Scale-invariance of Simpson's rule approximations to logIf I was trapped on a desert island and needed to compute log ⁡ ( 2 ), the natural logaritm of 2, one thing I could do is use the equality log ⁡ ( 2 ) = ∫ 1 2 1 x d xand approximate the definite integral with Simpson's rule. Say I want to partition [ 1 , 2 ] into 10 subintervals to do this. But, maybe I don't like working with fracftions very much. Well, then I could use instead log ⁡ ( 2 ) = ∫ 10 20 1 x d xsince it is easier to divide [ 10 , 20 ] into 10 subintervals.Now, it happens that using Simpon's rule approximation with n = 10 gives the same approximation to both of these definite integrals.

Question

Scale-invariance of Simpson&#039;s rule approximations to logIf I was trapped on a desert island and needed to compute   log  ⁡  (  2  ), the natural logaritm of   2, one thing I could do is use the equality  log  ⁡  (  2  )  =      ∫    1    2        1    x       d  xand approximate the definite integral with Simpson&#039;s rule. Say I want to partition   [  1  ,  2  ] into   10 subintervals to do this. But, maybe I don&#039;t like working with fracftions very much. Well, then I could use instead  log  ⁡  (  2  )  =      ∫          10              20            1    x       d  xsince it is easier to divide   [  10  ,  20  ] into   10 subintervals.Now, it happens that using Simpon&#039;s rule approximation with   n  =  10 gives the same approximation to both of these definite integrals.

Xzavier Shelton · Accepted Answer

This happens because   x is linear in   x. An approximation of the first integral could be   (  1      /    1  +  1      /    1.1  +  ⋯  +  1      /    2  )  ×  1      /    11 where the corresponding approximation of the second integral would be   (  1      /    10  +  1      /    11  +  ⋯  +  1      /    20  )  ×  10      /    11

Scale-invariance of Simpson's rule approximations to log If I was trapped on a desert island and ne

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