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[Skyrunner]

X^0 = 1 for all X. Therefore lim[n->0 f(n)^n]=1 for all f(n). The nature of the convergence depends on f(n), but the limit itself doesn't.

Actually, 0^0 is not defined. Doesn't matter in this case, but just saying

[ZetaX]

Every sensible mathematican defines 0^0 as 1. The simple fact that this "disagrees" (better: something is simply not continuous) with some limits is completely irrelevant for that.

[Skyrunner]

0^0=1 seems pretty natural and consistent with all other x^0, but my textbook and all the math teachers I have disagree. :c
A quick internet search tells me that this argument has gone on for some time o_O

[Shinziril]

You can think of it this way:

For any number n^x, multiplying this number by n produces a number equal to n^(x+1).  Similarly, dividing by n produces n^(x-1).  n^1 has been defined as just n.  n^0 is therefore n/n, which is 1 for any nonzero n.  However, 0/0 does not equal 1, but is instead undefined, therefore 0^0 is also undefined. 

0/0 has to be left undefined because of the way we set up the system of mathematics in the first place (ref).

[RedWarrior0]

Every sensible mathematican defines 0^0 as 1.

You say defines like it means something other than what it is.

0^0=1 seems pretty natural and consistent with all other x^0, but my textbook and all the math teachers I have disagree. :c

0^0=0 seems pretty natural and consistent with all other 0^x. Well, except the negative values of x.

Also, I'm pretty sure it's possible to break logic if 0⁰=1, and I'll avoid using calculus or the non-existent zeroth root.
edit: no wait it isn't because 0^0 is an indeterminate form. if lim x->n f(x) = 0 and lim x->n g(x) = 0, then lim x->n f(x)^g(x) could be any value.

[ZetaX]

Also, I'm pretty sure it's possible to break logic if 0⁰=1, and I'll avoid using calculus or the non-existent zeroth root.
edit: no wait it isn't because 0^0 is an indeterminate form. if lim x->n f(x) = 0 and lim x->n g(x) = 0, then lim x->n f(x)^g(x) could be any value.

No you can't break logic using that (you wouldn't even be able to do that with 0^0=0). 0^0=1 is the only definition that makes sense all over the areas of mathematics. As I already said, limits are not even an argument here, as it is of no use to wish for some property (continuity of something) which cannot be fulfilled. If you want a simple example where every mathematician uses 0^0=1 without even thinking about it:

Power series or polynomials: You often write a sufficiently nice function as f(x) = sum_{i=0...inf} a_n x^n. This implicitely assumes that 0^0=1 as otherwise you will need to add an exception or f(0).

For any number n^x, multiplying this number by n produces a number equal to n^(x+1).  Similarly, dividing by n produces n^(x-1).  n^1 has been defined as just n.  n^0 is therefore n/n, which is 1 for any nonzero n.  However, 0/0 does not equal 1, but is instead undefined, therefore 0^0 is also undefined. 

n^0 is not n/n. You can't just claim some rule to be true, as it isn't (this is the same as for the continuity arguments). This is the same as the argument that 1 = (1²)^(1/2) = ((-1)²)^(1/2) = -1, which is also an abusive use of formulas. Just because something is true in other cases says often nothing about the exempt ones.
But your post is somewhat true as well: e.g. the power series version wants a^0=1 for all a, which is a good start for the convenient inductive definition a^n := a·a^(n-1).

0^0=1 seems pretty natural and consistent with all other x^0, but my textbook and all the math teachers I have disagree. :c

Well, most serious math text book (graduate ones, for example) won't, as will professors. Teachers are probably never told about this kind of things anyway.

Every sensible mathematican defines 0^0 as 1.

You say defines like it means something other than what it is.

No I don't. There is a huge difference between an arbitrary definition (0^0=5) and a useful one (0^0=1). This is a standard phenomenon in mathematics. Another classical one is 1 (or more generally, a unit) not being a prime, as this simply makes stating theorems less tedious.

[Another]

It may not be mathematically strict but I think that a*x^0 in the usual power series expression is assumed to actually be reduced to a*1 before substituting any x there. Otherwise your formalism would produce a*0*x^(-1) term for differentiation and you would not be able to dismiss it at hand for all x because of 0/0 possibility.

[MagmaMcFry]

Just look at the product formula, x^n is defined as prod[k=1...n, x]. The empty product is always defined as 1.

[Another]

Deleted.

[Aptus]

Do you advocate accepting 0^0 by definition, MagmaMcFry? Because if you do, then (0^0)^0=1^0=1 but on the other hand (0^0)^0=0^(0^0)=0^1=0.

You are doing some illegal shit with those paranthesis. (0^0)^0=0^(0^0) is not a valid expression.

See for example 5^(2^3) = 390625, and (2^3)^5 = 32768.

You can't just move paranthesis around all willy nilly when you are dealing with exponents.

[ZetaX]

It may not be mathematically strict but I think that a*x^0 in the usual power series expression is assumed to actually be reduced to a*1 before substituting any x there. Otherwise your formalism would produce a*0*x^(-1) term for differentiation and you would not be able to dismiss it at hand for all x because of 0/0 possibility.

If x^0 is always 1 as in my definition, then you would just substitute that anyway. There is one fault here: you use that d/dx (x^n) = n x^(n-1), but this rule is not true for n=0 by exactly your reasoning; or it at least needs a convention like "0 as a constant has more zeroness than any function that is not constantly 0" (this could e.g. be put into a valid formal definition in some contexts like complex analysis). Again, this example is just a continuation of already existing rules you would like to be true, but which cannot be made true by _any_ definition of 0^0 (as for 0^0:=a \neq 1, the result wouldn't even be differentiable), so there is no point in trying or using it as a reason that 0^0=1 is bad.

