Simplify $1 - \sum_{k=0}^{n} {n \choose k} q_{1}^{k}(1-q_{1})^{n-k} \log(1+e^{2\beta(n-k)})$

Question

I am trying to find a simplification for:

$$1 - \sum_{k=0}^{n} {n \choose k} q_{1}^{k}(1-q_{1})^{n-k} \log\left(1+e^{2\beta(n-k)}\right)$$

so far I've tried this by using taylor expansions for $\log(1+x)$ and $e^{ax}$ however it did not seem to work. I've also tried out stirlings formula which did not help a lot either.

Any tips/tricks to tackle this problem?

Answer 1

you can use the $$ \log(1+x)\approx \log(x) $$ for high x

Answer 2

I am skeptical that the expression has an exact closed form. But one may provide an asymptotic formula which is valid for large $n$. Write

$$ S_n = S_{n,q_1,\beta} = 1 - \sum_{k=0}^{n} \binom{n}{k}q_1^k (1-q_1)^{n-k} \log\big( 1 + e^{2\beta(n-k)} \big). $$

If $q_1 \in (0, 1)$ and $\beta > 0$, then we can obtain an asymptotic formula. Indeed, introduce a random variable $N \sim \operatorname{Binomial}(n, 1-p_1)$. Then

\begin{align*} S_n &= 1 - \mathbf{E}\left[ \log\big(1 + e^{2\beta N} \big) \right] \\ &= 1 - \mathbf{E}[2\beta N] - \mathbf{E}\left[\log\big( 1+e^{-2\beta N} \big)\right]. \end{align*}

The second term is easily computed as $\mathbf{E}[2\beta N] = 2\beta n(1-q_1)$. For the last term, by utilizing the inequality $\log(1+x) \leq x$,

$$ \mathbf{E}\left[\log\big( 1+e^{-2\beta N} \big)\right] \leq \mathbf{E}[e^{-2\beta N}] = (q_1 + (1-q_1)e^{-2\beta})^n, $$

which decays exponentially as $r := q_1 + (1-q_1)e^{-2\beta} \in (0, 1)$. So

$$ S_n = 1 - 2\beta n(1-q_1) + \mathcal{O}(r^n) $$

as $n\to\infty$.