Textkit Logo

Calling All Math Folk

Philosophers and rhetoricians, Welcome!

Calling All Math Folk

Postby Geoff » Mon Sep 13, 2004 11:37 pm

I know there are some folks here who read math Theory (if not just for the Greek).

So help me out...

Why is X raised to the 0 power = 1 and not 0 ?
User avatar
Geoff
Textkit Fan
 
Posts: 345
Joined: Fri Aug 22, 2003 2:30 pm

Re: Calling All Math Folk

Postby Democritus » Tue Sep 14, 2004 3:20 am

Geoff wrote:Why is X raised to the 0 power = 1 and not 0 ?



Well, you know that fractional exponents are roots. So x^(1/2) (that is, x raised to the power of one half) is the square root of x, and x^(1/3) is the cubed root. So:

16^(1/2) = 4
16^(1/3) = 2.519842
16^(1/4) = 2
16^(1/5) = 1.741101
16^(1/6) = 1.587401
16^(1/10) = 1.319508
16^(1/15) = 1.203025
16^(1/25) = 1.117287

As the exponent gets smaller and smaller (closer to 0), the value of the expression approaches 1.

This makes intuitive sense. If the number y multipled by itself 25 times equals 16:

16^(1/25) = y --> y^25 = 16

then it is clear that y must be at least a little larger than one. The same is also true for all numbers z larger than 25.

16^(1/z) = y --> y^z = 16

As z increases into infinity, the exponent 1/z approaches 0, and the root y approaches 1.

So it makes sense to define 16^0 = 1.



You also know that negative exponents are the reciprocals, in other words, x^(-y) = 1/(x^y)

16^(-1) = 1/16 = 0.0625
16^(-2) = 1/16^2 = 0.003906
16^(-3) = 1/16^3 = 0.000244
16^(-4) = 1/16^4 = 0.000015

So as the exponent becomes more negative, the value of x^y approaches zero.

But as the exponent approaches zero from the negative direction, the value of x^y approaches one:

16^(-1/2) = 0.25
16^(-1/3) = 0.396850
16^(-1/4) = 0.5
16^(-1/5) = 0.574349
16^(-1/6) = 0.629961
16^(-1/10) = 0.757858
16^(-1/15) = 0.831238
16^(-1/25) = 0.895025

And that's another good reason to define 16^0 = 1.
Democritus
Textkit Fan
 
Posts: 331
Joined: Fri May 07, 2004 12:14 am
Location: California

Postby Titus Marius Crispus » Tue Sep 14, 2004 3:26 am

x^(y+z) = x^y * x^z

x^(y+0) = x^y = x^y * x^0

x^0 = 1
Titus Marius Crispus
Textkit Neophyte
 
Posts: 78
Joined: Wed Jul 14, 2004 2:00 am
Location: San Antonio, TX, USA

Postby Dacicus » Fri Sep 17, 2004 5:59 pm

Take the number x^y, with y an integer greater than 0. Every time you divide x^y by x, you decrease the value of the exponent by 1 because (a^b)/(a^c) = a^(b-c). Eventually, you'll reach x^1, which is x. If you divide this by x, you get 1, which is x^0.
Dacicus
Textkit Neophyte
 
Posts: 81
Joined: Fri Apr 09, 2004 1:13 am
Location: Los Angeles, California

Postby Emma_85 » Sat Sep 25, 2004 12:03 pm

can I call on you all too? what on earth do you do with 0^0? I mean 1 to the power of zero is certainly 1, but zero to the power of zero? My calculator tells me: domain error. lol, I need to know though, is it 1 too? I mean I could say that zero is like just getting very very close to 0, but I'm not too sure. I have a function and I have to know if it equals zero or not and it contains 0^0, and if that is 1 then the function in some cases makes more sense in others in makes not sense at all if it does equal 1. I'm confused :( .
phpbb
User avatar
Emma_85
Global Moderator
Global Moderator
 
Posts: 1564
Joined: Thu Jul 03, 2003 8:01 pm
Location: London

Postby Episcopus » Sat Sep 25, 2004 1:31 pm

You fools why are you complicating it. You have to know but the rule of dividing indices. When dividing you take each away, so 10^3/10^3 = 10^0 right, ye follow? Let us convert that to numbers. 1000/1000 = 1. Hence observe 10^0 to equal 1. boosh in yo eye.
phpbb
User avatar
Episcopus
Textkit Zealot
 
Posts: 2563
Joined: Sat Jun 14, 2003 8:57 pm

Postby Democritus » Sat Sep 25, 2004 6:20 pm

Have a look at this explanation:

http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

If you are thinking in terms of limits, then this describes the problem:

Notice that 0^0 is a discontinuity of the function f(x,y) = x^y, because no matter what number you assign to 0^0, you can't make x^y continuous at (0,0), since the limit along the line x=0 is 0, and the limit along the line y=0 is 1.


