Central Differences

The most common way of computing numerical derivative of a function $f(x)$ at any point $x^*$ is to approximate $f(x)$ by some polynomial $P_m(x)$ in the neighborhood of $x^*$ . It is expected that if selected neighborhood of $x^*$ is sufficiently small then $P_m(x)$ approximates $f(x)$ near $x^*$ well and we can assume that $f'(x^*)\approx P'_m(x^*)$ .

Let’s consider this approach in details (or go directly to the table of formulas).

At first, we sample $f(x)$ at the $N$ ( $N$ is odd) equidistant points around $x^*$ :

$f_k = f(x_k),\: x_k = x^*+kh,\: k=-\frac{N-1}{2},\dots,\frac{N-1}{2}$ ,

where $h$ is some step.

Then we interpolate points $\{(x_k,f_k)\}$ by polynomial of $(N-1)th$ degree: $P_{N-1}(x)=\sum_{j=0}^{N-1}{a_jx^j}$ . Its coefficients $\{a_j\}$ are found as a solution of system of linear equations:

$\left\{ P_{N-1}(x_k) = f_k\right\},\quad k=-\frac{N-1}{2},\dots,\frac{N-1}{2}$

By our assumption $f'(x^*)$ can be approximated by the derivative of the constructed interpolating polynomial:

$f'(x^*)\approx P'_{N-1}(x^*)$

To illustrate the process let’s consider case when $N=3$ . Here interpolating parabola are defined by the system (we assume $x^*=0$ for simplicity):

$\left\{\begin{matrix} P_2(-h)&=&f_{-1}\\ P_2(0)&=&f_0\\ P_2(h)&=&f_1 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} a_0-a_1h+a_2h^2&=&f_{-1}\\ a_0&=&f_0\\ a_0+a_1h+a_2h^2&=&f_1 \end{matrix}\right.$

After subtracting first equation from the last we get final result:

$f'(0)\approx P'_2(0)=a_1=\displaystyle\frac{f_{1}-f_{-1}}{2h}$

In the same way we can obtain expressions for any $N$ . Formulas for $N = 3,5,7,9$ are listed below:

$\begin{tabular}{|r|c|} \hline $N$& $N$-point stencil Central Differences \\ \hline &\\ 3&$\displaystyle\frac{f_{1}-f_{-1}}{2h}$\\ &\\ 5&$\displaystyle\frac{f_{-2}-8f_{-1}+8f_1-f_2}{12h}$\\ &\\ 7&$\displaystyle\frac{-f_{-3}+9f_{-2}-45f_{-1}+45f_1-9f_2+f_3}{60h}$\\ &\\ 9&$\displaystyle\frac{3f_{-4}-32f_{-3}+168f_{-2}-672f_{-1}+672f_{1}-168f_{2}+32f_{3}-3f_{4}}{840h}$\\ &\\ \hline \end{tabular}$

These expressions are very widely used in numerical analysis and commonly refered as central(finite) differences. As it can be clearly seen they have simple anti-symmetric structure and in general difference of $N$ -th order can be written as:

${f'(x^*)\approx \frac{1}{h}\sum_{k=1}^{(N-1)/2}{c_k(f_k-f_{-k})}}$ ,

where $c_k$ are coefficients derived by procedure described above.

It is easy to see that if $f(x)$ is a polynomial of a degree $N-1$ , then central differences of order $\ge N$ give precise values for $f(x)$ derivative at any point. This follows from the fact that central differences are result of approximating $f(x)$ by polynomial. If $f(x)$ is a polynomial itself then approximation is exact and differences give absolutely precise answer.

To differentiate a digital signal $u_n$ we need to use h=1/SamplingRate and replace $f_k$ by $u_{n+k}$ in the expressions above. In this case derivative of a signal $u_n$ is found by

$d_n = \sum_{k=1}^{(N-1)/2}{c_k(u_{n+k}-u_{n-k})}$

Frequency response of central differences is:

$H(\omega)=2i\sum_{k=1}^{(N-1)/2}{c_k\sin(k\omega)}$

Magnitude responses for $N = 3,5,7,9$ are drawn below:

Red dashed line is the magnitude response of an ideal differentiator $H_d(\omega)=i\omega$ . As $N$ grows $H(\omega)$ moves closer to ideal differentiator response (by increasing tangency order with $H_d(\omega)$ at $\omega=0$ ).

