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# Is 0^0 defined or not?

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My HP48GX calculator sais it's 1 Maple sais it's 1. But at school I learned it's undefined, and so sais my TI40x calculator. However, since Maple is made by mathematicians (I hope so ), who like defenitions and should shudder to program something that's mathematically incorrect, why did they say 0^0 = 1? Trying some things myself, I found that lim_x-->0 x^x = 1, as well as lim_x-->0 -x^-x lim_x-->0 -x^x lim_x-->0 x^-x lim_x-->0 x^0 But the only limitthat becomes 0 instead of 1 is: lim_x-->0 0^x So is that the only one limit that sais the opposite of all the other ones, and is that why they say it's undefined, but because all others say it's 1, they defined it as 1 in Maple? Anyway, personally I'd like if 0^0 was 1, it would make life so much easier EDIT: Hmm I just found this: http://home.att.net/~numericana/answer/algebra.htm, look at the second topic, some guy there is claiming that 0^0=1. Is he right, or is he going to be flamed by any mathematician that sees him? [edited by - Boops on January 1, 2004 6:13:26 PM]

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x^a / x^b = x^(a-b)
0^n / 0^n = 0^(n-n) = 0^0
0^n = 0
.: 0^0 = 0/0 = undefined.

HOWEVER

a / a = 1
let a = 0^n
.: 0^0 = 0^n/0^n = a/a = 1.

I dunno, I suspect that the programs in question might be being lazy... checking for ^0 first and automatically returning 1, without checking the 0 that you''re raising.

Richard "Superpig" Fine
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but it doesnt make sense in practice

0 pies raised 0 times = 1 pie?

[edited by - bigbadboo on January 1, 2004 7:02:20 PM]

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0^0 can be defined to be 1. It doesn't violate any other rules or lead to contradiction, so it's a definition such as the axiom a+(b+c) = (a+b)+c. It could've been defined to be 0, or even 42. But 0^0=1 is actually a pretty useful definition that takes some special cases away from certain formulas (e.g. Taylor series). On the other hand, there are some formulas where e.g. 0^0=0 would've been a better definition and such definition would've removed those special cases, but those formulas are in the minority.

[edited by - civguy on January 1, 2004 7:24:53 PM]

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When in practice would you raise zero pies to the zeroth power?

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Thanks for the replies

superpig: your 0^0 = 0/0 proof convinced me that it''s indeed undifined

nice explanation civguy.

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quote:
Original post by Boops
So is that the only one limit that sais the opposite of all the other ones, and is that why they say it''s undefined, but because all others say it''s 1, they defined it as 1 in Maple?
Strictly speaking, and AFAIK, it''s undefined.

Another related example is sin(x)/x. It''s undefined at x=0. But since this function is used so often, and its limit is 1, mathematicians have called it sinc(x), which is sin(x)/x if x!=0 and 1 if x = 0.

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quote:
Original post by Cedric
quote:
Original post by Boops
So is that the only one limit that sais the opposite of all the other ones, and is that why they say it''s undefined, but because all others say it''s 1, they defined it as 1 in Maple?
Strictly speaking, and AFAIK, it''s undefined.

Another related example is sin(x)/x. It''s undefined at x=0. But since this function is used so often, and its limit is 1, mathematicians have called it sinc(x), which is sin(x)/x if x!=0 and 1 if x = 0.

I may be incorrect, but I remember my dad (Ph.D. in Mathematics) telling me that it is 1.

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quote:
telling me that it is 1.

... if x->0

[edited by - DerekSaw on January 1, 2004 8:37:18 PM]

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I believe that it is one of the "indeterminate forms." It isn''t one...

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I believe, IMHO, the "anything" to the zero power is 1 means that the "anything" is any positive number larger than 0. 0^0 itself is just 0 since you are manipulating zero in the first place in multiplitive cases.

That is just comming from someone in Pre-Cal.

