Numerical Methods with MATLAB

1. sayısalyöntemlervematlab
uygulamalar
Numerical Methods And MATLAB
Applications
“interpolation method
PrepearedbyAliAbdullah.yousif

2. Content:
 Newton's forward interpolation method
 Newton's backward interpolation method
 central difference interpolation
 gauss forward difference interpolation formula
 Gauss backward difference interpolation formula
 Stirling interpolation formula

4. Newton's forward interpolation
method
Forward difference operator ∆𝑓 𝑥 = 𝑓(𝑥 + ℎ) − 𝑓(𝑥) are shown. wherein the first table. a difference
table as shown is prepared. 𝑥𝑜 − 𝑥1 The difference between (𝑓 𝑥𝑜 , 𝑓 𝑥1 ) the values
corresponding to the data is represented. Overall, ∆𝑓(𝑥)with the of the differences and higher-
order differences ∆²𝑓 are calculated as the difference of the previous differences. difference values.
∆𝑓 𝑥𝑜 = 𝑓(𝑥𝑜 + ℎ) − 𝑓(𝑥𝑜)
∆²𝑓(xo)= ∆ ∆𝑓 𝑥𝑜 = ∆𝑓 𝑥𝑜 + ℎ − ∆𝑓 𝑥𝑜
∆↑ 𝑟 + 1 𝑓 𝑥𝑜 = ∆ʳ𝑓 𝑥𝑜 + ℎ − ∆ʳ𝑓 𝑥𝑜
𝑓(𝑥𝑜 − 𝑥1) =
𝑓(𝑥1)−𝑓(𝑥𝑜)
ℎ
=
∆𝑓(𝑥𝑜)
ℎ
& 𝑓(𝑥𝑜, 𝑥1, … … . 𝑥𝑟) =
1
𝑟!
∆ ʳ 𝑓(𝑥0)
ℎʳ
forward finite difference table
∆ ⁴𝑓∆ ³𝑓
∆ ²𝑓∆𝒇𝒇𝒊xi
∆𝑓𝑜𝑓1𝑥1
∆ ⁴𝑓o
∆ ³𝑓o
∆ ²𝑓o∆𝑓1𝑓2𝑥2
∆ ³𝑓1
∆ ²𝑓1
∆𝑓2
𝑓3𝑥3
∆ ²𝑓2∆ 𝑓3
⋮
𝑓4
⋮
𝑥4
⋮

5. 𝑓(𝑥) = 𝑓(𝑥𝑜) + (𝑥 − 𝑥𝑜)
∆𝑓0
1!ℎ
+ (𝑥 − 𝑥𝑜)(𝑥 − 𝑥1)
∆²𝑓
2!ℎ²
+ ⋯ + +(𝑥 − 𝑥𝑜)(𝑥 − 𝑥1) ⋯ (𝑥 − 𝑥 𝑚−1)
∆ᵐ 𝑓𝑜
𝑚!ℎᵐ
+ 𝑅 ⋯ (6.42).
−1 < 𝛼 < 1 𝑥 = 𝑥𝑜 + 𝛼ℎ 𝑎𝑛𝑑 𝑥𝜅=𝑥𝑜 + 𝜅ℎ
𝑥 − 𝑥𝜅 = ℎ 𝛼 − 𝜅
𝒇 𝒙 = 𝒇 𝒙𝒐 + 𝜶𝒉 = 𝒇𝒐 + 𝜶∆𝒇𝒐 +
𝜶 𝜶−𝟏
𝟐!
∆ ²𝒇𝒐 + ⋯ +
𝜶 𝜶−𝟏 𝜶−𝟐 ⋯ 𝜶−𝒎+𝟏
𝒎!
∆ᵐ 𝒇𝒐 + 𝑹 ⋯(6.47)
The formula given in Eq.(6.47) is called the Newton’s forward interpolation
formula. This formula is used to interpolate the values of y near the beginning of
a set of equally spaced tabular values. This formula can also be used for
extrapolating the values of y a little backward .

