Question: The truncation error in a numerical algorithm is 0(n) where n denotes the number of iterations. It is found that after 3 iterations the

The truncation error in a numerical algorithm is 0(n) where n denotes the number of iterations. It is found that after 3 iterations the truncation error is 2.5%. What is the truncation error expected to be after 4, 5 and 6 iterations? Note: Having a truncation error of order 0(n) implies that the error reduces as the 4th power of number of iterations i.e. 6= Cn where Cis a constant.

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