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videogames101

Diamond Member
Aug 24, 2005
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I remember this topic being discussed awhile back around here, but i neither remember the outcome nor the arguments. To add to that, i was arguing the point on whether it 1 or 0 with some engineers, and we can't agree, at all. Can someone explain this again? I was under the impression that it did equal 1, but that may be depending on limits... Whatever, can someone please give me the correct answer?
 
H

hellokeith

Golden Member
Nov 12, 2004
1,664
0
0
(X^n) / (X^n-1) = X

{n=2}, (X^2)/(X^1)= (X^2)/X = X
{n=1}, (X^1)/(X^0) = X/1 = X
{n=0}, (X^0)/(X^-1) = 1/(1/x) = X
{n=-1}, (X^-1)/(X^-2) = (1/x)/(1/x^2) = (X^2)/(X^1) = X

etc. etc.
 
C

CycloWizard

Lifer
Sep 10, 2001
12,348
1
81
Originally posted by: CSMR
0^0 is defined to be 1 in the context of polynomials.
Click to expand...
If the polynomial is taken as a Maclaurin series and includes x=0. Of course, since 0^0 is "undefined," it can only have a value if we define one (as shown in the link I posted above). One can just as easily change the form of the polynomial to avoid this definition from
\sum_{i=0}^n{c_i x^i}
to
c_0+\sum_{i=1}^n{c_i x^i}.
 
S

shodan21

Junior Member
Jan 20, 2008
6
0
0
Originally posted by: BrownTown
Originally posted by: hellokeith
Originally posted by: Nathelion
If I read you correctly, you're dividing by 0.
Click to expand...

Incorrect.
Click to expand...

Well no, you are definitely dividing by 0 there...
Click to expand...

Actually he isn't. Yes, there is a 1/x in there when x = 0, but he has 1/(1/x), which simplifies to just x without dividing anything. It looks like he's dividing by 0 at first glance, but he's actually just simplifying an expression.

However, I don't see how his explanation answers the question, since with his statement:

{n=0}, (X^0)/(X^-1) = 1/(1/x) = X

he's assuming that 0^0 follows the law of anything powered to the 0, which would be 1. The statement still complies with the X^0/X^-1 = X theorem even if 0^0 = 0. If 0^0 equaling 0 caused the statement to not equal X, there'd be a valid argument there, but that's not the case.
 
H

hellokeith

Golden Member
Nov 12, 2004
1,664
0
0
Originally posted by: shodan21
However, I don't see how his explanation answers the question, since with his statement:

{n=0}, (X^0)/(X^-1) = 1/(1/x) = X

he's assuming that 0^0 follows the law of anything powered to the 0, which would be 1. The statement still complies with the X^0/X^-1 = X theorem even if 0^0 = 0. If 0^0 equaling 0 caused the statement to not equal X, there'd be a valid argument there, but that's not the case.
Click to expand...

(X^n)/(X^n-1) = X

or written another way

(X^n+1)/(X^n) = X

are used in proofs that X^0 = 1, not the other way around (i.e. someone didn't just arbitrarily say X^0 = 1 because they felt like it)

Watch..

(X^n+1) / (X^n) = X

now give values X = 3, n = 0, and C = X^n

(3^0+1) / C = 3
(3^1) / C = 3
3 / C = 3
C = 1
so C = X^n = 3^0 = 1

And this works for all values of X, even X = 0.
 
M

Mday

Lifer
Oct 14, 1999
18,647
1
81
Originally posted by: hellokeith
Originally posted by: shodan21
However, I don't see how his explanation answers the question, since with his statement:

{n=0}, (X^0)/(X^-1) = 1/(1/x) = X

he's assuming that 0^0 follows the law of anything powered to the 0, which would be 1. The statement still complies with the X^0/X^-1 = X theorem even if 0^0 = 0. If 0^0 equaling 0 caused the statement to not equal X, there'd be a valid argument there, but that's not the case.
Click to expand...

(X^n)/(X^n-1) = X

or written another way

(X^n+1)/(X^n) = X

are used in proofs that X^0 = 1, not the other way around (i.e. someone didn't just arbitrarily say X^0 = 1 because they felt like it)

Watch..

(X^n+1) / (X^n) = X

now give values X = 3, n = 0, and C = X^n

(3^0+1) / C = 3
(3^1) / C = 3
3 / C = 3
C = 1
so C = X^n = 3^0 = 1

And this works for all values of X, even X = 0.
Click to expand...

Proofs for X^0=1 all caveat for X != 0.
For n > 0, the statement "(X^n+1) / (X^n) = X" is undefined for X = 0 since you are dividing by 0.
 
