1. Numerical Differentiation
Sec:4.1
Numerical Differentiation

2. Forward-difference Formula Backward-difference Formula
Sec:4.1 Numerical Differentiation
𝒇 ′ 𝒙 𝟎 Approximation
Forward-difference Formula
Backward-difference Formula
𝑓 ′ 𝒙 𝟎 = 𝑓 𝒙 𝟎 +𝒉 −𝑓( 𝒙 𝟎 ) ℎ − ℎ 2 𝑓′′(𝜉)
𝑓 ′ 𝒙 𝟎 = 𝑓 𝒙 𝟎 −𝑓( 𝒙 𝟎 −𝒉) ℎ + ℎ 2 𝑓′′(𝜉)
where 𝜉 between 𝒙 𝟎 and 𝒙 𝟎 +𝐡
where 𝜉 between 𝒙 𝟎 and 𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 −𝐡
𝒙 𝟎
Two-Point Formula
Example
Use table 4.2 and forward-difference formula to approximate 𝑓 ′ 2 with ℎ=0.1
𝒉=𝟎.𝟏
1 (0.1) 𝑓 2.1 −𝑓(2)
𝑓′(2)≈
Use table 4.2 and backward-difference formula to approximate 𝑓 ′ 2 with ℎ= 0.2
1 (0.2) 𝑓 2 −𝑓(1.8)
𝑓′(2)≈
𝒉=𝟎.𝟐

3. Sec:4.1 Numerical Differentiation
𝒇 ′ 𝒙 𝟎 Approximation
Three-Point Midpoint Formula
Three-Point Endpoint Formula
𝒙 𝟎
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
Five-Point Midpoint Formula
Five-Point Endpoint Formula
𝒙 𝟎
𝒙 𝟎 +𝟐𝐡
𝒙 𝟎 −𝟐𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟑𝐡
𝒙 𝟎 +𝟒𝐡
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡

4. Three-Point Midpoint Formula Three-Point Endpoint Formula
Sec:4.1 Numerical Differentiation
𝒇 ′ 𝒙 𝟎 Approximation
Three-Point Midpoint Formula
Three-Point Endpoint Formula
𝒙 𝟎
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
Example
Use table 4.2 and three-point endpoint formula to approximate 𝑓 ′ 2 with ℎ=0.1
𝑓′(2)≈ = =

5. Five-Point Midpoint Formula Five-Point Endpoint Formula
Sec:4.1 Numerical Differentiation
𝒇 ′ 𝒙 𝟎 Approximation
Five-Point Midpoint Formula
Five-Point Endpoint Formula
𝒙 𝟎 −𝟐𝐡
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
𝒙 𝟎 +𝟑𝐡
𝒙 𝟎 +𝟒𝐡
Example
Use table 4.2 and five-point midpoint formula to approximate 𝑓 ′ 2 with ℎ=0.1
𝑓′(2)≈
The only five-point formula for which the table gives sufficient data is the midpoint formula =

6. Five-Point Midpoint Formula Five-Point Endpoint Formula
Sec:4.1 Numerical Differentiation
𝒇 ′ 𝒙 𝟎 Approximation
Five-Point Midpoint Formula
Five-Point Endpoint Formula
𝒙 𝟎 −𝟐𝐡
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
𝒙 𝟎 +𝟑𝐡
𝒙 𝟎 +𝟒𝐡
Example
Use table 4.2 and five-point midpoint formula to approximate 𝑓 ′ 2 with ℎ=0.1
𝑓′(2)≈ =
Compute the actual error
The data given in Table 4.2 are taken from the function 𝑓 𝑥 = 𝑥𝑒 𝑥
The actual error = −𝑓′(1.8) = −

7. Five-Point Midpoint Formula Five-Point Endpoint Formula
Sec:4.1 Numerical Differentiation
𝒇 ′ 𝒙 𝟎 Approximation
Five-Point Midpoint Formula
Five-Point Endpoint Formula
𝒙 𝟎 −𝟐𝐡
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
𝒙 𝟎 +𝟑𝐡
𝒙 𝟎 +𝟒𝐡
Example
Use table 4.2 and five-point midpoint formula to approximate 𝑓 ′ 2 with ℎ=0.1
𝑓′(2)≈ =
Compute the actual error
The data given in Table 4.2 are taken from the function 𝑓 𝑥 = 𝑥𝑒 𝑥
The actual error = −𝑓′(1.8) = −

