CSE 541 - Differentiation

Roger Crawfis

April 20, 2023 OSU/CIS 541 2

Numerical Differentiation

• The mathematical definition:

• Can also be thought of as the tangent line.0

( ) ( )'( ) lim

h

f x h f xf x

h

x x+h

April 20, 2023 OSU/CIS 541 3

Numerical Differentiation

• We can not calculate the limit as h goes to zero, so we need to approximate it.

• Apply directly for a non-zero h leads to the slope of the secant curve.

x x+h

April 20, 2023 OSU/CIS 541 4

Numerical Differentiation

• This is called Forward Differences and can be derived using Taylor’s Series:

2

2

( ) ( ) ( ) ( )2!

( ) ( ) ( ) ( )2!

( ) ( )( ) ( )

2!( ) ( )

( ) 0

hf x h f x f x h f

hf x h f x f x h f

f x h f x hf x f

hf x h f x

f x as hh

Theoretically speaking

April 20, 2023 OSU/CIS 541 5

Truncation Errors

• Let f(x) = a+e, and f(x+h) = a+f.• Then, as h approaches zero, e<<a and f<<a.• With limited precision on our computer, our

representation of f(x) a f(x+h).• We can easily get a random round-off bit as

the most significant digit in the subtraction.• Dividing by h, leads to a very wrong answer

for f’(x).

April 20, 2023 OSU/CIS 541 6

Error Tradeoff

• Using a smaller step size reduces truncation error.

• However, it increases the round-off error.

• Trade off/diminishing returns occurs: Always think and test!

Log error

Log step size

Truncation error

Round off error

Total errorPoint of

diminishingreturns

April 20, 2023 OSU/CIS 541 7

Numerical Differentiation

• This formula favors (or biases towards) the right-hand side of the curve.

• Why not use the left?

x x+hx-h

April 20, 2023 OSU/CIS 541 8

Numerical Differentiation

• This leads to the Backward Differences formula.

2

( ) ( ) ( ) ( )2!

( ) ( )( ) ( )

2!( ) ( )

( ) 0

hf x h f x f x h f

f x f x h hf x f

hf x f x h

f x as hh

April 20, 2023 OSU/CIS 541 9

Numerical Differentiation

• Can we do better?

• Let’s average the two:

• This is called the Central Difference formula.

1 ( ) ( ) ( ) ( ) ( ) ( )( )

2 2

f x h f x f x f x h f x h f x hf x

h h h

Forward difference Backward difference

April 20, 2023 OSU/CIS 541 10

Central Differences

• This formula does not seem very good.– It does not follow the calculus formula.

– It takes the slope of the secant with width 2h.

– The actual point we are interested in is not even evaluated.

x x+hx-h

April 20, 2023 OSU/CIS 541 11

Numerical Differentiation

• Is this any better?• Let’s use Taylor’s Series to examine the error:

2 3

2 3

3 3

( ) ( ) ( ) ( ) ( )2 3!

( ) ( ) ( ) ( ) ( )2 3!

( ) ( ) 2 ( ) ( ) ( )3! 3!

h hf x h f x f x h f x f

h hf x h f x f x h f x f

subtracting

h hf x h f x h f x h f f

April 20, 2023 OSU/CIS 541 12

Central Differences

• The central differences formula has much better convergence.

• Approaches the derivative as h2 goes to zero!!

2( ) ( ) 1( ) ( ) , ,

2 6

f x h f x hf x f h x h x h

h

2( ) ( )( )

2

f x h f x hf x O h

h

April 20, 2023 OSU/CIS 541 13

Warning

• Still have truncation error problem.

• Consider the case of:

• Build a table withsmaller values of h.

• What about largevalues of h for thisfunction?

( )100

100 100( )

21, 0.000333, 6

0.0100033 0.0099966( ) 0.010050

0.000666666Relative error:

0.01-0.0100500.5%

0.01

xf x

x h x h

f xh

at x h with significant digits

f x

April 20, 2023 OSU/CIS 541 14

Richardson Extrapolation

• Can we do better?

