hadron23
Nov23-10, 02:35 PM
Hello,
I am trying to come up with an expression for a bound on the sum of higher order terms, above second order. Consider the following Taylor expansion of a function f(x) around a point a,
f(x) = f(a) + \frac{f^{(1)}(a)}{1!}(x-a) + \frac{f^{(2)}(a)}{2!}(x-a)^2+ \frac{f^{(3)}(a)}{3!}(x-a)^3+...
Is it possible to come up with a value M such that
\begin{align}
&\left|\left|f(x) - [f(a) + \frac{f^{(1)}(a)}{1!}(x-a)]\right|\right|_2^2 \le M\\
&\Rightarrow \left|\left|\frac{f^{(2)}(a)}{2!}(x-a)^2+ \frac{f^{(3)}(a)}{3!}(x-a)^3+...\right|\right|_2^2 \le M
\end{align}
That is, come up with an upper bound on the error of the higher order terms.
Thanks
I am trying to come up with an expression for a bound on the sum of higher order terms, above second order. Consider the following Taylor expansion of a function f(x) around a point a,
f(x) = f(a) + \frac{f^{(1)}(a)}{1!}(x-a) + \frac{f^{(2)}(a)}{2!}(x-a)^2+ \frac{f^{(3)}(a)}{3!}(x-a)^3+...
Is it possible to come up with a value M such that
\begin{align}
&\left|\left|f(x) - [f(a) + \frac{f^{(1)}(a)}{1!}(x-a)]\right|\right|_2^2 \le M\\
&\Rightarrow \left|\left|\frac{f^{(2)}(a)}{2!}(x-a)^2+ \frac{f^{(3)}(a)}{3!}(x-a)^3+...\right|\right|_2^2 \le M
\end{align}
That is, come up with an upper bound on the error of the higher order terms.
Thanks