analysisnumber theory
Let $x \in \mathbb{R}$. Then I have the following
\begin{align} e^x (12 – 6x + x^2) &= \left((1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \mathcal{O}(x^5)\right)(12 – 6x + x^2) \\ &= 12 – 6x + 6x^2 + \mathcal{O}(x^5).
\end{align}
However, I don't get the last equation. Isn't $ x \mathcal{O}(x^n) = \mathcal{O}(x^{n+1})$?
I will present the full development as I have solved it.
Lemma. Let $a\in\mathbb{R}^{+}, a>1$ be fixed and $r=\frac{m}{n}\in\mathbb{Q},\, m\in\mathbb{Z},\, n\in\mathbb{N}$. Define $S_{r}(a)=\{x\in\mathbb{R}\mid 0\leq x^{n}\leq a^{m}\}$. $S_{r}(a)$ has supremum.
Proof. Let $m>0$. Define $S_{r}(a)=\{x\in\mathbb{R}\mid 0\leq x^{n}\leq a^{m}\}$. Because $1<a$ then $1=1^{n}=1^{m}=a^{0}<a^{m}$, that is $1\in S_{r}(a)$, thus $S_{r}(a)\neq\emptyset$. On the other hand, because $a>1$, $a^{m}>1$ and $x\geq a^{m}\longleftrightarrow x^{n}\geq a^{mn}>a^{m}$. Then $a^{m}$ is upper bound for $S_{r}(a)$.
Now, $m<0$. Let $S_{r}(a)$ as it was defined before. Because $1<a$ then $1=a^{0}>a^{m}\geq x^{n}>0$, and therefore $1$ is upper bound for $S_{r}(a)$. Also, because $a>1$, $0<a^{m}<1$ and $a^{mn}<a^{m}$. Thus $a^{m}\in S_{r}(a)$ and it follows that $S_{r}(a)\neq\emptyset$.
Then, by supremum axiom, $S_{r}(a)$ has supremum for each $r\in\mathbb{Q}$ and $a>1$.
The reason for taking the set $S_{r}(a)$ needs to be exposed. Intuitively the $x\in S_{r}(a)$ are such that $x<a^{r}$. But, since this is not already defined, we took the set of numbers that are such that $x^{n}<a^{m}$. Therefore, the former lemma says $a^{r}=\sup S_{r}(a)$. Thus, we can make the following
Definition. Let $a\in\mathbb{R}^{+},\, a>1$ fixed and $r\in\mathbb{Q}$. Define $\exp_{a}(r):\mathbb{Q}\longrightarrow\mathbb{R}$ as the function given by \begin{align*} r=\frac{m}{n}\neq 0 &\mapsto a^{r}=\sup S_{r}(a)=\sup\{x\in\mathbb{R}\mid 0\leq x^{n}\leq a^{m}\}\\ 0 &\mapsto 1 \end{align*}
We need to show that this is a good definition. With the former definition we are setting positive solution to the equation $x^{n}=a^{m}$. It could be possible for other positive numbers to be solutions. We must, then, prove that this equation has one and only one rational solution.
Lemma. Let $a\in\mathbb{R}^{+},\, a>1$ fixed and $r\in\mathbb{Q}$. $\rho=\sup S_{r}(a)$ is the unique positive solution of $x^{n}=a^{m}$.
Proof. Let $0<\epsilon<1$, then \begin{align*} (1+\epsilon)^{n} &=\sum_{k=0}^{n}\dbinom{n}{k}\epsilon^{k}\\ &=1+\sum_{k=1}^{n}\dbinom{n}{k}\epsilon^{k}\\ &<1+\sum_{k=1}^{n}\dbinom{n}{k}\epsilon=1+K_{n}\epsilon \end{align*} with $K_{n}$ constant for each $n$.
