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问个专业问题，请帮助看看# Computation - 科学计算

K*N2002-09-23 07:09

1 楼

关于算法order和精度的问题.
算法order越高,为什么精度不一定越高? 是不是精度和其他一些factor有关,比如stepzie.
numerical integration method中有一种叫bulirsch-stoer的算法,大家知道它的order
是什么,怎么推得呢?
xiexie.

t*n2002-09-23 07:09

2 楼

The order of algorithm means the order of your truncation error. Say a
second order algorithm means the error is O(dx^2), where dx is your step size.
So a higher order algorithm does not necessarily give out more accurate
solution than a lower order one.

【在 K*****N 的大作中提到】

: 关于算法order和精度的问题.
: 算法order越高,为什么精度不一定越高? 是不是精度和其他一些factor有关,比如stepzie.
: numerical integration method中有一种叫bulirsch-stoer的算法,大家知道它的order
: 是什么,怎么推得呢?
: xiexie.

w*g2002-09-23 07:09

3 楼

size.
~~~~~ This does not sound consistent logically to me... If your dx is the
same, high order scheme should render more accurate results, I think, provide
the algorithm is the same.
是不是精度和其他一些factor有关,比如stepzie.
method中有一种叫bulirsch-stoer的算法,大家知道它的order

【在 t****n 的大作中提到】

: The order of algorithm means the order of your truncation error. Say a
: second order algorithm means the error is O(dx^2), where dx is your step size.
: So a higher order algorithm does not necessarily give out more accurate
: solution than a lower order one.

K*N2002-09-23 07:09

4 楼

very nice. that's exactly what i am considering.
but, if i say, by the same stepsize, does mean that higher order algorithm
give higher accuracy??

【在 t****n 的大作中提到】

h*o2002-09-23 07:09

5 楼

f(x+dx) - f(x) = SIGMA{f_n(x)dx^n/n!}, n from 1 to infinity
if f_n(x)dx^n/n! > 0 for any n, higher order calculation has better
accuracy.
otherwise, it's hard to say.
However, if you increase the the order significantly, it's relatively safe
to say that you also get better accuracy.

【在 K*****N 的大作中提到】

: very nice. that's exactly what i am considering.
: but, if i say, by the same stepsize, does mean that higher order algorithm
: give higher accuracy??

t*n2002-09-23 07:09

6 楼

Let's consider a numerical differentiation of f'(x).
Assume we use a uniform grid system, which means grid space is same for
every node. A first order upwind approximation is
f'(x) = ( f(x+h) - f(x) ) / h - h * f''(csi) / 2
The central difference scheme, which is second order, has the form
f'(x) = ( f(x+h) - f(x-h) ) / 2 / h - h^2 * [ f'''(csi_1) + f'''(csi_2)]/3!
The first term in the above equations is the term we use to approximate
the derivative, while the rest of the equations are the

【在 K*****N 的大作中提到】

: very nice. that's exactly what i am considering.
: but, if i say, by the same stepsize, does mean that higher order algorithm
: give higher accuracy??

b*y2002-09-23 07:09

7 楼

The analysis using Taylor's series expansion assumes that function f(x)
is continuously differentiable up to desired order, which is not always
true in most cases. That's what I think why higher order approximation
fails.
If f(x) satisfies the continuous differentiable condition upto desired
order. One can always render higher accuracy under higher order scheme
for certain approximation, when stepsize is taken small enough.
Moreover, there is a trade off between stepsize and machine epsilon,
th

【在 K*****N 的大作中提到】

: very nice. that's exactly what i am considering.
: but, if i say, by the same stepsize, does mean that higher order algorithm
: give higher accuracy??