Question

Find the Maclaurin series for \(f(x)\) by using the definition of a Maclaurin series and also the radius of the convergence.

\(f(x) = \sinh x\)

Short answer

The Maclaurin series is \(f(x) = \sum\limits_{n = 0}^\infty {\frac{{{x^{2n + 1}}}}{{(2n + 1)!}}} \) and radius of convergence is\(\infty \).

Step-by-step solution

Concept used

(1) The expansion of the Maclaurin series\(f(x) = \sum\limits_{n = 0}^\infty {\frac{{{f^{(n)}}(0)}}{{n!}}} \)is,

\(f(0) + \frac{{{f^\prime }0}}{{1!}}x + \frac{{{f^{\prime \prime }}(0)}}{{2!}}{x^2} + \frac{{{f^{\prime \prime \prime }}(0)}}{{3!}}{x^3} + \cdots \)

(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 several derivatives of \(f(x)\) 

As given, the function is \(f(x) = \sinh x\)

Maclaurin series for \(f(x)\) is\(\sum\limits_{n = 0}^\infty {\frac{{{f^n}(0) \cdot {x^n}}}{{n!}}} \)

\(\begin{aligned}{l}f(x) &= \sinh x \Rightarrow f(0) &= \sinh (0) &= 0\\{f^\prime }(x) &= \cosh x \Rightarrow {f^\prime }(0) &= \cosh (0) &= 1\\{f^{\prime \prime }}(x) &= \sinh x \Rightarrow {f^{\prime \prime }}(0) &= \sinh (0) &= 0\\{f^3}(x) &= \cosh x \Rightarrow {f^3}(0) &= \cosh (0) &= 1\\{f^4}(x) &= \sinh x \Rightarrow {f^4}(0) &= \sinh (0) &= 0\end{aligned}\)

We see the pattern that

\(\begin{aligned}{l}{f^{2n}}(0) &= 0\\{f^{2n + 1}}(0) &= 1\end{aligned}\)

Maclaurin series for \(f(x)\)

Maclaurin series for \(f(x)\) is

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

Find the radius of convergence 

By ratio test, a series converges if

\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| < 1\)

Here,

\(\begin{aligned}{l}{a_n} = \frac{{{x^{2n + 1}}}}{{(2n + 1)!}}\\{a_{n + 1}} = \frac{{{x^{2n + 3}}}}{{(2n + 3)!}}\end{aligned}\)

We have

\(\begin{aligned}{l}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {{a_{n + 1}} \times \frac{1}{{{a_n}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{2n + 3}}}}{{(2n + 3)!}} \times \frac{{(2n + 1)!}}{{{x^{2n + 1}}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^2}}}{{(2n + 2)(2n + 3)}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \left| {\frac{{{x^2}}}{\infty }} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = 0\end{aligned}\)

It's worth noting that the limit is always 0, regardless of the value of\(x\).

As a result, for all values of\(x\), the Maclaurin series converges.

The Maclaurin series is \(f(x) = \sum\limits_{n = 0}^\infty {\frac{{{x^{2n + 1}}}}{{(2n + 1)!}}} \) .

Hence, interval of convergence is \(\mathbb{R}\), Radius of convergence is \(\infty \).