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chevron_right03-31-2003
x^x mathematics
The area of the generally fairly simple sounding formula y=x^x is quite interesting. To explain it for laymen, there is not much to it at first: 1^1=1, 2^2=4, 3^3=27, 4^4=256 and so on. However, how could one find a reversal for this arithmetic operation? How could one get from 256 to 4, from 27 to 3, and so on? So far, I have found only one way to do this, which can be done on almost any calculator:
This formula gives arbitrarily exact results, unless the x-value is larger than the irrational number e, which starts with 2.718281828. As soon as the x, which you don't know before, is greater than e, you get not only one result, but actually two - funnily enough, the calculator then always switches back and forth between the two results. You can see if this will be the case, if the number y, which you know before, is larger than about 15.15, because e^e = 15.15426..., which is so to say the limit for this approximation to x. Now, as already mentioned, we get not only one result, but two - the larger y is, whose x we want to calculate, the more the one value approaches 1 and the other value approaches y. Let's label these two new numbers a and b, in order to keep track of them. Now these two numbers have interesting properties, because a^b gives exactly the same result as b^a, i.e. exactly y. So mathematically we can say:
Now some more die-hard people will say: hold on, what if x is EXACTLY e? The problem in this case is: this is not possible at all, because e has infinitely many decimal places, it is irrational. In finite time one can only calculate finitely many decimal places of e, so that one can only speak of more or less than e, but never EXACTLY of e - one of the confusing properties of irrational numbers. The fact that we call the number "e" does not mean that we know it completely - we can only say about a completely known number that it is greater than, less than or equal to another number. There is always some uncertainty about a number that is not completely known - with an infinite number of decimal places, this uncertainty is infinite, so mathematically there can be no number that is exactly as large as what we call Euler's number. Eventually, the case x=e will be derivable in some completely different way - at present, however, it seems impossible to determine whether the case x=e belongs to the category x>e as well, or whether something completely unexpected is the case for this special case with infinitely many decimal places. In any case e represents the limit above which a and b exist.
An example of a and b are the only integers that occur in this subrange: a=2, b=4. 2^4 and 4^2 both add up to 16, and the x is somewhere around 2.745368024. As far as I know at the moment, this can only be solved with arbitrary precision using the "try-and-approximate" method if x is greater than e - of course, this can also be done using an algorithm if you want to program one.
Here is a graph in which x, y, a and b can be seen up to y=18.5:
The upward white line stands for the x, which is necessary in each case, so that x^x results in exactly y, which is marked by the white line from left to right. As soon as x is larger than the number e (2.718...) or as soon as y is larger than e^e (approx. 15.154), the two variables a and b appear, a is larger than x and converges to y while b is smaller than x and converges to 1. For these numbers, a^b and b^a both give the same result, namely y, i.e. exactly the same result as x^x. For each y there is only one single combination of a and b, all other numbers would perhaps yield the correct result y for a^b, but not for b^a - or vice versa. The value of a is marked in the graph by the uppermost point of the blue area - the lowest point of the red area represents b.
I suspect that a and b also exist if x is smaller than e - only that they then do not belong to the real numbers, but to the complex numbers. If anyone feels like it, feel free to try proving it. :)
X0 > 0
X1 = X0√y
(The X0-th root of y)
X2 = X1√y
X3 = X2√y
...
X1 = X0√y
(The X0-th root of y)
X2 = X1√y
X3 = X2√y
...
This formula gives arbitrarily exact results, unless the x-value is larger than the irrational number e, which starts with 2.718281828. As soon as the x, which you don't know before, is greater than e, you get not only one result, but actually two - funnily enough, the calculator then always switches back and forth between the two results. You can see if this will be the case, if the number y, which you know before, is larger than about 15.15, because e^e = 15.15426..., which is so to say the limit for this approximation to x. Now, as already mentioned, we get not only one result, but two - the larger y is, whose x we want to calculate, the more the one value approaches 1 and the other value approaches y. Let's label these two new numbers a and b, in order to keep track of them. Now these two numbers have interesting properties, because a^b gives exactly the same result as b^a, i.e. exactly y. So mathematically we can say:
x^x = y = a^b = b^a für x>e
x = a = b für x<e
x = a = b für x<e
Now some more die-hard people will say: hold on, what if x is EXACTLY e? The problem in this case is: this is not possible at all, because e has infinitely many decimal places, it is irrational. In finite time one can only calculate finitely many decimal places of e, so that one can only speak of more or less than e, but never EXACTLY of e - one of the confusing properties of irrational numbers. The fact that we call the number "e" does not mean that we know it completely - we can only say about a completely known number that it is greater than, less than or equal to another number. There is always some uncertainty about a number that is not completely known - with an infinite number of decimal places, this uncertainty is infinite, so mathematically there can be no number that is exactly as large as what we call Euler's number. Eventually, the case x=e will be derivable in some completely different way - at present, however, it seems impossible to determine whether the case x=e belongs to the category x>e as well, or whether something completely unexpected is the case for this special case with infinitely many decimal places. In any case e represents the limit above which a and b exist.
An example of a and b are the only integers that occur in this subrange: a=2, b=4. 2^4 and 4^2 both add up to 16, and the x is somewhere around 2.745368024. As far as I know at the moment, this can only be solved with arbitrary precision using the "try-and-approximate" method if x is greater than e - of course, this can also be done using an algorithm if you want to program one.
Here is a graph in which x, y, a and b can be seen up to y=18.5:
The upward white line stands for the x, which is necessary in each case, so that x^x results in exactly y, which is marked by the white line from left to right. As soon as x is larger than the number e (2.718...) or as soon as y is larger than e^e (approx. 15.154), the two variables a and b appear, a is larger than x and converges to y while b is smaller than x and converges to 1. For these numbers, a^b and b^a both give the same result, namely y, i.e. exactly the same result as x^x. For each y there is only one single combination of a and b, all other numbers would perhaps yield the correct result y for a^b, but not for b^a - or vice versa. The value of a is marked in the graph by the uppermost point of the blue area - the lowest point of the red area represents b.
I suspect that a and b also exist if x is smaller than e - only that they then do not belong to the real numbers, but to the complex numbers. If anyone feels like it, feel free to try proving it. :)