1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Does one raised to an undefined power equal anything?

  1. May 8, 2010 #1
    Does one raised to an undefined power "equal" anything?

    A commonly heard property of unity is "One raised to any power equals one", or:


    Another well known fact is that "One raised to the power of infinity is an indeterminate form", or:


    But what happens if the exponent of unity is "undefined", "nonsensical", "non-existent" and "disallowed", as in the expression:


    Can such an expression ever mean anything?
  2. jcsd
  3. May 8, 2010 #2
    Re: Does one raised to an undefined power "equal" anything?

    If you want it to. It can mean "I ate apple pie for desert", or anything you like. It's just a string of symbols.
  4. May 8, 2010 #3
    Re: Does one raised to an undefined power "equal" anything?


    I agree!

    But here's the problem...

    I know someone who absolutely insists that 1^(2/0) is an "indeterminate form".

    How could I convince him otherwise?

    What should I tell him?
    Last edited: May 8, 2010
  5. May 8, 2010 #4


    User Avatar
    Homework Helper

    Re: Does one raised to an undefined power "equal" anything?

    What exactly are you trying to do when convincing him? When you say 1 raised to any power is still 1 of course is restricted to any real number. Undefined numbers such as 1/0 are not real numbers. Indeterminate form MEANS the value [tex]1^\infty[/tex] can be anything, 0, 1, 10, or infinite or anything.

    Take the limit as x approaches infinite of [tex]\left(1+\frac{1}{x}\right)^x[/tex]

    This makes the indeterminate form [tex]1^\infty[/tex] but it actually equals e.

    Take [tex]\left(1+\frac{2}{x}\right)^x[/tex], this is the same form but equals e2.

    [tex]\left(1+\frac{1}{x}\right)^{2x}[/tex]... this is infinite.

    So, are you still going to prove your friend wrong?
  6. May 8, 2010 #5


    User Avatar
    Homework Helper

    Re: Does one raised to an undefined power "equal" anything?

    Oh sorry, you seem to agree that [tex]1^\infty[/tex] is indeterminate, but want to understand more about 1x where x is undefined.

    Undefined can be considered the same thing as infinite in terms of limits.

    Instead of taking the limit as x goes to infinite of 1x, take the limit as x goes to 0 of 11/x
  7. May 8, 2010 #6
    Re: Does one raised to an undefined power "equal" anything?

    Of course you have to use mathematical rigor. That includes saying what type of number x is. And you have to specify what type of exponentiation you want to use.
    With complex numbers 1^(1/2) is multivalued +1 or -1 and so on. For most x it will be multivalued.
    You can probably pick a definition of exponentiation which always choses 1. So yes, for all real numbers x we have 1^x=1.

    You are making the big mistake overinterpretation popular maths sayings. What they rather mean is that
    [tex]\lim_{n\to\infty} a_n=1[/tex]
    [tex]\lim_{n\to\infty} b_n=\infty[/tex]
    [tex]\lim_{n\to\infty} a_n^{b_n}=\text{indet. (with information given)}[/tex]
    Here at no point we are raising 1 to a power!
    Don't forget that [itex]\infty[/itex] is not something that you are allowed to use like normal variables. Infinity (in the calculus sense) is an algorithm, rather than a number.

    This expression is undefined. "Undefined" is not part of the real numbers, and we agreed that 1^x only works if x is a real number.

    No, 1/0 is indefined.
    Infinity is something completely different. It is the concept of limits with epsilons and all this.
  8. May 8, 2010 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Does one raised to an undefined power "equal" anything?

    First things first -- the notion of an "indeterminate form", and the notion that some expression is "undefined" are different concepts.

    A form is a string of symbols. Pay attention to the fact that the form "1+2" is not equal to "3". Of course, if we view them as arithmetic expressions, then the numbers they represent are equal.

    When we are using forms to denote arithmetic expressions, we might use the term "undefined" to refer to strings that are not arithmetic expressions. e.g. "1/0" is not an arithmetic expression of real numbers, because (1,0) is not in the domain of /.

    When doing calculus, we might talk about "limit forms", where we replace every atom in a form with its limit.

