Usenet 1994 October/usenetsourcesnewsgroupsinfomagicoctober1994disk2.iso/misc/volume24/gnucalc/part36

Text File | 1991-10-31 | 55.5 KB | 1,750 lines

Newsgroups: comp.sources.misc
From: [e-mail] (David Gillespie)
Subject:  v24i084:  gnucalc - GNU Emacs Calculator, v2.00, Part36/56
Message-ID: <[e-mail]>
X-Md4-Signature: 6bec56d735e7b86dd2319265b709ef85
Date: Thu, 31 Oct 1991 21:46:16 GMT
Approved: [e-mail]

Submitted-by: [e-mail] (David Gillespie)
Posting-number: Volume 24, Issue 84
Archive-name: gnucalc/part36
Environment: Emacs
Supersedes: gmcalc: Volume 13, Issue 27-45

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by four, except that they don't occur in years divisible by 100, except
that they @emph{do} in years divisible by 400.  We could work out the
answer by carefully counting the years divisible by four and the
exceptions, but there is a much simpler way that works even if we
don't know the leap year rule.
X
Let's assume the present year is 1991.  Years have 365 days, except
that leap years (whenever they occur) have 366 days.  So let's count
the number of days between now and then, and compare that to the
number of years times 365.  The number of extra days we find must be
equal to the number of leap years there were.
X
@group
@smallexample
1:  <Mon Jan 1, 10001>     2:  <Mon Jan 1, 10001>     1:  2925593
X    .                      1:  <Tue Jan 1, 1991>          .
X                               .
X
X  ' <jan 1 10001> RET         ' <jan 1 1991> RET          -
X
@end smallexample
@end group
@noindent
@group
@smallexample
3:  2925593       2:  2925593     2:  2925593     1:  1943
2:  10001         1:  8010        1:  2923650         .
1:  1991              .               .
X    .
X
X  10001 RET 1991      -               365 *           -
@end smallexample
@end group
X
@c [fix-ref Date Forms]
@noindent
There will be 1943 leap years before the year 10001.  (Assuming,
of course, that the algorithm for computing leap years remains
unchanged for that long.  @xref{Date Forms}, for some interesting
background information in that regard.)
X
@node Types Answer 7, Types Answer 8, Types Answer 6, Answers to Exercises
@subsection Types Tutorial Exercise 7
X
@noindent
The relative errors must be converted to absolute errors so that
@samp{+/-} notation may be used.
X
@group
@smallexample
1:  1.              2:  1.
X    .               1:  0.2
X                        .
X
X    20 RET .05 *        4 RET .05 *
@end smallexample
@end group
X
Now we simply chug through the formula.
X
@group
@smallexample
1:  19.7392088022    1:  394.78 +/- 19.739    1:  6316.5 +/- 706.21
X    .                    .                        .
X
X    2 P 2 ^ *            20 p 1 *                 4 p .2 RET 2 ^ *
@end smallexample
@end group
X
It turns out the @kbd{v u} command will unpack an error form as
well as a vector.  This saves us some retyping of numbers.
X
@group
@smallexample
3:  6316.5 +/- 706.21     2:  6316.5+/- 706.21
2:  6316.5                1:  0.1118
1:  706.21                    .
X    .
X
X    RET v u                   TAB /
@end smallexample
@end group
X
@noindent
Thus the volume is 6316 cubic centimeters, within about 11 percent.
X
@node Types Answer 8, Types Answer 9, Types Answer 7, Answers to Exercises
@subsection Types Tutorial Exercise 8
X
@noindent
The first answer is pretty simple:  @samp{1 / (0 .. 10) = (0.1 .. inf)}.
Since a number in the interval @samp{(0 .. 10)} can get arbitrarily
close to zero, its reciprocal can get arbitrarily large, so the answer
is an interval that effectively means, ``any number greater than 0.1''
but with no upper bound.
X
The second answer, similarly, is @samp{1 / (-10 .. 0) = (-inf .. -0.1)}.
X
Calc normally treats division by zero as an error, so that the formula
@w{@samp{1 / 0}} is left unsimplified.  Our third problem,
@w{@samp{1 / [0 .. 10]}}, also (potentially) divides by zero because zero
is now a member of the interval.  So Calc leaves this one unevaluated, too.
