Question

1) Calcule o somatório

∑ sen[2^(− k)] · cos[3 · 2^(− k)]
k de 0 a n

$\mathsf{\displaystyle\sum_{k=0}^n sen(2^{-k})\cdot cos(3\cdot 2^{-k})}$

e expresse a sua fórmula fechada em termos de n.

2) Mostre que a série

∑ sen[2^(− k)] · cos[3 · 2^(− k)]
k de 0 a ∞

$\mathsf{\displaystyle\sum_{k=0}^{\infty} sen(2^{-k})\cdot cos(3\cdot 2^{-k})}$

converge, e determine o seu valor.

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Obs.: Todos os arcos envolvidos estão em radianos.

Answer 1

Olá Lukyo.



Propriedades utilizadas


$\star~~\boxed{\boxed{\mathsf{\Delta a_k=a_{k+1}-a_k}}}\\\\\\\\\star~~\boxed{\boxed{\mathsf{\displaystyle\sum_{k=p}^{n} \Delta a_k=a_{n+1}-a_p}}}\\\\\\\\\star~~\boxed{\boxed{\mathsf{cos(\alpha+\beta)=cos~\alpha\cdot cos~\beta\mp sen~\alpha\cdot sen~\beta}}}\\\\\\\\\star~~\boxed{\boxed{\mathsf{sen(\alpha+\beta)=sen~\alpha\cdot cos~\beta\pm cos~\alpha\cdot sen~\beta}}}$

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1 - 


Chamaremos $\mathsf{2^{-k}}$ de x para facilitar os cálculos.

Vamos escrever aquela igualdade utilizando a identidade, seno da soma e da diferença de dois arcos.


$\mathsf{sen(3x+x)=sen(3x)\cdot cos~x+sen~x\cdot cos(3x)}\\\\\\\mathsf{sen(3x-x)=sen(3x)\cdot cos~x-sen~x\cdot cos(3x)}$


Fazendo a diferença entre as duas igualdades


$\mathsf{sen(3x+x)-sen(3x-x)=2sen~x\cdot cos(3x)}\\\\\\\mathsf{\dfrac{1}{2}\cdot\Big[sen(4x)-sen(2x)\Big]=sen~x\cdot cos(3x)}$


Substituindo


$\mathsf{x=2^{-k}}\\\\\\\mathsf{sen(x)\cdot cos(3x)=}\\\\\\\mathsf{\dfrac{sen(4\cdot2^{-k})-sen(2\cdot2^{-k})}{2}}\\\\\\\mathsf{\dfrac{sen(2^{2-k})-sen(2^{1-k})}{2}}\\\\\\\mathsf{\dfrac{sen(2^{-(k-2)})-sen(2^{-(k-1)})}{2}}\\\\\\\mathsf{\dfrac{-[sen(2^{-(k-1)})-sen(2^{-(k-2)})]}{2}}$

Tomando $\mathsf{a_k=\dfrac{sen(2^{-(k-2)})}{2}}$

$\mathsf{\Delta a_k=\dfrac{sen(2^{-[(k+1)-2]})}{2}-\dfrac{sen(2^{-(k-2)})}{2}}\\\\\\\mathsf{\Delta a_k=\dfrac{sen(2^{-(k-1)})-sen(2^{-(k-2)})}{2}}$

Pela propriedade telescópica

$\mathsf{\displaystyle\sum_{k=p}^n \Delta b_k=b_{n+1}-b_p}\\\\\\\\\mathsf{\displaystyle\sum_{k=0}^nsen(2^{-k})\cdot cos(3\cdot2^{-k})=-\sum_{k=0}^n\Delta a_k=-(a_{n+1}-a_0)=}\\\\\\\mathsf{-\Big[\dfrac{sen(2^{-[(n+1)-2]})-sen(2^{-(0-2)})}{2}\Big]}\\\\\\\\\boxed{\mathsf{\dfrac{1}{2}\cdot\Big[sen(4)-sen(2^{-(n-1)})}\Big]}}$


2 - 


$\mathsf{\displaystyle\sum_{k=0}^{\infty} sen(2^{-k})\cdot cos(3\cdot2^{-k})}$


$\mathsf{\displaystyle\lim_{n\to\infty}\displaystyle\sum_{k=0}^n sen(2^{-k})\cdot cos(3\cdot2^{-k})=}\\\\\\\mathsf{\displaystyle\lim_{n\to\infty} \dfrac{1}{2}\cdot\Big[sen(4)-sen(2^{-(n-1)})\Big]=}\\\\\\\mathsf{\displaystyle\lim_{n\to\infty}\dfrac{1}{2}\cdot\Big[sen(4)-sen\Big(\dfrac{2}{2^n}\Big)\Big]=}\\\\\\\\\mathsf{\dfrac{1}{2}\cdot\Big[sen(4)-sen(0)\Big]}\\\\\\\\\mathsf{\displaystyle\sum_{n=0}^{\infty}sen(2^{-k})\cdot cos(3\cdot2^{-k})=\dfrac{sen(4)}{2}}$



Dúvidas? comente.