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In my textbook the Lagrange's remainder which is associated with the Taylor's formula is defined as:

$R_{n}(x)= \frac{(x-a)^n}{n!} f^{(n)}(a + \vartheta (x-a))$, for some $\vartheta$ $\in$ <0 ,1>

1. I don't understand what this $\vartheta$ represents, why is it here what it means

Then is said that some function can be represented by Taylor series only if its Lagrange's remainder which is associated with its Taylor's formula goes to 0 as n goes to infinity (limit).

2. Why is that, what would be the intuition behind it.

Then to show that f(x)=$e^x$ can be represented by Taylor series it only says:

$R_{n}(x)=\frac{x^n}{n!}e^{\vartheta x}$, for some $\vartheta$ in <0, 1>

3. Shouldn't there be $e^{(a+\vartheta (x-a))}$ instead of $e^{\vartheta x}$ according to the definition?

$\lim_{x \to \infty}|R_{n}(x)|\leq e^{\vartheta x} \lim_{x \to \infty}\frac{|x|^n}{n!}=0$

4. Same as the third question and why is there $\leq$ sign? Isn't the left hand sind exactly equal to the right hand side that is shouldn't there be a $=$ sign?

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    $\begingroup$ The reason for the missing $a$ is that they are expanding about $a=0$. The $\le$ part should be equality. Then to get inequality we should replace $\vartheta x$ by $x$ if $x$ is positive, and by $1$ if $x$ is negative. $\endgroup$ Commented Aug 4, 2014 at 15:44

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This is notationally a little different from the most common version of the Lagrange form of the remainder. For one thing, $R_n(x)$ is usually the remainder when we truncate immediately just after the $(x-a)^n$ term. In this version, $R_n(x)$ seems to denote the remainder when we truncate just after the $(x-a)^{n-1}$ term. No problem, but it may lead to confusion if one looks at other sources.

You ask about the $\vartheta$. Using plain $\vartheta$ is useful, but potentially misleading. Actually, $\vartheta$ is a function of $x$ and $n$. Typically, however, we know little about $\vartheta(n,x)$ apart from the fact that it is between $0$ and $1$. Note that to say that $\vartheta$ is between $0$ and $1$ is precisely the same as saying that $a+\vartheta(x-a)$ is between $a$ and $x$. A more common version uses $f^{(n)}(\xi)$ where $\xi=\xi(n,x)$ is between $a$ and $x$.

As to the intuition, I will not say anything except to note that in the case $n=1$ (so we are truncating at the constant term) the error is $(x-a)f'(\xi)$ for some $\xi$ between $a$ and $x$. This is just the Mean Value Theorem. The Lagrange formula for the remainder is an extended version of the Mean Value Theorem, providing, for $n\gt 1$, a refined estimate of $f(x)$ that takes higher derivatives into account.

As I mentioned in a comment, in the discussion of $e^x$ they are taking $a=0$, so they are using the Maclaurin polynomials for $e^x$. This accounts for the seeming disappearance of $a$.

The limit assertion is less precise than it ought to be. Here is what should be said. If $x$ is positive, then $e^{\vartheta x}\le e^x$, since $\vartheta\lt 1$. If $x\lt 0$, then $e^{\vartheta x}\le 1$.

So if we let $m_x=\max(e^x,1)$, we have $$|R_n(x)|\le m_x \frac{|x|^n}{n!}.$$ Finally, let $n\to\infty$. For any fixed $x$, we have $\displaystyle\lim_{n\to\infty}\frac{|x|^n}{n!}=0$.

So for any fixed $x$, the error $R_n(x)$ approaches $0$ as $n\to\infty$. Put in other terms, this says that the Maclaurin series for $e^x$ converges to $e^x$ for all $x$.

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