In practice there is no need for ideal differentiators because usually signals contain noise at high frequencies which should be suppresed. This means that response of practical differentiator should be close to $H_d(\omega)$ on some passband interval $\omega\in[0,\omega_c]$ and should tend to zero in the upper part of Nyquist interval $\omega\in[\omega_c,\pi]$ .

From the plot we can see that central differences don’t resemble such behavior, all they care about is to get as closer as possible to the response of ideal differentiator, without supression of noisy high frequencies. As a consequence they perform well only on exact values, which contain no noise. Different technique is needed for robust derivative estimation of noisy signals. Such approaches are considered in the next sections:

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47 Comments

Kem

Posted February 19, 2009 at 8:32 pm | #

Hi Pavel, great tutorial!

How can this be done for the second derivatives f”(x) and also for the 2D case , especially how can we approximate d^2 f(x,y) / dxdy with different orders of approximation

Thank you!

Reply
Pavel Holoborodko

Posted February 20, 2009 at 5:42 pm | #

Thanks, Kem.

1. Second order central difference is simple to derive. We use the same interpolating polynomial and assume that $f^{\prime\prime}(x^*)\approx P^{\prime\prime}_m(x^*)$ . Final formulas are:

$\begin{tabular}{|r|c|} \hline $N$& Second order Central Differences \\ \hline &\\ 3&$\displaystyle\frac{f_{-1}-2f_{0}+f_1}{h^2}$\\ &\\ 5&$\displaystyle\frac{-f_{-2}+16f_{-1}-30f_0+16f_1-f_2}{12h^2}$\\ &\\ 7&$\displaystyle\frac{2f_{-3}-27f_{-2}+270f_{-1}-490f_0+270f_1-27f_2+2f_3}{180h^2}$\\ &\\ \hline \end{tabular}$

3. Third order central differences are:

$\begin{tabular}{|r|c|c|} \hline $N$&Third order Central Differences&Error\\ \hline &&\\ 5&$\displaystyle\frac{-f_{-2}+2f_{-1}-2f_1+f_2}{2h^3}$&$O(h^2)$\\ &&\\ 7&$\displaystyle\frac{f_{-3}-8f_{-2}+13f_{-1}-13f_1+8f_2-f_3}{8h^3}$&$O(h^4)$\\ &&\\ \hline \end{tabular}$

2. Estimation of the mixed second order derivative $\textstyle{\frac{\partial^2{f}}{\partial{x}\partial{y}}}$ is a little more elaborate but still follows the same idea. In this case we should build interpolating polynomial over the 2D grid $\{f_{k,l}=f(x^*+kh,y^*+lh)\}$ and find its mixed derivative at the point $(x^*,y^*)$ assuming that it is approximately equal to $\textstyle{\frac{\partial^2{f}}{\partial{x}\partial{y}}}$ .

Here are reference formulas for mixed second order differences:

For the $O(h^2)$ order of approximation:

$\displaystyle{{\frac{\partial^2{f}}{\partial{x}\partial{y}}}\approx \frac{1}{4\,h^2}\left[f_{-1,-1}+f_{1,1}-f_{1,-1}-f_{-1,1}\right] }$

For the $O(h^4)$ order of approximation:

$\displaystyle{{\frac{\partial^2{f}}{\partial{x}\partial{y}}}\approx \frac{1}{600\,h^2} \left[\begin{matrix} -63(f_{1,-2}+f_{2,-1}+f_{-2,1}+f_{-1,2})+\\ 63(f_{-1,-2}+f_{-2,-1}+f_{1,2}+f_{2,1})+\\ 44(f_{2,-2}+f_{-2,2}-f_{-2,-2}-f_{2,2})+\\ 74(f_{-1,-1}+f_{1,1}-f_{1,-1}-f_{-1,1}) \end{matrix}\right] }$

Reply
- Thomas Cameron
  
  Posted April 12, 2011 at 10:27 am | #
  
  Pavel,
  
  I just wanted to say how much i enjoyed finding this resource as i am taking my first course in numerical differential equations. I am having some confusion based on the definitions for the central difference operator that i am given and the one you are using. Could you post some of the steps taken to arrive at the finite difference approximation of: d^2 f(x,y) / dxdy for O(h^2).
  