6.  Example: find y= 𝑒3𝑥 𝑓or 𝑥 = 0.05 using the following table.
 Solution:
 The difference table is shown in below:
 We have 𝑥𝑜 = 0.0 , 𝑥 = 0.05, ℎ = 0.1
 Hence 𝛼 =
𝑥−𝑥𝑜
ℎ
=
0.05−0.0
0.1
= 0.5
 Using Newton s forward formula
𝒇 𝒙 = 𝒇𝒐 + 𝜶∆𝒇𝒐 +
𝜶 𝜶−𝟏
𝟐!
∆ ²𝒇𝒐 +
𝜶(𝜶−𝟏)(𝜶−𝟐)
𝟑!
∆ ³𝒇𝒐 +
𝜶(𝜶−𝟏)(𝜶−𝟐)(𝜶−𝟑)
𝟒!
+ ⋯
𝑓 0.05 = 1.0 + 0.5 0.3499 +
0.5 0.5 − 1
2
0.1224 +
0.5 0.5 − 1 0.5 − 2
6
0.0428 +
0.5 0.5 − 1 (0.5 − 2)(0.5 − 3)
24
= 1.16172
0.40.30.20.10𝒙
3.32012.45961.82211.34991y=𝑒3𝑥
∆⁴𝒇∆³𝒇∆²𝒇∆𝒇𝒇(𝒙) = 𝒆 𝟑𝒙𝒙
1.00000
0.12240.34991.34990.1
0.01500.04280.16520.47231.82210.2
0.05780.22300.63752.45960.3
0.86053.32010.4

7. Eample:
newton to forward data in the table with the correct interpolation method.
A) second degree.
B)-Find the third degree polynomial.
C) same method for (0.34) to calculate the reputation value.
D) for third-degree polynomial interpolation (0.34) Develop a MATLAB program to
calculate the reputation value.
Solution:
 A) table data given in the question.
 that is the function of a cubic function. the real value of the polynomial at the point, as
shown below 1.271
 𝑓 𝑥 = 𝑥3
+ 2𝑥2
+ 1 𝑓 0.34 = (0.34)3
+2(0.34)2
+1 = 1.271
 ℎ = 𝑥𝑖+1 − 𝑥𝑖 = 0.1
 Using newton eq.
 𝑓 𝑥 = 𝑓𝑜 + 𝛼∆𝑓𝑜 +
𝛼(𝛼−1)
2!
∆²𝑓𝑜 …..
 𝑓 𝑥 = 1.021 + 𝛼0.067 +
𝛼(𝛼−1)
2
 𝑓 𝑥 = 0.026𝛼2 + 0.041𝛼 + 1.021
 B)
0.60.50.40.30.20.1𝒙
1.9361.6251.3841.2071.0881.021𝑦
∆⁴𝒇∆³𝒇∆²𝒇∆𝒇𝒇𝒊𝒙𝒊
0.0671.0210.1
0.0520.1191.0880.2
0.0000.0060.0580.1771.2070.3
0.0000.0060.0640.2411.3840.4
0.0060.0700.3111.6250.5
1.9360.6

