# Taylor Series Expansion of an eqn

• Mar 5th 2009, 09:19 AM
markrvr
Taylor Series Expansion of an eqn
Hi, I am doing some work on density estimation using kernels and am looking at bias and variance in MISE to find optimal bandwidth. I have a book which deals with this but only gives an approximation to a taylor expansion. The book gives
int K(t){f(x-ht)-f(x)} dt
I need a taylor expansion of the form f(x-ht)= .....
I think it starts as f(x) - ht f'(x) + 1/2 h^2t^2f''(x)+...
Thanks
• Mar 6th 2009, 08:33 AM
markrvr
Can anybody tell me about the remainder term in a taylor series expansion and derivatives of the function f(x-ht) and how this relates to the order of taylor expansion?
• Mar 6th 2009, 09:15 AM
meymathis
Taylor series expansion of any function f around x at $x-h$ is

$f(x-h)=\sum_{n=0}^\infty (-h)^n f^{(n)}(x)/n!$

where $f^{(n)}(x)$ is the $n^{th}$ derivative of f evaluated at x.

See Wikipedia's Taylor series where their a is our x, and their x is our $x-h$.
• Mar 6th 2009, 11:01 AM
markrvr
yea thanks but if the function has derivatives only of finite order, then it can be approximated by Taylor expansion i.e. by a polynomial. So if a function has only 3 derivatives it can be approximated by a cubic polynomial. Im not quite sure how far to go with this expansion and the remainder term?
• Mar 6th 2009, 11:35 AM
meymathis
It depends on the function, how small ht is, and how good of an approximation you want. I think that the question you are asking is too general. How good of an approximation do you need? There isn't a "one-size fits all answer". For example, if you are looking at

$\frac{f(x)-f(x-ht)}{ht}$

and you are going to let $ht \rightarrow 0$, then there is no need to approximate beyond the linear term. Since the higher order terms are going to have an ht term (after dividing the ht in the denominator into each term) which is going to zero as ht does.

As for the size of the remainder term see Wikipedia's Taylor's theorem

If
$
f(x-ht) = f(x) + \frac{f'(x)}{1!}(- ht) + \frac{f^{(2)}(x)}{2!}(- ht)^2 + \cdots + \frac{f^{(n)}(x)}{n!}(- ht)^n + R_n(x-ht)$

So $R_n(x-ht)$ is the remainder term. Then there exists a number $\xi \in [x-ht,x]$ such that
$R_n(x-ht) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (-ht)^{n+1}$

assuming that f is n times differentiable in $[x-ht,x]$ and $n+1$ times differentiable in $(x-ht,x)$