Edit:
#include <iostream>#include <stdlib.h>int main(){    float answer=0^0;    std::cout << "0^0="<< answer << std::endl;    system("PAUSE");    return 0;}

Edit, correct way (time corrodes my memory):
#include <iostream>#include <stdlib.h>#include <math.h>int main(){    //Thank you furby100    std::cout << "0^0:" << pow(0,0) << std::endl;    system("PAUSE");    return 0;}

[edited by - Xiachunyi on January 1, 2004 11:05:15 PM]

[edited by - Xiachunyi on January 2, 2004 4:30:50 PM]

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The function x^0 is 1 everywhere except where x=0, where it is undefined. Asking what 0^0 doesn''t make much sense, but the limit as x->0 of x^0 is 1. Perhaps Maple used limits for indeterminite forms.

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By definition, 0^0 is defined to be "undefined".

Kami no Itte ga ore ni zettai naru!

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0^0 = 1 is (apparently) just plain useful in more situations than 0^0 = anything else. Strictly speaking it''s undefined, but in most situations it is assumed to be 1. Just leave it at that. Why is this topic so damn popular?

quote:
but it doesnt make sense in practice

0 pies raised 0 times = 1 pie?

Don''t start that shit again.

The following statement is true. The previous statement is false.
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Sometimes in the Blizzard forums over at "www.battle.net" they start .99~ == 1 threads. I guess every forum must have some disagreement.

Hmmm, maybe "OMG!!!111.99~.99~"?

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Usually, it is 1, according to this, but sometimes it can be 0. So either can be correct in certain circumstances.

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quote:
Original post by Xiachunyi
Edit:
#include <iostream>#include <stdlib.h>int main(){    float answer=0^0;    std::cout << "0^0="<< answer << std::endl;    system("PAUSE");    return 0;}

(uncontrollable laughter)

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http://mathworld.wolfram.com/ExponentLaws.html
quote:
Special cases include
(8) x1 = x
and
(9) x0 = 1
for x != 0 . The definition 00 = 1 is sometimes used to simplify formulas, but it should be kept in mind that this equality is a definition and not a fundamental mathematical truth (Knuth 1992; Knuth 1997, p. 56).

[edited by - extrarius on January 2, 2004 2:07:01 AM]

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"Definition" versus "truth." 1+1 is 2. There's no getting around that given the definition of addition and the definition of 1. That's the truth. Given our definitions, there is no truth about 0^0. So you either ignore it, or you define it.

0^0 is *defined* as being 1 in most cases for the same reason 0! is defined as being 1. It makes things easier. The discovery of zero screwed things up to the point that shortcuts are taken whenever possible. Trying to use logic to come to a conclusion because there is no logic behind laziness.

CM

*edit: PPS: yay Extrarius

[edited by - Conner McCloud on January 2, 2004 2:11:15 AM]

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For the LAST AND FINAL TIME

0^0 is defined as 1 , but the true result of the operation is undefined. so

0^0 = 1

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I have another question, kind of pretaining to this.

What does the the calculator mean when it says this:

"Warning: 0^0 replaced by 1"

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quote:
Original post by tHiSiSbOb
quote:
Original post by Cedric
quote:
Original post by Boops
So is that the only one limit that sais the opposite of all the other ones, and is that why they say it''s undefined, but because all others say it''s 1, they defined it as 1 in Maple?
Strictly speaking, and AFAIK, it''s undefined.

Another related example is sin(x)/x. It''s undefined at x=0. But since this function is used so often, and its limit is 1, mathematicians have called it sinc(x), which is sin(x)/x if x!=0 and 1 if x = 0.
I may be incorrect, but I remember my dad (Ph.D. in Mathematics) telling me that it is 1.
I don''t hold diplomas in very high esteem. Anyway, Mathworld is on my side. Not that it really matters.

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quote:
but it doesnt make sense in practice

0 pies raised 0 times = 1 pie?

[edited by - bigbadboo on January 1, 2004 7:02:20 PM]

Complex numbers don''t make sense in the sense of pies, but I don''t see anyone complaining about those.

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