Question

How do you find a power series converging to $f(x) = \dfrac{{\sin x}}{x}$ and determine the radius of convergence?

Answer 1

Hint : First we need to find the power series of $\dfrac{{\sin x}}{x}$ . A power series about ‘a’, or just power series, is any series that can be written in the form $\sum\limits_{n = 0}^\infty {{c_n}{{(x - a)}^n}}$ . Where ‘a’ and ${c_n}$ are numbers. The ${c_n}$ are often called the coefficients of the series. We can find the radius of convergence using the ratio test.

Complete step-by-step answer :
Given, $f(x) = \dfrac{{\sin x}}{x}$ .
Now let’s use the maclaurin expansion.
We know the formula of the maclaurin series: first we find the maclaurin series of $f(x) = \sin x$ .
$\sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(0)}}{{n!}}{{(x - 0)}^k} = f(0) + \dfrac{{f'(0)}}{{1!}}x + } \dfrac{{f''(0)}}{{2!}}{x^2} + \dfrac{{f'''(0)}}{{3!}}{x^3} + \dfrac{{{f^{(4)}}(0)}}{{4!}}{x^4} + \dfrac{{{f^{(5)}}(0)}}{{5!}}{x^5} + .....$
We need $f(0) = \sin (0) = 0$
Differentiating $\sin x$ with respect to ‘x’. $f'(x) = \cos x$
$f'(0) = \cos (0) = 1$
Again differentiate with respect to ‘x’. $f''(x) = - \sin x$
$f''(0) = - \sin (0) = 0$
Again differentiate with respect to ‘x’. $f'''(x) = - \cos x$
$f'''(0) = - \cos (0) = - 1$
Again differentiate with respect to ‘x’. ${f^{(4)}}(0) = \sin x$
${f^{(4)}}(0) = \sin (0) = 0$
Again differentiate with respect to ‘x’. ${f^{(5)}}(0) = \cos x$
${f^{(5)}}(0) = \cos (0) = 1$
Substituting we have,
$\Rightarrow \sin x = 0 + \dfrac{{(1)}}{{1!}}x + \dfrac{{(0)}}{{2!}}{x^2} + \dfrac{{( - 1)}}{{3!}}{x^3} + \dfrac{{(0)}}{{4!}}{x^4} + \dfrac{{(1)}}{{5!}}{x^5} + ......$
$\Rightarrow \sin x = 0 + x + 0 - \dfrac{{{x^3}}}{{3!}} + 0 + \dfrac{{{x^5}}}{{5!}} - ......$
$\Rightarrow \sin x = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - .....$
Expressing this in the sigma notation we have,
$\Rightarrow \sin x = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\dfrac{{{x^{2n + 1}}}}{{(2n + 1)!}}}$
But we need $\dfrac{{\sin x}}{x}$ . So divide the above equation by x
$\Rightarrow \dfrac{{\sin x}}{x} = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\dfrac{{{x^{2n + 1}}}}{{(2n + 1)!}}} .\dfrac{1}{x}$
$\Rightarrow \dfrac{{\sin x}}{x} = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\dfrac{{{x^{2n}}}}{{(2n + 1)!}}}$ . This is the required power series.
Now for checking the radius of convergence, we have ratio test sate that if we have a series $\sum {{a_n}}$ then $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L$ . If L is less than 1 then the series is convergent.
Here ${a_n} = {( - 1)^n}\dfrac{{{x^{2n}}}}{{(2n + 1)!}}$
Lets find $\dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{{\left( {{{( - 1)}^{n + 1}}\dfrac{{{x^{2(n + 1)}}}}{{(2(n + 1) + 1)!}}} \right)}}{{\left( {{{( - 1)}^n}\dfrac{{{x^{2n}}}}{{(2n + 1)!}}} \right)}}$
We have $\dfrac{{{{( - 1)}^{n + 1}}}}{{{{( - 1)}^n}}} = {( - 1)^{n + 1 - n}}$ above becomes
$\Rightarrow \dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{{\left( {{{( - 1)}^{n + 1 - n}}\dfrac{{{x^{2(n + 1)}}}}{{(2n + 2 + 1)!}}} \right)}}{{\left( {\dfrac{{{x^{2n}}}}{{(2n + 1)!}}} \right)}}$
We can rewrite it has,
$\Rightarrow \dfrac{{{a_{n + 1}}}}{{{a_n}}} = - \dfrac{{{x^{2n + 2}}}}{{(2n + 3)!}} \times \dfrac{{(2n + 1)!}}{{{x^{2n}}}}$
Since we have, $\dfrac{{{x^{2n + 2}}}}{{{x^{2n}}}} = {x^{2n + 2 - 2n}} = {x^2}$
$\Rightarrow \dfrac{{{a_{n + 1}}}}{{{a_n}}} = - \dfrac{{{x^2}(2n + 1)!}}{{(2n + 3)!}}$
Now substituting in the limit we have
$\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| { - \dfrac{{{x^2}(2n + 1)!}}{{(2n + 3)!}}} \right|$
Since the limit is for ‘n’ we can treat x as constant and removing outside the limit,
$= {x^2}\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{(2n + 1)!}}{{(2n + 3)!}}} \right|$
$= {x^2}\mathop {\lim }\limits_{n \to \infty } \dfrac{{(2n + 1)!}}{{(2n + 3)!}}$
This can be written as
$= {x^2}\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{(2n + 3)(2n + 2)}}$
$= 0$
The radius of convergence is 0.
So, the correct answer is “0”.

Note : Since it has a long calculation part be careful in each step. In the above problem we have, $\dfrac{{(2n + 1)!}}{{(2n + 3)!}}$ .
We know that $n! = n(n - 1)(n - 2).....$ . Hence we can write $(2n + 3)! = (2n + 3)(2n + 2)(2n + 1)!$ . Hence the numerator terms cancel out we will have $\dfrac{1}{{(2n + 3)(2n + 2)}}$ .