    The definition of continuity tells us that if the limit form represents an arithmetic expression at a point where it's continuous, then the value of the limit is equal to the value of the limit form. This is a very convenient computational device.

    When the limit form does not have that property, we say it is "indeterminate".
  9. May 8, 2010 #8
    Re: Does one raised to an undefined power "equal" anything?

    I thank you all for your replies.

    That was very kind of you.

    Here is a summary of your comments regarding the expression: 1^(2/0).

    Quoting Gerenuk:
    Quoting Hurkyl:
    Quoting Mentallic:
    Quoting Werg 22:
    Thus, we all agree that the expression 1^(2/0) is not possible
    if consistency in logic is to be maintained.

    I will direct my aquaintance to this topic, and allow him to decide for himself.
    Hopefully, he will think it through and change his mind.

    In case you are interested, here is his most recent comment. When I wrote...

    Quoting Myself:
    His response was:

    Any more comments?

  10. May 8, 2010 #9
    Re: Does one raised to an undefined power "equal" anything?

    You are fully correct with your explanation.

    I can strongly advice you one thing: this guys is a "i have not the faintest idea, but i will always play smartass and occationally quote a random line." I know a couple of those.
    I learned that discussing with such kind of guys is like discussion with mormon missionaries about physics. They are completely blind for what you say, so don't bother discussing this particular topic with that person.

    As you see all the guy could basically say to you was "No. You are wrong. I am right". He contributed no content whatsoever proofing his knowledge.
  11. May 9, 2010 #10
    Re: Does one raised to an undefined power "equal" anything?

    To: Gerenuk.

    Thank you.

  12. May 9, 2010 #11
    Re: Does one raised to an undefined power "equal" anything?

    You have to be careful with Don. He only understands things at the most superficial level, and has already reached his conclusions before he listens to your reasons. So he takes the part of what you say that superficially agrees with his conclusions, and ignores the rest that doesn't.

    For example, we were not talking about an expression with just constants, like 1^(2/0). We were talking about an expression that is a function of an unknown, and seems to evaluate to 1^(2/0) at one specific value of the variable. Like f(x)=(1+x)^(2/x) at x=0. That can be defined; f(0)=e. Which is what Mentallic said, but Don ignored. But he can only discuss it as 1^(2/0), because in that form it agrees with his pre-formed conclusion.

    Don does not agree that such an expression is indeterminate at x=0. He thinks that any expression like 1^g(x) is "prohibited" at every value in the domain of x, if g(x) can be zero at any one value in that domain. Which is specifically what we were discussing.

    He also does not understand that the literal expression 1/0 is undefined because it is infinite, but that the literal expression 1^(2/0) is undefined because it is indeterminate. In fact, he does not acknowledge that the expression 1^(2/x) is a form of [tex]1^\infty[/tex] at x=0, because he takes the statement "cannot be equal to infinity" at the superficial level. That infinity is a value that can't be equal to anything. Infinity is a quality, not a value, so 1/0 is infinity (it would be more accurate to say "is infinite" but Don refuses to grasp the difference) but is not equal to infinity. His quote was taken out of its context, which tries to explain that difference to him.

    Specifically, Don thinks he has proven Beal's Conjecture because he transformed c^z into (sorry, I don't know latex, and that thread was closed due to Don's pigheadedness) T*(c/T)[ (z*ln(c)/ln(T)-1)) / (ln(c)/ln(T)-1) ]. At c=T, this expression is indeteminate, but it is defined and is identically equal to c^z. But Don thinks the apparent division by zero makes the expression "prohibited," which he takes to mean c^z is "disallowed" under his assumptions that z>2 and so proves Beal's conjecture.

    He needs to be told, in plain language, and by as many people as possible, that he is wrong and has come nowhere near proving Beal's conjecture. One reason is that his expression is indeterminate, which he refuses to acknowledge. There are many other flaws as well - about the only thing that is right in his argument is that his expression does equal c^z.
  13. May 9, 2010 #12
    Re: Does one raised to an undefined power "equal" anything?

    Sorry guys.

