## Taylor's Inequality Theorem and Proof

The following theorem, known as Taylor's Inequality, is a necessary condition to show that a given function is

**e****quivalent**to its Taylor polynomial expansion. In essence, the theorem states the following.If the magnitude of the \((n + 1)^{st}\) derivative of \(f\) is bounded on a closed interval centered at \(a\), then the \(n^{th}\) remainder of the Taylor series expansion must be bounded. Moreover, this bound is directly proportional to the \((n + 1)^{st}\) power of \(|x - a|\).

**THEOREM | TAYLOR'S INEQUALITY**

If \(|f^{n + 1}(x)| \leq M \text{ } \forall x \in [a - r,a+r]\), where \(r > 0\), then the remainder, \(R_n(x)\), of the Taylor series satisfies \[|R_n(x)| \leq \frac{M}{(n + 1)!}|x - a|^{n + 1} \quad \text{for } x \in [a - r,a + r].\]

**PROOF | TAYLOR'S INEQUALITY**

Recall that, given a function \(f\) and a center \(a\), \[T_n(x) = \sum_{i = 0}^n{\frac{f^{(i)}(a)}{i!}(x - a)^i}\]and \[R_n(x) = f(x) - T_n(x).\]Given this notation, we can rewrite the conclusion of the theorem as\[|f(x) - T_n(x)| \leq \frac{M}{(n + 1)!}|x - a|^{n + 1}.\]We shall now argue inductively on \(n \in \{0,1,2,\ldots\}\) with the single assumption that \(x \in [a,a + r]\) (the proof does not change much for \(x \in [a - r,a]\)).

Let \(n = 0\) and suppose that \(|f^{(0 + 1)}(x)| = |f'(x)| \leq M \text{ } \forall x \in [a,a + r]\). Since \(f\) is differentiable, the Mean Value Theorem for derivatives states \(\exists c \in (a,a + r)\) such that \(f(x) - f(a) = f'(c) (x - a)\). More specifically, for our needs, this states that \(|f(x) - f(a)| = |f'(c) (x - a)|\). By our assumption, \(|f'(c)| \leq M\). Hence,\[|R_0(x)| = |f(x) - T_0(x)| = |f(x) - f(a)| \leq M |x - a|\]making the theorem true for \(n = 0\).

Assume the theorem is true for \(n = k\). That is, as long as \(|f^{(k)}(x)| \leq M \text{ } \forall x \in [a,a + r]\), then \[|R_k(x)| = |f(x) - T_k(x)| \leq \frac{M}{(k + 1)!}|x - a|^{k + 1}.\]

Let \(n = k + 1\) and suppose that \(f^{(k + 1)}(x) \leq M \text{ } \forall x \in [a,a + r]\). Let \(g(x) = f'(x)\). Then \(g^{(n)}(x) \leq M \text{ } \forall x \in [a,a + r]\). Let \(\overline{R_n}(x)\) be the remainder for the \(n^{th}\)-degree Taylor polynomial for \(g\), \(\overline{T_n}(x)\). By the induction hypothesis, this implies that \[|\overline{R_k}(x)| = |g(x) - \overline{T_k}(x)| \leq \frac{M}{(k + 1)!}|x - a|^{k + 1}.\]This implies that\[\overline{T_k}(x) - \frac{M}{(k + 1)!}|x - a|^{k + 1} \leq g(x) \leq \overline{T_k}(x) + \frac{M}{(k + 1)!}|x - a|^{k + 1}.\]Rewriting this inequality in terms of the original function, we get\[\sum_{i = 0}^k{\frac{f^{(i + 1)}(a)}{i!}(x - a)^i} -\frac{M}{(k + 1)!}|x - a|^{k + 1} \leq f'(x) \leq \sum_{i = 0}^k{\frac{f^{(i + 1)}(a)}{i!}(x - a)^i} + \frac{M}{(k + 1)!}|x - a|^{k + 1}.\]Finally, rewriting all three sides of the inequality in terms of \(t\) and integrating with respect to \(t\) from \(t = a\) to \(t = x\), we get\[\sum_{i = 0}^{k}{\frac{f^{(i + 1)}(a)}{i!}\cdot\frac{(x - a)^{i + 1}}{i + 1}} - \frac{M}{(k + 1)!}\cdot\frac{|x - a|^{k + 2}}{k + 2} \leq f(x) - f(a) \leq \sum_{i = 0}^{k}{\frac{f^{(i + 1)}(a)}{i!}\cdot \frac{(x - a)^{i + 1}}{i + 1}} + \frac{M}{(k + 1)!} \cdot \frac{|x - a|^{k + 2}}{k + 2}.\]This may look like a mess, but it cleans up very nicely. Notice that the left-hand side is equivalent to \(\sum_{i = 1}^{k + 1}{\frac{f^{(i)}(a)}{i!}(x - a)^i} - \frac{M}{(k + 2)!}|x - a|^{k + 2}\). The right-hand side cleans up just as nicely. Adding \(f(a)\) to all three sides, we get the re-indexed inequality\[\sum_{i = 0}^{k + 1}{\frac{f^{(i)}(a)}{i!}(x - a)^i} - \frac{M}{(k + 2)!}|x - a|^{k + 2} \leq f(x) \leq \sum_{i = 0}^{k + 1}{\frac{f^{(i)}(a)}{i!}(x - a)^i} + \frac{M}{(k + 2)!}|x - a|^{k + 2}.\]Noting that those sums are just \(T_{k + 1}(x)\) and subtracting the sums from both sides, we get\[-\frac{M}{(k + 2)!}|x - a|^{k + 2} \leq f(x) - T_{k + 1}(x) \leq \frac{M}{(k + 2)!}|x - a|^{k + 2},\]which gives the desired result of\[|R_{k + 1}| = |f(x) - T_{k + 1}(x)| \leq \frac{M}{(k + 2)!} |x - a|^{k + 2}.\]Hence, the claim must be true for all \(n \in \{0, 1, 2, \ldots\}\) by mathematical induction.