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GNU Emacs Calc 2.02 Manual

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.