They also mention this:

The discussion of 0^0 is very old. Euler argues for 0^0 = 1 since a^0 = 1 for a not equal to 0 . The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitshrift.


I will have to search my cellar for the older copies of Schlomilch's Zeitschrift. Ha ha. ;)
Democritus
Textkit Fan
 
Posts: 331
Joined: Fri May 07, 2004 12:14 am
Location: California

Postby Dacicus » Sat Sep 25, 2004 6:24 pm

I have a function and I have to know if it equals zero or not and it contains 0^0, and if that is 1 then the function in some cases makes more sense in others in makes not sense at all if it does equal 1.
Sounds like a calculus limit problem with indeterminate forms. You use logarithms to solve the 0^0 ones.

This page has some info about 0^0.

EDIT:
Democritus beat me to the explanation.

BTW, Episcopus, your explanation is the same as mine except with numbers instead of variables.
Dacicus
Textkit Neophyte
 
Posts: 81
Joined: Fri Apr 09, 2004 1:13 am
Location: Los Angeles, California

Postby Emma_85 » Sat Sep 25, 2004 9:39 pm

Thanks! It's too late for me to think of maths right now, better to think of going to bed :lol: . I'll take a look at your link in the morning Democritus, thank you.
phpbb
User avatar
Emma_85
Global Moderator
Global Moderator
 
Posts: 1564
Joined: Thu Jul 03, 2003 8:01 pm
Location: London

Postby Episcopus » Sun Sep 26, 2004 11:37 am

But Dacicus your explanation makes no sense it's just ^^^^^^exponential x d y √(x+z)√10^3^

That's how it seems anyhow for it loses people, when the explanation is quite simple.
phpbb
User avatar
Episcopus
Textkit Zealot
 
Posts: 2563
Joined: Sat Jun 14, 2003 8:57 pm

Postby Dacicus » Sun Sep 26, 2004 8:15 pm

I was trying to explain it in an informal proof using the simple properties of exponents. Examples are great, but if you want to show that it holds for all real numbers (or some other set), you need to use variables more.
Dacicus
Textkit Neophyte
 
Posts: 81
Joined: Fri Apr 09, 2004 1:13 am
Location: Los Angeles, California

Postby Emma_85 » Sat Oct 02, 2004 9:17 am

I think his proof is quite a good one actually, Episcopus. And now I know more about 0^0 than my maths teacher :wink: .
phpbb
User avatar
Emma_85
Global Moderator
Global Moderator
 
Posts: 1564
Joined: Thu Jul 03, 2003 8:01 pm
Location: London

Postby Episcopus » Sat Oct 02, 2004 10:37 am

Yeah sorry I'm just seriously ignorant when it comes to maths. Here at school the standard of people's maths is shockingly high! They seem to understand all this x d y √(x+z)√10^3^ with shocking ease; but you move on to any language work, english or especially other languages, and they are shockingly terrible. Shocking is an understatement. At x d y √(x+z)√10^3^ sincostan3x/^[2dfgsdfgs(b-x^7)]/2a I tend to lean back on my chair stare at a wall and then a girl and stare at the board again and laugh. I suppose I can't fathom how this would be of any use in real life. You're better off doing weights.
phpbb
User avatar
Episcopus
Textkit Zealot
 
Posts: 2563
Joined: Sat Jun 14, 2003 8:57 pm

Postby FerrariusVerborum » Wed Oct 06, 2004 3:33 am

The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitshrift.


I just love how the math controversy 'raged.' It makes me wish math professors actually got into fist-fights over this stuff.

Euler and Newton, duking it out. Beautiful.

-FV
User avatar
FerrariusVerborum
Textkit Neophyte
 
Posts: 15
Joined: Fri Jul 30, 2004 3:29 am
Location: Yuba City, CA

Postby ThomasGR » Thu Nov 04, 2004 9:53 pm

Very well so far, 0^0=1, but what happened with 0^x?
I mean 0^1=0, 0^2=0, etc.
and when we get 0^0, than suddenly we have a value 1?

Does it go also for negative numbers?
ThomasGR
Textkit Enthusiast
 
Posts: 444
Joined: Mon Aug 23, 2004 8:49 pm

Postby Eureka » Thu Nov 04, 2004 10:58 pm

ThomasGR wrote:Very well so far, 0^0=1, but what happened with 0^x?
I mean 0^1=0, 0^2=0, etc.
and when we get 0^0, than suddenly we have a value 1?

True, a number only has a value if all the limits used to evaluate it give the same result.

For 0^0 we have:

0^x as x-->0 = 0

and

x^0 as x-->0 = 1

Therefore 0^0 has no true value.
ThomasGR wrote:Does it go also for negative numbers?

I'm not sure what you mean by this, but in these two cases it doesn't matter whether 0 is approached from above or bellow. (I presume that's what you're talking about.)

Episcopus wrote: I suppose I can't fathom how this would be of any use in real life. You're better off doing weights.

