#1718 Failure of Taylor expansion

open
nobody
5
2012-11-18
2009-07-28
Anonymous
No

When I apply the function 'taylor(function,z,..., ...)' to a function containing the d/dz operator, it works correctly. However if the differential is of another variable it fails. Thus:

(%i1) G : diff(a(x,z),z,1);
d
(%o1) -- (a(x, z))
dz
(%i2) H : taylor(G,z,0,1);
!
! 2 !
d ! d !
(%o2)/T/ -- (a(x, z))! + (--- (a(x, z))! ) z + . . .
dz ! 2 !
!z = 0 dz !
!z = 0
=================================

works, but:

=================================
(%i3) G : diff(a(x,z),x,1);
d
(%o3) -- (a(x, z))
dx
(%i4) H : taylor(G,z,0,1);
d
(%o4)/T/ -- (a(x, z)) + . . .
dx
=================================

fails.

Regards,

David Webb.

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