Reduced Exponential: Work & Formulae

[ianbell]

"Reduced Exponential"

I am interested in what I call the "reduced exponential"
Sum_i=1 to infinity x^(i-1) / i!
where x is a general element in an algebra of interest.

Only when x is invertible is the reduced exponential equivalent to (exp(x)-1) /x .

Obviously we have a "reduced log", the inverse of the reduced exponential.

Does anybody of any work or formulae involving this construct? TIA.

 

[CompuChip]

How do you mean: "if x is invertible"?
If x is a number the series is always equal to (exp(x) - 1)/x, unless x= 0 in which case it converges to zero. If not, the notation with the division doesn't make sense.

Where did you encounter this function?

 

[ianbell]

"How do you mean: "if x is invertible"?"

x is an element of a general algebra, not merely a real or complex number but a multivector or matrix or similar such object that can be raised to integer powers and summed and so exponentiated. I've encountered this in quantum mechanics ..

 

[CompuChip]

OK, it is possible to define the exponential in such cases, but then I would write
(\exp(x) - 1) x^{-1} (or x^{-1} (\exp(x) - 1), though I think there is no difference here) instead of the division.

 

[D H]

Writing x^{-1} instead of dividing by x doesn't help. What if x is not invertible? For example, the matrix
x = \bmatrix 0 & 1 \\ 0 & 0 \endbmatrix
has no inverse but certainly has a "reduced exponential" as defined in the OP: \sum_{i=1}^{\infty} \frac {x^{i-1}}{i!} = \bmatrix 1 & 1/2 \\ 0 & 1 \endbmatrix

 

[CompuChip]

Doesn't help for what? I was just pointing out a notational inconvenience in

Only when x is invertible is the reduced exponential equivalent to (exp(x)-1) /x .

which still holds, as in the example you gave the sum evaluates to the identity which is not even close to exp(x) - 1 = x.

The question was

Does anybody of any work or formulae involving this construct? TIA.

which I must admit, I can't recall having seen or used anywhere.