Question

Question\(\sum\limits_{n = 0}^\infty {\frac{{{x^{2n}}}}{{(2n)!}}} {\rm{ }}\)

Prove the series expansion represents \(\cosh x\) for all \(x.\)

Short answer

The proving is stated below.

Step-by-step solution

The 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 squeeze theorem for calculation.

The derivatives of the function, \(f(x) = \cosh x\)is tabulated below.

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

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

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

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

Thus, by the 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.

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

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

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

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

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

Hence, the given series expansion represents \(\cosh x\) for all \(x.\)