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Taylor's Theorem (Taylor polynomial)
12-27-2010, 03:16 PM
Post: #1
Taylor's Theorem (Taylor polynomial)
Consider $\displaystyle{\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n dt}$. Apply integral by parts to get
$\displaystyle{-\frac{f^{(n)}(a)}{n!}(x-a)^n+\int_a^x\frac{f^{(n)}(t)}{(n-1)!}(x-t)^{(n-1)}dt=\cdots = -\sum_{k=1}^n \frac{f^{(k)}(a)}{k!}(x-a)^k+\int_a^x f^{\;'}(t)dt}$
Therefore $\quad (1)\quad\displaystyle{f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k+\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n dt}$
Since $(x-t)^n$ will not change sign when $t \in [a,x]$, we have
$\displaystyle{\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n dt = \frac{f^{(n+1)}(\xi )}{n!}\int_a^x (x-t)^n dt = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}}$ and so
$\quad\quad\quad\quad\quad (2)\quad \displaystyle{f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}}$
These expressions form the Taylor's theorem for single variable functions
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