Problems

Solution: If \(g(0) \in \mathbb{Z} \Rightarrow b \in \mathbb{Z}\). If \(g(1) \in \mathbb{Z} \Rightarrow a+b \in \mathbb{Z} \Rightarrow a \in \mathbb{Z}\), therefore \(a \cdot n + b \in \mathbb{Z}\), in particular \(g(n) \in \mathbb{Z}\) for all integers \(n\). \(f(0) \in \mathbb{Z} \Rightarrow C \in \mathbb{Z}\), \(f(1) \in \mathbb{Z} = A+ B + C \in \mathbb{Z} \Rightarrow A+ B \in \mathbb{Z}\) \(f(-1) \in \mathbb{Z} = A- B + C \in \mathbb{Z} \Rightarrow A- B \in \mathbb{Z}\) \(\Rightarrow 2A, 2B \in \mathbb{Z}\) \begin{align*} f(n) &= An^2 + Bn + C \\ &= An^2-An + An+Bn + C \\ &= 2A \frac{n(n-1)}2 + (A+B)n + C \\ &\in \mathbb{Z} \end{align*} Consider \(g(x) = f(x + \alpha)\), therefore \(g(0), g(1), g(-1) \in \mathbb{Z} \Rightarrow g(n) \in \mathbb{Z} \Rightarrow f(n+\alpha) \in \mathbb{Z}\)

Solution: Claim: If \(f(x) = c\) then \(\lozenge f(x) = 0\) Proof: Consider \(g(x) = x\) then \begin{align*} (1) && \lozenge(f(x)g(x)) &= g(x) \lozenge f(x) + f(x) \lozenge g(x) \\ \Rightarrow && \lozenge(c x) &= x \lozenge f(x) + c \lozenge x \\ (4) && \lozenge(c x) &= c \lozenge x \\ \Rightarrow && 0 &= x \lozenge f(x) \\ \Rightarrow && \lozenge f(x) &= 0 \end{align*} \begin{align*} (1) && \lozenge(x^2) &= x \lozenge x + x \lozenge x \\ (3) &&&= 2 x \cdot 1 \\ &&&= 2x \\ \\ (1) && \lozenge (x^3) &= x^2 \lozenge x + x \lozenge (x^2) \\ &&&= x^2 \cdot \underbrace{1}_{(3)} + x \cdot\underbrace{ 2x}_{\text{previous part}} \\ &&&= 3x^2 \end{align*} Claim: \(\lozenge h(x) = \frac{\d }{\d x} ( h(x))\) for any polynomial \(h\). Proof: (By (strong) induction on the degree of \(h\)). Base case: True, we proved this in the first part of the question. Inductive step: Assume true for all polynomials of degree less than or equal to \(k\). Then consider \(n = k+1\). We can write \(h(x) = ax^{k+1} + h_k(x)\) where \(h_k(x)\) is a polynomial of degree less than or equal to \(k\). Then notice: \begin{align*} && \lozenge (h(x)) &= \lozenge (ax^{k+1} + h_k(x)) \\ (2) &&&= \lozenge (ax^{k+1})+ \lozenge (h_k(x)) \\ &&&=\underbrace{a\lozenge (x^{k+1})}_{(4)}+ \underbrace{\frac{\d}{\d x} (h_k(x))}_{\text{inductive hypothesis}}\\ &&&= a \underbrace{\left (x \lozenge x^k + x^k \lozenge x \right)}_{(1)} + \frac{\d}{\d x} (h_k(x)) \\ &&&= a \left ( x \cdot \underbrace{k x^{k-1}}_{\text{inductive hyp.}} + x^k \cdot \underbrace{1}_{(3)} \right) + \frac{\d}{\d x} (h_k(x)) \\ &&&= (k+1)a x^k + \frac{\d}{\d x} (h_k(x)) \\ &&&= \frac{\d }{\d x} \left ( ax^{k+1} + h_k(x) \right) \\ &&&= \frac{\d }{\d x} (h(x)) \end{align*} Therefore since our statement is true for \(n=0\) and if it is true for \(n=k\) it is true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 0\)

Solution: \begin{align*} x^4 - 2x^3+x^2 &= x^2(x^2-2x+1) \\ &= x^2(x-1)^2 > 0 \end{align*} Since \(x\)and \(x-1\) can't both be zero, and square cannot be negative. \begin{align*} x^2 - 8x+17 &= (x-4)^2 +1 \geq 1 > 0 \end{align*} If \(x \leq 2\) then \(x^4 - 2x^3+x^2 > 0\) and \(17-8x \geq 1\) so \(x^4-2x^3+x^2-8x+17 > 0\) If \(x > 2\) then \(x^4-2x^3 = x^3(x-2) \geq 0\) and \(x^2-8x+17 > 0\) so \(x^4-2x^3+x^2-8x+17 > 0\), so for all real numbers our polynomial is positive and therefore cannot have any roots. Note that: \(x^4-x^3+x^2 = x^2(x^2-x+1) > 0\) and \(x^2-4x+4 =(x-2)^2 \geq 0\) If \(x < 1\) then \(x^4-x^3+x^2 > 0\) and \(4(1-x) > 0\) so \(x^4-x^3+x^2-4x+4 > 0\). If \(x > 1\) then \(x^4-x^3 = x^3(x-1) > 0\) and \(x^2-4x+4 \geq 0\) therefore \(x^4-x^3+x^2-4x+4 > 0\). Therefore \(x^4-x^3+x^2-4x+4 > 0\) for all real \(x\) and hence there are no real roots.