🚀 go-pugleaf

In article <1992Aug28.181224.14463@nas.nasa.gov> asimov@nas.nasa.gov (Daniel A. Asimov) writes:
>One way to generalize the exponential of a matrix to arbitrary
>continuous functions f: R^n -> R^n is as follows:
>
>exp(f): R^n -> R^n via
>
>	exp(f)(x) = x + f(x) + f(f(x))/2! + f(f(f(x)))/3! +...
>

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Comment: this is an unusual generalization of the exponential to nonlinear 
functions f.  I'd be curious as to its motivation.

The usual idea, of course, is to regard exp(tA)x, where A is a linear map,
as the solution of the differential equation
              dy/dt = Ay(t),  y(0) = x.

The standard generalization is the solution of
              dy/dt = f(y(t)), y(0) = x.

There is no "explicit" form of the solution, but there is a highly developed
theory, e.g. the theory of nonlinear semigroups.

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-- 
# Paul R. Chernoff                       chernoff@math.berkeley.edu     #
# Department of Mathematics              ucbvax!math!chernoff           #
# University of California               chernoff%math@ucbvax.bitnet    #
# Berkeley, CA  94720                                                   #