Thus, take $\frac{\rho}{2}<\rho_{2,\epsilon}<\rho<\rho_{1,\epsilon}<2\rho$ tal que $\rho_{1,\epsilon}=\rho(1+\epsilon)$ y $\rho_{2,\epsilon}=\frac{\rho}{1+\epsilon}$. Suppose that $\rho^{n}<a^{m}$, then exists $\epsilon$ such that $\rho^{n}(1+K_{n}\epsilon)<a^{m}$. Therefore \begin{align*} \rho_{1,\epsilon}^{n} &=\rho^{n}(1+\epsilon)^{n}\\ &<\rho^{n}(1+K_{n}\epsilon)\\ &<a^{m} \end{align*} then $\rho$, which is lesser than $\rho_{1,\epsilon}\in S_{r}(a)$, would not be even a upper bound for $S_{r}(a)$. But $\rho$ is the supremum of this set. Therefore $\rho^{n}\geq a^{m}$. Now, suppose that $\rho^{n}>a^{m}$, then exists $\epsilon$ such that $\frac{\rho^{n}}{1+K_{n}\epsilon}>a^{m}$ and it follows that \begin{align*} \rho_{2,\epsilon}^{n} &=\frac{\rho^{n}}{(1+\epsilon)^{n}}\\ &>\frac{\rho^{n}}{1+K_{n}\epsilon}\\ &>a^{m} \end{align*} thus $\rho$, which is greater than $\rho_{2,\epsilon}$, would not be the least upper bound of $S_{r}(a)$, that is the supremum, as there would be a lesser upper bound. But $\rho$ is the supremum, by hypothesis. Then, it remains not choice but to $\rho^{n}=a^{m}$ as we wanted to show.
On the other hand, suppose that exists $0<\rho_{1}\neq\rho$ such that $\rho_{1}^{n}=a^{m}$. The condition $\rho_{1}\neq\rho$ implies $\rho_{1}<\rho$ or $\rho<\rho_{1}$, and by exponential properties for integers $\rho_{1}<\rho\longleftrightarrow \rho_{1}^{n}<\rho^{n}$ and in a similar way for the other option. But both are equal to $a^{m}$, then $\rho_{1}=\rho$ and the positive root of the equation $x^{n}=a^{m}$ is only $\rho$.
This makes formal the intuitive notion of fractional exponents or roots.
Let us make a lemma for handle some expressions more easily
Lemma. Let $a\in\mathbb{R}^{+},\, a>1$ fixed and $0<r=\frac{m}{n}\in\mathbb{Q}$. Then $\exp_{a}(-r)=\frac{1}{\exp_{a}(r)}$
Proof. We are stating that $\sup S_{-r}(a)=\frac{1}{\sup S_{r}(a)}$ or, with the notation of the previous lemma \begin{align*} \rho_{-r}=\frac{1}{\rho_{r}} \end{align*}
We now by the previous lemma that $\rho_{r}$ is the unique positive solution of $x^{n}=a^{m}$ \begin{align*} \rho_{r}^{n} &=a^{m}\therefore\\ \frac{1}{\rho_{r}^{n}} &=\frac{1}{a^{m}}\\ &=a^{-m} \end{align*} and, also we know that $\rho_{-r}$ es the unique positive solution to $x^{n}=a^{-m}$ then \begin{align*} \rho_{-r}^{n} &=\frac{1}{\rho_{r}^{n}}\\ &=\left(\frac{1}{\rho_{r}}\right)^{n} \end{align*} thus, we conclude \begin{align*} \rho_{-r}=\frac{1}{\rho_{r}} \end{align*} which is the result we wanted to prove.
We need to show three additional results for this definition to be a good one: Function does not depend on the representation for a given rational, it is the same as the one for integers and is well defined for $0<a<1$.
