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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”.

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”.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
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Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
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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”.

```python def harmonic(n): harmonic = 1.00 c = 0 for i in range(2, n + 1): harmonic += 1 / i 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 += harmonic(i)[1] return c # Implementation of harmonic_sim(n) def harmonic_sim(n): c = 0 for i in range(2, n + 1): c += coupon(i) return c # Testcase print('harmonic_sim(100):', harmonic_sim(100)) ``` **Explanation:** 1. **Function `harmonic(n)`:** - Initializes a variable `harmonic` to 1.00 to start calculating the harmonic number. - Initializes a counter `c` to 0. - Iterates from 2 to `n` and calculates the harmonic number by adding the reciprocal of each number to `harmonic`. - Increments the counter `c` in each iteration. - Returns the rounded harmonic number (rounded to 2 decimal places) and the counter `c`. 2. **Function `harmonic_all(n)`:** - Initializes a counter `c` to 0. - Iterates from 2 to `n` and adds the counter from `harmonic(i)` to `c`. - Returns the total count `c`. 3. **Function `harmonic_sim(n)`:** - Initializes a counter `c` to 0. - Iterates from 2 to `n` and adds the result from a function `coupon(i)` to `c`. - Returns the total count `c`. 4. **Test Case:** - Calls `harmonic_sim(100)` and prints the result. **Note:** The function `coupon(i)` is used in `harmonic_sim(n)`, but its implementation is not provided in this snippet.
Transcribed Image Text:```python def harmonic(n): harmonic = 1.00 c = 0 for i in range(2, n + 1): harmonic += 1 / i 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 += harmonic(i)[1] return c # Implementation of harmonic_sim(n) def harmonic_sim(n): c = 0 for i in range(2, n + 1): c += coupon(i) return c # Testcase print('harmonic_sim(100):', harmonic_sim(100)) ``` **Explanation:** 1. **Function `harmonic(n)`:** - Initializes a variable `harmonic` to 1.00 to start calculating the harmonic number. - Initializes a counter `c` to 0. - Iterates from 2 to `n` and calculates the harmonic number by adding the reciprocal of each number to `harmonic`. - Increments the counter `c` in each iteration. - Returns the rounded harmonic number (rounded to 2 decimal places) and the counter `c`. 2. **Function `harmonic_all(n)`:** - Initializes a counter `c` to 0. - Iterates from 2 to `n` and adds the counter from `harmonic(i)` to `c`. - Returns the total count `c`. 3. **Function `harmonic_sim(n)`:** - Initializes a counter `c` to 0. - Iterates from 2 to `n` and adds the result from a function `coupon(i)` to `c`. - Returns the total count `c`. 4. **Test Case:** - Calls `harmonic_sim(100)` and prints the result. **Note:** The function `coupon(i)` is used in `harmonic_sim(n)`, but its implementation is not provided in this snippet.
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  • Using the given set of values of n from 2 to 1000 :
    • show that harmonic_sim(n) approaches n*H(n).
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