UPDATE and 7th POSTING:
1) Magee:
Right, it's a little clearer to use the terminology you propose:
x1(k+1) = sqrt(x1(k))
x2(k+1) = x1(k+1) - x2(k)
produces a sequence of points (x1(k), x2(k)), where the x1(k) rapidly converge towards 1-, while the x2(k) gravitates to becoming an alternation between two values. As you suggest, this ultimate behavior depends on your starting point (x1(1), x2(1)):
If you start at x1(1) = 0, x2(1) = a, then x1(k) = 0, and x2(k) = (-1)^n * a; and if a = 0, you get a fixed point in the mapping. The exploration of this behavior looks like the stuff I reference below: They discuss stability points, etc.
But it's not difficult to see the oscillation: Let's consider the case that x(1) is very close to 1:
x(1) = 1 - a ; typically 0 < a < 0.0000001
and
x2(1) = 0.5 - b.
Then:
x1(2) = sqrt(x1(1)) = sqrt(1 - a) = 1 - a/2.
x2(2) = x1(2) - x2(1)
= (1 - a/2) - (0.5 - b)
= 0.5 + (b - a/2)
So if a is very small, x1 jumps from (1 - a) to (1 - a/2), which is barely a change; but x2 jumps from (0.5 - b) to (0.5 + (b - a/2) which is about (0.5 + b); from my test cases, I can see that when a is extremely small (0.0000001) b is still typically 0.03. So x2 is oscillating about 0.5 with amplitude b.
2) jeredwm:
Yes, we have started focusing on the iterative mapping. The relevance to the original question (the limit of F(x) as x => 1-) is when the starting point (x1(1), x2(1)) is set equal to (x1(1), F(x1(1)). For the example shown in Posting 6, I arbitrarily CHOSE:
x(1) = x_1 = 0.8
and I CALCULATED:
x2(1) = F(x_1) = F(0.8) = 0.429183683755702
from my spreadsheet calculation that does the explicit calculation:
F(y) = y - y^2 + y^4 - y^8 + y^16 - ...
for 100 terms.
My original intent in focusing on using the iterative equation was to avoid having to accumulate a lot of numerical noise; but it also makes it a lot faster to generate trajectories, since I don't have to cut & paste stuff.
6th POSTING:
A new computational method: Since
F(x) + F(x^2) = x
by substituting y = sqrt(x), this can also be written
F(sqrt(y)) + F(y) = sqrt(y)
or
F(sqrt(y) = sqrt(y) - F(y)
Then define the sequence:
x_1 = 0.8; F(x_1) = 0.429183683755702
x_2 = sqrt(x_1) ;
F(x_2) = sqrt(x_1) - F(x_1)
= x_2 - F(x_1)
x_3 = sqrt(x_2) ;
F(x_3) = x_3 - F(x_2)
The sequence x_n goes climbs quite quickly up towards 1-, and the F(x_n) are quickly calculated without any more sums or concerns for inaccuracy. Here are the results:
0.80.429183684
0.8944271910.465243507
0.9457416090.480498102
0.9724924720.491994371
0.986150330.494155959
0.9930510210.498895062
0.9965194530.497624391
0.998258210.500633818
0.9991287250.498494907
0.9995642680.501069361
0.999782110.498712749
0.9998910490.5011783
0.9999455230.498767223
0.9999727610.501205538
0.999986380.498780843
0.999993190.501212348
0.9999965950.498784248
0.9999982980.50121405
0.9999991490.498785099
0.9999995740.501214476
0.9999997870.498785312
0.9999998940.501214582
0.9999999470.498785365
0.9999999730.501214609
0.9999999870.498785378
0.9999999930.501214615
0.9999999970.498785381
0.9999999980.501214617
0.9999999990.498785382
10.501214617
10.498785382
10.501214617
10.498785383
10.501214617
10.498785383
10.501214617
It seems that this sequence has fallen into something like a cycle oscillating between 0.4987 and 0.5012, and it doesn't look as though this is going away. Actually, the equations defining the sequence look like those systems of equations people used to study for chaos/bifurcation theory.
I've tried out a couple of other values, and the same thing happens: when the x_n's get very close to 1, the value of F(x_n) essentially oscillates between two values that are about 0.003 from 0.5.
OK, Magee, I have to agree with you: There is NO limit as x => 1-, so F(x) is not continuous at 1.
