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- Homework Statement
- Find the Maclaurin polynomial of order ##2n## for ##\cosh{x}##

- Relevant Equations
- Theorem 13: If ##f(x)=Q_{n}(x)+\Big((x-a)^{n+1}\Big)## as ##x\rightarrow{a}## where ##Q_{n}(x)## is a polynomial of degree at most ##n##, then ##Q_{n}(x)=P_{n}(x)##, that is, ##Q_{n}(x)## is the Taylor polynomial for ##f(x)## at ##x=a##.

Maclaurin formula for ##e^x## with errors in Big O form as ##x\rightarrow{0}##:

##e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots+\dfrac{x^n}{n!}+O\Big(x^{n+1}\Big)##

**Solution**Write the Taylor formula for ##e^x## at ##x=0##, with ##n## replaced by ##2n+1##, and then rewrite that with ##x## replaced with ##-x##. We get:

$$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots+\dfrac{x^{2n}}{(2n)!}+\dfrac{x^{2n+1}}{(2n+1)!}+O(x^{2n+2})$$

$$e^{-x}=1-x+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\cdots+\dfrac{x^{2n}}{(2n)!}-\dfrac{x^{2n+1}}{(2n+1)!}+O(x^{2n+2})$$

as ##x\rightarrow{0}##. Now average these two to get

$$\cosh{x}=\dfrac{e^x+e^{-x}}{2}=1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\cdots+\dfrac{x^{2n}}{(2n)!}+O(x^{2n+2})$$

as ##x\rightarrow{0}##. By Theorem 13 the Maclaurin polynomial ##P_{2n}(x)## for ##\cosh{x}## is

$$P_{2n}(x)=1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\cdots+\dfrac{x^{2n}}{(2n)!}$$

**Question**Isn't Theorem 13 stated for polynomials no more than ##n##-th-order? Or is there also understood that works as far as error term is of higher degree than the polynomial obtained: thus, we can mention theorem 13 at the end of the exercise?

¡Greetings!

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