jordan Posted August 2, 2004 Share Posted August 2, 2004 I was doing some summer calc work and was asked to find the range of the function sqr{x-1}. The book said the answer was infinity but I thought that an argument could be made for sqr{inf-1}. Am I missing something or is this worth getting into? Can't find the LaTeX symbols for sqr and inf either. Link to comment Share on other sites More sharing options...

pulkit Posted August 2, 2004 Share Posted August 2, 2004 What are sqr and inf supposed to mean ? Check LaTex symbols here Link to comment Share on other sites More sharing options...

jordan Posted August 2, 2004 Author Share Posted August 2, 2004 That's the site I was looking for. Ok then, it should look like this: I was doing some summer calc work and was asked to find the range of the function [math]\sqrt{x-1}[/math]. The book said the answer was [math]\infty[/math] but I thought that an argument could be made for [math]\sqrt{\infty-1}[/math]. Am I missing something or is this worth getting into? Link to comment Share on other sites More sharing options...

pulkit Posted August 2, 2004 Share Posted August 2, 2004 First of all [MATH]\infty[/MATH] is a property / tendancy and not a number. So there is nothing such as [MATH]x=\infty[/MATH] or [MATH]\sqrt{\infty-1}[/MATH]. And also Range is an interval, not a number so the range of the function you mention is acctually correctly written as [MATH][0,\infty)[/MATH] Link to comment Share on other sites More sharing options...

jordan Posted August 2, 2004 Author Share Posted August 2, 2004 I am aware that the you stated the correct range at the end there. I didn't simply because the topic did involve the low end of the range. I only wanted to discuss [math]\infty[/math]. As for the first part, I was also aware that [math]\sqrt{\infty-1}[/math] isn't a real funtion. I only intended to point out that it is more accurate than [math]\infty[/math] by inteself. By saying the function tends to infinity negates the fact that when y tends to [math]\infty[/math], x tends to an even greater [math]\infty[/math]. Therefore, would it be more accurate to give [math]\sqrt{\infty-1}[/math] as the upper bound? Link to comment Share on other sites More sharing options...

pulkit Posted August 2, 2004 Share Posted August 2, 2004 No infact it would be totally incorrect because as I pointed out [MATH]\infty[/MATH] is not a number and writing [MATH]\sqrt{\infty-1}[/MATH] would mean you are treating it as a number. Because its a property that a function approaches you ALWAYS say that the function goes to [MATH]\infty[/MATH] and not anything else. To a mathematician [MATH]\sqrt{\infty-1}[/MATH] is just rubbish. Link to comment Share on other sites More sharing options...

jordan Posted August 2, 2004 Author Share Posted August 2, 2004 Because its a property that a function approaches you ALWAYS say that the function goes to [MATH]\infty[/MATH'] and not anything else. Yes, but it's difficult to comprehend that. In order to make the function approach [math]\infty[/math] you need to have a number bigger than [math]\infty[/math], thus negating the initial infinity. To a mathematician [MATH]\sqrt{\infty-1}[/MATH] is just rubbish. Perhaps some other notation should be used then. Then again, mathmeticians have been getting along fine without my help for years. Link to comment Share on other sites More sharing options...

pulkit Posted August 2, 2004 Share Posted August 2, 2004 I do not understand what you find so difficult to comprehend. In order to make the function approach [MATH]\infty[/MATH] you need to have a number bigger than[MATH]\infty[/MATH] , thus negating the initial infinity. This makes no sense to me. [MATH]\infty[/MATH] may be defined as the tendancy to get larger than any numerical value that one can think of. View this problem in the light of this definition. Hence by saying that a function [MATH]f(x)[/MATH] has a range of[MATH](0,\infty)[/MATH] you have conveyed all the knowledge you ever need to. When you initially start with calculus, the concept of [MATH]\infty[/MATH] is rather an confusing one to grasp, but I am sure that as you are exposed to more problems and concepts you will find what I have said to be completely correct. Link to comment Share on other sites More sharing options...

jordan Posted August 2, 2004 Author Share Posted August 2, 2004 Let's say I believe numbers only go up to 5. You say no because 5+1=6. You have proven there is a number greater than five. Now, let's say we want to find the biggest number possible, and we'll call that concept infinity. Now, if I tell you than in order to find that biggest number (called infinity) you need to use a bigger one, you'd be confused. That's about where I stand. How can we say the function tends to the greatest possible number when you always have to use a number that is bigger to get there? Link to comment Share on other sites More sharing options...

