RE: Derivative of a List
- To: mathgroup at smc.vnet.net
- Subject: [mg42281] RE: [mg42277] Derivative of a List
- From: "Florian Jaccard" <jaccardf at eicn.ch>
- Date: Fri, 27 Jun 2003 06:31:13 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Hello !
You can use the numerical differentiation formulas :
v'(t)=(v(t+h)-v(t))/h + 0(h)
This is a good approximation if differnces of t's are small and irregular.
1) If the list of t is not regular :
liste = {{0., 2.}, {1., 2.5}, {2., 3.2}, {2.8, 3.6}};
numder[i_] := (liste[[i + 1,2]] - liste[[i,2]])/
(liste[[i + 1,1]] - liste[[i,1]])
vprime = Join[Table[numder[i], {i, 1, Length[liste] - 1}],
{numder[Length[liste] - 1]}];
TableForm[Transpose[Join[Transpose[liste], {vprime}]],
TableHeadings -> {{}, {"t", "v(t)", "v'(t)"}}]
If the intervals of t are regular, it is better tu use :
v'(t)=(v(t+h)-v(t-h))/(2h)+ 0(h^2)
2) If the list of t is regular :
In[7]:=
liste = {{0., 2.}, {1., 2.5}, {2., 3.2}, {3., 3.6}};
In[8]:=
betterNumder[i_] := (liste[[i + 1,2]] -
liste[[i - 1,2]])/(liste[[i + 1,1]] -
liste[[i - 1,1]])
In[9]:=
vprime = Join[{numder[1]}, Table[betterNumder[i],
{i, 2, Length[liste] - 1}],
{numder[Length[liste] - 1]}];
In[10]:=
TableForm[Transpose[Join[Transpose[liste], {vprime}]],
TableHeadings -> {{}, {"t", "v(t)", "v'(t)"}}]
3) If you have a lot of regular values, you can use the very good
Richardson formula :
v'(t)=(8*(v(t+h/2)-v(t-h/2))-v(t+h)+v(t-h))/(6h) + 0(h^4)
Greetings
F.Jaccard
-----Message d'origine-----
De : ch akpovo [mailto:akpovo at cennas.nhmfl.gov]
Envoyé : jeu., 26. juin 2003 11:36
À : mathgroup at smc.vnet.net
Objet : [mg42277] Derivative of a List
Hello,
I created a two columns table from experimental data say v (t) and would
like to compute the derivatives. How do I proceed.
Thanks