C

CSMR

Golden Member
Apr 24, 2004
1,376
2
81
Originally posted by: CycloWizard
Originally posted by: CSMR
0^0 is defined to be 1 in the context of polynomials.
Click to expand...
If the polynomial is taken as a Maclaurin series and includes x=0. Of course, since 0^0 is "undefined," it can only have a value if we define one (as shown in the link I posted above). One can just as easily change the form of the polynomial to avoid this definition from
\sum_{i=0}^n{c_i x^i}
to
c_0+\sum_{i=1}^n{c_i x^i}.
Click to expand...
Everything is undefined before you define it.
The change in form is not beneficial because each term in the series has a single form - c_i multiplied by x i times - and separating out one term leaves it up to the reader to work out this form. It is also more cumbersome and suggests that you are going to do something special with c_0.
As long as you only remove a finite number of these definitions it is a matter of style, but a definite matter of style nevertheless.
 
C

CycloWizard

Lifer
Sep 10, 2001
12,348
1
81
Originally posted by: CSMR
Everything is undefined before you define it.
The change in form is not beneficial because each term in the series has a single form - c_i multiplied by x i times - and separating out one term leaves it up to the reader to work out this form. It is also more cumbersome and suggests that you are going to do something special with c_0.
As long as you only remove a finite number of these definitions it is a matter of style, but a definite matter of style nevertheless.
Click to expand...
The forms are not the same. In the first, if the domain of interest includes x=0, then one must define x^0==1 for all x. In the second, this definition is unnecessary since no x^0 term is included. Thus, while you can call it whatever you want, you're simply framing the question in terms that necessitates your previous statement always being correct. I simply showed an alternative way to formulate a power series that does not require the definition that you supplied.
 
N

Nathelion

Senior member
Jan 30, 2006
697
1
0
Originally posted by: CSMR
Originally posted by: CycloWizard
Originally posted by: CSMR
0^0 is defined to be 1 in the context of polynomials.
Click to expand...
If the polynomial is taken as a Maclaurin series and includes x=0. Of course, since 0^0 is "undefined," it can only have a value if we define one (as shown in the link I posted above). One can just as easily change the form of the polynomial to avoid this definition from
\sum_{i=0}^n{c_i x^i}
to
c_0+\sum_{i=1}^n{c_i x^i}.
Click to expand...
Everything is undefined before you define it.
The change in form is not beneficial because each term in the series has a single form - c_i multiplied by x i times - and separating out one term leaves it up to the reader to work out this form. It is also more cumbersome and suggests that you are going to do something special with c_0.
As long as you only remove a finite number of these definitions it is a matter of style, but a definite matter of style nevertheless.
Click to expand...

Well thank you captain obvious.
Defining 0^0 other than in very special contexts would be very detrimental to math as we know it. To take an example that was brought up earlier:

1/(1/x) = ? when x = 0

Consider that using middle school algebra, I can rewrite this as x/1 = 0. Obviously.
But using that same set of rules I can also write 1/(1/x)=1/(1/0) OMGZORZ I'M DIVIDING BY 0!!! This result is undefined. We just basically "broke" algebra. Since we don't want to change all of algebra to avoid this problem, we just say it's undefined and leave it at that.
 
C

CSMR

Golden Member
Apr 24, 2004
1,376
2
81
Originally posted by: CycloWizard
Thus, while you can call it whatever you want, you're simply framing the question in terms that necessitates your previous statement always being correct. I simply showed an alternative way to formulate a power series that does not require the definition that you supplied.
Click to expand...
The limited definition (x^i over the integers i>0) is no more simple than the definition of x^i over the integers i>=0 and using the first rather than the second adds a little bit of complexity to the formula. I wasn't saying that your definition doesn't work just that it is not the best.
 
C

CSMR

Golden Member
Apr 24, 2004
1,376
2
81
Originally posted by: Nathelion
Defining 0^0 other than in very special contexts would be very detrimental to math as we know it.
Click to expand...
How so?
To take an example that was brought up earlier:
Click to expand...
With what connection to 0^0?
1/(1/x) = ? when x = 0
Consider that using middle school algebra, I can rewrite this as x/1 = 0. Obviously.[/quote]
Then your algebra book is wrong. Your book should have said that zero has no multiplicative inverse.
 
C

CycloWizard

Lifer
Sep 10, 2001
12,348
1
81
Originally posted by: CSMR
The limited definition (x^i over the integers i>0) is no more simple than the definition of x^i over the integers i>=0 and using the first rather than the second adds a little bit of complexity to the formula. I wasn't saying that your definition doesn't work just that it is not the best.
Click to expand...
What's best depends on the situation. In any case, it doesn't matter. :p
 
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