2
$\begingroup$

Let the polynomial $P(x)$ of odd degree $d$ be given by $$x^d+a_{d-1}x^{d-1}+...+a_0$$

Let $A = |a_0|+ ... + |a_{d-1}| + 1$

Then why is $$|a_{d-1}(-A)^{d-1} + a_{d-2}(-A)^{d-2} +...+ a_0| ≤ (A-1)A^{d-1}$$

I can understand why $$|a_{d-1}A^{d-1} + a_{d-2}A^{d-2} +...+ a_0| ≤ (A-1)A^{d-1}$$

Using the reasoning that $$(A-1)A^{d-1}$$ $$=|a_{d-1}|A^{d-1} + |a_{d-2}|A^{d-1}+...+|a_0|A^{d-1}$$ $$≥ |a_{d-1}|A^{d-1} + |a_{d-2}|A^{d-2}+...+|a_0|$$ $$≥ |a_{d-1}A^{d-1} + a_{d-2}A^{d-2}+...+a_0|$$

However I fail to see a similar reasoning for the other inequality.

New contributor
Gunnar is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
4
  • $\begingroup$ $|a_{d-1}(-A)^{d-1} + a_{d-2}(-A)^{d-2} +\cdots+ a_0| \le |a_{d-1}|A^{d-1} + |a_{d-2}|A^{d-2}+ \cdots+|a_0| $, so the same argument works. $\endgroup$
    – Martin R
    Commented yesterday
  • $\begingroup$ Sorry I still don't see immediately why this is true the reasons I found $|𝑎_{𝑑−1}(𝐴)^{𝑑−1}+𝑎_{𝑑−2}(𝐴)^{𝑑−2}+⋯+𝑎_{0}|≤|𝑎_{𝑑−1}|𝐴^{𝑑−1}+|𝑎_{𝑑−2}|𝐴^{𝑑−2}+⋯+|𝑎_{0}|$ is because everything is positive and |a| + |b| + |c| ≤ |a +b +c| $\endgroup$
    – Gunnar
    Commented yesterday
  • $\begingroup$ It is just the triangle inequality (and $A$ is positive). $\endgroup$
    – Martin R
    Commented yesterday
  • $\begingroup$ ok so I am assuming the triangle inequality is |a| + |b| ≥ |a + b| and trying to rewrite $|𝑎_{𝑑−1}(−𝐴)^{𝑑−1}+𝑎_{𝑑−2}(−𝐴)^{𝑑−2}+⋯+𝑎_{0}|$= $|𝑎_{𝑑−1}(𝐴)^{𝑑−1}+(-𝑎_{𝑑−2}(𝐴)^{𝑑−2})+⋯+𝑎_{0}|$ which would be less than or equal to $|𝑎_{𝑑−1}|(𝐴)^{𝑑−1}+|𝑎_{𝑑−2}|𝐴^{𝑑−2}+⋯+|𝑎_{0}|$ right? $\endgroup$
    – Gunnar
    Commented yesterday

1 Answer 1

3
$\begingroup$

First apply the triangle inequality: $$ |a_{d-1}(-A)^{d-1} + a_{d-2}(-A)^{d-2} +\cdots+ a_0| \le |a_{d-1}(-A)^{d-1}| + |a_{d-2}(-A)^{d-2}| +\cdots+ |a_0| $$ Now simplify the right-hand side, using $A > 0$: $$ |a_{d-1}(-A)^{d-1}| + |a_{d-2}(-A)^{d-2}| +\cdots+ |a_0| = |a_{d-1}| \cdot |{-A}|^{d-1} + |a_{d-2}| \cdot|{-A}|^{d-2} +\cdots+ |a_0| \\ = |a_{d-1}| A^{d-1} + |a_{d-2}| A^{d-2} + \cdots + |a_0| \, . $$ And this is (because $A \ge 1$) $$ \le |a_{d-1}| A^{d-1} + |a_{d-2}| A^{d-1} + \cdots + |a_0| A^{d-1} = (A-1) A^{d-1} \, . $$

Remark: This estimate holds for all polynomials of degree $d \ge 1$, the fact that $d$ is odd plays no role.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .