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Numerical Analysis with Python
Truncation errors and Tylor’s Series
LECTURE 03
• Truncation Errors
• Taylor’s Series
• Taylor’s Series remainder
• Maclaurin’s series
• Function approximation using Taylors Series
• Numerical differentiation
Truncation Errors
• ‘Truncation errors are those that result from using anapproximation in place of an exact mathematical procedure’.
• Example: Using finite-divided-difference equation to approximatethe derivative of function. E.g. velocity of a falling object undergravitational influence.
𝑑𝑣
𝑑𝑡≅Δ𝑣
Δ𝑡=𝑣 𝑡𝑖+1 − 𝑣 𝑡𝑖
𝑡𝑖+1 − 𝑡𝑖(1.1)
Truncation Errors
• A truncation error was introduced into the numerical solutionbecause the difference equation only approximates the true value ofthe derivative.
• One of the most widely used mathematical formulation in numericalmethods to express functions in an approximate fashion is Taylorseries.
• Taylor series provides a means to predict a function value at onepoint in terms of the function value and its derivatives at anotherpoint.
Tylor’s Series
Theorem:
“If the function f and its first (𝑛 + 1) derivatives are continuous on aninterval containing a and x, then the value of the function at x is givenby:”
Where 𝑅𝑛 is the remainder term is included to account for all termsfrom n+1 to infinity:
𝑓 𝑥 = 𝑓 𝑎 + 𝑓′ 𝑎 𝑥 − 𝑎 +𝑓′′ 𝑎
2!(𝑥 − 𝑎)2+
𝑓(3) 𝑎
3!(𝑥 − 𝑎)3. . . . +
𝑓(𝑛) 𝑎
𝑛!(𝑥 − 𝑎)𝑛+𝑅𝑛
Tylor’s Series Remainder
Remainder:
𝑅𝑛 is the remainder term used in Taylors series to account forTruncation Errors in all terms from (𝑛 + 1) to infinity
• Where 𝜉 is a point between a and x
𝑅𝑛 =𝑓(𝑛+1) 𝜉
(𝑛 + 1)!(𝑥 − 𝑎)𝑛+1
Maclaurin: special case of Taylor
Maclaurin series:
It is a special case of Taylor series with an exponential functionstarting at origin i.e. a=0
Proof:
• The Taylor series reduces to:
• Simplifying and truncating the remainder 𝑅𝑛
𝑓𝑜𝑟 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙, 𝑙𝑒𝑡 𝑓 𝑥 = 𝑒𝑥 𝑎𝑛𝑑 𝑎 = 0
𝑓 𝑥 = 𝑓 0 + 𝑓′ 0 𝑥 − 0 +𝑓′′ 0
2!(𝑥 − 0)2+
𝑓(3) 0
3!(𝑥 − 0)3. . . . +
𝑓(𝑛) 0
𝑛!(𝑥 − 0)𝑛+𝑅𝑛
𝑓 𝑥 = 𝑓 0 + 𝑓′ 0 𝑥 +𝑓′′ 0
2!(𝑥)2+
𝑓(3) 0
3!(𝑥)3. . . . +
𝑓(𝑛) 0
𝑛!(𝑥)𝑛
Maclaurin: special case of Taylor
Proof:
• The Taylor series reduces to:
• Hence, 𝑓 𝑥 = 0 = 𝑒0 = 1 , 𝑓′ 0 = 𝑒0 = 1 , 𝑓′′ 0 = 𝑒0 = 1
• Therefore on substitution above:
𝑓𝑜𝑟 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙, 𝑙𝑒𝑡 𝑓 𝑥 = 𝑒𝑥
𝑓 𝑥 = 𝑓 0 + 𝑓′ 0 𝑥 +𝑓′′ 0
2!(𝑥)2+
𝑓(3) 0
3!(𝑥)3. . . . +
𝑓(𝑛) 0
𝑛!(𝑥)𝑛
𝑑𝑓(𝑥)
𝑑𝑥= 𝑓′ 𝑥 = 𝑒𝑥 𝑓′′ 𝑥 = 𝑒𝑥 𝑓′′′ 𝑥 = 𝑒𝑥 𝑎𝑛𝑑 𝑠𝑜 𝑜𝑛 …
𝑒𝑥 = 1 + 1 ∗ 𝑥 +1
2!(𝑥)2+
1
3!(𝑥)3. . . . +
1
𝑛!(𝑥)𝑛
Maclaurin: special case of Taylor
Proof:
• Hence, Maclaurin Series is a special case of Tylor’s series forexponential function 𝒇 𝒙 = 𝒆𝒙 over the interval 𝟎 𝒙 𝒊. 𝒆. 𝒂𝒕 𝒂 =𝟎
𝒆𝒙 = 𝟏 + 𝒙 +𝒙𝟐
𝟐!+𝒙𝟑
𝟑!+𝒙𝟒
𝟒!+ . . . . +
𝒙𝒏
𝒏!
