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2d
comment Why do we assign values to divergent series?
@Ethan: I think "why do we say it equals -1/12" led to the duplicate marking. Your first question seemed to be asking, "What is the motivation for assigning values to divergent series?" But all other commentary indicates that you wanted to know, "Why do we assign the specific values we assign to particular divergent series?" Now that I re-read your question, I'm not sure which you were asking. Regardless of the intended question, the answers are in GH Hardy's book "Divergent Series"!
2d
answered Why do we assign values to divergent series?
Feb
15
comment Probability of tossing a coin twice & MATLAB?
In the formula you gave (for Bernoulli trials), the parameter $p$ is the probability of a single outcome. For tossing a coin, this should be 1/2. In your code, it should be 1/outcome.
Feb
1
comment What's wrong with $cos(t) /t$ and $e^{-t^2}$?
This has been proven. The answers to a similar question contain good references: math.stackexchange.com/questions/155/…
Jan
28
comment Provocations on the existence of mathematical objects
What is your criterion for the existence of objects of type 2? I'm happy to say that no object satisfying a list of properties that leads to a contradiction exists. If we are confident this doesn't happen, does that suffice for existence?
Jan
28
comment What function can be differentiated twice, but not 3 times?
Briefly, if you replace x by z, you have to extend the function to the complex plane, or at least to some open set in the plane. But no matter how you extend the function, it won't be complex-differentiable. Otherwise, the functions you see below are infinitely real-differentiable. For example, if $f(z) = z^3$ for $Re z \geq 0$ and $f(z) = -z^3$ otherwise, then $f(z)$ is not complex-differentiable. The piece-wise break occurs along the imaginary axis, and the calculation that worked for $x = 0$ below doesn't work for nonzero values along the axis. You really should ask a separate question!
Jan
26
comment simple question on POKER HAND probiblity {full house}
The reason why order matters is that Aces over Kings (3 aces and 2 kings) is a different hand (in poker) than Kings over aces (3 kings and 2 aces).
Jan
2
comment product is twice a square
This is nice. I gave a bad reason for why there should be fewer than $2n$ prime factors amongst the integers in the interval. The requirement that $[1,0,\ldots,0]$ be in the span unfortunately makes it so that knowing (or estimating) the dimension is not enough.
Dec
29
comment Why the largest eigenvalue is the bound?
I'm suspicious of the $1 + x^T A x$ in the denominator, because that would lead to an upper bound of 1, regardless of how small the largest eigenvalue is. I suspect $x^T x$ is supposed to be in the denominator instead.
Dec
27
comment An equivalence relation in primitive permutation group
Reflexivity does appear to be a problem. For an arbitrary $x \in X$, there need not be a $gB$ containing $x$. However, we can imagine non-transitive actions with blocks $B$ where this is an equivalence relation. If the given block $B$ satisfies $X = \cup gB$, then everything works. If the action is transitive, this always happens, but it is not hard to construct reasonable non-transitive actions where this equivalence relation makes sense.
Dec
20
comment yes, why does a negative times a negative make a positive?
"Does this attempted construction of the integers from the naturals work?" It appears your title worked against you.
Dec
20
comment yes, why does a negative times a negative make a positive?
I've looked at a lot of the details, but I think this could be made cleaner. In universal algebra, there is the general notion of "congruence relation" that applies to any algebraic structure. You've defined a very nice + and * operation on $\mathbb{N} \times \mathbb{N}$. I conjecture that if you define $\equiv$ by $(a_1,a_2) \equiv (b_1,b_2) \iff a_2 + b_1 = a_1 + b_2$ and mod out by that equivalence relation, you've got the integers. This is either what you are saying or very close. Regarding hardmath's comments: I do think your title is misleading!!
Dec
19
comment Concrete bases for functions on $S_n$
Is the intended addition and scalar multiplication point-wise addition and scalar multiplication of functions? If so, then the functions $\psi_{i,j}$ do not span the space of functions from $S_n$ to $\mathbb{R}$. A natural basis would consist of Kronecker deltas of permutations. (In other words, $\delta_{\sigma} (\tau) = 1$ if and only if $\sigma=\tau$.) Such Kronecker deltas would be orthogonal relative to your inner product, and you could make an orthonormal basis by scaling by $n!$. But maybe I'm misunderstanding the linear space you have in mind?
Dec
14
comment Is there any good reason not to define $0^0=1$ , such as contradictions in algebra or arithmetic?
The continuity requirement is not such a strange one when you consider that the real numbers are defined to be limit-like objects. So asserting $0^0=1$ over the reals does seem questionable - it will depend upon how you define exponentiation. Rudin has $a^x$ being a supremum of all $a^q$ with $q$ rational and $q<x$. With $a=0$ allowed, this would suggest $0^0=0$. (He requires a>0.) So, it's not a small offense to violate the continuity of $x^y$. But, in many contexts, it is reasonable to advocate $0^0 = 1$, because it can be interpreted reasonably and is very convenient.
Dec
14
comment Is there any good reason not to define $0^0=1$ , such as contradictions in algebra or arithmetic?
Thanks - I'd actually be surprised if even in this deeper "where does $0^0 = 1$ break down?" type question if it wasn't some sort of duplicate.
Dec
14
revised Is there any good reason not to define $0^0=1$ , such as contradictions in algebra or arithmetic?
deleted 481 characters in body
Dec
14
answered Is there any good reason not to define $0^0=1$ , such as contradictions in algebra or arithmetic?
Dec
13
comment How can I formalise a proof of this?
Yes. We can associate any ternary representation with 2's and 0's to a sequence of 1's and 0's, or equivalently, a subset of the natural numbers. (The number $n$ is in the associated subset if there is a 2 in the $n$-th place.) However, unlike the Cantor set, we'll be omitting anything that ends in an infinite string of 0's or 2's. The left-over set subtracts a countable set from the Cantor set. Hence, it is uncountable.
Dec
13
comment How can I formalise a proof of this?
I avoided writing it up as an answer because it's somewhat hard to formalize the answer :) With effort it can be made clean.
Dec
13
comment How can I formalise a proof of this?
Though this isn't the complement of the Cantor set, it does seem to be understandable through the ternary representation of numbers deleted from the Cantor set. In constructing the Cantor set, we delete any number that contains a 1 in its ternary representation. But here, all we're doing is deleting anything with a 1 and anything that terminates in a string of 0's at the end, e.g. 0.20000... Thus, I contend that 3/4 = 0.202020... in ternary is a counterexample for a = 0 and b = 1.