If the quantity, ${{X}_{r}}=\cos \dfrac{\pi }{{{2}^{r}}}+i\sin \dfrac{\pi }{{{2}^{r}}}$, prove that ${{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty =-1$.
Answer
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Hint: We have been given${{X}_{r}}=\cos \dfrac{\pi }{{{2}^{r}}}+i\sin \dfrac{\pi }{{{2}^{r}}}$, solve it for$ r=1,2,3$. Then use$\cos \theta +i\sin \theta ={{e}^{i\theta }}$to convert and then use the formula for the sum of infinite G.P, which is${{s}_{\infty }}=\dfrac{a}{1-r}$. Solve it, you will get the answer.
Complete step-by-step answer: A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence $2,6,18,....$ is a geometric progression with a common ratio $3$. Similarly, $10,5,2.5,......$ is a geometric sequence with common ratio $\dfrac{1}{2}$. Examples of a geometric sequence are powers of a fixed number ${{r}^{k}}$, such as ${{2}^{k}}$ and ${{3}^{k}}$. The general form of a geometric sequence is, $a,ar,a{{r}^{2}},....$ The ${{n}^{th}}$ term of a geometric sequence with initial value$a={{a}_{0}}$ and the common ratio $r$ is given by, ${{a}_{n}}=a{{r}^{n-1}}$ Such a geometric sequence also follows the recursive relation. ${{a}_{n}}={{a}_{^{n-1}}}r$ for every integer $n\ge 1$. Generally, to check whether a given sequence is geometric, one simply checks whether successive entries in the sequence all have the same ratio. The common ratio of a geometric sequence may be negative, resulting in an alternating sequence, with numbers alternating between positive and negative. The behavior of a geometric sequence depends on the value of the common ratio. If the common ratio is: $*$ Positive, the terms will all be the same sign as the initial term. $*$ Negative, the terms will alternate between positive and negative. $*$ Greater than 1, there will be exponential growth towards positive or negative infinity (depending on the sign of the initial term). $*$ 1, the progression is a constant sequence. $*$ Between −1 and 1 but not zero, there will be exponential decay towards zero. $*$ −1, the progression is an alternating sequence. $*$ Less than −1, for the absolute values there is exponential growth towards (unsigned) infinity, due to the alternating sign. Now we have been given that, ${{X}_{r}}=\cos \dfrac{\pi }{{{2}^{r}}}+i\sin \dfrac{\pi }{{{2}^{r}}}$ We know $\cos \theta +i\sin \theta ={{e}^{i\theta }}$. Now put $r=1$. ${{X}_{1}}=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}={{e}^{i\dfrac{\pi }{2}}}$
Now put $r=2$. ${{X}_{2}}=\cos \dfrac{\pi }{{{2}^{2}}}+i\sin \dfrac{\pi }{{{2}^{2}}}=\cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4}={{e}^{i\dfrac{\pi }{4}}}$ Also, now put $r=3$. ${{X}_{3}}=\cos \dfrac{\pi }{{{2}^{3}}}+i\sin \dfrac{\pi }{{{2}^{3}}}=\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}={{e}^{i\dfrac{\pi }{8}}}$
So for\[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty ={{e}^{i\dfrac{\pi }{2}}}.{{e}^{i\dfrac{\pi }{4}}}.{{e}^{i\dfrac{\pi }{8}}}......={{e}^{\pi i\left( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+... \right)}}\] Here we can see that\[\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.....\]is a series of G.P. Because we can see the common ratio $r=\dfrac{1}{2}$. Finding the sum of terms in a geometric progression is easily obtained by applying the formula : ${{n}^{th}}$partial sum of a geometric sequence, ${{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}$ So for$\infty $, ${{S}_{\infty }}=\dfrac{a}{1-r}$ Here the first term$=a=\dfrac{1}{2}$ and $r=\dfrac{1}{2}$. So using the formula we get,
${{S}_{\infty }}=\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}}=1$ So \[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty ={{e}^{i\dfrac{\pi }{2}}}.{{e}^{i\dfrac{\pi }{4}}}.{{e}^{i\dfrac{\pi }{8}}}......={{e}^{\pi i\left( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+... \right)}}={{e}^{\pi i(1)}}={{e}^{\pi i}}\] So we get, \[{{e}^{\pi i}}=\cos \pi +i\sin \pi \] We know that $\cos \pi =-1$ and $\sin \pi =0$. Substituting these values in the above equation, we get, \[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty ={{e}^{\pi i}}=\cos \pi +i\sin \pi =-1+0=-1\] So we get the final answer \[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty =-1\]. Hence proved.
Note: Read the question in a careful manner. You should be familiar with the concept of G.P. You must know the sum of $n$ terms which are ${{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}$ and also, for $\infty $ which is ${{S}_{\infty }}=\dfrac{a}{1-r}$. Don’t make a mistake while simplifying. Most students make silly mistakes.
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