format for :
64 bit ("double") float:
(sign)(11 bit exponent)(52 bit mantissa)
1.0= 0x3FF0000000000000 = 00 ten 1 rest 0
2.0= 0x4000000000000000 = 0 one 1 rest 0
0.5= 0x3FE0000000000000 = 00 nine 1 rest 0

find sqrt of a number stored in RAM
find 3 mistakes!

load r0 r1	1101x01 load opcode  #r1 holds S
address that holds number
mov r1,r2	0011x12 mov opcode
sub r2,r0,r2	0111202 sub opcode
2^52
clr r3		0001xx3 clr opcode
inc r3 r3	0011x33 inc opcode
asr r2 r3 r2	0111232 asr opcode
add r2 r0 r2	1111202 add opcode  #r2 holds the first approx: x
2^61
mov r0 r5	1011x05 mov opcode
2.0
mov r0 r6	1011x06 mov opcode
small
mov r0 r9	1011x09 mov opcode
2
mov rF rE	0011xFE mov opcode  #current PC=rF saved to rE
add r9 rE rE	01119EE add opcode  #note (rE) is address of following instruction
mov r2 r4	0011x24 mov opcode  #r4 is the starting x
fdiv r1 r2 r3	0111123 fdiv opcode
fadd r2 r3 r2	0111232 fadd opcode
fdiv r2 r5 r2	0111252 fdiv opcode #r2 is the updated x
fsub r2 r4 r4	0111232 fsub opcode
fdiv r4 r2 r4	0111123 fdiv opcode
fabs r4 r4	0011044 fabs opcode
fcmp r4 r6	011046x 2 fcmp opcode = fsub opcode
bpl rE		0011xEF conditional version of mov rE rF opcode
mov r0 rE	1011x0E mov opcode
address to output x
store r2 rE	0110CEx store opcode
halt




example: (all iterations shown)
take sqrt of 2.
bit pattern of 2.=4000000000000000 (hex)
bit pattern of x0=3FF8000000000000 (hex)
   1.5000000000000000     
   1.4166666666666665     
   1.4142156862745097     
   1.4142135623746899     
   1.4142135623730949     

bit pattern of xN=3FF6A09E667F3BCC (hex)

   1.4142135623730951 (actual)
   
take sqrt of pi
3.14159265358979323d0
bit pattern of pi= 400921FB54442D18 (hex)
bit pattern of x0= 3FFC90FDAA22168C (hex)
   1.7853981633974483     
   1.7725007746756094     
   1.7724538515266270     
   1.7724538509055159     
bit pattern of xN=3FFC5BF891B4EF6A (hex)

   1.7724538509055159 (actual)