Maclaurin Series
Numerical form of Taylor series
Numerical Form:
The numerical equivalent of Taylor series expansion of a function isas follow
• Letting (𝑥𝑖+1−𝑥𝑖) = ℎ , which is the incremental difference knownas the step size, the series can be rewritten:
• And the remainder
𝑓 𝑥𝑖+1 = 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 𝑥𝑖+1 − 𝑥𝑖 +𝑓′′ 𝑥𝑖2!
(𝑥𝑖+1 − 𝑥𝑖)2+
𝑓(3) 𝑥𝑖
3!(𝑥𝑖+1 − 𝑥𝑖)
3. . . . +𝑓𝑛 𝑥𝑖
𝑛!(𝑥𝑖+1 − 𝑥𝑖)
𝑛+𝑅𝑛
𝑓 𝑥𝑖+1 = 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 ℎ +𝑓′′ 𝑥𝑖2!
ℎ2 +𝑓(3) 𝑥𝑖
3!ℎ3. . . . +
𝑓𝑛 𝑥𝑖𝑛!
ℎ𝑛 + 𝑅𝑛
𝑅𝑛 =𝑓(𝑛+1) 𝜉
(𝑛 + 1)!ℎ𝑛+1
Taylor series: Examples
Problem statement 1:
Use zero- through fourth-order Taylor series expansions toapproximate the function at𝑥𝑖+1 =1 starting at 𝑥𝑖 = 0
Solution:
𝑓 𝑥 = −0.1𝑥4 − 0.15𝑥3 − 0.5𝑥2 − 0.25𝑥 + 1.2
Step size ℎ = 𝑥𝑖+1 − 𝑥𝑖 = 1,
exact value at 𝑥𝑖+1 = 1, 𝑓 𝑥𝑖+1 = 1 =0.2
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑎𝑡 𝑛 = 0;
𝑓 𝑥𝑖+1 = 𝑓 𝑥𝑖
𝑓 𝑥𝑖+1 = −0.1 ∗ 04 −0.15 ∗ 03 − 0.5 ∗ 02 − 0.25 ∗ 0 + 1.2 = 1.2
Taylor series: Examples
𝐹𝑜𝑟 𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑎𝑡 𝑛 = 1
𝑓 𝑥𝑖+1 = 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 ℎ
𝑓′ 𝑥𝑖 = −0.4𝑥3 − 0.45𝑥2 − 1 ∗ 𝑥 − 0.25
𝑓′ 𝑥𝑖 = 0 = −0.4 ∗ 03 −0.45 ∗ 02 − 1 ∗ 0 − 0.25 = −0.25
𝑓 𝑥𝑖+1 = 1.2 + −0.25 ∗ ℎ = 𝟎. 𝟗𝟓
Taylor series: Examples
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑎𝑡 𝑛 = 2
𝑓 𝑥𝑖+1 = 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 ℎ +𝑓′′ 𝑥𝑖2!
