Numerical techniques

Recursion (including iteration) combines well with Scheme’s mathematical primitive procedures to implement various numerical techniques. As an example, let’s implement Simpson’s rule, a procedure for finding an approximation for a definite integral.

The definite integral of a function *f*(*x*) within an
interval of integration [*a*,*b*] can be viewed as the
*area under the curve* representing *f*(*x*) from the
lower limit *x* = *a* to the upper limit *x* = *b*.
In other words, we consider the graph of the curve for
*f*(*x*) on the *x*,*y*-plane, and find the area enclosed
between that curve, the *x*-axis, and the *ordinates* of *f*(*x*) at *x* = *a* and *x* = *b*.

According to Simpson’s rule, we divide
the interval of integration [*a*,*b*]
into *n* evenly spaced
intervals, where *n* is even. (The larger *n*
is, the better the approximation.)
The interval boundaries constitute
*n* + 1 points on the *x*-axis, viz, *x*_{0}, *x*_{1},
`...`, *x*_{i}, *x*_{i+1}, `...`, *x*_{n}, where
*x*_{0} = *a* and *x*_{n} = *b*.
The length of each
interval is *h* = (*b* − *a*)/*n*, so each *x*_{i} = *a* + *i**h*. We then
calculate the ordinates of *f*(*x*) at the interval
boundaries. There are *n* + 1 such ordinates, viz,
*y*_{0}, `...`, *y*_{i}, `...`, *y*_{n}, where *y*_{i} = *f*(*x*_{i})
= *f*(*a* + *i**h*).
Simpson’s rule
approximates the definite integral of *f*(*x*) between
*a* and *b* with the value^{1}:

We define
the procedure `integrate‑simpson`

to take four arguments: the integrand `f`

; the
*x*-values at the limits `a`

and `b`

; and the
number of intervals `n`

.

(define integrate-simpson (lambda (f a b n) ;...

The first thing we do in
`integrate‑simpson`

’s body is ensure that
`n`

is even — if it isn’t, we simply bump its
value by 1.

;... (unless (even? n) (set! n (+ n 1))) ;...

Next, we put in the local variable `h`

the length of the interval. We introduce two more local variables
`h*2`

and `n/2`

to store the values of twice `h`

and half `n`

respectively, as we expect to
use these values often in the ensuing calculations.

;... (let* ((h (/ (- b a) n)) (h*2 (* h 2)) (n/2 (/ n 2)) ;...

We note that the sums *y*_{1} + *y*_{3} + `···` + *y*_{n−1}
and *y*_{2} + *y*_{4} + `···` + *y*_{n−2} both involve adding
every other ordinate. So let’s define a local procedure
`sum‑every‑other‑ordinate‑starting‑from`

that
captures this common iteration. By abstracting this
iteration into a procedure, we avoid having to repeat
the iteration textually. This not only reduces
clutter, but reduces the chance of error, since we have
only one textual occurrence of the iteration to debug.

`sum‑every‑other‑ordinate‑starting‑from`

takes two arguments:
the starting ordinate and the number of ordinates to be summed.

;... (sum-every-other-ordinate-starting-from (lambda (x0 num-ordinates) (let loop ((x x0) (i 0) (r 0)) (if (>= i num-ordinates) r (loop (+ x h*2) (+ i 1) (+ r (f x))))))) ;...

We can now calculate the three ordinate sums, and
combine them to produce the final answer. Note
that there are *n*/2 terms in *y*_{1} + *y*_{3} + `···` +
*y*_{n−1}, and (*n*/2) − 1 terms in *y*_{2} + *y*_{4} + `···`
+ *y*_{n−2}.

;... (y0+yn (+ (f a) (f b))) (y1+y3+...+y.n-1 (sum-every-other-ordinate-starting-from (+ a h) n/2)) (y2+y4+...+y.n-2 (sum-every-other-ordinate-starting-from (+ a h*2) (- n/2 1)))) (* 1/3 h (+ y0+yn (* 4.0 y1+y3+...+y.n-1) (* 2.0 y2+y4+...+y.n-2))))))

Let’s use `integrate‑simpson`

to find the definite
integral of the function

We first define in Scheme’s prefix
notation.^{2}

(define *pi* (* 4 (atan 1))) (define phi (lambda (x) (* (/ 1 (sqrt (* 2 *pi*))) (exp (- (* 1/2 (* x x)))))))

Note that we exploit the fact that in order to define `*pi*`

.^{3}

The following calls calculate the definite integrals of
`phi`

from 0 to 1, 2, and 3 respectively. They
all use 10 intervals.