• Is my choice of h a good one?

• Let’s subtract the two Taylor Series expansions again:

2 3 4 54 5

2 3 4 54 5

553 3

( ) ( ) ( ) ( ) ( ) ( ) ( )2 3! 4! 5!

( ) ( ) ( ) ( ) ( ) ( ) ( )2 3! 4! 5!

( ) ( )( ) ( ) 2 ( ) 2 2 2 ( )

3! 3! 5!

h h h hf x h f x f x h f x f x f x f x

h h h hf x h f x f x h f x f f x f x

subtracting

f x f x hf x h f x h f x h h h f x

April 20, 2023 OSU/CIS 541 15

Richardson Extrapolation

• Assuming the higher derivatives exist, we can hold x fixed (which also fixes the values of f(x)), to obtain the following formula.

• Richardson Extrapolation examines the operator below as a function of h.

2 4 62 4 6

1( ) ( ) ( )

2f x f x h f x h a h a h a h

h

1( ) ( ) ( )

2h f x h f x h

h

April 20, 2023 OSU/CIS 541 16

Richardson Extrapolation

• This function approximates f’(x) to O(h2) as we saw earlier.

• Let’s look at the operator as h goes to zero.2 4 6

2 4 6

2 4 6

2 4 6

( ) ( )

( ) ( )2 2 2 2

h f x a h a h a h

h h h hf x a a a

Same leading constants

April 20, 2023 OSU/CIS 541 17

Richardson Extrapolation

• Using these two formula’s, we can come up with another estimate for the derivative that cancels out the h2 terms.

4 64 6

4

3 15( ) 4 ( ) 3 ( )

2 4 16

1( ) ( ) ( ) ( )

2 3 2

hh f x a h a h

or

h hf x h O h

new estimate difference between old and new estimates

Extrapolates by assuming the new

estimate undershot.

April 20, 2023 OSU/CIS 541 18

Richardson Extrapolation

• If h is small (h<<1), then h4 goes to zero much faster than h2.

• Cool!!!

• Can we cancel out the h6 term?– Yes, by using h/4 to estimate the derivative.

April 20, 2023 OSU/CIS 541 19

Richardson Extrapolation

• Consider the following property:

• where L is unknown,

• as are the coefficients, a2k.

22

1

22

1

( ) ( ) kk

k

kk

k

h f x a h

L a h

0

lim ( )h

L h f x

April 20, 2023 OSU/CIS 541 20

Richardson Extrapolation

• Do not forget the formal definition is simply the central-differences formula:

• New symbology (is this a word?):

1( ) ( ) ( )

2h f x h f x h

h

2

1

,02

,02

n

k

nk

hD n

hL A k

From previous slide

April 20, 2023 OSU/CIS 541 21

Richardson Extrapolation

• D(n,0) is just the central differences operator for different values of h.

• Okay, so we proceed by computing D(n,0) for several values of n.

• Recalling our cancellation of the h2 term.

4

4

1( ) ( ) ( ) ( )

2 3 2

1(1,0) (1,0) (0,0)

4 1

h hf x h O h

D D D O h

April 20, 2023 OSU/CIS 541 22

Richardson Extrapolation

• If we let hh/2, then in general, we can write:

• Let’s denote this operator as:

4

1( ) ( ,0) ( ,0) ( 1,0)

4 1 2n

hf x D n D n D n O

1

1,1 ( ,0) ,0 1,0

4 1D n D n D n D n

April 20, 2023 OSU/CIS 541 23

Richardson Extrapolation

• Now, we can formally define Richardson’s extrapolation operator as:

• or

4 1, ( , 1) 1, 1 , 1

4 1 4 1

m

m mD n m D n m D n m m n

new estimate old estimate

1, ( , 1) , 1 1, 1

4 1mD n m D n m D n m D n m

April 20, 2023 OSU/CIS 541 24

Richardson Extrapolation

• Now, we can formally define Richardson’s extrapolation operator as:

Memorize me!!!!

