### List Tutorial Exercise 8

The first step is to build a list of values of x.

1:  [1, 2, 3, ..., 21]  1:  [0, 1, 2, ..., 20]  1:  [0, 0.25, 0.5, ..., 5]
.                       .                       .

v x 21 RET              1 -                     4 /  s 1


Next, we compute the Bessel function values.

1:  [0., 0.124, 0.242, ..., -0.328]
.

V M ' besJ(1,$) RET  (Another way to do this would be 1 TAB V M f j.) A way to isolate the maximum value is to compute the maximum using V R X, then compare all the Bessel values with that maximum. 2: [0., 0.124, 0.242, ... ] 1: [0, 0, 0, ... ] 2: [0, 0, 0, ... ] 1: 0.5801562 . 1: 1 . . RET V R X V M a = RET V R + DEL  It's a good idea to verify, as in the last step above, that only one value is equal to the maximum. (After all, a plot of @c{$\sin x\$} sin(x) might have many points all equal to the maximum value, 1.)

The vector we have now has a single 1 in the position that indicates the maximum value of x. Now it is a simple matter to convert this back into the corresponding value itself.

2:  [0, 0, 0, ... ]         1:  [0, 0., 0., ... ]    1:  1.75
1:  [0, 0.25, 0.5, ... ]        .                        .
.

r 1                         V M *                    V R +


If a = had produced more than one 1 value, this method would have given the sum of all maximum x values; not very useful! In this case we could have used v m (calc-mask-vector) instead. This command deletes all elements of a "data" vector that correspond to zeros in a "mask" vector, leaving us with, in this example, a vector of maximum x values.

The built-in a X command maximizes a function using more efficient methods. Just for illustration, let's use a X to maximize besJ(1,x)' over this same interval.

2:  besJ(1, x)                 1:  [1.84115, 0.581865]
1:  [0 .. 5]                       .
.

' besJ(1,x), [0..5] RET            a X x RET
`

The output from a X is a vector containing the value of x that maximizes the function, and the function's value at that maximum. As you can see, our simple search got quite close to the right answer.