Let f ( x ) = a r c t a n ( x ). Use the derivative approximation: f ′ ( x ) = 8 f ( x + h ) − 8 f ( x − h ) − f ( x + 2 h ) + f ( x − 2 h ) 12 h to approximate f ′ ( 1 4 π ) using h − 1 = 2 , 4 , 8 . Try to take h small enough that the rounding error effect begins to dominate the mathematical error. For what value of h does this begin to occur?

Question

Let   f  (  x  )  =  a  r  c  t  a  n  (  x  ). Use the derivative approximation:      f    ′    (  x  )  =            8      f      (      x      +      h      )      −      8      f      (      x      −      h      )      −      f      (      x      +      2      h      )      +      f      (      x      −      2      h      )              12      h        to approximate       f    ′    (      1    4    π  ) using       h    −    1 =   2  ,  4  ,  8 . Try to take   h small enough that the rounding error effect begins to dominate the mathematical error. For what value of h does this begin to occur?

Tristin Case · Accepted Answer

Partial answerTo get an idea how good the approximation is, you can calculate the result for the special functions   f  (  x  )  =  1  ,  x  ,      x    2    ,      x    3    ⋯. Here, we get the exact result for   f  (  x  )  =  1  ,  x  ,      x    2    ,      x    3    ,      x    4  , so the formula is exact upto degree   4 (Polynomials upto degree   4 are differentiated exactly by the formula)For   f  (  x  )  =      x    5  , you get       f    ′    (  x  )  =  5      x    4    −  4      h    4  , so you have an error of   4      h    4