Question

If ${x_n} = \cos \left( {\dfrac{\pi }{{{2^n}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^n}}}} \right),$ then ${x_1}.{x_2}.{x_3}..........\infty$ is equal to_________

Answer 1

Hint : We solve this by putting the values of n in the given equation. We use the relations between cosine, sine and exponential function ${e^{i\theta }} = \cos \theta + i\sin \theta$ . Which is also known as Euler’s formula in complex analysis. Using this find the value of ${x_1}$ , ${x_2}$ , ${x_3}$ …. and substituting in ${x_1}.{x_2}.{x_3}..........\infty$ we get the required value.

Complete step-by-step answer :
Now, given ${x_n} = \cos \left( {\dfrac{\pi }{{{2^n}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^n}}}} \right)$ . ---- (1) and ${x_1}.{x_2}.{x_3} - - - - \infty$ ----- (2)
To find the values of ${x_1}$ , ${x_2}$ , ${x_3}$ ….
Put $n = 1$ in equation (1) we get:
$\Rightarrow {x_1} = \cos \left( {\dfrac{\pi }{{{2^1}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^1}}}} \right)$
On the right hand side, Comparing with ${e^{i\theta }} = \cos \theta + i\sin \theta$ .
$\Rightarrow {x_1} = {e^{i\left( {\dfrac{\pi }{2}} \right)}}$
Put $n = 2$ in equation (1) we get:
$\Rightarrow {x_2} = \cos \left( {\dfrac{\pi }{{{2^2}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^2}}}} \right)$
$\Rightarrow {x_2} = \cos \left( {\dfrac{\pi }{4}} \right) + i\sin \left( {\dfrac{\pi }{4}} \right)$
On the right hand side, Comparing with ${e^{i\theta }} = \cos \theta + i\sin \theta$ .
$\Rightarrow {x_2} = {e^{i\left( {\dfrac{\pi }{4}} \right)}}$
Put $n = 3$ in equation (3) we get:
$\Rightarrow {x_3} = \cos \left( {\dfrac{\pi }{{{2^3}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^3}}}} \right)$
$\Rightarrow {x_3} = \cos \left( {\dfrac{\pi }{8}} \right) + i\sin \left( {\dfrac{\pi }{8}} \right)$
On the right hand side, Comparing with ${e^{i\theta }} = \cos \theta + i\sin \theta$ .
$\Rightarrow {x_3} = {e^{i\left( {\dfrac{\pi }{8}} \right)}}$
Continuing like this and substituting in equation (2), we get:
${x_1}.{x_2}.{x_3}..........\infty = {e^{i\left( {\dfrac{\pi }{2}} \right)}}.{e^{i\left( {\dfrac{\pi }{4}} \right)}}.{e^{i\left( {\dfrac{\pi }{8}} \right)}} - - - - \infty$
Since the base is same on the right hand side we can add the exponent’s term, we get:
Taking $i$ as a common, we get:
$= {e^{i\left( {\left( {\dfrac{\pi }{2}} \right) + \left( {\dfrac{\pi }{4}} \right) + \left( {\dfrac{\pi }{8}} \right) + - - - - \infty } \right)}}$
Taking $\pi$ as common, we get:
$= {e^{i\pi \left( {\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + - - - - - - } \right)}}$
We know that $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + - - - - - = \sum\limits_{n = 1}^\infty {{{\left( {\dfrac{1}{2}} \right)}^n}}$
Also,
$\sum\limits_{n = 1}^\infty {{{\left( {\dfrac{1}{2}} \right)}^n}} = \dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{1}{2}}} = 1$ (Taking L.C.M and simplifying)
$= {e^{i\pi \left[ {\dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{1}{2}}}} \right] }}$
$= {e^{i\pi }}$
We know from Euler’s formula that ${e^{i\theta }} = \cos \theta + i\sin \theta$ . Comparing and we get,
$= \cos \pi + i\sin \pi$
We know the values of $\sin \pi = 0$ and $\cos \pi = - 1$ .
Substituting in above we get,
$= - 1$
Hence, the value of ${x_1}.{x_2}.{x_3} - - - - \infty = - 1$ .
So, the correct answer is “-1”.

Note : All we did is find the values of ${x_1}.{x_2}$ and ${x_3}$ using the Euler’s formula. Substitute this in equation in (2), we get the required value. Remember this ${e^{i\theta }} = \cos \theta + i\sin \theta$ well. We used the basic exponent and power formula ${e^a}.{e^b}.{e^c} = {e^{a + b + c}}$ . Since the base is the same we can add the exponent. We express cosine and sine in terms of $e$ . Just be careful in the substituting and calculation part. The calculation part might be difficult.