X
If you turn on ``infinite'' mode by pressing @kbd{m i}, you will
instead get the answer @samp{[0.1 .. inf]}, which includes infinity
as a possible value.
X
The fourth calculation, @samp{1 / (-10 .. 10)}, has the same problem.
Zero is buried inside the interval, but it's still a possible value.
It's not hard to see that the actual result of @samp{1 / (-10 .. 10)}
will be either greater than @i{0.1}, or less than @i{-0.1}.  Thus
the interval goes from minus infinity to plus infinity, with a ``hole''
in it from @i{-0.1} to @i{0.1}.  Calc doesn't have any way to
represent this, so it just reports @samp{[-inf .. inf]} as the answer.
It may be disappointing to hear ``the answer lies somewhere between
minus infinity and plus infinity, inclusive,'' but that's the best
that interval arithmetic can do in this case.
X
@node Types Answer 9, Types Answer 10, Types Answer 8, Answers to Exercises
@subsection Types Tutorial Exercise 9
X
@group
@smallexample
1:  [-3 .. 3]       2:  [-3 .. 3]     2:  [0 .. 9]
X    .               1:  [0 .. 9]      1:  [-9 .. 9]
X                        .                 .
X
X    [ 3 n .. 3 ]        RET 2 ^           TAB RET *
@end smallexample
@end group
X
@noindent
In the first case the result says, ``if a number is between @i{-3} and
3, its square is between 0 and 9.''  The second case says, ``the product
of two numbers each between @i{-3} and 3 is between @i{-9} and 9.''
X
An interval form is not a number; it is a symbol that can stand for
many different numbers.  Two identical-looking interval forms can stand
for different numbers.
X
The same issue arises when you try to square an error form.
X
@node Types Answer 10, Types Answer 11, Types Answer 9, Answers to Exercises
@subsection Types Tutorial Exercise 10
X
@noindent
Testing the first number, we might arbitrarily choose 17 for @cite{x}.
X
@group
@smallexample
1:  17 mod 811749613   2:  17 mod 811749613   1:  533694123 mod 811749613
X    .                      811749612              .
X                           .
X
X    17 M 811749613 RET     811749612              ^
@end smallexample
@end group
X
@noindent
Since 533694123 is (considerably) different from 1, the number 811749613
must not be prime.
X
It's awkward to type the number in twice as we did above.  There are
various ways to avoid this, and algebraic entry is one.  In fact, using
a vector mapping operation we can perform several tests at once.  Let's
use this method to test the second number.
X
@group
@smallexample
2:  [17, 42, 100000]               1:  [1 mod 15485863, 1 mod ... ]
1:  15485863                           .
X    .
X
X [17 42 100000] 15485863 RET           V M ' ($$ mod $)^($-1) RET
@end smallexample
@end group
X
@noindent
The result is three ones (modulo @cite{n}), so it's very probable that
15485863 is prime.  (In fact, this number is the millionth prime.)
X
Note that the functions @samp{($$^($-1)) mod $} or @samp{$$^($-1) % $}
would have been hopelessly inefficient, since they would have calculated
the power using full integer arithmetic.
X
Calc has a @kbd{k p} command that does primality testing.  For small
numbers it does an exact test; for large numbers it uses a variant
of the Fermat test we used here.  You can use @kbd{k p} repeatedly
to prove that a large integer is prime with any desired probability.
X
@node Types Answer 11, Types Answer 12, Types Answer 10, Answers to Exercises
@subsection Types Tutorial Exercise 11
X
@noindent
There are several ways to insert a calculated number into an HMS form.
One way to convert a number of seconds to an HMS form is simply to
multiply the number by an HMS form representing one second:
X
@group
@smallexample
1:  31415926.5359     2:  31415926.5359     1:  8726@@ 38' 46.5359"
X    .                 1:  0@@ 0' 1"              .
X                          .
X
X    P 1e7 *               0@@ 0' 1"              *
X
@end smallexample
@end group
@noindent
@group
@smallexample
2:  8726@@ 38' 46.5359"             1:  6@@ 6' 2.5359" mod 24@@ 0' 0"
1:  15@@ 27' 16" mod 24@@ 0' 0"          .
X    .
X
X    x time RET                         +
@end smallexample
@end group
X
@noindent
It will be just after six in the morning.
X
The algebraic @code{hms} function can also be used to build an
HMS form:
X
@group
@smallexample
1:  hms(0, 0, 10000000. pi)       1:  8726@@ 38' 46.5359"
X    .                                 .
