Constant Exponent

The equation in question is: ex=x{ e }^{ x }=x. It would be interesting to find a constant point in a function like exponent. Obviously there are no real solutions. But what if we consider complex numbers, are there any constant exponents?

We are looking for a complex number zz such as ez=z{ e }^{ z }=z. Let z=x+iyz=x+iy, where xx - real part, yy - imaginary part. Following the expanded definition of the exponent:

ex+iy=ex(cos(y)+isin(y))=x+iy{ e }^{ x+iy }={ e }^{ x }\cdot (cos(y)+i\cdot sin(y))=x+iy

Thus we get a system of two equations:

{x=excos(y)y=exsin(y)\begin{cases} { x=e }^{ x }\cdot cos(y) \\ y={ e }^{ x }\cdot sin(y) \end{cases}

Since there are no real solutions, y0y\neq 0 which also means sin(y)0sin(y)\neq 0. That means we can safely divide by yy and express xx:

xy=cos(y)sin(y)\frac { x }{ y } =\frac { cos(y) }{ sin(y) }

x=ycot(y)x=y\cdot cot(y)

Using this expression we get an equation that should help us get the value of yy:

y=eycot(y)sin(y)y={ e }^{ y\cdot cot(y) }\cdot sin(y)

Unfortunately, I wasn't able to derive a concrete root of this equation. However, let's consider these two functions:

f(y)=ysin(y)g(y)=eycot(y)f(y)=\frac { y }{ sin(y) } \\ g(y)={ e }^{ y\cdot cot(y) }

Both are even functions. If we build their graphs we get:

(f(y)f(y) - green, g(y)g(y) - red)

As we can see there is an infinite number of intersections occurring roughly once in every 2π2\pi interval. That means there is an infinite number of fixed complex points who are equal to their exponent which I find fascinating.

If you have any idea how to derive zz or found an error, feel free to comment.

Note by Nick Kent
3 weeks, 1 day ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Interesting investigation. Starting with your system of two equations, I used multi-variate Newton Raphson iteration to find xx and yy .
Here is the zz value with the smallest magnitude. Of course, there are other solutions as well:

z0.3181±1.3372iz \approx 0.3181 \pm 1.3372 i

Python code is attached:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
import math
import numpy as np
import random

S = 5.0

# Initial guesses

x = -S + 2.0*S*random.random()
y = -S + 2.0*S*random.random()

# Iterative solution

for j in range(0,100):

    f1 = math.exp(x) * math.cos(y) - x   # functions from system
    f2 = math.exp(x) * math.sin(y) - y

    fvec = np.array([f1,f2])

    J11 = math.exp(x) * math.cos(y) - 1.0   # Jacobian matrix entries
    J12 = -math.exp(x) * math.sin(y)  

    J21 = math.exp(x) * math.sin(y)  
    J22 = math.exp(x) * math.cos(y) - 1.0 

    J = np.array([[J11,J12],[J21,J22]])
    vec = np.array([x,y])

    right = np.dot(J,vec) - fvec    

    Sol = np.linalg.solve(J, right)   

    x = Sol[0]
    y = Sol[1]

print x
print y
print ""

print ""
print "###############################"
print ""

Steven Chase - 3 weeks, 1 day ago

Log in to reply

@Nick Kent Would you like to post this as a problem?

Steven Chase - 3 weeks, 1 day ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...