  Thank you,
  Thomas
  
  Reply
shinji

Posted August 12, 2009 at 11:30 pm | #

anyone know about the derivation of the third order central differences formula?

Reply
Chris Willy

Posted June 12, 2010 at 9:33 pm | #

Hi Pavel –
Do you have a published reference for the O(h^4) approximation for the mixed 2nd partial derivatives? Thanks.

Reply
- Pavel Holoborodko
  
  Posted June 13, 2010 at 10:32 pm | #
  
  Here is formula for the $O(h^4)$ order of approximation:
  
  $\displaystyle{{\frac{\partial^2{f}}{\partial{x}\partial{y}}}\approx \frac{1}{600\,h^2} \left[\begin{matrix} -63(f_{1,-2}+f_{2,-1}+f_{-2,1}+f_{-1,2})+\\ 63(f_{-1,-2}+f_{-2,-1}+f_{1,2}+f_{2,1})+\\ 44(f_{2,-2}+f_{-2,2}-f_{-2,-2}-f_{2,2})+\\ 74(f_{-1,-1}+f_{1,1}-f_{1,-1}-f_{-1,1}) \end{matrix}\right] }$
  
  Reply
  - Chris Willy
    
    Posted June 13, 2010 at 11:46 pm | #
    
    Hi Pavel -
    Thanks for your quick reply. I read that you had posted that same formula earlier … in fact, it’s the earlier post that prompted me to write. I’m looking for a published reference (book or journal article) that contains the formula so that I can cite the reference in a paper I’m writing. Thanks.
    Best regards,
    Chris Willy
    
    Reply
    - Pavel Holoborodko
      
      Posted June 14, 2010 at 9:26 pm | #
      
      Well, I’ve derived this formula by myself. I have no idea is it published somewhere or not.
      
      You can reference my site – it would be of huge support for me. Several published papers already include references to my site – so I think there is no problem.
      
      You can tell me more about your task, maybe I can derive more suitable filters for your conditions.
      
      Reply
Mercè Bruned

Posted July 1, 2010 at 7:15 pm | #

Hello Pavel,
Congratulations for your good and well organised work. I’m not mathematic and I’m writing an algorithm to derivate a discrete function. My set of points are not equidistant, so they have x values that are no constant. I read your advanced work about derivatives for noisy functions and probably I will use it in a future.
For the moment, I will derivate with central differences method. I wonder if it’s easy to extend the method when points are not equidistant. Is it ok to divide by the x increment?
Thanks for your blog

Reply
- Pavel Holoborodko
  
  Posted July 2, 2010 at 3:41 pm | #
  
  Thank you for your comment!
  
  Central difference extension for irregular spaced data is:
  
  $\displaystyle {f'(x^*)\approx \sum_{k=1}^{(N-1)/2}{c_k\cdot\frac{f_{k}-f_{-k}}{ x_{k}-x_{-k}}\cdot 2k}}$
  
  So, yes, difference of function values should be divided by the $x$ increment with $2k$ weight. Coefficients $\{c_k\}$ are from formula for uniform spacing. For instance, for $N=3,\,c_1=1/2$ and for $N=5,\,c_1=2/3,\,c_2=-1/12$ .
  
  Actually it is better to use $\{c_k\}$ from smooth noise-robust differentiators, since they are much more numerically stable even in the absence of noise. Plus they have similar approximation properties to central differences – so you won’t lose any precision and get more stability as a bonus.
  
  Coefficients are for $N=5,\,c_1=1/4,\,c_2=1/8$ and for $N=7,\,c_1=13/32,\,$ $c_2=1/8,\,c_3=-5/96$ . See referenced page for other $\{c_k\}$ .
  
  Let me know about the results please.
  
  Reply
Mercè Bruned

Posted July 2, 2010 at 8:40 pm | #

Hello again Pavel,

Thank you very much for your fast and good answer.

