/sci/ - Science & Math

I've heard about blockchain a few years ago when my network tech interest friend got into Dogecoin and then again when /biz/ was created here. The hurdle to make an account on an exchange turned me off and the idea that it's too late was also lingering in my mind. Now after coming back to a normal routine after finishing my PhD in May I couldn't miss the news about Ethereum blowing up.

So I eventually signed up and quickly made tens of k gains, which drew me more in, obviously.

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I'm currently toying around with a bot who makes some money on Bittrex and I've also downloaded the NEO testnet block headers and will support the guy who writes the Python software developement kit on git. If you're interested in either of those two, I'm happy for collaborators. I have a friend who studies math and stats and I'm currently evaluating the ticker data, running moving avarage algorithms on it and so on. The buy and sell conditions are tricky to improve, but it's real math, which is nice.

Here the API which I too use in my routine
https://github.com/ericsomdahl/python-bittrex
Here the western NEO dev community, which are pretty cool folks
https://github.com/CityOfZion
You find the Slack chat at the bottom of this page
https://cityofzion.io/

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Regarding opinions, I have a lot, and I surely a better technical few on things than /biz/

First off, it must be noted that it's all currently unregulated (taxes and IPC (aka ICO) regulations suck balls in the US, though, which is thankfully not where I live) and low market cap (money involved), which is both good. I think even moreso than in the dot-com area, it's an unprecedented situation in that every 14 year old can make an account and start trading stocks, essentially. Both play into it being extremely volatile, and this is also the reason I can make money with a bot I only scripted 2 weeks ago. It will not stay like this forever.
The global situation is interesting.

cont.

finally got the PhD but now I can't keep up with my projects.

It's of course related to doing a 40 hour job now, but there's also social media I think I should do to break out of basic wage life, and that crypto shit which magically seems to be a money machine. I try to combine that with dependently typed memes as good as I can, but I must somehow externalize some capacities. Maybe I'll find a schedule that works for me and then I'd like to do time series and probability, as this relates to everything.

This may a little bit opaque, but maybe it's of some value.

All questions of "why the function E(x):=e^x" come down to
>because it's the group homomorphism between the two operations in the ring:
E(x+y) = E(x) · E[y]

Consider the collection of all complex numbers of norm 1.
This is e^{iu} for u in [0, 2pi].
It's a representation of U(1), which lives in the category of locally compact abelian groups.

For any group G, the set Hom(G,U(1)) inherits a group structure from U(1). (If you can't see what this means, ask.)
This is called the dual group and its elements are called "characters" X.

For example, consider R with addition. We'll see that this group is actually self-dual. Fix a real number p and then e.g. [math] x \mapsto e^{ipx} [/math] is an element [math] X_p [/math] of Hom(R,U(1)). It's a homomorphism, because [math] e^{ipa} · e^{ipb}= e^{ip(a+b)} [/math] . We can use the multiplication of phases in U(1) to define a group structure * on these X's by
[math]X_p*X_q \equiv (x \mapsto e^{ipx}) · (x\mapsto e^{iqx}) := x \mapsto e^{i(p+q)x} [/math].
We have a X_p for all real numbers p and so, in this case, as promised, we find Hom(R,U(1)) = R.
(That "new" R from constructing the dual is the domain of Fourier space!)

Now since we see Hom(-,U(1)) maps groups to groups, we can consider it as an auto-functor in the category of locally compact abelian groups.
It now turns out that the twice applied functor [math] Hom(Hom(-,U(1)),U(1)) [/math] is naturally isomorphic to the identity functor. Actually: If X in Hom(G,U(1)), then we can map g\in G to [math]g\mapsto (X\mapsto X(g))[/math].

Since locally compact groups have the Haar measure, one can form the space of integrable functions G -> C and carry it along with these functors. This procedure is the Fourier transform and it really works for all groups in this category! E.g. for U(1) itself, i.e. the periodic interval, we find that the dual group is Z and the associated transform is the Fourier series.

This may a little bit opaque, but maybe it's of some value.

All questions of "why the function E(x):=e^x" come down to
>because it's the group homomorphism between the two operations in the ring:
E(x+y) = E(x) · E[y]

Consider the collection of all complex numbers of norm 1.
This is e^{iu} for u in [0, 2pi].
It's a representation of U(1), which lives in the category of locally compact abelian groups.

For any group G, the set Hom(G,U(1)) inherits a group structure from U(1). (If you can't see what this means, ask.)
This is called the dual group and its elements are called "characters" X.

For example, consider R with addition. We'll see that this group is actually self-dual. Fix a real number p and then e.g. [math] x \mapsto e^{ipx} [/math] is an element [math] X_p [/math] of Hom(R,U(1)). It's a homomorphism, because e^{ipa} · e^{ipb}= e^{ip(a+b)}. We can use the multiplication of phases in U(1) to define a group structure * on these X's by
[math] X_p * X_q \equiv (x \mapsto e^{ipx}) · (x\mapsto e^{iqx}) := x \mapsto e^{i(p+q)x} [/math].
We have a X_p for all real numbers p and so, in this case, as promised, we find Hom(R,U(1)) = R.
(That "new" R from constructing the dual is the domain of Fourier space!)

Now since we see Hom(-,U(1)) maps groups to groups, we can consider it as an auto-functor in the category of locally compact abelian groups.
It now turns out that the twice applied functor [math] Hom(Hom(-,U(1)),U(1)) [/math] is naturally isomorphic to the identity functor. Actually: If X in Hom(G,U(1)), then we can map g\in G to [math] g\mapsto (X\mapstoX(g)) [/math].

Since locally compact groups have the Haar measure, one can form the space of integrable functions G -> C and carry it along with these functors. This procedure is the Fourier transform and it really works for all groups in this category! E.g. for U(1) itself, i.e. the periodic interval, we find that the dual group is Z and the associated transform is the Fourier series.

>>8418127
You do examples, try to seek a pattern, try to rewrite it or connect it to something already known.

For your sum, I'd first try to write down what you really want to compute.

It's

[math] \sum_{k=1}^n k^{n-(k-1)} [/math]

Sanity check: it's of the form a^b and for the lower bound k=1 you get 1^(n-(1-1)) = 1^n and for the upper bound k=n you get n^(n-(n-1)) = n^(-(-1)) = n.

The sum is also

[math] \sum_{k=1}^n k^{n+1-k} [/math]

or

[math] \sum_{k=1}^n k^{n+1} \cdot k^{-k} [/math]

Now just

[math] \sum_{k=1}^n k^{n+1} [/math]

is something well studied, namely under the name Generalized Harmonic numbers:
https://en.wikipedia.org/wiki/Harmonic_number#Generalization
That sum is [math] H_{n, \, -n-1} [/math]
It's some integer, but not trivial to compute.

What you have is something rather more complicated. Each term of the above sum, you weight with a small number 1/k^k. I.e. your sum is bound by that H, except the higher numbers are strongly suppressed.

The sum over 1/k^k alone is also known and goes by
https://en.wikipedia.org/wiki/Sophomore%27s_dream

Given the variations that are already soft or tricky, there will likely be no simple form for your sum.

I have colleges from school who live off of poker and one of my classmates is a millionaire. In fact, he was a millionaire at 17 even before he was legally allowed to have an account at the software platform he started playing for money.
It's not a cool job. But it makes more than most other jobs. It's a job, though.