MaplePrimes - answers and comments on Question, Combining functions,

Defining functions

Robert Israel — Wed, 03 Feb 2010 20:42:48 Z

I hope you defined your functions something like this, not as you wrote them:

> N40 := xi -> piecewise(0<xi and xi<=1, a, b);
  N50 := xi -> piecewise(1 < xi and xi <= 2, c, d);

Then to combine them:

> N41 := xi -> (xi-1)*N40(xi) + (2-xi)*N50(xi);

Or you could even do this, with the help of an "identity function":

> id:= x -> x:
  N41 := (id - 1)*N40 + (2-id)*N50;

Thanks that worked, yea I

Søren Mikkel Andersen — Wed, 03 Feb 2010 21:03:20 Z

Thanks that worked,

yea I forgot the actual values of the piecwise function in my initial question..

What is the difference between the notations

N40(xi):=piecewise(0<xi and xi<=1, a, b);
N40 :=xi -> piecewise(0<xi and xi<=1, a, b);

Hi, I now have made the

Søren Mikkel Andersen — Thu, 04 Feb 2010 00:37:14 Z

Hi, I now have made the generic functions

for instance, I have

i:=3: p := 1:

N31:=xi->N40(xi)*((kn[i+p+1]-xi)/(kn[i+p+1]-kn[i]+1])):

Now N31 works as intended, but I want to make another generic function

i:=4: p := 1:

N41:=xi->N50(xi)*((kn[i+p+1]-xi)/(kn[i+p+1]-kn[i]+1])):

But then that changes the behaviour of N31,

how can I make N31 remain constant with respect to changing my counting variables i and p ?

instantiation

pagan — Thu, 04 Feb 2010 02:50:31 Z

> unassign('i'):unassign('p'):

> N:=xi->N40(xi)*((kn[i+p+1]-xi)/(kn[i+p+1]-kn[i]+1));
      N40(xi) (kn[i + p + 1] - xi)
xi -> ----------------------------
       kn[i + p + 1] - kn[i] + 1  

> N31:=subs([i=3,p=1],eval(N));
      N40(xi) (kn[5] - xi)
xi -> --------------------
       kn[5] - kn[3] + 1  

> N41:=subs([i=4,p=1],eval(N));
      N40(xi) (kn[6] - xi)
xi -> --------------------
       kn[6] - kn[4] + 1  

> N31(3.2);
                           N40(3.2) (kn[5] - 3.2)
                           ----------------------
                             kn[5] - kn[3] + 1   

> i,p := 177,-20:

> N31(3.2);
                           N40(3.2) (kn[5] - 3.2)
                           ----------------------
                             kn[5] - kn[3] + 1

Defining functions

Robert Israel — Wed, 03 Feb 2010 22:43:24 Z

The normal way to define a function is using ->. With N40(xi):= ... you are literally defining the value of the procedure N40 only on the literal input xi (technically, you are assigning an entry in the remember table of N40), and this will not affect N40(anything else).

If you are using 2D input, N40(xi):= ... will cause a popup window asking you to clarify what you mean, a function definition or a remember table assignment. If you choose function definition, the result will be as if you used ->. But you might as well use -> in the first place.