Taylor

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### Taylor's Theorem Taylor's theorem provides an approximation of a $k$-times differentiable function around a given point by a polynomial of degree $k$, called the Taylor polynomial. It also provides a remainder term that quantifies the error in this approximation. #### Statement If $f$ is a real function such that $f^{(n-1)}(x)$ is continuous on the closed interval $[a, x_0]$ (or $[x_0, a]$) and $f^{(n)}(x)$ exists on the open interval $(a, x_0)$ (or $(x_0, a)$), then there exists at least one point $c$ strictly between $a$ and $x_0$ such that: $$f(x_0) = f(a) + (x_0-a)f'(a) + \frac{(x_0-a)^2}{2!}f''(a) + \dots + \frac{(x_0-a)^{n-1}}{(n-1)!}f^{(n-1)}(a) + R_n$$ Where $R_n$ is the remainder term, representing the error of the approximation. #### Schlömilch and Roche Form of Remainder The remainder term $R_n$ can be expressed as: $$R_n = \frac{(x_0-a)^n (1-\theta)^{n-p}}{(n-1)!p} f^{(n)}(a+\theta(x_0-a))$$ for some $\theta \in (0,1)$ and any positive integer $p$. Letting $x_0 - a = h$, the form becomes: $$R_n = \frac{h^n (1-\theta)^{n-p}}{(n-1)!p} f^{(n)}(a+\theta h)$$ **Special Cases:** - **Cauchy's Form (p=1):** $R_n = \frac{h^n (1-\theta)^{n-1}}{(n-1)!} f^{(n)}(a+\theta h)$ - **Lagrange's Form (p=n):** $R_n = \frac{h^n}{n!} f^{(n)}(a+\theta h)$ #### Proof Let $F(x)$ be an auxiliary function defined as: $$F(x) = f(x) + (x_0-x)f'(x) + \frac{(x_0-x)^2}{2!}f''(x) + \dots + \frac{(x_0-x)^{n-1}}{(n-1)!}f^{(n-1)}(x)$$ We want to show that $f(x_0) - F(a) = R_n$. Consider another auxiliary function $G(x)$ defined on $[a, x_0]$: $$G(x) = F(x) + K(x_0-x)^p$$ where $K$ is a constant chosen such that $G(a) = G(x_0)$. First, evaluate $G(x_0)$: $$G(x_0) = F(x_0) = f(x_0) + (x_0-x_0)f'(x_0) + \dots = f(x_0)$$ Now, set $G(a) = G(x_0)$: $$F(a) + K(x_0-a)^p = f(x_0)$$ This allows us to determine $K$: $$K = \frac{f(x_0) - F(a)}{(x_0-a)^p}$$ Since $G(a) = G(x_0)$ and $G(x)$ is continuous on $[a, x_0]$ and differentiable on $(a, x_0)$ (as $f^{(n)}(x)$ exists), by Rolle's Theorem, there exists a point $c \in (a, x_0)$ such that $G'(c) = 0$. Let's compute $F'(x)$: $$F'(x) = f'(x) + [-f'(x) + (x_0-x)f''(x)] + \left[-\frac{2(x_0-x)}{2!}f''(x) + \frac{(x_0-x)^2}{2!}f'''(x)\right] + \dots$$ $$ + \left[-\frac{(n-1)(x_0-x)^{n-2}}{(n-1)!}f^{(n-1)}(x) + \frac{(x_0-x)^{n-1}}{(n-1)!}f^{(n)}(x)\right]$$ All intermediate terms cancel out in pairs. For example, $(x_0-x)f''(x)$ from the second term cancels with $-\frac{2(x_0-x)}{2!}f''(x)$ from the third term. So, $F'(x)$ simplifies to: $$F'(x) = \frac{(x_0-x)^{n-1}}{(n-1)!}f^{(n)}(x)$$ Now, compute $G'(x)$: $$G'(x) = F'(x) - K p (x_0-x)^{p-1}$$ Setting $G'(c) = 0$: $$\frac{(x_0-c)^{n-1}}{(n-1)!}f^{(n)}(c) - K p (x_0-c)^{p-1} = 0$$ Assuming $x_0-c \neq 0$: $$K = \frac{(x_0-c)^{n-1}}{(n-1)! p (x_0-c)^{p-1}} f^{(n)}(c) = \frac{(x_0-c)^{n-p}}{(n-1)! p} f^{(n)}(c)$$ Equating the two expressions for $K$: $$\frac{f(x_0) - F(a)}{(x_0-a)^p} = \frac{(x_0-c)^{n-p}}{(n-1)! p} f^{(n)}(c)$$ Rearranging to find $f(x_0) - F(a)$: $$f(x_0) - F(a) = \frac{(x_0-a)^p (x_0-c)^{n-p}}{(n-1)! p} f^{(n)}(c)$$ This term $f(x_0) - F(a)$ is exactly the remainder $R_n$. Since $c \in (a, x_0)$, we can write $c = a + \theta(x_0-a)$ for some $\theta \in (0,1)$. Then, $x_0-c = x_0 - (a+\theta(x_0-a)) = (x_0-a) - \theta(x_0-a) = (x_0-a)(1-\theta)$. Substituting these into the remainder expression: $$R_n = \frac{(x_0-a)^p ((x_0-a)(1-\theta))^{n-p}}{(n-1)! p} f^{(n)}(a+\theta(x_0-a))$$ $$R_n = \frac{(x_0-a)^p (x_0-a)^{n-p} (1-\theta)^{n-p}}{(n-1)! p} f^{(n)}(a+\theta(x_0-a))$$ $$R_n = \frac{(x_0-a)^n (1-\theta)^{n-p}}{(n-1)! p} f^{(n)}(a+\theta(x_0-a))$$ This is the Schlömilch and Roche form of the remainder. Letting $x_0-a = h$, we get the more common form: $$R_n = \frac{h^n (1-\theta)^{n-p}}{(n-1)! p} f^{(n)}(a+\theta h)$$