8. Three-Point Midpoint Formula Three-Point Endpoint Formula
Sec:4.1 Numerical Differentiation
𝒇 ′ 𝒙 𝟎 Approximation
Three-Point Midpoint Formula
Three-Point Endpoint Formula
𝒙 𝟎
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
Example
Use table 4.2 and three-point midpoint formula to approximate 𝑓 ′ 2 (consider all possible cases)
𝒉=𝟎.𝟏
1 2(0.1) 𝑓 2.1 −𝑓(1.9)
𝑓′(2)≈
1 2(0.2) 𝑓 2.2 −𝑓(1.8)
𝑓′(2)≈
𝒉=𝟎.𝟐

9. Three-Point Midpoint Formula Three-Point Endpoint Formula
Sec:4.1 Numerical Differentiation
𝒇 ′ 𝒙 𝟎 Approximation
Three-Point Midpoint Formula
Three-Point Endpoint Formula
𝒙 𝟎
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
Example
Use table 4.2 and three-point endpoint formula to approximate 𝑓 ′ (consider all possible cases)
𝒉=−𝟎.𝟏
1 2(−0.1) −3𝑓 𝑓 2.1 −𝑓(2)
𝑓′(2.2)≈
𝒉=−𝟎.𝟐
1 2(−0.2) −3𝑓 𝑓 2 −𝑓(1.8)
𝑓′(2.2)≈

10. Sec:4.1 Numerical Differentiation
𝒇 ′ 𝒙 𝟎 Approximation
Three-Point Midpoint Formula
Three-Point Endpoint Formula
𝒙 𝟎
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
Five-Point Midpoint Formula
Five-Point Endpoint Formula
𝒙 𝟎
𝒙 𝟎 +𝟐𝐡
𝒙 𝟎 −𝟐𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟑𝐡
𝒙 𝟎 +𝟒𝐡
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡

11. Sec:4.1 Numerical Differentiation
Comparison Study
Example
𝑓′(2.0)≈
True Error
The true value
Three-point endpoint with h = 0.1
1.35 × e−1
Three-point endpoint with h = −0.1
1.13 e−1
Three-point midpoint with h = 0.1
−6.16 e-2
Three-point midpoint with h = 0.2
−2.47 e-1
Five-point midpoint with h = 0.1
1.69 e-4
Table 4.2 represents values of 𝑓 𝑥 =𝑥 𝑒 𝑥
If we had no other information we would accept the five-point midpoint approximation using h = 0.1 as the most accurate, and expect the true value to be between that approximation and the three-point mid-point approximation that is in the interval [22.166, ].

12. Second Derivative Midpoint Formula
Sec:4.1 Numerical Differentiation
𝒇 ′ ′ 𝒙 𝟎 Approximation
Second Derivative Midpoint Formula
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
Example
Use table 4.2 and Second Derivative Midpoint Formula to approximate 𝑓 ′ ′ 2 with ℎ=0.1,0.2
𝑓′′(2)≈
𝑓′′(2)≈ = = = =
 𝑓′′(2.0)≈
True Error
The true value
midpoint with h = 0.1
−3.70 e-2
midpoint with h = 0.2
−1.48 e-1

13. Sec:4.1 Numerical Differentiation
𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒐𝒏𝒔
𝐿𝑎𝑔𝑟𝑎𝑛𝑔𝑒 𝐼𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙
𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒐𝒏𝒔
𝑻𝒉𝒆𝒔𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂𝒔
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠
Product of two functions
𝑫 𝒙 𝒇𝒈 =
𝒇 ′ 𝒈+𝒇𝒈′
𝑫 𝒙 𝒙− 𝒙 𝟏 𝒙− 𝒙 𝟐 =
𝒙− 𝒙 𝟐 + 𝒙− 𝒙 𝟏
Product of three functions
𝑫 𝒙 𝒇𝒈𝒌 =
𝒇 ′ 𝒈𝒌+𝒇 𝒈 ′ 𝒌+𝒇𝒈𝒌′
𝑫 𝒙 𝒙− 𝒙 𝟏 𝒙− 𝒙 𝟐 𝒙− 𝒙 𝟑 =
𝒙− 𝒙 𝟐 𝒙− 𝒙 𝟑 + 𝒙− 𝒙 𝟏 𝒙− 𝒙 𝟑
+ 𝒙− 𝒙 𝟏 𝒙− 𝒙 𝟐

