Question

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

Answer 1

Step 1: Simplifying the limit expression:

Find the value of $\underset{n\to \infty }{\lim }\left(\frac{2}{n}\right)\left[f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)..............f\left(1\right)\right]$

Given data

$f:\left(0,2\right)\to R$

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

Consider the given Equation as,

$$\begin{gathered} I=\underset{n\to \infty }{\lim }\left(\frac{2}{n}\right)\left[f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)..............f\left(1\right)\right] \\ I=\underset{n\to \infty }{\lim }\left(\frac{2}{n}\right)\left[f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)..............f\left(\frac{n}{n}\right)\right] \\ I=\underset{n\to \infty }{\lim }\left(\frac{2}{n}\right)\sum _{r=1}^{n}f\left(\frac{r}{n}\right) \\ I=\underset{n\to \infty }{\lim }\left(\frac{2}{n}\right)\sum _{r=1}^n{\log }_2\left[1+\tan \frac{\mathrm{\pi }}{4}\times \frac{r}{n}\right] \end{gathered}$$

Step 2: Forming the Integral function:

Let us assume that

$$\begin{gathered} \frac{r}{n}=t \\ \frac{1}{n}dr=dt \\ \left\{\text{When}\begin{array}{ll}r=1,& t=o\\ r=n,& t=1\end{array}\right. \end{gathered}$$

Then the integral becomes,

$I=2{\int }_0^1{\log }_2\left(1+\tan \frac{\mathrm{\pi }}{4}t\right)dt$

Let us assume that

$$\begin{gathered} \frac{\mathrm{\pi }}{4}=x \\ \frac{\mathrm{\pi }}{4}dt=dx \\ dt=\frac{4}{\mathrm{\pi }}dx \\ \left\{\text{When}\begin{array}{ll}t=0,& x=o\\ t=1& x=\frac{\mathrm{\pi }}{4}\end{array}\right. \end{gathered}$$

Then the integral becomes,

$I=2\times \frac{4}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\log }_2\left(1+\tan n\right)dn......\left(1\right)$

We know that

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$

Then the integral becomes,

$$\begin{gathered} I=\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\log }_2\left(1+\tan \left(0+\frac{\mathrm{\pi }}{4}-x\right)\right)dx \\ I=\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\log }_2\left(1+\tan \left(\frac{\mathrm{\pi }}{4}-x\right)\right)dx \end{gathered}$$

We know that

$$\begin{gathered} \tan \left(A+B\right)=\frac{\tan A+\tan B}{1+\tan A\tan B} \\ \tan \frac{\mathrm{\pi }}{4}=1 \end{gathered}$$

Then the integral becomes,

$$\begin{gathered} I=\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\log }_2\left(1+\frac{1-\tan x}{1+\tan x}\right)dx \\ I=\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\log }_2\left(\frac{2}{1+\tan x}\right)dx \end{gathered}$$

We know that

$\log \left(\frac{m}{n}\right)=\log m-\log n$

Then the integral becomes,

$$\begin{gathered} I=\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\log }_2\left(2\right)-{\log }_2\left(1+\tan x\right)dx \\ I=\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\log }_2\left(2\right)dx-\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\log }_2\left(1+\tan x\right)dx......\left(2\right) \end{gathered}$$

Step 3: Adding the equations $\left(1\right)$ and $\left(2\right)$

$$\begin{gathered} 2I=\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}1.dx-\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\log }_2\left(1+\tan x\right)dx+\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}{\log }_2\left(1+\tan x\right)dx \\ 2I=\frac{8}{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{4}}1.dx \\ 2I=\frac{8}{\mathrm{\pi }}{\left[x\right]}_0^{\frac{\mathrm{\pi }}{4}} \\ 2I=\frac{8}{\mathrm{\pi }}\times \frac{\mathrm{\pi }}{4} \\ 2I=2 \\ I=1 \end{gathered}$$

Hence, ,$\underset{n\to \infty }{\lim }\left(\frac{2}{n}\right)\left[f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)..............f\left(1\right)\right]$ is equal to $1$.