Introduction to Numerical Analysis I
Handout 9
1
Example 1.3.
Numerical Differentiation and Integration
f (x) ≈ P2 (x) = f [a] + f [a, a + h] (x − a)
We developed a tool (interpolation) which will serve
us in future as following: Given Pn (x) is a polynomial interpolation of f (x), for each linear operator L
we have
Lf (x) ≈ LPn (x)
f 0 (x) ≈
f (3) (c̃)
(x − a) (x − a − h)
6
Thus,
and the error given by
f (a + h) − f (a) h (2)
− f (c)
h
2
Example 1.4. We previously found sin x ≈ P3 (x) =
4
4 2
d
4
π x − π 2 x , therefore cos x sin’ x ≈ dx P3 (x) = π −
8
π 2 x ≈ 1.27 − 0.81x
f 0 (a) ≈
e (x) = L [f (x) − Pn (x)] = Lf {[x0 , . . . , xn , x] p (x)}
where p (x) =
n
Q
(x − xk ).
k=0
1.1
f (a + h) − f (a)
f (2) (c)
+
((x − a − h) + (x − a)) +
h
2
Numerical Differentiation/Finite Differences
Definition 1.1 (First Numerical Derivative).
Let Pn (x) be polynomial interpolation of f (x) at inn
terpolation points {xk }k=0 , then
f 0 (x) = Pn0 (x) +
d
{f [x0 , . . . , xn , x] p (x)}
dx
Definition 1.5 (Higher order Numerical Differentiation). Let Pn (x) be polynomial interpolation
n
of f (x) at interpolation points {xk }k=0 . Suppose
that f (x) have n + m + 1 derivatives (m ≤ n), then
The error is given by
d
{f [x0 , . . . , xn , x] p (x)} =
dx
d
= f [x0 , . . . , xn , x] p0 (x) + p (x) f [x0 , . . . , xn , x]
dx
e(x) = f 0 (x) − Pn0 (x) =
dm
{f [x0 , . . . , xk , x] p (x)}
dxm
And the error is given by
f (m) (x) = Pn(m) (x) +
dm
{f [x0 , . . . , xk , x] p (x)} =
dxm
f [x0 , . . . , xn , x] p0 (x) + f [x0 , . . . , xn , x, x] p (x) =
d
From the term dx
f [x0 , . . . , xn , x] we learn that we
need to differentiate divided differences.
(m)
e (x) = f (m) (x) − Pn
dm−1
dxm−1
dm−2 f [x0 , . . . , xn , x] p00 (x) + 2f [x0 , . . . , xn , x, x] p0 (x) +
dxm−2
f [x0 , . . . , xn , x, x, x] p (x)} = · · ·
Definition 1.2 (Derivative of DD).
d
f [x0 , . . . , xk , x] =
dx
f [x0 , . . . , xk , x + h] − f [x0 , . . . , xk , x]
lim
=
h→0
h
lim f [x0 , . . . , xk , x, x + h] = f [x0 , . . . , xk , x, x]
Definition 1.6 (Forward (Backward) Differences:).
(partially) uses interpolation points {a + jh}nj=0 {a − jh}nj=0
Definition 1.7 (Central Differences:). (partially)
uses interpolation points {a + jh}nj=0 ∪{a − jh}nj=0 symmetrically.
h→0
Finally, given that f (x) have n + 2 derivatives,
the error formula is
Theorem 1.8. Let m be an integer and let n +
1 = 2m. Given interpolation points x0 ≤ ... ≤ xn
symmetrically distributed around a point a, that is
a − x(m−1)−k = − (a − xm+k ) or alternatively
e (x) = f 0 (x) − Pn0 (x) =
f [x0 , . . . , xn , x] p0 (x) + p (x) f [x0 , . . . , xn , x, x] =
=
(x) =
f (n+2) (c̃)
f (n+1) (c) 0
p (x) +
p (x)
(n + 1)!
(n + 2)!
a=
Note that in the simplest case, one of the polynomial parts p(x) or p0 (x) vanishes.
for any k, then
1
d
dx
x(m−1)−k + xm+k
2
n
Q
(x − xk )
= 0.
k=0
x=a
1.1.1
1.1.2
Developing FD schemes using Polynomial Basis
Consider f 0 (a) ≈ c1 f (a) + c2 f (a − h) + c3 f (a + h)
The Taylor expansion gives
Definition 1.9 (The order of approximation).
The order of approximation is defined to be the minimal p that satisfies the following inequality
p
f (a) = f (a)
f (a ± h) = f (a) ± hf 0 (a) +
p
|e(x)| ≤ ch = O(h ),

