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.