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Zero to the zeroth power
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videogames101
Diamond Member
For the win.
Um...I think the point is that the calculator simply defines 0^0 as 1 if it ever comes across it, because any other definition would screw things up where it shouldn't. Forget the ti-89. If a supercomputer tried to do the same thing, it would have the same problems we are having. 0 is weird- it doesn't follow the rules that any other number does. We have to make up definitions for expressions that use 0, since we essentially made up the number 0. Examples of this are everywhere: what other number, when multiplied or divided by any number, is always equal to that same number again? Same with dividing by 0. Some would say this is always infinity, but this doesn't work at all. So we say it's impossible; an undefined value.
@ inspire
and why cant something be defined by its limit? is that not how derivatives are defined in the classical sense? Imagine a hyperbola. y=1/x. As x approaches o from the positve side the value approaches +inf. As x approaches 0 from the neg side the value approaches -inf. The hyperbola is undefined at 0 not because it results in infinity but because it has 2 values. Consider another example, the circle x^2 + y^2 =1. At x=0, y can have 2 values, this time they are finite.
In conclusion, I suggest the value of 0^0 has 2 values - 0 and 1.
and why cant something be defined by its limit? is that not how derivatives are defined in the classical sense? Imagine a hyperbola. y=1/x. As x approaches o from the positve side the value approaches +inf. As x approaches 0 from the neg side the value approaches -inf. The hyperbola is undefined at 0 not because it results in infinity but because it has 2 values. Consider another example, the circle x^2 + y^2 =1. At x=0, y can have 2 values, this time they are finite.
In conclusion, I suggest the value of 0^0 has 2 values - 0 and 1.
imported_inspire
Senior member
You're conclusion proves my point, 0^0 is not well-defined in the mathematical sense. The example of a circle falls in the same trap - a circle is not a function. The reason you can't define something at it's limit is because the existence of a limit does not imply continuity. True, it implies a removeable discontiuity, which is why limits makes useful approximations for inference. However, algebraically, an expression does not evaluate to its limit unless it is continuous at that point.
@CSMR - I'll have to look into that some more - I don't want to give you some half-baked answer, so give me a while to think about it and I'll try to answer that as best I can. I have found so far that Cauchy considered 0^0 undefined, but I'll have some reading to do on all this.
@Qriz - This problem that you bring up actually comes up a lot in abstract algebra with additive iidentities.
imported_inspire
Senior member
Okay, CSMR, I'm not sure if this will anser your question. Consider evaluating the limit of an expression x^y where x and y both tend to zero.
For example:
Lim (4x)^{3/ln(x)}
x->0+
Now, we can see that as x->0, 4x goes to zero, as well as 3/ln(x) - keep that in mind.
Now, let's simplify things and define z as z=(4x)^{3/ln(x)}. Now, since we're approaching zero from the right we can look at the ln(z) since it is 1-1 with z. We now have:
Lim [3 ln(4x)]/[ln(x)]
x->0+
This, however gives us another indeterminate form, so we apply L'Hopital's Rule to obtain 3.
Thus, our original limit, by reversing the earlier transformation is e^3.
By this example, I mean now to say that the limit of x^y as x and y approach zero is undefined since there is no one value that it evaluates to.
For example:
Lim (4x)^{3/ln(x)}
x->0+
Now, we can see that as x->0, 4x goes to zero, as well as 3/ln(x) - keep that in mind.
Now, let's simplify things and define z as z=(4x)^{3/ln(x)}. Now, since we're approaching zero from the right we can look at the ln(z) since it is 1-1 with z. We now have:
Lim [3 ln(4x)]/[ln(x)]
x->0+
This, however gives us another indeterminate form, so we apply L'Hopital's Rule to obtain 3.
Thus, our original limit, by reversing the earlier transformation is e^3.
By this example, I mean now to say that the limit of x^y as x and y approach zero is undefined since there is no one value that it evaluates to.
Yes that shows that the limit set of x^y as (x,y)->0 is not unique. But it would not normally be taken as axiomatic that the function "raised to the power of" has to be continuous on the domain it is defined on.
But Cauchy was very interested in continuity and real numbers. Cauchy sequences are a very nice way to define the real numbers and then define operations like multiplication and addition on them in a continuous way. So it that might explain why he preferred not to define 0^0. Although he also worked a lot in complex analysis where taylor series are quite important, and for these one would want to take 0^0=1.
But Cauchy was very interested in continuity and real numbers. Cauchy sequences are a very nice way to define the real numbers and then define operations like multiplication and addition on them in a continuous way. So it that might explain why he preferred not to define 0^0. Although he also worked a lot in complex analysis where taylor series are quite important, and for these one would want to take 0^0=1.
0^0 is 1 by definition. It "makes sense" since every other number divided by itself equals 1 even if you get a division by zero error for 0.
Another example: You can't extract the root of a negative number because every negative number multiplied by itself becomes positive. But the introduction of a definition of "non existing" surd numbers allows working with "insolvable" equations and the results are correct after transforming the surd numbers back to rational ones.
Another example: You can't extract the root of a negative number because every negative number multiplied by itself becomes positive. But the introduction of a definition of "non existing" surd numbers allows working with "insolvable" equations and the results are correct after transforming the surd numbers back to rational ones.
imported_inspire
Senior member
@chcarnanage
Defining 0^0 as one runs into problems. I've presented one.
You can 'extract' roots of negative numbers. You can't 'extract' even roots of negative numbers and expect to get a Real (much less Rational) answer.
