+2004-05-16 Steve Kargl <sgk@troutmask.apl.washington.edu>
+
+ * arith.c: Fix comment typos.
+
2004-05-15 Tobias Schlueter <tobias.schlueter@physik.uni-muenchen.de>
PR fortran/13742
We first get the argument into the range 0.5 to 1.5 by successive
multiplications or divisions by e. Then we use the series:
- ln(x) = (x-1) - (x-1)^/2 + (x-1)^3/3 - (x-1)^4/4 + ...
+ ln(x) = (x-1) - (x-1)^2/2 + (x-1)^3/3 - (x-1)^4/4 + ...
Because we are expanding in powers of (x-1), and 0.5 < x < 1.5, we
have -0.5 < (x-1) < 0.5. Ignoring the harmonic term, this means
x = Nln2 + r
- Then we obtain exp(r) from the McLaurin series.
+ Then we obtain exp(r) from the Maclaurin series.
exp(x) is then recovered from the identity
exp(x) = 2^N*exp(r). */
x= N*2pi + r
- Then we obtain sin(r) from the McLaurin series. */
+ Then we obtain sin(r) from the Maclaurin series. */
void
sine (mpf_t * arg, mpf_t * result)
/* Make sure a constant numeric expression is within the range for
- it's type and kind. Note that there's also a gfc_check_range(),
+ its type and kind. Note that there's also a gfc_check_range(),
but that one deals with the intrinsic RANGE function. */
arith