14. Three-Point Midpoint Formula Three-Point Endpoint Formula
Sec:4.1 Numerical Differentiation
𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒐𝒏𝒔
Three-Point Midpoint Formula
Three-Point Endpoint Formula
𝒙 𝟎
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
the second Lagrange polynomial
𝑓 𝑥 =𝑓 𝑥 0 −ℎ 𝑥− 𝑥 0 𝑥− 𝑥 0 −ℎ −ℎ −2ℎ +𝑓 𝑥 0 𝑥− 𝑥 0 +ℎ 𝑥− 𝑥 0 −ℎ ℎ −ℎ +𝑓 𝑥 0 +ℎ 𝑥− 𝑥 0 +ℎ 𝑥− 𝑥 ℎ ℎ
+ 𝒇′′′(𝝃(𝒙)) 𝟑! (𝒙− 𝒙 𝟎 +𝒉)(𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉)
Differentiate both sides
𝑓′ 𝑥 =𝑓 𝑥 0 −ℎ 𝑥− 𝑥 0 + 𝑥− 𝑥 0 −ℎ −ℎ −2ℎ +𝑓 𝑥 0 𝑥− 𝑥 0 +ℎ + 𝑥− 𝑥 0 −ℎ ℎ −ℎ +𝑓 𝑥 0 +ℎ 𝑥− 𝑥 0 +ℎ + 𝑥− 𝑥 ℎ ℎ
+ 𝒇′′′(𝝃(𝒙)) 𝟑! 𝑫 (𝒙− 𝒙 𝟎 +𝒉)(𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉) +𝑫 𝒇′′′(𝝃(𝒙)) 𝟑! (𝒙− 𝒙 𝟎 +𝒉)(𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉)
𝑫 (𝒙− 𝒙 𝟎 +𝒉)(𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉) = (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉)+ (𝒙− 𝒙 𝟎 +𝒉)(𝒙− 𝒙 𝟎 −𝒉)+ (𝒙− 𝒙 𝟎 +𝒉)(𝒙− 𝒙 𝟎 )
Set 𝑥= 𝑥 0
𝑓′ 𝑥 0 =𝑓 𝑥 0 −ℎ −ℎ −ℎ −2ℎ +𝑓 𝑥 0 ℎ + −ℎ ℎ −ℎ +𝑓 𝑥 0 +ℎ ℎ 2ℎ ℎ
+ 𝒇 ′′′ 𝝃 𝒙 𝟑! (− 𝒉 𝟐 )

15. Three-Point Midpoint Formula Three-Point Endpoint Formula
Sec:4.1 Numerical Differentiation
Three-Point Midpoint Formula
Three-Point Endpoint Formula
𝒙 𝟎
𝒙 𝟎 −𝐡
𝒙 𝟎
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝐡
𝒙 𝟎 +𝟐𝐡
the second Lagrange polynomial
𝑓 𝑥 =𝑓 𝑥 0 𝑥− 𝑥 0 −ℎ 𝑥− 𝑥 0 −2ℎ −ℎ −2ℎ +𝑓 𝑥 0 +ℎ 𝑥− 𝑥 0 𝑥− 𝑥 0 −2ℎ ℎ −ℎ +𝑓 𝑥 0 +2ℎ 𝑥− 𝑥 0 𝑥− 𝑥 0 −ℎ 2ℎ ℎ
+ 𝒇 ′′′ 𝝃 𝒙 𝟑! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉)(𝒙− 𝒙 𝟎 −𝟐𝒉)
Differentiate both sides
𝑓′ 𝑥 =𝑓 𝑥 0 𝑥− 𝑥 0 −ℎ + 𝑥− 𝑥 0 −2ℎ −ℎ −2ℎ +𝑓 𝑥 0 +ℎ 𝑥− 𝑥 0 + 𝑥− 𝑥 0 −2ℎ ℎ −ℎ +𝑓 𝑥 0 +2ℎ 𝑥− 𝑥 0 + 𝑥− 𝑥 0 −ℎ 2ℎ ℎ
+𝐷 (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉)(𝒙− 𝒙 𝟎 −𝟐𝒉) 𝒇 ′′′ 𝝃 𝒙 𝟑! +𝐷 𝒇 ′′′ 𝝃 𝒙 𝟑! (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉)(𝒙− 𝒙 𝟎 −𝟐𝒉)
𝐷 (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉)(𝒙− 𝒙 𝟎 −𝟐𝒉) = (𝒙− 𝒙 𝟎 −𝒉)(𝒙− 𝒙 𝟎 −𝟐𝒉) + (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝟐𝒉) + (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉)
Set 𝑥= 𝑥 0
𝑓′ 𝑥 0 =𝑓 𝑥 0 −ℎ + −2ℎ −ℎ −2ℎ +𝑓 𝑥 0 +ℎ −2ℎ ℎ −ℎ +𝑓 𝑥 0 +2ℎ −ℎ 2ℎ ℎ
+𝐷 (𝒙− 𝒙 𝟎 )(𝒙− 𝒙 𝟎 −𝒉)(𝒙− 𝒙 𝟎 −𝟐𝒉) 𝒇 ′′′ 𝝃 𝒙 𝟑!

