Quoting James Holton <[email protected]>:
Now the coefficients of a Taylor polynomial are themselves values of the
derivatives of the function being approximated. Each time you take a
derivative of "f(x)", you divide by the units (and therefore dimensions)
of "x". So, Pete's coefficients below: 1, -1/6, and 1/120 have
dimension of [X]^-1, [X]^-2, [X]^-3, respectively.
James,
The the factors 1, 1/6, 1/120, etc. in the Taylor series of a funcion
f(x) do not come from the derivatives of that function. They simply
come from the coefficients 1/(n!) that pre-multiply each term (each
derivative) in the series. They are, of course, dimensionless (note
that n is just an integer number).
Marc
