Test 2 Solutions

Test 2 Solutions
MA 522 Fall 2008
r(x)
be the (n, m)-Padé approximant to the function f (x). Show that the
t(x)
xk r(x)
.
(n + k, m)-Padé approximant to xk f (x) is
t(x)
1. (20 points) Let
Solution: From our assumptions we have that
r
≡f
t
x 6 | t,
mod xn+m , deg(r) < n,
deg(t) ≤ m.
Therefore, xn+m |r − f t, which implies that xn+m+k |(xk r) − (xk f )t. Furthermore, deg(xk r) <
k r(x)
k + n, deg(t) ≤ m, and x 6 | t. Thus x t(x)
is the (n + k, m)-Padé approximant to xk f (x).
2. Discriminants
Let f = ad xd + ad−1 xd−1 + . . . + a0 ∈ C[x] with ad 6= 0. The discriminant of f is defined as
disc(f ) :=
1
Res(f, f 0 , x)
ad
where Res(f, f 0 , x) is the determinant of the Sylvester resultant matrix of f and f 0 (the
derivative of f ).
(a) (20 points) Prove that f has a multiple factor (i.e. f is divisible by h2 for some
h ∈ C[x] of positive degree) if and only if disc(f ) = 0.
Hint: (You can see the hint for -5 points)Q
prove that in this case
gcd(f, f 0 ) 6= 1 by differentiatP
r
ing the product form of f , i.e. f = ad i=1 (x − αi )ei where ri=1 ei = d.
Solution: If f = ad
Qr
i=1 (x
− αi )ei then
0
f = ad
r
X
ei (x − αi )ei −1
i=1
Y
(x − αj )ej .
j6=i
Then there exists i such that ei ≥ 2 if and only if each of the summands of f 0 is divisible
by x − αi , thus f 0 (αi ) = 0. In other words, f has a multiple factor if and only if f and f 0
has at least one common root, which is equivalent to gcd(f, f 0 ) 6= 1. But that is equivalent
to the Sylvester matrix of f and f 0 being singular, i.e. Res(f, f 0 , x) = 0, thus disc(f ) = 0.
(b) (20 points) Compute the discriminant of the quadratic polynomial f = ax2 + bx + c.
Explain how your answer relates to the quadratic formula, and, without using part (a),
prove that f has a multiple root if and only if the discriminant vanishes.
1
Solution: f = ax2 + bx + c, f 0 = 2ax + b. Then the Sylvester matrix of f and f 0 is the 3 × 3
matrix:

T
a b c
Syl(f, f 0 ) =  0 2a b 
2a b 0
with determinant Res(f, f 0 , x) = ab2 − 4a2 c = a (b2 − 4ac) , which gives
disc(f ) = b2 − 4ac.
The quadratic formula for the solution is
x1,2 =
−b ±
p
disc(f )
,
2a
and the two roots collide if and only if disc(f ) = 0.
3. (a) (20 points) Prove that every polynomials f ∈ C[x, y, z] can be written as
f = h1 (y − x2 ) + h2 (z − x3 ) + r
where r is a polynomial in x alone.
Hint: (you can see the hint for -5 points) Use the division with remainder algorithm for multivariate polynomials, and specify carefully the monomial ordering to be used.
Solution: Let the monomial ordering be the lexicographic ordering with x < y < z.
Then lt(y − x2 ) = y and lt(z − x3 ) = z, and {y − x2 , z − x3 } forms a Göbner basis for
I = hy − x2 , z − x3 i ⊂ C[x, y, z]. Thus for every f ∈ C[x, y, z] there is a unique r such that
none of the terms in r are divisible by y or z, and f − r ∈ I. But this implies that r is a
polynomial in x alone, and satisfies the conditions of the problem.
(b) (20 points) The curve C := {(t, t2 , t3 ) : t ∈ C} ⊂ C3 is called the twisted cubic curve.
Show that I(C) = hy − x2 , z − x3 i. (Recall that I(V ) = {f ∈ C[x, y, z] : ∀v ∈ V f (v) = 0}).
Hint: (you can see the hint for -5 points) For f ∈ I(C) use that f (t, t2 , t3 ) = 0, together with
the expression (1) in part (a), to prove that f ∈ hy − x2 , z − x3 i.
Solution: To simplify the notation let p1 := y − x2 and p2 := z − x3 . Since p1 (t, t2 , t3 ) =
t2 − t2 = 0, and p2 (t, t2 , t3 ) = t3 − t3 = 0 for all t ∈ C, therefore p1 , p2 ∈ I(C), so
hp1 , p2 i ⊂ I(C).
On the other hand, let f ∈ I(C). Then by part (a) we have f = g + r where g ∈ hp1 , p2 i
and r ∈ C[x]. Therefore r ∈ I(C), which implies that r(t) = 0 for all t ∈ C, thus r ≡ 0.
Therefore f ∈ hp1 , p2 i. This proves that I(C) = hy − x2 , z − x3 i.
(c) (10 points) Let V ⊂ C3 be the curve parametrized by (t, tm , tn ), n, m ≥ 2. Determine
I(V ).
Solution: I(V ) = hy − xm , z − xn i.
2

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