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About LinkBacksHow come 4 to the power of 0 = 1? I just don't get it.
September 9th, 2006, 04:36 PM
How come 4 to the power of 0 = 1? I just don't get it.
What do you think it should be?
September 9th, 2006, 06:37 PM
Why 4 specifically?Originally Posted by The P-manator
It's part of the definition of integer exponents. If x is a non-zero real number, we define x^0=1, and x^n recursively by x^n=x*x^(n-1) for all integers n>0. If n is a negative integer, we define x^(n)=1/(x^(-n)).
That's all, it's just the definition that 4^0=1. That's all there is to 'get'.
If you want to ask *why* someone would consider this is a sensible definition for x^0, you like the exponent law (x^n)*(x^m)=x^(n+m) don't you? What if you defined 4^0 to be something other than 1?
I am sorry, english please? Is there any other method of explaining this in a simpler way?
x^0 is defined to be 1 when x is non-zero.
Simple enough?
Definition. That's the easiest way to explain it. It's a definition that comes from how we define exponents.
maybe we can put it this way.
1) x^n/x^m=x^(n-m)...........by definition
2)a/a=1........of course a is non-zero
hence, if n=m
we have x^n/x^n
=x^(n-n)
=x^0
and x^n/x^n=1
thus x^0=1
hehe
September 10th, 2006, 01:46 AM
There are lots of reasons but one reason is for consistency with the rules of algebra.
if 4^0 = a
then by the rule of doing the same thing to both sides of an equation
4(4^0) = 4 a
then by the rule of adding exponents when multiplying powers with the same base
4^(0+1) = 4 a
or 4^1 = 4 a
so 4 = 4 a
which means that a must be 1
Another is for consistency with other mathematical functions like the logarithm.
4^0 = a
should be equivalent to log base 4 of a = 0
but the log base 4 of a should be the same as (log a)/(log 4)
but (log a)/(log 4) can only equal 0 if log a = 0
which means that a must be 1
Or using the rules of algebra again
then log(4^0) = log a
then 0 log(4) = log a
which means that log a = 0
and so a must be 1
what about factorial operation.
0!=1
apparently this is for consistancy with the index laws?
September 10th, 2006, 02:45 AM
OK 4^1 = 4
4^ 0.5 = 2
4^ 1/3 = 1.58
4^ 0.25 = 1.414
4^0.1 = 1.14
Now as the power approaches 0 so the result approaches 1
If you have a piece of log/lin graph paper you can plot it you will easily see where it's going. Whoops.. do they still make graph paper....
If you are into calculus you will then see n^0 = 1
I think there is no mathematical proof for this question other than the fact that it follows by axioms on indices :-?.
i think this is a valid proof
x^1=x
x^2=xx
x^n=x times itself n times
x^n/x^m=x times itself n times divided with x times itself m times, and since you can remove m x´s then it turns to be x^(n-m)
and when n=m we have the same amount both above and under the divition. and we all know something divided with itself is 1
0!=1 also by definition.
It's a sensible choice for a few reasons. It makes the formula for "n choose k" work when k=0 or k=n, this keeps us from having to make the binomial theorem icky by having two terms floating outside the sum.
factorial is defined recursively using n=n*((n-1)!), so even if you decided to 'start' this definition at 1 by definining 1!=1, this recursive definition suggests that 0! should satisfy 1!=1*(0!) so you'd need 0!=1. This is also suggesting that you'd need 1=0!=0*((-1)!), so there is no way to define factorial for negaitve integers if we wanted this recursion to hold and they are left undefined (these turns up as poles in the gamma function).
The gamma function would still be regarded as "the"* extension of the factorial to the postitive reals even if we had left 0! undefined, and gamma would tell us we should have 0!=1 if it's anything at all.
*there are many ways to extend it, but gamma is the unique log convex extension, and easily the most useful.
0!=1 is a tricky question since factorial is property of natural number, that do not includes 0.
what aboutr (-1)!?
-1 or for that matter -200 do not count, these are relative not absolute values. There is NO such physical quantity as -1.
Ask your teacher.
yes i know, but that dont answer my question
what if i take (-1)! and if that doesnt work why?
What's tricky about it being defined as 0!=1? Every definition of factorial I've ever seen included 0.
ps. 'natrual numbers' sometimes include 0 as well, there's no universal convention.
exactly what I said before. If you wanted the recursive definition to hold you'd need 0!=0*((-1)!). In other words (-1)! when multiplied by 0 would have to give 1, there's no real number that satisfies this.
