Tschirnhaus Transformation

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Theorem

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

$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0$


Then the substitution: $y = x + \dfrac {a_{n-1}} {n a_n}$

converts $P_n$ into a depressed polynomial:

$b_n y^n + b_{n-1} y^{n-1} + \cdots + b_1 y + b_0 = 0$

where $b_{n-1} = 0$.

Such a substitution is called a Tschirnhaus transformation.


Proof

Substituting $y = x + \dfrac {a_{n-1}} {n a_n}$ gives us $x = y - \dfrac {a_{n-1}} {n a_n}$.

By the Binomial Theorem:

$a_n x^n = a_n \left({y^n - \dfrac {a_{n-1}} {a_n} y^{n-1} + P'_{n-2} \left({y}\right)}\right)$

where $P'_{n-2} \left({y}\right)$ is a polynomial in $y$ of order $n-2$.

Now we note that:

$a_{n-1} x^{n-1} = a_{n-1} y^{n-1} - P''_{n-2} \left({y}\right)$

where $P''_{n-2} \left({y}\right)$ is another polynomial in $y$ of order $n-2$.

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

Hence the result.

$\blacksquare$


Source of Name

This entry was named for Ehrenfried Walther von Tschirnhaus.