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We have the following inequality:

$$2^x \leq \sum_{i=0}^m{{x \choose i}\lambda^i}$$

All the variables are in $\mathbb{N}_{>0}$

I need to find a tight upper bound for $x$ using $m,\lambda$.

In the case of $\lambda = 1$ we can use the binomial theorem to show $x \leq m$. However for $\lambda>1$ I have no idea how to find a tight upper bound for this.

It can be shown that: $$2^x \leq \sum_{i=0}^m{{x \choose i}\lambda^i} \leq \left(\frac{\lambda e x}{m}\right)^m$$

And then we can use the solution from here: Upper bound $2^x \leq (ax)^c$

But I need a tighter bound than this. Is there any way to bound $x$ directly from this partial binomial theorem sum?

I thought of maybe doing something like this:

$$2^x = (1 + \lambda)^{x\log_{1 + \lambda}(2)}=(1 + \lambda)^{\frac{x}{\log_2(1 + \lambda)}}=\\ \sum_{i=0}^{{\frac{x}{\log_2(1 + \lambda)}}}{{{\frac{x}{\log_2(1 + \lambda)}} \choose i}\lambda^i} \leq \sum_{i=0}^m{{x \choose i}\lambda^i}$$

But I'm not sure how to continue from here (or if it even helps).

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  • $\begingroup$ I suppose that you face an hypergeometric function for the rhs. $\endgroup$ – Claude Leibovici May 16 '20 at 15:33
  • $\begingroup$ @ClaudeLeibovici I'm not sure what that means, or how can I find an upper bound for x with that. $\endgroup$ – Tomer Wolberg May 16 '20 at 15:50
  • $\begingroup$ You could start with a simple upper bound of $m(x\lambda)^m$ for the RHS. $\endgroup$ – Aravind May 16 '20 at 15:59
  • $\begingroup$ @Aravind but I need to find an upper bound for $x$ not for the sum. $\endgroup$ – Tomer Wolberg May 16 '20 at 16:05
  • $\begingroup$ @Aravind what does rhs mean? $\endgroup$ – Tomer Wolberg May 16 '20 at 16:08
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This is more of a long comment than an answer, but I don't get the same upper bound that you get in the $\lambda \leq 1$ case.

Assuming that $\lambda$ (and hence everything) is positive, it seems to me that:

$$\sum_{i=0}^m{{x \choose i}\lambda^i} \leq \sum_{i=0}^x{{x \choose i}\lambda^i} $$

with equality if and only if $m \geq x$.

But the right hand side of this new inequality equals $(1 + \lambda)^x$, by the binomial theorem.

So substituting this back into the original inequality we obtain:

$$2^x \leq (1 + \lambda)^x$$

When $\lambda > 1$ we get this inequality for free and so we don't learn anything new about $x$, which is similar to the problem you experienced.

When $\lambda = 1$ we have equality in the last inequality I typed, which means that we also need equality in the first equality I typed which impies $x \leq m$ as you also found.

But if $\lambda < 1$ then this inequality puts a rather strong restriction on $x$, namely:

$$x = 0$$

For any $x > 0$ the above inequality with $\lambda < 1$ is violated.

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    $\begingroup$ Yes, you're right. In the case it's smaller than 1 then x=0. But I'm more interested in the case it's bigger than 1. In my specific algorithm $\lambda$ is some integer constant that is bigger than 2. $\endgroup$ – Tomer Wolberg May 16 '20 at 22:06

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