Question

Find the Taylor series for \(f(x)\) centered at \(\frac{\pi }{2}\)and its radius of the convergence.

\(f(x) = \sin x,a = \frac{\pi }{2}\)

Short answer

The Taylor series is \(f(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n} \cdot {{(x - \pi /2)}^{2n}}}}{{(2n)!}}} \)and radius of convergence is \(\infty \).

Step-by-step solution

Concept used

(1) If\(f\)has a power series expansion at\(a,f(x) = \sum\limits_{n = 0}^\infty {\frac{{{f^{(n)}}(a)}}{{n!}}} {(x - a)^n}\), \(f(x) = f(a) + \frac{{{f^\prime }(a)}}{{1!}}(x - a) + \frac{{{f^{\prime \prime }}(a)}}{{2!}}{(x - a)^2} + \frac{{{f^{\prime \prime \prime }}(a)}}{{3!}}{(x - a)^3} + \cdots \)

(2) The Ratio Test:

(i) If\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L < 1\), then the series\(\sum\limits_{n = 1}^\infty {{a_n}} \)is unconditionally convergent (and it is also convergent).

(ii) If\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L > 1\)or\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \infty \), the series\(\sum\limits_{n = 1}^\infty {{a_n}} \)then becomes divergent.

(iii) If\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = 1\), the Ratio Test inconclusive; that is, no conclusion can be drawn about the convergence or divergence of\(\sum\limits_{n = 1}^\infty {{a_n}} \).

Find the Taylor series 

Taylor series for the function \(f(x)\) at \(x = a\) is\(\sum\limits_{k = 0}^\infty {\frac{{{f^k}(a) \cdot {{(x - a)}^k}}}{{k!}}} \)

\(\begin{aligned}{l}f(x) &= \sin x\\f(\pi /2) &= \sin (\pi /2)\\f(\pi /2) &= 1\end{aligned}\)

\(\begin{aligned}{l}{f^\prime }(x) &= \cos x\\{f^\prime }(\pi /2) &= \cos (\pi /2)\\{f^\prime }(\pi /2) &= 0\end{aligned}\)

\(\begin{aligned}{l}{f^{\prime \prime }}(x) &= - \sin x\\\mathop \smallint \nolimits^{\prime \prime } (\pi /2) &= - \sin (\pi /2)\\\mathop \smallint \nolimits^{\prime \prime } (\pi /2) &= - 1\end{aligned}\)

\(\begin{aligned}{l}{f^3}(x) &= - \cos x\\{f^3}(\pi /2) &= - \cos (\pi /2)\\{f^3}(\pi /2) &= 0\end{aligned}\)

\(\begin{aligned}{l}{f^4}(x) &= \sin x\\{f^5}(\pi /2) &= \sin (\pi /2)\\{f^5}(\pi /2) &= 1\end{aligned}\)

\(\begin{aligned}{l}{f^5}(x) &= \cos x\\{f^5}(\pi /2) &= \cos (\pi /2)\\{f^5}(\pi /2) &= 0\end{aligned}\)

We see the pattern that

\(\left\{ {\begin{aligned}{*{20}{l}}{{f^{2n}}(\pi /2) = {{( - 1)}^n}}\\{{f^{2n + 1}}(\pi /2) = 0}\end{aligned}} \right.\)

Since every integer \(k\) is in the form 2 n or \(2n + 1\), we can write

\(\sum\limits_{k = 0}^\infty {\frac{{{f^k}(\pi /2) \cdot {{(x - \pi /2)}^k}}}{{k!}}} = \sum\limits_{n = 0}^\infty {\frac{{{f^{2n}}(\pi /2) \cdot {{(x - \pi /2)}^{2n}}}}{{(2n)!}}} + \sum\limits_{n = 0}^\infty {\frac{{{f^{2n + 1}}(\pi /2) \cdot {{(x - \pi /2)}^{2n + 1}}}}{{(2n + 1)!}}} \)

Substitute \({f^{2n}}(\pi /2) = {( - 1)^n}\) and \({f^{2n + 1}}(\pi /2) = 0\)

\(\begin{aligned}{l}\sum\limits_{k = 0}^\infty {\frac{{{f^k}(\pi /2) \cdot {{(x - \pi /2)}^k}}}{{k!}}} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n} \cdot {{(x - \pi /2)}^{2n}}}}{{(2n)!}}} + \sum\limits_{n = 0}^\infty {\frac{{0 \cdot {{(x - \pi /2)}^{2n + 1}}}}{{(2n + 1)!}}} \\\sum\limits_{k = 0}^\infty {\frac{{{f^k}(\pi /2) \cdot {{(x - \pi /2)}^k}}}{{k!}}} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n} \cdot {{(x - \pi /2)}^{2n}}}}{{(2n)!}}} + 0\end{aligned}\)

\(f(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n} \cdot {{(x - \pi /2)}^{2n}}}}{{(2n)!}}} \)

Find the radius of convergence 

Use the Ratio Test

The absolute value bars allow us to remove \({( - 1)^{n + 1}}\) and \({( - 1)^n}\)

\(\begin{aligned}{l}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( - 1)}^{n + 1}} \cdot {{(x - \pi /2)}^{2n + 2}}}}{{(2n + 2)!}} \div \frac{{{{( - 1)}^n} \cdot {{(x - \pi /2)}^{2n}}}}{{(2n)!}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{(x - \pi /2)}^{2n + 2}}}}{{(2n + 2)!}} \cdot \frac{{(2n)!}}{{{{(x - \pi /2)}^{2n}}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{(x - \pi /2)}^{2n + 2}}}}{{{{(x - \pi /2)}^{2n}}}}} \right| \cdot \frac{{(2n)!}}{{(2n + 2)!}}\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{{(x - \pi /2)}^{2n + 2 - 2n}}} \right| \cdot \frac{1}{{(2n + 2)(2n + 1)}}\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \frac{{{{(x - \pi /2)}^2}}}{{(2n + 2)(2n + 1)}}\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \frac{{{{(x - \pi /2)}^2}}}{\infty }\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= 0\end{aligned}\)

Since, the limit is always zero, irrespective to the value of \(x\), the Taylor series converges for all values of \(x\).

Hence, the Taylor series is \(f(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n} \cdot {{(x - \pi /2)}^{2n}}}}{{(2n)!}}} \).

The radius of convergence is \(\infty \).