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2h
comment Can I solve this integral with a squared sum in it?
Yes, it's possible... it's a sum of exponentials of negative definite quadratics. Each term is a Gaussian integral.
Aug
25
comment Help find the MacLaurin series for $\frac{1}{e^x+1}$
The series $1/(1+y)=1-y+y^2-y^3+y^4-\ldots$ only converges when $|y|<1$, and it converges fastest when $|y|$ is close to zero. So applying it to $1/(1+e^x)$ (i.e., setting $y=e^x$), where $x$ is close to zero, is not a good idea. Since $e^x$ is close to $1$ when $x$ is close to zero, that series converges slowly, if at all, for small $x$.
Aug
19
awarded  real-analysis
Aug
12
answered Why does $\sqrt{3x} \left(\dfrac {x}{2} \right)^x$ approximate $x!$ pretty well?
Aug
11
answered Is ths FOL structure necessarily infinite?
Aug
9
answered Infinite expected value?
Aug
9
comment Infinite expected value?
Intuitively it seems like every day the kid adds $0.6$ days to his sentence on average; i.e., he stretches out his grounding by a factor of $5/2$. So I'm guessing the EV should be $10-20$ days?
Aug
8
comment Average of a list of numbers, read one at a time
If you know how many numbers are in the set already, then you can update the average as follows: $\mu \rightarrow (n \mu + x) / (n + 1)$. This works because $n\mu$ is the old sum, $n\mu + x$ is the new sum, and so $n\mu + x$ divided by $n+1$ is the new mean.
Aug
7
comment Smallest number consisting of only ones and zeroes, divisible by a given number.
Yes... note that if $X$ and $Y$ are equal to $i$ modulo $N$, then $XZ$ and $YZ$ are also equal modulo $N$, where $Z$ is any string of additional digits. So you can safely ignore the larger of $X$ and $Y$; i.e., you don't need to visit the state $i$ more than once.
Aug
7
comment Is $\exists x(P(x)\rightarrow\forall y P(y))$ a tautology?
There is a man such that if he is awake, all men are awake.
Aug
6
answered Find the value of this infinite term
Aug
6
comment Where does this argument showing there are uncountably many TMs fail?
"Construct a Cantor TM which runs every $n$-th TM up to the $n$-th bit and outputs the reversed bit." How? How would you construct this Cantor TM that knows how to simulate each of the TMs on a given infinite list? If you try to make this part precise, you'll see that you can't do it. Note that for the proof by contradiction to show that there are uncountably many TMs (of a particular type), the initial list has to be arbitrary (except for the restriction to that type)... it's not enough to choose a particular list, like TMs generating 1, 1/2, 1/3, etc.
Jul
14
comment Can I have a logical explanation for why this number is so ridiculously close to a whole number?
It's more impressive that it's very close to a whole number that is very close to a perfect cube.
Jul
13
answered An “apparent” contradiction for eigenvalues signs of $A=\left( \begin{array}{cc} a & -a \\ a-1 & 1-a \\ \end{array} \right)$.
Jul
9
comment Is it possible to have random number chain relation?
You're adding a bunch of stuff to $I$ to get $38$, and subtracting the same bunch of stuff from $I$ to get $2$. That is, $I+x=38$ and $I-x=2$, where $x=5+7+2+4$ (the sum of your four random numbers). From this you can deduce that $I=20$ and that $x=18$, but that's it... there's no way to break $x$ back into its components.
Jul
9
revised Is it possible to have random number chain relation?
deleted 3 characters in body
Jul
7
comment Does the Russell Set exist?
If you have a predicate $\phi(x)$ that is true or false for every set, then $\{x : \phi(x) \}$ is a class (and may or may not also be a set). The "universal set" (really, a proper class) comes from letting $\phi(x)$ be true for all $x$. There's really no question as to whether a class "exists"... if $\phi(x)$ is a well-formed formula in the language of set theory, then the associated class is meaningful, and moreover can be intersected with sets to yield sets.
Jul
7
revised Another kind of derangement?
added 494 characters in body
Jul
7
answered Another kind of derangement?
Jul
7
comment Another kind of derangement?
Agreed. If the first reading is correct, then it's irrelevant how big the teams are... that's the only reason I think the second reading may be intended. But OP is going to have to weigh in.