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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-x1c-this> - 2012-07-31 20:06 +0000
    Re: What is the command to do a power of a value "bugbear" <bugbear@1:261/38.remove-x1c-this> - 2012-07-31 20:07 +0000
    Re: What is the command to do a power of a value "glen herrmannsfeldt" <glen.herrmannsfeldt@1:261/38.remove-x1c-this> - 2012-07-31 20:07 +0000
      Re: What is the command to do a power of a value "xvictoryeohx" <xvictoryeohx@1:261/38.remove-x1c-this> - 2012-07-31 20:07 +0000
    Re: What is the command to do a power of a value "Roedy Green" <roedy.green@1:261/38.remove-x1c-this> - 2012-07-31 20:07 +0000
      Re: What is the command to do a power of a value "Andreas Leitgeb" <andreas.leitgeb@1:261/38.remove-x1c-this> - 2012-07-31 20:07 +0000
        Re: What is the command to do a power of a value "Patricia Shanahan" <patricia.shanahan@1:261/38.remove-x1c-this> - 2012-07-31 20:07 +0000
          Re: What is the command to do a power of a value "Patricia Shanahan" <patricia.shanahan@1:261/38.remove-x1c-this> - 2012-07-31 20:07 +0000
          Re: What is the command to do a power of a value "glen herrmannsfeldt" <glen.herrmannsfeldt@1:261/38.remove-x1c-this> - 2012-07-31 20:07 +0000
            Re: What is the command to do a power of a value "Patricia Shanahan" <patricia.shanahan@1:261/38.remove-x1c-this> - 2012-07-31 20:07 +0000
            Re: What is the command to do a power of a value "Andreas Leitgeb" <andreas.leitgeb@1:261/38.remove-x1c-this> - 2012-07-31 20:07 +0000

#16850 — What is the command to do a power of a value

From: "xvictoryeohx" <xvictoryeohx@1:261/38.remove-dpk-this>

From: "xvictoryeohx" <xvictoryeohx@1:261/38.remove-rf4-this>

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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#16851

  To: xvictoryeohx
From: "bugbear" <bugbear@1:261/38.remove-dpk-this>

  To: xvictoryeohx
From: "bugbear" <bugbear@1:261/38.remove-rf4-this>

  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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#16852

  To: xvictoryeohx
From: "glen herrmannsfeldt" <glen.herrmannsfeldt@1:261/38.remove-dpk-this>

  To: xvictoryeohx
From: "glen herrmannsfeldt" <glen.herrmannsfeldt@1:261/38.remove-rf4-this>

  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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#16853

  To: glen herrmannsfeldt
From: "xvictoryeohx" <xvictoryeohx@1:261/38.remove-dpk-this>

  To: glen herrmannsfeldt
From: "xvictoryeohx" <xvictoryeohx@1:261/38.remove-rf4-this>

  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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#16856

  To: xvictoryeohx
From: "Roedy Green" <roedy.green@1:261/38.remove-dpk-this>

  To: xvictoryeohx
From: "Roedy Green" <roedy.green@1:261/38.remove-rf4-this>

  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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#16862

  To: Roedy Green
From: "Andreas Leitgeb" <andreas.leitgeb@1:261/38.remove-dpk-this>

  To: Roedy Green
From: "Andreas Leitgeb" <andreas.leitgeb@1:261/38.remove-rf4-this>

  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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#16863

  To: Andreas Leitgeb
From: "Patricia Shanahan" <patricia.shanahan@1:261/38.remove-dpk-this>

  To: Andreas Leitgeb
From: "Patricia Shanahan" <patricia.shanahan@1:261/38.remove-rf4-this>

  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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#16866

  To: Patricia Shanahan
From: "Patricia Shanahan" <patricia.shanahan@1:261/38.remove-dpk-this>

  To: Patricia Shanahan
From: Patricia Shanahan <pats@acm.org>

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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#16868

  To: Patricia Shanahan
From: "glen herrmannsfeldt" <glen.herrmannsfeldt@1:261/38.remove-dpk-this>

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

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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#16874

  To: glen herrmannsfeldt
From: "Patricia Shanahan" <patricia.shanahan@1:261/38.remove-dpk-this>

  To: glen herrmannsfeldt
From: Patricia Shanahan <pats@acm.org>

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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#16875

  To: glen herrmannsfeldt
From: "Andreas Leitgeb" <andreas.leitgeb@1:261/38.remove-dpk-this>

  To: glen herrmannsfeldt
From: Andreas Leitgeb <avl@gamma.logic.tuwien.ac.at>

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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