How to calculate the optimal zipf distribution of word frequencies in a text

for a homework assignment i have to plot the word frequencies of a text and compare it to an optimal zipf distribution.

Plotting the counted word frequencies of the text according to their rank in a log log graph seems to work fine.

But i have trubble with calculating the optimal zipf distribution. The result should look something like this:

I dont understand what the equation would look like to calculate the straight zipf line.

On the german wikipedia page of the zipf law I found an equation that seems to work

but there are no sources cited, so i dont understand where the constant of 1.78 comes from.

#tokenizes the file 
tokens = word_tokenize(raw)
tokensNLTK = Text(tokens)

#calculates the FreqDist of all words - all words in lower case
freq_list = FreqDist([w.lower() for w in tokensNLTK]).most_common()

#Data for X- and Y-Axis plot
values=[]
for item in (freq_list):
    value = (list(item)[1]) / len([w.lower() for w in tokensNLTK])
    values.append(value)

#graph of counted frequencies gets plotted
plt.yscale('log')
plt.xscale('log')
plt.plot(np.array(list(range(1, (len(values)+1)))), np.array(values))

#graph of optimal zipf distribution is plotted
optimal_zipf = 1/(np.array(list(range(1, (len(values)+1))))* np.log(1.78*len(values)))###1.78
plt.plot(np.array(list(range(1, (len(values)+1)))), optimal_zipf)
plt.show()

My results with this script look like this:

but i am just not sure if the optimal zipf distribution is calculated right. If so, shouldnt the optimal zipf distribution cross the X-axis at one point?

EDIT: if it helps, my text has 2440400 tokens and 27491 types

Take a look at this research paper by Andrew William Chisholm. Specifically page #22.

H(N) ≈ ln(N) + γ

Where γ is the Euler-Mascheroni constant with approximate value 0.57721. Noting that exp(γ) ≈ 1.78, equation <...> can be re-written to become for large N (N must be greater than 1,000 for this to be accurate to one part in a thousand).

pr ≈ 1 / [r*ln(1.78*N)]