# How and why does it mention Theorem 13?

• mcastillo356
In summary, the conversation discusses rewriting the Taylor formula for e^x at x=0 with n replaced by 2n+1 and x replaced with -x. The average of the new formulas is then used to find the Maclaurin polynomial for cosh x. The question is raised about the use of Theorem 13, which states that if f(x) = Q_n(x) + O((x-a)^{n+1}) then Q_n(x) = P_n(x). The discussion concludes that Theorem 13 can be applied to polynomials of any degree, as long as the error term is no larger than O((x-a)^{n+1}).

#### mcastillo356

Gold Member
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!

Last edited:
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

topsquark and mcastillo356
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 $O((x-a)^{n+1})$.

topsquark and mcastillo356
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!