ℎ2
𝑓′′ 𝑥𝑖 = −1.2𝑥2 − 0.9𝑥 − 1
𝑓′′ 𝑥𝑖 = 0 = −1.2 ∗ 02 −0.9 ∗ 0 − 1 = −1
𝑓 𝑥𝑖+1 = 1.2 − 0.25ℎ −1
2 ∗ 1ℎ2 = 𝟎. 𝟒𝟓
Taylor series: Examples
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑎𝑡 𝑛 = 3
𝑓 𝑥𝑖+1 = 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 ℎ +𝑓′′ 𝑥𝑖2!
ℎ2 +𝑓(3) 𝑥𝑖
3!ℎ3
𝑓(3) 𝑥𝑖 = −2.4𝑥 − 0.9
𝑓(3) 𝑥𝑖 = 0 = −2.4 ∗ 0 − 0.9 = −0.9
𝑓 𝑥𝑖+1 = 1.2 − 0.25ℎ −1
2 ∗ 1ℎ2 −
0.9
3 ∗ 2 ∗ 1ℎ2 = 𝟎. 𝟑
Taylor series: Examples
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑎𝑡 𝑛 = 4
𝑓 𝑥𝑖+1 = 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 ℎ +𝑓′′ 𝑥𝑖2!
ℎ2 +𝑓(3) 𝑥𝑖
3!ℎ3 +
𝑓(4) 𝑥𝑖4!
ℎ4
𝑓(4) 𝑥𝑖 = −2.4 𝑓 4 𝑥𝑖 = 0 = −2.4
𝑓 𝑥𝑖+1
= 1.2 − 0.25ℎ −1
2 ∗ 1ℎ2 −
0.9
3 ∗ 2 ∗ 1ℎ3 −
2.4
4 ∗ 3 ∗ 2 ∗ 1ℎ4
= 𝟎. 𝟐
Taylor series: Examples
𝐻𝑒𝑛𝑐𝑒 fourth-order Taylor series expansions
𝑓 𝑥 = −0.1𝑥4 − 0.15𝑥3 − 0.5𝑥2 − 0.25𝑥 + 1.2
𝑓 𝑥𝑖+1 = 1 = 1.2 − 0.25 ∗ 1 −1
2 ∗ 1∗ 12
−0.9
3∗2∗1∗ 13 −
2.4
4∗3∗2∗1∗ 14
𝑓 1 = 𝟎. 𝟐
Taylor series: Examples
Problem statement 2:
Use Taylor series expansions with n=0 to 6 to approximate 𝑓 𝑥 =𝑐𝑜𝑠𝑥 𝑎𝑡 𝑥𝑖+1 =
𝜋
3on the basis of the value of f(x) and its derivatives at
𝑥𝑖 =𝜋
4.
Solution:
Step size ℎ = 𝑥𝑖+1 − 𝑥𝑖 =𝜋
3−
𝜋
4=
𝜋
12
exact value at 𝑥𝑖+1 =𝜋
3, 𝑓
𝜋
3= cos
𝜋
3= 0.5
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑎𝑡 𝑛 = 0;
𝑓 𝑥𝑖+1 ≅ 𝑓 𝑥𝑖
𝑓 𝑥𝑖+1 ≅ cos𝜋
4= 𝟎. 𝟕𝟎𝟕𝟏𝟎𝟔
Taylor series: Examples
Solution:
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑎𝑡 𝑛 = 1
𝑓 𝑥𝑖+1 ≅ 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 ℎ
𝑓(1) 𝑥𝑖 = −sin 𝑥𝑖 , 𝑓 1 𝑥𝑖 =𝜋
4= −sin
𝜋
4= −0.707106
𝑓 𝑥𝑖+1 ≅ 0.707106 + −0.707106 ∗𝜋
12= 0.521986659
Taylor series: Examples
Solution:
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑎𝑡 𝑛 = 2
𝑓 𝑥𝑖+1 ≅ 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 ℎ + 𝑓′ 𝑥𝑖 ℎ +𝑓′′ 𝑥𝑖2!
