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| bio | website | |
|---|---|---|
| location | ||
| age | 17 | |
| visits | member for | 1 year, 11 months |
| seen | Mar 3 at 6:37 | |
| stats | profile views | 19 |
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Jan 4 |
awarded | Yearling |
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Nov 13 |
revised |
Linear Programming-minimize cost escaped $ symbol so it does not trigger latex |
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Nov 13 |
suggested | suggested edit on Linear Programming-minimize cost |
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Nov 2 |
comment |
Find the limit of $\frac{\bar{z}}{z}$ as $z$ goes to $0$. @DavidThompson For a limit to exist, it has to have the same value, no matter which direction you approach the limit from. This shows that the function approaches a different value depending on $t$. |
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Oct 29 |
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How can I find the limiting value of a time-dependent PDE? Perhaps you can take the laplace transform and then use the Final value theorem to find $f(\infty)$ |
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Oct 29 |
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Is $0^0=1$ postulate independent of all other axioms of complex numbers? hmm... If you are referring to $g^0=e$ in that pdf, I guess that would be considered an axiom. However that is a statement for all $g\in G$. In that context, $0^0=1$ is a true statement derived from that axiom by substituting $g\to0$ so $0^0=1$ is not itself an axiom. |
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Oct 29 |
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Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Anixx I believe $E(x)$ a unary operator that is similar to $f: y\to e^y$ and if you define the binary operator $f: (x,y)\to x^y$ and evaluate it at $(0,0)$ will will run into the same $\log{0}$ problem. $E(0)$ is just $e^0=1$. |
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Oct 29 |
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Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Anixx I see that they explicitly define it in that article but I would not consider it an axiom because it can be derived from other definitions such as the recursive hyper operator or cardinality of sets. |
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Oct 29 |
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Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Anixx HagenVonEitzen's derivation is based on the cardinality of sets. I don't think you can get any closer to the "standard axioms" than that. |
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Oct 29 |
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Taking inverse Fourier transform of complicated multipart equation just out of curiosity, do these equations represent anything such as a physical system? |
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Oct 29 |
revised |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? added 102 characters in body |
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Oct 29 |
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Is $0^0=1$ postulate independent of all other axioms of complex numbers? Re definitions: $0^0$ is still not an axiom by itself though. It can be derived from the definitions/axioms for the natural number case. For reference, I interpreted your question as "is $0^0=1$ {independent axiom} or {non-independent axiom}" |
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Oct 29 |
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Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Anixx well alright the limit may not be necessary but the part about $\log{0}$ still holds. |
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Oct 29 |
revised |
Is $0^0=1$ postulate independent of all other axioms of complex numbers? deleted 1 characters in body |
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Oct 29 |
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Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Anixx because usually these edge cases where the original definition may not apply are defined as their limits. |
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Oct 29 |
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Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Anixx If you use the hyper operator definition, it is defined in the recursion. See the $n\ge 3, b=0$ case |
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Oct 29 |
answered | Is $0^0=1$ postulate independent of all other axioms of complex numbers? |
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Oct 28 |
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Is $0^0=1$ postulate independent of all other axioms of complex numbers? @Graphth Testing on all my calculators: results. |
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Oct 27 |
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Is $0^0=1$ postulate independent of all other axioms of complex numbers? This is one case where it is $1$ but if you evaluate the limit $$\lim_{x\to0, y\to0}x^y$$ in some directions you will not get 1 |
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Oct 26 |
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DCT and Inverse DCT Formulas @whynot If that answered your question, don't forget to accept the answer :) |
