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What is the command to do a power of a value

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  What is the command to do a power of a value "xvictoryeohx" <xvictoryeohx@1:261/38.remove-rf4-this> - 2012-07-30 19:00 +0000
    Re: What is the command to do a power of a value "bugbear" <bugbear@1:261/38.remove-rf4-this> - 2012-07-30 19:00 +0000
    Re: What is the command to do a power of a value "glen herrmannsfeldt" <glen.herrmannsfeldt@1:261/38.remove-rf4-this> - 2012-07-30 19:00 +0000
      Re: What is the command to do a power of a value "xvictoryeohx" <xvictoryeohx@1:261/38.remove-rf4-this> - 2012-07-30 19:00 +0000
    Re: What is the command to do a power of a value "Roedy Green" <roedy.green@1:261/38.remove-rf4-this> - 2012-07-30 19:00 +0000
      Re: What is the command to do a power of a value "Andreas Leitgeb" <andreas.leitgeb@1:261/38.remove-rf4-this> - 2012-07-30 19:00 +0000
        Re: What is the command to do a power of a value "Patricia Shanahan" <patricia.shanahan@1:261/38.remove-rf4-this> - 2012-07-30 19:00 +0000
          Re: What is the command to do a power of a value Patricia Shanahan <pats@acm.org> - 2012-07-30 13:15 -0700
          Re: What is the command to do a power of a value glen herrmannsfeldt <gah@ugcs.caltech.edu> - 2012-07-30 22:26 +0000
            Re: What is the command to do a power of a value Patricia Shanahan <pats@acm.org> - 2012-07-30 16:01 -0700
            Re: What is the command to do a power of a value Andreas Leitgeb <avl@gamma.logic.tuwien.ac.at> - 2012-07-30 23:11 +0000
          Re: What is the command to do a power of a value "Andreas Leitgeb" <andreas.leitgeb@1:261/38.remove-dpk-this> - 2012-07-31 18:02 +0000

#16665 — What is the command to do a power of a value

From: xvictoryeohx@gmail.com

C=L(1+i/100)power of n

i am stuck here
for example square root is Math.sqrt(x) How do i do Power of a value?

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#16666

  To: xvictoryeohx
From: bugbear <bugbear@trim_papermule.co.uk_trim>

xvictoryeohx@gmail.com wrote:
> C=L(1+i/100)power of n
>
> i am stuck here
> for example square root is Math.sqrt(x)
> How do i do Power of a value?

I have a dim memory from Numerical Analysis that the "obvious" way to evaluate 
that is inaccurate for small "i" and large "n".

  BugBear

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#16667

  To: xvictoryeohx
From: glen herrmannsfeldt <gah@ugcs.caltech.edu>

xvictoryeohx@gmail.com wrote:
> C=L(1+i/100)power of n

> i am stuck here
> for example square root is Math.sqrt(x)
> How do i do Power of a value?

Math.pow().

-- glen

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#16668

  To: glen herrmannsfeldt
From: xvictoryeohx@gmail.com

On Monday, July 30, 2012 12:31:35 PM UTC+8, glen herrmannsfeldt wrote:
> xvictoryeohx@gmail.com wrote:
>
> > C=L(1+i/100)power of n
>
>
>
> > i am stuck here
>
> > for example square root is Math.sqrt(x)
>
> > How do i do Power of a value?
>
>
>
> Math.pow().
>
>
>
> -- glen

Thanks!

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#16671

  To: xvictoryeohx
From: Roedy Green <see_website@mindprod.com.invalid>

On Sun, 29 Jul 2012 21:28:11 -0700 (PDT), xvictoryeohx@gmail.com wrote, quoted 
or indirectly quoted someone who said :

>C=L(1+i/100)power of n
>
>i am stuck here
>for example square root is Math.sqrt(x)
>How do i do Power of a value?

see http://mindprod.com/jgloss/power.html
--
Roedy Green Canadian Mind Products
http://mindprod.com
The greatest shortcoming of the human race is our inability to understand the 
exponential function.
 ~ Dr. Albert A. Bartlett (born: 1923-03-21 age: 89)
http://www.youtube.com/watch?v=F-QA2rkpBSY

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#16676

  To: Roedy Green
From: Andreas Leitgeb <avl@gamma.logic.tuwien.ac.at>

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%29

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#16678

  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.

Patricia

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#16681

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

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#16685

Patricia Shanahan <patricia.shanahan@1:261/38.remove-rf4-this> wrote:

(snip, someone wrote)
>>   x ^ n  =  exp ( log(x) * n )    |  x = (1 + i/100)
>>          = exp ( log( 1 + i/100 ) * n)
>>          = exp ( log1p ( i/100 ) * n)

(snip)

> 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?

I didn't know about the log1p function. (Does Java have one?)

As x gets small log(1+x) loses precision. Consider:

      log(1+1e-60)

The x87 instruction set has an instruction for evaluating log(1+x)
and one for evaluating exp(x-1). Most HLLs don't.

-- glen

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#16688

On 7/30/2012 3:26 PM, glen herrmannsfeldt wrote:
...
> I didn't know about the log1p function. (Does Java have one?)

It's in java.lang.Math, since 1.5.

Patricia

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#16691

glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:
> and one for evaluating exp(x-1).

That is exp(x)-1, not exp(x-1), just for the record ;-)

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#16782

  To: Patricia Shanahan
From: Andreas Leitgeb <avl@gamma.logic.tuwien.ac.at>

Patricia Shanahan <pats@acm.org> wrote:
> On 7/30/2012 6:22 AM, Andreas Leitgeb wrote:
>> 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%
29
>
> 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. ..."

Well, one ulp of i/100 is likely smaller than one ulp of 1+i/100 (at least it 
is for 0 <= i <= 100, the typical range for interest rates). It's like 
calculating sin(0.0) versus sin(Math.PI), where sin() makes the same promise 
wrt precision up to an ulp.

If the OP had been interested in the interest value alone, i.e. in
  I = L*( (1+i/100)^n ) - L
then using  log1p() and expm1() probably would beat the precision of pow() by, 
um, a few decimal digits, depending of course on the values of i and n.

Anyway, I think it's good to know that log1p() and expm1() exist, even if the 
example at hand doesn't now seem to cry out for them as loudly as I thought it 
did on first read.

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