Polinom monic de gradul 4 cu |f(i)|=1
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Polinom monic de gradul 4 cu |f(i)|=1
Fie polinomul \( f=x^4+ax^3+bx^2+cx+d \) cu coeficienti reali si radacinile reale astfel incat \( |f(i)|=1 \). Sa se arate ca \( a=b=c=d=0 \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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pohoatza
Fie \( x_{1}, x_{2}, x_{3}, x_{4} \) cele patru radacini.
Cum insa polinomul se mai poate scrie \( f(x)=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4}) \), deducem deci ca
\( |f(i)|=|i-x_{1}| \cdot |i-x_{2}| \cdot |i-x_{3}| \cdot |i-x_{4}|=\prod_{k=1}^{4}{\sqrt{1+x_{k}^{2}}} \geq 1 \), deci \( x_{k}=0 \), cu \( k=\overline{1,4} \), rezultand astfel ca \( f(x)=x^{4} \)
Cum insa polinomul se mai poate scrie \( f(x)=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4}) \), deducem deci ca
\( |f(i)|=|i-x_{1}| \cdot |i-x_{2}| \cdot |i-x_{3}| \cdot |i-x_{4}|=\prod_{k=1}^{4}{\sqrt{1+x_{k}^{2}}} \geq 1 \), deci \( x_{k}=0 \), cu \( k=\overline{1,4} \), rezultand astfel ca \( f(x)=x^{4} \)