Problem 2

Let

a_1,a_2,\ldots,a_n

be real constants and let

x

be real. Define

f(x)=\cos(a_1+x)+\frac12\cos(a_2+x)+\cdots+\frac1{2^{n-1}}\cos(a_n+x)

. Given that

f(x_1)=f(x_2)=0

, prove that

x_2-x_1=m\pi

for some integer

m

Step 1 of 4: Collect the sine and cosine coefficients

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In plain words

Every term has the same frequency, so changing the phases only changes the two coefficients, not the frequency.

f(x)=B\cos x-C\sin x,\quad B=\sum_{j=1}^n2^{1-j}\cos a_j,\quad C=\sum_{j=1}^n2^{1-j}\sin a_j

Detailed analysis

Use $\cos(a_j+x)=\cos a_j\cos x-\sin a_j\sin x$ term by term. The weighted sum therefore has the form $f(x)=B\cos x-C\sin x$ , where $B,C$ are constants independent of $x$ .