16. Sec:4.1 Numerical Differentiation
𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒐𝒏𝒔
𝐿𝑎𝑔𝑟𝑎𝑛𝑔𝑒 𝐼𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙
𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒐𝒏𝒔
𝑻𝒉𝒆𝒔𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂𝒔
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠
Use Lagrange polynomial and the given nodes to express 𝑓(𝑥) as 𝑃(𝑥)+𝑒𝑟𝑟𝑜𝑟 then differentiate both sides. Next set 𝑥 = 𝑥 0

17. Sec:4.1 Numerical Differentiation
𝑻𝒂𝒚𝒍𝒐𝒓 𝒔𝒆𝒓𝒊𝒆𝒔
𝑓 𝑥 0 +𝑘 =𝑓 𝑥 0 +𝑘𝑓′ 𝑥 𝑘 2 𝑓′′ 𝑥 𝑘 3 𝑓′′′ 𝑥 𝑘 4 𝑓′′′′ 𝑥 0 +⋯
𝒔𝒆𝒕 𝒌=𝒉
𝑓 𝑥 0 +ℎ =𝑓 𝑥 0 +ℎ𝑓′ 𝑥 ℎ 2 𝑓′′ 𝑥 ℎ 3 𝑓′′′ 𝑥 ℎ 4 𝑓′′′′ 𝑥 0 +⋯
𝒔𝒆𝒕 𝒌=𝟐𝒉
𝑓 𝑥 0 +2ℎ =𝑓 𝑥 0 +2ℎ𝑓′ 𝑥 ℎ 2 𝑓′′ 𝑥 ℎ 3 𝑓′′′ 𝑥 ℎ 4 𝑓′′′′ 𝑥 0 +⋯
𝒔𝒆𝒕 𝒌=−𝒉
𝑓 𝑥 0 −ℎ =𝑓 𝑥 0 −ℎ𝑓′ 𝑥 ℎ 2 𝑓′′ 𝑥 0 − 1 6 ℎ 3 𝑓′′′ 𝑥 ℎ 4 𝑓′′′′ 𝑥 0 +⋯
𝒔𝒆𝒕 𝒌=−𝟐𝒉
𝑓 𝑥 0 −2ℎ =𝑓 𝑥 0 −2ℎ𝑓′ 𝑥 ℎ 2 𝑓′′ 𝑥 0 − 4 3 ℎ 3 𝑓′′′ 𝑥 ℎ 4 𝑓′′′′ 𝑥 0 +⋯