Really, I have a noisy discrete function (points come from peak calculation of a laser stripe, acquired by a digital camera. I use the laser stripe to enhance profile of objects).

I need to smooth the function for display purpose and I need also to derivate the function in order to find inflexion points, local maxima and other features that I use to do measurements of the object that I’ve enhanced the profile.

At this moment I’m smoothing the function using Savitzky-Golay filter and I apply central differences to the smoothed function. I read your work and I realized that smooth noise-robust differentiators would be really useful to my work, but I don’t know how to use it to smooth the function for display purposes.

Reply
- Pavel Holoborodko
  
  Posted July 5, 2010 at 10:21 am | #
  
  Thank you for the description of your project – it is really interesting.
  
  I think your application could benefit from using SNRD filters. I will help you to apply them for smoothing (for visual output) and for finding derivative.
  
  Just let me know what polynomial degree and number of points you are using now in Savitzky-Golay filter. Knowing that info I can build SNRD filters with better properties for your task.
  
  Reply
  - Mercè Bruned
    
    Posted July 12, 2010 at 7:45 pm | #
    
    Hello Pavel, I appreciate your help very much.
    
    I’m smoothing my function with a third degree polynomial and I’m using 7 points. Once smoothed I’m applying central differences to calculate first and second derivative.
    
    I’ve tryed to calculate first derivative by using SNRD filters on the Savitzy-Golay smoothed functions, and I’m really loosing important function detail, I’m smoothing too much, I think this is not a good way to proceed.
    
    Thanks again for your interest and best regards from Spain.
    
    Reply
    - Pavel Holoborodko
      
      Posted July 13, 2010 at 12:30 pm | #
      
      Congratulations on the Victory!!
      
      Thank you for the information about your test results.
      I have several comments though.
      
      I’ve checked properties of 3rd order 7 point SG and derive some special filters with better properties. I would appreciate if you would try them and give me feedback.
      
      1. For your application I recommend to use following filters for smoothing (for visualization):
      
      $\displaystyle {\hat{f}(x_0)\approx a_0f_0+\sum_{k=1}^{(N-1)/2}{2a_k\cdot\frac{f_{k}(x_k-x_0)+f_{-k}(x_0-x_{-k})}{x_{k}-x_{-k}}}}$
      
      where coefficients $\{a_k\}$ are:
      
      $\scriptsize{ \begin{tabular}{|l|l|} \hline &\\ N = 7&$a_0=1/2,\,a_1 = 9/32,\,a_2=0,\,a_3 = -1/32$\\ &\\ N = 9&$a_0=55/128,\,a_1= 9/32,\,a_2 = 3/64,\,a_3 = -1/32,\,a_4 = -3/256$\\ &\\ \hline \end{tabular} }$
      
      Both of them preserve important details better than SG and have soft and guaranteed noise suppression ( $N=9$ is stronger).
      
      2. For derivative estimation I recommend to use
      
      $\displaystyle {f'(x^*)\approx \sum_{k=1}^{(N-1)/2}{c_k\cdot\frac{f_{k}-f_{-k}}{ x_{k}-x_{-k}}\cdot 2k}}$
      
      with
      
      $\scriptsize{ \begin{tabular}{|l|l|} \hline &\\ N = 7&$c_1=13/32,\,c_2=1/8,\,c_3=-5/96$\\ &\\ N = 9&$c_1= 9/32,\,c_2 = 1/6,\,c_3 = -1/96,\,c_4 = -1/48$\\ &\\ \hline \end{tabular} }$
      
      Important: do not apply differentiation filters after smoothing filters. This could hurt approximation order (as you experienced with SNRD after SG smoothing).
      
      Apply derivative filter directly on noisy non-smoothed data.
      
      Reply
      - Susan
        
        Posted April 6, 2011 at 2:58 pm | #
        
        Thank you for your info on central differencing!! For a homework assignment I am given a table (similar to yours) which gives us 3 and 5 point central difference (along with 2,3, and 5 point forward and backward) formulas, but then we are asked to determine the 4 point central difference formula from the table. I’m confused because I cannot find information on this anywhere online, is it possible to do a 4 point central divided difference approximation for a first derivative?? Any help would be greatly appreciated.
Jan Kadlec

Posted July 20, 2010 at 10:22 pm | #

Awesome tutorial.

