python this is connected to the last problem - the second part of the question is added. my attempt on this problem shows that the part b) (approaches N*NH) is not really wokring..  a) (answered) with a function “harmonic(n)” that computes the n-th harmonic number, write a function “harmonic_all(n)” that returns the number of values generated until all values are obtained as a function of the range of possible values n, then write a function “harmonic_sim(n)” that repeats “harmonic_all(n)” a total of n_sim = 100 times. (Attaching the code from the answer for a)    d) Show that as n increases (e.g., with a doubling experiment), from n = 2 to n_max = 1,000, the value of “coupon_sim(n)” approaches “n * Hn”.

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this is connected to the last problem - the second part of the question is added. my attempt on this problem shows that the part b) (approaches N*NH) is not really wokring.. 

a) (answered) with a function “harmonic(n)” that computes the n-th harmonic number, write a function “harmonic_all(n)” that returns the number of values generated until all values are obtained as a function of the range of possible values n, then write a function “harmonic_sim(n)” that repeats “harmonic_all(n)” a total of n_sim = 100 times.

(Attaching the code from the answer for a) 

 

d) Show that as n increases (e.g., with a doubling experiment), from n = 2 to n_max = 1,000, the value of “coupon_sim(n)” approaches “n * Hn”.

: def harmonic(n) :
harmonic = 1.00
c = 0
for i in range (2, n + 1) :
harmonic += 1 / i
C = c + 1
return round (harmonic,2),c
# Implementation of harmonic_all(n)
def harmonic_all(n):
c = 0
for i in range (2, n+1):
c = c + harmonic(i) [1]
return c
# Implementation of harmonic_sim(n)
def harmonic_sim(n):
c = 0
for i in range (2, n+1):
c = c + coupon(i)
return c
# Testcase
print('harmonic_sim(100):',harmonic_sim(100))
coupon_sim(100): 166650
Transcribed Image Text:: def harmonic(n) : harmonic = 1.00 c = 0 for i in range (2, n + 1) : harmonic += 1 / i C = c + 1 return round (harmonic,2),c # Implementation of harmonic_all(n) def harmonic_all(n): c = 0 for i in range (2, n+1): c = c + harmonic(i) [1] return c # Implementation of harmonic_sim(n) def harmonic_sim(n): c = 0 for i in range (2, n+1): c = c + coupon(i) return c # Testcase print('harmonic_sim(100):',harmonic_sim(100)) coupon_sim(100): 166650
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