Difference between revisions of "1969 IMO Problems/Problem 2"

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==Problem==
 
Let <math>a_1, a_2,\cdots, a_n</math> be real constants, <math>x</math> a real variable, and <cmath>f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).</cmath> Given that <math>f(x_1)=f(x_2)=0,</math> prove that <math>x_2-x_1=m\pi</math> for some integer <math>m.</math>
 
Let <math>a_1, a_2,\cdots, a_n</math> be real constants, <math>x</math> a real variable, and <cmath>f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).</cmath> Given that <math>f(x_1)=f(x_2)=0,</math> prove that <math>x_2-x_1=m\pi</math> for some integer <math>m.</math>
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Revision as of 14:50, 17 February 2018

Problem

Let $a_1, a_2,\cdots, a_n$ be real constants, $x$ a real variable, and \[f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).\] Given that $f(x_1)=f(x_2)=0,$ prove that $x_2-x_1=m\pi$ for some integer $m.$