(integrate-simpson phi 0 1 10) (integrate-simpson phi 0 2 10) (integrate-simpson phi 0 3 10)

To four decimal places, these values should be
0.3413, 0.4772, and 0.4987 respectively [2, Table
26.1]. Check to see that our implementation
of Simpson’s rule does indeed produce comparable
values!^{4}

It is not always convenient to specify the number `n`

of intervals. A number that is good enough for
one integrand may be woefully inadequate for another. In
such cases, it is better to specify the amount of
*tolerance* `e`

we are willing to grant the final answer, and let
the program figure out how many intervals are needed. A
typical way to accomplish this is to have the program
try increasingly better answers by steadily increasing
`n`

, and stop when two successive sums differ within
`e`

. Thus:

(define integrate-adaptive-simpson-first-try (lambda (f a b e) (let loop ((n 4) (iprev (integrate-simpson f a b 2))) (let ((icurr (integrate-simpson f a b n))) (if (<= (abs (- icurr iprev)) e) icurr (loop (+ n 2)))))))

Here we calculate successive Simpson integrals (using
our original procedure `integrate‑simpson`

) for
`n`

= 2, 4, `...`. (Remember that `n`

must be even.)
When the integral `icurr`

for the current `n`

differs within `e`

from the integral `iprev`

for
the immediately preceding `n`

, we return `icurr`

.

One problem with this approach is that we don’t take
into account that only some *segments* of the
function benefit from the addition of intervals. For
the other segments, the addition of intervals merely
increases the computation without contributing to a
better overall answer. For an improved adaptation, we
could split the integral into adjacent segments, and
improve each segment separately.

(define integrate-adaptive-simpson-second-try (lambda (f a b e) (let integrate-segment ((a a) (b b) (e e)) (let ((i2 (integrate-simpson f a b 2)) (i4 (integrate-simpson f a b 4))) (if (<= (abs (- i2 i4)) e) i4 (let ((c (/ (+ a b) 2)) (e (/ e 2))) (+ (integrate-segment a c e) (integrate-segment c b e))))))))

The initial segment is from `a`

to `b`

. To find
the integral for a segment, we calculate the Simpson
integrals `i2`

and `i4`

with the two smallest
interval numbers 2 and 4. If these are within `e`

of
each other, we return `i4`

. If not we split the
segment in half, recursively calculate the integral
separately for each segment, and add. In
general, different segments at the same level converge
at their own pace. Note that when we integrate a half
of a segment, we take care to also halve the tolerance,
so that the precision of the eventual sum does not
decay.

There are still some inefficiencies in this
procedure: The integral `i4`

recalculates three
ordinates already determined by `i2`

, and the integral
of each half-segment recalculates three ordinates
already determined by `i2`

and `i4`

. We avoid
these inefficiencies by making explicit the sums used
for `i2`

and `i4`

, and by transmitting more parameters
in the named-`let`

`integrate‑segment`

. This makes for
more sharing, both within the body of `integrate‑segment`

and across successive calls to `integrate‑segment`

:

(define integrate-adaptive-simpson (lambda (f a b e) (let* ((h (/ (- b a) 4)) (mid.a.b (+ a (* 2 h)))) (let integrate-segment ((x0 a) (x2 mid.a.b) (x4 b) (y0 (f a)) (y2 (f mid.a.b)) (y4 (f b)) (h h) (e e)) (let* ((x1 (+ x0 h)) (x3 (+ x2 h)) (y1 (f x1)) (y3 (f x3)) (i2 (* 2/3 h (+ y0 y4 (* 4.0 y2)))) (i4 (* 1/3 h (+ y0 y4 (* 4.0 (+ y1 y3)) (* 2.0 y2))))) (if (<= (abs (- i2 i4)) e) i4 (let ((h (/ h 2)) (e (/ e 2))) (+ (integrate-segment x0 x1 x2 y0 y1 y2 h e) (integrate-segment x2 x3 x4 y2 y3 y4 h e)))))))))

`integrate‑segment`

now explicitly sets four
intervals of size `h`

, giving five ordinates `y0`

,
`y1`

, `y2`

, `y3`

, and `y4`

. The integral
`i4`

uses all of these ordinates, while the integral
`i2`

uses just `y0`

, `y2`

, and `y4`

, with an
interval size of twice `h`

. It is easy to verify that
the explicit sums used for `i2`

and `i4`

do correspond
to Simpson sums.

Compare the following approximations of
_{0}^{20} *e*^{x} *d**x*:

(integrate-simpson exp 0 20 10) (integrate-simpson exp 0 20 20) (integrate-simpson exp 0 20 40) (integrate-adaptive-simpson exp 0 20 .001) (- (exp 20) 1)

The last one is the analytically correct answer. See
if you can figure out the smallest `n`

(overshooting is
expensive!)
such that `(integrate‑simpson exp 0 20 n)`

yields a result
comparable to that returned by the `integrate‑adaptive‑simpson`

call.

