Question

Answer it please with proper explanation !! ​

Answer 1

Question : (x) denote the sum of the infinite trigonometric, f(x) = $\sf{\displaystyle{ \sum^{ \infty }_{ \sf \: n = 1} }} \: \: sin \dfrac{2x}{ {3}^{n} } \: \: sin \dfrac{x}{ {3}^{n} }$

Find f(x) (independent of n) also evaluate the sum of the solution of the equation f(x) = 0 lying in the intervals ( 0. 629

Solution

$\sf{ \bf \: f(x) = \displaystyle \sum^{ \infty } _{n = 1} \: \sf \: sin} \: \dfrac{2x}{ {3}^{n} } \: \: \: sin \dfrac{x}{ {3}^{n} }$

$\sf{ \: \qquad \qquad =\dfrac{1}{2} \displaystyle \sum^{ \infty } _{ \sf \: n = 1} \sf2sin \: \dfrac{2x}{ {3}^{n} }\: sin \: \: \dfrac{x}{ {3}^{n} } }$

$\qquad \qquad \qquad \sf = \dfrac{1}{2} \displaystyle \sum^{ \infty } _{ \sf \: n = 1} \bigg \lgroup \sf \: cos \dfrac{x}{ {3}^ n} - cos \frac{x}{ {3}^{n - 1} } \bigg \rgroup$

Now substituting n = 1 , 2, 3 , 4 ....

$\bf \: f(x) = \dfrac{1}{2} \bigg \lgroup \: cos \dfrac{x}{3} - cos \: x \bigg \rgroup + \dfrac{1}{2} \bigg \lgroup \: cos \dfrac{x}{ {3}^{2} } - cos \dfrac{x}{3} \bigg \rgroup$

$\qquad \qquad \sf + \dfrac{1}{2} \bigg \lgroup \: cos \dfrac{x}{3 {}^{3} } - cos \: \dfrac{x}{ {3}^{2} } \bigg \rgroup... + \dfrac{1}{2} \bigg \lgroup \: cos \dfrac{x}{ {3}^{n} } - cos \dfrac{x}{3 {}^{n \: \: 1} } \bigg \rgroup$

$\bf \: f(x) = \displaystyle \: \lim_{ \sf \: n \longrightarrow \infty } \frac{1}{2} \bigg \lgroup \sf- cos \frac{x}{ { 3}^{n} } - cos \: n \bigg \rgroup$

$\bf \: f(x) = 0$

$\qquad \: = - cos \: x \: + 1 = 0 \\ \\ \qquad = \cancel{cos} \: x = 1 = \cancel{cos}2 \: n \: \pi \\ \\ \large \implies\boxed{\pink{x = 2n \pi}}$

$\bf \qquad \qquad \qquad \: where \: n \in \: i$

Answer 2

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