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visits member for 2 years, 10 months
seen Apr 13 at 2:21

Apr
10
comment Visually deceptive “proofs” which are mathematically wrong
-_- Now that's just silly.
Apr
8
comment Visually deceptive “proofs” which are mathematically wrong
@Oliver They are not being serious, right?
Mar
28
comment Can a piece of A4 paper be folded so that it's thick enough to reach the moon?
Somehow I just knew someone here would brute-force 2^x :)
Mar
20
comment How does the exponent of a function effect the result?
What are you smoking? $x^\frac{m}{n}$ is perfectly defined!
Mar
17
comment Definition of convolution?
The Cross-correlation does use x+y.
Jan
23
comment How do I convince my students that the choice of variable of integration is irrelevant?
I just knew that someone would write a program to show this :P
Jan
4
awarded  Yearling
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 $t$.
Oct
29
comment 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 $f(\infty)$
Oct
29
comment 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.
Oct
29
comment 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 $e^0=1$.
Oct
29
comment 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.
Oct
29
comment 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.
Oct
29
comment Taking inverse Fourier transform of complicated multipart equation
just out of curiosity, do these equations represent anything such as a physical system?
Oct
29
revised Is $0^0=1$ postulate independent of all other axioms of complex numbers?
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Oct
29
comment 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}"
Oct
29
comment 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.
Oct
29
revised Is $0^0=1$ postulate independent of all other axioms of complex numbers?
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Oct
29
comment 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.
Oct
29
comment 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