05/03/2009, 09:45 PM
(This post was last modified: 05/03/2009, 09:51 PM by Base-Acid Tetration.)
Quote:If we take the equations as recursive definitions then \( \operatorname{It} \) is only defined for functions in the second argument that are constant. Thatswhy I proposed the different version.
Why? If one defined the It operator for functions in the second argument the definition would not be a "recursive" definition? then say that the functional equations are true for the It operator for all functions, but the equations are not "recursive". ok.
