bc and exponents containing decimals and fractions.

Dogs · 09-15-2009, 12:52 AM

I need bc to allow me to use expressions such as 2^4.7

Currently, I get the error "Runtime warning (func=(main), adr=9): non-zero scale in exponent"

I downloaded the source awhile ago and located the error message, but I'm unsure as to how to phrase it so it will actually work.

I need this to work for my math class I'm in. Any help would be appreciated, and any help that would allow me to understand more about how this process works would be even moreso appreciated.

estabroo · 09-15-2009, 08:47 AM

use octave instead, I'm guessing it would require serious changes to get bc to do decimal/fractional exponents

octave:1> 2^4
ans = 16
octave:2> 2^4.7
ans = 25.992
octave:3> 2^(2/3)
ans = 1.5874

Dogs · 09-15-2009, 05:56 PM

Quote:

Originally Posted by estabroo

use octave instead, I'm guessing it would require serious changes to get bc to do decimal/fractional exponents

octave:1> 2^4
ans = 16
octave:2> 2^4.7
ans = 25.992
octave:3> 2^(2/3)
ans = 1.5874

Hey, thanks a lot. I'll give Octave a try and report back. I still want to change bc, though, as my main focus right now is being a darn good programmer. Part of this shall include ridiculous and unnecessary challenges, like adding functionality to bc through it's source code.

jlinkels · 09-15-2009, 10:13 PM

I wonder what you kids learn in school nowadays.

x=a^^b
ln(x)=ln(a^^b)
ln(x)=b*ln(a)
exp(ln(x))=exp(b*ln(a))
x=(exp(b*ln(a))

and of course this works in lc as well:

Code:

echo "e(4.7*l(2))" | bc -l
25.99207668339953672244

jlinkels

gmbastos · 09-15-2009, 10:55 PM

Quote:

Originally Posted by jlinkels

I wonder what you kids learn in school nowadays.

x=a^^b
ln(x)=ln(a^^b)
ln(x)=b*ln(a)
exp(ln(x))=exp(b*ln(a))
x=(exp(b*ln(a))

and of course this works in lc as well:

Code:

echo "e(4.7*l(2))" | bc -l
25.99207668339953672244

jlinkels

Good point, jlinkels, but remember it looses some precision due to introducing more rounding off.

One possible workaround would be to increase precision by two digits, performing the calculation and rounding back to the desired precision (still looses precision due to rounding but way less).

p.s.: don't trust that much in high school curriculum: there is much more distraction for kids nowadays (TV and internet for instance...).

jlinkels · 09-16-2009, 03:59 AM

I assume (but not sure) that ln and exp are calculated internally as Taylor series. I also think that the a^^b is calculated as Taylor series. That means in the first case two Taylor approximations and in the second case only one.

The question is how many Taylor terms are being used and how fast the series converge for ln, exp and ^^. The so called rest term of an n-length Taylor can be calculated. And how this is realated to the internal precision of bc.

Though the question is academic, it would be nice to look into that (by someone who has more time available than me)

jlinkels

sxdw · 03-12-2013, 04:27 PM

Quote:

Originally Posted by jlinkels

I wonder what you kids learn in school nowadays.

x=a^^b
ln(x)=ln(a^^b)
ln(x)=b*ln(a)
exp(ln(x))=exp(b*ln(a))
x=(exp(b*ln(a))

and of course this works in lc as well:

Code:

echo "e(4.7*l(2))" | bc -l
25.99207668339953672244

jlinkels

A calculator is supposed to solve expressions. You are not supposed to make life easier for the calculator, it is supposed to be the other way around...