MathLabs

Problem 3

Let f(x)=anxn+an−1xn−1+⋯+a0f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 and g(x)=cn+1xn+1+cnxn+⋯+c0g(x)=c_{n+1}x^{n+1}+c_nx^n+\cdots+c_0 be non-zero real polynomials such that g(x)=(x+r)f(x)g(x)=(x+r)f(x) for some real rr. If a=max⁡(∣an∣,…,∣a0∣)a=\max(|a_n|,\ldots,|a_0|) and c=max⁡(∣cn+1∣,…,∣c0∣)c=\max(|c_{n+1}|,\ldots,|c_0|), prove that ac≤n+1\frac{a}{c}\le n+1.
Step 3 of 5: Use forward induction when ∣r∣≥1|r|\ge1
∣r∣≥1⟹∣a0∣≤c,∣a1∣≤2c,∣ak∣≤(k+1)c|r|\ge1\Longrightarrow |a_0|\le c,\quad |a_1|\le2c,\quad |a_k|\le(k+1)c
Detailed analysis

For ∣r∣≥1|r|\ge1, ∣a0∣=∣c0/r∣≤c|a_0|=|c_0/r|\le c. Also ∣a1∣=∣(c1−a0)/r∣≤∣c1∣+∣a0∣≤2c|a_1|=|(c_1-a_0)/r|\le|c_1|+|a_0|\le2c. If ∣ak∣≤(k+1)c|a_k|\le(k+1)c, then ∣ak+1∣=∣(ck+1−ak)/r∣≤∣ck+1∣+∣ak∣≤(k+2)c|a_{k+1}|=|(c_{k+1}-a_k)/r|\le|c_{k+1}|+|a_k|\le(k+2)c. Thus ∣ak∣≤(k+1)c≤(n+1)c|a_k|\le(k+1)c\le(n+1)c for every kk.