Look up the gamma function:
http://mathworld.wolfram.com/GammaFunction.html
It has poles that would correspond to the "negative factorials", it 'blows up' as you approach these poles.
Sorry to everyone who attempted to help me, but it is Billco's definition which I understand. I'm only in Grade 9 enriched math, so all that fance log stuff I don;t get. What does ^ mean?
I never know that -1 is a natural number
Eh? I certainly didn't say that. -1 is never considered a natural number. 0 sometimes is, sometimes isn't.I never know that -1 is a natural number
He didn't give anything that was a 'definition', just a motivation for 4^0 to be 1 if you want 4^x to be continuous at x=0.
What would you think if I suggested 0^0 should be 0 based on:
0^1=0
0^0.5=0
0^0.1=0
0^0.001=0
and then went on to suggest 0^0 should be 1 based on:
1^0=1
0.5^0=1
0.1^0=1
0.001^0=1
a^b is "a to the power b". It's pretty standard for ascii notation, and is used in many/most/(maybe all) computer maths programs, and things like LaTeX. You wouldn't see it used in a text book where superscripts are possible.
How could you have understood anything in Bilco's post without knowing what the ^ meant?
Shmoe is giving an example of function of two variables which has two different limits at the origin from two different directions, f(x,y) = x^y at (x,y)=(0,0). But I don't see that algebra or consistency with the logarithm will work on this one either.
taking the logarithm of both sides of 0^0 = a
0 log 0 = log a
But log 0 is undefined except as the limit of negative infinity.
Also the other approach of (log a)/log 0 = 0 doesn't help, even when you use the limit of log 0 -> negative infinity for any a solves this.
Therefore I think we must conclude that 0^0 is undefined
Hmmm.... what about a limit from a different angle
like the 45 degree angle using the limit of x^x as x approaches 0
emacs lisp gives me the following progression
(expt .1 .1) 0.7943282347242815
(expt .01 .01) 0.954992586021436
(expt .001 .001) 0.9931160484209338
(expt .0001 .0001) 0.9990793899844618
(expt .00001 .00001) 0.9998848773724686
(expt .000001 .000001) 0.9999861845848758
(expt .0000001 .0000001) 0.9999983881917339
which suggests that the limit is 1.0
I think this means that the limit is 1.0 from all directions except along x=0 where f(x,y) = 0^y = 0
I don't know that might be a case for 1.0 being the best answer if there is any answer at all, other than undefined that is.
He didn't give anything that was a 'definition', just a motivation for 4^0 to be 1 if you want 4^x to be continuous at x=0.
What would you think if I suggested 0^0 should be 0 based on:
0^1=0
0^0.5=0
0^0.1=0
0^0.001=0
and then went on to suggest 0^0 should be 1 based on:
1^0=1
0.5^0=1
0.1^0=1
0.001^0=1
a^b is "a to the power b". It's pretty standard for ascii notation, and is used in many/most/(maybe all) computer maths programs, and things like LaTeX. You wouldn't see it used in a text book where superscripts are possible.
How could you have understood anything in Bilco's post without knowing what the ^ meant?
Was I wrong? I pointed out (effectively I thought ) that Y was depedent upon X.
That y=4^X
Your attempt which starts "What would you think...."etc Does not appear to have an independant variable it is not a function just a series of equates.
I thought I was doing OK for a guy who learnt Calculus 55 years ago and last practiced it about 25 years ago.
It is not necessary to understand the character ' ^ ' in my post, I said "as the power approaches 0 so the result = 1"
Finally y=4^x produces a curve which is valid for all values of x so why should it NOT be valid to explain it from first principles?.
I didn't say anything was wrong with your post, just that it's not a definition.
It was just to get him to think about what can happen if you try to use continuity as your motivation for definitions.
Your post also had a series of equations in essentially the same form, but you are ok with that in your post but not mine? I find that odd.
If you take out the parts with '^', your post begins "Now as the power approaches 0 so the result approaches 1" and I would be left to wonder, power of what?
This isn't a criticism of your post, and there is nothing at all wrong with someone not knowing what the "^" meant. I just find it hard to believe anything useful could have been gleaned (from any of the responses) without knowing what '^' means, and would highly suggest that P-manator go back and read them with this new knowledge.
That is NOT first principles. You don't even have a curve y=4^x until you have defined what these powers mean.
