Editing Taylor's theorem (section)

=== Alternate proof for Taylor's theorem in one real variable ===

Let <math>f(x)</math> be any real-valued continuous function to be approximated by the Taylor polynomial.

Step 1: Let <math display="inline">F</math> and <math display="inline">G</math> be functions. Set <math display="inline">F</math> and <math display="inline">G</math> to be

<math display="block">\begin{align}
F(x) = f(x) - \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k} 
\end{align}</math>

<math display="block">\begin{align}
G(x) = (x-a)^{n}
\end{align}</math>

Step 2: Properties of <math display="inline">F</math> and <math display="inline">G</math>:

<math display="block">\begin{align}
F(a) & = f(a) - f(a) - f'(a)(a - a) - ... - \frac{f^{(n-1)}(a)}{(n-1)!}(a-a)^{n-1} = 0 \\
G(a) & = (a-a)^n = 0
\end{align}</math>

Similarly, 

<math display="block">\begin{align}
F'(a) = f'(a) - f'(a) - \frac{f''(a)}{(2-1)!}(a-a)^{(2-1)} - ... - \frac{f^{(n-1)}(a)}{(n-2)!}(a-a)^{n-2} = 0
\end{align}</math>

<math display="block">\begin{align}
G'(a) &= n(a-a)^{n-1} = 0\\
&\qquad \vdots\\
G^{(n-1)}(a) &= F^{(n-1)}(a) = 0
\end{align}</math>

Step 3: Use Cauchy Mean Value Theorem

Let <math>f_{1}</math> and <math>g_{1}</math> be continuous functions on <math>[a, b]</math>. Since <math>a < x < b</math> so we can work with the interval <math>[a, x]</math>. Let <math>f_{1}</math> and <math>g_{1}</math> be differentiable on <math>(a, x)</math>. Assume <math>g_{1}'(x) \neq 0</math> for all <math>x \in (a, b)</math>.
Then there exists <math>c_{1} \in (a, x)</math> such that

<math display="block">\begin{align}
\frac{f_{1}(x) - f_{1}(a)}{g_{1}(x) - g_{1}(a)} = \frac{f_{1}'(c_{1})}{g_{1}'(c_{1})}
\end{align}</math>

Note: <math>G'(x) \neq 0</math> in <math>(a, b)</math> and <math>F(a), G(a) = 0</math> so

<math display="block">\begin{align}
\frac{F(x)}{G(x)} = \frac{F(x) - F(a)}{G(x) - G(a)} = \frac{F'(c_{1})}{G'(c_{1})}
\end{align}</math>

for some <math>c_{1} \in (a, x)</math>.

This can also be performed for <math>(a, c_{1})</math>:

<math display="block">\begin{align}
\frac{F'(c_{1})}{G'(c_{1})} = \frac{F'(c_{1}) - F'(a)}{G'(c_{1}) - G'(a)} = \frac{F''(c_{2})}{G''(c_{2})}
\end{align}</math>

for some <math>c_{2} \in (a, c_{1})</math>.
This can be continued to <math>c_{n}</math>.

This gives a partition in <math>(a, b)</math>:

<math display="block">a < c_{n} < c_{n-1} < \dots < c_{1} < x</math>

with

<math display="block">\frac{F(x)}{G(x)} = \frac{F'(c_{1})}{G'(c_{1})} = \dots = \frac{F^{(n)}(c_{n})}{G^{(n)}(c_{n})} .</math>

Set <math>c = c_{n}</math>:

<math display="block">\frac{F(x)}{G(x)} = \frac{F^{(n)}(c)}{G^{(n)}(c)}</math>

Step 4: Substitute back

<math display="block">\begin{align}
\frac{F(x)}{G(x)} = \frac{f(x) - \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k}}{(x-a)^{n}} = \frac{F^{(n)}(c)}{G^{(n)}(c)}
\end{align}</math>

By the Power Rule, repeated derivatives of <math>(x - a)^{n}</math>, <math>G^{(n)}(c) = n(n-1)...1</math>, so:

<math display="block">\frac{F^{(n)}(c)}{G^{(n)}(c)} = \frac{f^{(n)}(c)}{n(n-1)\cdots1} = \frac{f^{(n)}(c)}{n!}.</math>

This leads to:

<math display="block">\begin{align}
f(x) - \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k} = \frac{f^{(n)}(c)}{n!}(x-a)^{n}
\end{align}.</math>

By rearranging, we get:

<math display="block">\begin{align}
f(x) = \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k} + \frac{f^{(n)}(c)}{n!}(x-a)^{n}
\end{align},</math>

or because <math>c_{n} = a</math> eventually:

<math display="block">f(x) = \sum^{n}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k}.</math>