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 4 of 5: Use backward induction when |r| is less than 1
0<∣r∣<1⟹∣an∣≤c,∣an−1∣<2c,∣an−k∣<(k+1)c0<|r|<1\Longrightarrow |a_n|\le c,\quad |a_{n-1}|<2c,\quad |a_{n-k}|<(k+1)c
Detailed analysis

For 0<∣r∣<10<|r|<1, ∣an∣=∣cn+1∣≤c|a_n|=|c_{n+1}|\le c and ∣an−1∣=∣cn−ran∣<c+c=2c|a_{n-1}|=|c_n-ra_n|<c+c=2c. If ∣an−k∣≤(k+1)c|a_{n-k}|\le(k+1)c, then ∣an−k−1∣=∣cn−k−ran−k∣<c+(k+1)c=(k+2)c|a_{n-k-1}|=|c_{n-k}-ra_{n-k}|<c+(k+1)c=(k+2)c. Hence every coefficient satisfies ∣aj∣≤(n+1)c|a_j|\le(n+1)c.