Question

Prove that the series expansion of \(f(x) = \sum\limits_{n = 0}^\infty {\frac{{{x^{2n + 1}}}}{{(2n + 1)!}}} {\rm{ }}\)represents \({\rm{s}}inhx\) for all\(x\).

Short answer

The proving is stated below.

Step-by-step solution

Taylor’s inequality and squeeze theorem.

1. Taylor’s inequality: If\(\left| {{f^{(n + 1)}}(x)} \right| \le M\)for\(|x - a| \le d\), then the remainder\({\rm{Rn(x)}}\)for the Taylor series satisfies the inequality\(\left| {{R_n}(x)} \right| \le \frac{M}{{(n + I)!}}|x - a{|^{n + I}}\)for\(|x - a| \le d\).
2. For every real number\(x,\mathop {\lim }\limits_{n \to \infty } \frac{{{x^n}}}{{n!}} = 0\).
3. Squeeze theorem: If\({x_n} \le {z_n} \le {y_n}{\rm{ for }}n \ge N\)and\(\mathop {\lim }\limits_{n \to \infty } {x_n} = \mathop {\lim }\limits_{n \to \infty } {y_n} = L,\)then the value of\(\mathop {\lim }\limits_{n \to \infty } {z_n}\)is\(L\).
4. If\({\rm{f(x) = Tn(x) + Rn(x)}}\), where\({\rm{Tn}}\)is\({\rm{nth}}\)degree Taylor polynomial for\({\rm{f}}\)at\({\rm{a}}\)and\(\mathop {\lim }\limits_{n \to \infty } \left| {{R_n}(x)} \right| = 0\)for\({\rm{|x - a| < R}}\), then\({\rm{f}}\)is equal to the sum of its Taylor series on the interval\({\rm{|x - a| < R}}\).

Use the Taylor’s inequality and the squeeze theorem for calculation. 

The derivatives of function, \(f(x) = \sinh x\) is tabulated below:

\(n\)	\({f^n}(x)\)	\({f^n}(0)\)
\(0\)	\(\sinh x\)	\(0\)
\(1\)	\(\cosh x\)	\(1\)
\(2\)	\(\sinh x\)	\(0\)
\(3\)	\(\cosh x\)	\(1\)
\(4\)	\(\sinh x\)	\(0\)
\(:\)	\(:\)	\(:\)

From the above table, it can be concluded that \({f^{(n + 1)}}(x) = \cosh x{\rm{ or }}\sinh x\)and\(a = 0\).

Since\(|\sinh x| < |\cosh x| = \cosh x{\rm{ for all }}x\), for each case \(\left| {{f^{(n + 1)}}(x)} \right| \le \cosh x\) for all \(n.\)

If \(d\) is any positive number and\({\rm{|x| < d}}\), then \(\cosh x \le \cosh d\).

That is, \(\left| {{f^{(n + 1)}}(x)} \right| \le \cosh x \le \cosh d\).

Here, \(M = \cosh d\).

Thus, by result (1) stated above\(\left| {{R_n}(x)} \right| \le \frac{{\cosh d}}{{(n + 1)!}}|x - 0{|^{n + 1}}\).

Since\(0 \le \left| {{R_n}(x)} \right| \le \frac{{\cosh d}}{{(n + 1)!}}|x{|^{n + 1}}\), apply the squeeze theorem and obtain the required relation.

\(\begin{aligned}{l}\mathop {\lim }\limits_{n \to \infty } 0 \le \mathop {\lim }\limits_{n \to \infty } 0\left| {{R_n}(x)} \right| \le \mathop {\lim }\limits_{n \to \infty } \frac{{\cosh d}}{{(n + 1)!}}|x{|^{n + 1}}\\0 \le \mathop {\lim }\limits_{n \to \infty } 0\left| {{R_n}(x)} \right| \le \mathop {\lim }\limits_{n \to \infty } \frac{{\cosh d}}{{(n + 1)!}}|x{|^{n + 1}}\end{aligned}\)

Obtain the limit \(\mathop {\lim }\limits_{n \to \infty } \frac{1}{{(n + 1)!}}|x{|^{n + 1}}\) by the result (2) stated above.

\(\mathop {\lim }\limits_{n \to \infty } \frac{{|x{|^{n + 1}}}}{{(n + 1)!}} = 0\)

Compute\(\mathop {\lim }\limits_{n \to \infty } \frac{{\cosh d}}{{(n + 1)!}}|x{|^{n + 1}}\).

\(\begin{aligned}{l}\mathop {\lim }\limits_{n \to \infty } \frac{{\cosh d}}{{(n + 1)!}}|x{|^{n + 1}} = \cosh d \cdot \mathop {\lim }\limits_{n \to \infty } \frac{{|x{|^{n + 1}}}}{{(n + 1)!}}\\ = \cosh d(0)\\ = 0\end{aligned}\)

Thus, by the squeeze theorem \(\mathop {\lim }\limits_{n \to \infty } \left| {{R_n}(x)} \right| = 0.\)

That is, \(\left| {{R_n}(x)} \right| \to 0{\rm{ as }}n \to \infty {\rm{ for }}|x| \le d,\) here \(d\)is an arbitrary positive number.

Since \(\mathop {\lim }\limits_{n \to \infty } \left| {{R_n}(x)} \right| = 0,\) it can be concluded that the sum is equal to its Taylor series.

Here, the given series expansion represents \(\sinh x\) for all \(x.\)