How does big-O notation relate to the actual error involved in a numerical differentiation?Suppose I have some position data x 1 , x 2 , . . . x n that was sampled at an interval h. If I wanted the velocity data, I could apply a finite difference scheme: v 1 = x 2 − x 1 h + O ( h )O(h) denotes that the error term is proportional to the step size. ..What exactly does this mean physically? For instance, say my x terms were taken at 5 samples/second (so h=0.2). Does this mean the error in my velocity data is ± 0.2? How do I interpret the error in a physical sense, and can I write this like an uncertainty to a measurement? E.g., v 1 = 10 ± 0.2?

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How does big-O notation relate to the actual error involved in a numerical differentiation?Suppose I have some position data             x      1        ,          x      2        ,    .    .    .          x      n       that was sampled at an interval h. If I wanted the velocity data, I could apply a finite difference scheme:      v    1    =                    x        2            −              x        1              h    +  O  (  h  )O(h) denotes that the error term is proportional to the step size. ..What exactly does this mean physically? For instance, say my x terms were taken at 5 samples/second (so h=0.2). Does this mean the error in my velocity data is   ± 0.2? How do I interpret the error in a physical sense, and can I write this like an uncertainty to a measurement? E.g.,       v    1    =  10  ±  0.2?

losing2mema · Accepted Answer

Look at the simplest functions x(t) and compute the exact expressions   v  (  h  ,  t  )  =            x      (      t      +      h      )      −      x      (      t      )        h  For x(t)=at you have   v  (  h  ,  t  )  =            x      (      t      +      h      )      −      x      (      t      )        h    =  a and therefore the error term O(h) is zero.For   x  (  t  )  =  a      t    2   you have   v  (  h  ,  t  )  =  2  a  t  +  a  h and O(h)=ah. Thus the error is constant in time, it only depends on a,h.For   x  (  t  )  =  a      t    3   you have   v  (  h  ,  t  )  =  3  a      t    2    +  3  a  t  h  +  a      h    2   and   O  (  h  )  =  3  a  t  h  +  a      h    2    .  . Once a again you have a constant term, here   a      h    2    ,  , but now the second error term 3ath is time-dependent ant its absolute value increases.In the two first cases you have an obvious error of the form ±x.y m/s but in the last case it is not so easy. You can give a maximum error for the considered t range; or if you have to bound the error you have maximum time interval for a valid approximation.

Avah Knapp · Accepted Answer

More generally, if x is twice continuously differentiable, there exists a   ξ  ∈  [  0  ,  1  ] for which             x      (      t      +      h      )      −      x      (      t      )        h    =      h          2      !            x    ″    (  t  +  ξ  h  ), so that you need to estimate the second derivative:       v    1    =  10  ±      0.2    2        v          m      a      x        ″

How does big-O notation relate to the actual error involved in a numerical differentiation? Suppose

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