Guys, like I implied before, you should stop feeding the troll. He will only get more fat and happy. His happiness is directly proportional to how many posts he can keep a crank thread going. :rolleyes:

[QUOTE=gd_barnes;261174]He will only get more fat and happy.[/QUOTE]

does he have artery damage if so we can cause a blockage by finding it.

I say that 1^n = 1^(0/0) = 1 .

"xilman" says:
[QUOTE]
Nope. 1^(0/0) is undefined.
[/QUOTE]

"Condor" says:
[QUOTE]
...it is [I]not[/I] true that "1^(0/0)=1."
[/QUOTE]

and the rest of you apparently [I][B]agree[/B][/I]
with "Condor" and "xilman". :missingteeth:

Don.

[QUOTE=Don Blazys;261214]and the rest of you apparently [I][B]agree[/B][/I]
with "Condor" and "xilman". :missingteeth: [/QUOTE]
Yeah, so? Which is more parsimonious: that you are wrong? Or that the rest of the world is wrong?

Because we know the continuous extension of [TEX]\({e^X}\)^{\({\frac{e^X-1}{X^2}}\)}[/TEX] is [TEX]e[/TEX] as [TEX]X\to 0[/TEX]:
[code]X e^X e^X-1 X^2 (e^x)^((e^X-1)/X^2)
1.0000 2.7183 1.7183 1.0000 5.5749
0.3333 1.3956 0.3956 0.1111 3.2767
0.1000 1.1052 0.1052 0.0100 2.8625
0.0333 1.0339 0.0339 0.0011 2.7645
0.0100 1.0101 0.0101 0.0001 2.7320
0.0033 1.0033 0.0033 0.0000 2.7228
0.0010 1.0010 0.0010 0.0000 2.7196
0.0003 1.0003 0.0003 0.0000 2.7187
0.0001 1.0001 0.0001 0.0000 2.7184
etc.[/code]
Which proves what I said before, and that Don is wrong. The indeterminate form 1^(0/0) could be any value, not just 1. For any particular instance of it, you need to take the limits to determine what value you can use.

[QUOTE=Condor;261219]Which proves what I said before, and that Don is wrong. The indeterminate form 1^(0/0) could be any value, not just 1. For any particular instance of it, you need to take the limits to determine what value you can use.[/QUOTE]

Wrong. This has NOTHING to do with limits. 1^(0/0) is undefined. It
is not 'any value'. It is undefined because (0/0) is [b]NOT A NUMBER[/b].
Asking about the value of 1^(0/0) makes as much sense (ZERO) as asking
about 1^blue or 1^envy or 1^sodomy. (0/0) is not a number and you
can't raise 1 to the power of something that isn't a number.

[QUOTE=akruppa;261057]Fourth powers are 0,1 (mod 5). If d^4 == 1 (mod 5), the LHS must consist of two summands that are 0 (mod 5) and one that is 1 (mod 5). If d^4 == 0 (mod 5), the only way that the LHS sums to 0 (mod 5) is by all three summands being 0 (mod 5). So for w=x=y=z=4, I think there will always be a pair of bases with common factor 5. With any exponent >4, there might be coprime solutions.[/QUOTE]

Yes. of course. Quite simple.

[QUOTE=akruppa;261065]Some of us kinda enjoy this. Admittedly it's on the same level as laughing at the handicapped kid during recess, but I'm not choosy about entertainment.[/QUOTE]

Allow me to quote/paraphrase Underwood Dudley:

"When asked what kind of advice he would give other researchers,
Mr. Dudley does not equivocate. 'You want to discourage the cranks as much as possible,' he says. 'Try to get them to quit, throw their stuff away. Tell
them there's a federal law against this stuff, and if they don't stop, they'll
be thrown in jail. Anything to get them to quit.'"

[QUOTE=R.D. Silverman;261221]Wrong. This has NOTHING to do with limits. 1^(0/0) is undefined. It
is not 'any value'. It is undefined because (0/0) is [b]NOT A NUMBER[/b].
Asking about the value of 1^(0/0) makes as much sense (ZERO) as asking
about 1^blue or 1^envy or 1^sodomy. (0/0) is not a number and you
can't raise 1 to the power of something that isn't a number.[/QUOTE]
Sorry, I keep trying to put it in language Don can understand in his warped version of reality. That way I can show him that his conclusion is invalid even when the assumptions and conditions he ignores are satisfied.

What I meant was:[list][*]Say you have an function of the form F(x)=f(x)^[g(x)/h(x)].[*]And that for some constant x1, f(x1)=1 and g(x1)=h(x1)=0.[*]Then F(x1) is indeed undefined, because it is 1^(0/0).[*]But, if you can assume F(x) is continuous at x=x1, and f(x), g(x), and h(x) satisfy some other critical conditions as x approaches x1, then you can define a value for F(x1) that is contingent on the assumption of continuity.[/list]Don is saying (without reference to any of them) that these conditions are always met, and that this value is always 1. As you point out, they aren't always met. I'm saying that when they are, the value you define - which is contingent on the assumption of continuity - isn't always 1. And I gave an example that proves it.

You might recognize that my example is equivalent to [tex]e^{\frac{e^X-1}{X}[/tex]. The exponent is a well-known indeterminate form that is 1 [I]under the assumption of continuity[/I] at x=0.

And my ultimate point is that if you start with a function that is well-defined and continuous at x=x1, and artifically introduce a division by zero into a part of it through manipulation, that you can make the assumption of continuity. The conditions required for the various parts should be met, and you can safely conclude that there is no need to "disallow" the offending value x=x1 [i]from the original function[/i]. It is not a part of the range of the modified function, but the original still works fine.

[QUOTE=Don Blazys;261214]I say that 1^n = 1^(0/0) = 1 .[/QUOTE]

You're quite wrong, as others have pointed out. If I had written this in my grade school algebra class I would have lost points.

It is clear to me that Don is saying that since n/0 is undefined that 1^(n/0) is undefined, yet since 0/0 is indeterminate, i.e. not undefined, that 1^(0/0) is determinate, i.e. 1^(0/0) = 1.

IIUC, 0/0 is both indeterminate and undefined, whereas n/0 for n != 0 is undefined but not indeterminate. At least this is what I understand based upon [url]http://en.wikipedia.org/wiki/Indeterminate_form[/url] and [url]http://en.wikipedia.org/wiki/Defined_and_undefined[/url].

My understanding is that since 0/0 is undefined, then 1^(0/0) is also undefined.

If someone wants to correct me, feel free to do so.