X
X  ' hms(0, 0, 1e7 pi) RET             =
@end smallexample
@end group
X
@noindent
The @kbd{=} key is necessary to evaluate the symbol @samp{pi} to
the actual number 3.14159...
X
@node Types Answer 12, Types Answer 13, Types Answer 11, Answers to Exercises
@subsection Types Tutorial Exercise 12
X
@noindent
As we recall, there are 17 songs of about 2 minutes and 47 seconds
each.
X
@group
@smallexample
2:  0@@ 2' 47"                    1:  [0@@ 3' 7" .. 0@@ 3' 47"]
1:  [0@@ 0' 20" .. 0@@ 1' 0"]          .
X    .
X
X    [ 0@@ 20" .. 0@@ 1' ]              +
X
@end smallexample
@end group
@noindent
@group
@smallexample
1:  [0@@ 52' 59." .. 1@@ 4' 19."]
X    .
X
X    17 *
@end smallexample
@end group
X
@noindent
No matter how long it is, the album will fit nicely on one CD.
X
@node Types Answer 13, Types Answer 14, Types Answer 12, Answers to Exercises
@subsection Types Tutorial Exercise 13
X
@noindent
Type @kbd{' 1 yr RET u c s RET}.  The answer is 31557600 seconds.
X
@node Types Answer 14, Types Answer 15, Types Answer 13, Answers to Exercises
@subsection Types Tutorial Exercise 14
X
@noindent
How long will it take for a signal to get from one end of the computer
to the other?
X
@group
@smallexample
1:  m / c         1:  3.3356 ns
X    .                 .
X
X ' 1 m / c RET        u c ns RET
@end smallexample
@end group
X
@noindent
(Recall, @samp{c} is a ``unit'' corresponding to the speed of light.)
X
@group
@smallexample
1:  3.3356 ns     1:  0.81356 ns / ns     1:  0.81356
2:  4.1 ns            .                       .
X    .
X
X  ' 4.1 ns RET        /                       u s
@end smallexample
@end group
X
@noindent
Thus a signal could take up to 81 percent of a clock cycle just to
go from one place to another inside the computer, assuming the signal
could actually attain the full speed of light.  Pretty tight!
X
@node Types Answer 15, Algebra Answer 1, Types Answer 14, Answers to Exercises
@subsection Types Tutorial Exercise 15
X
@noindent
The speed limit is 55 miles per hour on most highways.  We want to
find the ratio of Sam's speed to the US speed limit.
X
@group
@smallexample
1:  55 mph         2:  55 mph           3:  11 hr mph / yd
X    .              1:  5 yd / hr            .
X                       .
X
X  ' 55 mph RET       ' 5 yd/hr RET          /
@end smallexample
@end group
X
The @kbd{u s} command doesn't figure out how to simplify this,
but @w{@kbd{u b}} (convert to base units) does.  Now we take the
logarithm base two to find the final answer, assuming that each
successive pill doubles his speed.
X
@group
@smallexample
1:  19360.       2:  19360.       1:  14.24
X    .            1:  2                .
X                     .
X
X    u b              2                B
@end smallexample
@end group
X
@noindent
Thus Sam can take up to 14 pills without a worry.
X
@node Algebra Answer 1, Algebra Answer 2, Types Answer 15, Answers to Exercises
@subsection Algebra Tutorial Exercise 1
X
@noindent
@c [fix-ref Declarations]
The result @samp{sqrt(x)^2} is simplified back to @cite{x} by the
Calculator, but @samp{sqrt(x^2)} is not.  (Consider what happens
if @w{@cite{x = -4}}.)  If @cite{x} is real, this formula could be
simplified to @samp{abs(x)}, but for general complex arguments even
that is not safe.  (@xref{Declarations}, for a way to tell Calc
that @cite{x} is known to be real.)
X
@node Algebra Answer 2, Algebra Answer 3, Algebra Answer 1, Answers to Exercises
@subsection Algebra Tutorial Exercise 2
X
@noindent
It is easiest to use @kbd{a P x RET} to get a list of solutions.
X
@group
@smallexample
1:  17 x^2 - 6 x^4 + 3 = 0   2:  17 x^2 - 6 x^4 + 3 = 0
2:  17 x^2 - 6 x^4 + 3 = 0   1:  [(0, 0.408), (0, -0.408), 1.732, -1.732]
X    .                            .