ℎ2
𝑓(2) 𝑥𝑖 = −cos 𝑥𝑖 , 𝑓 2 𝑥𝑖 =𝜋
4= −𝑐𝑜𝑠
𝜋
4= −0.707106
𝑓 𝑥𝑖+1 ≅ cos𝜋
4−sin
𝜋
4
𝜋
12−𝑐𝑜𝑠
𝜋4
2!
𝜋
12
2
= 0.497754491
Taylor series: Examples
Solution:𝐴𝑓𝑡𝑒𝑟 𝑐𝑜𝑚𝑝𝑢𝑡𝑖𝑛𝑔 𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑢𝑝 𝑡𝑜 𝑛 = 6 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑟𝑒𝑠𝑢𝑙𝑡𝑠 𝑎𝑟𝑒 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑
Taylor series: Examples with python
Problem statement 3:
Implement a program in python to solve the problem statement 1 and 2and display the results both in tabular and graphical form. Hint usethe symbolic expression modules (sympy) to compute the derivativesterms in the series.
Clue: to compute and evaluate derivative of
Truncation error in Taylor Series
• For a simple monomial with a finite number of derivatives. Thispermits a complete determination of the Taylor series remainder.
Problem statement 4:
Employ the zero-order to first-order Taylor series to approximate thefunction 𝑓 𝑥 = 𝑥4 at 𝑥𝑖+1=2 when 𝑥𝑖 = 1. Compute the remainderof the Taylor series approximation.
Solution
Step size ℎ = 𝑥𝑖+1 − 𝑥𝑖 = 1,
exact value at 𝑥𝑖+1 = 1, 𝑓 𝑥𝑖+1 = 2 = 24 =16
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑎𝑡 𝑛 = 0;
𝑓 𝑥𝑖+1 ≅ 𝑓 𝑥𝑖 𝑓 𝑥𝑖+1 ≅ 14 = 𝟏
Truncation error in Taylor Series
𝑎𝑡 𝑛 = 0
𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟, 𝑅𝑛 = 𝑓′ 𝑥𝑖 ℎ +𝑓′′ 𝑥𝑖2!
ℎ2 +𝑓(3) 𝑥𝑖
3!ℎ3. . . . +
𝑓𝑛 𝑥𝑖𝑛!
ℎ𝑛
Since 𝑓′ 𝑥 = 4𝑥3, 𝑓(2) 𝑥 = 12𝑥2, 𝑓(3) 𝑥 = 24𝑥, 𝑓(4) 𝑥 = 24
Any higher derivatives are zero i.e. 𝑓(5) 𝑥 = 0 and so on---
𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟, 𝑅𝑛 = 4𝑥3ℎ +12𝑥2
2!ℎ2 +
24𝑥
3!ℎ3 +
24
4!ℎ4 + 0 + 0 +⋯
Truncation error in Taylor Series
Solution
At 𝑥 = 1
𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟, 𝑅𝑛
= 4 ∗ 13 1 +12 ∗ 12
2!12 +
24 ∗ 1
3!13 +
24
4!14 + 0 + 0 = 𝟏𝟓
𝑇𝑎𝑦𝑙𝑜𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑎𝑡 𝑛 = 1
𝑓 𝑥𝑖+1 ≅ 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 ℎ= 14+ 4 ∗ 13=5
𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟, 𝑅𝑛 =12 ∗ 12
2!12 +
24 ∗ 1
3!13 +
24
4!14 + 0 + 0 = 𝟏𝟏
Numerical Differentiation
• Taylor Series Expansion can be used to derive numericalapproximations of first-order derivatives of a function.
