3.7.51 Algebra Tutorial Exercise 4

The hard part is that V R + is no longer sufficient to add up all the contributions from the slices, since the slices have varying coefficients. So first we must come up with a vector of these coefficients. Here's one way:

     2:  -1                 2:  3                    1:  [4, 2, ..., 4]
     1:  [1, 2, ..., 9]     1:  [-1, 1, ..., -1]         .
         .                      .

         1 n v x 9 <RET>          V M ^  3 <TAB>             -

     1:  [4, 2, ..., 4, 1]      1:  [1, 4, 2, ..., 4, 1]
         .                          .

         1 |                        1 <TAB> |

Now we compute the function values. Note that for this method we need eleven values, including both endpoints of the desired interval.

     2:  [1, 4, 2, ..., 4, 1]
     1:  [1, 1.1, 1.2,  ...  , 1.8, 1.9, 2.]
         .

      11 <RET> 1 <RET> .1 <RET>  C-u v x

     2:  [1, 4, 2, ..., 4, 1]
     1:  [0., 0.084941, 0.16993, ... ]
         .

         ' sin(x) ln(x) <RET>   m r  p 5 <RET>   V M $ <RET>

Once again this calls for V M * V R +; a simple * does the same thing.

     1:  11.22      1:  1.122      1:  0.374
         .              .              .

         *              .1 *           3 /

Wow! That's even better than the result from the Taylor series method.