Question

Prove that the series of expansion\(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{(2n + 1)!}}} {(x - \pi )^{2n + 1}}\) represent\(\sin \pi x\) 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 inequality and squeeze theorem for proving.

The derivatives of the function, \(f(x) = \sin \pi x\)is tabulated below:

\(n\)	\({f^n}(x)\)
\(0\)	\(\sin \pi x\)
\(1\)	\(\pi \cos (\pi x)\)
\(2\)	\( - {\pi ^2}\sin (\pi x)\)
\(3\)	\( - {\pi ^3}\cos (\pi x)\)
\(4\)	\( - {\pi ^4}\sin (\pi x)\)
\(:\)	\(:\)

From the above table, it can be concluded that\({f^{(n + 1)}}(x) = \left( { \pm {\pi ^{n + 1}}\sin \pi x} \right)\) or\(\left( { \pm {\pi ^{n + 1}}\cos \pi x} \right)\)and\(a = 0\).

For each case\(\left| {{f^{(n + 1)}}(x)} \right| \le {\pi ^{n + 1}}\) …… (1)

Here, \(M = {\pi ^{n + 1}}\).

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

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

\(\begin{aligned}{l}0 \le \left| {{R_n}(x)} \right| \le \frac{{{\pi ^{n + 1}}}}{{(n + 1)!}}\left| {{x^{n + 1}}} \right|\\0 \le \left| {{R_n}(x)} \right| \le \frac{{\left| {\pi {x^{n + 1}}} \right|}}{{(n + 1)!}}\end{aligned}\)

Obtain the limit\(\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {\pi {x^{n + 1}}} \right|}}{{(n + 1)!}}\).

By the result (2) stated above \(\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {\pi {x^{n + 1}}} \right|}}{{(n + 1)!}} = 0\).

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

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\(\sin \pi x\)for all\(x.\)