How and why does it mention Theorem 13?

[mcastillo356]

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!

 

[Mark44]

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

 

[pasmith]

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

 

[mcastillo356]

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!. $n\in{\mathbb{N}}$?

 

[FactChecker]

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

 

[Mark44]

¡The same as about the $n$ at most order polynomial at Theorem 13!. $n\in{\mathbb{N}}$?

Yes, of course...

 

[mcastillo356]

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!