[math-fun] Proof that 0^0 = 1
I hear 0^0 should be defined as 1 for convenience, but perhaps there are proofs this has to be so (the *value* zero raised to the *value* 0, not the indeterminate limits of the form 0^0). What's your favorite proof that 0^0 = 1? Here's one. 0^0 = (1 - 1)^0 = \sum_{k=0}^0 \binom{0}{k} 1^{0-k} (-1)^k = \binom{0}{0} * 1 * 1 = \frac{0!}{0! \cdot 0!} * 1 = 1 * 1 = 1 Andres.
There are b^n words of length n words over an alphabet with b letters. There is one empty word over any alphabet, including the empty alphabet. Also: every formula in combinatorics where 0^0 may occur (seriously, I'd love to see a single counter-example!). Your example says 0^0 = binomial(0,0) = number of ways to choose zero apples from a bag with zero apples = 1. All this of course over the integers. IIRC there was a discussion regarding 0^0 on math-fun a while ago. That discussion may have been discouraged at some point, though. Best regards, jj * Andres Valloud <ten@smallinteger.com> [Jun 13. 2020 15:28]:
I hear 0^0 should be defined as 1 for convenience, but perhaps there are proofs this has to be so (the *value* zero raised to the *value* 0, not the indeterminate limits of the form 0^0). What's your favorite proof that 0^0 = 1? Here's one.
0^0 = (1 - 1)^0 = \sum_{k=0}^0 \binom{0}{k} 1^{0-k} (-1)^k = \binom{0}{0} * 1 * 1 = \frac{0!}{0! \cdot 0!} * 1 = 1 * 1 = 1
Andres.
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Let's not start that discussion again. 0^0 is undefined. End of story. Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Sat, Jun 13, 2020 at 10:03 AM Joerg Arndt <arndt@jjj.de> wrote:
There are b^n words of length n words over an alphabet with b letters. There is one empty word over any alphabet, including the empty alphabet.
Also: every formula in combinatorics where 0^0 may occur (seriously, I'd love to see a single counter-example!).
Your example says 0^0 = binomial(0,0) = number of ways to choose zero apples from a bag with zero apples = 1.
All this of course over the integers.
IIRC there was a discussion regarding 0^0 on math-fun a while ago. That discussion may have been discouraged at some point, though.
Best regards, jj
* Andres Valloud <ten@smallinteger.com> [Jun 13. 2020 15:28]:
I hear 0^0 should be defined as 1 for convenience, but perhaps there are proofs this has to be so (the *value* zero raised to the *value* 0, not the indeterminate limits of the form 0^0). What's your favorite proof that 0^0 = 1? Here's one.
0^0 = (1 - 1)^0 = \sum_{k=0}^0 \binom{0}{k} 1^{0-k} (-1)^k = \binom{0}{0} * 1 * 1 = \frac{0!}{0! \cdot 0!} * 1 = 1 * 1 = 1
Andres.
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Your example says 0^0 = binomial(0,0) = number of ways to choose zero apples from a bag with zero apples = 1.
All this of course over the integers.
It came in really handy when proving that \sum_{k \geq 0} \binom{2k}{k} x^k = \frac{1}{\sqrt{1-4x}} for | x | < 1/4, via complex calculus. I would have had to excuse the case x = 0, but with 0^0 defined as 1 it's simpler. Andres.
Lisp (expt 0 0) returns 1. Brent On 6/13/2020 12:27 PM, Andres Valloud wrote:
Your example says 0^0 = binomial(0,0) = number of ways to choose zero apples from a bag with zero apples = 1.
All this of course over the integers.
It came in really handy when proving that
\sum_{k \geq 0} \binom{2k}{k} x^k = \frac{1}{\sqrt{1-4x}}
for | x | < 1/4, via complex calculus. I would have had to excuse the case x = 0, but with 0^0 defined as 1 it's simpler.
Andres.
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(The real number 0) ^ (The real number 0) I think is best regarded as undefined, since x^y has an essential singularity at (0,0). But (The cardinal 0) ^ (The cardinal 0) is clearly 1. X^Y is the number of functions from a set of size Y to a set of size X, and there is exactly one function with range and domain the null set. Andy On Sat, Jun 13, 2020 at 7:27 AM Andres Valloud <ten@smallinteger.com> wrote:
I hear 0^0 should be defined as 1 for convenience, but perhaps there are proofs this has to be so (the *value* zero raised to the *value* 0, not the indeterminate limits of the form 0^0). What's your favorite proof that 0^0 = 1? Here's one.
0^0 = (1 - 1)^0 = \sum_{k=0}^0 \binom{0}{k} 1^{0-k} (-1)^k = \binom{0}{0} * 1 * 1 = \frac{0!}{0! \cdot 0!} * 1 = 1 * 1 = 1
Andres.
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participants (5)
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Andres Valloud -
Andy Latto -
Brent Meeker -
Joerg Arndt -
Neil Sloane