Matematické Fórum

stuart clark · 26. 03. 2011 03:40

Let $f(x) = cos\left(a_{1}+x\right)+\frac{1}{2}cos\left(a_{2}+x\right)+\frac{1}{2^2}cos\left(a_{3}+x\right)+.............+\frac{1}{2^{n-1}}cos\left(a_{n}+x\right)$ . Where $a_{1},a_{2},...........................a_{n}\in\mathbb{R}$ .

If $f(x_{1}) = f(x_{2}) = 0$ . Then $\left|x_{2}-x_{1}\right| =$

musixx · 28. 03. 2011 10:36 — Editoval musixx (28. 03. 2011 11:05)

Firstly, we must realize that $f(x)$ is not constant. But $\sum_{i=1}^\infty\frac1{2^i}=1$ , so therefore any finite sum $\sum_{i=1}^{n-1}\frac1{2^i}$ is smaller than 1. For sure, there is such $x_{pos}$ that $\cos(a_1+x_{pos})=1$ . Since $\sum_{i=1}^{n-1}\frac1{2^i}<1$ , then also $\sum_{i=1}^{n-1}\left(\frac1{2^i}\cdot k_i\right)<1$ for any sequence $\{k_i;\ k_i\leq1\}$ , so specially for a sequence $\{\cos(a_{i+1}+x_{pos})\}$ . Therefore $f(x_{pos})>0$ . Analogically, we can show that there exists $x_{neg}$ such that $f(x_{neg})<0$ . So our function $f(x)$ is not a constant one. EDIT: A constant function would degenerate the following...

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Futher (an extension via the induction): a linear combination
$a\cos(b+x)+c\cos(d+x)$
is equal to
$a\cos b\cos x-a\sin b\sin x+c\cos d\cos x-c\sin d\sin x$ , so also to
$A\cos x+B\sin x$ , so also to
$C\cos(D+x)$ (for suitable constants)
mentioned e.g. here (in Czech), so your whole function $f(x)$ is nothing more than $U\cos(V+x)$ for suitable constants $U$ and $V$ , where $U$ is nonzero because of the previous paragraph. Therefore your $|x_2-x_1|$ is an integral multiple of $\pi$ , since there is -- in the end -- nothing more than one single cosinus.

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A remark: We used strongly a fact how $x$ was used in all the arguments. A general counter-example:
$a(x)=\sin(10x)-\sin(x)$
$b(x)=\sin(x)$ .
Minimal periods:
$a(x)\ \rightarrow\ 2\pi$
$b(x)\ \rightarrow\ 2\pi$
$a(x)+b(x)\ \rightarrow\ \pi/5$

Matematické Fórum

#1 26. 03. 2011 03:40

Trigonometric equation

#2 28. 03. 2011 10:36 — Editoval musixx (28. 03. 2011 11:05)

Re: Trigonometric equation

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