Partial Solutions to Homework XII
Y. Zhou
Section 6.5
6.
k
Proof. It is sufficient to show that Bik (x) = Bi−1
(x − 1). The claim of the problem follows from the
iterative application of this equality untill getting B0k :
k
k
Bik (x) = Bi−1
(x − 1) = Bi−2
(x − 2) = · · · = B0k (x − i) = B0k (x − ti ).
k
(x − 1).
We use mathmatical induction to prove Bik (x) = Bi−1
k
k
For k = 0, Bi (x) = Bi−1 (x − 1) follows directly from the definition of Bi0 (x). Assume that
k−1
Bik−1 (x) = Bi−1
(x − 1), we have
Bik (x)
x − ti
ti+k+1 − x k−1
B k−1 (x) +
B
(x)
ti+k − ti i
ti+k+1 − ti+1 i+1
x − ti
ti+k+1 − x k−1
k−1
=
Bi−1
(x − 1) +
B
(x − 1)
ti+k − ti
ti+k+1 − ti+1 i
ti+k − (x − 1) k−1
(x − 1) − ti−1 k−1
B
(x − 1) +
Bi (x − 1)
=
ti−1+k − ti−1 i−1
ti+k − ti
= Bi−1 (x − 1),
=
where the first and the last equalities follow from the iterative relation of B-splines; the second
eqnality follows from the assumption about Bik−1 (x), and the third equality holds true since ti = i
and x − ti = (x − 1) − (ti − 1) = (x1 ) − ti−1 .
8.
Proof. Since Bik (x) is compactly supported in (ti , ti+k+1 ), we know
∞
X
0=
ci Bik (x) =
i=−∞
j
X
ci Bik (x), ∀x ∈ (tj , tj+1 ).
i=j−k
k
This equation indicates that cj−k , · · · , cj are all zeros because Bj−k
(x), · · · , Bj are linearly independent on (ti , ti+k+1 ) (Lemma 8). Letting j vary over all integers we will get that ci = 0 for all
i.
16.
Proof.
∞
X
k
ci Bi (x) dx
−∞ Z
∞
Z
∞
≤
∞
X
ci Bik (x) dx
−∞ −∞
∞
∞ X
−∞
Z
=
|ci |Bik (x)dx
−∞ −∞
=
∞
X
Z
=
|ci |
−∞
≤
∞
X
Bik (x)dx
−∞
−∞
∞
X
∞
|ci |
ti+k+1 − ti
k+1
|ci |m
−∞
∞
X
= c
|ci |,
−∞
where we use the fact that Bik (x) ≥ 0 in deriving the first inequality, and use the conclusion of
problem 7 in deriving the third equality. In deriving the second inequality we use the relation that
ti+k+1 − ti = (ti+k+1 − ti+k ) + (ti+k − ti+k−1 ) + · · · + (ti+1 − ti ), with m terms in total and each
term less than m according to supi |ti+1 − ti | ≤ m.
1
2
6.6 2.
Proof. For any x ∈ R, we can always find j such that x ∈ [tj , tj+1 ), and
Sf
=
∞
X
f (xi )Bi2 (x)
i=−∞
=
j
X
f (xi )Bi2 (x)
i=j−2
By using the definition
Sf
=
of xi and Bi2 (x),
2
f (xj−2 )Bj−2
(x)
we have
2
+ f (xj−1 )Bj−1
(x) + f (xj )Bj2 (x)
x − tj−2 1
tj+1 − x
1
Bj−2 (x) +
Bj−1
(x) +
tj − tj−2
tj+1 − tj−1
x − tj−1
tj+2 − x 1
1
Bj−1
(x) +
Bj (x) +
f (xj−1 )
tj+1 − tj−1
tj+2 − tj
x − tj
tj+3 − x
1
Bj1 (x) +
Bj+1
(x)
f (xj )
tj+2 − tj
tj+3 − tj+1
x − tj−1
tj+1 − x
1
= Bj−1
(x) f (xj−2 )
+ f (xj−1 )
+
tj+1 − tj−1
tj+1 − tj−1
tj+2 − x
x − tj
Bj1 (x) f (xj−1 )
+ f (xj )
tj+2 − tj
tj+2 − tj
tj + tj+1 x − tj−1
tj−1 + tj tj+1 − x
1
+
= Bj−1
(x)
+
2
tj+1 − tj−1
2
tj+1 − tj−1
tj+1 + tj+2 x − tj
tj + tj+1 tj+2 − x
+
Bj1 (x)
2
tj+2 − tj
2
tj+2 − tj
tj+1 − x tj−1 + tj tj+1 − x
tj + tj+1 x − tj−1
=
+
+
tj+1 − tj
2
tj+1 − tj−1
2
tj+1 − tj−1
tj + tj+1 tj+2 − x
x − tj
tj+1 + tj+2 x − tj
+
.
tj+1 − tj
2
tj+2 − tj
2
tj+2 − tj
= f (xj−2 )
This is a quadratic equation. The coefficient of x2 is
tj−1 + tj
tj + tj+1
1
tj + tj+1
tj+1 + tj+2
1
−
+
+
+
= 0,
−
tj+1 − tj
2(tj+1 − tj−1 ) 2(tj+1 − tj−1 )
tj+1 − tj 2(tj+2 − tj ) 2(tj+2 − tj )
the constant term is
tj+1 [(tj−1 + tj )tj+1 − (tj + tj+1 )tj−1 ] tj [(tj + tj+1 )tj+2 − (tj+1 + tj+2 )tj ]
−
2(tj+1 − tj )(tj+1 − tj−1 )
2(tj+1 − tj )(tj+2 − tj )
tj+1 tj
tj+1 tj
−
=
2(tj+1 − tj ) 2(tj+1 − tj )
=0,
and it can be checked that coefficient of x is 1.
