W.9. Fie \( \varphi=\frac{1+\sqrt{5}}{2} \) si functia \( f(x)=1+\varphi\cos x+A\cos 2x+B\cos 3x \) astfel incat \( f(x)\geq 0 \), pentru orice \( x\in [0,\pi] \). Sa se determine \( A \) si \( B \).
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functie pozitiva si numarul de aur
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Cezar Lupu
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functie pozitiva si numarul de aur
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lasamasatelas
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In principiu problema poate fi atacata cu "artilerie grea". Intrucat \( \cos2x=2\cos^2-1 \) si \( \cos3x=4\cos^3x-3\cos{x} \) avem \( f(x)=P(\cos{x}) \) unde \( P(X)=4BX^3+2AX^2+(\phi-3)X+1-A \). Deci trebuie sa gasim \( A,B \) a.i. \( P(X)\geq0 \) \( {\forall}X\in[-1,1] \). Asta ar merge cu multa munca daca folosim teorema lui Sturm.
Cum insa stim ca e problema de concurs ar trebui sa ne asteptam la o solutie "frumosa". Acel \( \sqrt5 \) din \( \phi \) ni-l sugereaza pe \( \sqrt5 \) care apare in cosinusurile \( \cos\frac{k\pi}5 \). Pt. cine nu stie, astea pot fi calculate asa: Daca \( \zeta=e^{\frac{2k\pi}5i} \) unde \( k=1,2,3,4 \) atunci \( \zeta^5=1 \) si \( \zeta\neq1 \) deci \( 1+\cdots\zeta^4=0 \). Rezulta ca \( 0=\zeta^2+\zeta^{-2}+\zeta+\zeta^{-1}+1=(\zeta+\zeta^{-1})^2+(\zeta+\zeta^{-1})-1 \). Dar \( \zeta+\zeta^{-1}=2\cos\frac{2k\pi}5 \).
Deci \( 2\cos\frac{2\pi}5 \) si \( 2\cos\frac{4\pi}5 \) sunt solutii ale ecuatiei \( X^2+X-1=0 \). Avem \( 2\cos\frac{2\pi}5=\frac{-1+\sqrt5}2=\phi-1=\phi^{-1} \) si \( 2\cos\frac{4\pi}5=\frac{-1-\sqrt5}2=-\phi \). Deci \( \cos\pm\frac{2\pi}5=\frac{\phi^{-1}}2 \) si \( \cos\pm\frac{4\pi}5=-\frac{\phi}2 \). Avem si \( \cos\pm\frac{\pi}5=-\cos\frac{4\pi}5=\frac{\phi}2 \) si \( \cos\pm\frac{3\pi}5=-\frac{2\pi}5=-\frac{\phi^{-1}}2 \).
Avem:
\( 0{\leq}f(\pi)=1+\phi\cos\pi+A\cos2\pi+B\cos3\pi=1-\phi+A-B \) si
\( 0{\leq}f(\frac{3\pi}5)=1+\phi\cos\frac{3\pi}5+A\cos\frac{6\pi}5+B\cos\frac{9\pi}5= \)
\( 1+\phi\cdot(-\frac{\phi^{-1}}2)-\frac\phi2A+\frac\phi2B= \)
\( \frac12-\frac\phi2A+\frac\phi2B \).
Daca inmultim a doua ineg. cu \( 2\phi^{-1}=2(\phi-1) \) obtinem \( 0\leq\phi-1-A+B \). Prin urmare trebuie sa avem egalitati adica \( f(\pi)=f(\frac{3\pi}5)=0 \) si \( A-B=\phi-1 \).
Observam ca \( f \) are minim local (si global) in \( \frac{3\pi}5 \). Rezulta ca \( f\prime(\frac{3\pi}5)=0 \). Cum \( \sin2x=2{\sin}x{\cos}x \) si \( sin3x={\sin}x(2\cos2x+1) \) avem \( f\prime(x)=-\phi{\sin}x-2A\sin2x-B\sin3x=-{\sin}x(\phi+2{\cos}xA+3(2\cos2x+1)B) \).
Prin urmare: \( 0=f\prime(\frac{3\pi}5)=-\sin\frac{3\pi}5(\phi+2\cos\frac{3\pi}5A+3(2\cos\frac{6\pi}5+1)B) \) \( =-\sin\frac{3\pi}5(\phi-\phi^{-1}A+(-\phi+1)B) \), care implica \( 0=\phi-\phi^{-1}A-\phi^{-1}B \), adica \( A+B=\phi^2=\phi+1 \). Impreuna cau \( A-B=\phi-1 \) asta implica \( A=\phi \) si \( B=1 \).
