Evaluate the product:
$$\Pi= \prod_{n=1}^{\infty} \frac{1}{1-\tan^2 2^{-n}}$$
Solution
Let $ \displaystyle a_n = \frac{1}{{1 - {{\tan }^2}\left( {{2^{ - n}}} \right)}} = \frac{{{{\cos }^2}\left( {{2^{ - n}}} \right)}}{{{{\cos }^2}\left( {{2^{ - n}}} \right) - {{\sin }^2}\left( {{2^{ - n}}} \right)}} = \frac{{{{\cos }^2}\left( {{2^{ - n}}} \right)}}{{{{\cos }}\left( {{2^{ - n + 1}}} \right)}}$.
Then:
$$\begin{aligned}
{a_1}{a_2} \cdots {a_n} &=\frac{{{{\cos }^2}\left( {{2^{ - 1}}} \right)}}{{\cos \left( {{2^0}} \right)}} \cdot \frac{{{{\cos }^2}\left( {{2^{ - 2}}} \right)}}{{\cos \left( {{2^{ - 1}}} \right)}} \cdots \frac{{{{\cos }^2}\left( {{2^{ - n}}} \right)}}{{\cos \left( {{2^{ - n + 1}}} \right)}} \\
&= \frac{1}{{\cos 1}}\cos \left( {\frac{1}{2}} \right)\cos \left( {\frac{1}{4}} \right) \cdots \cos \left( {\frac{1}{{{2^n}}}} \right)\\
&= \frac{1}{{\cos 1}}\frac{{\cos \left( {\frac{1}{2}} \right)\cos \left( {\frac{1}{4}} \right) \cdots 2\sin \left( {\frac{1}{{{2^n}}}} \right)\cos \left( {\frac{1}{{{2^n}}}} \right)}}{{2\sin \left( {\frac{1}{{{2^n}}}} \right)}}\\
&= \frac{1}{{\cos 1}}\frac{{\cos \left( {\frac{1}{2}} \right)\cos \left( {\frac{1}{4}} \right) \cdots \cos \left( {\frac{1}{{{2^{n - 1}}}}} \right)\sin \left( {\frac{1}{{{2^{n - 1}}}}} \right)}}{{2\sin \left( {\frac{1}{{{2^n}}}} \right)}}\\
&= \frac{1}{{\cos 1}} \cdot \frac{{\sin 1}}{{{2^n}\sin \left( {\frac{1}{{{2^n}}}} \right)}} = \tan 1 \cdot \frac{1}{{{2^n}\sin \left( {\frac{1}{{{2^n}}}} \right)}}
\end{aligned}$$
Hence $\displaystyle \prod_{n=1}^{\infty}\frac{1}{1-\tan^2 2^{-n}}= \tan 1$ and the evaluations are complete.
The exercise can also be found in mathematica.gr
$$\Pi= \prod_{n=1}^{\infty} \frac{1}{1-\tan^2 2^{-n}}$$
Solution
Let $ \displaystyle a_n = \frac{1}{{1 - {{\tan }^2}\left( {{2^{ - n}}} \right)}} = \frac{{{{\cos }^2}\left( {{2^{ - n}}} \right)}}{{{{\cos }^2}\left( {{2^{ - n}}} \right) - {{\sin }^2}\left( {{2^{ - n}}} \right)}} = \frac{{{{\cos }^2}\left( {{2^{ - n}}} \right)}}{{{{\cos }}\left( {{2^{ - n + 1}}} \right)}}$.
Then:
$$\begin{aligned}
{a_1}{a_2} \cdots {a_n} &=\frac{{{{\cos }^2}\left( {{2^{ - 1}}} \right)}}{{\cos \left( {{2^0}} \right)}} \cdot \frac{{{{\cos }^2}\left( {{2^{ - 2}}} \right)}}{{\cos \left( {{2^{ - 1}}} \right)}} \cdots \frac{{{{\cos }^2}\left( {{2^{ - n}}} \right)}}{{\cos \left( {{2^{ - n + 1}}} \right)}} \\
&= \frac{1}{{\cos 1}}\cos \left( {\frac{1}{2}} \right)\cos \left( {\frac{1}{4}} \right) \cdots \cos \left( {\frac{1}{{{2^n}}}} \right)\\
&= \frac{1}{{\cos 1}}\frac{{\cos \left( {\frac{1}{2}} \right)\cos \left( {\frac{1}{4}} \right) \cdots 2\sin \left( {\frac{1}{{{2^n}}}} \right)\cos \left( {\frac{1}{{{2^n}}}} \right)}}{{2\sin \left( {\frac{1}{{{2^n}}}} \right)}}\\
&= \frac{1}{{\cos 1}}\frac{{\cos \left( {\frac{1}{2}} \right)\cos \left( {\frac{1}{4}} \right) \cdots \cos \left( {\frac{1}{{{2^{n - 1}}}}} \right)\sin \left( {\frac{1}{{{2^{n - 1}}}}} \right)}}{{2\sin \left( {\frac{1}{{{2^n}}}} \right)}}\\
&= \frac{1}{{\cos 1}} \cdot \frac{{\sin 1}}{{{2^n}\sin \left( {\frac{1}{{{2^n}}}} \right)}} = \tan 1 \cdot \frac{1}{{{2^n}\sin \left( {\frac{1}{{{2^n}}}} \right)}}
\end{aligned}$$
Hence $\displaystyle \prod_{n=1}^{\infty}\frac{1}{1-\tan^2 2^{-n}}= \tan 1$ and the evaluations are complete.
The exercise can also be found in mathematica.gr
Now that we know that value of the product we can get the value of a series involving $\log$. Taking $\log $ at both sides , we have:
ReplyDelete$$\ln \Pi = \ln \left ( \prod_{n=1}^{\infty} \frac{1}{1-\tan^2 2^{-n}} \right )= \sum_{n=1}^{\infty}\ln \left ( \frac{1}{1-\tan^2 2^{-n}} \right )= \\ - \sum_{n=1}^{\infty}\ln \left ( 1-\tan^2 2^{-n} \right )$$
Hence:
$$ \sum_{n=1}^{\infty}\ln \left ( 1-\tan^2 2^{-n} \right )=-\ln \tan 1 $$