I have proposed a classifier whose complexity can be written as

\begin{align*} \mathcal{O}\left(f(m)\right)&=\mathcal{O}\left(\log\left(\max_{x\in(0,1)}\left\{g(x,m)\right\}\right)\right)\\ g(x,m)&=\frac{(1+x)^m(1-x)}{1-x^{m+1}} \end{align*}

where $m\in\{0,1,2,3,...\},x\in(0,1)$.

There exist an evident upper bound for $g(x,m)$ but it is quite large:

\begin{align} (1+x)^m=\sum_{i=0}^m{\binom{m}{i}x^i}&<\binom{m}{{\lceil}\frac{m}{2}{\rceil}}\sum_{i=0}^{m}{}x^i\\ &<\binom{m}{{\lceil}\frac{m}{2}{\rceil}}\frac{1-x^{m+1}}{1-x} \end{align} \begin{equation} \longrightarrow \forall x: \frac{(1+x)^m(1-x)}{1-x^{m+1}}<\binom{m}{{\lceil}\frac{m}{2}{\rceil}} \end{equation} As you see $\binom{m}{{\lceil}\frac{m}{2}{\rceil}}$ is a generous upper bound.

Translating my question to Combinatorics language, I am seeking for a technique that provides the smallest or a small enough $U_m$ in the below formula. \begin{equation} \sum_{i=0}^m{\binom{m}{i}x^i}<U_m\sum_{i=0}^{m}{}x^i \end{equation} Can anyone provide any hint, please?


Hint. Note that the function $$f(x)=\frac{(1+x)^m(1-x)}{1-x^{m+1}}$$ is increasing in $(0,1)$ and $$\lim_{x\to 1^-} f(x)=\frac{2^m}{m+1}.$$

  • $\begingroup$ Thanks Robert!! $\endgroup$ – hossayni Mar 22 '18 at 8:57

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