• Truncating the series after the first derivative term:
• On rearranging
• Where 𝑓′ 𝑥𝑖 is the first-order derivatives
𝑓 𝑥𝑖+1 = 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 ℎ + 𝑅𝑛 1.2
𝑓′ 𝑥𝑖 =𝑓 𝑥𝑖+1 − 𝑓 𝑥𝑖
ℎ−𝑅𝑛ℎ
𝑓 𝑥𝑖+1 − 𝑓 𝑥𝑖ℎ
= 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 − 𝑜𝑟𝑑𝑒𝑟 𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛
𝑅𝑛ℎ
= 𝑖𝑠 𝑡ℎ𝑒 𝑇𝑟𝑢𝑛𝑐𝑎𝑡𝑖𝑜𝑛 𝐸𝑟𝑟𝑜𝑟
Finite-Divided-Difference: Forward difference
• When Taylor series is expanded forward to calculate a next value onthe basis of a present value.
• Truncating the remainder and using first-order term only from theseries we have:
𝜵𝒇𝒊 is referred to as the first forward difference
𝑓 𝑥𝑖+1 ≅ 𝑓 𝑥𝑖 + 𝑓′ 𝑥𝑖 ℎ
𝑓′ 𝑥𝑖 ≅𝑓 𝑥𝑖+1 − 𝑓 𝑥𝑖
ℎ=∇𝑓𝑖ℎ
Finite-Divided-Difference: Backward difference
• When Taylor series is expanded backward to calculate a previousvalue on the basis of a present value.
• Truncating the remainder and using first-order term only from theseries we have:
𝜵𝒇𝒊 is referred to as the first backward difference
𝑓 𝑥𝑖 ≅ 𝑓 𝑥𝑖−1 + 𝑓′ 𝑥𝑖 ℎ
𝑓′ 𝑥𝑖 ≅𝑓 𝑥𝑖 − 𝑓 𝑥𝑖−1
ℎ=∇𝑓𝑖ℎ
Finite-Divided-Difference: Centered Difference
• A third way to approximate the first derivative is to subtract Eq.(4.19) from the forward Taylor series expansion.
• The result yields
• 𝜵𝒇𝒊 is referred to as the first centered difference
𝑓′ 𝑥𝑖 ≅𝑓 𝑥𝑖+1 − 𝑓 𝑥𝑖−1
2ℎ=∇𝑓𝑖2ℎ
Finite-Divided-Difference: examples
Problem statement 5:
a) Use forward, backward and centered difference approximation toestimate the first derivative of the function below at x=0.5 using astep size h=0.5. Repeat the computation using h=0.25.
b) Estimate the truncation error in each case.
Solution:
𝑓 𝑥 = −0.1𝑥4 − 0.15𝑥3 − 0.5𝑥2 − 0.25𝑥 + 1.25
𝒖𝒔𝒊𝒏𝒈 𝒆𝒙𝒂𝒄𝒕 𝒎𝒆𝒕𝒉𝒐𝒅
𝑓′ 𝑥𝑖 = −0.4𝑥3 − 0.45𝑥2 − 1 ∗ 𝑥 − 0.25
𝑎𝑡 x=0.5
𝑓′ 0.5 = −0.4 ∗ 0.53 − 0.45 ∗ 0.52 − 1 ∗ 0.5 − 0.25 = −0.9125.
