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Subject: Re: math function for calculator

Re: math function for calculator

From: BlueChip <cs_bluechip_at_webtribe.net>
Date: Mon, 03 May 2004 18:14:35 +0100

>Woo, really thanks. I use your algorithm and it works like a charm.
>
>Your algorithm is very efficient especially for my program.
>I am using scientific number format, the number needed to
>find root is always between 0 and 10. I take the midpoint of
>operand and 1 as the initial guess. Most of time, it takes less
>than 10 repeat. Average is about 5 times! Math is beautiful!

Thank you, and you are most welcome :)

>Can you tell me the name of the algorithm?

Sorry, I have no idea if it has a name - I based it on an estimation theory
I found when trying to calcualte PI for use in a random number generator a
few years ago.

>And do you know the algorithms for other functions like sin,cos,
>especially log?

Sin & cos can be done in code with the "cordic" algorithm - I'm sure google
will aid you with some source code - as this is what is in most calculators
- if not give me a shout and I will dig some out for you - it is FAR too
horrific to try and describe here, and the method I could describe here
is really too chunky and time-consuming for use in this scenario - check
out cordic, it's really quite nice :)

If I ever find out why I should care what a log is, you will be the first
to know - LOL
I'm sure you should be able to use it to work out how many bits are
required to store a specified number - but I've never worked out how to
actually do it
/me bows his head in shame
Anyway, the point is, sorry, no I have never understood logs let alone
worked on theories for them :(

pihCeulB

>Isaac
>
>------------------ Original Message ------------------
>
> >Try this:
> >
> >
> >V/G = N ... (G+N)/2 = G
> >repeat until G=N
> >
> >
> >V = Value (for which we are seeking the root)
> >G = most recent Guess
> >N = New guess
> >
> >
> >Example for root of V=12 with an arbitrary intial guess of 2
> >
> >
> >find _/12 ... guess 2.000
> >12/2.000 = 6.000 ... (2.000+6.000)/2 = 4.000
> >12/4.000 = 3.000 ... (4.000+3.000)/2 = 3.500
> >12/3.500 = 3.429 ... (3.500+3.429)/2 = 3.465
> >12/3.465 = 3.463 ... (3.465+3.463)/2 = 3.464
> >12/3.464 = 3.464
> >
> >
> >3.464^2 = 11.999
> >
> >
> >_/12 is irrational, so to get 12 back out of the other end you will need an
> >infinite number of digits after the decimal place, hence the result above
> >being 11.999, that is because I decided to offer a result to only 3 decimal
> >places.
> >
> >
> >Obviously faster methods can be found, but generally you will need to rely
> >on the existence of other mathematical functions (such as log-n) or big
> >memory-hungry lookup tables - Although for this application where our app
> >is 32K end-of-story, a lookup table of 1..2K would not be harmful.
> >
> >
> >One use of a lookup table might be to get a good initial guess...
> >With a table of all known integer squares:
> >
> >
> >_/105 lies somewhere between _/100 (=10) and _/121 (=11)
> >121 - 100 = 21
> >21 / (105-100) = 4.200
> >(1) / 4.200 = .238
> >
> >
> >So an initial guess of 10.238 would be reasonable
> >The answer is actually 10.247
> >
> >