0
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Background, this seems to work (This is manually transcribed from the program, beware of errors):

J1[f_, i_] := DifferenceDelta[f[i], i]^2
D[J1[f + t v, i], t]
2 (-v[i] + v[i+1]) (-(f + t v)[i] + (f + t v)[i+1])

What I would like to do is replace the ()^2 with a generic function JJ, and later specify

JJ[x_] := ...

Here's what happens with the generic JJ:

J2[f_, i_] := JJ[DifferenceDelta[f[i], i]]
D[J2[f + t v, i],t]
0

Why can't it return a derivative expression in terms of a generic derivative JJ', and how could I modify my code to make it work?

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  • 1
    $\begingroup$ I get (-v[i] + v[1 + i]) Derivative[1][JJ][-(f + t v)[i] + (f + t v)[1 + i]], not 0. It may still not be what you want. If you're getting 0, then you might want to quit the kernel and try again. Perhaps v has the value 0. $\endgroup$ – Michael E2 Jun 7 '14 at 11:58
  • $\begingroup$ I think the issue was caused by a typo in the original post (tv was used instead of t v in D[J2[...]]); and now, with Michael's edit, it is fixed :) $\endgroup$ – kglr Jun 7 '14 at 13:22
1
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Like this?

JJ[x_] := x^2

J2[f_, i_] := JJ[ DifferenceDelta[ f[i], i]]

D[ J2[ f + t v, i], t]
2 (-v[i] + v[1 + i]) (-(f + t v)[i] + (f + t v)[1 + i])
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