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?<br>
The Abel-Ruffini theorem states that general functions of degree 5 or higher cannot be solved using a finite number of additions, subtractions, multiplications, and divisions. Now, that isn't to say that *no* function of 5 or higher can be solved -- quite the contrary, actually: x<sup>5</sup> = 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 ±<sub>1</sub> and ±<sub>2</sub> are independent of the other but related to itself (thus both ±<sub>1</sub> are the same sign always), producing five solutions, four of which are complex.<br>
But there also exist equations with no exact solutions, such as x<sup>5</sup> - 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.<br><br>
While they cannot be solved generally, quintic functions can be *reduced* significantly. Take, for example, the function x<sup>5</sup> + a<sub>4</sub>x<sup>4</sup> + a<sub>3</sub>x<sup>3</sup> a<sub>2</sub>x<sup>2</sup> + a<sub>1</sub>x + a<sub>0</sub> = 0. Making the substitution x = v - a<sub>4</sub>/4 produces the new equation v<sup>5</sup> + b<sub>3</sub>v<sup>3</sup> + b<sub>2</sub>v<sup>2</sup> + b<sub>1</sub>v + b<sub>0</sub> = 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 4<sup>th</sup> power term.<br>
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][1] 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 y<sup>3</sup> + 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 $y^{3} = 9q^{3}\alpha /p^{2}+\frac{4}{3}\alpha pq-\frac{8}{27}p^{3}-2q^{2}$<br>.
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.<br>
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.<br><br>
Thank you in advance.<br><br>
Gabriel Benamy<br><br><br><br>
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).<br>
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?
[1]: http://www.sigsam.org/bulletin/articles/145/Adamchik.pdf