Q: I have defined the operator in the following way:
When I type in
I get the correct result. But when I try to compute
the input is returned. How can I specify the operator in a general way?
A: Hartmut Wolf (email@example.com) writes: If you consult Section A.2.7 of The Mathematica Book, you will find that the operator is fully associative. Definitions respecting associativity will work without problem. Noting that , an associative definition that works for any number of arguments is
Another natural extension of the original definition is the sequence
This will, in general, lead to different results than the definition above. An associative definition for an arbitrary number of arguments is immediate.
David Park (firstname.lastname@example.org) writes: Either use parentheses, since your definition only accommodates two arguments,
or extend the definition of to more than two arguments by adding the (right-association) rule
Now the desired result is obtained without using parentheses.
Note that is the addition formula for capacitors in series or resistors in parallel.
As another example, the following approach makes it straightforward to implement the velocity addition formula of special relativity for an arbitrary number of velocities, where is the speed of light.
Relativistic addition of to gives .
Ordinary addition of the following three velocities exceeds .
However, relativistic addition leads to a result less than .
In special relativity, the rapidity parameter is defined by , where , and is the relative velocity between the moving reference frames. The benefit of the rapidity representation is the simple formula for the addition of rapidities (in the same spatial direction).
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