \(h\) to ever larger values of \(x\), it must produce values that go to infinity, keeping the behavior of \(f\) and \(g\) in sync.

By maintaining the condition \(h(x) \to \infty\), function composition will preserve these equivalences, which is important in analysis and calculus when managing complex functions and their behaviors at infinity.