MaplePrimes - answers and comments on Question, Problem with convert to - 2-D math input

Defining a function

Robert Israel — Fri, 19 Feb 2010 02:03:10 Z

You are defining phi[1](z) rather than defining the function phi[1]. What you probably meant is

> phi[1] := z -> A[1]*z^(lambda+I*epsilon)+B[1]*z^(lambda-I*epsilon);

The way you have it, phi[1](z) is defined (technically, it's an entry in the remember table of phi[1]), but this would not define phi[1] of anything else.
The 2-D math parser is asking you whether you really meant what you wrote (the remember table entry) or the function definition as I wrote it.

Hi Robert Thank for your

BrianB — Fri, 19 Feb 2010 02:17:34 Z

Hi Robert

Thank for your answer.

Is it correctly understood then that it is only unknown functions that figures in an expression that you write as phi[0](z) so that diff(phi[0](z), z) is understood and not just 0? When a funtion on the other hand is declared you do not need this because it follows from the expression what it is a funtion of, right?.

Hi Thank you for your

BrianB — Tue, 09 Mar 2010 17:16:45 Z

Thank you for your answer. I can now do the following which was what I wanted.

> unassign(phi[0])

> diff(phi[0](z), z)

> phi[0] := z -> a*z

> diff(phi[0](z), z)

Please explain

Robert Israel — Fri, 19 Feb 2010 02:41:03 Z

Please explain the question. I'm not sure what you mean. Whether you define phi[0] or phi[0](z), diff(phi[0](z), z) should work. On the other hand, if you define phi[0](z), Maple would not know what phi[0](t) or diff(phi[0](t),t) is.