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On one can find an hint at how to derive Taylor expansions from the mean value theorem.

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

(equation 4)

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

enter image description here

Repeating the process indefinitely yields

enter image description here

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

enter image description here (equation 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 $\xi_1 = x + \frac{\Delta}{2}$. Doesn't it mean 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?

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1 Answer 1

up vote 3 down vote accepted

(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:'s_Theorem/One_Variable

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