You used the term 'surd' to talk about complex numbers when a surd is actually an irrational number. You also cannot transform an irrational number to a rational number - the two are mutually exclusive.
@CSMR - Good point. I didn't think my example would answer your question, but I thought it was interesting enough to discuss. I haven't been able to think of an axiomatic discrepancy, so I'll concede that point 🙂.
Defining 0^0 as one runs into problems. I've presented one.
You can 'extract' roots of negative numbers. You can't 'extract' even roots of negative numbers and expect to get a Real (much less Rational) answer.
You used the term 'surd' to talk about complex numbers when a surd is actually an irrational number. You also cannot transform an irrational number to a rational number - the two are mutually exclusive.
@CSMR - Good point. I didn't think my example would answer your question, but I thought it was interesting enough to discuss. I haven't been able to think of an axiomatic discrepancy, so I'll concede that point 🙂.
beansbaxter
Senior member
0^0 doesn't equal anything, It's undefined and so it doesnt actually exist. The interesting question would be why exactly is it undefined?
Biftheunderstudy
Senior member
Its obvious from our definitions of taylor series that 0^0 = 1
We can all agree that e^0 = 1
e^x = sum(x^n/n!)
e^0 = 0^0/0! + 0^1/1! + ....
since all the terms other than the leading term ->0 the first term has to ->1
the gamma function tells us that 0! = 1 so 0^0 has to =1
P.S.
come to think of it I think the proof of 0!=1 relies on 0^0 as well, can't remember though
We can all agree that e^0 = 1
e^x = sum(x^n/n!)
e^0 = 0^0/0! + 0^1/1! + ....
since all the terms other than the leading term ->0 the first term has to ->1
the gamma function tells us that 0! = 1 so 0^0 has to =1
P.S.
come to think of it I think the proof of 0!=1 relies on 0^0 as well, can't remember though
When you take something like
(5^5) / (5^5) you simply follow the rules of exponents here and subtract them leave 5/5 which is equal to 1.
If you take (5^7) / (5^5) you end up with 5^2 = 25 if you take (5^5) / (5^7)= 5^-2 = 1/25
So knowing this and applying this lets put (0^5) / (0^5) you get 0/0 which is undefined.
(5^5) / (5^5) you simply follow the rules of exponents here and subtract them leave 5/5 which is equal to 1.
If you take (5^7) / (5^5) you end up with 5^2 = 25 if you take (5^5) / (5^7)= 5^-2 = 1/25
So knowing this and applying this lets put (0^5) / (0^5) you get 0/0 which is undefined.
imported_inspire
Senior member
I'm lost in the wording of your first sentece there, but the second one is even more confusing. I read it to say that: if we define 'any nonzero x to the zeroth power is 1' then, that implies 'that x to the zeroth power is 1 when x is zero'. Which, on its own, doesn't make sense. We usually go Major Premise + Minor Premise = Conclusion.
I'm guessing I missed something in the first paragraph, but I can't make heads or tails of it.
@ Bif - I don't have a book handy right now, so put up with me for a minute.
On what interval is the expansion sum(x^n/n!) valid as being equal to e^x (does it include 0? Does n start at 0?)
I don't follow the thing about 'every other term other than the leading term ->0' They don't approach zero - they're constants - they are what they are. You may be thinking of sequences... True, these terms will 'ringdown', but that doesn't mean their sum is neglible by any means.
I've never thought of what the gamma function evaluates to at 0, but I do know that the gamma function is not analytic at 0 - which makes me think of residues.
If you find that proof, shoot me a link.
Biftheunderstudy
Senior member
The taylor series of the exponential is from 0 to +infinity.
In e^0 only the first term in the expansion is non-zero the rest all have powers of zero in them....so I guess ->0 is the wrong terminology, what I mean is they are 0
for example: the next term is x^1/1! = 0^1/1! = 0*0/1 = 0, the third term in the series is x^2/2! = 0^2/2 = 0 and so on
A friend has stolen my mathematical physics textbook so the proof is eluding me, however its in the same way that you can define factorials of fractional number and wierd things like that, your bang on about it having to due with complex analysis and residues....so evil.
In e^0 only the first term in the expansion is non-zero the rest all have powers of zero in them....so I guess ->0 is the wrong terminology, what I mean is they are 0
for example: the next term is x^1/1! = 0^1/1! = 0*0/1 = 0, the third term in the series is x^2/2! = 0^2/2 = 0 and so on
A friend has stolen my mathematical physics textbook so the proof is eluding me, however its in the same way that you can define factorials of fractional number and wierd things like that, your bang on about it having to due with complex analysis and residues....so evil.
As far as I know, 0^0 is considered an indeterminate form that can evaluate to anything, but it's often convenient (just in terms of making a lot of expressions look nicer) to assume that the two zeros are of same "strength" and take 0^0=1. The issue with the exponential power series has to do with the way it's written and is nothing inherent in the series, but it provides an example of this. The actual series is 1+x+... so you could write it as 1 + sum[x^k/k!, {k,1,infinity}] or sum[x^k/k!, {k,0,infinity}]; they're the same thing when x is not 0, but the second one looks neater and you can often use that if you take 0^0=1 in this specific case.
It has a simple pole there, so it's just infinity.
Evil? That stuff is way cool. 🙂 You can even use basically the same ideas to get noninteger order derivatives and a host of other stuff.
It has a simple pole there, so it's just infinity.
Evil? That stuff is way cool. 🙂 You can even use basically the same ideas to get noninteger order derivatives and a host of other stuff.
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