18. Sec:4.1 Numerical Differentiation
Example
Derive a three-point formula to approximate f’(x0) that uses f (x0), f (x0 + h), f (x0 + 2h). Then find the error term [what is the order of the error O( ℎ 𝑚 ) ]
𝒇 ′ 𝒙 𝟎 =𝑨 𝒇 𝒙 𝟎 +𝑩𝒇 𝒙 𝟎 +𝒉 +𝑪𝒇 𝒙 𝟎 +𝟐𝒉 +𝑬𝒓𝒓𝒐𝒓 𝑨
𝑓 𝑥 0 =𝑓 𝑥 0 𝑩
𝑓 𝑥 0 +ℎ =𝑓 𝑥 0 +ℎ 𝑓 ′ 𝑥 ℎ 2 𝑓 ′′ 𝑥 ℎ 3 𝑓 ′′′ 𝑥 0 +⋯ 𝑪
𝑓 𝑥 0 +2ℎ =𝑓 𝑥 0 +2ℎ 𝑓 ′ 𝑥 ℎ 2 𝑓 ′′ 𝑥 ℎ 3 𝑓 ′′′ 𝑥 0 +⋯
+ ℎ 𝐵+2𝐶 𝑓 ′′ 𝑥 0
𝑨 𝒇 𝒙 𝟎
+𝑩𝒇 𝒙 𝟎 +𝒉
+𝑪𝒇 𝒙 𝟎 +𝟐𝒉
𝐴+𝐵+𝐶 𝑓 𝑥 0 +ℎ 𝐵+2𝐶 𝑓 ′ 𝑥 0 =
+ ( 1 6 𝐵𝑓′′′ 𝑥 𝐶𝑓′′′ 𝑥 0 )ℎ 3
We have 3 constants to determine (A, B, C). So we need three equations
𝐴+𝐵+𝐶 =0
𝑨= −𝟑 𝟐𝒉
𝑩= 𝟐 𝒉
𝑪= −𝟏 𝟐𝒉
ℎ 𝐵+2𝐶 =0
ℎ 𝐵+2𝐶 =1
𝒇 ′ 𝒙 𝟎 ≈− 𝟑 𝟐𝒉 𝒇 𝒙 𝟎 + 𝟐 𝒉 𝒇 𝒙 𝟎 +𝒉 − 𝟏 𝟐𝒉 𝒇 𝒙 𝟎 +𝟐𝒉

19. Sec:4.1 Numerical Differentiation
Example
Derive a three-point formula to approximate f’(x0) that uses f (x0), f (x0 + h), f (x0 + 2h). Then find the error term [what is the order of the error O( ℎ 𝑚 ) ]
𝒇 ′ 𝒙 𝟎 =𝑨 𝒇 𝒙 𝟎 +𝑩𝒇 𝒙 𝟎 +𝒉 +𝑪𝒇 𝒙 𝟎 +𝟐𝒉 +𝑬𝒓𝒓𝒐𝒓 𝑨
𝑓 𝑥 0 =𝑓 𝑥 0 𝑩
𝑓 𝑥 0 +ℎ =𝑓 𝑥 0 +ℎ 𝑓 ′ 𝑥 ℎ 2 𝑓 ′′ 𝑥 ℎ 3 𝑓 ′′′ 𝑥 0 +⋯ 𝑪
𝑓 𝑥 0 +2ℎ =𝑓 𝑥 0 +2ℎ 𝑓 ′ 𝑥 ℎ 2 𝑓 ′′ 𝑥 ℎ 3 𝑓 ′′′ 𝑥 0 +⋯
+ ℎ 𝐵+2𝐶 𝑓 ′′ 𝑥 0
𝑨 𝒇 𝒙 𝟎
+𝑩𝒇 𝒙 𝟎 +𝒉
+𝑪𝒇 𝒙 𝟎 +𝟐𝒉
𝐴+𝐵+𝐶 𝑓 𝑥 0 +ℎ 𝐵+2𝐶 𝑓 ′ 𝑥 0 =
+ ( 1 6 𝐵𝑓′′′ 𝑥 𝐶𝑓′′′ 𝑥 0 )ℎ 3
find the error term
𝒇 ′ 𝒙 𝟎 ≈− 𝟑 𝟐𝒉 𝒇 𝒙 𝟎 + 𝟐 𝒉 𝒇 𝒙 𝟎 +𝒉 − 𝟏 𝟐𝒉 𝒇 𝒙 𝟎 +𝟐𝒉
𝑨= −𝟑 𝟐𝒉
𝑩= 𝟐 𝒉
𝑪= −𝟏 𝟐𝒉
Leading
error term
=( 1 6 𝐵𝑓′′′ 𝑥 𝐶𝑓′′′ 𝑥 0 )ℎ 3
= ( 𝟏 𝟑𝒉 𝒇 ′′′ 𝒙 𝟎 − 𝟐 𝟑𝒉 𝒇′′′ 𝒙 𝟎 )𝒉 𝟑
= − 𝟏 𝟑 𝒇′′′ 𝒙 𝟎 𝒉 𝟐
𝑬𝒓𝒓𝒐𝒓=𝑶( 𝒉 𝟐 )
𝒇 ′ 𝒙 𝟎 =− 𝟑 𝟐𝒉 𝒇 𝒙 𝟎 + 𝟐 𝒉 𝒇 𝒙 𝟎 +𝒉 − 𝟏 𝟐𝒉 𝒇 𝒙 𝟎 +𝟐𝒉 − 𝟏 𝟑 𝒇′′′ 𝝃 𝒉 𝟐