19.
Proof. Lemma 2 in Section 6.5 says that Bik (x) > 0 for any x ∈ (ti , ti+k+1 ). Thus for any x ∈ (t0 , t1 ),
Bik (x) > 0 for −k ≤ i ≤ 0, i.e., k +1 nonzeros spline functions in total. Similarly, for any x ∈ (t1 , t2 ),
Bik (x) > 0 for −k + 1 ≤ i ≤ 1, adding one more nonzero function B1k (x). Repeating the argument
for each of the remaining subintervals (t2 , t3 ), · · · , (tn−1 , tn ), we can have total n + k nonzero splines
k
B−k
(x), · · · , Bnk (x).
6.8 3.
Q
Proof. Let g(x) be the best approximation of f (x) in n , we need to show that g(x) = g(−x).
According to the definition of best approximation, we know
Z 1
(f (x) − g(x))xk dx = 0, ∀0 ≤ k ≤ n,
−1
3
Q
as {xk }, k = 0, · · · , n are the basis functions of n . Since f (x) is an even function, we have for k
even
Z 1
0 =
(f (x) − g(x))xk dx
−1
1
Z
(f (−x) − g(x))(−x)k dx
=
−1
1
Z
(f (−x) − g(x))(−x)k dx
= −
−1
1
Z
(f (−x) − g(x))(−x)k d(−x)
=
−1
Z −1
=
(f (x) − g(−x))xk dx
1
−1
Z
(f (x) − g(−x))xk dx
= −
1
Z
1
(f (x) − g(−x))xk dx,
=
−1
suggesting that
Z
1
(g(x) − g(−x))xk dx = 0, ∀k even.
−1
Following the similarly procedure we can show that
Z 1
(g(x) − g(−x))xk dx = 0, ∀k odd.
−1
Q
Q
As g(x) ∈ n , g(−x) ∈ n , g(x) − g(−x) ∈ n and {xk } are the basis functions of n , it follows
that g(x) − g(−x) = 0 and therefore g(x) is an even function.
Q
Q
8.
Proof. By definition,
an = hxPn−1 , Pn−1 i/hPn−1 , Pn−1 i,
hence we only need to show that
Z
a
hxPn−1 , Pn−1 i =
2
xPn−1
wdx = 0.
−a
2
Note that Pn−1
is nonnegative and xw is odd, so a sufficient condition for above equality is that
2
Pn−1 is even. In other words, it is sufficient to show that Pn is an even or an odd function. We will
prove this by using mathematical induction.
By definition P0 = 1 is an even function, and thus a1 = 0 and P1 = x is an odd function. It
follow that a2 = 0 and
P2 = xP1 + b2 P0
is an even function. But then a3 = 0 and
P3 = xP2 + b3 P1
is an odd function. Let Pn−2 be an odd function, Pn−1 be an even function, we have an = 0,
Pn = xPn−1 + bn Pn−2
being an odd function, an+1 = 0, and
Pn+1 = xPn + bn+1 Pn−1
being an odd function. Hence we proved that Pn is an even function if n is even and is an odd
function if n is odd, as required.
11.
4
Proof. By defintion,
bn = hxPn−1 , Pn−2 i/hPn−2 , Pn−2 i = hxPn−1 , Pn−2 i/kPn−2 k2 ,
thus it suffices to show that
hxPn−1 , Pn−2 i = kPn−1 k2 .
Note that
hxPn−1 , Pn−2 i = hxPn−2 , Pn−1 i
and
Pn−1 = (x − an−1 )Pn−2 + bn−1 Pn−3 , or xPn−2 = Pn−1 + an−1 Pn−2 − bn−1 Pn−3
we conclude that
hxPn−1 , Pn−2 i = hxPn−2 , Pn−1 i
= hPn−1 + an−1 Pn−2 − bn−1 Pn−3 , Pn−1 i
= hPn−1 , Pn−1 i + an−1 hPn−2 , Pn−1 i + bn−1 hPn−3 , Pn−1 i
= hPn−1 , Pn−1 i = kPn−1 k2
as hPn−2 , Pn−1 i = 0, hPn−3 , Pn−1 i = 0 due to the orthogonality. This finishes the proof.
16.
Proof. We need to show that hf1 , f2 i = 0 for any f1 , f2 are such that Af1 = λ1 f1 , Af2 = λ2 f2 with
λ1 6= λ2 . For such f1 , f2 and self-adjoint A we have
λ1 hf1 , f2 i = hλ1 f1 , f2 i
=
hAf1 , f2 i
=
hf1 , Af2 i
=
hf1 , λ2 f2 i
=
λ2 hf1 , f2 i.
Since λ1 6= λ2 , this equality holds ture if and only if hf1 , f2 i = 0.