NA I, Hw 7 (due 7/6)
Example 1 3.3/1(b) List the Chebyshev interpolation nodes x1 , . . . , xn in the
interval [−2, 2] with n = 4.
solution: The general formula from Chebyshev nodes on [a, b] is
xi =
b−a
(2i − 1)π a + b
cos
+
,
2
n
2
i = 1, . . . , n.
The four nodes on [−2, 2] are therefore
xi = 2 cos
(2i − 1)π
,
8
i = 1, 2, 3, 4.
Example 2 3.3/2 Find the upper bound for |(x−x1 ) . . . (x−xn )| on the interval
and Chebyshev nodes used in Exercise 1.
solution: Because our interval is not [−1, 1] we have to use the more general
inequality, see p. 166 eq. (3.20), which is valid on an interval [a, b] and it is
given by
n
( b−a
2 )
|(x − x1 ) . . . (x − xn )| ≤ n−1
.
2
Setting a = −2 and b = 2 the right-hand side evaluates to 2.
Example 3 3.3/3 Assume that Chebyshev interpolation is used to find a fifth
degree interpolating polynomial Q5 (x) on the interval [−1, 1] for the function
f (x) = ex . Use the interpolation error formula to find a worst-case estimate
for the error |ex − Q5 (x)| that is valid for x throughout the interval [−1, 1].
How many digits after the decimal point will be correct when Q5 (x) is used to
approximate ex ?
solution: To find a degree 5 interpolating polynomial, we need (at least) n = 6
nodes. By Theorem 3.3, the interpolation error is
ex − Q5 (x) =
(x − x1 ) . . . (x − x6 ) (vi)
f (c),
6!
where c ∈ [−1, 1]. Using eq. (3.20), and similarly to Ex. 3.11, we get
|ex − Q5 (x)| ≤
|f (vi) (c)|
e
= 5 ≈ 1.1798 × 10−4 .
25 6!
2 6!
Since 0.11798 × 10−3 < 0.5 × 10−3 , Q5 (x) and ex will agree to 3 decimal places
for x ∈ [−1, 1].
1
Example 4 (Our conjecture) Show whether the Chebyshev polynomials
Tn+1 (x) = 2xTn (x) − Tn−1 (x),
are odd functions with odd n and even with n even.
solution: A function is odd if
f (−x) = −f (x)
and even if
f (−x) = f (x).
The first Chebyshev polynomials are
T0 (x) = 1
T1 (x) = x
T2 (x) = 2x2 − 1
T3 (x) = 4x3 − 3x
...
With n = 0, 1 (and respective T0 , T1 even and odd), we have
T2 (−x) = 2(−x)T1 (−x) − T0 (−x) = 2xT1 (x) − T0 (x) = T2 (x)
T3 (−x) = 2(−x)T2 (−x) − T1 (−x) = −2xT2 (x) + T1 (x) = −T3 (x).
Next assume that for some n > 1, the corresponding Tn−1 and Tn are even and
odd, respectively. We have
Tn+1 (−x) = 2(−x)Tn (−x) − Tn−1 (−x) = 2xTn − Tn−1 (x) = Tn+1 (x),
Tn+2 (−x) = 2(−x)Tn+1 (−x) − Tn (−x) = −2xTn (x) + Tn (x) = −Tn+2 (x),
and so with Tn−1 and Tn even and odd, resp., we get that Tn+1 and Tn+2 are
even and odd, respectively.
2