# Is it possible to algebraically prove that the $n$th degree Taylor remainder of $f(x)$ is less than $K|\Delta x|^{n+1}$ for $K \in \mathbb{R^+}$?

I found a purely algebraic proof, given below, that for a mononomial $f(x) = x^n$ the magnitude of the error of its linear approximation $| f(x) - [f(a) + f'(a)(x-a)] |$ is less than $K(x-a)^2$ for $K \in \mathbb{R^+}$. I can generalize this result to polynomials, and hence any Taylor series.

I was wondering if it's possible, using algebraic means*, to generalize this proof for the error of any kind of approximation. For example, I'm looking to prove that the magnitude of the error of the quadratic approximation of a monomial, $| f(x) - [f(a) + f'(a)(x-a) + f''(a)(x - a)^2 / 2! ] |$, is less than or equal to $K(x-a)^3$ .

Furthermore, is it possible to prove that it is impossible to find $K$ such that $| f(x) - [f(a) + f'(a)(x-a)] | < K(x-a)^n$ for $n > 2$? Or to find $K$ such that the $n$th degree Taylor remainder is less than $K(x-a)^{n+2}$?

I'm asking all of this mainly to attain a better understanding of the Taylor series, and the limitations of algebra as opposed to tools such as the bisection algorithm.

*(I'm aware that what I'm asking can be proven using, for example, the Lagrange form of the Taylor remainder.)

Proposition 1 (What I've proved). Let $f(x) = x^n$. Then for all $a, x \in [-I,I]$, there exists $K \in \mathbb{R^+}$ such that the error of the linear approximation of $f(x)$ at $a$ is less than $K(x-a)^2$: $$|f(x) - f(a) - f'(a)(x-a)| \leq K(x-a)^2$$

Proof. \begin{align*} f(x) - f(a) - f'(a)(x-a)&= (x-a)\sum_{i=1}^n x^{n-i}a^{i-1} - (x-a)na^{n-1} \\ &= (x-a)\sum_{i=1}^n(x^{n-i}a^{i-1}-a^{n-1}) \\ &= (x-a)\sum_{i=1}^n(x^{n-i}-a^{n-i})a^{i-1}.\\ &= (x-a)\sum_{i=1}^{n-1}(x^{n-i}-a^{n-i})a^{i-1}\\ &= (x-a)\sum_{i=1}^{n-1}(x-a)\sum_{j=1}^{n-i}(x^{n-i-j}-a^{j-1})a^{i-1} \\ &= (x-a)^2\sum_{i=1}^{n-1}\sum_{j=1}^{n-i}x^{n-i-j}a^{i+j-2}.\\ \therefore \quad |f(x) - f(a) - f'(a)(x-a)| &= (x-a)^2\left|\sum_{i=1}^{n-1}\sum_{j=1}^{n-i}a^{n-i-j}x^{i+j-2}\right|\\ &\leq (x-a)^2\left|\sum_{i=1}^{n-1}\sum_{j=1}^{n-i}I^{n-i-j}I^{i+j-2}\right|.\\ \end{align*}

Proposition 2 (An attempt to generalize Proposition 1). Let $f(x) = x^n$. Then for all $a, x \in [-I,I]$, there exists $K \in \mathbb{R^+}$ such that the error of the quadratic approximation of $f(x)$ at $a$ is less than $K(x-a)^3$: $$|f(x) - f(a) - f'(a)(x-a) - f''(a)(x-a)^2| \leq K(x-a)^3$$

Proof (Incomplete). The right hand side of this proof is identical to the proof for proposition 1, save for the final set of equations after the $\therefore$ sign. \begin{array}{l} f(x) - f(a) - f'(a)(x-a) - \frac{f''(a)}{2!}(x-a)^2 + \frac{n(n-1)a^{n-2}}{2}(x-a)^2\\ = x^n - a^n - (x-a)na^{n-1} \\ = (x-a)\sum_{i=1}^n x^{n-i}a^{i-1} - (x-a)na^{n-1}\\ = (x-a)\sum_{i=1}^n(x^{n-i}a^{i-1}-a^{n-1})\\ = (x-a)\sum_{i=1}^n(x^{n-i}-a^{n-i})a^{i-1}\\ = (x-a)\sum_{i=1}^{n-1}(x^{n-i}-a^{n-i})a^{i-1} \\ = (x-a)\sum_{i=1}^{n-1}(x-a)\sum_{j=1}^{n-i}(x^{n-i-j}-a^{j-1})a^{i-1} \\ = (x-a)^2\sum_{i=1}^{n-1}\sum_{j=1}^{n-i}x^{n-i-j}a^{i+j-2}\\ \therefore\quad f(x) - f(a) - f'(a)(x-a) - \frac{f''(a)}{2!}(x-a)^2 \\ = (x-a)^2\sum_{i=1}^{n-1}\sum_{j=1}^{n-i}x^{n-i-j}a^{i+j-2} - \frac{n(n-1)a^{n-2}}{2}(x-a)^2\\ = (x-a)^2\left[\sum_{i=1}^{n-1}\left(\sum_{j=1}^{n-i}x^{n-i-j}a^{i+j-2} - \frac{na^{n-2}}{2}\right)\right]\\ \end{array}

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Not only I am sure that this can be done but also I think I have seen it somewhere. Unfortunately I absolutely cannot remember where. Probably in some introductory book in algebraic geometry. –  Giuseppe Negro Dec 14 '12 at 2:58