Do there exist $a_k$ and $b_k$ so the equation $\sum\limits_{k=1}^{n} (a_k \sin(kx) + b_k \cos(kx)) = 0$ has no roots?

Do there exist real numbers $a_1, a_2, ..., a_n$ and $b_1, b_2, ..., b_n$ such that the equation $$\sum\limits_{k=1}^{n} (a_k \sin(kx) + b_k \cos(kx)) = 0$$ has no solutions?

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Using complex numbers $c_k=a_k+ib_k$, $z=\cos x+i\sin x$, the question is whether $f(z)=\sum c_kz^k$ intersects the real axis for some $z$ on the unit circle. – Hagen von Eitzen May 13 '14 at 6:49
The integral of this trigonometric polynomial over any interval of length $2\pi$ is zero. Since this function is continuous and real-valued, it must possess a zero within that interval. – Sangchul Lee May 13 '14 at 7:12

1 Answer

It is as sos440 says: $$\int_0^{2\pi} \left[ \sum\limits_{k=1}^{n} (a_k \sin(kx) + b_k \cos(kx)) \right] \; dx = 0$$ A strictly positive or strictly negative function cannot integrate to zero. Thus $f(x) = \sum\limits_{k=1}^{n} (a_k \sin(kx) + b_k \cos(kx))$ is neither. So there exist real numbers $a$ and $b$ with $f(a) > 0$, $f(b) < 0$. By intermediate value theorem, $f(x) = 0$ for some $x$.

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thanks @Goos and @sos440!!! – PandaMan May 13 '14 at 8:50