求解函数$f_n(x)$在区间$(0,\pi)$内的最大值点及相关方程

Let's break down how to find the maximum points of this function over the interval $(0, \pi)$:
$$f_n(x)=\frac{1}{n+1}\left(\frac{\sin\left((n+1)\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right)^2$$

Core Equation for Maximum Points #

To locate the maxima, we start by taking the derivative of $f_n(x)$, setting it to zero, and simplifying using trigonometric identities. This leads us to the key transcendental equation that all maximum points must satisfy:
$$\tan\left((n+1)\frac{x}{2}\right) = (n+1)\tan\left(\frac{x}{2}\right)$$

Polynomial Rewrite #

We can also transform this equation into a polynomial form in terms of $\tan^2\frac{x}{2}$—a useful alternative for numerical solving or algebraic manipulation:
$$\sum_{k=1}^{{\left\lfloor\frac{n+1}{2}\right\rfloor}(-1)}k k {n+1\choose{2k+1}}\tan^{2k}\frac{x}{2}=0$$

Additional Equivalent Forms #

There are other valid representations of this equation waiting to be derived. You could explore using more advanced multiple-angle trigonometric identities or algebraic substitutions to expand it into different, potentially more convenient forms.

内容的提问来源于stack exchange，提问作者Birendra Singh