Newton forward-difference and backward-forward

Question

Approximate f(0.05) use the following data and the Newton forward-difference formula

 

x                0.0             0.2             0.4             0.6             0.8

f(x)       1.00000         1.22140      1.49182      1.82212     2.22554

 

b. use the Newton backward-difference formula to approximate f(0.65)

Answer 1

 

 The linear function is  f0 +  (x-x0)(f1-f0)/(x1-x0) =

 

                                1 +  (x-0)(1.22140-1)/(0.2) =

                                

                                 1 + (0.2214)x/(0.2) =

 

                                1 + 1.107x

 

So f(0.05) is approximately 1 + 1.107(0.05) = 1.105535

 

 

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Likewise the backwards difference results in :

 

 f1 - (x-x1)(f1-f0)/(x1-x0) =

   

1.82212 - (x-0.6)( 2.22554-1.82212)/(-0.2)=

 

1.82212 + (x-0.6)(2.0171)

 

 

So for x=0.65

1.922975