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PDF Editor FAQ
Are there quick, efficient methods to do matrix multiplications in machine learning, like the way FFT is used to do efficient fourier transform in DSP?
NOTE This used to be a completely different question. This is BY FAR not the fastest - they were looking for FFT’s and Matrix math in pythonTrying to think of the correlation. So here is both.You get Add, Subtract and Divide only :) This is using Python and .# add(), subtract() and divide() # importing numpy for matrix operations import numpy # initializing matrices x = numpy.array([[1, 2], [4, 5]]) y = numpy.array([[7, 8], [9, 10]]) # using add() to add matrices print ("The element wise addition of matrix is : ") print (numpy.add(x,y)) # using subtract() to subtract matrices print ("The element wise subtraction of matrix is : ") print (numpy.subtract(x,y)) # using divide() to divide matrices print ("The element wise division of matrix is : ") print (numpy.divide(x,y)) And a FFT you can plot!%matplotlib inline import numpy as np import matplotlib.pyplot as plt import scipy.fftpack # Number of samplepoints N = 600 # sample spacing T = 1.0 / 800.0 x = np.linspace(0.0, N*T, N) y = np.sin(50.0 * 2.0*np.pi*x) + 0.5*np.sin(80.0 * 2.0*np.pi*x) yf = scipy.fftpack.fft(y) xf = np.linspace(0.0, 1.0/(2.0*T), N/2) fig, ax = plt.subplots() ax.plot(xf, 2.0/N * np.abs(yf[:N//2])) plt.show()
Why do some people still wear wristwatches?
Its just the awesome feeling that it gives ..Interestingly there is also a research study which tells that it is good to wear watchesIn an exploratory sample (N > 100) and a confirmatory sample (N > 600), we compared big-five personality traits between individuals who do or do not regularly wear a standard wristwatch. Significantly higher levels of conscientiousness were observed in participants who wore a watch. In a third study (N = 85), watch wearers arrived significantly earlier to appointments in comparison to controls.Source: Watch-wearing as a marker of conscientiousnessYou look awesome with a watch..Look even President Obama wears a watch :)
A .529 sample of gas occupies 125 mL at 60 cm of Hg and 25 C. What is the molar mass of the gas?
First use the ideal gas law to calculate the moles of the gas. Then divide the mass by the moles to get molar mass. I am going on the assumption that 0.529 is the mass in grams.Ideal gas lawPV = nRTP = pressure = 60 cmHg × 10 mmHg/1 cmHg = 600 mmHgV = 125 mL × 1 L/1000 mL = 0.125 Ln = mole = ? molR = gas constant = 62.3638 L·mmHg/K·molT = temperature (K) = 25°C + 273.15 = 298 KSolve for n.n = PV/RTn = (600 mmHg)(0.125 L)/(62.3638 L·mmHg/K·mol)(298 K) = 0.0040356 molmolar mass = 0.529 g/0.0040356 mol = 131 g/mol = 100 g/mol to one significant figure (600 mmHg)Note: If 0.529 is not grams, you need to convert it to grams.
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