Question

57. By using Newton's backward difference table form the following data: f(30) =
0.5000, f (35) = 0.5736, f (40) = 0.6428, f (45) = 0.7071. What is the value of V3yn?
-0.0049
3-1.872
-0.0005
-0.0469​

Answer 1

Given : f(30) = 0.5000, f (35) = 0.5736, f (40) = 0.6428, f (45) = 0.7071

To find :  ▽³yₙ using Newton's backward difference table

-0.0049

3-1.872

-0.0005

-0.0469​

Solution:

 

▽yₙ= yₙ - yₙ₋₁ is called the first backward difference

x           f(x)                   ▽y            ▽²y               ▽³y

30        0.5000

                                  0.0736

35        0.5736                            -0.0044  

                                 0.0692                        -0.0005

40         0.6428                           -0.0049    

                                 0.0643                  

45        0.7071

▽³yₙ   = -0.0005

 

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