X
X    RET                          a P x RET
@end smallexample
@end group
X
@noindent
Now we can substitute these into the original formula.
X
@group
@smallexample
4:  17 x^2 - 6 x^4 + 3 = 0
3:  [(0., 0.408), (0., -0.408), 1.732, -1.732]
2:  [(0., 0.408), (0., -0.408), 1.732, -1.732]
1:  17 x^2 - 6 x^4 + 3
X    .
X
X    M-2 RET TAB   v u DEL
@end smallexample
@end group
X
@noindent
The @kbd{v u} command unpacks @samp{a = 0} to @samp{a} and 0; we then
delete the 0.
X
@group
@smallexample
3:  17 x^2 - 6 x^4 + 3 = 0
2:  [(0., 0.408), (0., -0.408), 1.732, -1.732]
1:  [1e-11, 1e-11, -1e-10, -1e-10]
X    .
X
X    V M $ RET
@end smallexample
@end group
X
@noindent
These numbers are reasonable approximations to zero, considering
the precision to which we computed the solutions.
X
@node Algebra Answer 3, Algebra Answer 4, Algebra Answer 2, Answers to Exercises
@subsection Algebra Tutorial Exercise 3
X
@group
@smallexample
1:  x sin(pi x)         1:  (sin(pi x) - pi x cos(pi x)) / pi^2
X    .                       .
X
X  ' x sin(pi x) RET   m r   a i x RET
X
@end smallexample
@end group
@noindent
@group
@smallexample
1:  [y, 1]
2:  (sin(pi x) - pi x cos(pi x)) / pi^2
X    .
X
X  ' [y,1] RET TAB
X
@end smallexample
@end group
@noindent
@group
@smallexample
1:  [(sin(pi y) - pi y cos(pi y)) / pi^2, (sin(pi) - pi cos(pi)) / pi^2]
X    .
X
X    V M $ RET
X
@end smallexample
@end group
@noindent
@group
@smallexample
1:  (sin(pi y) - pi y cos(pi y)) / pi^2 + (pi cos(pi) + sin(pi)) / pi^2
X    .
X
X    V R -
X
@end smallexample
@end group
@noindent
@group
@smallexample
1:  (sin(3.14159 y) - 3.14159 y cos(3.14159 y)) / 9.8696 - 0.3183
X    .
X
X    =
X
@end smallexample
@end group
@noindent
@group
@smallexample
1:  [0., -0.95493, 0.63662, -1.5915, 1.2732]
X    .
X
X    v x 5 RET  TAB  V M $ RET
@end smallexample
@end group
X
@node Algebra Answer 4, Rewrites Answer 1, Algebra Answer 3, Answers to Exercises
@subsection Algebra Tutorial Exercise 4
X
@noindent
The hard part is that @kbd{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:
X
@group
@smallexample
2:  -1                 2:  3                    1:  [4, 2, ..., 4]
1:  [1, 2, ..., 9]     1:  [-1, 1, ..., -1]         .
X    .                      .
X
X    1 n v x 9 RET          V M ^  3 TAB             -
X
@end smallexample
@end group
@noindent
@group
@smallexample
1:  [4, 2, ..., 4, 1]      1:  [1, 4, 2, ..., 4, 1]
X    .                          .
X
X    1 |                        1 TAB |
@end smallexample
@end group
X
@noindent
Now we compute the function values.  Note that for this method we need
eleven values, including both endpoints of the desired interval.
X
@group
@smallexample
2:  [1, 4, 2, ..., 4, 1]
1:  [1, 1.1, 1.2,  ...  , 1.8, 1.9, 2.]
X    .
X
X 11 RET 1 RET .1 RET  C-u v x
X
@end smallexample
@end group
@noindent
@group
@smallexample
2:  [1, 4, 2, ..., 4, 1]
1:  [0., 0.084941, 0.16993, ... ]
X    .
X
X    ' sin(x) ln(x) RET   m r  p 5 RET   V M $ RET
@end smallexample
@end group
X
@noindent
Once again this calls for @kbd{V M * V R +}; a simple @kbd{*} does the
same thing.
X
@group
@smallexample
1:  11.22      1:  1.122      1:  0.374
X    .              .              .
X
X    *              .1 *           3 /
@end smallexample
@end group
X
@noindent
Wow!  That's even better than the result from the Taylor series method.