1
 1
1
min |xi − xj | 6 h 6 max |xi − xj | .
i6=j
Definition 1.11 (Algebraic Order of Exactness).
An approximation method has an algebraic order of
exactness p if p is the maximal degree of a polynomial for which the approximation provides an exact
solution.
1.1.3
Example 1.12. For example consider Polynomial
basis {1, x, x2 , x3 , ...} The forward formula gives a
first order
f
f
f
h3 (3)
f
6
(a) + O h4
0
1
−1


 
f (a)
0
f (a)
0
1   hf (a)  =  f (a + h)  +O h3
h2 00
1
f (a − h)
f (a)
2
Subtract between the (±) equations
to get f (a + h)−
f (a − h) = 2hf 0 (a) + O h3 then solve for f 0 (a) to
get the central formula. Similarly, add between the
equations to get
f (a + h) − 2f (a) + f (a − h) = h2 f 00 (a) + O h4
f (a + h) − 2f (a) + f (a − h)
⇒ f 00 (a) =
+ O h2
h2
f (a + h) − f (a − h)
+ O(h2 )
f 0 (a) =
2h
f
(a) ±
We don’t need to solve the linear system. One writes
2
f (a ± h) − f (a) = ±hf 0 (a) + h2 f 00 (a) + O h3
Example 1.10. For example the central difference
formula for the first derivative has a first order, that
is of second order, since
0
h2 00
f
2
Rewrite in the matrix form
where e(x) is the error of approximation and
i6=j
Developing FD schemes using Taylor
Expansion
Sensitivity to error
Let f˜ (x) = f (x) + ẽ (x) and assume |ẽ (xj )| 6 ε
and f 3 (x) 6 M . Use the central formula for first
derivative f 0 (a) =
Then
f (a + h) − f (a)
(a) ≈
h
1−1
=1⇒
= 0 = f0
h
x+h−x
= 1 = f0
=x⇒
h
(x + h)2 − x2
2xh + h2
= x2 ⇒
=
= 2x + O (h)
h
h
f (a+h)−f (a−h)
2h
−
f (3) (c) 2
h .
6
f (a + h) − f (a − h) f (3) (c) 2
f˜0 (a) =
−
h +
2h
6
ẽ (a + h) − ẽ (a − h)
2h
Thus, the error is
(3)
f (c) 2 ẽ (a + h) − ẽ (a − h) 6
e (x) = −
h +
6
2h
M 2 |ẽ (a + h)| + |ẽ (a − h)| M 2 ε
h +
→ ∞
= h +
6 2h
6
h h→0
The latest example provide an algorithm to create FD schemes. Let D be differential operator. To
p
P
approximate D at point a by Df (a) =
cj f (a+jh)
j=1
one solves the following linear system of equation
To find the optimal value of h one solves





D1
Dx
..
.
D xp





c1
c2
..
.
cp


 
 
=
 
D1
Dx
..
.
D xp

e0 (x) =




to get h =
Example 1.13. Let f 0 (a) ≈ c1 f (a) + c2 f (a − h).
To find c1 , c2 solve
f = 1 ⇒ c1 + c2 = 0 ⇒ c2 = −c1
f = x ⇒ c1 a + c2 (a − h) = 1
To get the backward formula f 0 (a) ≈
f (a)−f (a−h)
h
2
q
3
3
M ε.
M
ε
h− 2 =0
3
h
See the graph of e(x) below.