Wish to solve exponential congruence equation

  • Thread starter jackmell
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In summary: I am happy to have been able to help you. Keep up the good work !In summary, the conversation discusses finding a value for ##n## in the equation $$e^{\frac{2n\pi i p}{q}}=e^{-\pi i p/q},\quad n=1,2,\cdots, q-1, \quad (p,q)\in \mathbb{N}\backslash 0$$ and how this relates to computing the analytically-continuous branch across a multi-valued function. It is shown that there can only be one solution for ##n## and this can be found using the formula $$n+1=k(q/p)$$ where ##k=p## and the ending
  • #1
jackmell
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Hi,

Is it possible to find a value of ##n## for the following expression other than by exhaustive search?

$$e^{\frac{2n\pi i p}{q}}=e^{-\pi i p/q},\quad n=1,2,\cdots, q-1, \quad (p,q)\in \mathbb{N}\backslash 0$$

I can write a short program to search for the value of n, but that would be infeasible if p and q were millions of digits long.
 
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  • #2
Your question is confusing. You ask for values of n, but you say n = 1,2,...,q-1. What are you looking for?
 
  • #3
mathman said:
Your question is confusing. You ask for values of n, but you say n = 1,2,...,q-1. What are you looking for?

Hi mathman. Sorry for the confusion. I'm looking for a positive integer ##n## in the range of one to ##q-1## which satisfies the equation. There should be only one value of n in that range which does this. This problem is in regards to computing the analytically-continuous branch across a multi-valued function.

For example, consider the function ##w=z^{-3/4}(1-z)^{-7/5}## which produces the equation for unknown n:

$$ e^{-14 n \pi i/5}=e^{14 \pi i/5}=e^{4\pi i/5}$$

such that n can be only 1, 2, 3, or 4.

Now:

For n=1:

$$e^{-14 \pi i/5}\neq e^{4\pi i/5}$$

For n=2:

$$e^{-28 \pi i/5}\neq e^{4\pi i/5}$$

For n=3:

$$e^{-42\pi i/5}\neq e^{4\pi i/5}$$

For n=4:

$$e^{-56\pi i/5}=e^{-6\pi i/5}=e^{4\pi i/5}$$

so n=4 is the solution. But that is my problem: is this the only way to solve these kinds of equations? Do I need to go through them one by one until I find the correct n or is there an algorithm or function that I could simply compute to find the answer?
 
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  • #4
Hi jackmell !
The relationship given in your first post :
exp(2*n*pi*i*p/q) = exp(-pi*i*p/q)
is not consistent with the exemple given in your second post :
exp(-14*n*pi*i/5) = exp(14*pi*i/5)
If the exemple is correct, then the first equation should be :
exp(n*pi*i*p/q) = exp(-pi*i*p/q)
Where is the mistake ?

Anyway, it seems rather easy to solve the problem :
if exp(2*n*pi*i*p/q) = exp(-pi*i*p/q) is correct then
exp(2*n*pi*i*p/q) = exp(-pi*i*p/q + 2*k*pi*i)
where k is any positive or negative integer.
2*n*pi*i*p/q = -pi*i*p/q + 2*k*pi*i
2*n = -1 + 2*k*q/p
Depending on p and q, there are some conditions for (-1 + 2*k*q/p) to be even. If these conditions are fullfiled, one can find the formula leading to couples (k, n).
 
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  • #5
JJacquelin said:
Hi jackmell !
The relationship given in your first post :
exp(2*n*pi*i*p/q) = exp(-pi*i*p/q)
is not consistent with the exemple given in your second post :
exp(-14*n*pi*i/5) = exp(14*pi*i/5)
If the exemple is correct, then the first equation should be :
exp(n*pi*i*p/q) = exp(-pi*i*p/q)
Where is the mistake ?

Anyway, it seems rather easy to solve the problem :
if exp(2*n*pi*i*p/q) = exp(-pi*i*p/q) is correct then
exp(2*n*pi*i*p/q) = exp(-pi*i*p/q + 2*k*pi*i)
where k is any positive or negative integer.
2*n*pi*i*p/q = -pi*i*p/q + 2*k*pi*i
2*n = -1 + 2*k*q/p
Depending on p and q, there are some conditions for (-1 + 2*k*q/p) to be even. If these conditions are fullfiled, one can find the formula leading to couples (k, n).

. . . Jesus, then I'm no better then those in the homework forum I criticize for not being clear. Hope they don't see this. Ok, let me try to be precisely clear and I think from your work JJacquelin, I have found a solution to the problem!

I wish to determine for a particular starting branch of the function ##w=z^{r/s}(1-z)^{p/q}##, which branch I will land on when I make one complete revolution around the singular point ##z=1##. If I begin on the principal branch at say ##z=1/2##, then by analyzing the argument change of the function around this singular point, I arrive at the expression:

$$e^{p/q i(\pi+2n\pi)}=e^{p/qi(-\pi)}$$
or:
$$e^{2n\pi i p/q}=e^{-2\pi ip/q}$$

Now following what you said above:

$$e^{2n\pi i p/q}=e^{-2\pi ip/q+2k\pi i},\quad n=1,2,\cdots,q-1 \quad \text{and }k\in \mathbb{Z}$$

or:

$$2\pi i(np/q)=2\pi i(k-p/q)$$

$$ np=kq-p$$
$$ (n+1)=k(q/p)$$

and this is the critical point: Since I know beforehand that there can be only one solution to this problem, one n only, I can always let ##k=p## which means the ending branch will always be ##q-1## and this is consistent with the experimental results above as well as other specific cases I've analyzed: the ending branch was always the last one. And likely by a similar argument, characterize the branching geometry likewise around the origin thereby completely characterizing the analytic continuation of the beta function via the pochhammer contour for rational exponents in that thread I'm working on in the Topology forum. Really a beautiful problem in my opinion guys if anyone reading this is interested and just so you know, it's really a struggle for me and I appreciate the help. :)

Thanks a bunch JJacquelin!
 
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  • #6
Don't mention it !
So, were was only a "2" missing in the equation. After correcting it, all's well that ends well.
Sure, it's an interresting problem, especially with the explanation of the context.
 

1. What is an exponential congruence equation?

An exponential congruence equation is a type of mathematical equation that involves an exponential function and a modular arithmetic statement. It is typically written in the form of a^x ≡ b (mod m), where a and b are integers, x is the variable, and m is the modulus.

2. How do you solve an exponential congruence equation?

To solve an exponential congruence equation, you can use the properties of modular arithmetic and algorithms such as the Chinese Remainder Theorem or Euler's Theorem. These methods involve manipulating the equation to reduce it to a simpler form that can be solved using basic algebraic techniques.

3. What are some common applications of exponential congruence equations?

Exponential congruence equations have many real-world applications, including cryptography, computer science, and engineering. They are also used in number theory to study the properties of integers and their relationships with prime numbers.

4. Can an exponential congruence equation have more than one solution?

Yes, an exponential congruence equation can have multiple solutions. This is because modular arithmetic involves finding the remainders of a number when divided by a modulus, and there can be multiple numbers that have the same remainder when divided by a specific modulus.

5. Are there any special cases of exponential congruence equations?

Yes, there are two special cases of exponential congruence equations: linear congruence equations and quadratic congruence equations. Linear congruence equations are of the form ax ≡ b (mod m), while quadratic congruence equations are of the form ax^2 + bx + c ≡ 0 (mod m).

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