    I didn't think that would happen.

    Please lock this thread.

    Thanks again for your help.

  14. May 9, 2010 #13


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Does one raised to an undefined power "equal" anything?

    Please take your personal squabbles elsewhere, and refrain from derailing threads.

    I do feel compelled to point out, for the sake of getting the little details correct, that the partial function
    is quite plainly undefined at zero. However, its continuous extension to the set of nonnegative reals is defined at zero, and has value e2.
  15. May 10, 2010 #14
    Re: Does one raised to an undefined power "equal" anything?

    I'm sorry it was brought here, but I didn't do it. I just wanted to make sure that whatever comments you made were not misquoted or misrepresented. So I am not "derailing the thread," I am trying to make sure it addresses the question Don thinks he was asking, but didn't. And that Don is told the answer to the correct question, with no reasonable way to misinterpret it. Don will do that anyway, but I want the record clear. Anyway, I'll just explain what I mean, and drop it.

    Hurkyl, you made a minor error of sematics, that is exactly what I feared Don would misunderstand. The partial function
    is quite plainly indeterminate at x=0. What that means, is that without further information, you can't determine a value for it. So it is undefineable in general, but that is not what Don wants undefined to mean to math. He wants "can't ever be defined, so must be expunged." Sometimes, this expression can be defined. Don ignored this:
    Emphasis added. (Oops, sorry about my typo where I said e1. I started with (1+x)(1/x) but changed it, but apparently not everywhere.)

    Here, you are saying that under specific conditions you can determine a value at x=0. Don quoted what you said only up to the point of saying "undefined," and concluded that you meant (1) it can never be defined, (2) that x=0 must be "disallowed" under all circumstances, and (3) if you can't disallow x=0, then one of the assumptions that led to the derivation of the equation must be wrong. Proof by reductio ad absurdum.

    But let's say that the equation
    I(r) = I0 (1-(1+r)(-1/r)) / (1-e-1)​
    is derived for a current in a system, where r is the resistance of a component in it. Don thinks that you said the current is undefined if you short that component. What you said was, the expression itself is undefined if you don't know what I and r represent; but that if the resistance can be represented by a nonnegative real, and everything should be continuous at r=0, I(0) is defined to be I0.

    Don further feels that since r=0 must be "disallowed" in every way, that the system cannot be built in the first place. Not that something will blow up (i.e., "is really not continuous"), that you can't get that equation legitimately. That "undefined" currents cannot exist, but there obviously will be a current, so something is wrong elsewhere. That is the meaning he placed on your statement where you used that word, and I just want you to say more clealry what you meant. That it can be defined sometimes, so you can't conclude the expression itself is invalid.

    What Don left out of this thread, is that the "specific circumstances" for the expression he wants to use, are that it isn't even indeterminate. He just made it look that way via algebraic manipulation that was a little more complicated than the following, but essentially the same in concept:
    f(x) = (x/k)x = (x/k)((x-k) * x/(x-k))
    = [(x/k)(x-k)](x/(x-k))
    From this he concludes that, since there is no reason x can't be equal to k, but we must "disallow" x=k, that something is wrong with the assumptions that define f(x).

    The problem is that Don won't listen to any of the reasons his argument is not correct, just one of which is that the partial expression (x/k)(x/(x-y)) is indeterminate. He is right that when it is expressed that way we can't, in general, define a value. But since it is indeterminate, and we derived it from an expression that had no such division by zero, it is in fact defined for positive k.
    Last edited: May 10, 2010
  16. May 10, 2010 #15


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Does one raised to an undefined power "equal" anything?

    Don didn't create a personal attack on himself -- you were the one who decided to do that, rather than address the relevant math.

    Anyways, I'm going to insist on making true statements in this thread. There is a time where imprecise, nearly-true statements are appropriate, and I'm pretty sure this is not it.

    The claim that the partial function defined by
    I(r) = I0 (1-(1+r)(-1/r)) / (1-e-1)​
    is defined at 0 is an example of one of these useful fictions. Most of the time, when defining functions like this, we don't mean what we literally wrote -- we really mean things like "the continuous extension of what we wrote" or "we do this in most places, and make the obvious adjustments where it doesn't work" or what-not. We're lazy and we don't want to go through the extra trouble of saying all these niggling details that everybody knows anyways.