I’m writing a GPU raytracer and need to estimate the gradient (∂/∂x, ∂/∂y, ∂/∂z) of a distance function (for surface normals). I assume that the function is expensive to evaluate.

Now I’m using central differences – these should get me O(h²) error with seven function samples (the center is always evaluated). I was wondering if I could get the same error order with fewer samples – some kind of sample sharing between the three axes (I’ve tried the tetrahedral arrangement, but it’s even worse than forward differences because of ∂x∂y-like terms).

Reply
- Pavel Holoborodko
  
  Posted July 21, 2010 at 11:06 am | #
  
  Thanks.
  
  Actually 7-points central difference has $O(h^6)$ . Three point stencil is enough to achieve $O(h^2)$ .
  
  To get the lowest error with the smallest number of points in 3D one should apply 1D stencil along the corresponding axis. Only points along the required axis play major role for approximation order, others can be used for other goals (like noise suppression).
  
  P.S.
  I like your site. Some of your projects made me nostalgic about old good DOS-hacking times. Trees were taller and grass was greener back then…….
  
  Reply
  - Jan Kadlec
    
    Posted July 21, 2010 at 7:02 pm | #
    
    Thanks for the quick answer and for visiting my site .
    
    What I originally meant was using the three-point stencil along every axis (which makes 6 evaluations in total + one in the center for other unrelated purposes).
    
    I’ve finally settled for 4-point forward differences (1+9 evaluations in total with error O(h³)): $-2 f_{-1} -3 f_0 + 6 f_1 - f_2 \over 6h$ .
    
    Reply
    - Pavel Holoborodko
      
      Posted July 22, 2010 at 1:45 pm | #
      
      Oh, sorry for misunderstanding. Glad you have found suitable solution.
      Maybe I should include one sided differences in the tutorial.
      
      Reply
Bob

Posted September 30, 2010 at 3:09 am | #

What would be the two dimensional 7 and 9 point formulas for (d^2/dx^2 + d^2/dy^2))f(x,y)?

Reply
- Pavel Holoborodko
  
  Posted September 30, 2010 at 10:42 am | #
  
  What do you need them for?
  
  Reply
dan

Posted October 28, 2010 at 4:10 pm | #

hey!
how can one prove that the third derivative of the central difference scheme of the function
f(x)=1/x with arguments a,b,c,d is given as
f”’(1/x)=(4ab(c-d)+4cd(a-b))/abcd

Reply
Catherine

Posted November 20, 2010 at 6:34 am | #

Thanks for putting up these solutions. Do you have any opinion on the usefulness of Dispersion-Relation-Preserving finite difference schemes* for time dependent problems? If so, do you know of any site where coefficients are listed for various N-point stencils?

* (ie C. K. W. Tam, and J. C. Webb, “Dispersion-Relation-Preserving Finite Difference Schemes for Computational Acoustics”, Journal of Computational Physics, Vol. 107, 1993, pp. 262-281)

Reply
- Pavel Holoborodko
  
  Posted November 20, 2010 at 11:14 am | #
  
  It seems that paper is not available online for free. Could you send me pdf file or scan if you have it?
  
  I found related papers of C.K.W. Tam:
  
  http://www.math.fsu.edu/~tam/Multi-size-mesh%20DRP.pdf
  
  http://www.math.fsu.edu/~tam/Tam_IJCFD_04.pdf
  
  They look interesting – I’ll read them first and then get back to your question.
  
  Thanks for the pointer.
  
  Reply
  - Pavel Holoborodko
    
    Posted November 29, 2010 at 2:57 pm | #
    
    Sorry for the late reply.
    
    After having read these papers I would say that DRP are almost the same to classical central differences.
    
    The criterion authors use to construct DRP (minimum deviation from frequency response of ideal differentiators) is actually (near-)equivalent to the exactness on polynomials condition (which is the basement of CD).
    
    I see no particular reason to choose DRP over CD. Actually I would suggest to use CD since they have rational coefficients, so rounding error is minimized.
    
