Making the Imports

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from fractions import Fraction
import plotly.graph_objects as go

1.

First thing I need to show is that for a small number $x$, the value of $e^x$ is approximately equal to $$ e^x = 1+x $$ To prove this, we'll first show that $$ 10^x = 1+cx$$ here c is constant which as we'll see is a scale factor. Which will turn out to be $\ln{10}$. To get this, we'll start out with trying to get the value of $10^{x}$ for a number x. We'll start by calculating the square root of $10$ and then keep doing this.

#showing 8 digits
pd.set_option("display.precision", 8)
#creating x
power = np.arange(0, 20, 1)
x = 1/2**power
#creating 10^x
ten_x = 10**x
#creating the DataFrame
df1 = pd.DataFrame(data = [x, ten_x], index = ['x', '10^x']).transpose()
#calculating (10^x-1)/x
df1[r'c = (10^x-1)/x'] = df1['x'].apply(lambda x: (10**x-1)/x)

df1

	x	10^x	c = (10^x-1)/x
0	1.00000000	10.00000000	9.00000000
1	0.50000000	3.16227766	4.32455532
2	0.25000000	1.77827941	3.11311764
3	0.12500000	1.33352143	2.66817146
4	0.06250000	1.15478198	2.47651176
5	0.03125000	1.07460783	2.38745051
6	0.01562500	1.03663293	2.34450742
7	0.00781250	1.01815172	2.32342038
8	0.00390625	1.00903504	2.31297148
9	0.00195312	1.00450736	2.30777050
10	0.00097656	1.00225115	2.30517585
11	0.00048828	1.00112494	2.30387999
12	0.00024414	1.00056231	2.30323242
13	0.00012207	1.00028112	2.30290873
14	0.00006104	1.00014055	2.30274690
15	0.00003052	1.00007027	2.30266600
16	0.00001526	1.00003514	2.30262554
17	0.00000763	1.00001757	2.30260532
18	0.00000381	1.00000878	2.30259521
19	0.00000191	1.00000439	2.30259015

#log10
np.log(10)

2.302585092994046

As we can see, the value of $c$ is coming out to be about 2.30259. This is approximately the value of $\ln{10} =2.302585092994046 $. Using this, we see that for samll value of x: $$ e^x = 1 + x$$

2.

Now we shall proceed to prove tha Euler's equation: $$ e^{ix} = \cos(x) + i\sin(x) $$ The only assumption we need to make is that for small value of x, we must have $$ e^{ix} = 1 + ix$$

We shall start by using: $$ e^{i\frac{1}{1024}} = 1 + i\times \frac{1}{1024} $$ and follow the exact opposite method as done above. We'll multiply the above value to itself to get the next values.
To get more accurate value, we'll start from $x = \frac{1}{2^{15}}$

# power
power = np.arange(0, 16, 1)
#Setting the base complex number
base_num = np.complex(1,1/2**15)
x = 2**power
# p = Fraction((1/1024)*2**power).limit_denominator()
#Getting the complex numbers
e_ix = base_num**x

df2 = pd.DataFrame(data = [x, e_ix], index = ['x', 'e^x']).transpose()
df2['power'] = df2['x'].apply(lambda x: int(x.real))
df2.drop(columns = ['x'], inplace = True)
df2['x'] = df2['power'].apply(lambda x: Fraction((1/2**15)*x).limit_denominator())
df2=df2[['power', 'x', 'e^x']]

df2

	power	x	e^x
0	1	1/32768	1.00000000+0.00003052j
1	2	1/16384	1.00000000+0.00006104j
2	4	1/8192	0.99999999+0.00012207j
3	8	1/4096	0.99999997+0.00024414j
4	16	1/2048	0.99999989+0.00048828j
5	32	1/1024	0.99999954+0.00097656j
6	64	1/512	0.99999812+0.00195312j
7	128	1/256	0.99999243+0.00390624j
8	256	1/128	0.99996960+0.00781242j
9	512	1/64	0.99987817+0.01562437j
10	1024	1/32	0.99951224+0.03124493j
11	2048	1/16	0.99804846+0.06245938j
12	4096	1/8	0.99219956+0.12467497j
13	8192	1/4	0.96891612+0.24740490j
14	16384	1/2	0.87758926+0.47942920j
15	32768	1	0.54031055+0.84148382j

#Saving th csv file
numbers = df2.drop(['power'], axis = 1)
numbers['Real Part'] = df2['e^x'].apply(lambda x: x.real)
numbers['Imaginary Part'] = df2['e^x'].apply(lambda x: x.imag)
numbers.drop(['e^x'], axis = 1).to_csv('numbers.csv')

It is obvious from the table above that the real value of $e^{ix}$ is decreasing while the imaginary part is increasing. To see the effects more clearly, we'll use $e^{i/8}$ as base number and repeat the procedure above.

