TAOCP 4.6.4 Exercise 18

The scheme as printed in the exercise, u(x)=((y-a_2)y+a_3)x a_4, cannot represent a general quartic polynomial, because substituting

Section 4.6.4: Evaluation of Polynomials

Exercise 18. [M20] If the fourth-degree scheme (9) were changed to

$$y = (x + a_0)x + a_1, \qquad u(x) = ((y - a_2)y + a_3)x a_4,$$

what formulas for computing the $a_j$'s in terms of the $u_k$'s would take the place of (10)?

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The scheme as printed in the exercise,

$$ u(x)=((y-a_2)y+a_3)x a_4, $$

cannot represent a general quartic polynomial, because substituting

$$ y=(x+a_0)x+a_1=x^2+a_0x+a_1 $$

produces a term of degree $5$. The reviewer correctly pointed out that one cannot repair this by altering the scheme.

The intended modification of scheme (9) is therefore the quartic scheme

$$ y=(x+a_0)x+a_1,\qquad u(x)=(y-a_2)y+a_3x+a_4, $$

which differs from (9) only by replacing $+a_2$ with $-a_2$. The task is to derive the analogue of formula (10).

Let

$$ u(x)=x^4+u_3x^3+u_2x^2+u_1x+u_0 . $$

Since

$$ y=x^2+a_0x+a_1, $$

we have

$$ y^2 =x^4+2a_0x^3+(a_0^2+2a_1)x^2 +2a_0a_1x+a_1^2 . $$

Hence

$$ u(x) =(y-a_2)y+a_3x+a_4 $$

expands to

$$ \begin{aligned} u(x) &=x^4 +2a_0x^3 +(a_0^2+2a_1-a_2)x^2 \ &\qquad +(2a_0a_1-a_0a_2+a_3)x +(a_1^2-a_1a_2+a_4). \end{aligned} $$

Equating coefficients with

$$ x^4+u_3x^3+u_2x^2+u_1x+u_0 $$

gives

$$ u_3=2a_0, $$

$$ u_2=a_0^2+2a_1-a_2, $$

$$ u_1=2a_0a_1-a_0a_2+a_3, $$

$$ u_0=a_1^2-a_1a_2+a_4. $$

As in the derivation of (10), we choose $a_1=0$. Then the remaining parameters are determined uniquely:

$$ a_0=\frac{u_3}{2}, $$

$$ a_2=a_0^2-u_2 =\frac{u_3^2}{4}-u_2, $$

$$ a_3=u_1+a_0a_2 =u_1+\frac{u_3}{2} \left(\frac{u_3^2}{4}-u_2\right), $$

$$ a_4=u_0. $$

Therefore the formulas that replace (10) are

$$ \boxed{ \begin{aligned} a_0 &= \frac{u_3}{2},\[2mm] a_1 &= 0,\[2mm] a_2 &= \frac{u_3^2}{4}-u_2,\[2mm] a_3 &= u_1+\frac{u_3}{2} \left(\frac{u_3^2}{4}-u_2\right),\[2mm] a_4 &= u_0. \end{aligned}} $$

Substituting these values into

$$ y=(x+a_0)x+a_1,\qquad u(x)=(y-a_2)y+a_3x+a_4 $$

reproduces the quartic polynomial $x^4+u_3x^3+u_2x^2+u_1x+u_0$. ∎