Can single variable equations always be solved using basic math principles?

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Discussion Overview

The discussion revolves around whether all equations with a single variable can be solved using basic math principles. Participants explore the nature of solvability, particularly focusing on specific examples and the implications of higher-level mathematics.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions if all single-variable equations can be solved, citing a specific example: 5 = (x)/(1+x^7).
  • Another participant asserts that not all equations can be solved, providing the example y=xe^x, which cannot be expressed in terms of elementary functions.
  • There is a suggestion that polynomials up to degree four can be solved in closed form, while higher-degree polynomials may not be solvable in general, depending on the specific case.
  • Numerical methods, such as the Newton-Raphson method, are proposed as potential solutions for equations that cannot be solved analytically.
  • One participant introduces the concept of complex numbers, stating that they allow for solutions that are not real, and notes that any polynomial can be factored into first and second-order factors with real coefficients.
  • Another participant discusses the ProductLog function as a way to express solutions for certain equations, questioning what constitutes a "solved" equation.

Areas of Agreement / Disagreement

Participants express differing views on the solvability of single-variable equations, with no consensus reached on whether all such equations can be solved or under what conditions. The discussion includes multiple competing perspectives on the nature of solutions and the role of numerical methods.

Contextual Notes

The discussion highlights limitations in the definitions of solvability and the conditions under which solutions may or may not exist, particularly regarding the use of complex numbers and the applicability of higher-level mathematical concepts.

Fascheue
Can all equations with a single variable be solved?

5 = (x)/(1+x^7)

How can something like this be solved for example. Sometimes I have equations with one variable and I can’t see any way to solve them with the knowledge of math that I currently have. Is it just that a higher level of math is required to solves problems like the one above or am I missing something?
 
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Fascheue said:
Can all equations with a single variable be solved?
No. ##y=xe^x## for example can't.
5 = (x)/(1+x^7)
Not sure about this one. Polynomials up to degree four can be solved in a closed form. Equations of higher degrees like this one can only sometimes be solved, depending on the given case. I haven't checked this one though.
How can something like this be solved for example.
Numerically by algorithms like the Newton-Raphson method.
Sometimes I have equations with one variable and I can’t see any way to solve them with the knowledge of math that I currently have. Is it just that a higher level of math is required to solves problems like the one above or am I missing something?
It is a bit of high level mathematics (Galois theory) to see why from degree five on, polynomials cannot be generally solved anymore, and it is not quite trivial to check a certain polynomial.
 
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It depends what you mean by "solved". If you mean, "write down the answer in terms of elementary functions", then the answer is no, as fresh_42 said. But let's take his example of y=x \exp(x). The solution is x=W_n(y), where Wn is the ProductLog function. This is a tabulated function where you can look up the values, so in this sense it is no different from sin(x). So do you consider the problem "solved"? If by "solved" you mean "find the number(s) for which the equation is satisfied to a sufficient degreee of accuracy", then the solutions for any equation of this type can be solved by numerical methods like Newton-Raphson, if a solution exists.
 
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Are you allowing complex numbers? Complex numbers allow solutions that are not real. Any polynomial can be factored to factors of order 1 and 2 with real numbers. The order 1 factors give a solution and order 2 factors give 2 conjugate complex solutions. Your example can be manipulated to an equivalent polynomial problem, 5 (1 + x⁷) - x = 0, with x=-1 (and some complex numbers) disallowed.
 

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