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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 2 years, 7 months |
| seen | 19 hours ago | |
| stats | profile views | 487 |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Brian M. Scott no, it is not true. |
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Oct 27 |
accepted | Limit $\lim_{x\to \infty } \, \frac{x}{\ln (x)-\ln \left(\frac{1}{x}\right)}$ |
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Oct 27 |
asked | Limit $\lim_{x\to \infty } \, \frac{x}{\ln (x)-\ln \left(\frac{1}{x}\right)}$ |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? You wrote: "For example, if $x_n \rightarrow 0, x_n \neq 0$, then $\lim_{n\rightarrow \infty}0^{x_n}=0$, since each term in the sequence is $0$." - This is NOT true. Actually, this limit does not exist. |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Mark Bennet is it definition or proof? The answer claims there are proofs. |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? you say "if you define it in terms of repeated products, you get 0^0=1" I suppose "you get" means "you can derive it"? |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? @MarkBennet the answer says "There are various proofs which prove either case." and also "It is most definitely not an axiom.". So I want to see the proofs that the answer claims to exist. |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? that empty product is 1 is an axiom or provable? |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Carl Mummert oh I see I confused it with another limit, $\lim x^x$, in that case you always get 1 at zero unless you are approaching zero by a spiral. |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? so $0^0=1$ is an independent axiom BOTH for natural exponentiation and for complex exponentiation? |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Navin if you do not switch the logarithm branch how can you get something other than 1 in that limit? It seems to me that to get non-1 result you have to switch the logarithm branches on fly in the process of approaching 0. |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? Are not natural numbers a subset of complex numbers? Can $0^0=1$ be independently postulated for complex numbers as an axiom? |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? Well if there are proofs that prove either of the contradicting cases, then the set of axioms is inconsistent. -1 |
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Oct 27 |
comment |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? Well I see but this does not answer the question directly. |
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Oct 27 |
asked | Is $0^0=1$ postulate independent of all other axioms of complex numbers? |
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Oct 14 |
awarded | Yearling |
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Sep 21 |
awarded | Custodian |
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Sep 5 |
revised |
Prove that $i^i$ is a real number typo fix...... |
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Sep 5 |
suggested | suggested edit on Prove that $i^i$ is a real number |
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Aug 30 |
asked | Is exponent of discrete-analytic function also discrete-analytic? |