    Reply
Jeramy Dickerson

Posted November 25, 2010 at 6:15 am | #

Hello,

I am trying to set up a FD equation of the form:
d/dx(a(x)*d/dx f(x))
I am looking for a five point stencil in particular.
This is basically the Poisson equation where the permitivity is not a constant.

Thanks for any help!

Reply
Daniel

Posted December 28, 2010 at 11:22 am | #

What is the 9 point formula for the second derivative in one dimension?

Reply
- Pavel Holoborodko
  
  Posted December 28, 2010 at 11:26 am | #
  
  You can try 9 point stencils from here Noise Robust Second Derivative.
  
  Reply
KS

Posted January 22, 2011 at 8:51 am | #

Hi Pavel,

Thanks for this post, it was very insightful. I have not seen an in depth discussion of this topic past the N = 3 case.

I am somewhat confused on the technique you used to solve for the coefficients {c_k}. You give a general expression for the central difference based derivative approximation in terms of {c_k} and say “where {c_k} are coefficients derived by the procedure described above.” But I do not see a general technique above for finding the coefficients. I can derive the coefficients for a specific h & N by setting up the linear system Ax = b and finding the 2nd row of Pinv(A). I was curious how you derived the general form in terms of the variable h.

Thanks

Reply
- Pavel Holoborodko
  
  Posted January 23, 2011 at 10:52 am | #
  
  Hi KS,
  
  Yes, “procedure described above” refers to the solution of Ax=b.
  Usually differentiators are considered to have anti-symmetric coefficients.
  That is what I meant under “general form” without mathematical rigor.
  
  There is a simple algorithm on how to derive $\{c_k\}$ without solving linear system.
  Generation of Finite Difference Formulas for arbitrary Spaced Grids. Bengt Fornberg
  
  I think you might find it interesting.
  
  Reply
  - KS
    
    Posted January 23, 2011 at 2:59 pm | #
    
    Thanks for the reference!
    
    Reply
Ed

Posted February 3, 2011 at 11:06 pm | #

Pavel,

thanks for the nice introduction!

For mixed second derivatives it may be more efficient to use an alternative formula:

$\frac{\partial^2{f}}{\partial{x}\partial{y}}}\approx \frac{ - f_{+1,-1} - f_{-1,+1} + f_{1,0} + f_{-1,0} + f_{0,1} + f_{0,-1} - 2 f_{0,0}}{2\,h^2}$

Its advantage is that evaluating the mixed derivative along with the “diagonal” derivatives (d^2 f / dx^2 and d^2 f / dy^2) requires only 2 extra function values instead of 4.

Same idea can be used to derive formula accurate to $h^3$ :

$\frac{\partial^2{f}}{\partial{x}\partial{y}}}\approx \frac{1}{24\,h^2} \left[\begin{matrix} 16 (f_{0, -1} + f_{0,1} + f_{-1,0} + f_{1,0})\\ -(f_{2,0} + f_{-2,0} + f_{0,2} + f_{0,-2})\\ -16 (f_{-1, 1} + f_{1, -1})\\ + f_{-2, 2} + f_{2, -2}\\ -30 f_{0, 0} \end{matrix}\right]$

Only 4 extra values are needed here.

Cheers!

Reply
- Pavel Holoborodko
  
  Posted February 4, 2011 at 2:31 pm | #
  
  Ed, thank you for the nice formulas!!
  
  Just checked, first one has $O(h^2)$ , second $O(h^4)$ .
  
  Interesting how they behave in frequency domain – I’ll take a look on the transfer functions to check isotropy (important for 2D formulas).
  
  Maybe I should start collecting all the formulas in one big compendium.
  
  P.S.
  You can insert math formulas into comments by using LaTeX syntax:
  $ ... $ – for inline formulas
  \[ ... \] – for displayed.
  
  Reply
Jono

Posted March 14, 2011 at 8:00 pm | #

Pavel

Thanks for this interesting intro. I am not mathematical in any way and am trying to argue the case over the advantages of a 5-point derivative method over say a 3-point method.

