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Taylor Polynomial

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The Taylor polynomial

$\displaystyle p_n(x) = f(a) + f'(a) (x-a) + \cdots + \frac{f^{(n)}(a)}{n!} (x-a)^n $

interpolates the derivatives of a function $ f$ at a point $ a$ up to order $ n$, i.e., $ p_n^{(k)}(a)=f^{(k)}(a)$ for $ k= 0,\ldots ,n$. If $ f$ is $ (n+1)$-times continuously differentiable,

$\displaystyle f(x) = p_n(x) + R,\quad R = \frac{f^{(n+1)}(t)}{(n+1)!} (x-a)^{n+1} \,, $

for some $ t$ between $ a$ and $ x$.

A straightforward computation shows that the derivatives of $ f$ and $ p_n$ match.

To check the expression for the remainder $ R$, we augment an additional term:

$\displaystyle q(y) = p_n(y) + c (y-a)^{n+1} \,, $

where the constant $ c$ is chosen so that $ q(x)=f(x)$. As a consequence, $ q-f$ has a zero of order $ n+1$ at $ y=a$ and an additional zero at $ y=x$, i.e., the error hat $ n+2$ zeros, counting multiplicities. By Rolle's theorem, the $ (n+1)^$st derivative of $ q-f$ must have at least one zero $ t$:

$\displaystyle 0 = q^{(n+1)}(t) - f^{(n+1)}(t) = c (n+1)! - f^{(n+1)}(t) \,. $

The asserted form of the remainder now follows from

$\displaystyle R = f(x) - p_n(x) = q(x) - p_n(x) = c (x-a)^{n+1} \,. $

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  automatisch erstellt am 20.  7. 2016