Lines Matching refs:LOG
6831 # 1.3 If |X| < 16380 log(2), go to Step 2. #
6842 # 16380 log(2) used in Step 1.3 is also in the compact #
6844 # |X| < 16380 log(2). There is no harm to have a small #
6846 # 16380 log(2) and the branch to Step 9 is taken. #
6862 # constant := single-precision( 64/log 2 ). #
6999 # 1.3 If |X| < 70 log(2), go to Step 2. #
7104 # Step 10. Calculate exp(X)-1 for |X| >= 70 log 2. #
8078 # slognp1(): computes the log(1+X) of a normalized input #
8079 # slognp1d(): computes the log(1+X) of a denormalized input #
8086 # fp0 = log(X) or log(1+X) #
8096 # Step 1. If |X-1| < 1/16, approximate log(X) by an odd #
8105 # Step 3. Define u = (Y-F)/F. Approximate log(1+u) by a #
8106 # polynomial in u, log(1+u) = poly. #
8109 # log(X) = log( 2**k * Y ) = k*log(2) + log(F) + log(1+u) #
8110 # by k*log(2) + (log(F) + poly). The values of log(F) are #
8114 # Step 1: If |X| < 1/16, approximate log(1+X) by an odd #
8120 # log(1+X) as k*log(2) + log(F) + poly where poly #
8121 # approximates log(1+u), u = (Y-F)/F. #
8125 # log(F)'s need to be tabulated. Moreover, the values of #
8327 #--ENTRY POINT FOR LOG(X) FOR X FINITE, NON-ZERO, NOT NAN'S
8344 blt.w LOGNEG # LOG OF NEGATIVE ARGUMENT IS INVALID
8356 #--THE IDEA IS THAT LOG(X) = K*LOG2 + LOG(Y)
8357 #-- = K*LOG2 + LOG(F) + LOG(1 + (Y-F)/F).
8359 #--LOG(1+U) CAN BE VERY EFFICIENT.
8368 lea LOGTBL(%pc),%a0 # BASE ADDRESS OF 1/F AND LOG(F)
8399 #--LOG(1+U) IS APPROXIMATED BY
8419 add.l &16,%a0 # ADDRESS OF LOG(F)
8425 fadd.x (%a0),%fp1 # LOG(F)+U*V*(A2+V*(A4+V*A6))
8427 fadd.x %fp1,%fp0 # FP0 IS LOG(F) + LOG(1+U)
8447 #--LOG(X) = LOG(1+U/2)-LOG(1-U/2) WHICH IS AN ODD POLYNOMIAL
8489 #--REGISTERS SAVED FPCR. LOG(-VE) IS INVALID
8495 #--ENTRY POINT FOR LOG(X) FOR DENORMALIZED INPUT
8530 bra.w LOGBGN # begin regular log(X)
8552 bra.w LOGBGN # begin regular log(X)
8555 #--ENTRY POINT FOR LOG(1+X) FOR X FINITE, NON-ZERO, NOT NAN'S
8575 ble.w LP1NEG0 # LOG OF ZERO OR -VE
8582 #--SIMPLY INVOKE LOG(X) FOR LOG(1+Z).
8592 #--EXP(-1/16) < X < EXP(1/16). LOG(1+Z) = LOG(1+U/2) - LOG(1-U/2)
8667 #--ENTRY POINT FOR LOG(1+Z) FOR DENORMALIZED INPUT
8788 # Step 1. Call slognd to obtain Y = log(X), the natural log of X. #
8789 # Notes: Even if X is denormalized, log(X) is always normalized. #
8791 # Step 2. Compute log_10(X) = log(X) * (1/log(10)). #
8802 # Step 1. Call sLogN to obtain Y = log(X), the natural log of X. #
8804 # Step 2. Compute log_10(X) = log(X) * (1/log(10)). #
8815 # Step 1. Call slognd to obtain Y = log(X), the natural log of X. #
8816 # Notes: Even if X is denormalized, log(X) is always normalized. #
8818 # Step 2. Compute log_10(X) = log(X) * (1/log(2)). #
8837 # Step 3. Call sLogN to obtain Y = log(X), the natural log of X. #
8839 # Step 4. Compute log_2(X) = log(X) * (1/log(2)). #
8862 bsr slogn # log(X), X normal.
8874 bsr slognd # log(X), X denorm.
8904 bsr slogn # log(X), X normal.
8919 bsr slognd # log(X), X denorm.
8959 # 1. If |X| > 16480*log_10(2) (base 10 log of 2), go to ExpBig. #
8963 # 3. Set y := X*log_2(10)*64 (base 2 log of 10). Set #
8970 # log_10(2)/64 and L10 is the natural log of 10. Then #
9215 #--USUAL CASE, 2^(-70) <= |X| <= 16480 LOG 2 / LOG 10
23493 # ILOG is the log base 10 of the input value. It is #
23674 # ILOG is the log base 10 of the input value. It is approx-
Indexes created Tue Oct 28 12:10:06 GMT 2025