Lemma. Let $a\in\mathbb{R}^{+},\, a>1$ fixed and $r=\frac{m}{n}=\frac{p}{q}\in\mathbb{Q}$ such that $p,q$ are relatively prime. Then \begin{align*} \sup S_{r}(a)=\sup\{x\in\mathbb{R}\mid 0\leq x^{n}\leq a^{m}\}=\sup\{x\in\mathbb{R}\mid 0\leq x^{q}\leq a^{p}\}=\sup T_{r}(a) \end{align*} Proof. From the definition of equality of rationals $mq=np$, then $p$ divides $mq$, but, since $p,q$ son relatively prime then $p$ divides $m$ that is $m=kp$, thus $kqp=np$ and since $q\neq 0$ then $n=kq$ or $q$ divides $n$ and the factor is the same for $m,n$. Notice that $k\in\mathbb{N}$, since $m, p$ must have the same sign. Thus $x\in T_{r}(a)\longleftrightarrow x^{q}\leq a^{p}\longleftrightarrow x^{kq}\leq a^{kp}\longleftrightarrow x^{n}\leq x^{m}\longleftrightarrow x\in S_{r}(a)$, by the exponential properties for naturals. Then $T_{r}(a)=S_{r}(a)$ and finally $\sup S_{r}(a)=\sup T_{r}(a)$.
This lemma solves the first point since every representation of a given rational is connected, via multiples, with the irreductible representation (which is unique, unless order, by the fundamental theorem of arithmetic applied to numerator and denominator).
Lemma. Let $a\in\mathbb{R}^{+},\, a>1$ fixed and $r=\frac{m}{1}\in\mathbb{Q}$. Then \begin{align*} \sup S_{m}(a)=\sup\{x\in\mathbb{R}\mid 0\leq x\leq a^{m}\}=a^{m} \end{align*}
Proof. Obviously we work with $S_{m}(a)=[0,a^{m}]$. It is clear that $a^{m}$ is an upper bound for this set. Moreover, there is no other which is strictly lesser, since if there would be one then we would construct$\sup S_{m}(a)<\sup S_{m}(a)+\frac{a^{m}-\sup S_{m}(a)}{2}<a^{m}$ and then we would see that $\sup S_{m}(a)$ is not an upper bound for $S_{m}(a)$, what is against the fact that it is a supremum. It follows $\sup S_{m}(a)=a^{m}$.
Then the exponential as we have defined reduces to the exponential for integers at integer exponents.
Now, we extend for $0<a<1$. We know how to calculate $\exp_{a}$ when $a>1$. But if $0<a<1$ we also know that $\frac{1}{a}>1$. Thus, in this case, for every $r$ \begin{align*} a^{r} &=\left(\frac{1}{a}\right)^{-r} \end{align*} and this definition is good since we are writing it in terms of something that is well defined, as we have proven. We, then, get.
Definition. Let $a\in\mathbb{R}^{+}$ fixed and $r\in\mathbb{Q}$. Define $\exp_{a}(r):\mathbb{Q}\longrightarrow\mathbb{R}$ as the function given by \begin{align*} r=\frac{m}{n} &\mapsto a^{r}=\sup S_{r}(a)=\sup\{x\in\mathbb{R}\mid 0\leq x^{n}\leq a^{m}\},\, a>1\\ 0 &\mapsto 1\\ r=\frac{m}{n} &\mapsto a^{r}=\left(\frac{1}{a}\right)^{-r},\, 0<a<1 \end{align*}
This definition must follow the same properties that the one for integers, that is
Theorem. Let $a,b\in\mathbb{R}^{+}$ fixed y $r,s\in\mathbb{Q}$ then
Proof.
Best Answer
Note that the coefficient of $x^5$ in $$ e^x (12 - 6x + x^2) $$
$$= \left(1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 +\frac{1}{120}x^5+...\right)(12 - 6x + x^2) \\ $$
$$= 12 - 6x + 6x^2 + \frac {1}{60}x^5+...$$
$$= 12 - 6x + 6x^2 + \mathcal{O}(x^5)$$