#Taking the base as 1/8
x = np.arange(0,80)
base_num = df2['e^x'][12]
e_x = base_num**x
df3 = pd.DataFrame(data = [x, e_x], index = ['x', 'e^x']).transpose()
df3['power'] = df3['x'].apply(lambda x: int(x.real))
df3.drop(columns = ['x'], inplace = True)
df3['x'] = df3['power'].apply(lambda x: Fraction((1/8)*x).limit_denominator())
df3=df3[['power', 'x', 'e^x']]

df3

	power	x	e^x
0	0	0	1.00000000+0.00000000j
1	1	1/8	0.99219956+0.12467497j
2	2	1/4	0.96891612+0.24740490j
3	3	3/8	0.93051295+0.36627462j
4	4	1/2	0.87758926+0.47942920j
...	...	...	...
75	75	75/8	-0.99890422+0.04976453j
76	76	19/2	-0.99731672-0.07516201j
77	77	77/8	-0.98016638-0.19891615j
78	78	39/4	-0.94772079-0.31956673j
79	79	79/8	-0.90048618-0.43523103j

80 rows × 3 columns

#Saving this dataframe
base_8 = df3.drop(['power'], axis = 1)
base_8['Real Part'] = df3['e^x'].apply(lambda x: x.real)
base_8['Imaginary Part'] = df3['e^x'].apply(lambda x: x.imag)
base_8.drop(['e^x'], axis = 1).to_csv('base_8.csv', index = False)

Plotting The Graphs

#Imaginary part
plt.figure(figsize=(8,6))
t = np.arange(0, len(df3), 1)
plt.scatter(t/8, df3['e^x'].values.imag, label='Imaginary Part')
plt.plot(t/8, np.sin(t/8), color='black', label='$\sinx$')
plt.plot(t/8, 0*t, '--', color='black', label='x-axis')
plt.xlabel('x', fontdict={'fontsize': 16})
plt.ylabel('Imaginary Part of $e^{ix}$', fontdict={'fontsize': 16})
plt.legend()
plt.ylim([-1,1])
plt.xlim([0,10]);

#Real part
plt.figure(figsize=(8,6))
t = np.arange(0, len(df3), 1)
plt.scatter(t/8, df3['e^x'].values.real, label='Real Part')
plt.plot(t/8, np.cos(t/8), color='black', label='$\cosx$')
plt.plot(t/8, 0*t, '--', color='black', label='x-axis')
plt.xlabel('x', fontdict={'fontsize': 16})
plt.ylabel('Real Part of $e^{ix}$', fontdict={'fontsize': 16})
plt.legend()
plt.ylim([-1,1])
plt.xlim([0,10]);

#Both the real and Imaginary Parts
plt.figure(figsize=(10,6))
t = np.arange(0, len(df3), 1)
plt.plot(t/8, df3['e^x'].values.imag, 'r', label = 'Imaginary')
plt.plot(t/8, df3['e^x'].values.real, 'b', label = 'Real')
plt.plot(t/8, 0*t, '--', color='black', label='x-axis')
plt.scatter(t/8, np.sin(t/8),label='$\sin(x)$', color = 'green')
plt.scatter(t/8, np.cos(t/8),color='black', label='$\cos(x)$')
plt.legend()
plt.ylim([-1,1])
plt.xlim([0,10]);

Creating an Interactive Plot

#creating the DataFrame
X= df3['power']/8
data  = pd.DataFrame(data = [X], index = ['x']).transpose()
data['real'] = df3['e^x'].values.real
data['imaginary'] = df3['e^x'].values.imag

data

	x	real	imaginary
0	0.000	1.00000000	0.00000000
1	0.125	0.99219956	0.12467497
2	0.250	0.96891612	0.24740490
3	0.375	0.93051295	0.36627462
4	0.500	0.87758926	0.47942920
...	...	...	...
75	9.375	-0.99890422	0.04976453
76	9.500	-0.99731672	-0.07516201
77	9.625	-0.98016638	-0.19891615
78	9.750	-0.94772079	-0.31956673
79	9.875	-0.90048618	-0.43523103

80 rows × 3 columns

#Making the plot
t = data['x']
y = np.cos(t)
y2= np.sin(t)
fig = go.Figure(data=go.Scatter(x=t, y=y,mode='lines', line_color='blue', name='cos(x)'))
fig.add_trace(go.Scatter(x=t, y=y2,mode='lines', line=dict(color="red"), name='sin(x)'))
fig.add_trace(go.Scatter(x=t, y=data['real'],mode='markers', line_color='red', name='Real Part'))
fig.add_trace(go.Scatter(x=t, y=data['imaginary'],mode='markers', line_color='blue', name='Imaginary Part'))
fig.update_layout(title="Plot of e^{ix}", xaxis_title='x', yaxis_title='e^{ix}')
fig.show()