11/18/2007, 10:26 AM
I have struck on the following:
We can define 0^0 =0 or 0^0 = 1 = depending on the scale of 0 we use in first 0 and exponent ( based on various orders of infinitesimals ). if we denote 0 as dx, dy, than:
1) dx^dy = 0 if dx is infinitely smaller than dy, but dx=0, dy=0 dx+dy=0, dx/dy = 0 or
2) dx^dy=1 if dx is infinitely greater than dy, but dx=0, dy=0, dx+dy=0, dx/dy=infinity.
3) The only problem arises when dx=dy=0 and dx+dy=0, dx/dy=finite.
Then the problem 0^0 is solved axiomatically to fit other math.
but let us leave this case open.
if we look at h(0)= 0^0^0^0^............. infinite times:
and take partial exponents:
if 0^0 = 1, 0^(0^0)=0, 0^(0^(0^0))=0^(0^1)=0^0=1,
so we get oscillating values 1,0,1,0,1,0,1,0........
if 0^0=0, 0^(0^0)=0^0 = 0, 0^(0^(0^0))=0^(0^0) =0^0 =0
We get value 0 all the time.
4) So it might be correct to suppose that value of 0^0 when 0/0=finite value lies somewhere between these 2 cases.
But how to evaluate h(0) when 0^0=1?
5) If we look at sum 1-1+1-1+1-1+1..............we see that partial sums develop like 0,1,0,1 .....similar to case h(0) when 0^0=1.
We know that Euler summation gives values to those series =1/2. We might expect that something similar can work in case h(0) when 0^0=1.
6) but there is a difference:
we can look at 1-1+1-1+1 ....... as sum of
1+1+1+1+1.........= + infinity
and -1-1-1-1-1- = - infinity
So value 1/2 is actually attributed to sum -infinity+infinity= 1/2.
7) In case with tetration, we can look at :
0^1^0^1^0^1.............. = 0, 0^1=0
0^-1^-1^-1^-1.........= infinity , 0^-1=infinity
0^(-1)^0^-1^0^-1^0.........= either infinity or 0 or 1 depending on 0/0 and 0^0.
0^-1 = infinity.
minus infinity is never as result in exponentation of 0, so we are looking at value that would describe multiplication
0^(-1) ^-1^-1........* 0^(1)^1^1....... = 0^(0) which is something like 0*infinity so could be finite.
the conjecture is, that there could exist a value atributable to the h(0) when 0^0=1 as a value of 0,1,0,1,0,1.............and that this value also gives the correct answer to 0^0 when 0/0 = 1.
h(0) when 0^0=0 is 0.
9) This value is not 1/2. What could it be? pi/2? pi/4? 1/e? i/2? sqrt(i)? something else?
The definition of h(z) by - W(-ln(0)/ln(0) does not seem very helpful here, but perhaps to ln(0) can be formally attributed some value as well?
We can define 0^0 =0 or 0^0 = 1 = depending on the scale of 0 we use in first 0 and exponent ( based on various orders of infinitesimals ). if we denote 0 as dx, dy, than:
1) dx^dy = 0 if dx is infinitely smaller than dy, but dx=0, dy=0 dx+dy=0, dx/dy = 0 or
2) dx^dy=1 if dx is infinitely greater than dy, but dx=0, dy=0, dx+dy=0, dx/dy=infinity.
3) The only problem arises when dx=dy=0 and dx+dy=0, dx/dy=finite.
Then the problem 0^0 is solved axiomatically to fit other math.
but let us leave this case open.
if we look at h(0)= 0^0^0^0^............. infinite times:
and take partial exponents:
if 0^0 = 1, 0^(0^0)=0, 0^(0^(0^0))=0^(0^1)=0^0=1,
so we get oscillating values 1,0,1,0,1,0,1,0........
if 0^0=0, 0^(0^0)=0^0 = 0, 0^(0^(0^0))=0^(0^0) =0^0 =0
We get value 0 all the time.
4) So it might be correct to suppose that value of 0^0 when 0/0=finite value lies somewhere between these 2 cases.
But how to evaluate h(0) when 0^0=1?
5) If we look at sum 1-1+1-1+1-1+1..............we see that partial sums develop like 0,1,0,1 .....similar to case h(0) when 0^0=1.
We know that Euler summation gives values to those series =1/2. We might expect that something similar can work in case h(0) when 0^0=1.
6) but there is a difference:
we can look at 1-1+1-1+1 ....... as sum of
1+1+1+1+1.........= + infinity
and -1-1-1-1-1- = - infinity
So value 1/2 is actually attributed to sum -infinity+infinity= 1/2.
7) In case with tetration, we can look at :
0^1^0^1^0^1.............. = 0, 0^1=0
0^-1^-1^-1^-1.........= infinity , 0^-1=infinity
0^(-1)^0^-1^0^-1^0.........= either infinity or 0 or 1 depending on 0/0 and 0^0.
0^-1 = infinity.
minus infinity is never as result in exponentation of 0, so we are looking at value that would describe multiplication
0^(-1) ^-1^-1........* 0^(1)^1^1....... = 0^(0) which is something like 0*infinity so could be finite.
the conjecture is, that there could exist a value atributable to the h(0) when 0^0=1 as a value of 0,1,0,1,0,1.............and that this value also gives the correct answer to 0^0 when 0/0 = 1.h(0) when 0^0=0 is 0.
9) This value is not 1/2. What could it be? pi/2? pi/4? 1/e? i/2? sqrt(i)? something else?
The definition of h(z) by - W(-ln(0)/ln(0) does not seem very helpful here, but perhaps to ln(0) can be formally attributed some value as well?