证明组合函数$f(k)={n\choose k}(1+e^{-k})^n$在k=0处取最大值

Great question! Let's break down how to rigorously prove that ( f(k) ) reaches its highest value when ( k=0 ) for all integers ( k \in {0,1,...,n} ). We'll use two intuitive, solid approaches to confirm this.

Approach 1: Show ( f(k) ) is strictly decreasing #

To prove ( f(0) ) is the maximum, we can show that each subsequent term is smaller than the previous one—meaning the function strictly decreases as ( k ) increases. We do this by analyzing the ratio of consecutive terms ( \frac{f(k+1)}{f(k)} ):

1. Calculate the combination ratio:
   The binomial coefficient ratio simplifies to:
   [
   \frac{\binom{n}{k+1}}{\binom{n}{k}} = \frac{n - k}{k + 1}
   ]

2. Calculate the exponential term ratio:
   Simplify the fraction inside the nth power:
   [
   \frac{1 + e^{-(k+1)}}{1 + e^{-k}} = \frac{e^{k+1} + 1}{e^{k+1} + e} = 1 - \frac{e - 1}{e(e^k + 1)}
   ]
   This value is strictly less than 1 for all ( k \geq 0 ), since we're subtracting a positive number from 1. Raising it to the nth power keeps it less than 1.

3. Verify the full ratio is less than 1:

   • For ( k=0 ): The ratio becomes ( n \cdot \left( \frac{1 + e^{-1}}{2} \right)^n ). The constant ( \frac{1+e^{-1}}{2} \approx 0.684 ), and the function ( g(n) = n \cdot (0.684)^n ) peaks at ( n \approx 2.6 ) with a maximum value ~0.96 (still less than 1). For all ( n \geq 1 ), this ratio stays below 1.
   • For ( k \geq 1 ): The combination ratio ( \frac{n -k}{k+1} ) is at most ( \frac{n-1}{2} ) (maximized when ( k=1 )), and the exponential ratio is at most ( \left( \frac{1+e^{-2}}{1+e{-1}} \right)^n \approx (0.83)^n ). Even for small ( n ), ( \frac{n-1}{2} \cdot (0.83)^n <1 ), and as ( n ) grows, the exponential term decays far faster than the linear term can grow.

Since ( \frac{f(k+1)}{f(k)} <1 ) for all valid ( k ), ( f(k) ) strictly decreases as ( k ) increases. This means ( f(0) ) is the largest value.

Approach 2: Direct comparison to ( f(0) ) #

We know ( f(0) = \binom{n}{0}(1+e⁰⁾n = 2^n ). For any ( k \geq1 ), we just need to show ( \binom{n}{k}(1+e^{-k})n < 2^n ), or equivalently:
[
\binom{n}{k} < \left( \frac{2}{1+e^{-k}} \right)^n = \left( \frac{2e^k}{ek +1} \right)^n
]

• For ( k=1 ): ( \frac{2e}{e+1} \approx1.462 ), and ( n < (1.462)^n ) holds for all ( n \geq1 ) (exponential growth always outpaces linear growth eventually, and it's true even for small ( n ): ( 1 <1.462 ), ( 2 <2.137 ), etc.).
• For ( k \geq2 ): ( \frac{2e^k}{ek+1} ) gets closer to 2 as ( k ) increases, so ( \left( \frac{2e^k}{ek+1} \right)^n ) becomes even larger relative to ( \binom{n}{k} ).
• For mid-range ( k ) (like ( k=n/2 )): The largest binomial coefficient ( \binom{n}{n/2} \approx \frac{2^n}{\sqrt{\pi n/2}} ). Since ( \left( \frac{2e^k}{ek+1} \right)^n > (1.462)^n ), and ( (1.462/2)^n = (0.731)^n ) decays exponentially, ( \frac{2^n}{\sqrt{\pi n/2}} < (1.462)^n ) holds for all ( n \geq1 ).

Every case confirms ( f(k) < f(0) ) when ( k \geq1 ).

内容的提问来源于stack exchange，提问作者Happy Mittal