1 years, 96 days

## @acer This seems to be the more complete...

@acer This seems to be the more complete answer, and seems to me faster for entry. Thanks!

## @Carl Love Exactly what I was looking fo...

@Carl Love Exactly what I was looking for, and your extra remark also helped. Do you know why I am not able to do this by the method I used in the OP?

## @Carl Love Fantastic! This actually...

Fantastic! This will help me a lot!

## doesn't seem to help to expand the expre...

It seems that it doesn't solve my issue; the expression is still not expanded:

expr := expand(Q(h));
expr :=

/                                       2    / 3\\
alpha[-2] \f(x) - 2 D(f)(x) h + 2 @@(D, 2)(f)(x) h  + O\h //
------------------------------------------------------------
h

alpha[0] f(x)
+ -------------
h

/                     9                 2    / 3\\
alpha[3] |f(x) + 3 D(f)(x) h + - @@(D, 2)(f)(x) h  + O\h /|
\                     2                          /
+ -----------------------------------------------------------
h

InertForm:-MakeInert(expr);
%+(%/(%*(alpha[-2], h(%f(x), 0, %*(-2, D(f)(x)), 1,

%*(2, @@(D, 2)(f)(x)), 2, %O(1), 3)), h),

%/(%*(alpha[0], %f(x)), h), %/(%*(alpha[3], h(%f(x), 0,

%*(3, D(f)(x)), 1, %*(%/(9, 2), @@(D, 2)(f)(x)), 2, %O(1), 3)),

h))

expand(((alpha[-2] %* h(%f(x), 0, (-2) %* D(f)(x), 1, 2 %* (D@@2)(f)(x), 2, %O(1), 3)) %/ h) %+ ((alpha[0] %* %f(x)) %/ h) %+ ((alpha[3] %* h(%f(x), 0, 3 %* D(f)(x), 1, (9 %/ 2) %* (D@@2)(f)(x), 2, %O(1), 3)) %/ h));
%+(%/(%*(alpha[-2], h(%f(x), 0, %*(-2, D(f)(x)), 1,

%*(2, @@(D, 2)(f)(x)), 2, %O(1), 3)), h),

%/(%*(alpha[0], %f(x)), h), %/(%*(alpha[3], h(%f(x), 0,

%*(3, D(f)(x)), 1, %*(%/(9, 2), @@(D, 2)(f)(x)), 2, %O(1), 3)),

h))

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