fr0mhell's math stash: Taylor series proof

Taylor series proof

Let x, t, a \in R, f (x) be a real valued function s.t. infinite times differentiable

and let

$D^{n} f (x) = \frac{d^{n} f}{d x^{n}}, j_{n} (t) = \frac{(- 1)^{n + 1} (x - t)^{n}}{n!}$

by solving this simple integral, we get

$\int_{a}^{x} (x - t) D^{2} f (t) d t = f (x) - f (a) - f^{'} (a) (x - a)$

and using integration-by-parts method, we are able to observe that

$\int_{a}^{x} (x - t) D^{2} f (t) d t$

$= {[j_{2} (t) D^{2} f (t)]}_{a}^{x} - \int_{a}^{x} j_{2} (t) D^{3} f (t) d t$

$= {[j_{2} (t) D^{2} f (t)]}_{a}^{x} - {[j_{3} (t) D^{3} f (t)]}_{a}^{x} + \int_{a}^{x} j_{3} (t) D^{4} f (t) d t$

$= {[j_{2} (t) D^{2} f (t)]}_{a}^{x} - {[j_{3} (t) D^{3} f (t)]}_{a}^{x} + {[j_{4} (t) D^{4} f (t)]}_{a}^{x} - \int_{a}^{x} j_{4} (t) D^{5} f (t) d t \dots$

since $j_{n} (x) = 0. j_{n} (a) = \frac{(x - a)^{n} (- 1)^{n + 1}}{n!}$ the equation above becomes

$= \frac{(x - a)^{2} D^{2} f (a)}{2!} + \frac{(x - a)^{3} D^{3} f (a)}{3!} + \frac{(x - a)^{4} D^{4} f (a)}{4!} \dots$

thus, if

$lim_{n \to \infty} \int_{a}^{x} j_{n} (t) D^{n + 1} f (t) d t = 0$

then,

$\int_{a}^{x} (x - t) D^{2} f (t) d t = f (x) - f (a) - f^{'} (a) (x - a)$

$= \frac{(x - a)^{2} D^{2} f (a)}{2!} + \frac{(x - a)^{3} D^{3} f (a)}{3!} + \frac{(x - a)^{4} D^{4} f (a)}{4!} \dots$

$∴ f (x) = f (a) + \frac{(x - a) D^{1} f (a)}{1!} + \frac{(x - a)^{2} D^{2} f (a)}{2!} + \frac{(x - a)^{3} D^{3} f (a)}{3!} \dots$

$∴ f (x) = lim_{n \to \infty} \sum_{k = 0}^{n} \frac{(x - a)^{k} D^{k} f (a)}{k!}$

fr0mhell's math stash

이 블로그 검색

Taylor series proof

Taylor series proof

댓글 2개:

all posts