8.  B)
 𝒇 𝒙 = 𝒇𝒐 + 𝜶∆𝒇𝒐 +
𝜶 𝜶−𝟏
𝟐!
∆ ²𝒇𝒐 +
𝜶(𝜶−𝟏)(𝜶−𝟐)
𝟑!
∆ ³𝒇𝒐..
 𝑓 𝑥 = 1.021 + 𝛼0.067 +
𝛼(𝛼−1)
2
0.052 +
𝛼(𝛼−1)(𝛼−2)
6
0.006
 𝑓 𝑥 = 0.001𝛼3
+ 0.023𝛼2
+ 0.043𝛼 + 1.021
 C) x=0.34 and 𝑥𝑜 = 0.3 𝑎𝑙𝑠𝑜 ℎ = 0.1 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑝𝑎𝑐𝑖𝑛𝑔
 𝑥𝑖=𝑥 𝑜+𝛼ℎ ≫ 𝛼
𝑥−𝑥𝑜
ℎ
→ 𝛼 =
0.34−0.3
0.1
= 0.4
 𝑓 𝑥 = 1.021 + 𝛼0.067 +
𝛼(𝛼−1)
2
0.064
 𝑓 0.34 = 1.021 + 0.4 0.067 +
0.4 0.4−1
2
0.064 = 1.270
 actual relative error for the second-degree polynomial.𝜀ℎ =
(1.271−1.270)
1.271
≅ 0.08%
 𝑓 𝑥 = 1.021 + 𝛼0.067 +
𝛼(𝛼−1)
2
0.052 +
𝛼(𝛼−1)(𝛼−2)
6
0.006
 𝑓 0.34 = 1.021 + 0.4 ∗ 0.067 +
0.4 0.4−1
2
0.052 +
0.4 0.4−1 0.4−2
6
0.006 = 1.271
 actual relative error for the third-degree polynomial.
 𝜀ℎ =
(1.271−1.271)
1.271
≅ 0.00%

9.  >> x=[0.1 0.2 0.3 0.4 0.5 0.6 ];
 y=[1.021 1.088 1.207 1.384 1.625 1.9361];
 x0=0.34;
 degree=3;
 n=size(x,2);
 f=zeros(n,n);
 for i=1:n-1
 f(i,1)=(y(i+1)-y(i));
 end;
 for j=2:n-1
 for i=1:n-j
 f(i,j)=(f(i+1,j-1)-f(i,j-1));
 end;
 end;
 d=abs(x-x0);
 ind=find(d==min(d));
 differencevalue =[y(ind) f(ind,:)];
 alpha=(x0-x(ind))/(x(2)-x(1))
 fx=differencevalue(1)+alpha+differencevalue(2);
 fori=i:degree
 product=alpha
 for j=i:i
 product=product*(alpha-j);
 end;
 fx=fx+product*differencevalue(i+2)/factorial(i+1);
 alpha = 0.4000 fori = 1 2 3 product = 0.4000

10. Newton'sbackwardinterpolationmethod
backward difference operator. 𝛻𝑓 𝑥 = 𝑓 𝑥 − 𝑓 𝑥 − ℎ are shown. wherein a difference in the table
as in the forward difference method is prepared. numerical values in the table is the same as the
forward difference table for recognizing the numerical values are different. (𝑥0 , 𝑥1) corresponding to
you efficiency (𝑓 𝑥0 , 𝑓(𝑥1)) the difference between the values 𝛻𝑓 𝑥 are represented. when asked
what happened to the money that the calculation of the values at the beginning of the use of table
functions .newton backward difference equation. Newton’s forward interpolation formula is not
suitable for interpolation values of y near the end of a table of values.
𝛻𝑓 𝑥𝑜 = 𝑓(𝑥𝑜) − 𝑓(𝑥𝑜 − ℎ)
𝛻²𝑓(xo)= 𝛻 𝛻𝑓 𝑥𝑜 = 𝛻𝑓(𝑥𝑜) − 𝛻𝑓(𝑥𝑜 − ℎ)
𝛻 𝑟+1
𝑓 𝑥 = 𝛻ʳ𝑓 𝑥𝑜 − 𝛻ʳ𝑓 𝑥𝑜 + ℎ
−1 < 𝛼 < 1 𝑥 = 𝑥 𝑛 + 𝛼ℎ 𝑎𝑛𝑑 𝑥 − 𝑥 𝑛−1 = 𝑥 𝑛 − 𝑥 𝑛−1 + 𝛼ℎ = ℎ + 𝛼ℎ = ℎ 𝛼 + 1
𝒇 𝒙 = 𝒇𝒏 +
𝜶
𝟏!
∆𝒇 𝒏−𝟏 +
𝜶(𝜶+𝟏)
𝟐!
∆ 𝟐 𝒇 𝒏−𝟐 + ⋯ +
𝜶 𝜶+𝟏 ⋯ 𝜶+𝒎−𝟏
𝒎!
∆ 𝒎 𝒇 𝒏−𝒎 + 𝑹 ⋯(6.55)
𝒇 𝒙 = 𝒇𝒏 +
𝜶
𝟏!
𝜵𝒇 𝒏 +
𝜶(𝜶+𝟏)
𝟐!
𝜵 𝟐 𝒇 𝒏 + ⋯ +
𝜶 𝜶+𝟏 ⋯ 𝜶+𝒎−𝟏
𝒎!
𝜵 𝒎 𝒇 𝒏 + 𝑹 ⋯ (6.56)
Estimated error value:
𝑅 =
ℎ 𝑚+1
(𝑚 + 1)!
𝛼 𝛼 + 1 ⋯ 𝛼 + 𝑚 𝑓 𝑚+1(𝜖)

