<Tr> <Td> f (x 0) = − 14.1014 (\ displaystyle f (x_ (0)) = - 14.1014) </Td> <Td> f (x 1) = − 0.931596 (\ displaystyle f (x_ (1)) = - 0.931596) </Td> <Td> f (x 2) = 0 (\ displaystyle f (x_ (2)) = 0) </Td> <Td> f (x 3) = 0.931596 (\ displaystyle f (x_ (3)) = 0.931596) </Td> <Td> f (x 4) = 14.1014 (\ displaystyle f (x_ (4)) = 14.1014) </Td> </Tr> <P> Using six digits of accuracy, we construct the table </P> <Dl> <Dd> − 3 2 − 14.1014 17.5597 − 3 4 − 0.931596 − 10.8784 1.24213 4.83484 0 0 0 0 1.24213 4.83484 3 4 0.931596 10.8784 17.5597 3 2 14.1014 (\ displaystyle (\ begin (matrix) - (\ tfrac (3) (2)) & - 14.1014&&&& \ \ &&17. 5597&&& \ \ - (\ tfrac (3) (4)) & - 0.931596&& - 10.8784&& \ \ &&1. 24213&&4. 83484& \ \ 0&0&&0&&0 \ \ &&1. 24213&&4. 83484& \ \ (\ tfrac (3) (4)) &0. 931596&&10. 8784&& \ \ &&17. 5597&&& \ \ (\ tfrac (3) (2)) &14. 1014&&&& \ \ \ end (matrix))) </Dd> </Dl> <Dd> − 3 2 − 14.1014 17.5597 − 3 4 − 0.931596 − 10.8784 1.24213 4.83484 0 0 0 0 1.24213 4.83484 3 4 0.931596 10.8784 17.5597 3 2 14.1014 (\ displaystyle (\ begin (matrix) - (\ tfrac (3) (2)) & - 14.1014&&&& \ \ &&17. 5597&&& \ \ - (\ tfrac (3) (4)) & - 0.931596&& - 10.8784&& \ \ &&1. 24213&&4. 83484& \ \ 0&0&&0&&0 \ \ &&1. 24213&&4. 83484& \ \ (\ tfrac (3) (4)) &0. 931596&&10. 8784&& \ \ &&17. 5597&&& \ \ (\ tfrac (3) (2)) &14. 1014&&&& \ \ \ end (matrix))) </Dd>

Newton divided difference method for interpolation can be used for