Maths has a million uses. :)
phpbb
User avatar
Eureka
Textkit Zealot
 
Posts: 741
Joined: Tue Feb 17, 2004 3:52 am
Location: Melbourne, Australia

Postby classicalclarinet » Fri Nov 05, 2004 12:14 am

just like Latin 8)
User avatar
classicalclarinet
Textkit Enthusiast
 
Posts: 400
Joined: Wed Aug 11, 2004 12:27 am
Location: Anc, AK, USA

Postby Eureka » Fri Nov 05, 2004 12:57 am

Emma_85 wrote:I think his proof is quite a good one actually, Episcopus. And now I know more about 0^0 than my maths teacher :wink: .

Your proof is quite good, it's just wrong, that's all. :P

Teachers, *pech* my primary school teachers used to tell me that a/0=0. I tried arguing with them, but that never works. Teachers know everything. :roll: :)


classicalclarinet wrote:just like Latin 8)

Even more so, if that's possible. :shock:
phpbb
User avatar
Eureka
Textkit Zealot
 
Posts: 741
Joined: Tue Feb 17, 2004 3:52 am
Location: Melbourne, Australia

Postby ThomasGR » Fri Nov 05, 2004 7:27 am

Eureka wrote:I'm not sure what you mean by this, but in these two cases it doesn't matter whether 0 is approached from above or bellow. (I presume that's what you're talking about.)

I think there is also a problem with negative numbers.

4*0^(-4)
=
4/(0^4)
4/(0)
=
0?


Teachers, *pech* my primary school teachers used to tell me that a/0=0. I tried arguing with them, but that never works. Teachers know everything. :Rolling Eyes: :Smile:

Maybe your teacher was correct, after all? :lol:
No wonder that number 0 was adopted with such difficulty! :wink: It does nothing but create problems! :lol:
ThomasGR
Textkit Enthusiast
 
Posts: 444
Joined: Mon Aug 23, 2004 8:49 pm

Postby Eureka » Fri Nov 05, 2004 9:17 am

ThomasGR wrote:I think there is also a problem with negative numbers.
4*0^(-4)
=
4/(0^4)
4/(0)
=
0?

hmmmmmm… 0^(-4):

x^(-4) as x-->0 = infinity
0^x as x-->-4 = 0

Yep, we have no solution.
ThomasGR wrote:Maybe your teacher was correct, after all? :lol:

Yes, I’m sure they were basing their comments on the concept of multiple limits. The scary thing is, you made me realise that they were partially correct. :shock: (I’m going to have to rethink my life now.)



…Still, for most purposes, a/0=infinity.
phpbb
User avatar
Eureka
Textkit Zealot
 
Posts: 741
Joined: Tue Feb 17, 2004 3:52 am
Location: Melbourne, Australia

Postby Democritus » Sat Nov 06, 2004 7:27 pm

ThomasGR wrote:I think there is also a problem with negative numbers.


I think you are right, and to put it another way, 4*0^(-4) is undefined, because 4/0 is undefined.


Eureka wrote:hmmmmmm… 0^(-4):

x^(-4) as x-->0 = infinity
0^x as x-->-4 = 0


I think the problem is that the expression "0^x as x-->-4" is itself invalid, because when x is near -4 it's negative, and that means we have zero in the denominator. In other words, 0^x is undefined, for all x<0.

Eureka wrote:…Still, for most purposes, a/0=infinity.


Maybe what you meant to say was, as b approaches zero, the expression a/b increases to infinity. Infinity is not a number, and the expression a/0 is not a valid expression. So we can't say that a/0=infinity. Both sides of this equation are not valid.
Democritus
Textkit Fan
 
Posts: 331
Joined: Fri May 07, 2004 12:14 am
Location: California

Postby Eureka » Sun Nov 07, 2004 12:59 am

Democritus wrote:Maybe what you meant to say was, as b approaches zero, the expression a/b increases to infinity.

Yes, I know infinity is not a number, but I wrote “=infinity” rather than “-->infinity” for simplicity.
Democritus wrote:
Eureka wrote:hmmmmmm… 0^(-4):

x^(-4) as x-->0 = infinity
0^x as x-->-4 = 0


I think the problem is that the expression "0^x as x-->-4" is itself invalid, because when x is near -4 it's negative, and that means we have zero in the denominator. In other words, 0^x is undefined, for all x<0.

I’ve consulted my old notes. If you want to evaluate a^b at a=0, b=-4:

first set b=-4
so, a^b = a^-4
evaluate the limit as a-->0:
a^-4-->infinity

then set a=0
so, a^b=0^b=0
evaluate the limit as b-->-4
0=0 regardless of the value of b

Because the methods do not agree, we have no solution. Both 0/x=0 and 0/x-->infinty are slightly correct, but technically incorrect.
phpbb
User avatar
Eureka
Textkit Zealot
 
Posts: 741
Joined: Tue Feb 17, 2004 3:52 am
Location: Melbourne, Australia


Return to The Academy

Who is online

Users browsing this forum: No registered users and 14 guests