   Chapter 11.10, Problem 28E

Chapter
Section
Textbook Problem

# Prove that the series obtained in Exercise 25 represents sin x for all x.

To determine

To prove: The series expansion of n=0(1)n+1(2n+1)!(xπ)2n+1 represents sinx for all x .

Explanation

Given:

Consider the series from Exercise 25, n=0(1)n+1(2n+1)!(xπ)2n+1 .

Result used:

(1)Taylor’s Inequality:  If |f(n+1)(x)|M for |xa|d , then the remainder Rn(x) of the Taylor series satisfies the inequality |Rn(x)|M(n+1)!|xa|n+1 for |xa|d  .

(2) For every real numbers x, limnxnn!=0 .

Theorem used:

(1) Squeeze Theorem: If xnznyn for nN and limnxn=limnyn=L , then the value of limnzn is L.

(2) If f(x)=Tn(x)+Rn(x) , where Tn is nth degree Taylor polynomial of f at a and limn|Rn(x)|=0 for |xa|<R , then f is equal to the sum of its Taylor series on the interval |xa|<R .

Proof:

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

 n fn(x) 0 sinx 1 cosx 2 −sinx 3 −cosx 4 sinx 5 −sinx ⋮ ⋮

From the above table, it can be concluded that f(n+1)(x)=±sinx or ±cosx and a=0

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