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)+i⋅sin(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=ex⋅cos(y)y=ex⋅sin(y)\begin{cases} { x=e }^{ x }\cdot cos(y) \\ y={ e }^{ x }\cdot sin(y) \end{cases}

Since there are no real solutions, y≠0y\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=y⋅cot(y)x=y\cdot cot(y)

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

y=ey⋅cot(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)=ey⋅cot(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.

#Algebra

Note by Nick Kent
1 year, 10 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 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 sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

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:

z≈0.3181±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 - 1 year, 10 months ago

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

Steven Chase - 1 year, 10 months ago
×

Problem Loading...

Note Loading...

Set Loading...