    But in those situations where we really are paying attention to these details, it is flat out wrong to make the claim that, for the function defined above, that I(0) is defined.

    And in your other example, the relation you wrote
    (x/k)x = (x/k)((x-k) * x/(x-k))
    is incorrect. (If is, of course, valid when restricted to the domain where x>0, k>0, and x and k are different)

    The function defined by "x/x" is undefined at 0. The limit form written "x/x" is indeterminate. There's a difference between what those two statements mean. It's not a difference that we often care paying attention to, but if we're going to do so, then we should make true statements.
    Last edited: May 10, 2010
  17. May 11, 2010 #16
    Re: Does one raised to an undefined power "equal" anything?

    I am truly sorry for that - but I felt it was necessary to make you see just how your words would be misinterpreted if I didn't. The issue really isn't the math, it is the seemingly purposeful misunderstanding of the math, in order to support invalid conclusions that have nothing to do with that math.
    I agree, and have tried to do so within the bounds of how it will be (mis)understood. And that's the problem. But I'd also like to simplify the "true statements" without adding "yes, but that also means...".
    1. Yes or no, is an expression of the form f(x) = g(x)1/h(x) an example of the indeterminate form [tex]1^\infty[/tex] at x1>0, if g(x1)=1 and h(x1)=0? I'm not stating whether or not something like "continuous extension" is appropriate. I'm asking if that form, with a division by zero in the exponent, is something the form [tex]1^\infty[/tex] intends when it does apply.
    2. Can you use a well-defined "continuous extension" value of f(x), when the domain of x is the integers? Or does that domain prevent the use of the "continuous extension?" I'm not really sure how this point applies to anything, because of the next one, but it came up.
    3. If you take a function f(x) that is well defined over all positive values of x (example: f(x) = x3), and (correctly) manipulate it into the form f(x) = x1*g(x)1/h(x) where g(x1)=1 and h(x1)=0 for x1>0, does that make f(x) undefined, or in any way invalid, at any positive values of x? (example: g(x)= (x/x1), h(x)=(ln(x)-ln(x1))/(3*ln(x)-ln(x1)). This is the reason we are asking if it is indeterminate. It's not because the equation started in that form, but because it was manipulated into it. So that labeling it "indeterminate" is not so it can be defined or evaluated, but so we can decide if the manipulation was artificial and has no bearing on whether f(x) is defined or can be evaluated.
    4. If you answered "yes" to the last question, does that mean those values of x must be removed from the domain of f(x)?
    Don answers these questions "no," "no," "yes," and "for some xi's." I'm sorry this all sounds like another attack, but these are the answers Don wants to support, and he is asking questions out of context to get the answers he prefers. He wants g(x)1/h(x) to not be indeterminate, because I claimed being indeterminate makea the relationship artificial, having no bearing on f(x). He is using your statement that it is "undefined" to claim that f(x) is also undefined, so some values of x must be removed from the domain of f(x).
    I did not mean that it was literally defined without other information. I meant that if you built a circuit with ideal components, derived that equation for it, and set r=0, that the current would be I0. You could determine that by using a variable resistor and dialing it to zero, which is the realization of your "continuous extension;" or just by shorting it, assuming that all of the components can handle the sudden change. Yes, I know there could be other issues like damping times, but I'm assuming they are negligible. It is possible that whatever interactions cause the exponential nature of this result means there are singularities in the circuit, which is why I assumed ideal components. To me, that means that they can handle those singularities.

    The point is, arriving at that equation alone is not enough to conclude that you can't short the resistor without destroying the electrical components, which is effectively what Don wants to do. Even worse, the equation he really had was a well-defined I(r), over all positive r, that he manipulated into the form I(r)=I0(t)*g(r,t)^(1/h(r,t)) for some completely made-up parameter t. And I intentionally left it out of the function I(r), because it does not affect that value in any way. Unless, of couse, it makes the right-hand side undefined. Which is where "indeterminate" comes in.

