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Numerical Differentiation

MACM 316

1/9

Numerical Differentiation
® Suppose we have a list of points
x0 < x1 < x2 < · · · < xn
and corresponding function values
f (x 0 ), f (x 1 ), f (x 2 ), . . . , f (x n )
® A natural question is whether we can use the data above to approximate f ′ (x) at some point x ∈ [x0, xn].
® The answer is easy provided the points xi are equally spaced so that xi − xi−1 = h (constant) and x.
The easiest way to motivate derivative formulas is using the definition of derivative f (x + h) − f (x )
f ′ (x) = lim
h→0
h
which suggests many possible difference formulas:
f ′ (x ) ≈

f (x + h) − f (x )

h
f (x ) − f (x − h)
f ′ (x ) ≈
h
f (x + h) − f (x − h)
f ′ (x ) ≈
2h
f (x + 2 h) − f (x − 2 h)

f (x ) ≈
4h

(forward difference)

(backward difference)

(centered difference)

(wide centered difference)

···
® These formulas are accurate only if h is “small enough.” October 30, 2008

c Steven Rauch and John Stockie

Numerical Differentiation

MACM 316

2/9

Example
Suppose we’re approximating the derivative of f (x) = 2 sin(3x) using the equally-spaced data
x 0.3000 0.3250 0.3500 0.3750 0.4000 0.4250 0.4500 0.4750 0.5000 f (x) 1.5667 1.6554 1.7348 1.8045 1.8641 1.9131 1.9514 1.9788 1.9950

The approximations of f ′ (0.4) with h = 0.1 are:
1. f ′ (x) ≈

f (x+h)−f (x)
h

f ′ (0.4) ≈

1.9950−1.8641
0. 1

= 1.3090

(40%)

2. f ′ (x) ≈

f (x)−f (x−h)
h

f ′ (0.4) ≈

1.8641−1.5667
0. 1

= 2.9740

(37%)

3. f ′ (x) ≈

f (x+h)−f (x−h)
2h

f ′ (0.4) ≈

1.9950−1.5667
0. 2

= 2.1415

(1.5%)

where the relative errors are computed using the exact value f ′ (0.4) = 6 cos(3 × 0.4) = 2.17414652686004
2.1
2

y

1.9
1.8
1.7
1.6
1.5
0.25

0.3

0.35

0.4

0.45

0.5

0.55

0.6

x
Figure 1: f (x) = 2 sin(3x)

October 30, 2008

c Steven Rauch and John Stockie

Numerical Differentiation

MACM 316

3/9

Example (cont’d)
Investigate what happens when h is decreased to 0.05:
x 0.3000 0.3250 0.3500 0.3750 0.4000 0.4250 0.4500 0.4750 0.5000 f (x) 1.5667 1.6554 1.7348 1.8045 1.8641 1.9131 1.9514 1.9788 1.9950

The approximations of f ′ (0.4) with h = 0.05 are:
1. f ′ (x) ≈

f (x+h)−f (x)
h

f ′ (0.4) ≈

1.9514−1.8641
0.05

= 1.7460

(20%)

2. f ′ (x) ≈

f (x)−f (x−h)
h

f ′ (0.4) ≈

1.8641−1.7348
0.05

= 2.5860

(19%)

3. f ′ (x) ≈

f (x+h)−f (x−h)
2h

f ′ (0.4) ≈

1.8641−1.7348
0. 1

= 2.1660

(0.4%)

4. f ′ (x) ≈

f (x+2h)−f (x−2h)
4h

f ′ (0.4) ≈

1.9950−1.5667
0. 2

= 2.1415

(1.5%)

Notice that:
• The forward and backward difference formulas (1 and 2) have similar accuracy.
• The centered difference (3) is much more accurate than the one-sided differences.
• Decreasing h increases the accuracy of the approximation. Question: Can this be explained?

October 30, 2008

c Steven Rauch and John Stockie

Numerical Differentiation

MACM 316

4/9

Error Analysis
To analyse the error in finite difference formulas, use Taylor series approximations. Example 1: Forward difference formula
• Write the Taylor polynomial of degree n = 1, with error term: ′

f (x + h) = f (x ) + f (x ) h +

f ′′ (c)
2

h2

• Then
f (x + h) − f (x )
h



= f (x ) +

f ′′ (c)
2

h

= f ′ (x ) + O (h)
• Decreasing h clearly reduces the error.
Example 2: Centered difference formula
• Taylor polynomials for f (x + h) and f (x − h) to O (h4): f ′′(x) 2 f ′′′(x) 3
f (4) (x) 4
f (5) (c1 ) 5
f (x + h) = f (x) + f (x) h +
h+
h+
h+
h
2
6
4!
5!


f (x − h) = f (x) − f ′(x) h +

f ′′ (x)
2

h2 −

f ′′′(x)
6

h3 +

f (4) (x)
4!

h4 −

f (5) (c2 )
5!

h5

• Subtract the second equation from the first and divide by 2h: f ′′′(x) 2
f (x + h) − f (x − h)

= f (x) +
h + O (h4 )
2h
6

• Error in centered formula is smaller (as expected)!
October 30, 2008

c Steven Rauch and John Stockie...
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