Menu Close

Un-ln-cos-1-2-n-show-that-Un-0-




Question Number 201533 by Rodier97 last updated on 08/Dec/23
                 Un = ln (cos (1/2^n ) )      show  that Un ≤0
$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{Un}\:=\:{ln}\:\left({cos}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\right) \\ $$$$\:\:\:\:{show}\:\:{that}\:{Un}\:\leqslant\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by AST last updated on 08/Dec/23
0≤cos((1/(2n)))≤1=e^0 ⇒U_n ≤ln(e^0 )=0
$$\mathrm{0}\leqslant{cos}\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\leqslant\mathrm{1}={e}^{\mathrm{0}} \Rightarrow{U}_{{n}} \leqslant{ln}\left({e}^{\mathrm{0}} \right)=\mathrm{0} \\ $$
Answered by Mathspace last updated on 08/Dec/23
we have 0≤(1/2^n )<(π/2)  ∀n≥1 ⇒  0<cos((1/2^n ))≤1 ⇒ln(cos(1/2^n ))≤0 ⇒  u_n ≤0
$${we}\:{have}\:\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{2}^{{n}} }<\frac{\pi}{\mathrm{2}}\:\:\forall{n}\geqslant\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{0}<{cos}\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\leqslant\mathrm{1}\:\Rightarrow{ln}\left({cos}\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\leqslant\mathrm{0}\:\Rightarrow \\ $$$${u}_{{n}} \leqslant\mathrm{0} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *