Homework: Numerical Differentiation1
Instructor: Prof. Hector D. Ceniceros
1. Let f(x) = e
x
.
(a) Compute the centered difference approximation of f
0
(1/2), i.e. D0
h
f(1/2), for
h = 0.1/2
n
, n = 0, 1, . . . , 10, and verify its quadratic rate of convergence.
(b) Determine approximately the optimal value h0 which gives the minimum total
error (the sum of the discretization error plus the round-off error) and verify
this numerically.
(c) Construct and compute a fourth order approximation to f
0
(1/2) by applying
Richardson extrapolation to D0
h
f(1/2). Verify the rate of convergence numerically.
What is the optimal h0 in this case?
(d) As seen in class, we can use Cauchy’s integral formula to express f
0
(x0) as
f
0
(x0) = 1
2πr Z 2π
0
f(x0 + reiθ)e
iθdθ. (1)
Use the composite trapezoidal rule applied to (1) to approximate f
0
(1/2) to
machine precision. You can choose r freely in this problem.
2. Use Taylor series expansions to derive the error term of the sided difference approximation
to f
0
(x0):
Dhf(x0) = f(x0 + 2h) + 4f(x0 + h) 3f(x0)
2h
. (2)
3. Consider the data points (x0, f0),(x1, f1), . . . ,(xn, fn), where the points x0, x1, . . . , xn
are distinct but otherwise arbitrary (they could be for example the Chebyshev
nodes). Then the derivative of the interpolating polynomial of these data is
(x)fj
, (3)
where the lj
’s are the elementary Lagrange polynomials:
lj (x) = 1
αj
Yn
k=0
k6=j
(x xk), αj =
Yn
k=0
k6=j
(xj xk). (4)
We can evaluate (3) at each of the nodes x0, x1, . . . , xn, which will give us an
approximation to the derivative of f at those points, i.e. f
0
(xi) ≈ P
0
n
(xi). We can
write this as
f
0 ≈ Dnf, (5)
where f = [f0 f1 . . . fn]
T
, f
0 = [f
0
(x0) f
0
(x1). . . f0
(xn)]T and Dn is the Differentiation
Matrix, (Dn)ij = l
(xi).
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(a) Prove that
l
0
j
(x) = lj (x)
Xn
k=0
k6=j
1
x xk
, (6)
Hint: differentiate log lj (x).
(b) Using (6) prove that
(Dn)ij =
, i 6= j, (7)
(Dn)ii =
Xn
k=0
k6=i
1
xi xk
. (8)
(c) Prove that
Xn
j=0
(Dn)ij = 0 for all i = 0, 1, . . . , n. (9)
(d) Obtain D2 for the Chebyshev points x0 = 1, x1 = 0, x2 = 1.
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