pulkit Posted August 2, 2004 Share Posted August 2, 2004 THere is no greates number possible....that is the very essence of infinity. We are not trying to look for numbers here. I will use this concept of infinity whenever I want to describe the behaviour of a function. I can then observe that given a set of numbers, what values can my function take. If I realise that there is no largest number in my set, then obviously no metter how large a number I think up, my function can always take a value (finite) more than this. THen I say that the function goes to infinity. You CAN Not call the biggest number possible infinity because if you are talking of a finite set of numbers, the concept of infinity itself is redundant. If you "believe" in numbers only upto 5, you will never evr encounter things that approach infinity. Link to comment Share on other sites More sharing options...

jordan Posted August 2, 2004 Author Share Posted August 2, 2004 I do understand the concept of infinity and why certain functions tend to infinity. What I'm still not understanding is this: How can you call the solution infinity when you had to put a number greater than it into the function? Doesn't that mean the solution isn't infinity anymore but rather that the input is infinity? I understand exactly what you're saying, pulkit. What I'm looking for is more of a reason than an explanation. I understand why they say this function tends to infinity but haven't gotten an explanation of why I should believe. Does that make sense? Link to comment Share on other sites More sharing options...

pulkit Posted August 2, 2004 Share Posted August 2, 2004 I did not need to put "a number greater than infiinity"(The mathematician in me dies a thousand deaths when I use Lingo like that ) to figure the range out. All I saw was that as I keep putting in larger and larger numbers the function keeps giving higher and higher outputs. In other words, the function is an increasing function, i.e., if x>y then f(x)>f(y). As there is no limit to the hugeness of x that I can choose over the domain of real numbers, there is also no limit to the values that I can get out of the function. Hence I conclude that the function must approach infinity. I do not acctualy need to plug in any numbers into the function, its a purely theoretical arguement. And whenevr you have to determine the range of a function (a very tricky thing to do by the way when you get to complex functions), you either do so graphically or by the type of theoretical arguements I mention. So the question of plugging in "numbers greater than infinity" (sigh !) never arise. I hope I resolved your issue, its bed time for now ! Link to comment Share on other sites More sharing options...

jordan Posted August 2, 2004 Author Share Posted August 2, 2004 Could this be solved with the talk I've heard recently about the greater and lesser degrees of infinity? Link to comment Share on other sites More sharing options...

MandrakeRoot Posted August 3, 2004 Share Posted August 3, 2004 It is pretty simple actually. [math][0,\infty)[/math] is just a notation to say that your function can take on any positive real value. In this sense infinity is just a notation and nothing more. The lesser and greater number of infinity is more something that arises in set theory. For exemple it is possible to show that [math]\mathbb{N}[/math] and [math]\mathbb{Q}[/math] contain just as much elements (though both contain an infinite number of them. On the contrairy [math]\mathbb{R}[/math] contains strictly more elements than either of these two sets mentioned above, and the set consisting of all subsets of [math]\mathbb{R}[/math] contains again strictly more elements and so on and so on, but that has nothing to do with numbers or calculating. You cannot calculate with infinite "as a number" except when you impose trivial rules as : [math]\infty + a = \infty \forall a \in \mathbb{R}[/math] and so on. Here [math]\infty - \infty[/math] will always remain undefined, because otherwise you can arrive to contradictions. So to make short the argument, infinity is a usefull concept or in the use [math][0,\infty)[/math] just a simple notation to say "all positive real numbers greater or equal to zero". I hope that makes it more clear now ? Mandrake Link to comment Share on other sites More sharing options...

Dapthar Posted August 3, 2004 Share Posted August 3, 2004 Could this be solved with the talk I've heard recently about the greater and lesser degrees of infinity?Nope. The range of a function doesn't have anything to do with "degrees of infinity". As far as I know, the concept of "degrees of infinity" is not a formal mathematical construct. The only concept that I have seen that would fit the "degrees of infinity" description you have supplied is the "big O" notation used in Computer Science classes, which is used to describe the asymptotic behavior of functions. Usually the functions of interest describe the execution time of various algorithms. Link to comment Share on other sites More sharing options...

jordan Posted August 3, 2004 Author Share Posted August 3, 2004 I'm still relatively unsure, but I'll take everyone's words for it. I'll probably just have a quick chat with the Calc teacher about it when school starts. Link to comment Share on other sites More sharing options...

