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I am studying from Spivak' Calculus, and he states Taylor's Theorem as follows:


Let $f',\cdots,f^{(n+1)}$ be defined on $[a,x]$ and let $R_{n,a}(x)$ be defined by

$$R_{n,a}(x)=f(x)-\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}x^k$$

Then, for some $t\in (a,x)$

$$\eqalign{ & {R_{n,a}}(x) = \frac{{{f^{\left( {n + 1} \right)}}\left( t \right)}}{{n!}}{\left( {x - t} \right)^n}\left( {x - a} \right) \cr & {R_{n,a}}(x) = \frac{{{f^{\left( {n + 1} \right)}}\left( t \right)}}{{\left( {n + 1} \right)!}}{\left( {x - a} \right)^{n + 1}} \cr} $$

Moreover, if $f^{(n+1)}$ is integrable over $[a,x]$; then

$${R_{n,a}}(x) = \int\limits_a^x {\frac{{{f^{\left( {n + 1} \right)}}\left( t \right)}}{{n!}}{{\left( {x - t} \right)}^n}} dt$$

On the other hand, Landau's older textbook states:

THEOREM Let $h>0$. Suppose $f^{(n)}$ continuous for $0\leq x\leq h$ and differentiable on $0<x<h$. Let

$$\Phi=f(h)-\sum_{k=0}^n \frac{f^{(k)}(0)}{k!}h^k$$

Then, there exists a $t$ in $(0,h)$ such that


and then uses this to prove.

THEOREM (Taylor's Theorem) Let $h>0$. Suppose $f^{(n)}$ continuous for $\mu \leq x\leq \mu+h$ and differentiable on $\mu <x<\mu+h$. Then there exists an $t$ such that $\mu<t<\mu +h $ and$$f(\mu+h)=\sum\limits_{v = 0}^n {\frac{{{f^{\left( v \right)}}\left( \mu \right)}}{{v!}}} {h^v} + \frac{{{h^{n+1}}}}{{(n+1)!}}{f^{\left( n+1 \right)}}\left( t \right)$$

Now: I see the hypothesis are both the same, and they both address the remainder, but I can't see why Landau fixes $h>0$ and then gives a formula for this fixed $h$ and the $t$ (he actually names this $x$, but I found it a little conflicting) , while Spivak gives the remainder as a function of $x$ and a fixed $t$. Maybe it is just to make his (Landau's) proof simpler? I see how to go from Spivak's result to Landau's, but not the other way around.

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some $x$ and $t$'s are mixed up. on the fourth formula you perhaps need $dt$. – Maesumi Oct 19 '12 at 19:42
@Maesumi Corrected. – Pedro Tamaroff Oct 19 '12 at 19:43
up vote 2 down vote accepted

The Landau's version of Taylor theorem is eq. 3 of Spivak's version.

They are actually defined in the same way, up to a translation to make the segment $[a,x]$ start from the origin. You can jump from Spivak version to Landau using the coordinate transformation $$s(p)=\frac{(p-a)h}{x-a},$$ where $p$ is the point in the Spivak coordinate system you're considering (in particular if $p=t_{\text{Spivak}}$ then $s(p)=t_{\text{Landau}}$).

I don't think you can conclude Spivak stronger result without requiring integrability.

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