How and why does it mention Theorem 13?

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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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It's unfortunate that n appears in both Thm. 13 and in your work. Instead of ##P_{2n}##, write this as ##P_m## (where m = 2n). Then what does Thm. 13 say about your degree m polynomial
 
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mcastillo356 said:
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?

Theorem 13 states that if f(x) = Q_n(x) + O((x-a)^{n+1}) then Q_n(x) = P_n(x). Thus it applies to non-polynomial f provided it can written as a polynomial of degree n plus an error term which is no larger than O((x-a)^{n+1}).
 
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Mark44 said:
It's unfortunate that n appears in both Thm. 13 and in your work. Instead of ##P_{2n}##, write this as ##P_m## (where m = 2n). Then what does Thm. 13 say about your degree m polynomial
¡The same as about the ##n## at most order polynomial at Theorem 13!. ##n\in{\mathbb{N}}##?
 
mcastillo356 said:
thus, we can mention theorem 13 at the end of the exercise?
Getting a polynomial is direct, but Theorem 13 is required to prove that the polynomial is the same as the Maclaurin polynomial
 
mcastillo356 said:
¡The same as about the ##n## at most order polynomial at Theorem 13!. ##n\in{\mathbb{N}}##?
Yes, of course...
 
Mark44 said:
It's unfortunate that n appears in both Thm. 13 and in your work. Instead of ##P_{2n}##, write this as ##P_m## (where m = 2n).
Definetely, thanks, @Mark44.
@pasmith, @FactChecker, Theorem 13 is been very difficult to understand, but I worked it out in a Spanish Maths Forum. I'd love to share with PF, if required to.
Peace and Love!
 
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