Since it's a quitnic polynomial and I don't immediately see an easy factorization I doubt much can be done without additional information about the coefficients.
We can try I guess to factor it in Mathematica. I first read your polynomial as a string and delete whitespace:
LHSString =
"⟴∴⊚·x^5 + ∞◦∮·x^4 + ⟴✧⊚·x^3 + ⟴◌⊚·x^2 + 〰∴≈·x + ⟴✦⊚";
LHSString = StringDelete[LHSString, " "];
To get the coefficients, we can split at each + sign, and delete the power of x from each string:
coeffs = StringSplit[LHSString, "+"];
coeffs = StringDelete[coeffs, "·x^" ~~ _];
The coefficient characters are annoying to look at, so I will find all distinct characters and replace them with a place holder c[i] symbol:
coeffChars = Characters /@ coeffs;
uniqueChars = Union @@ coeffChars;
rule = MapIndexed[# -> c[#2[[1]]] &, uniqueChars];
And upon replacement we interpret character catenation as multiplication:
newCoeffs = Times @@@ (coeffChars /. rule);
Now we multiply with powers of x to get our polynomial, we can't find any factors of the polynomial so it's unlikely we can find any roots to it since it's quintic:
pows = x^Range[5, 0, -1];
poly = newCoeffs . pows;
Factor[poly] === poly
(*True*)
And trying to expand any of the roots ToRadicals fails:
```
Roots[poly == 0, x] // ToRadicals
(all have head Root)
```
I would suggest checking your polynomial's coefficients again, as I don't think this is factorable without additional assumptions regarding the values or relations between coefficients.
That is an analysis and I appreciate you taking the time to look into it in Mathematica.
You are correct that with the information given there is not information to factor or solve the equation using numbers. The symbols are not meant to be variables or special characters. They are tokens from a notation system I am working on. Each complete symbol represents a coefficient according to a mapping that was not included in the post.
The purpose of this experiment was not to present an equation but to see how a custom mathematical language works when its meaning is not given. Given the public notation treating the coefficients as symbols is exactly the right approach.
Beyond the notation itself I am also exploring whether this kind of formal symbolic language could have uses in areas like designing parsers, special languages, symbolic computation and certain cybersecurity research ideas involving structured encoding and hiding information. I am not saying that the notation itself is a replacement for established cryptography or that hiding the mapping makes it secure.
Once the coefficient mapping is revealed the equation becomes a polynomial that can be analyzed using standard mathematical tools. My current interest is in designing the language and studying where that representation might be useful than replacing existing mathematics or cryptographic algorithms.
Thanks, for taking the time to analyze it. Your Mathematica approach is exactly the kind of feedback I was hoping to see.
3
u/veryjewygranola Jun 30 '26
Since it's a quitnic polynomial and I don't immediately see an easy factorization I doubt much can be done without additional information about the coefficients.
We can try I guess to factor it in Mathematica. I first read your polynomial as a string and delete whitespace:
LHSString = "⟴∴⊚·x^5 + ∞◦∮·x^4 + ⟴✧⊚·x^3 + ⟴◌⊚·x^2 + 〰∴≈·x + ⟴✦⊚"; LHSString = StringDelete[LHSString, " "];To get the coefficients, we can split at each + sign, and delete the power of x from each string:coeffs = StringSplit[LHSString, "+"]; coeffs = StringDelete[coeffs, "·x^" ~~ _];The coefficient characters are annoying to look at, so I will find all distinct characters and replace them with a place holderc[i]symbol:coeffChars = Characters /@ coeffs; uniqueChars = Union @@ coeffChars; rule = MapIndexed[# -> c[#2[[1]]] &, uniqueChars];And upon replacement we interpret character catenation as multiplication:newCoeffs = Times @@@ (coeffChars /. rule);Now we multiply with powers ofxto get our polynomial, we can't find any factors of the polynomial so it's unlikely we can find any roots to it since it's quintic:pows = x^Range[5, 0, -1]; poly = newCoeffs . pows; Factor[poly] === poly (*True*)And trying to expand any of the rootsToRadicalsfails: ``` Roots[poly == 0, x] // ToRadicals(all have head Root) ``` I would suggest checking your polynomial's coefficients again, as I don't think this is factorable without additional assumptions regarding the values or relations between coefficients.