CODEWITHSUNDEEP
maximadifferentiation
Karl
Karl

Reputation: 263

maxima gradef differentiation

I try to define the derivative of the standard normal pdf in terms of the function:

φ(x) := exp (-x^2/2)/sqrt(2 * %pi); gradef(φ(x),-x*φ(x));

but if I type then:

diff(φ(x),x);

I get:

-(x*%e^(-x^2/2))/(sqrt(2)*sqrt(%pi))`

not as I want -x*φ(x).

What I am doing wrong?

Thanks do not

Karl

EDiT :

Unfortunately both suggestions do not work.

Upvotes: 2

Views: 413

Answers (1)

Robert Dodier
Robert Dodier

Reputation: 17573

I think there's nothing wrong; Maxima is just evaluating phi according to the definition you gave when you call gradef.

I can think of a couple of things to try. (1) Call gradef before defining phi. Then maybe you'll get phi in the output when you call diff. Not sure if that will work.

(2) Define the gradef using a noun expression, i.e., gradef(φ(x),-x*'φ(x)). Notice the single quote mark ' before φ(x); that makes a so-called noun expression in which the argument x may be evaluated but the function φ is not called. Later on in order to evaluate the function, when you want to, you can say ev(someexpression, nouns) to evaluate all noun expressions in someexpression.

EDIT: Here's another idea. This works for me. The previous ideas didn't work because φ gets evaluated too soon; this new idea goes to a greater length to prevent evaluation. Note that the gradef is defined for 'φ(x), so you have to write diff('φ(x), x) in order to apply the gradef.

(%i12) gradef('φ(x), -x*'φ(x));
(%o12)                        φ(x)
(%i13) diff('φ(x), x);
(%o13)                      - x φ(x)

The gradef produces a noun expression -x*'φ(x), so to verbify it you can say:

(%i14) ev(%, nouns);
                                    2
                                   x
                                 - --
                                   2
                             x %e
(%o14)                 - -----------------
                         sqrt(2) sqrt(%pi)

Looks like the chain rule is applied as expected:

(%i15) diff('φ(x/a), x);
                                  x
                              x φ(-)
                                  a
(%o15)                      - ------
                                 2
                                a
(%i16) ev(%, nouns);
                                    2
                                   x
                                - ----
                                     2
                                  2 a
                            x %e
(%o16)               - --------------------
                                          2
                       sqrt(2) sqrt(%pi) a

Upvotes: 0

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