11. Eample:
From the following table estimate the number of student who obtained mark in computer
programming between 75 and 80 .
Solution:
The cumulative frequency table is shown in the table .
To find the number of student with mark less than 80
Let 𝑥 𝑛 = 85 , 𝑥 = 80 , ℎ = 10 , 𝛼 =
𝑥−𝑥 𝑛
ℎ
=
(80−850
10
=-0.5
then using NEWTON backward formula we obtain
𝒇 𝒙 = 𝒇𝒏 +
𝜶
𝟏!
𝜵𝒇 𝒏 +
𝜶(𝜶 + 𝟏)
𝟐!
𝜵 𝟐
𝒇 𝒏 + ⋯ +
𝜶 𝜶 + 𝟏 ⋯ 𝜶 + 𝒎 − 𝟏
𝒎!
𝜵 𝒎
𝒇 𝒏 +
= 200 + −0.5 20 +
−0.5 −0.5+1
2
−40 +
−0.5 −0.5+1 −0.5+2
6
−40 +
−0.5 −0.5+1 −0.5+2 −0.5+3
24
−20 = 198.2813
so number of student getting marks in computer programming between 75 and 80
=198-180=18
75-8565-7555-6545-5535-45Mark
2060604020No.of student
𝛁 𝟒
𝒚𝛁 𝟑
𝒚𝛁 𝟐
𝒚𝛁𝒚No.of
student
Mark
lessthan(x)
2045
406055
206012065
-2006018075
-20-40-402020085

12.  Example: using the data given in 6.4, with Newton's backward difference
formula.
 a) Find a polynomial of second and third degree.
 b) the same method to calculate x = 0:34 reputation value.
 c) the polynomial interpolation from 3.derece f = (0.39) to develop a program
that will calculate .
Solution:
prepared for the difference in the table can be used to calculate .method point but according to this
example the function ( 𝒙 𝒏 > 𝒙) must be selected by considering, for example, x = 0.5. step size value .(𝒉 =
𝒙𝒊+𝟏 − 𝒙 𝟏 = 𝟎. 𝟏 )so has the second-degree polynomial as follows using the first and second order forward
difference terms.
𝒇 𝒙 = 𝟏. 𝟔𝟐𝟓 + 𝜶𝟎. 𝟐𝟒𝟏 +
𝜶(𝜶 + 𝟏)
𝟐
𝟎. 𝟎𝟔𝟒
𝒇 𝒙 = 𝟎. 𝟎𝟑𝟐𝜶 𝟐
+ 𝟎. 𝟐𝟕𝟑𝜶 + 𝟏. 𝟔𝟐𝟓
third degree polynomial interpolation from 3.derece
𝒇 𝒙 = 𝟏. 𝟔𝟐𝟓 + 𝜶𝟎. 𝟐𝟒𝟏 +
𝜶(𝜶 + 𝟏)
𝟐
𝟎. 𝟎𝟔𝟒 +
𝜶(𝜶 + 𝟏)(𝜶 + 𝟐)
𝟔
𝟎. 𝟎𝟎𝟔
𝒇 𝒙 = 𝟎. 𝟎𝟎𝟏𝜶 𝟑
𝟎. 𝟎𝟑𝟓𝜶 𝟐
+ 𝟎. 𝟐𝟕𝟓𝜶 + 𝟏. 𝟔𝟐𝟓
𝛁 𝟑 𝒇𝛁 𝟐 𝒇𝛁𝒇𝒇𝒊𝒙𝒊
1.0210.1
0.0520.0671.0880.2
0.0060.0580.1191.2070.3
0.0060.0640.1771.3840.4
0.0060.0700.2411.6250.5
0.3111.9360.6

