The Remainder of Taylor Polynomial

Question

In the Lang's Calculus bok says that

$R_n (x)=\dfrac {f^{n}\left( c\right) } {n!}(x-a)^{n}$ where $c$ is between $x$ and $a$.

But, I saw that in the a lecture note,

$R_n (x)=\dfrac {f^{n+1}\left( c\right) } {(n+1)!}(x-a)^{n+1}$ where $c$ is between $x$ and $a$.

Which is the true? So, for an example, $f(x)=e^x$. The remainder ,as the lecture note, is $R_4=\left| \dfrac {e^{c}} {120}\left( x-1\right) ^{5}\right| $ but as Lang's the book formula, $R_4=\left| \dfrac {e^{c}} {24}\left( x-1\right) ^{4}\right|$.

Answer 1

Langrange's form of remainder goes as follows:

Let $f$ posess atleast $n+1$ derivatives over an open interval $I$ and let $c\in I$.

Let $$P_n(x)=\sum_0^n\dfrac{f^{(i)}(c)(x-c)^i}{i!}$$ and let $$R_n(x)=f(x)-P_n(x)$$ Then for each $x\in I$ there exists $z$ between $x$ and $c$ such that $$R_n(x)=\dfrac{f^{(n+1)}(z)(x-c)^{(n+1)}}{(n+1)!}$$

In your example, $e^x$ is infinitely differentiable over $R$. So, for a the Taylor polynomial of degree $4$, the remainder should be of the second form you mentioned.