# 0 to the power 0

• B
Mentor
There is an excellent math teacher out here in Aus (he won young Australian of the year) who makes good videos. Here is one about the old conundrum 0^0:

As you can see in my comment, I'm afraid I have to disagree with Eddie here. 0^0 is undefined because you get different answers depending on how you take the double limit. It shows why one must be really careful with what it means by powers. You really need calculus to define them properly. You will find it in real analysis or honours calculus textbooks, but sketch it define log to base e, Ln(x) as integral 1 to x 1/y dy. We see a problem immediately at x =0. Differentiating Ln(xy), you get 1/x. Thus Ln(xy) = ln (x) + C. If x = 1 we see C = Ln(y) or Ln(xy) = Ln(x) + Ln(y). Let e^x be the inverse of Ln(x). Let a = e^x, b = e^y, and we have e^(x+y) = e^(Ln(a) +Ln(b)) = e^Ln(a*b) = a*b = e^x*e^y. We can, from these relations, work out all the other fundamental relations of logs and powers. As I explained in the comments, this greatly helps clarify 0^0. Basically, Eddie was a bit sloppy and an example of why you need real analysis.

Thanks
Bill

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PeroK
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The convention that ##0^0 = 1## turns up implicity in power series, where we have for example:
$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$ And, if we set ##x = 0## we have: $$e^0 = \frac{0^0}{0!} + \sum_{n=1}^{\infty} \frac{0^n}{n!} = 0^0$$ And we need the convention that ##0^0 = 1##.

• bhobba
fresh_42
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$$0^0=\prod_{\iota\in\{\}}0=\prod_{\iota\in\{\}}c=1$$

jedishrfu
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Curiously, I did a plot on Desmos and got that ##0^0 = 1##

The curve drawn was interesting as well exhibiting the limit of ##y = x^x## drawing a curve akin to a parabola (but of course not a parabola) like x^2 with a minimum at x= 0.368 and y= 0.692.

Also, nothing is graphed in the x<0 quadrants on the Desmos plot.

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etotheipi
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$$0^0=\prod_{\iota\in\{\}}0=\prod_{\iota\in\{\}}c=1$$
Could you explain? fresh_42
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Could you explain? The power function is an abbreviation for a certain multiplication. The neutral element is the necessary result of a void multiplication, as is ##\sum_{k\in \emptyset} a_k=0.## So whatever we multiply over zero many factors has to be ##1.##

• etotheipi
I always like the videos of blackpenredpen, and here is one on this topic, claiming it is undefined. Also walframalpha claims it is undefined, giving pretty much the same point as blackpenredpen. Lastly numberphyle claims the same.

So ##\lim_{x \to 0^+} x^x = 1##, also ##\lim_{x \to 0^+} x^0 = 1##, but ##\lim_{x \to 0^+} 0^x = 0##. This means you get a contradiction based on which limit you use. Therefore it is undefined, as @bhobba already said.

PeroK
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Well, it's undefined until ... you define it!

• • cormsby and jedishrfu
fresh_42
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I always like the videos of blackpenredpen, and here is one on this topic, claiming it is undefined. Also walframalpha claims it is undefined, giving pretty much the same point as blackpenredpen. Lastly numberphyle claims the same.

So ##\lim_{x \to 0^+} x^x = 1##, also ##\lim_{x \to 0^+} x^0 = 1##, but ##\lim_{x \to 0^+} 0^x = 0##. This means you get a contradiction based on which limit you use. Therefore it is undefined, as @bhobba already said.
Who defined that continuity had to play a role? This is already the first unspoken assumption. My argument is based on logic.

My argument is based on logic.
So is mine "The fact that all Mathematics is Symbolic Logic is one of the greatest discoveries of our age"
Bertrand Russell

fresh_42
Mentor
So is mine "The fact that all Mathematics is Symbolic Logic is one of the greatest discoveries of our age"
Bertrand Russell
No, you assumed continuity of certain functions and concluded, that the fact, that this cannot be achieved for these two functions will make the symbol undefined. But the requirement of continuity, let alone of both functions simultaneously, is out of thin air. My argument uses symbolism only.

WWGD
Gold Member
Curiously, I did a plot on Desmos and got that ##0^0 = 1##

The curve drawn was interesting as well exhibiting the limit of ##y = x^x## drawing a curve akin to a parabola (but of course not a parabola) like x^2 with a minimum at x= 0.368 and y= 0.692.

Also, nothing is graphed in the x<0 quadrants on the Desmos plot.
It becomes needed to use Complex numbers since negative roots , e.g. ## (-1/2)^{1/2} ## are not defined in/for Real numbers.

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• jedishrfu
wrobel
the whole this thread will be another argument that there is no correct definition

• dextercioby, jedishrfu and Arjan82
jedishrfu
Mentor
If we go with many worlds then in one the mathematicians say it’s one and in the other they say it’s zero.

Usually the definition that wins out is the one that leads to new mathematics.

Having said that, it’s perhaps a good time to close this thread and go on to other matters.

• fresh_42