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+--- Thread: After 6 years I still don't understand this (/showthread.php?tid=39280)
After 6 years I still don't understand this - JoeTang - 2011-03-16
2147483647 Wrote:You stated a few yourself. I'll even add in a bonus. Here is the plot of f(x,y) = x^y
The point (0,0) is clearly on 1.
![[Image: msp526919egh73abi38f26g.gif]](http://img196.imageshack.us/img196/4999/msp526919egh73abi38f26g.gif)
Nothing about that image is clear. The fact that as x approaches 0 and y approaches negative values obscures the point (0,0) alone makes the image unclear.
After 6 years I still don't understand this - ChaosCorpse - 2011-03-16
Jumping into similar content, I've been curious as to why !0=1 myself. Care to enlighten, oh math majors?
After 6 years I still don't understand this - Locked - 2011-03-16
LongShotJohn Wrote:Jumping into similar content, I've been curious as to why !0=1 myself. Care to enlighten, oh math majors?
n!=n(n-1)!
n=1
1! = 1(0)!
1!=0!
0! = 1.
After 6 years I still don't understand this - thinbear - 2011-03-16
LongShotJohn Wrote:Jumping into similar content, I've been curious as to why !0=1 myself. Care to enlighten, oh math majors?
from wiki: 0! is a special case that is explicitly defined to be 1.[1]
ie. rules are rules
After 6 years I still don't understand this - Noah - 2011-03-16
2147483647 Wrote:You stated a few yourself.
As far as I know, I haven't. :O
2147483647 Wrote:I'll even add in a bonus. Here is the plot of f(x,y) = x^y
The point (0,0) is clearly on 1.
From that graph, we can only see that its limit goes towards 1. However, we cannot see or evaluate its actual value by the graph itself.
A plot of the function 0^(x^2) would say its limit goes towards 0. Same rule applies here.
Noah
After 6 years I still don't understand this - shouri - 2011-03-16
2147483647 Wrote:Can you read? I'm saying that your statement (bolded) would be true if there had existed valid proofs supporting both sides of the argument. In this case, there is only valid proof supporting one side in the 0^0 problem. It's not the same mathematically, logically, or any other way.
Unless you're saying that mathematics is actually religion, which would be absolutely moronic.
My apologies, I apparently haven't learned my lesson either.
After 6 years I still don't understand this - 2147483647 - 2011-03-16
Noah Wrote:As far as I know, I haven't. :O
From that graph, we can only see that its limit goes towards 1. However, we cannot see or evaluate its actual value by the graph itself.
A plot of the function 0^(x^2) would say its limit goes towards 0. Same rule applies here.
Noah
![[Image: capture2to.jpg]](http://img198.imageshack.us/img198/5596/capture2to.jpg)
After 6 years I still don't understand this - JoeTang - 2011-03-16
2147483647 Wrote:
http://www.wolframalpha.com/input/?i=y+%3D+x%2Fx
x/x at x = 0 must equal 1 then.
After 6 years I still don't understand this - 2147483647 - 2011-03-16
JoeTang Wrote:http://www.wolframalpha.com/input/?i=y+%3D+x%2Fx
x/x at x = 0 must equal 1 then.
0/0 is an undeterminate form. You're comparing apples to oranges.
After 6 years I still don't understand this - JoeTang - 2011-03-16
2147483647 Wrote:0/0 is an undeterminate form. You're comparing apples to oranges.
0^0 is an indeterminate form. You're comparing apples to apples.
After 6 years I still don't understand this - hadriel - 2011-03-17
A limit exists doesn't mean a value at that point exists. I note that some of what is written here is true because x^x = 1 at x = 0, but that doesn't establish a proof that x^x = 0. The most accurate one to me is still the x^(2-2) proof, although strictly speaking cancelling variables isn't the most correct thing to do. A change of variables would stamp that.
Stop insulting each other - you all sound like little kids quarrelling. Can't we have a civilised argument?
Hadriel