Lines Matching refs:X

6 |	denormalized number. slognp1 computes log(1+X), and slognp1d
7 |	computes log(1+X) for denormalized X.
12 |	Output:	log(X) or log(1+X) returned in floating-point register Fp0.
20 |		argument X such that |X-1| >= 1/16, which is the usual
27 |	Step 1. If |X-1| < 1/16, approximate log(X) by an odd polynomial in
28 |		u, where u = 2(X-1)/(X+1). Otherwise, move on to Step 2.
30 |	Step 2. X = 2**k * Y where 1 <= Y < 2. Define F to be the first seven
37 |	Step 4. Reconstruct log(X) = log( 2**k * Y ) = k*log(2) + log(F) + log(1+u)
42 |	Step 1: If |X| < 1/16, approximate log(1+X) by an odd polynomial in
43 |		u where u = 2X/(2+X). Otherwise, move on to Step 2.
45 |	Step 2: Let 1+X = 2**k * Y, where 1 <= Y < 2. Define F as done in Step 2
46 |		of the algorithm for LOGN and compute log(1+X) as
330 	movel	(%a0),X(%a6)
331 	movel	4(%a0),X+4(%a6)
332 	movel	8(%a0),X+8(%a6)
334 	cmpil	#0,%d0		| ...CHECK IF X IS NEGATIVE
336 	cmp2l	BOUNDS1,%d0	| ...X IS POSITIVE, CHECK IF X IS NEAR 1
340 |--THIS SHOULD BE THE USUAL CASE, X NOT VERY CLOSE TO 1
342 |--X = 2^(K) * Y, 1 <= Y < 2. THUS, Y = 1.XXXXXXXX....XX IN BINARY.
344 |--THE IDEA IS THAT LOG(X) = K*LOG2 + LOG(Y)
353 	asrl	#8,%d0		| ...SHIFTED 16 BITS, BIASED EXPO. OF X
360 	movel	#0x3FFF0000,X(%a6)	| ...X IS NOW Y, I.E. 2^(-K)*X
371 	fmovex	X(%a6),%fp0
425 	fsubs	one,%fp1		| ...FP1 IS X-1
426 	fadds	one,%fp0		| ...FP0 IS X+1
427 	faddx	%fp1,%fp1		| ...FP1 IS 2(X-1)
428 |--LOG(X) = LOG(1+U/2)-LOG(1-U/2) WHICH IS AN ODD POLYNOMIAL
429 |--IN U, U = 2(X-1)/(X+1) = FP1/FP0
484 |--ENTRY POINT FOR LOG(1+X) FOR X FINITE, NON-ZERO, NOT NAN'S