Finite-Divided-Difference: examples
Solution:
𝑢𝑠𝑖𝑛𝑔 𝒇𝒐𝒓𝒘𝒂𝒓𝒅 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑡 ℎ = 0.5, 𝑥𝑖 =0.5
𝑓′ 𝑥𝑖 ≅𝑓 𝑥𝑖+1 − 𝑓 𝑥𝑖
ℎ
𝑥𝑖+1 − 𝑥𝑖 = ℎ 𝑥𝑖+1 = 1
𝑓 0.5 = −0.1 ∗ 0.54 − 0.15 ∗ 0.53 − 0.5 ∗ 0.52 − 0.25 ∗ 0.5 + 1.25= 0.975𝑥𝑖+1=1𝑓 1 = 0.25
𝑓′ 0.5 =0.25 − 0.975
0.5= −1.45
Finite-Divided-Difference: examples
𝑢𝑠𝑖𝑛𝑔 𝒇𝒐𝒓𝒘𝒂𝒓𝒅 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑡 ℎ = 0.25, 𝑥𝑖+1 =0.5
𝑓 0.5 = 0.975
𝑓 0.75 = 0.686 𝑓′ 0.5 =0.686 − 0.975
0.25= −1.156
𝜀𝑡 =−0.9125− (−1.156)
−0.9125∗ 100 = 26.68%
𝜀𝑡 =−0.9125− (−1.45)
−0.9125∗ 100 = 58.9%
Finite-Divided-Difference: examples
𝑢𝑠𝑖𝑛𝑔 𝒃𝒂𝒄𝒌𝒘𝒂𝒓𝒅 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑡 ℎ = 0.5, 𝑥𝑖+1 =0.5
𝑓′ 𝑥𝑖 ≅𝑓 𝑥𝑖 − 𝑓 𝑥𝑖−1
ℎ
𝑥𝑖 − 𝑥𝑖−1 = ℎ 𝑥𝑖−1 = 0
𝑓(𝑥𝑖−1) = −0.1 ∗ 04 − 0.15 ∗ 03 − 0.5 ∗ 02 − 0.25 ∗ 0 + 1.2 = 𝟏. 𝟐𝟓
𝑓(𝑥𝑖) =𝑓 0.5 = −0.1 ∗ 0.54 − 0.15 ∗ 0.53 − 0.5 ∗ 0.52 − 0.25 ∗ 0.5 + 1.25= 0.975
𝑓′ 0.5 =0.975 − 1.25
0.5= −0.55
Finite-Divided-Difference: examples
𝑢𝑠𝑖𝑛𝑔 𝒃𝒂𝒄𝒌𝒘𝒂𝒓𝒅 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑡 ℎ = 0.25, 𝑥𝑖 =0.5
𝑓 0.25 = 1.153
𝑓 0.5 = 0.975
𝑓′ 0.5 =0.975 − 1.153
0.25= −0.712
𝜀𝑡 =−0.9125− (−0.712)
−0.9125∗ 100 = 22%
𝜀𝑡 =−0.9125− (−0.55)
−0.9125∗ 100 = 39.7%
Finite-Divided-Difference: examples
𝑢𝑠𝑖𝑛𝑔 𝒄𝒆𝒏𝒕𝒆𝒓𝒆𝒅 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑡 ℎ = 0.5, 𝑥𝑖 =0.5
𝑓′ 𝑥𝑖 ≅𝑓 𝑥𝑖+1 − 𝑓 𝑥𝑖−1
2ℎ
𝑓 0 = 1.25 𝑓 𝑥𝑖−1
𝑓 0.5 = 0.975 𝑓 𝑥𝑖
𝑓 1 = 0.25 𝑓 𝑥𝑖+1
𝑓′ 0.5 =0.25 − 1.25
2 ∗ 0.5= −1
𝜀𝑡 =−0.9125− (−1)
−0.9125∗ 100 = 9.59%
Finite-Divided-Difference: examples
𝑢𝑠𝑖𝑛𝑔 𝒄𝒆𝒏𝒕𝒆𝒓𝒆𝒅 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑎𝑡 ℎ = 0.25, 𝑥𝑖+1 =0.5
𝑓′ 𝑥𝑖 ≅𝑓 𝑥𝑖+1 − 𝑓 𝑥𝑖−1
2ℎ
𝑓 0.25 = 1.153 𝑓 𝑥𝑖−1
𝑓 0.5 = 0.975 𝑓 𝑥𝑖
𝑓 0.75 = 0.686 𝑓 𝑥𝑖+1
𝑓′ 0.5 =0.686 − 1.153
2 ∗ 0.25= −0.934
𝜀𝑡 =−0.9125− (−0.934)
−0.9125∗ 100 = 2.36%