-] Problem 5 Full Width at Half Maximum(FWHM). Distribution width is a very important quantity. It tells you how wide a the distribution is. One way to do that is to use a quantity called full-width half max This is the width of the distribution at half maximum. For example for distribution shown below the maximum is at x=0.0 and y=1.0 The maximum of this distribution is 1.0. So we are interested in the width of the distribution when the value of the distribution is 1.0/2=0.5 The width of the distribution is shown in black line. In this case it's about 2.35 import math def demo(x): 1 2 3 4 x = np.linspace (-3,3) 5 return math.exp(-x**2/2.0) y = [demo(xx) for xx in x] 6 plt.plot(x,y) 7 plt.axhline (0.5, color='red') 8 9 plt.axvline (2.355/2,color='red',linestyle= 'dashed') 10 plt.axvline (-2.355/2, color='red', linestyle= 'dashed') 11 12 plt.grid() 13 plt.annotate( 14 15 16 '', xy=(-2.355/2, 0.4), xycoords='data', xytext=(2.355/2, 0.4), textcoords='data', arrowprops={'arrowstyle': '<->'}) 17 plt.text(0,0.3, 'FWHM',horizontalalignment ='center') 10- 0.8 0.6 0.4 0.2 0.0 -2 FWHM ✓ ✓ 5.1)Find FWHM of the following function. Make sure the bound on error of the FWHM is less than 10-4. 1 import math mu = 2.345 return gamma/((x-mu)**2+gamma**2) 2 def f(x, gamma=3.1): 3 4 5 6 y = [f(xx) for xx in x] x = np.linspace (-10,10,200) [25] 7 plt.plot(x,y) ... ... [] 0.30- 0.25 0.20 0.15 0.10 0.05- -10.0 -7.5 -5.0 -2.5 0.0 2.5 5.0 7.5 10.0 5.2) Plot FWHM(y-axis) as a function of (gamma) from y = 0.5... 3. Make sure you have at least 30 points. 1 print(f(1.0, gamma-0.2)) # you may find this useful [26] ... 0.108165114047 [ ] 5.3) (Optional) Prove the relation you found in 5.2) 1

def findOccurrences(s, ch):     lst = []     for i in range(0, len(s)):         if a==s[i]:             lst.append(i)     return lst Use the code above instead of enumerate in the code posted below. n=int(input("Number of rounds of Hangman to be played:")) for i in range(0,n):     word = input("welcome!")     guesses = ''     turns = int(input("Enter the number of failed attempts allowed:")) def hangman(word):            secrete_word = "-" * len(word)     print(" the secrete word " + secrete_word)         user_input = input("Guess a letter: ")         if user_input in word:         occurences = findOccurrences(word, user_input)         for index in occurences:             secrete_word = secrete_word[:index] + user_input + secrete_word[index + 1:]         print(secrete_word)     else:         user_input = input("Sorry that letter was not found, please try again: ")   def findOccurrences(s, ch):     return [i for i, letter in enumerate(s) if letter == ch]  *** enumerate not discussed in…

2. Problem 2: In this problem you will test our "in class" version of Gaussian elimination against the version in numpy. Use both codes to solve some randomly chosen linear systems of size N 10, 20, 40, 80, 160, 320, 640, 1280, 2560, and 5120. Depending on your computer, the last two may or may not work. If they don't, just say so in your report. If you computer can do N- 10240 then try this as well. Note that in cach case N 10 * 2k for k = 1,..., 10. That is, each time we are doubling the size of the matrix. abone Plot the results as two curves. Our code in red and numPy in bluc. (Or whatever colors you prefer). Can you guess the function which governs the runtime? Based on this guess, how long would it take to do N 40960 using numPy? N=81920?

int sum, k, i, j; int x[4] [4]={1,2,3,4}, {5,6,7,8},{9,8,7,3},{2, 1,7,1}; sum=x[0] [0]; for (k=1; k<=3;k++) sum+=x[k] [k]; Give the value in sum after the statements are executed: