2
$\begingroup$

Let $f(x)=(1-x)^b (1+x)^{(n-b)}= \sum_{i=0}^n a_ix^i$, where $n$ is a positive integer and $b$ is a non-negative integer less than $n$. I want to find an upper bound on $\sum_{i=0}^n |a_i|$ other than the trivial upper bound $2^n$. Also for $b=0,1,\frac{n}{2}$, it is easy. Is there any integration type of approach for this problem?

$\endgroup$
4
  • $\begingroup$ a. Typeset your question. b. the symbol $e$ stands for a frequently used constant, and using it in this context is very confusing. $\endgroup$
    – Igor Rivin
    Nov 21, 2012 at 17:59
  • 7
    $\begingroup$ The coefficient $a_i$ is a value of a Krawtchouk polynomial. A lot is known about their asymptotics. See for instance www6.cityu.edu.hk/rcms/publications/preprint21.pdf and references [11] and [15] therein. $\endgroup$ Nov 21, 2012 at 18:51
  • $\begingroup$ @Stanley: Thank you very much for your kind help. I have tried to understand main results from these papers. But there is no such closed form. I need closed form as a function of $b,n$ so that I can calculate upper bound on $\sum_{i=0}^n |a_i|$. Thank you again for your help. $\endgroup$
    – user29295
    Nov 22, 2012 at 1:00
  • $\begingroup$ You DO NOT need a closed form to get estimates. See my answer. $\endgroup$
    – Igor Rivin
    Nov 22, 2012 at 4:04

2 Answers 2

2
$\begingroup$

Assume without loss of generality that $b\leq n/2$. Writing $f(x)=(1-x^2)^b(1+x)^{n-2b}$ shows that an upper bound is $2^{n-b}$, but this is very crude.

$\endgroup$
1
  • $\begingroup$ @Stanley: Thank you very much. For $b\leq n/3$, I have tried similar approach but do not get anything better than $2^{n-b}$. $\endgroup$
    – user29295
    Nov 22, 2012 at 16:32
3
$\begingroup$

This is analyzed exhaustively (also exhaustingly) by Domenici in this 2005 preprint.

$\endgroup$
1
  • $\begingroup$ @Rivin: Thank you very much for the paper. In the paper author tried to estimate different $a_i$. From these estimates $\sum_{i=0}^n|a_i|$ is not easily followed for general $b$. Thank you again for your kind help. $\endgroup$
    – user29295
    Nov 22, 2012 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.