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Given a polynomial of any degree $\sum_{i=0}^n a_ix^i$ can it be proven that the substitution $x=t-$${a_{n-1}}\over{na_n}$ will convert the equation to depressed form $b_nt^n$ + $\sum_{i=0}^{n-2} b_it^i$, for some new coefficients $b_i$ (i.e. remove the $n-1$ term)?

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According to wikipedia this transformation is known as a Tschirnhaus transformation: – user11393 Jan 24 '12 at 14:20
Yes, it can. ${}{}$ – Peter Taylor Jan 24 '12 at 15:53
up vote 0 down vote accepted

We have \begin{align*} P(x)&=\sum_{p=0}^na_p\sum_{j=0}^pt^j\binom pj\left(-\frac{a_{n-1}}{na_n}\right)^{p-j}\\ &=\sum_{0\leq j\leq p\leq n}a_pt^j\binom pj\left(-\frac{a_{n-1}}{na_n}\right)^{p-j}\\ &=\sum_{j=0}^n\sum_{p=j}^nt^ja_p\binom pj\left(-\frac{a_{n-1}}{na_n}\right)^{p-j}\\ &=\sum_{j=0}^n\left(\sum_{p=j}^na_p\binom pj\left(-\frac{a_{n-1}}{na_n}\right)^{p-j}\right)t^j, \end{align*} in particular, the coefficient of $n-1$ in $t$ is $$\sum_{p=n-1}^na_p\binom p{n-1}\left(-\frac{a_{n-1}}{na_n}\right)^{p-(n-1)}=a_{n-1}+a_nn\left(-\frac{a_{n-1}}{na_n}\right)=0.$$

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It is straightforward to derive this substitution from scratch. Suppose $f$ has degree $n > 0\:.$

$$\begin{eqnarray}{}\quad\ \ f(t+c) &=&\ a_n (t^n + n c\ t^{n-1} +\ \cdots\ )\ +\ a_{n-1} (t^{n-1}\ +\ \cdots\ )\ +\ \cdots \\ \\ &=&\ a_n\ t^n + (n c\ a_n + a_{n-1})\ t^{n-1} +\ \cdots \end{eqnarray} $$

Thus the coefficient of $t^{n-1}$ is zero iff $\ n c\ a_n + a_{n-1} = 0\ $ iff $\ c = -\dfrac{a_{n-1}}{n a_n}$.

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