At 01:11 PM 10/15/2005 -0400, you wrote:
>At 09:34 AM 10/15/2005 -0700, you wrote:
>> > How would you calculate e^(-at) in 32 bit integer math?
>> >
>> > Why does it need to be integer math?  From your previous post
>> > I got the impression this calculation was going to be done at
>> > assembly time.
>>
>>The assembler I'm using only supports integer math. No fractions. No
>>floating point. Just integers. So if you put in .36 it becomes 0. 92/256 =
>>0. Integer math.
>
>Okay so you imagine a radix point at an appropriate place. You can move
>it around after each intermediate calculation with shifts, provided you
>don't overflow. Unfortunately, you've got to deal with "C-like" integer
>math and don't have access to the full 64-bits of a multiply, only the
>least significant 32 bits.
>
>For example, 0.36 might be represented as 0x0002E14 (0.36 * 32768). The
>radix point is between the MS 2 bytes and the LS 2 bytes.
>
>>How do you calculate e^(-x) where x is between 1/256 and 192/256 using only
>>integer math.
>
>A Taylor series might do it acceptably well.


James:

Here is an example in c, where xf, yf are 32-bit integers.

   xf = x * 128;
   yf = 32768 - xf  + (xf* xf)/65536 - (((xf * xf)/32768 )*xf)/(6*32768) + 
(((xf*xf)/32768)* ((xf*xf)/32768))/(32768 * 24)
    - (((((xf*xf)/32768)* ((xf*xf)/32768))/32768) * xf)/(32768 * 120)  ;

The result is ~= exp(-x/256) * 32768 over the range
x = 1 to 192.

yf/32768 agrees with exp(-x/256) within 0.001 over that range.


You can easily extend this to additional terms, although it gets a bit 
tedious.

It's based on the textbook Taylor/Maclaurin series

            infinity
            ---
            \
exp(x) =    >  (x^n)/n!
            /
            ---
           n= 0


Best regards,

----
Tria


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