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I am attempting to understand the connection between the Mean Value Theorem and Taylor's Theorem. For the 2nd order expansion, I was able to use the MVT to derive a remainder that looks very similar to Taylor's remainder. However, I find that Taylor's theorem sets the coefficient on the remainder to be $\frac{1}{2}$ $(\frac{1}{n !}$ for general case in $\mathbb{R}^n$), while the MVT-based derivation yields a coefficient that falls somewhere in the interval $(0,1)$. Why does Taylor's theorem set the coefficient to a specific value in the $(0,1)$ interval? Below is the derivation I have been referring to.

Let $f:U \subseteq \mathbb{R} \mapsto \mathbb{R}$, where $f \in C^2$ on $\mathbb{R}$. Let $x,y \in \mathbb{R}$, and let $x<y$. Given this, we can represent $f(y)$ as follows:$$f(y)=f(x) + f'(x)(y-x)+R_2(y)$$ Isolating the remainder term from above eq., and applying the Mean Value Theorem (MVT) twice, I can show the following: \begin{align} R_2(y)&=f(y)-f(x) - f'(x)(y-x) \\ &= f'(z)(y-x) - f'(x)(y-x) \ \ \text{ where} \ z \in (x,y) \ \ \ \ \ \ \text{[By MVT on }f(y)-f(x)] \\ &= (y-x) (f'(z)- f'(x)) \\ &= (y-x) (f''(z')(z-x)) \ \ \text{where} \ z \in (x,z) \subseteq(x,y) \ \ \ \ \ \ \text{[By MVT on }f'(z)-f'(x)] \\ &= (y-x) (f''(z')(\overbrace{x+\xi(y-x)}^{\text{implied by} \ z \in (x,y) }-x)) \ \ \text{ where} \ \xi \in (0,1) \\ &= (y-x) f''(z')\xi(y-x) \\ &= \xi f''(z')(y-x)^2 \end{align}

Now, Taylor's Theorem states that $$R_2(y)=\frac{1}{2}f''(z')(y-x)^2 \text{ for some} \ \ z'\in(x,y) $$

The remainder from Taylor's theorem is identical to the remainder I derived, except for the $\xi$ term which has been set to $\xi=\frac{1}{2}$ in Taylor's Theorem, while $\xi \in (0,1)$ in the MVT-based derivation above. Why does Taylor's theorem fix $\xi=\frac{1}{2}$?

lasoon
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    Looking superficially: it's not the same $z'$ in both cases, which your notation may hide. – Clement C. Nov 15 '17 at 22:18
  • To Clement's point, if you want to write $f(x)=a_0+a_1x+a_2\frac{x^2}{2}$, then while Taylor's theorem will give you a possible set of $a_i$, these might not be optimal, in the sense that say, you would like to minimize $|f(x)-a_0-a_1x+a_2\frac{x^2}{2}|$ on some interval $[m,n]$ – Alex R. Nov 15 '17 at 22:35

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I know this is an older thread, but let me try to give the derivation for which the coefficient is 1/2, using the notation in the original question.

Assume that $f'$ is continuous and differentiable over the closed interval [x, y]. $$ R_{2}(y) = f(y) - f(x) - f'(x)(y-x) $$ Now, taking the derivative with respect to $y$, we have: $$ R'_{2}(y) = f'(y) - 0 - f'(x)\\ \text{or } \frac{R'_{2}(y)}{y-x} = \frac{f'(y) - f'(x)}{y-x} $$ The quantity $\frac{f'(y) - f'(x)}{y-x}$ represents the average rate of change in the value of the function $f'(t)$ as $t$ goes from $x$ to $y$. So by Mean Value Theorem, $\exists z \in [x, y]$ s.t. $$ f''(z) = \frac{f'(y) - f'(x)}{y-x} = \frac{R'_{2}(y)}{y-x}\\ \text{or } R'_{2}(y) = f''(z)(y-x) $$ Taking anti-derivative with respect to $y$ ( where $f''(z)$ is just a constant here ), we have: $$ R_{2}(y) = \frac{1}{2}f''(z)(y-x)^{2} $$ which gives the result you may be looking for.

abcd
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    Here is an application of the (generalized) mean value theorem in the real line to obtain the Taylor polynomial of order $n$ when a function is $n-1$ times differentiable in a neighborhood of a point $a$, and admits $n$-derivatives at $a$. https://math.stackexchange.com/a/3406636/121671 – Mittens Aug 07 '20 at 04:50
  • Shouldn't the antiderivative include a constant? – user599310 Apr 09 '23 at 11:01
  • Furthermore, why $z$ is a constant when it clearly depends on the values of $x$ and $y$? – user599310 Apr 09 '23 at 16:37