- #1
hadron23
- 28
- 1
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,
[tex]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+...
[/tex]
Is it possible to come up with a value [tex]M[/tex] such that
[tex]
\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}
[/tex]
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,
[tex]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+...
[/tex]
Is it possible to come up with a value [tex]M[/tex] such that
[tex]
\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}
[/tex]
That is, come up with an upper bound on the error of the higher order terms.
Thanks