Simpson’s rule cannot be directly applied to *improper integrals* (integrals such that either
the value of the integrand is unbounded somewhere
within the interval of integration, or the interval of
integration is itself unbounded). However, the rule can still be
applied for a *part* of the integral, with the
remaining being approximated by other means. For
example, consider the function.
For *n* > 0, (*n*) is defined as the
following integral with unbounded upper limit:

From this, it follows that (a) (1) = 1, and (b)
for *n* > 0, (*n* + 1) = *n*(*n*). This implies
that if we know the value of in the interval
(1, 2), we can find (*n*) for any *real* *n* > 0.
Indeed, if we relax the condition *n* > 0, we can use
result (b) to extend the domain of (*n*) to
include *n* ≤ 0, with the understanding that the function
will diverge for *integer* *n* ≤ 0.^{5}

We first implement a Scheme procedure `gamma‑1‑to‑2`

that requires its argument `n`

to be within the
interval (1, 2). `gamma‑1‑to‑2`

takes a
second argument `e`

for the tolerance.

(define gamma-1-to-2 (lambda (n e) (unless (< 1 n 2) (error 'gamma-1-to-2 "argument outside (1, 2)")) ;...

We introduce a local variable `gamma‑integrand`

to hold
the -integrand *g*(*x*) = *x*^{n−1}*e*^{x}:

;... (let ((gamma-integrand (let ((n-1 (- n 1))) (lambda (x) (* (expt x n-1) (exp (- x)))))) ;...

We now need to integrate *g*(*x*) from 0 to .
Clearly we cannot deal with an infinite number of
intervals; we therefore use Simpson’s rule for only a
portion of the interval [0, ), say [0, *x*_{c}]
(*c* for “cut-off”). For the remaining, “tail”,
interval [*x*_{c}, ), we use a tail-integrand
*t*(*x*) that reasonably approximates *g*(*x*), but has the
advantage of being more tractable to analytic solution.
Indeed, it is easy to see that for sufficiently large
*x*_{c}, we can replace *g*(*x*) by an exponential decay
function *t*(*x*) = *y*_{c} *e*^{−(x − xc )}, where *y*_{c}
= *g*(*x*_{c}). Thus:

The first integral can be solved using Simpson’s rule,
and the second integral is just *y*_{c}. To find *x*_{c},
we start with a low-ball value (say 4), and then refine
it by successively doubling it until the ordinate at
2*x*_{c} (ie, *g*(2*x*_{c})) is within a certain
tolerance of the ordinate predicted by the tail-integrand
(ie, *t*(2*x*_{c})). For both the Simpson
integral and the tail-integrand calculation, we will
require a tolerance of `e/100`

, an order of 2 less than
the given tolerance `e`

, so the overall
tolerance is not affected:

;... (e/100 (/ e 100))) (let loop ((xc 4) (yc (gamma-integrand 4))) (let* ((tail-integrand (lambda (x) (* yc (exp (- (- x xc)))))) (x1 (* 2 xc)) (y1 (gamma-integrand x1)) (y1-estimated (tail-integrand x1))) (if (<= (abs (- y1 y1-estimated)) e/100) (+ (integrate-adaptive-simpson gamma-integrand 0 xc e/100) yc) (loop x1 y1)))))))

We can now write a more general procedure `gamma`

that returns (*n*) for any real *n*:

(define gamma (lambda (n e) (cond ((< n 1) (/ (gamma (+ n 1) e) n)) ((= n 1) 1) ((< 1 n 2) (gamma-1-to-2 n e)) (else (let ((n-1 (- n 1))) (* n-1 (gamma n-1 e)))))))

Let us now calculate (3/2).

(gamma 3/2 .001) (* 1/2 (sqrt *pi*))

The second value is the analytically correct answer.
(This is because (3/2) =
(1/2)(1/2), and (1/2) is known to be
^{1/2}.)
You can modify `gamma`

’s second
argument (the tolerance) to get as close an
approximation as you desire.

^{1} Consult any elementary
text on the calculus for an explanation of why this
approximation is reasonable.

^{2}
is the probability density of a
random variable with a *normal* or *Gaussian*
distribution, with mean = 0 and standard deviation = 1.
The definite integral _{0}^{z} (*x*) *d**x*
is the probability that the random
variable assumes a value between 0 and *z*.
However, you don’t need to know all this in
order to understand the example!

^{3} If Scheme didn’t
have the `atan`

procedure, we could use our
numerical-integration procedure to get an approximation
for _{0}^{1} (1 + *x*^{2})^{−1} *d**x*, which is /4.

^{4} By pulling constant factors — such as ```
(/
1 (sqrt (* 2 *pi*)))
```

in `phi`

— out of the
integrand, we could speed up the ordinate calculations
within `integrate‑simpson`

.

^{5} (*n*) for real *n* > 0 is
itself an extension of the “decrement-then-factorial” function
that maps *integer* *n* > 0 to (*n*−1)!.