okay, they're definitely there and let me tell you why. Because when we dream we effectively simulate these other worlds and take measurements of points in those abstract mathemetical spaces with our bwains!
If we could calculate far enough we could take measurements of any abstract world and show that there's life inside, if the world is conducive to life like us
@EnjoysMath what kind of capabilities are we missing that could allow is to get a glimpse of these alternate universes? could we ever physically reach them?
I'm so thankful to God that such amazingly beautiful things are possible. They give a meaning to my life :-) $$\sum_{n=1}^{\infty} \frac{\sin(n)}{n 2^n}=\frac{1}{2} + \arctan\left(3 \tan\left(\frac{1}{2}\right)\right)$$
@Victor yes, but if you always have to use the chain rule while differentiating, that should give you some information about what you'll always have to do on the other side
I think everyone in this channel is arguably a druggie because pure math is the ultimate mind altering substance for effectively abstracting your consciousness away from reality
@MickLH Maybe, but keep in mind it's also about giving a lot of pleasure, it's a way of eating a good quality chocolate with your mind (if I can say that). :-)
But I do not believe that makes the vast majority of us druggies, I see the math crowd specifically as such because they are interested in their drug, study it religiously, and optimize the mental effects of it at all costs
For $\prod_0^\infty (1+x^{2^n})$, although it's possible to multiply both sides by $1-x$, it seems nicer to solve it by noting that each nonnegative integer has a unique binary representation.
@DanielFischer Let $\varphi: \mathfrak A \rightarrow \mathfrak B$ be a $^*$-isomorphism (not necessarily isometric) and $A \in \mathfrak A$, and $\sigma(A)$ the spectrum of $A$. Then $\sigma(A) = \sigma(\varphi(A))$, correct?
Because if $A-\lambda I$ is invertible iff $\varphi(A-\lambda I) = \varphi(A) - \lambda(\varphi I)= \varphi(A) - \lambda I_{\mathfrak B}$ is (coming from the fact that $\varphi$ is a $\ast$-isomorphism)
(i.e., if the latter had an inverse, it would give me an inverse of $A - \lambda I$, and vice versa)
I don't see anything wrong with this argument but I don't trust myself to even the most obvious facts
At least twice, then, @DanielFischer. Thanks. I'm overly cautious when I'm working with stuff that's fundamentally new to me, since I'm prone to making errors, and I don't want my formative moments to give me the wrong ideas.
@Mike Being cautious is good. Asking for confirmation too. But when you looked over your argument a couple of times, and it still looks good, start thinking it is probably right.
I kind of like how much the author leaves to the reader in this book (C* algebras by example). It's not too much, but oftentimes his equalities leave some explanation to be desired; and it's not all too difficult to explain them.
Let our memory slots be represented by elements of $\Bbb{Z}_p$ for a prime $p$. $k$ memory slots would be $k$ copies of the ring: $R = (\Bbb{Z}_p)^k$. Suppose that for a problem $f : X \to Y$, there are $X \xrightarrow{\phi} R \xrightarrow{\psi} Y$, each map computable in $O(|x|)$ time for eac...
Any one could give a reference to the formula in an answer? "The series (in $\log n$ and $\log\log n$) of $\pi(n)/n$ is known, although I believe it's a divergent expansion. It's been known since the 18th century."
@Victor I'm not gonna look at it, but you might check the wikipedia and MathWorld articles for the prime number theorem - both certainly have better asymptotics than $n/\log n$
@Chris'ssis yes, it was very intuitive to do so because you choose $1/2^n$ and $1/2^{n+1}-/2^n$ happens to not depend on $n$!!! And it also happens that $\pi-x) /2$ has a nice Fourier series
@Chris'ssis $\zeta(n)$ is monotonically decreasing. $\frac1n$ is monotonically decreasing. $\sum\sin(n)$ is bounded. Use Dirichlet to show that it converges.
@DanielFischer Let me quote you: "I don't remember. I don't recall. I have no memory of anything at all." No, I don't remember Chris's opinion on the matter. :-)
I'm trying to make an algorithm for finding $log_a(b)$ where $a, b\in\mathbb{Q}$ as a value in $\mathbb{Q}$, if there's no such number, abort to an approximation algorithm.
Anyone knows an algorithm that can find answers in \mathbb{Q}, the one I found only finds answers in \mathbb{N}
factor $a$. the exponents in the prime factors will tell you which $n$th roots are rational. one such $n$ will be maximal; store $n$ and take the $n$th root: call it $r$. the perform your algorithm for $\log_r(b)$
if you don't want to factor your rationals then yolo
What algorithm should I implement, General number field sieve is only the fastest for numbers larger than 100 digits. What's the fastest for smaller numbers?
@Darksonn In terms of what's most efficient, I definitely don't know. The thing I described would probably end up getting you something in polynomial time.
For numbers that small? You might honestly just factor.
@DanielFischer I'm reviewing my notes and I see the claim that it is possible in a metric space for the unit open ball to have multiple centers and multiple radius (ie different centers/radius may define the same unit ball). Isee examples of different radius that yield the same ball (discrete metric) but I cannot find for different centers...
@GabrielR. In an ultrametric space $B_r(x) = B_r(y)$ for all $y\in B_r(x)$. If you take an ultrametric space where the set of possible distances has only $0$ as an accumulation point - say $\mathbb{Z}$ with a $p$-adic metric - you get both.
@Darksson Let's say you have an algorithm that checks if $a$ is a perfect $n$th power (for some fixed $n$). Use this algorithm for all $2 \leq n \leq \log_2(a)$. For whichever $n$ is greatest, take the $n$th root of $a$ call it $r$. Thn if $\log_a(b)$ is rational, then $\log_r(b)$ is an integer; so use your log algorithm that checks for integer powers.
Let our memory slots be represented by elements of $\Bbb{Z}_p$ for a prime $p$. $k$ memory slots would be $k$ copies of the ring: $R = (\Bbb{Z}_p)^k$. Suppose that for a problem $f : X \to Y$, there are $X \xrightarrow{\phi} R \xrightarrow{\psi} Y$, each map computable in $O(|x|)$ time for eac...
