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Posted: Thu May 12, 2005 3:15 pm
by Psychophanta
MrMat understands it. And he can explain, but looks like he doesn't want

Posted: Thu May 12, 2005 4:20 pm
by MrMat
Sorry i wasn't around to answer. The mathematical definition of a limit probably won't help describe it but basically it's the result of a convergent sequence. Here's a couple of examples:
e.g. take the sequence given by 2-1/x. Then as x goes to infinity the 1/x term gets smaller and smaller (it is converging), leaving the 2 term behind and so lim x->inf 2-1/x = 2.
Another example, take the limit of sin(x)/x as x goes to zero. Clearly sin(x)/x is undefined (we can't divide by zero) but if you look at a graph you can see the value it is approaching is 1. To show this mathematically you can use an expansion of sin(x)=x-(x^3)/6+(x^5)/120+... Then divide that by x to get 1-(x^2)/6+(x^4)/120+... and so as x goes to zero the only term left is 1. Then lim x->0 sin(x)/x = 1.
So what it's saying is that a function may have a limit even though the function itself doesn't exist at that point. An interesting point is that the limit may be different depending on whether you're approaching from above or below so the + or - sign is often used to show from which side you were approaching the limit:
e.g. lim x->0 x-[x] where [x] means the integer part of x (i.e. [5.678]=5). If you look at the graph of that it slopes up from 0 to 1 then jumps back down to 0 and slopes up to 1 and so on. So lim x->0- x-[x] = 1 (that's the limit as x goes to 0 from below, i.e. going up a slope) and lim x->0+ x-[x] = 0 (the limit as x goes to 0 from above, i.e. going down a slope). Then lim x->0 x-[x] doesn't exist because the limits from each direction are different.
So onto the question (at last). Lim x^x as x->0+ means you consider the graph of x to the power of x and look at what it does as x gets smaller and smaller (but starting from a positive x). Why do that? Perhaps because it would give a useful value for 0^0, but from earlier examples we've seen that lim 0^x as x->0 gives a different value (zero), so the value for 0^0 depends on what problem you're solving in the first place. For most people the answer 1 seems to be the useful answer so that's why it is often used but for pedantic people like myself i'd say 0^0 is indeterminate. It really doesn't matter though as long as you know what the value given by the programming language is and if for some (probably unlikely) reason you need it to be a different value then you code around it.
If you've read all that and not fallen to sleep then well done

Posted: Fri May 13, 2005 9:34 am
by Psychophanta
MrMat, that certain humbleness i see in explanations like that makes me to trust.
I wish there were more humble people.
However, i can't see why lim x->0- x-[x] = 1. Are not that wrong?
Posted: Fri May 13, 2005 9:36 am
by thefool
so im not humble? i just want evidence for what i hear. Also again, it was NOT an huge bug, as its the correct result it returns. But again i do belive that its NOT = 1 but its normally defined to be 1!
Posted: Fri May 13, 2005 10:10 am
by Psychophanta
thefool wrote:so im not humble? i just want evidence for what i hear. Also again, it was NOT an huge bug, as its the correct result it returns. But again i do belive that its NOT = 1 but its normally defined to be 1!
I recognize the title of this thread is not humble, but it's just a quest to put onto the "General Discussion" table.
Posted: Fri May 13, 2005 10:21 am
by thefool
okay good. but you didnt answer me, am i the most un-humble you ever met?
Posted: Fri May 13, 2005 10:35 am
by LarsG
Psychophanta:
why can't you let the subject rest...
I haven't read any arguments from you as to why you think the result is wrong?
Other in this thread have been argumenting with ref. to books and people..
The fact is that the result is unresolvable, *but* it is accepted as the result of "1" by almost all the higher mathematicians, professors etc. etc...
(and as you can see, it is even embedded into the calculators)
If you don't like it, then I suggest you use a programatic way around it (easily done), or drop the whole calculation, and fiddle with something else..
Posted: Fri May 13, 2005 10:36 am
by Psychophanta
thefool wrote:okay good. but you didnt answer me, am i the most un-humble you ever met?
hahaha! you must know it, not me!
Anyway, that's not the matter in this thread

Posted: Fri May 13, 2005 10:45 am
by Psychophanta
LarsG wrote:I haven't read any arguments from you as to why you think the result is wrong?
Result as 1 is wrong because it is not an absolute true result, that's all.
If 0^0=1 then why 0/0 is not 0 or 1 too? (with your arguments: 0 divided by anything is 0, don't? , or anything divided by samething is 1, don't?)
LarsG wrote:The fact is that the result is unresolvable, *but* it is accepted as the result of "1" by almost all the higher mathematicians, professors etc. etc...
No, no. That's not so clear.
LarsG wrote:(and as you can see, it is even embedded into the calculators)
FALSE
LarsG wrote:If you don't like it, then I suggest you use a programatic way around it (easily done), or drop the whole calculation, and fiddle with something else..
Thans, i did it yet :roll:
Posted: Fri May 13, 2005 10:46 am
by thefool
Lars is right! you really havent been argumenting for your case!
we other have. Mrmat has explained it, but you still bring this topic up and call all other that tries to prove that its 1, for unhumble persons..!
Lars explained it the simplest way. You havent been agumenting! whenever we said something you asked if we were 11 years old or said that "What would niels bohr think if he heard this" etc.. I tried to say it was = 1 using math books and calculators. Now i know its not = 1 but it should return 1 because, as lars said, its the result that almost all proffersors or mathematicans accept it as!
Even my programmable graph-calculator wich can solve equations or whatever you would like it to, returned 1.
so instead of keeping insulting people you should know that this was not too easy to understand because _YOU_ didnt come with any prove that pb should fail...
edit: And it IS included in every calculator that arent from [-∞ ; 1930] blah
and why do you bring in division with zero? you think its the same thing?
Posted: Fri May 13, 2005 10:50 am
by Psychophanta
PB do it well done, it is Intel FPU which should be programmable to return 0, 1 or indetermined

Posted: Fri May 13, 2005 10:52 am
by thefool
good you see it! and amd's processors too

well was about to say that you werent humble, just impolite but ill take it back if you can see what you have done in this thread :roll:
Posted: Fri May 13, 2005 10:59 am
by Psychophanta
thefool wrote:you werent humble, just impolite
Could be, but pease, come on, here we are not analising and discussing about people, but about facts

Posted: Fri May 13, 2005 11:05 am
by thefool
why do you bring people in then?
Posted: Fri May 13, 2005 1:05 pm
by DoubleDutch
Code: Select all
If you don't like it, then I suggest you use a programatic way around it (easily done), or drop the whole calculation, and fiddle with something else..
Thats almost exactly what I said to him at least 10 messages ago, he is obviously not listening.
The one thing that I think should change is the rating your given (eg Psychophanta is rated as an expert, myself as an Apprentice). He is obviously not an expert & I don't really consider myself his appentice!
It needs to be another system rather than how many questions you have asked. The current system is quite insulting actually!
LarsG wrote:
(and as you can see, it is even embedded into the calculators)
FALSE
It is indeed true! Just disassemble the code from a HP calculator. The rom for the calculators is available on the HP website. What calculator do you have where is not pre-calculated? The code I've seen appear to show x^0=0 and x^1=x, others are calculated.
-Anthony