X
@node Rewrites Answer 1, Rewrites Answer 2, Algebra Answer 4, Answers to Exercises
@subsection Rewrites Tutorial Exercise 1
X
@noindent
We'll use Big mode to make the formulas more readable.
X
@group
@smallexample
X                                               ___
X                                          2 + V 2
1:  (2 + sqrt(2)) / (1 + sqrt(2))     1:  --------
X    .                                          ___
X                                          1 + V 2
X
X                                          .
X
X  ' (2+sqrt(2)) / (1+sqrt(2)) RET         d B
@end smallexample
@end group
X
@noindent
Multiplying by the conjugate helps because @cite{(a+b) (a-b) = a^2 - b^2}.
X
@group
@smallexample
X          ___    ___
1:  (2 + V 2 ) (V 2  - 1)
X    .
X
X  a r a/(b+c) := a*(b-c) / (b^2-c^2) RET
X
@end smallexample
@end group
@noindent
@group
@smallexample
X         ___                         ___
1:  2 + V 2  - 2                1:  V 2
X    .                               .
X
X  a r a*(b+c) := a*b + a*c          a s
@end smallexample
@end group
X
@noindent
(We could have used @kbd{a x} instead of a rewrite rule for the
second step.)
X
The multiply-by-conjugate rule turns out to be useful in many
different circumstances, such as when the denominator involves
sines and cosines or the imaginary constant @code{i}.
X
@node Rewrites Answer 2, Rewrites Answer 3, Rewrites Answer 1, Answers to Exercises
@subsection Rewrites Tutorial Exercise 2
X
@noindent
Here is the rule set:
X
@group
@smallexample
[ fib(n) := fib(n, 1, 1) :: integer(n) :: n >= 1,
X  fib(1, x, y) := x,
X  fib(n, x, y) := fib(n-1, y, x+y) ]
@end smallexample
@end group
X
@noindent
The first rule turns a one-argument @code{fib} that people like to write
into a three-argument @code{fib} that makes computation easier.  The
second rule converts back from three-argument form once the computation
is done.  The third rule does the computation itself.  It basically
says that if @cite{x} and @cite{y} are two consecutive Fibonacci numbers,
then @cite{y} and @cite{x+y} are the next (overlapping) pair of Fibonacci
numbers.
X
Notice that because the number @cite{n} was ``validated'' by the
conditions on the first rule, there is no need to put conditions on
the other rules because the rule set would never get that far unless
the input were valid.  That further speeds computation, since no
extra conditions need to be checked at every step.
X
Actually, a user with a nasty sense of humor could enter a bad
three-argument @code{fib} call directly, say, @samp{fib(0, 1, 1)},
which would get the rules into an infinite loop.  One thing that would
help keep this from happening by accident would be to use something like
@samp{ZzFib} instead of @code{fib} as the name of the three-argument
function.
X
@node Rewrites Answer 3, Rewrites Answer 4, Rewrites Answer 2, Answers to Exercises
@subsection Rewrites Tutorial Exercise 3
X
@noindent
He got an infinite loop.  First, Calc did as expected and rewrote
@w{@samp{2 + 3 x}} to @samp{f(2, 3, x)}.  Then it looked for ways to
apply the rule again, and found that @samp{f(2, 3, x)} looks like
@samp{a + b x} with @w{@samp{a = 0}} and @samp{b = 1}, so it rewrote to
@samp{f(0, 1, f(2, 3, x))}.  It then wrapped another @samp{f(0, 1, ...)}
around that, and so on, ad infinitum.  Joe should have used @kbd{M-1 a r}
to make sure the rule applied only once.
X
(Actually, even the first step didn't work as he expected.  What Calc
really gives for @kbd{M-1 a r} in this situation is @samp{f(3 x, 1, 2)},
treating 2 as the ``variable,'' and @samp{3 x} as a constant being added
to it.  While this may seem odd, it's just as valid a solution as the
``obvious'' one.  One way to fix this would be to add the condition
@samp{:: variable(x)} to the rule, to make sure the thing that matches
@samp{x} is indeed a variable, or to change @samp{x} to @samp{quote(x)}
on the lefthand side, so that the rule matches the actual variable
@samp{x} rather than letting @samp{x} stand for something else.)