    It turns out this manipulation can be done for any positive integer r, and that g(r,r)=1 and h(r,r)=0. From that he concludes that the r's he wants to get rid of are "disallowed" for I(r), but the ones he does not choose this way are "allowed." And that is the ultimate question he wants answered - is are those r's "disallowed" ?
    Last edited: May 11, 2010
  18. May 11, 2010 #17


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Does one raised to an undefined power "equal" anything?

    This hypothesis is vacuous -- the following three statements are collectively contradictory:
    • The domain of f(x) includes all positive real numbers
    • h(x) is zero for some positive real number
    • f(x) = g(x)1/h(x) is true on the domain of f

    e.g. a simpler, similar example: 1 = x/x (where x is a real variable) is not true, and any calculation that assumes that equation is suspect. Now, making such substitutions is a commonly useful thing to do -- but it requires breaking a problem into two parts: one part where x is assumed to be nonzero (so that 1=x/x is identically true, and may be correctly used), and one part where x is assumed to be zero, which must be handled in another fashion.
  19. May 11, 2010 #18
    Re: Does one raised to an undefined power "equal" anything?

    I didn't say f(x) = g(x)1/h(x) is true on the domain of f. It has a problem at x=x1, which is the one Don is trying to take advantage of. But other than that, the manipulation was correct. He is trying to claim that because of that substitution, the domain of f(x) is restricted the same way the domain of g(x)1/h(x) is. One clue that it is not true, is that the form of g(x)1/h(x) matches the indeterminate form [tex]1^\infty[/tex] (see question #1 for caveats) at that point, which is what led Don here to try to claim it didn't.

    But you have now been helpful enough in finding the problem. Thank you.
    Last edited: May 12, 2010
  20. May 11, 2010 #19
    Re: Does one raised to an undefined power "equal" anything?

    While often toted as a fact by calculus teachers, the notion of an "indeterminate form" is ill-defined. It only survives because it's a helpful pedagogical tool for students who can't distinguish between a function and evaluations of that function.

    Division by zero isn't defined.

    Exponentiation of zero by zero isn't defined.

    Infinity isn't a real number.

    Talking about "2/0" is like talking about the "shape of blue". It's nonsense. It's not well-formed. Similarly "0^0" and "1^∞". Those are "the color of nothing" and "the sound of silence" respectively They. Are. Nonsense.

    What ISN'T nonsense are limits. Limits allow you to give reasonable meanings to these undefined values. The reason they are able to this (while the expressions alone can't) is because they carry more information. Think about it like this. With an expression of two variables, there's only one possible representation if we substitute 0 for those values.

    1) f(0, 0)

    But with a limit, there are several ways to do it:

    1) f(0, 0)
    2) lim x->0: f(x, 0)
    3) lim x->0: f(0, x)
    4) lim x->0: f(x, x)

    See how we have a more expressive "language" with limits?

    Now let f(x, y) = x^y. How many of these possibilities are legitimate still?

    1) 0^0 (illegitimate, because as we said above exponentiation of 0 by 0 is not defined)
    2) lim x->0: x^0
    3) lim x->0: 0^x
    4) lim x->0: x^x

    So, with one case invalid, we still have three different possible limits.

    Naively, we can map all three of these back to the illegitimate expression 0^0. Even though it's a bastardization (because remember 0^0 is absolute nonsense), we accept that even if it was legitimate, it would be ambiguous. Seeing 0^0, we naturally want to map it back to a limit so we can evaluate it -- but oh no, there's not enough information to decide which kind of limit we meant.

    This is the reason for introducing "indeterminate forms". The language of expressions is less "expressive" than the language of limits. But since the former is more compact, we often use it instead. This leads to ambiguities, and we call that little screw-up in our mathematical history by a funny name: "indeterminate form."
  21. Sep 17, 2011 #20
    Re: Does one raised to an undefined power "equal" anything?

    your explanation about 1^infinity is very clear thanks,but if 1/0 is undefined then why we were taught like " anything divided by o is infinity" ?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook