Prove Taylor expansion with mean value theorem

Question

http://vapour-trail.blogspot.com/2006/03/brief-explanation-of-taylor-series-via.html provides an intuitive derivation of Taylor expansions from the mean value theorem that confuses me.

The derivation is described as follow. By the mean value theorem we have (assuming that $f$ has the differentiability properties requires for an infinite Taylor expansion)

$$f(x+\Delta)=f(x)+\Delta\cdot\frac{df}{dx}(\xi_1),\quad x<\xi_1<x+\Delta \tag4$$

Then one can reapply the mean value theorem to the first derivative in equation 4 to get

\begin{array} { l } \displaystyle { \frac { d f } { d x } \left( \xi _ { 1 } \right) = \frac { d f } { d x } ( x ) + \left( \xi _ { 1 } - x \right) \cdot \frac { d ^ { 2 } f } { d x ^ { 2 } } \left( \xi _ { 2 } \right) \quad ; x < \xi _ { 2 } < \xi _ { 1 } } \\ \displaystyle{ \Delta _ { 2 } \equiv \left( \xi _ { 1 } - x \right) } \\ \displaystyle{ \frac { d f } { d x } \left( \xi _ { 1 } \right) = \frac { d f } { d x } ( x ) + \Delta _ { 2 } \cdot \frac { d ^ { 2 } f } { d x ^ { 2 } } \left( \xi _ { 2 } \right) } \\[2ex] \displaystyle f ( x + \Delta ) = f ( x ) + \Delta \cdot \frac { d f } { d x } \left( \xi _ { 1 } \right) \,\,\text{ from equation (4)}\\ \displaystyle { f ( x + \Delta ) = f ( x ) + \Delta \left[ \frac { d f } { d x } ( x ) + \Delta _ { 2 } \cdot \frac { d ^ { 2 } f } { d x ^ { 2 } } \left( \xi _ { 2 } \right) \right] } \\ \displaystyle { f ( x + \Delta ) = f ( x ) + \Delta \cdot \frac { d f } { d x } ( x ) + \Delta \cdot \Delta _ { 2 } \cdot \frac { d ^ { 2 } f } { d x ^ { 2 } } \left( \xi _ { 2 } \right) } \end{array}

Repeating the process indefinitely yields

$$ f ( x + \Delta ) = f ( x ) + \Delta \cdot \frac { d f } { d x } ( x ) + \Delta \cdot \Delta _ { 2 } \cdot \frac { d ^ { 2 } f } { d x ^ { 2 } } ( x ) + \Delta \cdot \Delta _ { 2 } \cdot \Delta _ { 3 } \cdot \frac { d ^ { 3 } f } { d x ^ { 3 } } ( x ) + \ldots, \\ f(x+\Delta)=\sum _ { n = 0 } ^ { \infty } \left[ \left( \prod _ { p = 1 } ^ { n } \Delta _ { p } \right) \cdot \frac { d ^ { n } f } { d x ^ { n } } ( x ) \right], \quad \Delta_0=1 \,\,\&\,\, \Delta_1=\Delta $$

Now, as noted by the author, the last equation would be equivalent to an infinite Taylor series if we had

$$\Delta_p=\frac{\Delta_1}{p},\quad \forall p\in N\tag 8.$$

The author writes "Unfortunately, equation 8 is not as easy to derive/proof as its simplicity may otherwise suggest. As such, this article would end here." Someone asked for hints in the comments but the author never answered.

Could anyone provide a proof of (equation 8)?

It seems weird to me. For instance we have $\xi_1 = x + \Delta_2$. Equation 8 says $\Delta_2 = \frac{\Delta}{2}$. So $\xi_1 = x + \frac{\Delta}{2}$. Wouldn't that imply that for every $f$ (again having the required properties) and for every interval $[x,x+\Delta]$ in the domain of $f$, the mean value is always half way through the interval (i.e. the derivative at the half point in the interval is equal the the mean derivative over the interval)? This is obviously false (right?) so where is my mistake (if any)?

Answer 1

Equation (8) is certainly not true. A simple counterexample is $f(t)=t^3, x=0, \Delta=1,$ because we get $f'(\xi_1)=3\xi_1^2=f(1)-f(0)=1, \xi_1={1\over \sqrt3}$ and hence $\Delta_2=\xi_1={1\over \sqrt3}\neq {\Delta \over 2}.$ Also Taylor's Theorem doesn't say that an infinitely differentiable function necessarily coincides with its Taylor series. All Taylor's Theorem does is give a measure for the error made by estimating a function with its $n^{th}$ Taylor polynomial.

If you want to learn about the connection between the Taylor expansion and the Mean Value Theorem I suggest:
http://en.wikipedia.org/wiki/Taylor%27s_theorem
http://www.proofwiki.org/wiki/Taylor%27s_Theorem/One_Variable

Answer 2

A non-rigorous approach can be writing the power series and finding the coefficients through reasoning...

$$ f(x) = a_0 + a_1(x-a) + a_2(x-a)^2 + \dots + a_n(x-a)^n + \dots $$

from there $a_0=f(a)$, $a_1= f'(a)$ and successive terms can be found by differentiating both sides further.

Answer 3

I'm not entirely sure what the exact proof is, but I would like to point something out. Let us take a look at:

$$\Delta_p = \frac{\Delta_1}{p}$$

I think on this one we have to think backwards. By using the mean value theorem we ended up with the intervals beyond $x$: $$x < x + \Delta_p + \Delta_{p-1} + \dots + \Delta_2 < x + \Delta_1$$ for all integers $p$. As each of these $\Delta_k$ is a individual limit; that is, we may choose how small the number beyond $x$ is for each $\Delta_k$(as long as it is follows the mean value theorem and it in fact stays in our bounded interval). Furthermore we may define: $$\Delta_k = \frac{\Delta_1}{p}$$ for $k\leq p$.

Now if it will work with other limits is beyond me, but I think this is a start to solving the problem. Perhaps if we used any other interval it would converge too fast and wouldn't be as accurate of an approximation? (I'm thinking in particularly of $\Delta_k = \frac{\Delta_1}{p^2}$ )

Answer 4

I'm not very experienced with math myself, by I belive that by comparing the expansion using the Mean Value approach and the Taylor expansion we can deduct the value of $Δ_p$ in terms of $Δ$.

From Mean Value Theorem: $$ f(x+Δ)=f(x)+Δ⋅{\frac{df}{dx}}(x)+Δ⋅Δ_2⋅{\frac{d^2f}{dx^2}}(x)+Δ⋅Δ_2⋅Δ_3⋅{\frac{d^3f}{dx^3}}(x)+ ... \tag{1}\label{eq1} $$

From Taylor Theorem: $$ f(x+Δ)=f(x)+Δ⋅{\frac{df}{dx}}(x)+Δ^2⋅{\frac{1}{2!}}⋅{\frac{d^2f}{dx^2}}(x)+Δ^3⋅{\frac{1}{3!}}⋅{\frac{d^3f}{dx^3}}(x)+ ... \tag{2}\label{eq2} $$

If \eqref{eq1} and \eqref{eq2} are valid, then: $$ Δ_p = {\frac{Δ_1}{p}} $$ Where \eqref{eq1} can be expanded in terms of Δ $$ f(x+Δ)=f(x)+Δ⋅{\frac{df}{dx}}(x)+Δ⋅{\frac{Δ}{2}}⋅{\frac{d^2f}{dx^2}}(x)+Δ⋅{\frac{Δ}{2}}⋅{\frac{Δ}{3}}⋅{\frac{d^3f}{dx^3}}(x)+ ... $$