My application is in using the first and second derivative of angular displacement data to yield velocity and acceleration. I have smoothed the displacement data, collected at 100Hz, with a forward and reverse 5th order butterworth filter (1/50 ‘low’) prior to calculating velocity and acceleration. However I am confused as to the advantages and disadvantages of the 3 and 5-point methods.

Thanks

Reply
- Pavel Holoborodko
  
  Posted March 15, 2011 at 11:05 am | #
  
  I think results of 3 and 5-points methods won’t differ very much after smoothing you’re using.
  
  Reply
Lee-Ping

Posted April 28, 2011 at 1:08 am | #

Hi there,

Your website was very helpful! Thank you. My scientific application is a least-squares minimization in a large parameter space where some of the parameteric derivatives are known analytically but others are unknown. I used finite difference derivatives to estimate the gradient and diagonal elements of the Hessian, and I fill in the rest of the Hessian elements using BFGS.

For my application, I checked the three-point difference result against the seven-point difference result and got agreement to within five decimal places. This means that the three-point difference is usable in the actual application.

Reply
Tal

Posted May 13, 2011 at 8:03 am | #

Hello all, i’ve wondered if anyone can help me with the following:
I have a poisson equation: (d^2/dx^2+d^2/dy^2)=-1
Let’s say that around point U(i,j) i have 8 points as follows:
U(i-1,j-1), U(i+1,j+1), U(i,j+1), U(i,j-1), U(i-1,j), U(i+1,j), U(i-1,j+1), U(i+1,j-1)
How would the solution for U(i,j) be if DeltaX and DeltaY (finite differences) are constant but different (DeltaX not equal DeltaY)?

Thanks for the help

Reply
razye

Posted May 26, 2011 at 1:36 pm | #

hello,what is the discretization of Hessian term ([(∂^2 f)/〖∂x〗^2 ][(∂^2 f)/〖∂y〗^2 ]-〖((∂^2 f)/∂x∂y)〗^2)with 9-point stencil?

Reply
BiLinear Approx

Posted June 17, 2011 at 12:30 am | #

great tutorial! It helps me with a better understanding of how an approx of derivative to an arbitrary accuracy is achieved.

Reply
Tharanga

Posted August 16, 2011 at 3:29 pm | #

Your tutorial is greadfull.I want to know how explicit central differance method is apply to the wave equation

Reply
ochi

Posted August 20, 2011 at 8:05 pm | #

please can you help with the fomular for the order of o(h)^3

thanks

Reply
Lutfi Mulyadi

Posted September 19, 2011 at 11:01 am | #

Dear Mr Pavel
Let me introduce myself. my name Lutfi mulyadi I come from Indonesia. I was very interested when I saw your blog. now I’m working on my BSc thesis in the field of nuclear reactor physics in Bandung Institut Of Technology, Indonesia. My thesis title is Fast Nuclear Reactor Burnup Analysis Using Fourth Order Runge Kutta Methods. in this thesis I solve the boundary value problem with the central finite difference and then the initial value problem with 4 Order Runge Kutta Methods. I have managed to create a program for the Runge Kutta method but I have not succeeded for the central difference to the neutron diffusion equation . if you can analyze the mistakes that I do? I really hope you can help me. a have sent to you my computer code in C programming language for solving neutron diffusion equations with central difference and I attach also my thesis to your email that include the numerical methods that I use in chapter 3 Simulation Methods (Numerical Methods) .

Thank you so much for your help

Reply
ashkan

Posted October 16, 2011 at 11:50 pm | #

HI
i want fifth order central difference (3 or 5 point).can anyone help me,my email:ghanbari.ashkan@yahoo.com

Reply
Mike

Posted November 18, 2011 at 1:21 pm | #

Hi. I am looking for a formulation of backward finite difference in 2d. Can anyone help?

Reply
Mitzey

Posted January 10, 2012 at 2:05 pm | #

Hi ! can the central difference formula be used even when the interval between x values is not constant? if not, what formula can be applied? thanks!

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- Pavel Holoborodko
  
  Posted January 10, 2012 at 2:54 pm | #
  
  Yes, it can, check this comment for formula.
  
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franzes

Posted February 16, 2012 at 12:24 am | #

do you have some kind of history or concept for this central difference method?

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Central Differences

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