A good reason for 0^0=1 being the wrong choice would be an example where it causes more drag and nuisance than another specific one.

[lordnincompoop]

I'm writing a mathematics essay on applied Markov chains, and I've got a Markov chain-style word predictor and text writer that I drafted for this project. Problem is, when I made these algorithms up for my program, I wasn't thinking mathematical terms but just in what I could easily write and compile and have it run decently fast. Not only that, but this is the first time I've ever tried to define something formally and tried to work through it using purely mathematical functions that describe each process. As far as I know, storage variables and side-effects (I don't know if this even makes sense, which goes to show how little I know).

I'm trying to handle the task of describing my algorithms and steps my program goes through as best I can. I've managed thus far, and my results are in the spoiler below.

Spoiler (click to show/hide)

Such a word predictor needs to do only one thing: Look at the current word, and predict the next; though other bots and programs, such as Swiftkey and M.V. Shaney have made use of context, but that is not essential. To do so, however, it needs to examine all the words that have followed the current word before and determine which one was the most common from frequency of occurrence. This process can easily be described as a Markov chain:

Let there be a sequence of words representing input prose called T. Let there be a finite state space S representing all the unique words in T. Let C be a matrix with as many rows and columns as there are states in S. If the number of times word i has transitioned immediately to j in text T can be represented as Ci,j, the ith column and jth row of C represents Ci,j. This will be the basis to creating the transition matrix.



To be able to predict words and add more input text, fixed probabilities were unfeasible. There were two methods of calculating transition probabilities that I created. The first was more suitable for large pieces of text in the manner of M.V. Shaney, as it gave greater variation in diction and virtually eliminated instances where the program would simply repeat a phrase over and over again. That first method is detailed below.

For a transition matrix P describing state space S, let Pi,j, representing the transition probability from state i in S to j in S, equal f(i,j).

   Where n equals the last element of S.

But now I'm stuck on something completely different. The second method for finding the transition probabilities looks like it'd be very difficult to translate, since it relies on iteration (kinda) and on if-then-else constructs, which I've never really seen or used yet. Below is the snippet of code directly pertaining to the second method.

Spoiler (click to show/hide)

std::string MarkovNode::FindLikeliest() {
       
        std::pair <std::string, unsigned long long> likeliest;
        if (linksForward.empty() != true) {
                likeliest = *linksForward.begin();
                for ( std::map <std::string, unsigned long long>::iterator it = linksForward.begin(); it != linksForward.end(); ++it ) {
                        if(it->second > likeliest.second) likeliest = *it;
                        else if((rand()%2) == 1) likeliest = *it;
                }
                return likeliest.first;
        }
               
        else return "There are no other words that follow this word.\n";
}

What this means is as follows: Set the highest count to be the first column element in the row corresponding to the current state. Iterating through the row corresponding to the current state in matrix of counts C (defined in the first spoiler), if the element in the next column has a higher count, set the highest count to be that. If it has the same count, there is a 50% chance that the highest count becomes that. Repeat to the last column and return the highest count thus calculated.

That means that for a word state where there is exactly one column with the highest count for the corresponding row in matrix C, there is a 100% chance of transitioning to the word corresponding to the column. If there are more with the same count, there is an equal chance for all of them. (Or at least it's supposed to - right now, it creates a situation where for a number of situations 2^n-1, where n is the number of equally most common next words, the last word in the set (arbitrarily ordered, unfortunately) is chosen in half of those situations, and the next last is a quarter, and so on dividing the chance in half until the two first elements each receive 1/(2^n-1) chance. I'll get around to fixing that.) The issue is that this is obviously not the kind of function I used for the first method or something that easily fits into a stochastic matrix - but I have absolutely no idea how to convert it into something that would.

Can anyone help?

[palsch]

X^0 = 1 for all X. Therefore lim[n->0 f(n)^n]=1 for all f(n). The nature of the convergence depends on f(n), but the limit itself doesn't.

To try to decrypt this.

X^0 = 1 for all X is a convenience as argued already by others. It's often assumed true even for undefined, infinite or zero values, even if it's not formally true. Not assuming it (especially for 0^0) breaks binomial theorem, and having binomial theorem work is more important than formal completeness at this particular level.

In this case, we have lim[n->0 f(n)^n]=1 where f(n) = 1/n. For all n>0, f(n)^n is continuous and real.* This allows us to simply take the limit as the value at n=0, so long as you are willing to accept the (1/0)^0 = 1. If you take 1/0 undefined then the entire operation is invalid, which is why the assumption of any X^0 being 1 and then ignoring ignoring the actual nature of X is the first step I'd take.

I fully understand that this could be taken as arguing backwards - that the X^0 = 1 is using the limit of X^0 as the assumed value - but the assumption of X^0 = 1 is more usually seen as the fundamental axiom from which the limit is taken rather than the other way around. Which kinda breaks the question, but still.

* Actually the entire function f(n)^n is continuous but complex everwhere but 0 with a limit of 1 (or 1 + 0i if you have a strict requirement about complex notation) from either approach. This is probably a little above the requested level of mathematics, but is the reason you can confidently say that the limit is 1 the way that the question was originally worded, rather than just the limit as n goes from +inf to 0.

[MagmaMcFry]

If you have f(n) = 2^n-1, what is lim[n->0, f(n)^n] then?

[Jim Groovester]

Can anyone help?

Programming help or representing a stochastic matrix for your second part?

If I'm understanding the logic of your algorithm as you intend it to run, for a given row in your stochastic matrix, you're ignoring all non-maximal entries, so just replace the stochastic matrix with a new one with all non-maximal entries in a row set to zero, and the remaining entries scaled by the appropriate factor.