Tschirnhaus Transformation

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Theorem

Let Pn(x)=0$P_n\left(x\right)=0$ be a polynomial equation of order n$n$:

anxn+an−1xn−1+⋯+a1x+a0=0$a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0=0$

Then the substitution: y=x+an−1nan$y=x+{\displaystyle \frac{a_{n-1}}{n{a}_n}}$

converts Pn$P_n$ into a depressed polynomial:

bnyn+bn−1yn−1+⋯+b1y+b0=0$b_n{y}^n+b_{n-1}{y}^{n-1}+\cdots +b_{1}y+b_0=0$

where bn−1=0$b_{n-1}=0$.

Such a substitution is called a Tschirnhaus transformation.

Proof

Substituting y=x+an−1nan$y=x+{\displaystyle \frac{a_{n-1}}{n{a}_n}}$ gives us x=y−an−1nan$x=y-{\displaystyle \frac{a_{n-1}}{n{a}_n}}$.

By the Binomial Theorem:

anxn=an(yn−an−1anyn−1+P′n−2(y))$a_n{x}^n=a_n\left(y^n-{\displaystyle \frac{a_{n-1}}{a_n}}{y}^{n-1}+P_{n-2}^{\prime }\left(y\right)\right)$

where P′n−2(y)$P_{n-2}^{\prime }\left(y\right)$ is a polynomial in y$y$ of order n−2$n-2$.

Now we note that:

an−1xn−1=an−1yn−1−P′′n−2(y)$a_{n-1}{x}^{n-1}=a_{n-1}{y}^{n-1}-P_{n-2}^{″}\left(y\right)$

where P′′n−2(y)$P_{n-2}^{″}\left(y\right)$ is another polynomial in y$y$ of order n−2$n-2$.

The terms in yn−1$y^{n-1}$ cancel out.

Hence the result.

■$◼$

Source of Name

This entry was named for Ehrenfried Walther von Tschirnhaus.