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Bwyard
Newbie
Joined: 28 Apr 2009 Posts: 17

Posted: 01 Dec 2009 01:12:13 pm Post subject: 


i need code to find limits on the calculator or some help on how to figure it out. i have symbolic and omnicalc so i could use that in the code i dont mind. I am writing a calculus program which is why i need the code. 

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ztrumpet
Active Member
Joined: 06 May 2009 Posts: 555

Posted: 01 Dec 2009 03:45:21 pm Post subject: 


Limits? Like what?
Do you mean something like how many digits of precision the calc has? It has 14, although commands like fpart only have 12 or 13.
Last edited by Guest on 01 Dec 2009 03:45:39 pm; edited 1 time in total 

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GloryMXE7 Puzzleman 3000
Active Member
Joined: 02 Nov 2008 Posts: 604

Posted: 01 Dec 2009 04:02:35 pm Post subject: 


based on the context I believe he means calculus limits 

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calcdude84se
Member
Joined: 09 Aug 2009 Posts: 207

Posted: 01 Dec 2009 05:44:18 pm Post subject: 


Limits are typically evaluated by getting really close to the limit or by simplifying symbolically. I must say that Symbolic's symbolic capabilities aren't that powerful. (I prove my point by asking my calculator to simplify (x^2+3x+2)/(x+2) and getting back what I put in) Though you won't get exacts, evaluating it by getting very close to the limit is the only practical way until someone writes a better symbolic manipulation program. The problem with this is that you can't evaluate limits with other variables in them. I am not in the place to write a CAS, so hope that someone else will... (or do it yourself ) 

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thornahawk μολών λαβέ
Active Member
Joined: 27 Mar 2005 Posts: 569

Posted: 01 Dec 2009 08:19:44 pm Post subject: 


Quote: Limits are typically evaluated by getting really close to the limit or by simplifying symbolically...
For a human evaluating limits, yes. For a computer, except in very special cases, the usual method for symbolic evaluation starts with a series expansion of the function(s) involved (just like in the application of L'Hospital's rule), as well as any number of rather complicated tricks (e.g. writing rational functions in terms of powers with negative exponent).
Programming all those, especially on a very little calculator, are none too easy.
There's the numerical approach (which just *estimates* limits and is easier to implement), but these methods aren't foolproof, especially for the "I got it from the calculator so it must be true" people.
thornahawk
Last edited by Guest on 01 Dec 2009 08:21:32 pm; edited 1 time in total 

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Bwyard
Newbie
Joined: 28 Apr 2009 Posts: 17

Posted: 03 Dec 2009 01:13:45 pm Post subject: 


well i know it can be done because a + calc does it but i dont know how it does it. so could you help me determine the code. 

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calcdude84se
Member
Joined: 09 Aug 2009 Posts: 207

Posted: 03 Dec 2009 04:29:11 pm Post subject: 


You mean a TI83/84+(SE)? I didn't think they could do that...
(Checks...)
Well, they can't as far as I know.... What are you thinking of? 

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thornahawk μολών λαβέ
Active Member
Joined: 27 Mar 2005 Posts: 569

Posted: 04 Dec 2009 09:07:19 am Post subject: 


I said, "none too easy", not "impossible". Unless you've perused fully up to the little nuances of symbolic computation, I cannot recommend any algorithm you can easily input into your calculator.
Just to give you a hint of one of the more general methods, it requires determining symbolically the first few derivatives of the function whose limit is being taken...
thornahawk 

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Builderboy2005
Advanced Newbie
Joined: 19 Apr 2009 Posts: 51

Posted: 04 Dec 2009 05:59:46 pm Post subject: 


A good way to find the actual number that is a limit, you can do something like this: If your limit is
Lim (x^24x5)/(X5)
X>5
You can input that straight into Y= and then plug in 5.000000001 or 4.9999999999
If your limit is Infinity, just plug in 1E99 or 1E99 for negative infinity. 

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thornahawk μολών λαβέ
Active Member
Joined: 27 Mar 2005 Posts: 569

Posted: 04 Dec 2009 07:37:14 pm Post subject: 


Substituting numbers just a bit near the actual point where the limit is to be evaluated is not a foolproof solution:
Code: e12→X:(1+Xֿ¹)^X
2.718281828
e99→X:(1+Xֿ¹)^X
1
e12→X:(e^(X)1)/X
1
e13→X:(e^(X)1)/X
0
But if you're going to try this one, try not to go past 10^12 or within ± 10^12 of a finite number...
To be fair, though, the numerical approach I was alluding to in an earlier post involves evaluating a sequence of numbers that keep approaching the limit before extrapolating; if you're lucky with these sort of algorithms, you already have a good answer after the first few sequence terms. :)
thornahawk 

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