13.  B)
 𝒇 𝒙 = 𝒇𝒐 + 𝜶𝛁𝒇𝒐 +
𝜶 𝜶+𝟏
𝟐!
𝛁²𝒇𝒐 +
𝜶(𝜶+𝟏)(𝜶+𝟐)
𝟑!
𝛁 ³𝒇𝒐..
 C) x=0.34 and 𝑥𝑛 = 0.5 𝑎𝑙𝑠𝑜 ℎ = 0.1 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑝𝑎𝑐𝑖𝑛𝑔
 𝑥=𝑥 𝑛+𝛼ℎ ≫ 𝛼 =
𝑥−𝑥𝑛
ℎ
→ 𝛼 =
0.34−0.5
0.1
= −1.6
𝒇 𝒙 = 𝟎. 𝟎𝟑𝟐𝜶 𝟐 + 𝟎. 𝟐𝟕𝟑𝜶 + 𝟏. 𝟔𝟐𝟓
𝒇 𝟎. 𝟑𝟒 = 𝟎. 𝟎𝟑𝟐(−𝟏. 𝟔) 𝟐+𝟎. 𝟐𝟕𝟑 −𝟏. 𝟔 + 𝟏. 𝟔𝟐𝟓 = 𝟏. 𝟐𝟕𝟎
𝒇 𝒙 = 𝟎. 𝟎𝟎𝟏𝜶 𝟑 + 𝟎. 𝟎𝟑𝟓𝜶 𝟐 + 𝟎. 𝟐𝟕𝟓𝜶 + 𝟏. 𝟔𝟐𝟓
𝒇 𝟎. 𝟑𝟒 = 𝟎. 𝟎𝟎𝟏(−𝟏. 𝟔) 𝟑
+𝟎. 𝟎𝟑𝟓(−𝟏. 𝟔) 𝟐
+𝟎. 𝟐𝟕𝟓 −𝟏. 𝟔 + 𝟏. 𝟔𝟐𝟓 = 𝟏. 𝟐𝟕𝟏

14.  x=[0.1 0.2 0.3 0.4 0.5 0.6 ];
 y=[1.021 1.088 1.207 1.625 1.936 ];
 xn=0.39;
 x=[0.1 0.2 0.3 0.4 0.5 0.6 ];
 y=[1.021 1.088 1.207 1.384 1.625 1.9361];
 xn=0.34;
 degree=3;
 n=size(x,2);
 f=zeros(n,n);
 for i=1:n-1
 f(i,1)=(y(i+1)-y(i));
 end;
 for j=2:n-1
 for i=1:n-j
 f(i,j)=(f(i+1,j-1)-f(i,j-1));
 end;
 end;
 ind=find(x>xn);
 ind=ind(1);
 xind=ind:-1:ind-degree+1;
 yind=1:i:degree;
 differencevalue=[y(ind+1) f(sub2ind(size(f),xind,yind))]
 alpha=(xn-x(ind+1))/(x(2)-x(1));
 fx=differencevalue(1)+alpha+differencevalue(2);
 for i=1:degree-1
 product=alpha;
 for j=1:i
 product=product*(alpha+j);
 end;
 fx=fx+product*differencevalue(i+2)/factorial(i+1);
 End ; differencevalue = 1.6250 0.2410 0.0640 0.0060

