crntaylor
Joined 17 years ago
Joined 17 years ago
We only had 23 years of Python interpreter development,
how would things look like when Python is 42, like C?
C, which I always think of as an ancient venerable systems language, is less than twice as old as Python, which I think of as a hot new kid on the block.In thirty years, when my career will probably be drawing to a close, Python will be 53 years old and C will be 72 years old. Barely any difference at all.
As humour goes, it's a few levels of indirection away from Seinfeld. But you're on a forum full of people who spend all day thinking of abstractions for their abstractions, so what do you expect?
Answer quickly - how much of your net worth would you risk on a bet with 1% chance of a 1000X payout? Now how much would you risk if you can hand off 90% of any loss you take to someone else?
The difference is what we call moral hazard: http://en.wikipedia.org/wiki/Moral_hazard
A great read if you're interested in learning more about the history and operating procedures of the sales & trading side of investment banking is Traders, Guns and Money by Satyajit Das. Its sections on credit default swaps and collateralized debt obligations are particularly interesting when you consider that they were written in 2006, pre-crisis (around the same time that Leveraged Sell Out was getting started, in fact).
In particular, dijit asserted that "Not ACID => Not secure" (which is debatable, but that doesn't matter here) from which you can also validly deduce the contrapositive "Secure => ACID". However, you then (sarcastically) asserted that dijit is saying "ACID => Secure", whereas in fact he said nothing like that.
Source: I used to work in high frequency trading.
The reason it works is that 9998 = 10^4 - 2. You can expand as
1 / (10^n - 2) = 1/10^n * 1/(1 - 2/10^n)
= 1/10^n * (1 + 2/10^n + 2^2 /10^2n + 2^3 /10^3n + ...)
which gives the observed pattern. It breaks down when 2^k has more than n digits, which happens approximately when 2^k > 10^n => k > n log(10) / log(2)
which comes out to 4 * log(10)/log(2) = 13.28 when n = 4.---
Another pattern can be generated from the power series expansion
x / (1 - x)^2 = x + 2x^2 + 3x^3 + 4x^4 + ...
setting x = 1/10^n gives the infinite series 1/10^n + 2/10^2n + 3/10^3n + ...
which leads to the neat fact that 1 / 998001 = 0.000 001 002 003 004 005 006 007...
---Another example is the fraction
1000 / 997002999 = 0.000 001 003 006 010 015 021 ...
which goes through the triangle numbers[0] in its expansion, or 1 / 998999 = 0.000 001 001 002 003 005 008 013 021 ...
which goes through the Fibonacci numbers[1].---
Getting the squares is harder, but you can do it with
1001000 / 997002999 = 0.001 004 009 016 025 036 049 ...
[0] http://en.wikipedia.org/wiki/Triangle_number more (portable and secure)
rather than (more portable) and secure
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