5th POSTING:
Jeredwm, that is an interesting idea: Just do the sums. I didn't think about it because I don't have a C compiler, but I can do this on a spreadsheet with some futzing about. Here are some results:
x____________Sum(up to n = 47)
0.5000000_____0.3086
0.7000000_____0.39576
0.9000000_____0.46777
0.9900000_____0.494099
0.9990000_____0.50012
0.9999000_____0.502025
0.9999900_____0.4973599
0.9999990_____0.50073949
0.9999999_____0.50185152
There is some oscillation near 1. This could be a precision issue in Excel, but on the other hand, when I do the calculation, I find quite accurately that:
F(x) + F(x^2) = x for x up to 0.99999999. So the spreadsheet seems quite reliable!
So, on the one hand, it is not incompatible with a limiting value of 0.5; but on the other hand, f(x) does not seem to be monotonic. I don't see any way to predict oscillatory behavior from the form of the series; or to rule it out.
4th POSTING:
Magee, I don't share your opinion: I think it is very likely that:
Lim(x => 1-) Lim(n => infinity) G(n, x) exists,
where:
G(n, x) = Sum(k = 1, n)[(-1)^k * x^(2^k) ]
I realize that
Lim(x => 1-) G(2N, x) = 0
and
Lim(x => 1-) G(2N+1, x) = 1
so
Lim(n => infinity) Lim(x => 1-)G(n, x) does not exist. However, that only forces me to conclude that I cannot interchange the order of taking limits in this case. At least we know that Lim(n => infinity) G(n, x) converges and is continuous on (0,1).
A related example:
1/(1+x) = 1 - x + x^2 - x^3 + x^4 - ...
shows the same problem as x => 1-:
- At x = 1, the sum turns into the indefinite:
1 - 1 + 1 - 1 + 1 - ...
- The nth partial sum H(n, x) = (1 - (-x)^n)/(1 + x)
so Lim(x => 1-) H(2N, x) = 0
but
Lim(x => 1-) H(2N+1, x) = 1
nonetheless
Lim(x => 1-) Lim(n => infinity) H(n, x) = 1/2
This is not a proof; but it shows the possibility. I suspect the function
G(x) = Lim(n => infinity) G(n, x) is better behaved than I know how to prove.
3rd POSTING:
We see from the initial equation that:
0 = f(0) + f(0^2) = f(0) + f(0) = 2f(0)
=> f(0) = 0
1 = f(1) + f(1^2) = f(1) + f(1) = 2f(1)
=> f(1) = 1/2
The formal solution is:
F(x) = x - x^2 + x^4 - x^8 + x^16 - x^32 + ...
which has been shown to be continuous on (-1, 1); we see immediately that F(0) = 0. However, F(1) cannot be evaluated from the series, because it becomes the meaningless:
1 - 1 + 1 - 1 + 1 - 1 + ...
We can just define F(1) = 1/2, and then it would satisfy the problem. But the remaining question is: Is F(x) continuous as thus defined? In other words, is
Lim(x=>1-) F(x) = 1/2 ?
If the limit exists, then it can have no other value, because it satisfies the equation on (0,1). But does the limit actually exist?
Since on [0,1):
1 > x = F(x) + F(x^2) > max(F(x), F(x^2)) > 0
Therefore, 1 > F(y) ≥ 0 on [0,1).
Hmm, I am very close to concluding that this limit exists (and must therefore be 1/2), but not quite there...
2nd POSTING:
A series that formally satisfies the equation is:
f(x) = x - x^2 + x^4 - x^8 + x^16 - ...
= Sum(n=0)[(-1)^n * x^(2^n) ]
By comparison to the geometric series for 1/(1-x ) which is:
1 + x + x^2 + x^3 + ...
this must also converge absolutely for x in (-1, 1), which includes [0,1] except for the point x = 1, where the series is divergent: 1 - 1 + 1 - 1 + ...
However, I do not see any problem with simply assigning the value of 1/2 to f(1), to satisfy our earlier requirement. f will then be continuous; however, I am no longer 100% sure that f is monotonically increasing in x. It probably is.
1st POSTING:
I can't construct a function that satisfies this equation, but I don't see any reason why such a function should not be continuous:
Assume f(x) is an increasing function of x; then f(x^2) is also an increasing function. Then f(x) + f(x^2) must also be increasing, but since this is just x, no problem. So f could be increasing on [0,1].
To meet the values at x = 0 and 1:
Set f(0) = 0
Set f(1) = 1/2
Heck, let's even assume f is differentiable:
f'(x) + 2x*f'(x^2) = 1
We would then conclude
f'(0) = 1
f'(1) = 1/3
I don't see any obvious problem. Why shouldn't f exist and be monotonically increasing, continuous and even differentiable on [0,1] ?