MandrakeRoot Posted August 4, 2004 Share Posted August 4, 2004 I do not agree with you Daphtar. The big O notation is something indicating the speed at which something tends to zero or infinity or whatever. Say for instance you have two algorithms to solve something with error resp. O(1/n) and O(1/n^2), the second would be preferable since it converges more quickly, but that has nothing to do with infinity. The only reference i see to different degrees of infinity is the different (infinite) cardinals. You can create sets which are all infinite and each time one is strictly bigger than another. (two sets are equal in size if you can create a bijection in between them). In the original question of jordan infinity is not an element, but a notation. Mandrake Link to comment Share on other sites More sharing options...

pulkit Posted August 4, 2004 Share Posted August 4, 2004 I do not agree with you Daphtar. The big O notation is something indicating the speed at which something tends to zero or infinity or whatever. Say for instance you have two algorithms to solve something with error resp. O(1/n) and O(1/n^2), the second would be preferable since it converges more quickly, but that has nothing to do with infinity. An algorithm's time complexity can nevr converge to 0. Orders like O(1/n) and O(1/n^2) are impossible. It would mean a greater problem size needs lesser time. Orders do indeed diverge to infinity and what you do is figure whcih one does how slowly, that'll be the better algorithm. Link to comment Share on other sites More sharing options...

MandrakeRoot Posted August 4, 2004 Share Posted August 4, 2004 I was talking about the precision of the algorithm not the time complexity (n being the number of iterations or something like that and not the size of the problem). Time complexities could ofcourse never converge to zero for obvious reasons, like you say. Still say having an O(n) (this time n = size of problem) alg. and one O(n^2), then both would be just as "infinite" since it would suffice to put an n^2 size problem to the first and an n-size problem to the second. Mandrake Link to comment Share on other sites More sharing options...

pulkit Posted August 4, 2004 Share Posted August 4, 2004 Still say having an O(n) (this time n = size of problem) alg. and one O(n^2), then both would be just as "infinite" since it would suffice to put an n^2 size problem to the first and an n-size problem to the second. Whats that suposed to mean ? The only place you ever encountered the big O notation is when you study the assymptotic time complexity and not "efficiency". And also, no. of iterations would never approach zero. Link to comment Share on other sites More sharing options...

MandrakeRoot Posted August 5, 2004 Share Posted August 5, 2004 No you are wrong the big "O" notation is also used in mathematics and is used there to indicate the behaviour of errors in function of the "step of discretization" for example. I was using this notation in that sense, so O(1/n) is the behaviour of the error the algorithm makes in function of the number of iterations ! (When we do more iterations, the error will become smaller), which means nothing on the number of iterations. I will write it down even more clear because otherwise you might misunderstand : Suppose the algorithm is supposed to calculate the quantity X, but in fact calculates X_bar, and we know by analysing the algorithm that |X - X_bar| = O(1/n), where n is the number of iterations. Then if we desire a precision of approx. 1/100, we will need to do something like 100 iterations ! Depending on the size of the problem m, the algorithm could easily have time complexity O(m^5). The goal of the O notation is to majorate some function of something, by another more easy function in order to be sure to have some knowledge of "worst" case behaviour. If your algorithm has time complexity O(m), (m size of problem), then that does not mean that for every instance of your problem the algorithm wil terminate in this time, it might as well terminate before, but the "worst" instance will generate this behaviour ! Mandrake Link to comment Share on other sites More sharing options...

pulkit Posted August 5, 2004 Share Posted August 5, 2004 Ah ! I know about the last part that you mention : worst case analysis : coz I have to do it all the time. Never acctualy used O notation in maths, I was merely referring to how you analyse algorithms in computer science using it...........its a very nice concept though, can be used to analyse those probablistic algos. Link to comment Share on other sites More sharing options...

MandrakeRoot Posted August 6, 2004 Share Posted August 6, 2004 Yeah it is that. There are some algorithms with O(exp(m)) behaviour, where m is the size of the problem, but in practice some of them are not that bad and actually terminate. It is just that the theoretical worst case behaviour is disastrous for such algorithms, though for some "real life" problems that might be a lot better. Mandrake Link to comment Share on other sites More sharing options...

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