2
$\begingroup$

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 $\xi=\frac{1}{2}$?

$\endgroup$
2
  • 3
    $\begingroup$ Looking superficially: it's not the same $z'$ in both cases, which your notation may hide. $\endgroup$
    – Clement C.
    Nov 15 '17 at 22:18
  • $\begingroup$ 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 $[m,n]$ $\endgroup$
    – Alex R.
    Nov 15 '17 at 22:35
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $a$. math.stackexchange.com/a/3406636/121671 $\endgroup$ Aug 7 '20 at 4:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.