Lines Matching refs:Y

30 |	Step 2. X = 2**k * Y where 1 <= Y < 2. Define F to be the first seven
31 |		significant bits of Y plus 2**(-7), i.e. F = 1.xxxxxx1 in base
32 |		2 where the six "x" match those of Y. Note that |Y-F| <= 2**(-7).
34 |	Step 3. Define u = (Y-F)/F. Approximate log(1+u) by a polynomial in u,
37 |	Step 4. Reconstruct log(X) = log( 2**k * Y ) = k*log(2) + log(F) + log(1+u)
45 |	Step 2: Let 1+X = 2**k * Y, where 1 <= Y < 2. Define F as done in Step 2
48 |		u = (Y-F)/F.
342 |--X = 2^(K) * Y, 1 <= Y < 2. THUS, Y = 1.XXXXXXXX....XX IN BINARY.
343 |--WE DEFINE F = 1.XXXXXX1, I.E. FIRST 7 BITS OF Y AND ATTACH A 1.
344 |--THE IDEA IS THAT LOG(X) = K*LOG2 + LOG(Y)
345 |--			 = K*LOG2 + LOG(F) + LOG(1 + (Y-F)/F).
346 |--NOTE THAT U = (Y-F)/F IS VERY SMALL AND THUS APPROXIMATING
349 |--DIVISION IS NEEDED TO CALCULATE (Y-F)/F.
351 |--GET K, Y, F, AND ADDRESS OF 1/F.
360 	movel	#0x3FFF0000,X(%a6)	| ...X IS NOW Y, I.E. 2^(-K)*X
362 	andil	#0xFE000000,FFRAC(%a6) | ...FIRST 7 BITS OF Y
374 	fsubx	F(%a6),%fp0		| ...Y-F
376 |--SUMMARY: FP0 IS Y-F, A0 IS ADDRESS OF 1/F, FP1 IS K
381 	fmulx	(%a0),%fp0	| ...FP0 IS U = (Y-F)/F