Chapter 10.1, Problem 25E

To determine

To find: The values of $x$ if the series ${\displaystyle \sum _{n=0}^{\infty }{\sin }^{n}x}$ converges and then also find the function of $x$ .

Answer to Problem 25E

The function of $x$ is $\frac{1}{1-\sin x}$ with $x\ne \left(2c+1\right)\frac{\pi }{2}$ if the series ${\displaystyle \sum _{n=0}^{\infty }{\sin }^{n}x}$ converges.

Explanation of Solution

Given:

The series is ${\displaystyle \sum _{n=0}^{\infty }{\sin }^{n}x}$ .

Concept used:

The mathematical expression of infinite geometric series in general is written as $a+ar+a{r}^2+a{r}^3+\cdots$ where the first term of the series is $a$ and the common ratio between two consecutive terms is $r$ .

The convergence test of series states that the geometric series converges if $\left|r\right|<1$ otherwise the series diverges where $r$ is the constant ratio.

Furthermore, the convergence test of geometric series also ensures that the convergent infinite geometric series ${\displaystyle \sum _{n=1}^{\infty }a}{r}^{n-1}$ converges to $\frac{a}{1-r}$ .

Calculation:

The series ${\displaystyle \sum _{n=0}^{\infty }{\sin }^{n}x}$ must be converging if the sequence ${\sin }^{n}x$ is 0.

The series ${\displaystyle \sum _{n=0}^{\infty }{\sin }^{n}x}$ can be written as ${\displaystyle \sum _{n=0}^{\infty }{\left(\sin x\right)}^n}$ .

Now in order to converge, the magnitude of rate for the function $\sin x$ must be less than 1 since $x^n$ converges to 0 if $0<x<1$ .

Therefore, the function $\left|\sin x\right|<1$ .

Now solve the absolute inequality $\left|\sin x\right|<1$ by the concept of inequality.

  $-1<\sin x<1$

It can be seen that the value of $\sin x$ lies between $-1\text{ and 1}$ for the convergence of the series.

The value of $\left|\sin x\right|=1$ at all odd multiples of $\frac{\pi }{2}$ . Therefore, the series ${\displaystyle \sum _{n=0}^{\infty }{\sin }^{n}x}$ will converge for all $x$ except odd multiples of $\frac{\pi }{2}$ .

It can be observed that the series ${\displaystyle \sum _{n=0}^{\infty }{\left(\sin x\right)}^n}$ is the infinite geometric series.

The given function fits the formula for infinite geometric series ${\displaystyle \sum _{n=1}^{\infty }a}{r}^{n-1}$ that converges to $\frac{a}{1-r}$ .

The initial term is $1$ and each sequential term is multiplied by $\left(\sin x\right)$ and the last term is the $n^{th}$ term that is equal to ${\left(\sin x\right)}^n$ .

The values are obtained as $a=1,r={\sin }^{n}x$ .

The series ${\displaystyle \sum _{n=0}^{\infty }{\left(\sin x\right)}^n}$ will converge to $\frac{1}{1-\sin x}$ with $x\ne \left(2c+1\right)\frac{\pi }{2}$ .

Conclusion:

Thus, the function of $x$ is $\frac{1}{1-\sin x}$ with $x\ne \left(2c+1\right)\frac{\pi }{2}$ if the series ${\displaystyle \sum _{n=0}^{\infty }{\sin }^{n}x}$ converges.