20. Sec:4.1 Numerical Differentiation
Example
Derive a three-point formula to approximate f’’(x0) that uses f (x0), f (x0 + h), f (x0 + 2h).
𝒇 ′′ 𝒙 𝟎 =𝑨 𝒇 𝒙 𝟎 +𝑩𝒇 𝒙 𝟎 +𝒉 +𝑪𝒇 𝒙 𝟎 +𝟐𝒉 +𝑬𝒓𝒓𝒐𝒓 𝑨
𝑓 𝑥 0 =𝑓 𝑥 0 𝑩
𝑓 𝑥 0 +ℎ =𝑓 𝑥 0 +ℎ 𝑓 ′ 𝑥 ℎ 2 𝑓 ′′ 𝑥 ℎ 3 𝑓 ′′′ 𝑥 0 +⋯ 𝑪
𝑓 𝑥 0 +2ℎ =𝑓 𝑥 0 +2ℎ 𝑓 ′ 𝑥 ℎ 2 𝑓 ′′ 𝑥 ℎ 3 𝑓 ′′′ 𝑥 0 +⋯
+ ℎ 𝐵+2𝐶 𝑓 ′′ 𝑥 0
𝑨 𝒇 𝒙 𝟎
+𝑩𝒇 𝒙 𝟎 +𝒉
+𝑪𝒇 𝒙 𝟎 +𝟐𝒉
𝐴+𝐵+𝐶 𝑓 𝑥 0 +ℎ 𝐵+2𝐶 𝑓 ′ 𝑥 0 =
+ ( 1 6 𝐵𝑓′′′ 𝑥 𝐶𝑓′′′ 𝑥 0 )ℎ 3
We have 3 constants to determine (A, B, C). So we need three equations
𝐴+𝐵+𝐶 =0
ℎ 𝐵+2𝐶 =𝟏
ℎ 𝐵+2𝐶 =𝟎

21. Sec:4.1 Numerical Differentiation
𝑓 ′ 𝑥 0 =𝑨𝑓 𝑥 0 −ℎ +𝑩 𝑓 𝑥 0 +𝑪𝑓 𝑥 0 +ℎ +𝑫𝑓 𝑥 0 +2ℎ +𝑬𝑓 𝑥 0 +3ℎ +𝑬𝒓𝒓𝒐𝒓 𝑨
𝑓 𝑥 0 −ℎ =𝑓 𝑥 0 −ℎ𝑓′ 𝑥 ℎ 2 𝑓′′ 𝑥 0 − 1 6 ℎ 3 𝑓′′′ 𝑥 ℎ 4 𝑓′′′′ 𝑥 0 +⋯ 𝑩
𝑓 𝑥 0 =𝑓 𝑥 0 𝑪
𝑓 𝑥 0 +ℎ =𝑓 𝑥 0 +ℎ𝑓′ 𝑥 ℎ 2 𝑓′′ 𝑥 ℎ 3 𝑓′′′ 𝑥 ℎ 4 𝑓′′′′ 𝑥 0 +⋯
𝑓 𝑥 0 +2ℎ =𝑓 𝑥 0 +2ℎ𝑓′ 𝑥 ℎ 2 𝑓′′ 𝑥 ℎ 3 𝑓′′′ 𝑥 ℎ 4 𝑓′′′′ 𝑥 0 +⋯ 𝑫 𝑬
𝑓 𝑥 0 +2ℎ =𝑓 𝑥 0 +2ℎ𝑓′ 𝑥 ℎ 2 𝑓′′ 𝑥 ℎ 3 𝑓′′′ 𝑥 ℎ 4 𝑓′′′′ 𝑥 0 +⋯
𝑳𝑯𝑺=
𝐴+𝐵+𝐶+𝐷+𝐸 𝑓 𝑥 0 +⋯𝑓′ 𝑥 0 +⋯𝑓′′ 𝑥 0
+⋯𝑓′′′ 𝑥 0 +⋯𝑓′′′′ 𝑥 0 +⋯

22. Sec:4.1 Numerical Differentiation