1, ( , 1) , 1 1, 1

4 1mD n m D n m D n m D n m

April 20, 2023 OSU/CIS 541 25

Richardson Extrapolation Theorem

• These terms approach f’(x) very quickly.

2

1

, ,2

k

nk m

hD n m L A k m

Order starts much higher!!!!

April 20, 2023 OSU/CIS 541 26

Richardson Extrapolation

• Since m n, this leads to a two-dimensional triangular array of values as follows:

• We must pick an initial value of h and a max iteration value N.

0,0

1,0 1,1

2,0 2,1 2,2

,0 ,1 ,2 ,

D

D D

D D D

D N D N D N D N N

April 20, 2023 OSU/CIS 541 27

Example

52

3

cos(100( )

11.3,

1280,0 16.696386

1,0 40.583393

2,0 109.322528

3,0 135.031747

4,0 142.068615

5,0 143.866937

xf x

x

x h

D

D

D

D

D

D

1

0.002444096

h

April 20, 2023 OSU/CIS 541 28

Example

0,0 16.696386

1,0 40.583393 1,1 48.583393

2,0 109.322528 2,1 132.235574

3,0 135.031747 3,1 143.601487

4,0 142.068615 4,1 144.414238

5,0 143.866937 5,1 144.466377

D

D D

D D

D D

D D

D D

1

3extrapolate

April 20, 2023 OSU/CIS 541 29

Example

0,0 16.696386

1,0 40.583393 1,1 48.583393

2,0 109.322528 2,1 132.235574 2,2 137.814897

3,0 135.031747 3,1 143.601487 3,2 144.359214

4,0 142.068615 4,1 144.414238 4,2 144.468421

5,0 143.866937 5,1 144

D

D D

D D D

D D D

D D D

D D

.466377 5,2 144.469853D

1

15extrapolate

April 20, 2023 OSU/CIS 541 30

Example

• Which converges up to eight decimal places.

• Is it accurate?

16.696386

40.583393 48.583393

109.322528 132.235574 137.814897

135.031747 143.601487 144.359214 144.463092

142.068615 144.414238 144.468421 144.470154 144.470182

143.866937 144.466377 144.469853 144.469876 144.469875 5,D 5 144.4698751

1023extrapolate1

255extrapolate

1

63extrapolate

1

15extrapolate

1

3extrapolate

April 20, 2023 OSU/CIS 541 31

Example

• We can look at the (theoretical) error term on this example.

• Taking the derivative:

2

55 1

12 2

57

5,5 ,52

11.3 (6,5) ,5

4096 2

k

k

k

k

hD L A k

hf A A k

2-144

(1.3) 144.469874253f Round-off error

April 20, 2023 OSU/CIS 541 32

Second Derivatives

• What if we need the second derivative?

• Any guesses?

2 3 4 54 5

2 3 4 54 5

( ) ( ) ( ) ( ) ( ) ( ) ( )2 3! 4! 5!

( ) ( ) ( ) ( ) ( ) ( ) ( )2 3! 4! 5!

h h h hf x h f x f x h f x f x f x f x

h h h hf x h f x f x h f x f f x f x

April 20, 2023 OSU/CIS 541 33

Second Derivatives

• Let’s cancel out the odd derivatives and double up the even ones:– Implies adding the terms together.

2 4

4( ) ( ) 2 ( ) 2 ( ) 2 ( )2 4!

h hf x h f x h f x f x f x

April 20, 2023 OSU/CIS 541 34

Second Derivatives

• Isolating the second derivative term yields:

• With an error term of:

2

( ) 2 ( ) ( )( )

f x h f x f x hf x

h

421

12E h f

April 20, 2023 OSU/CIS 541 35

Partial Derivatives

• Remember: Nothing special about partial derivatives:

, ,,

2, ,

,2

f x h y f x h yfx y

x hf x y h f x y hf

x yy h

April 20, 2023 OSU/CIS 541 36

Calculating the Gradient

• For lab 2, you need to calculate the gradient.

• Just use central differences for each partial derivative.

• Remember to normalize it (divide by its length).