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 \[\dfrac{1}{{(2n + 3)(2n + 2)}}\] .