   Chapter 11.10, Problem 30E

Chapter
Section
Textbook Problem

# Prove that the series obtained in Exercise 18 represents cosh x for all x.

To determine

To prove: The series expansion n=0x2n(2n)! represents coshx for all x .

Explanation

Given:

The series is n=0x2n(2n)! .

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)=coshx as shown below table.

 n fn(x) fn(0) 0 coshx 1 1 sinhx 0 2 coshx 1 3 sinhx 0 4 coshx 1 ⋮ ⋮ ⋮

In the above table, it can be concluded that f(n+1)(x)=coshx or sinhx .

Since |sinhx|<|coshx|=coshx for all x , for each case |f(n+1)(x)|coshx for all n

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