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If we have a degree $n$ polynomial $$p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x+a_0$$with coefficients in a field, say $\Bbb C$, for concreteness, it is well known that the substitution $y= x + \frac{a_{n-1}}{na_n}$ "kills" the $a_{n-1}x^{n-1}$ term, that is, we get something of the form $$b_ny^n +b_{n-2}y^{n-2}+\cdots+ b_1y+b_0.$$ I think of that substitution as "translating" the polynomial such that the mean of the roots "in the new variable $y$" is zero. Everything seems to work fine since the arithmetic mean of $n$ numbers is a linear function of them, in some sense.

Question: is there a suitable substitution that eliminates an arbitrary term $a_kx^k$, $0\leq k < n $, of our choice?

Working with $n=3$ around, it seems that there are two possibilities for "depressing" the $a_1x$ term (given by the quadratic formula), and these expressions are very close to the closed expressions for the critical points of a third degree polynomial. I don't know if this is a coincidence or not, but I digress.

Also, thinking of Vieta's formulae doesn't seem to lead anywhere, because of the non-linearity of the expressions involved. I googled around a bit, and this might be related to Tschirnhaus transformations, but my background in algebra is weak and I don't know enough Galois theory to digest some of that information (and that's probably overkill).

I would like some intuition behind such substitutions instead of the obvious "write $y = x+t$, plug in, put the desired coefficient equal to zero and solve for $t$", if said substitutions do exist - I'd like some intuition that avoids computations as most as possible (something geometric, maybe?)

Thanks, and sorry for the rant. (You can edit in tags that might be also relevant, I'm a bit lost here)

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    $\begingroup$ (1) .A polynomial substitution $x=f(y)$ in $p(x)$ that eliminates the term $a_0 $ would yield a polynomial $y q(y)=p(x),$ which means that $f(0)$ is a solution to $p(x)=0,$ so finding a suitable $f$ requires finding a zero of $p$ to begin. (2). Eliminating $a_kx^k$ by letting $x=y+c,$ when deg $(p)= n$ and $1\leq k \leq n-2$ will usually require solving $r(c)=0$ where $r$ is a polynomial with $deg (r)=n-k.$ $\endgroup$ – DanielWainfleet Mar 5 '16 at 7:22
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Yes, if you want to eliminate only one term, then the linear substitution $x = y+t$ is enough. For example, given the cubic,

$$P(x): = ax^3+bx^2+cx+d\tag1$$

Let $x = y+t$, collect the new variable $y$, and you get,

$$P(y): = a y^3 + (b + 3 a t) y^2 + (c + 2 b t + 3 a t^2) y + (d + c t + b t^2 + a t^3)$$

You can eliminate the non-leading terms by equating any of the coefficients to zero and solving for $t$. I'm sure you can see this easily generalizes to higher powers. However, once you go above the quintic, getting rid of some terms may involve $t$ that is no longer in radicals.

P.S. If you want to get rid of two or three terms simultaneously, then you need a Tschirnhausen transformation which is just a generalization, though you have to equate your polynomial to zero. See this post for an easy explanation on how to do this as applied to the quintic.

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    $\begingroup$ This will always work over an algebraically closed field like op's suggestion of $\mathbb{C}$, but note that it doesn't necessarily imply that there's a real $t$ that gets the job done, and indeed, that's not always possible - consider the polynomial $p(x)=x^2+1$; then no substitution of the form $x=y+t$ with $t\in\mathbb{R}$ can eliminate the constant term - otherwise, we'd have $p(y+t)=ay^2+by$, implying that $p(0+t)=a\cdot 0^2+b\cdot 0=0$ and thus that $t$ is a (non-existant) root of the original equation. $\endgroup$ – Steven Stadnicki Mar 5 '16 at 6:45
  • $\begingroup$ I see, thanks. What I was looking for, though, it was some way to see geometrically what $t$ should be, since for eliminating the $n-1$ degree term we can actually guess it without having to solve for $t$, thinking of means of the roots. Your answer sort of indicates that there is no such way if the degree of $p$ is too high :(. $\endgroup$ – Ivo Terek Mar 5 '16 at 15:15
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    $\begingroup$ @IvoTerek: For the case when both the $x^{n-1}$ and $x^{n-2}$ terms are eliminated simultaneously, a geometric interpretation would be the sum of the roots and the sum of their squares are equal to zero. But that would need a quadratic Tschirnhausen transformation whose exact form one can't "guess" a priori unless you're Ramanujan. $\endgroup$ – Tito Piezas III Mar 5 '16 at 16:00
  • $\begingroup$ @StevenStadnicki: Agreed. However, I was more focused on whether the OP wanted $t$ in radicals. That why I mentioned quintics (since eliminating any of its intermediate terms would not involve anything higher than a quartic). $\endgroup$ – Tito Piezas III Mar 5 '16 at 16:06

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