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I want to calculate a finite difference (something like this SO Post). My data is as follows: I have x-values that are powers of two (4, 8, 16, 32 and 64). Corresponding to them are y-values, such that

$$Y=f(X)$$

where $f$ is a monotonic function. My interest is in calculating finite difference such as

$$\text{Diff}(x) = (y_2-y_1) /(x_2-x_1)$$ at different values of X, where X is in the domain (i.e. $4\le x \le 64$). Now my problem is that when I take piecewise linear approximation, Diff(x) changes a lot between 4-to-8 and 8-to-16. This is undesirable. If I try to fit quadratic equation, by taking three points at a time, and then calculate finite difference, it becomes negative at some points, which is not physically acceptable, since $f$ is monotonic function. In other words, I fitted $ax^2+bx+c$ and thus $\text{Diff}(X)=2ax+b$. So, the problem is that $2ax+b$ is not positive at all points in the domain.

Can someone point-out a method to solve this problem. I need to use C/C++ for solve this, I cannot use matlab. I just have five X values, so solution of any complexity is acceptable. Thanks for the help.

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  • $\begingroup$ Is there any specific functional form you're looking for? $\endgroup$ – Geoff Oxberry Feb 14 '12 at 15:24
  • $\begingroup$ @GeoffOxberry Any low-order polynomial should suffice. $\endgroup$ – user984260 Feb 14 '12 at 15:29
  • $\begingroup$ Do you seek a continuous function from the data given and determine the derivative of this continuous function using finite differences? $\endgroup$ – Paul Feb 14 '12 at 15:31
  • $\begingroup$ @Paul Thanks. Yes. The only thing is that: I don't prefer a single continuous function which spans entire domain. Multiple functions, which model part of the domain are OK and preferable. $\endgroup$ – user984260 Feb 14 '12 at 15:45
  • $\begingroup$ @user984260: Perhaps something like piece-wise continuous splines? $\endgroup$ – Paul Feb 14 '12 at 16:20
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Instead of fitting a general polynomial, you way want to consider using piece-wise cubic splines to interpolate the function in the interval between each gridpoint. one good property of cubic splines is that they minimize the bending energy of the fitted curve, which may lead to fewer radical oscillations (which may be the cause of the negative values you experience with the quadratic approach you previously tried).

You can obtain a piecewise cubic spline by proposing a separate cubic polynomial of the form

$f_i(x)=a_ix^3+b_ix^2+c_ix+d_i$ in each interval $i$

whose endpoints are the gridpoints you selected. Then, you can impose the following conditions: 1. All functions must agree with the data at each of the nodes 2. The functions must be continuous at each of the node points 3. The first and second derivatives of each function must continuous at each of the node points

You can get two extra equations by imposing the so-called "natural spline" condition, that is the second derivative must be zero at the first and last node. This should give you enough information to solve for each of the coefficients of each cubic polynomial in each interval.

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A simple and accurate approach in this case is to use standard differences in $\log$ space.

$$ \begin{align*} s(x) &= \log x \\ \partial_x f(x) &= \partial_s f(e^{s(x)}) \partial_x s(x) \\ &= \frac 1 x \partial_s f(e^{s(x)}) \\ \end{align*} $$

Since $s(x)$ is equally spaced, it is easy to approximate the derivative on the bottom line using any standard differencing.

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  • $\begingroup$ Thanks. It is very good idea. The issue is that, still I get large difference in Diff(X). E.g. for X={4,8,16}, Y={23363,44528,78328}, I get using piece-wise linear approximation Diff(7)=3023 (using Y1 and Y2) and Diff(8)=4225 (using Y2 and Y3). May be nature of my problem itself is like that. $\endgroup$ – user984260 Feb 14 '12 at 16:13
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Fitting by a parametrized family of functions (like degree $n$-polynomials, where there are $n+1$ free parameters corresponds to minimizing an error term on a parameter space. If the error term is a least squares approach, then this leads to a quadratic program.

Quadratic Programming

If you want your function to be non-negative, or to be a monotonic function (it is not clear what you mean), then you arrive a contrained minimization, which is another topic on its own.

Constraint programming

In your case, a low order polynomial should suffice. The first wikipedia article should lead to plenty of ways to solve your problem.

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  • $\begingroup$ Thanks for reply. I fitted ax^2+bx+c and thus Diff(x)=2ax+b. So, the problem is that 2ax+b is not positive at all points in the domain. Can you suggest something? Constraint programming does not help at all. $\endgroup$ – user984260 Feb 14 '12 at 15:25

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