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Taylor's theorem

Let $I \subset \mathbb{R}$ be an interval, $x_0 \in I$ and $n \geq 1$. Further let $f: I \to \mathbb{R}$ be a function $n$ times differentiable.

First part of the theorem[edit]

For the funtion $R: I \to \mathbb{R}$ defined by

[math]f \left( x \right) = \left( \sum_{k=0}^{n} \frac{ f^{ \left( k \right) } \left( x_0 \right) }{k!} \cdot \left( x - x_0 \right)^{k} \right) + R \left(x \right)[/math]

it is true that

[math]\lim_{x \to x_0} \frac{ R\left( x \right) }{\left( x - x_0 \right)^{n}} = 0[/math]

We call $\sum_{k=0}^{n} \frac{ f^{ \left( k \right) } \left( x_0 \right) }{k!} \cdot \left( x - x_0 \right)^{k}$ the Taylor polynomial (also Taylor's polynomial) of $n$th degree in $x_0$ and R the corresponding remainder.

Second part of the theorem[edit]

If $f$ is even $n+1$ times differentiable, we will find that for every $x \in I \setminus \{ x_0 \}$ a $\xi$ between $x$ and $x_0$ exists, with

[math]R \left( x \right) = \frac{ f^{ \left( n+1 \right)} \left( \xi \right) }{ \left( n+1 \right)! } \cdot \left( x - x_0 \right)^{n+1}[/math].

This is called the Lagrange form of the remainder.

The proof[edit]

Prerequisite information[edit]

First part[edit]

We will name the Taylor polynomial of $n$th degree $T$:

[math]T \left( x \right) = \sum_{k=0}^{n} \frac{ f^{ \left( k \right) } \left( x_0 \right) }{k!} \cdot \left( x - x_0 \right)^{k}[/math]

From that we find:

[math]\frac{ R \left( x \right) }{ \left( x - x_0 \right)^{n}} = \frac{ f \left( x \right) - T \left( x \right) }{ \left( x - x_0 \right)^{n} }[/math]

We will now use the rule of Bernoulli-De L’Hospital to prove that $\lim_{x \to x_0} \frac{ R \left( x \right) }{ \left( x - x_0 \right)^{n}} = 0$. It is evident that in $ I \setminus \{ x_0 \}$ the first $n$ derivates of $\left( x - x_0 \right)^{n}$ are non zero and that from the construction follows that

$T \left( x_0 \right) = f \left( x_0 \right) $
$T' \left( x_0 \right) = f' \left( x_0 \right) $
$\vdots$
$T^{\left( n \right)} \left( x_0 \right) = f^{\left( n \right)} \left( x_0 \right)$

Repeatin the rule of Bernoulli-De L’Hospital, we deduce that:

[math]\begin{eqnarray}\lim_{x \to x_0} \frac{ R\left( x \right) }{\left( x - x_0 \right)^{n}} &=& \lim_{x \to x_0} \frac{ f \left( x \right) - T \left( x \right) }{ \left( x - x_0 \right)^{n} } \\ &=& \lim_{x \to x_0} \frac{ f' \left( x \right) - T' \left( x \right) }{ n \cdot \left( x - x_0 \right)^{n-1} } \\ &\vdots& \\ &=& \lim_{x \to x_0} \frac{ f^{\left( n-1 \right) } \left( x \right) - T^{\left( n-1 \right) } \left( x \right) }{ n! \cdot \left( x - x_0 \right) } \\ &=& \lim_{x \to x_0} \left( \frac{ f^{\left( n-1 \right) } \left( x \right) - f^{\left( n-1 \right) } \left( x_0 \right) }{ n! \cdot \left( x - x_0 \right) } - \frac{ T^{\left( n-1 \right) } \left( x \right) - T^{\left( n-1 \right) } \left( x_0 \right) }{ n! \cdot \left( x - x_0 \right) }\right) \\ &=& \frac{1}{n!} \cdot \left( f^{ \left( n \right) } \left( x_0 \right) - T^{ \left( n \right) } \left( x_0 \right) \right) = 0\end{eqnarray}[/math]

Second part[edit]

We will look at

[math]h \left( t \right) = \frac{ f \left( t \right) - T \left( t \right) }{ \left( n+1 \right)! } \cdot \left( x- x_0 \right)^{ n+1 } - R \left( x \right) \cdot \frac{ \left( t- x_0 \right)^{ n+1} }{ \left( n+1 \right)! }[/math]

We know that $T^{ \left( n+1 \right)} = 0$ and therefore

[math]h^{ \left( n+1 \right) } \left( t \right) = \frac{ f^{ \left( n+1 \right) } \left( t \right) }{ \left( n+1 \right)! } \cdot \left( x- x_0 \right)^{ n+1 } - R \left( x \right)[/math]

Now all that is left to do ist to show that $h^{ \left( n+1 \right) }$ has a zero $ \xi $ between $x$ and $x_0$.

Ofcourse $h \left( x_0 \right) = h' \left( x_0 \right) = \dots = h^{ \left( n \right) } \left( x_0 \right)$ and [math]h \left( x \right) = \left( f \left( x \right) - T \left( x \right) - R \left( x \right) \right) \cdot \frac{ \left( x - x_0 \right)^{ n+1} }{ \left( n+1 \right)! } = 0[/math]

And then there has to be a $\xi_1$ between $x$ and $x_0$ with $h' \left( \xi_1 \right) = 0$. And then there has to be a $\xi_2$ between $x$ and $x_0$ with $h \left( \xi_2 \right) = 0$. And so on and so forth...

In the end we receive out $\xi = \xi_{n+1}$ between $x$ and $x_0$ with $h^{\left( n+1 \right) } \left( \xi \right) = 0$.