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$f(x) = {\log _2}\left( {1 + \tan \left( {\frac{{\pi x}}{4}} \right)} \right)$

$\mathop {\lim }\limits_{n \to \infty } \frac{2}{n}\left( {f\left( {\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + ............ + f(1)} \right)$=?

Let $f: (0, 2) \to R $ be defined as $f(x) = {\log _2}\left( {1 + \tan \left( {\frac{{\pi x}}{4}} \right)} \right)$. Then $\mathop {\lim }\limits_{n \to \infty } \frac{2}{n}\left( {f\left( {\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + ............ + f(1)} \right)$ is equal to _ _ _ _ .

Solution

The given limit $\mathop {\lim }\limits_{n \to \infty } \frac{2}{n}\left( {f\left( {\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + ............ + f(1)} \right)$

$ = \mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{r = 1}^n {f\left( {\frac{r}{n}} \right)} } \right).\frac{2}{n}$

$ = \int\limits_0^1 {f(x).2dx = I} $

$I = 2\int\limits_0^1 {{{\log }_2}\left( {1 + \tan \frac{{\pi x}}{4}} \right)} dx$

$ \Rightarrow I = 2\int\limits_0^1 {{{\log }_2}\left( {1 + \tan \frac{{\pi (1 - x)}}{4}} \right)} dx = 2\int\limits_0^1 {{{\log }_2}\left( {1 + \tan \left( {\frac{\pi }{4} - \frac{{\pi x}}{4}} \right)} \right)} dx$

$ \Rightarrow I = 2\int\limits_0^1 {{{\log }_2}\left( {1 + \frac{{1 - \tan \frac{{\pi x}}{4}}}{{1 + \tan \frac{{\pi x}}{4}}}} \right)} dx = 2\int\limits_0^1 {{{\log }_2}\left( {\frac{2}{{1 + \tan \frac{{\pi x}}{4}}}} \right)} dx$

$ \Rightarrow I = 2\int\limits_0^1 {\left[ {\log {}_22 - {{\log }_2}\left( {1 + \tan \frac{{\pi x}}{4}} \right)} \right]} dx = 2\int\limits_0^1 {\left[ {1 - {{\log }_2}\left( {1 + \tan \frac{{\pi x}}{4}} \right)} \right]} dx$

$ \Rightarrow I = 2 - I$

$ \Rightarrow I = 1$