Question

What is the radius of convergence of the Maclaurin’s series expansion for the function $f\left(x\right)=\sinh x$?

Answer 1

Hint: We solve this problem by finding the Maclaurin’s expansion of the given function. The Maclaurin;s expansion of a function $p\left(x\right)$ is given as,
$p\left(x\right)=p\left(0\right)+{\displaystyle \frac{x}{1!}}{p}^{\prime }\left(0\right)+{\displaystyle \frac{x^2}{2!}}{p}^{″}\left(0\right)+{\displaystyle \frac{x^3}{3!}}{p}^{‴}\left(0\right)+........$
We use the standard formula of hyperbolic sine trigonometric function that is,
$\sinh x={\displaystyle \frac{e^x-e^{-x}}{2}}$
After finding the Maclaurin’s series expansion we find the values of ${}^{\prime }{x}^{\prime }$ for which the series is convergent using the ratio test which is given as if $a_n$ is the $n^{th}$ term of the series and,
$l={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{a_{n+1}}{a_n}}\right|}$
If $l<1$ then the series is convergent and if $l\ge 1$ then the series is divergent.

Complete step-by-step solution:
We are given that the function as,
$f\left(x\right)=\sinh x$
We know that the formula of hyperbolic sine function is given as,
$\sinh x={\displaystyle \frac{e^x-e^{-x}}{2}}$
By using this formula we get the given function as,
$\Rightarrow f\left(x\right)={\displaystyle \frac{e^x-e^{-x}}{2}}$
Now, let us find the Maclaurin’s expansion for the above function.
Let us substitute $x=0$ in above function then,
$\Rightarrow f\left(0\right)={\displaystyle \frac{e^0-e^{-\left(0\right)}}{2}}=0$
Now, let us differentiate the given function and substitute $x=0$ then we get,
$\begin{array}{rl}& \Rightarrow f^{\prime }\left(x\right)={\displaystyle \frac{e^x+e^{-x}}{2}}\\ & \Rightarrow f^{\prime }\left(0\right)={\displaystyle \frac{e^0+e^{-0}}{2}}=1\end{array}$
Now, let us differentiate the function again and substitute $x=0$ then we get,
$\begin{array}{rl}& \Rightarrow f^{″}\left(x\right)={\displaystyle \frac{e^x-e^{-x}}{2}}\\ & \Rightarrow f^{″}\left(0\right)={\displaystyle \frac{e^0-e^{-0}}{2}}=0\end{array}$
Now, let us differentiate the function again and substitute $x=0$ then we get,
$\begin{array}{rl}& \Rightarrow f^{‴}\left(x\right)={\displaystyle \frac{e^x+e^{-x}}{2}}\\ & \Rightarrow f^{‴}\left(0\right)={\displaystyle \frac{e^0+e^{-0}}{2}}=1\end{array}$
We know that the Maclaurin’s expansion for a function $p\left(x\right)$ is given as,
$p\left(x\right)=p\left(0\right)+{\displaystyle \frac{x}{1!}}{p}^{\prime }\left(0\right)+{\displaystyle \frac{x^2}{2!}}{p}^{″}\left(0\right)+{\displaystyle \frac{x^3}{3!}}{p}^{‴}\left(0\right)+........$
By using this formula to above given function then we get the Maclaurin expansion as,
$\begin{array}{rl}& \Rightarrow f\left(x\right)=0+{\displaystyle \frac{x}{1!}}\left(1\right)+{\displaystyle \frac{x^2}{2!}}\left(0\right)+{\displaystyle \frac{x^3}{3!}}\left(1\right)+........\\ & \Rightarrow f\left(x\right)=x+{\displaystyle \frac{x^3}{3!}}+{\displaystyle \frac{x^5}{5!}}+......\end{array}$
Now, let us represent the above series in summation then we get,
$\Rightarrow f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{x^{2n+1}}{(2n+1)!}}$
Here, we can see that the $n^{th}$ term of above series is given as,
$a_n={\displaystyle \frac{x^{2n+1}}{(2n+1)!}}$
Now, let us use the ratio test.
We know that the ratio test which is given as if $a_n$ is the $n^{th}$ term of the series and,
$l={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{a_{n+1}}{a_n}}\right|}$
If $l<1$ then the series is convergent and if $l\ge 1$ then the series is divergent.
Now, let us find the value of $l$ for the given function then we get,
$$\begin{array}{rl}& \Rightarrow l={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{\left[{\displaystyle \frac{x^{2(n+1)+1}}{(2(n+1)+1)!}}\right]}{\left[{\displaystyle \frac{x^{2n+1}}{(2n+1)!}}\right]}}\right|}\\ & \Rightarrow l={\displaystyle \underset{n\to \mathrm{\infty }}{lim}|{\displaystyle \frac{x^{2n+3}}{(2n+3)!}}\times {\displaystyle \frac{(2n+1)!}{x^{2n+1}}}|}\end{array}$$
Here, we know that the term $(2n+3)!$ can be represented in terms of $(2n+1)!$ as,
$\Rightarrow (2n+3)!=\left[(2n+3)(2n+2)\right](2n+1)!$
By replacing this value in above equation then we get,
$$\begin{array}{rl}& \Rightarrow l={\displaystyle \underset{n\to \mathrm{\infty }}{lim}|{\displaystyle \frac{x^{2n+1}\times x^2}{\left[(2n+3)(2n+2)\right](2n+1)!}}\times {\displaystyle \frac{(2n+1)!}{x^{2n+1}}}|}\\ & \Rightarrow l=\left|x^2\right|{\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{1}{(2n+3)(2n+2)}}\right|}\end{array}$$
Here, we can see that, as the value $n\to \mathrm{\infty }$ the denominator in the limit tends to $\mathrm{\infty }$ so that fraction inside the limit tends to ‘0’.
So, we can take the value of ${}^{\prime }{l}^{\prime }$ as,
$\Rightarrow l=0<1$
Here, we can say that the given function is convergent.
So, we can say that for any value of ${}^{\prime }{x}^{\prime }$ the series is convergent.
We know that the radius of convergence is given as the value of ${}^{\prime }{x}^{\prime }$ for which the function is convergent.
So, the radius of convergence is given as real numbers $\mathbb{R}$
Therefore, we can conclude that the radius of given function $f\left(x\right)=\sinh x$ is given as,
$\therefore x\in \mathbb{R}$

Note: Students may make mistakes by not giving the radius of convergence.
To find the radius of convergence we need to find the value of ${}^{\prime }{x}^{\prime }$ where the given function or series is convergent so that the answer for this problem should be $x\in \mathbb{R}$
But students may make mistakes by concluding that the given function is convergent. Given function may or may not be convergent for any value of ${}^{\prime }{x}^{\prime }$ so we need to find those values. Bit concluding the question directly mentioning that the given function is convergent is not the correct answer for the question.
We need to give the answer which is relevant to the question asked.