Max-estimate version of Lagrange remainder formula: Difference between revisions

Latest revision as of 20:17, 12 July 2012

Statement

About a general point

Suppose $f$ is a function of one variable and $x_{0}$ is a point in the domain such that $f$ is $(n + 1)$ times differentiable at $x_{0}$ . Denote by $R_{n} (f; x_{0})$ the function of $x$ given by $x \mapsto f (x) - P_{n} (f; x_{0}) (x)$ , i.e., $R_{n} (f; x_{0})$ is the remainder when we subtract from $f$ its $n^{t h}$ Taylor polynomial at $x_{0}$ .

For any $x$ , let $J$ is the interval between $x$ and $x_{0}$ (it might be the interval $[x, x_{0}]$ or $[x_{0}, x]$ depending on whether $x < x_{0}$ or $x > x_{0}$ ). If $f$ is $n + 1$ times differentiable everywhere on $J$ , then we have:

$| R_{n} (f; x_{0}) (x) | \leq ({sup}_{t \in J} | f^{(n + 1)} (t) |) \frac{| x - x_{0} |^{n + 1}}{(n + 1)!}$

If $f^{(n + 1)})$ is continuous on $J$ , the $sup$ can be replaced by $max$ :

$| R_{n} (f; x_{0}) (x) | \leq ({max}_{t \in J} | f^{(n + 1)} (t) |) \frac{| x - x_{0} |^{n + 1}}{(n + 1)!}$

About the point 0

Suppose $f$ is a function of one variable such that $f$ is $(n + 1)$ times differentiable at $0$ . Denote by $R_{n} (f; 0)$ the function of $x$ given by $x \mapsto f (x) - P_{n} (f; 0) (x)$ , i.e., $R_{n} (f; 0)$ is the remainder when we subtract from $f$ its $n^{t h}$ Taylor polynomial at $0$ .

For any $x$ , let $J$ is the interval between $x$ and $0$ (it might be the interval $[x, 0]$ or $[0, x]$ depending on whether $x < 0$ or $x > 0$ ). If $f$ is $n + 1$ times differentiable everywhere on $J$ , then we have:

$| R_{n} (f; 0) (x) | \leq ({sup}_{t \in J} | f^{(n + 1)} (t) |) \frac{| x |^{n + 1}}{(n + 1)!}$

If $f^{(n + 1)})$ is continuous on $J$ , the $sup$ can be replaced by $max$ :

$| R_{n} (f; 0) (x) | \leq ({max}_{t \in J} | f^{(n + 1)} (t) |) \frac{| x |^{n + 1}}{(n + 1)!}$

@@ Line 11: / Line 11: @@
 If <math>f^{(n+1)})</math> is continuous on <math>J</math>, the <math>\sup</math> can be replaced by <math>\max</math>:
-<math>|R_n(f;x_0)(x)| \le \left( \sup_{t \in J} |f^{(n+1)}(t)|\right) \frac{|x - x_0|^{n+1}}{(n+1)!}</math>
+<math>|R_n(f;x_0)(x)| \le \left( \max_{t \in J} |f^{(n+1)}(t)|\right) \frac{|x - x_0|^{n+1}}{(n+1)!}</math>
 ===About the point 0===
@@ Line 17: / Line 17: @@
 For any <math>x</math>, let <math>J</math> is the interval between <math>x</math> and <math>0</math> (it might be the interval <math>[x,0]</math> or <math>[0,x]</math> depending on whether <math>x < 0</math> or <math>x > 0</math>). If <math>f</math> is <math>n + 1</math> times differentiable everywhere on <math>J</math>, then we have:
+<math>|R_n(f;0)(x)| \le \left( \sup_{t \in J} |f^{(n+1)}(t)|\right) \frac{|x|^{n+1}}{(n+1)!}</math>
+If <math>f^{(n+1)})</math> is continuous on <math>J</math>, the <math>\sup</math> can be replaced by <math>\max</math>:
 <math>|R_n(f;0)(x)| \le \left( \max_{t \in J} |f^{(n+1)}(t)|\right) \frac{|x|^{n+1}}{(n+1)!}</math>