有限傅里叶和的最小值求解咨询

Hey there! Great call suspecting that ( m_N = 0 ) for all ( N \geq 2 )—you’re absolutely correct, and we can break this down with some handy trigonometric identities to prove it clearly.

First, let’s start by simplifying the given function ( f_N(x) ). The key insight here is recognizing that this weighted cosine sum can be rewritten using a well-known trigonometric identity for such finite series. Let’s work through the steps:

1. Trigonometric Identity for Weighted Cosine Sums
   The function ( f_N(x) = \frac{N}{2} + \sum_{k=1}^{N-1} (N-k)\cos(kx) ) is actually equivalent to:
   [
   f_N(x) = \frac{\sin^{2\left(\frac{Nx}{2}\right)}{2\sin}2\left(\frac{x}{2}\right)}
   ]
   Let’s verify this with small ( N ) to be sure:

   • For ( N=2 ): ( f_2(x) = 1 + \cos x = 2\cos^2\left(\frac{x}{2}\right) ), and ( \frac{\sin^2(x)}{2\sin2\left(\frac{x}{2}\right)} = \frac{4\sin^{2\left(\frac{x}{2}\right)\cos}2\left(\frac{x}{2}\right)}{2\sin^2\left(\frac{x}{2}\right)} = 2\cos^2\left(\frac{x}{2}\right) ), which matches perfectly.
   • For ( N=3 ): ( f_3(x) = \frac{3}{2} + 2\cos x + \cos 2x ). Using the identity, ( \frac{\sin^{2\left(\frac{3x}{2}\right)}{2\sin}2\left(\frac{x}{2}\right)} ) expands to ( 2\cos^2x + 2\cos x + \frac{1}{2} )—exactly the simplified form of ( f_3(x) ).

2. Proving Non-negativity and Zero Attainment

   • Non-negativity: Since squares of real numbers are always non-negative, ( \sin^2\left(\frac{Nx}{2}\right) \geq 0 ) and ( \sin^2\left(\frac{x}{2}\right) \geq 0 ) for all ( x \in [0, 2\pi] ). The denominator ( 2\sin^2\left(\frac{x}{2}\right) ) is positive except when ( x=0 ) or ( x=2\pi ), where taking the limit gives ( \frac{N^2}{2} > 0 ). So ( f_N(x) \geq 0 ) everywhere in the interval.
   • Zero Attainment: We just need to find an ( x ) where the numerator is zero (and the denominator isn’t). Set ( \sin\left(\frac{Nx}{2}\right) = 0 )—this happens when ( \frac{Nx}{2} = k\pi ) for integer ( k ). Choosing ( k=1,2,...,N-1 ), we get ( x = \frac{2k\pi}{N} ). For these ( x ), ( \sin\left(\frac{x}{2}\right) = \sin\left(\frac{k\pi}{N}\right) \neq 0 ) (since ( k < N )), so the denominator is non-zero, making ( f_N(x) = 0 ).

Putting it all together: since ( f_N(x) ) is never negative and we can explicitly find points where it equals 0, the minimum ( m_N = 0 ) for all ( N \geq 2 ).

备注：内容来源于stack exchange，提问作者cowsin