X
@node Rewrites Answer 4, Rewrites Answer 5, Rewrites Answer 3, Answers to Exercises
@subsection Rewrites Tutorial Exercise 4
X
@noindent
@tindex seq
Here is a suitable set of rules to solve the first part of the problem:
X
@group
@smallexample
[ seq(n, c) := seq(n/2,  c+1) :: n%2 = 0,
X  seq(n, c) := seq(3n+1, c+1) :: n%2 = 1 :: n > 1 ]
@end smallexample
@end group
X
Given the initial formula @samp{seq(6, 0)}, application of these
rules produces the following sequence of formulas:
X
@example
seq( 3, 1)
seq(10, 2)
seq( 5, 3)
seq(16, 4)
seq( 8, 5)
seq( 4, 6)
seq( 2, 7)
seq( 1, 8)
@end example
X
@noindent
whereupon neither of the rules match, and rewriting stops.
X
We can pretty this up a bit with a couple more rules:
X
@group
@smallexample
[ seq(n) := seq(n, 0),
X  seq(1, c) := c,
X  ... ]
@end smallexample
@end group
X
@noindent
Now, given @samp{seq(6)} as the starting configuration, we get 8
as the result.
X
The change to return a vector is quite simple:
X
@group
@smallexample
[ seq(n) := seq(n, []) :: integer(n) :: n > 0,
X  seq(1, v) := v | 1,
X  seq(n, v) := seq(n/2,  v | n) :: n%2 = 0,
X  seq(n, v) := seq(3n+1, v | n) :: n%2 = 1 ]
@end smallexample
@end group
X
@noindent
Given @samp{seq(6)}, the result is @samp{[6, 3, 10, 5, 16, 8, 4, 2, 1]}.
X
Notice that the @cite{n > 1} guard is no longer necessary on the last
rule since the @cite{n = 1} case is now detected by another rule.
But a guard has been added to the initial rule to make sure the
initial value is suitable before the computation begins.
X
While still a good idea, this guard is not as vitally important as it
was for the @code{fib} function, since calling, say, @samp{seq(x, [])}
will not get into an infinite loop.  Calc will not be able to prove
the symbol @samp{x} is either even or odd, so none of the rules will
apply and the rewrites will stop right away.
X
@node Rewrites Answer 5, Rewrites Answer 6, Rewrites Answer 4, Answers to Exercises
@subsection Rewrites Tutorial Exercise 5
X
@noindent
@tindex nterms
If @cite{x} is the sum @cite{a + b}, then `@t{nterms(}@i{x}@t{)}' must
be `@t{nterms(}@i{a}@t{)}' plus `@t{nterms(}@i{b}@t{)}'.  If @cite{x}
is not a sum, then `@t{nterms(}@i{x}@t{)}' = 1.
X
@group
@smallexample
[ nterms(a + b) := nterms(a) + nterms(b),
X  nterms(x)     := 1 ]
@end smallexample
@end group
X
@noindent
Here we have taken advantage of the fact that earlier rules always
match before later rules; @samp{nterms(x)} will only be tried if we
already know that @samp{x} is not a sum.
X
@node Rewrites Answer 6, Programming Answer 1, Rewrites Answer 5, Answers to Exercises
@subsection Rewrites Tutorial Exercise 6
X
@noindent
Here is a rule set that will do the job:
X
@group
@smallexample
[ a*(b + c) := a*b + a*c,
X  opt(a) O(x^n) + opt(b) O(x^m) := O(x^n) :: n <= m
X     :: constant(a) :: constant(b),
X  opt(a) O(x^n) + opt(b) x^m := O(x^n) :: n <= m
X     :: constant(a) :: constant(b),
X  a O(x^n) := O(x^n) :: constant(a),
X  x^opt(m) O(x^n) := O(x^(n+m)),
X  O(x^n) O(x^m) := O(x^(n+m)) ]
@end smallexample
@end group
X
If we really want the @kbd{+} and @kbd{*} keys to operate naturally
on power series, we should put these rules in @code{EvalRules}.  For
testing purposes, it is better to put them in a different variable,
say, @code{O}, first.
X
The first rule just expands products of sums so that the rest of the
rules can assume they have an expanded-out polynomial to work with.
Note that this rule does not mention @samp{O} at all, so it will
apply to any product-of-sum it encounters---this rule may surprise
you if you put it into @code{EvalRules}!
X
In the second rule, the sum of two O's is changed to the smaller O.