Pt. suficienta, avem \( f(x)=P({\cos}x) \) unde \( P \) are gradul 3 si primul coeficient \( 4B=4 \). Cum \( f(\pi)=f(\frac{3\pi}5)=f\prime(\frac{3\pi}5) \) deci \( P(\cos\pi)=P(\cos\frac{3\pi}5)=P\prime(\cos\frac{3\pi}5)=0 \). Rezulta ca \( P(X)=4(X+1)(X+\frac{\phi^{-1}}2)^2 \), care e \( \geq0 \) pe intervalul \( [-1,1] \).
Cum insa stim ca e problema de concurs ar trebui sa ne asteptam la o solutie "frumosa". Acel \( \sqrt5 \) din \( \phi \) ni-l sugereaza pe \( \sqrt5 \) care apare in cosinusurile \( \cos\frac{k\pi}5 \). Pt. cine nu stie, astea pot fi calculate asa: Daca \( \zeta=e^{\frac{2k\pi}5i} \) unde \( k=1,2,3,4 \) atunci \( \zeta^5=1 \) si \( \zeta\neq1 \) deci \( 1+\cdots\zeta^4=0 \). Rezulta ca \( 0=\zeta^2+\zeta^{-2}+\zeta+\zeta^{-1}+1=(\zeta+\zeta^{-1})^2+(\zeta+\zeta^{-1})-1 \). Dar \( \zeta+\zeta^{-1}=2\cos\frac{2k\pi}5 \).
Deci \( 2\cos\frac{2\pi}5 \) si \( 2\cos\frac{4\pi}5 \) sunt solutii ale ecuatiei \( X^2+X-1=0 \). Avem \( 2\cos\frac{2\pi}5=\frac{-1+\sqrt5}2=\phi-1=\phi^{-1} \) si \( 2\cos\frac{4\pi}5=\frac{-1-\sqrt5}2=-\phi \). Deci \( \cos\pm\frac{2\pi}5=\frac{\phi^{-1}}2 \) si \( \cos\pm\frac{4\pi}5=-\frac{\phi}2 \). Avem si \( \cos\pm\frac{\pi}5=-\cos\frac{4\pi}5=\frac{\phi}2 \) si \( \cos\pm\frac{3\pi}5=-\frac{2\pi}5=-\frac{\phi^{-1}}2 \).
Avem:
\( 0{\leq}f(\pi)=1+\phi\cos\pi+A\cos2\pi+B\cos3\pi=1-\phi+A-B \) si
\( 0{\leq}f(\frac{3\pi}5)=1+\phi\cos\frac{3\pi}5+A\cos\frac{6\pi}5+B\cos\frac{9\pi}5= \)
\( 1+\phi\cdot(-\frac{\phi^{-1}}2)-\frac\phi2A+\frac\phi2B= \)
\( \frac12-\frac\phi2A+\frac\phi2B \).
Daca inmultim a doua ineg. cu \( 2\phi^{-1}=2(\phi-1) \) obtinem \( 0\leq\phi-1-A+B \). Prin urmare trebuie sa avem egalitati adica \( f(\pi)=f(\frac{3\pi}5)=0 \) si \( A-B=\phi-1 \).
Observam ca \( f \) are minim local (si global) in \( \frac{3\pi}5 \). Rezulta ca \( f\prime(\frac{3\pi}5)=0 \). Cum \( \sin2x=2{\sin}x{\cos}x \) si \( sin3x={\sin}x(2\cos2x+1) \) avem \( f\prime(x)=-\phi{\sin}x-2A\sin2x-B\sin3x=-{\sin}x(\phi+2{\cos}xA+3(2\cos2x+1)B) \).
Prin urmare: \( 0=f\prime(\frac{3\pi}5)=-\sin\frac{3\pi}5(\phi+2\cos\frac{3\pi}5A+3(2\cos\frac{6\pi}5+1)B) \) \( =-\sin\frac{3\pi}5(\phi-\phi^{-1}A+(-\phi+1)B) \), care implica \( 0=\phi-\phi^{-1}A-\phi^{-1}B \), adica \( A+B=\phi^2=\phi+1 \). Impreuna cau \( A-B=\phi-1 \) asta implica \( A=\phi \) si \( B=1 \).
Pt. suficienta, avem \( f(x)=P({\cos}x) \) unde \( P \) are gradul 3 si primul coeficient \( 4B=4 \). Cum \( f(\pi)=f(\frac{3\pi}5)=f\prime(\frac{3\pi}5) \) deci \( P(\cos\pi)=P(\cos\frac{3\pi}5)=P\prime(\cos\frac{3\pi}5)=0 \). Rezulta ca \( P(X)=4(X+1)(X+\frac{\phi^{-1}}2)^2 \), care e \( \geq0 \) pe intervalul \( [-1,1] \).