English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Assume a function F = F1(x) F2(y)
where x = X + x', y = Y + y'. X is a mean value and x' the fluctuation about mean.
F ~ [F1(X) + (dF1/dx) x'] [F1(Y) + (dF2/dy) y']; this uses 0th and 1st order approximation of Taylor series expansion; derivatives are expressed at mean value X and Y.
So F ~ F1(X) F2(Y) + (dF1/dx) (dF2/dy) x'y'; here I took the average of F, so that the average of x' and y' is zero.
My question is that I'm dealing with large fluctuations about the mean and I found that F cannot be approximate by a Taylor series expansion even when I take higher order terms. Should I give up on this, or is there a better way of expanding functions with large fluctuations about mean?
Any help is appreciated, and please contact me if you need clarifications.

2007-01-16 09:03:53 · 2 answers · asked by ben 1 in Science & Mathematics Mathematics

Thanks for Scarlet time and answer. Here are few more details about this problem:
1- function F (and F1, F2) are know mathematically. Say F1 = x^2 g(x), where g has a bit complicated form to write here, but I can consider it as a constant (one) for the purpose of solving this problem.
F2 = y^1.5
Here, of course, x and y are periodic functions of time (I don't know these functions but I can compute it by solving transient PDE's of a multiphase flow). The whole purpose of doing this is to avoid running transient and time-consuming simulations, so we do some kind of RANS approach for this problem.
The time average x'y' is not small; and higher orders such as x'x'y'y' that come from high order Taylor series are even larger in mag. So it doesn't matter if derivatives are decreasing because higher order fluctuations such as x'y'^2 and x'^2y'^2 are much larger than for e.g. x'y'.

2007-01-17 01:53:38 · update #1

2 answers

If the derivatives after a reasonable number of orders are not small over a significant fraction of the region in which your fluctuations occur, you won't be able to get good approximations through Taylor expansion.

You probably need to adopt an interpolative approach so that you're not trying to establish behaviour of the whole function from derivatives at a single point (how reliable are those derivatives anyway?)

The first thing to do is to make sure your assumptions are correct: the function really is continuous and differentiable, and your variables are scaled appropriately - you may find that converting one to a logarithmic or exponential scale works better, but you should be able to tell from the nature of the variables.

Assuming that's all OK, you can approximate the function in an adaptive way so that reasonably linear areas of the function are represented by few points, while highly curved areas are represented by many points. This minimises the number of function evaluations required. (I assume evaluation is difficult or computationally expensive, so you can't just evaluate it at every point of interest.) An outline for this is given below.

Start off with four evaluations, one at each corner of the region of interest. [Note: All evaluations are stored and used for future interpolation.] Interpolate these to predict the value at the centre, then evaluate at the centre [keep this evaluation too, no point wasting it] and compare. If they match to within the required tolerance (a note on this later), you're done. Otherwise, divide the region into quarters and apply the same process to each quarter, continuing until you have sufficient evaluations everywhere to interpolate within the given tolerance.

Note that the tolerance used for comparison can be bigger than your desired end tolerance (I suggest twice as large) since every comparison will have that central point evaluated and we'll keep that evaluation for interpolation when we've finished the above process. That means x and y distances from the nearest evaluation point will be halved, so the maximum error should be roughly quartered from the error at the last comparison.

2007-01-16 19:51:15 · answer #1 · answered by Scarlet Manuka 7 · 0 0

Way to much to read and I DO NOT understand any word in that LONGGGG paragraph :))

2007-01-23 10:42:00 · answer #2 · answered by ramos_chellie 2 · 0 0

fedest.com, questions and answers