Originally posted by loop quantum gravity
so what other solutions are there?
For those of you who don't care to read below, I put these approximate solutions first (generated by a program I built using a method described below). Because the program I built stores the numbers in a relatively small memory space the answers aren't as good as they could be.
(1.5, 7.40876)
(1.13, 32.0162)
(2.5, 2.97029)
(12.1, 1.30982)
(8.4343, 1.43866)
I neglected to mention any specific numbers earlier because I didn't know how to exactly solve ln(a)/a = ln(b)/b for either a or b. Remember, all we have to do to find a solution to a
b = b
a where a != b is find a solution to ln(a)/a = ln(b)/b where a != b because a solution to one of these equations is a solution to other as well. The latter is easier to solve numerically, so we'll go with that (the following requires basic calculus).
assume a > 0 is given (we want to find b):
ln(a)/a = ln(b)/b
b ln(a)/a = ln(b)
b ln(a)/a - ln(b) = 0
Now from here we could simply go on the solve the equation numerically for b by applying Newton's Method. Unfortunantly, not all given values of a will produce a solution. In addition, Newton's Method requires an initial guess to get the ball rolling and if we guess inappropriately we might just end up with b = a (which we already know would solve the equation but we ruled out by a != b because it was too trivial).
In order to get a better idea of what's going on we will define a function f(x) which will equal the left-hand side of the equation that is zeroed above and analyze when this function has roots, and if it does, where they exist because these roots are the solutions we are looking for.
b ln(a)/a - ln(b) = 0; a,b > 0 (from above)
f(x) = x ln(a)/a - ln x, x > 0
To get an idea of what f(x) looks like, we plug in some values and analyze the derivative:
f'(x) = ln(a)/a - 1/x, x > 0
f'(x) = ( x ln a - a )/( ax ) (possible extrema at x = a/ln a, if a > 1, otherwise a/ln a would be negative and outside the domain of f(x))
f'(x < a/ln a) < 0
f'(x > a/ln a) > 0 (last two confirm x = a/ln a is an absolute minumum if a > 1)
Limit[x->0, f(x)] = +inf
Limit[x->+inf, f(x)] = +inf if a > 1
Limit[x->+inf, f(x)] = -inf if a < 1
These limits inform us that if an absolute minumum exists (a > 1) and is also negative, then f(x) will cut the x-axis twice and therefore have two roots. One of the roots would just be x=a as we mentioned earlier. The other would be the one we're looking for. On the other hand, the absolute minumum being zero would imply only one root at the trivial x=a (the abs. min can't be positive because the trivial root always exists). If a < 1 and there is no absolute minumun then the limits inform us only one root exists (which would be the trivial x=a). So let's assume a > 1 and find when the abs. min is negative.
Let us see what the absolute minumun is:
f(a/ln a) = (a/ln a)(ln(a)/a) - ln(a/ln a)
f(a/ln a) = 1 - ln(a/ln a)
When is this absolute minumun less than zero:
1 - ln(a/ln a) < 0
1 < ln(a/ln a)
e < a/ln a (e
x is an increasing function)
The last is true whenever a > 1 and a != e.
Okay, so far we have established if we are given a > 1 and a != e, then f(x) will have two roots and one of these will be the solution for b that we are in search for. To actually find this root we will use Newton's method, but which root should we go after? The one to the left or the right of the minumum? One is a "dud" and the other is the actual solution. We know the dud is at x = a. This will be to the left of the minumum (at x = a/ln a) if ln a < 1 (or 1 < a < e) and to the right of the minumum if ln a > 1 (or 1 < e < a). So if a < e, our initial guess to Newton's method should be to the right of the minumum and vice-versa for a > e.
We have always talked about 'a' being given, but it since the original equation is "symmetrical", we could just as well have been talking as if 'b' were given. To be fair to both sides, well say the given is x. So in summary, given any number x > 1, x != e, we can find another number b for which x
b = b
x is true by applying Newton's method as follows:
if x < e, start with b
0 = 2 x/ln x (right of minumum)
if x > e, start with b
0 = x/(2 ln x) (left of minumum)
b
n+1 = x b
n(ln b
n - 1) / (b
n ln x - x)