15. CENTRAL DIFFERENCE INTERPOLATION
 a step size equal to f(x) function ( 𝛿𝑓𝑘+1/2) center at the point of difference, 𝛿𝑓𝑘+1/2
= 𝑓𝑘+1 − 𝑓𝑘will be recognized. The central difference operator, indicated by 𝛿𝑓(𝑥) Various
central difference formula for the intermediate values ​​can be used in prepared
statements difference. about the central difference
 correlations are shown below.
 𝛿𝑓 𝑥 = 𝑓 𝑥 +
ℎ
2
− 𝑓 𝑥 −
ℎ
2
 𝛿2
𝑓 𝑥 𝑘 = 𝛿𝑓 𝑥 𝑘 +
ℎ
2
− 𝛿𝑓(𝑥 𝑘 −
ℎ
2
)
 𝛿 𝑟+1
𝑓 𝑥 𝑘 = 𝛿 𝑟
𝑓 𝑥 𝑘 +
ℎ
2
− 𝛿 𝑟
𝑓(𝑥 𝑘 −
ℎ
2
)
 at 𝑥𝑜 < 𝑥 < 𝑥1 𝛼 = 𝑥 − 𝑥𝑜/ℎ
 𝑓 𝑥 = 𝑓𝑜 + 𝑥 − 𝑥𝑜 𝛿𝑓1
2
+ x − 𝑥0 𝑥 − 𝑥1
𝛿2 𝑓𝑜
2!ℎ2 + 𝑥 − 𝑥0 𝑥 − 𝑥0 𝑥 − 𝑥−1
𝛿3 𝑓1
2
3!ℎ3 + ( 𝑥

18. Example: using the data given in 6.4, the central difference interpolation method, the first and.
finding a second and third degree polynomial x = 0:34 calculate the values: Prepared difference table
used for this example. According to the method for the function to be calculated in 0.34 point x = (xo <
x < x1) must be chosen as the x = 0.3 and x = 0.4.
step size value h = x-x1 = 0.1
𝑥 = 0.34 → 𝑓 0.34 = 1.207 + 0.34 − 0.3 0.177 = 1.214
Second degree central difference interpolation
𝑓 𝑥 = 𝑓𝑜 + 𝑥 − 𝑥𝑜 𝛿𝑓1/2 + 𝑥 − 𝑥𝑜 𝑥 − 𝑥𝑖
𝛿2 𝑓𝑜
2! ℎ2
𝑓 𝑥 = 1.207 + 𝑥 − 0.3 0.177 +
𝑥 − 0.3 𝑥 − 0.4 0.058
2 0.1 2
= 2.9𝑥2
− 1.853𝑥 + 1.502
At x=0.34 then 𝑓 𝑥 = 1.2077
Third degree central difference interpolation
𝑓 𝑥 = 2.9𝑥2
− 1.853𝑥 + 1.502 + x − 0.3 x − 0.4
𝑥 − 0.2 0.006
6 0.1 2
𝑓 𝑥 = 𝑥3
+ 2𝑥2
− 1.593𝑥 + 1.478
𝑓 0.34 = 1.206
𝑓 𝑥 = 𝑓𝑜 + 𝑥 − 𝑥 𝑜 𝛿𝑓1
2
→ 𝑓 𝑥 = 1.207 + 𝑥 − 0.3 0.177
𝜹 𝟒
𝒇𝜹 𝟑
𝒇𝜹 𝟐
𝒇𝜹𝒇𝒇𝒊𝒙𝒊
0.0671.0210.1
0.0520.1191.08
8
0.2
0.0060.0580.1771.2070.3
0.0000.0060.0640.2411.3840.4
0.0000.0060.0700.3111.6250.5
1.9360.6