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NNTP-Posting-Date: Mon, 30 Jul 2012 15:15:20 -0500
Date: Mon, 30 Jul 2012 13:15:16 -0700
From: Patricia Shanahan <pats@acm.org>
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Subject: Re: What is the command to do a power of a value
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On 7/30/2012 12:00 PM, Patricia Shanahan wrote:
>    To: Andreas Leitgeb
> From: Patricia Shanahan <pats@acm.org>
>
> On 7/30/2012 6:22 AM, Andreas Leitgeb wrote:
>> On Sun, 29 Jul 2012 21:28:11 -0700 (PDT), xvictoryeohx@gmail.com wrote:
>>> C=L(1+i/100)power of n
>>
>>    x ^ n  =  exp ( log(x) * n )    |  x = (1 + i/100)
>>           = exp ( log( 1 + i/100 ) * n)
>>           = exp ( log1p ( i/100 ) * n)
>>
>> If you're doing more calculations with same interest-rate but
>> different periods, then you may want to calculate
>>     double logBase = Math.log1p( i / 100 );
>> once, and use that for the individual calculations:
>>     C = L * Math.exp( logBase * n )
>>
>> The gist of this response is, that for the kind of base (1+i/100),
>> you better separate the pow operation out into log and exp, and
>> actually use log1p on (i/100) instead of log on (1+i/100) for
>> efficiency's and precision's sake.
>>
>> For "Math.log1p" see:
>> http://docs.oracle.com/javase/6/docs/api/java/lang/Math.html#log1p%28double%2
> 9
>>
>
> I am curious about why you expect this to be more precise than: "The computed
> result must be within 1 ulp of the exact result. Results must be
> semi-monotonic." (From the pow description). Or do you know of cases where
> Math.pow gets an over-large rounding error?
>
> Note that I am not disagreeing with your method for calculating a power of a
> number slightly greater than 1, just questioning whether doing it explicitly
> gets more precise answers than using Math.pow.

I've thought about this some more, and I think I get it now.

If i is reasonably small compared to 100, the 1 + i/100 addition loses
several bits of precision.

Patricia