The optional constant coefficients are there mostly so that
@samp{O(x^2) - O(x^3)} and @samp{O(x^3) - O(x^2)} are handled
as well as @samp{O(x^2) + O(x^3)}.
X
The third rule absorbs higher powers of @samp{x} into O's.
X
The fourth rule says that a constant times a negligible quantity
is still negligible.  (This rule will also match @samp{O(x^3) / 4},
with @samp{a = 1/4}.)
X
The fifth rule rewrites, for example, @samp{x^2 O(x^3)} to @samp{O(x^5)}.
(It is easy to see that if one of these forms is negligible, the other
is, too.)  Notice the @samp{x^opt(m)} to pick up terms like
@w{@samp{x O(x^3)}}.  Optional powers will match @samp{x} as @samp{x^1}
but not 1 as @samp{x^0}.  This turns out to be exactly what we want here.
X
The sixth rule is the corresponding rule for products of two O's.
X
Another way to solve this problem would be to create a new ``data type''
that represents truncated power series.  We might represent these as
function calls @samp{series(@var{coefs}, @var{x})} where @var{coefs} is
a vector of coefficients for @cite{x^0}, @cite{x^1}, @cite{x^2}, and so
on.  Rules would exist for sums and products of such @code{series}
objects, and as an optional convenience could also know how to combine a
@code{series} object with a normal polynomial.  (With this, and with a
rule that rewrites @samp{O(x^n)} to the equivalent @code{series} form,
you could still enter power series in exactly the same notation as
before.)  Operations on such objects would probably be more efficient,
although the objects would be a bit harder to read.
X
@c [fix-ref Compositions]
Some other symbolic math programs provide a power series data type
similar to this.  Mathematica, for example, has an object that looks
like @samp{PowerSeries[@var{x}, @var{x0}, @var{coefs}, @var{nmin},
@var{nmax}, @var{den}]}, where @var{x0} is the point about which the
power series is taken (we've been assuming this was always zero),
and @var{nmin}, @var{nmax}, and @var{den} allow pseudo-power-series
with fractional or negative powers.  Also, the @code{PowerSeries}
objects have a special display format that makes them look like
@samp{2 x^2 + O(x^4)} when they are printed out.  (@xref{Compositions},
for a way to do this in Calc, although for something as involved as
this it would probably be better to write the formatting routine
in Lisp.)
X
@node Programming Answer 1, Programming Answer 2, Rewrites Answer 6, Answers to Exercises
@subsection Programming Tutorial Exercise 1
X
Just enter the formula @samp{ninteg(cos(pi t^2/2), t, 0, x)}, type
@kbd{Z F}, and answer the questions.  Since this formula contains two
variables, the default argument list will be @samp{(t x)}.  We want to
change this to @samp{(x)} since @cite{t} is really a dummy variable
to be used within @code{ninteg}.
X
The exact keystrokes are @kbd{Z F c fresC RET RET C-b C-b DEL DEL RET y}.
(The @kbd{C-b C-b DEL DEL} are what fix the argument list.)
X
While this definition works, you'll find the @code{erf}-based definition
described afterwards to be much more efficient; @code{erf} is also defined
by an integral (involving @samp{exp(-t^2)}), but Calc has good algorithms
for computing it that don't rely on brute-force numerical integration.
The @kbd{Z F} formula would be @samp{re((i+1) erf(sqrt(pi) (1-i) x / 2) / 2)},
where @code{re} is the function that extracts the real part of a complex
number.  Actually, you'd want to type @kbd{=} on this formula first to
evaluate the @code{pi} and @code{i} variables in it; otherwise your
@kbd{z c} function will return a formula that needs to have @kbd{=}
typed on it each time.
X
@node Programming Answer 2, Programming Answer 3, Programming Answer 1, Answers to Exercises
@subsection Programming Tutorial Exercise 2
X
@noindent
One way is to move the number to the top of the stack, operate on
it, then move it back:  @kbd{C-x ( M-TAB n M-TAB M-TAB C-x )}.
X
Another way is to negate the top three stack entries, then negate
again the top two stack entries:  @kbd{C-x ( M-3 n M-2 n C-x )}.
X
Finally, it turns out that a negative prefix argument causes a
command like @kbd{n} to operate on just the specified stack entry,
which is just what we want:  @kbd{C-x ( M-- 3 n C-x )}.
