 # Jan Hudde’s Second Letter: On Maxima and Minima - Mathematical Background: Double Roots

Author(s):
Daniel J. Curtin (Northern Kentucky University)

Building on the work of Fermat and Descartes [Curtin, pp. 263-264], Hudde considered algebraic expressions set equal to $z$ and sought the maximum value for $z.$ Strictly speaking his procedure finds a local maximum, but I will say “maximum” as he does. I will also use functional notation in this section, though Hudde did not.

In our notation Hudde wanted to find the maximum value $z_{\rm{max}}$ of $f(x) = z.$ This will occur when $f(x) = z$ has a double root. For smaller values of $z,$ $f(x) = z$ will have two (or more) solutions (see Figure 1). Figure 1. Near its maximum value, $f(x) = z$ has two (or more) solutions; at the maximum value $z_{\rm{max}},$ $f(x) = z$ has a double root.

For a minimum, invert the picture. Hudde recognized that the procedure for finding maxima and minima was the same. His theorem statements refer to both, though his examples use only the word “maximum.”

### Double Roots

For algebraic functions the problem of finding a double root is purely algebraic. No limit or infinitesimal quantity is needed.

Hudde’s letter begins with a theorem telling how to do this [Hudde, p. 507]. For a polynomial

$p(x) = c_0 + c_1x + c_2x^2 + \cdots + c_nx^n$

and any arithmetic progression $a, a+b, a+2b, a+3b,\dots,$ combine the two to get

${\hat p}(x) = ac_0 + (a+b)c_1x + (a+2b)c_2x^2 + \cdots + (a+nb)c_nx^n.\quad (1)$ In our notation, we can see

${\hat p}(x) = ap(x)+bxp^{\prime}(x).\quad\quad\quad\quad\quad\quad (2)$

Hudde claimed that if $p(x) = 0$ has a double root at $x=x_0,$ then ${\hat p}(x) = 0$ has a single root there. The converse, which is what Hudde really needed, is also true, though Hudde left this implicit.

To prove this, we might proceed as follows. Assume $p(x) = 0$ has a double root $x=x_0.$ Thus $p(x) = {{(x-{x_0})}^2}{q(x)},$ where $q(x)$ is a polynomial that is not zero at $x_0.$ From (2),

${\hat p}(x) = a{{(x-{x_0})}^2}q(x) + bx[2(x-{x_0})q(x) + {{(x-{x_0})}^2}{q^{\prime}(x)}]$

$=(x-{x_0})[ a{(x-{x_0})}q(x) + 2bxq(x) + {(x-{x_0})}{q^{\prime}(x)}].$

Since $q(x_0)\not=0,$ the quantity in the square brackets will not be zero, unless $x_0 = 0,$ in which case identifying a double root would be quite easy.

How did Hudde argue? He looked at the simplest example with a double root, namely $x=y$:

$p(x)=x^2-2xy+y^2={(x-y)}^2=0.$

Apply the progression to obtain

${\hat p}(x) = (a+2b)x^2-2(a+b)xy+ay^2=0.$

Hudde considered it obvious that $y$ is a single root. We might elaborate

$0=(a+2b)x^2-2(a+b)xy+ay^2=a{(x-y})^2+2b(x^2-xy)$

$=(x-y)[a(x-y)+2bx],$

and this expression has $y$ as a single root (again excepting the easy case $y=0$).

For the general case, Hudde gave an example that easily generalizes. He considered the following equation, which has $y$ as a double root:

${(x-y)}^2(x^3+px^2+qx+r)=0.$

$(x^2-2xy+y^2)x^3+p(x^2-2xy+y^2)x^2+q(x^2-2xy+y^2)x+r(x^2-2xy+y^2)=0.$

If the left-hand side is expanded into powers of $x$ and each term $x^k$ multiplied by the corresponding term $a+kb$ of an arithmetic progression, $k=0,1,2,3,4,5,$ then each of the occurrences of $x^2-2xy+y^2$ is multiplied by a different progression. For example,

$(x^2-2xy+y^2)x^3=x^5-2x^4y+x^3y^2$

becomes

$[(a+5b)x^2-(a+4b)2xy+(a+3b)y^2]x^3.$

Thus $y$ is a single root of this piece, by the argument for $x^2-2xy+y^2=0$ and any arithmetic progression. Since this works for each piece, it follows that $y$ is a single root of the entire equation.

Clearly, if the cubic factor is replaced by any polynomial or rational expression (that isn’t $0$ at $y$), the same argument applies. Hudde also observed that if the expression starts with a triple root, then the transformed expression has a double root, etc.

Daniel J. Curtin (Northern Kentucky University), "Jan Hudde’s Second Letter: On Maxima and Minima - Mathematical Background: Double Roots," Convergence (June 2015)

## Dummy View - NOT TO BE DELETED

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