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RE:? D[f,{x,n}]
*To*: mathgroup at smc.vnet.net
*Subject*: [mg25559] RE:[mg25495] ? D[f,{x,n}]
*From*: "Ersek, Ted R" <ErsekTR at navair.navy.mil>
*Date*: Sat, 7 Oct 2000 03:36:07 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
Jack Goldberg had trouble making it so
D[f, x] and D[f, {x, n}]
perform some "simplification" on (f)
before the built in D is used.
I give some code below that does this.
Notice if you do these examples without
changing D you get factored polynomials.
Since I don't give a rule for D[f, {x, n}] this
seems to prove that D[f, {x, n}] evaluates
Nest[ D[#, x]&, f, n ]
or something equivalent.
-----------------------------
In[1]:=
Unprotect[D];
flag=True;
D[f_,x_]/;flag:=
Block[{flag}, D[Expand[f],x] ]
In[4]:=
D[(x+2)^3,x]
Out[4]=
12 + 12*x + 3*x^2
In[5]:=
D[(x+2)^4,{x,2}]
Out[5]=
48 + 48*x + 12*x^2
--------------------
Notice, inside Block I didn't have to set (flag=False)
We only care that flag is not True.
--------------------
On a related note, the Help Browser says
"Derivative is generated when you apply D to a function
whose derivative Mathematica does not know." We see Derivative
in the FullForm below.
In[6]:=
f'[x] //FullForm
Out[6]=
Derivative[1][f][x]
In[7]:=
f''[x] //FullForm
Out[7]=
Derivative[2][f][x]
--------------------
Regards,
Ted Ersek
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http://www.verbeia.com/mathematica/tips/Tricks.html
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