Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Practice Exercises - Page 553: 59

$\Sigma_{n=0}^\infty \dfrac{(-1)^n (\pi)^{2n+1} (x)^{2n+1}}{(2n+1)!}$

Work Step by Step

Since, we know that the Taylor Series for $\sin x$ is defined as: $\sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$ Now, $\sin (\pi x)=\Sigma_{n=0}^\infty \dfrac{(-1)^n (\pi x)^{2n+1}}{(2n+1)!}=(\pi x)-\dfrac{(\pi x)^3}{3!}+\dfrac{(\pi x)^5}{5!}-\dfrac{(\pi x)^7}{7!}+...$ or, $=\Sigma_{n=0}^\infty \dfrac{(-1)^n (\pi)^{2n+1} (x)^{2n+1}}{(2n+1)!}$

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