Hi
I’m working on a python script which uses a cascaded triangular arbitrash in order to multiply money.
But so far the calculations seem to be all unsatisfying.
What I’m doing so far is fetching and filtering and interpolating the values in order to estimate the trade development:
#!/usr/bin/python
import sys
import os
import copy
import math
# delays:
from time import sleep
# web api
from Cryptsy import Api
#plotting the stuff
import matplotlib
matplotlib.use("TkAgg")
from matplotlib.pyplot import plot
from matplotlib.pyplot import figure
from matplotlib.pyplot import close
from matplotlib.backends.backend_tkagg import FigureCanvasTkAgg, NavigationToolbar2TkAgg
from IPython import display
#numpy stuff:
from numpy import delete
from numpy import array
from numpy import linspace
from numpy import polyfit
from numpy import poly1d
from numpy import newaxis
from numpy import ones
#scipy stuff
from scipy.signal import wiener
from scipy.optimize import curve_fit
from scipy.ndimage.filters import convolve1d
from scipy.interpolate import interp1d
from scipy.interpolate import InterpolatedUnivariateSpline
from scipy.signal import wiener
from scipy.signal import gaussian
from scipy.signal import savgol_filter
#sklearn
from sklearn.gaussian_process import GaussianProcess
from sklearn.linear_model import LinearRegression
from sklearn.isotonic import IsotonicRegression
from sklearn.cross_validation import cross_val_predict
from sklearn.preprocessing import PolynomialFeatures
from sklearn.pipeline import Pipeline
from sklearn.utils import check_random_state
#pyqt fit
from pyqt_fit import npr_methods
import pyqt_fit.nonparam_regression as smooth
A friend asked me a question on how feasible it would be to use gamma rays as an energy source. So let’s crunch some numbers: The wavelength of gamma rays is
\[\lambda=\frac{c}{f}=0.005 nm\]
Since the optimum length l for an antenna is
\[l=\frac{\lambda}{4}=0.00125 nm=1.25 pm\]
We are a bit in trouble here since the radius of the atom Sodium for instance is approximately 160 pm.
Physically it’s just not possible to build such a small antenna.
Today… Let’s just say: I’ve found the StarBucks.
It’s exactly as the StarBucks in Switzerland. I mean exactly the same.
Only that an Americano middle sized cup (for take away) is only 3.50 CHFr and not 5.00 CHFr.! ^_^
Agricultural paradise
Well… I didn’t find it in time… Also as far as I could see… It’s not really as interesting as some travelling blogs have described it.
Lovers road
Well. I didn’t go straight there either because I was stuck at the bubble tea and steamed bread corner where I’ve found fresh Manju (with red bean stuffing! *_*) as you can find a recipe for on my blog.
Also I’ve found Banh Bao… After this dinner I was so stuffed that I decided to roll back home.
Ahh! I talked Chinese a lot more today ^_^
Actually just one of so many parks in 珠海. I passed by this place on my way to the 客车68号 (Bus number 68) to the office of Allwinner Technology.
My way to Allwinner Technology
Since the security guard at the entrance already looked a bit confused when I sat down and tried to make clear to him with my dictionary that I was two hours too early for the meeting I’ve let it be to make pictures of the entrance.
Just be told, that the company has a beautiful modern new building now with leather seats and mings vases and two big red Chinese lanterns at the entrance.
A beautiful office building! *_*
Dinner (sponsored by Allwinner)
Best food of my life (2.0)! ^^
我不想回来瑞士! (I don’t wanna go back to Switzerland) *_*
Ah and btw, maybe I don’t have to.
I seems as if the dinner has brought up some additional business ideas which would justify my employment at their company =D
Thanks to the Russian duty and tax office my clothes are still all in Moscow, which forced me to buy new ones at the stores around the corner (first pic). Actually it’s not such a problem I guess, I think I look quiet good in these 0.30$ clothes ^^ (second pic)
Food
After having had some trouble making myself understood I managed to get my favourite food after all.
Given, a complicated darlington circuitry:
We now replace the darlington pairs with single NPN transistors by defining: $$I_C=I_B \cdot a$$ with $$a=(\beta+1)\cdot\beta$$ and $$I_E=I_B \cdot b$$ with $$b=(\beta+1)^2$$
Now we set:
$$I_{KD} = I_{B1} + I_{C3} = I_{B1} + I_{B3} \cdot a$$ and $$I_{RL}=I_{C4}=I_{B4}\cdot a$$ and $$I_{B2}=I_{B5}$$ and $$I_{KD} + I_{C1} + I_{RL} = I_{E5} + I_{E2} $$
$$I_{KD} + I_{B1} \cdot a + I_{RL} = I_{B5} \cdot b + I_{B2} \cdot b = 2\cdot I_{B5}\cdot b = 2\cdot I_{B2}\cdot b$$
$$\Rightarrow I_{RL}=- a \cdot I_{KD} – I_{KD} + a^2 \cdot I_{B3} + 2 \cdot b \cdot I_{B2}$$
And resolve the current equation:
$$I_{B3} \cdot b = I_{E3} = I_{C5}+I_{B2}+I_{B5}$$
$$=I_{C5}+2 \cdot I_{B2}$$
$$=I_{C5}+2 \cdot I_{B5}$$
$$=a \cdot I_{B5}+2 \cdot I_{B5}=(a+2)\cdot I_{B5}=(a+2)\cdot I_{B2}$$