What if you approach along the curve (x,log(a)/log(x)), where a>0 and x>0 and x->0?
It's a moot point.
And perhaps one minor mis-understanding .
I had mis-understood a point in your post.
But I think we are both on a similar wavelength.
All right all right, I get it, your the big shot mathematician that vents himself with big formulas to a 4rth year highschooler.L.E.A.P. wrote:
I am sorry, english please? Is there any other method of explaining this in a simpler way?
x^0 is defined to be 1 when x is non-zero.
Simple enough?
I maent that big formula that you typed was very hard to understand to me!
It's not that I enjoy it, it's just that the only thing I'm good at is confusing highschoolers. I'm of the opinion that if someone is only good at one thing, they are doing a disservice to humanity if they don't cultivate that ability to it's full potential, even if they don't enjoy it.
(Nothing I've said in this thread, or anywhere in this forum should indicate in any way that I am anything resembling a "big shot mathematician".)
You *can * understand it if you are willing to think about it carefully. I'm sure of this.
The basic idea is as simple as the integers themselves. Let x be a non-zero number. We are going to start with the definition:
x^0=1.
We would like to define powers to an arbitrary integer, so let's start with the positive ones. We'll make these definitions:
x^1=x
x^2=x * x
x^3=x * x * x
x^4=x * x * x * x
I'm getting tired. You can see how to go from one line to the next clearly enough right? We just multiply by another x each time. Let me rewrite the above with this in mind, and I'm going to add in our earlier x^0=1 definition:
x^0=1
x^1=x * x^0
x^2=x * x^1
x^3=x * x^2
x^4=x * x^3
This is saving a little time, there are less x's to write this way. It's not really necessary for me to keep adding more lines if I can just tell you how to get from one line to the next is it? You know we will have the line:
x^100=x * x^99
somewhere, so I don't really need to write it. It would save me alot of time (that's an understatement) if I could just describe how to get from one line to the next. The relationship for the nth line, where n>=1 is (the x^0=1 line is the 0th line) x^n=x * x^(n-1). So I'm going to save myself alot of time and define x^n for non-negative integers this way:
x^0=1
x^n= x *x^(n-1) when n>=1
This lets you know in no uncertain terms how x^7, of x^102 are defined. It would take a bit of work, but you can find x^102 by working backwards from this definition:
x^102=x * x^101 = x * (x * x^100)=x * (x * (x * x^99))=......
(you didn't think I was actually going to go all the way back to x^0 did you?) This is how you'd compute powers by hand, just start multiplying a number by itself a bunch of times. Ok this isn't a smart way, but shockingly few people seem aware of the repeated squaring approach, but that's another topic, so let's just call it a way.
You can now go out and show the usual exponent law (x^n)*(x^m)=x^(n+m) holds for all n,m integers >=0, and we are happy. So very happy. What about the negative integers? We'd like the exponent law to hold for them as well, and it turns out if we were to just define x^(-n) as 1/x^n when n<0 (for example x^(-3)=1/x^3) we will be able to prove this exponent law does in fact hold. (I'm not going to prove here that these definitions will make the exponent law hold)
That's the way the definitions are often organized. This is an example of a recursive or inductive proof that you'll see more of later. They aren't really avoidable (see the factorial definition I gave). A thing called "the principle of mathematical induction" is a fundamental part of the natural numbers (sometimes thinly disguised as "well ordering"), this is something you'll come to love if you carry on in maths.
Shmoe is suggesting this because on this curve our two variable function f(x,y) = x^y becomes x^(log base x of a) = a, which means that the limit as x goes to 0 on this curve is a.What if you approach along the curve (x,log(a)/log(x)), where a>0 and x>0 and x->0?
For this limit to work (without resorting to analytic continuations of our function) I think that a must not only be greater than 0 but also less than 1.
Now the question I immediately had was at what angle does this approach the origin and as you would expect the tangent to this curve at the origin is the line x=0.
As a result I was correct in my conclusion that approaching the origin at any angle except the line x=0 does give the limit of 1. It is only that when you approach the origin from the angle of the line x=0 that you can get any limit from 0.0 to 1.0 depending on how you take the limit (that is what curve you follow).
Thank you shmoe, I nderstand now, I wil try tu do the reverse when I get back from school. I just wanted you to know that I WILL carry on maths for the simple fact that I need them.
He explained it using something called words, which most non-mathematical experts understand.
Thanks again, Billco.
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