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MTH603 (Spring 2012) - Divided Differences & Derivatives, Exercises of Mathematics

Institute of Mathematical SciencesMathematics

The solution to assignment #2 for the mth603 course taken in spring 2012. The assignment includes constructing divided difference tables and using newton's interpolation polynomial to find function values, as well as finding derivatives using forward and backward differences. The document also includes the steps and calculations for each question.

Typology: Exercises

2011/2012

Uploaded on 08/03/2012

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Solution File Assignment # 2
MTH603 (Spring 2012)
Total marks: 10
Lecture # 23-28
Due date: 20-06-2012
Dear students as it was told there are 3 questions in the assignment but only one question
will be graded. Question 3 will be graded.
Question#1 Marks 10
A function y=f(x) is given by the following table.
x 321.0 322.8 324.2 325.0
y=f(x) 2.50651 2.50893 2.51081 2.51188
a) Construct the divided differences table for the above data.
b) Find f (323.5) by Newton’s interpolation polynomial.
Solution:
The divided difference table for the given data is constructed as;
x f(x) 1st D.D 2nd D.D 3rd D.D
321.0 2.50651
322.8 2.50893 0.001344
324.2 2.51081 0.001343 -3.125*10-7
325.0 2.51188 0.001338 -2.273*10-6 -4.901*10-7
Now, using Newton’s divided difference formula, we have
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Solution File Assignment # 2

MTH603 (Spring 2012)

Total marks: 10 Lecture # 23- Due date: 20-06-

Dear students as it was told there are 3 questions in the assignment but only one question will be graded. Question 3 will be graded.

Question#1 Marks 10

A function y=f(x) is given by the following table.

x 321.0 322.8 324.2 325. y=f(x) 2.50651 2.50893 2.51081 2.

a) Construct the divided differences table for the above data. b) Find f (323.5) by Newton’s interpolation polynomial.

Solution:

The divided difference table for the given data is constructed as;

x f(x) 1 st^ D.D 2 nd^ D.D 3 rd^ D.D 321.0 2. 322.8 2.50893 0. 324.2 2.51081 0.001343 -3.12510- 325.0 2.51188 0.001338 -2.27310-6^ -4.901*10 -

Now, using Newton’s divided difference formula, we have

0 0 0 1 0 1 0 1 2 0 1 2 0 1 2 3

7 7

( ) [ , ]

( )( ) [ , , ]

( )( )( ) [ , , , ]

y y x x y x x x x x x y x x x x x x x x x y x x x x y x x x x x x y x

 

7 2 4 7 3 4 2

4 7 2

x x x x x

x x x x

   

 

7 3 2 7 3

x x x x x

  

Hence, the value of f(323.5)

2 7 3

y f

Question#2 Marks 10

Find / / /

y (0.6)^ and y (0.6)from the following table,

x 0.4 0.5 0.6 0.7 0. y 1.5836494 1.7974426 2.0442376 2.3275054 2.

Solution:

'(0.8) 1 0.3235764 0.0403086^ 0.0038358^ 0.

y (^) n yn y^ n^ yn^ yn h

y

y

y

y

 ^   ^  ^  

 ^    

2 3 4 2

2

2

2

y (^) n (^) h yn yn yn

y

y

y

y

 ^      

 ^   