Recently in my Intro to Proofs class, we've been talking about the fundamental theorem of algebra, which states that all polynomials of degree n always have n, not necessarily distinct, not necessarily real, solutions. Every high school student is taught the formula $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ for solving quadratic equations, and there exist solutions to the general cubic (Cardano's formula) and general quartic (Ferrari's solution). But what about higher functions?
The Abel-Ruffini theorem states that general functions of degree 5 or higher cannot be solved using a finite number of additions, subtractions, multiplications, divisions, and root extractions. Now, that isn't to say that no function of 5 or higher can be solved — quite the contrary, actually: x5 = 1 has solutions $x = \sqrt[5]{1}\in\{1,-\frac{1\pm_1\sqrt{5}}{4}\pm_2 i\sqrt{\frac{5\pm_1\sqrt{5}}{8}}\}$ where ±1 and ±2 are independent of the other but related to itself (thus both ±1 are the same sign always), producing five solutions, four of which are complex.
But there also exist equations with no exact solutions, such as x5 – x – 1 = 0. Sure, it has a real solution at x≈1.1673, but that's only an approximation, good to 4 decimal places. However, this is not really what my question is about.
While they cannot be solved generally, quintic functions can be reduced significantly. Take, for example, the function x5 + a4x4 + a3x3 a2x2 + a1x + a0 = 0. Making the substitution x = v – a4/4 produces the new equation v5 + b3v3 + b2v2 + b1v + b0 = 0, where the b coefficients are in terms of the a coefficients. Realize that this is just a linear shift to the side of the previous equation, but it becomes simpler to manage, as it's missing the 4th power term.
Tschirnhaus took it a step further, though. He used a method, which is now called the Tschirnhaus Transformation (the subject of my question) to solve the general cubic in a manner separate to Cardano's solution, and proposed that it could be used to solve any polynomial (and was mistaken). My question lies in what exactly it was that he did. Here is a paper which explains his transformation. I've followed it up to (7), and I understand that we're then solving for alpha in order to reduce the equation to a simple y3 + g, but then I am completely lost as to where the resulting equation came from. (7) is $res_{x}(P_{3},T) = y^3 + (p\alpha^{2}-\frac{1}{3}p^2 +3\alpha q)y + \frac{2}{3}\alpha^{2}p^{2}-\alpha^{3}q+q^{2}+\alpha pq+\frac{2}{27}p^3$, and the result is $y^{3} = 9q^{3}\alpha /p^{2}+\frac{4}{3}\alpha pq-\frac{8}{27}p^{3}-2q^{2}$
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Additionally, I followed the separate method for reducing (but not solving) the quintic through to (12), but I don't understand how (13) and (14) are derived.
If anybody here would be kind enough to explain where they came from, then thank you very much. I realize that there may be typos here, but I can't for the life of me understand what's going on anymore.
Thank you in advance.
Gabriel Benamy
PS. If any syntax is off, it's because the preview, auto preview, and tag look-ahead prompt are all down and I had to write everything on Wikipedia (which has somewhat different syntax, anyway).
PPS. Apparently, my notify email address is invalid, but I didn't have the option of changing it, so I had to uncheck "notify me" in order to post. Am I doing something wrong?
EDIT: Because apparently, I am unable to comment (as Mathoverflow functionality has completely died for me), I have to add this as an edit. Qiaochu Yuan said that I've misquoted both theorems. From my understanding, having "n complex roots" is the same as n, not necessarily real, roots. After all, 1 + i is "not necessarily real" but is a root of x3 – 3x2 + 4x – 2 = 0. 1 is also a real root of this (and is also complex, because all real values are complex).
Additionally, he says that the Abel-Ruffini theorem says that "polynomials of degree 5 or higher cannot be solved using the arithmetic operations and root extractions." This is not true, as demonstrated with x^5 – 1 = 0, which is a polynomial of degree 5 and clearly can be solved. The AR theorem states that not all can be solved in such a way, not that none can be solved in such a way.
Best Answer
Equation 7 is derived by using the resultant of the two polynomials in 5 and 6. There is a wikipedia article on the resultant. It can be computed using the Sylvester matrix which again has a wikipedia article.
Steps 13 and 14 are derived using the Tschirnhaus transformation to get an equation for the power sums of $y_k$ for 1 and 2 in terms of the power sums of the $x_k$. $y_k$ for 1 and 2 are known and substituting the known power sums for the $x_k$ for 1 and 2 after using the Tschirnhaus transformation for the power sum $y_k$ for 1 give equation 13 . Similarly substituting the known power sums for $x_k$ for 1 through 4 in the Tschirnhaus transformation of the power sum $y_k$ for 2 give equation 14.