X
Just for kicks, let's also do it algebraically:
@w{@kbd{C-x ( ' -$$$, $$, $ RET C-x )}}.
X
@node Programming Answer 3, Programming Answer 4, Programming Answer 2, Answers to Exercises
@subsection Programming Tutorial Exercise 3
X
@noindent
Each of these functions can be computed using the stack, or using
algebraic entry, whichever way you prefer:
X
@noindent
Computing @c{$\displaystyle{\sin x \over x}$}
@cite{sin(x) / x}:
X
Using the stack:  @kbd{C-x (  RET S TAB /  C-x )}.
X
Using algebraic entry:  @kbd{C-x (  ' sin($)/$ RET  C-x )}.
X
@noindent
Computing the logarithm:
X
Using the stack:  @kbd{C-x (  TAB B  C-x )}
X
Using algebraic entry:  @kbd{C-x (  ' log($,$$) RET  C-x )}.
X
@noindent
Computing the vector of integers:
X
Using the stack:  @kbd{C-x (  1 RET 1  C-u v x  C-x )}.  (Recall that
@kbd{C-u v x} takes the vector size, starting value, and increment
from the stack.)
X
Alternatively:  @kbd{C-x (  ~ v x  C-x )}.  (The @kbd{~} key pops a
number from the stack and uses it as the prefix argument for the
next command.)
X
Using algebraic entry:  @kbd{C-x (  ' index($) RET  C-x )}.
X
@node Programming Answer 4, Programming Answer 5, Programming Answer 3, Answers to Exercises
@subsection Programming Tutorial Exercise 4
X
@noindent
Here's one way:  @kbd{C-x ( RET V R + TAB v l / C-x )}.
X
@node Programming Answer 5, Programming Answer 6, Programming Answer 4, Answers to Exercises
@subsection Programming Tutorial Exercise 5
X
@group
@smallexample
2:  1              1:  1.61803398502         2:  1.61803398502
1:  20                 .                     1:  1.61803398875
X    .                                            .
X
X   1 RET 20         Z < & 1 + Z >                I H P
@end smallexample
@end group
X
@noindent
This answer is quite accurate.
X
@node Programming Answer 6, Programming Answer 7, Programming Answer 5, Answers to Exercises
@subsection Programming Tutorial Exercise 6
X
@noindent
Here is the matrix:
X
@example
[ [ 0, 1 ]   * [a, b] = [b, a + b]
X  [ 1, 1 ] ]
@end example
X
@noindent
Thus @samp{[0, 1; 1, 1]^n * [1, 1]} computes Fibonacci numbers @cite{n+1}
and @cite{n+2}.  Here's one program that does the job:
X
@example
C-x ( ' [0, 1; 1, 1] ^ ($-1) * [1, 1] RET v u DEL C-x )
@end example
X
@noindent
This program is quite efficient because Calc knows how to raise a
matrix (or other value) to the power @cite{n} in only @c{$\log_2 n$}
@cite{log(n,2)}
steps.  For example, this program can compute the 1000th Fibonacci
number (a 209-digit integer!) in about 10 steps; even though the
@kbd{Z < ... Z >} solution had much simpler steps, it would have
required so many steps that it would not have been practical.
X
@node Programming Answer 7, Programming Answer 8, Programming Answer 6, Answers to Exercises
@subsection Programming Tutorial Exercise 7
X
@noindent
The trick here is to compute the harmonic numbers differently, so that
the loop counter itself accumulates the sum of reciprocals.  We use
a separate variable to hold the integer counter.
X
@group
@smallexample
1:  1          2:  1       1:  .
X    .          1:  4
X                   .
X
X    1 t 1       1 RET 4      Z ( t 2 r 1 1 + s 1 & Z )
@end smallexample
@end group
X
@noindent
The body of the loop goes as follows:  First save the harmonic sum
so far in variable 2.  Then delete it from the stack; the for loop
itself will take care of remembering it for us.  Next, recall the
count from variable 1, add one to it, and feed its reciprocal to
the for loop to use as the step value.  The for loop will increase
the ``loop counter'' by that amount and keep going until the
loop counter exceeds 4.
X
@group
@smallexample
2:  31                  3:  31
1:  3.99498713092       2:  3.99498713092
X    .                   1:  4.02724519544
X                            .
X
X    r 1 r 2                 RET 31 & +
@end smallexample
@end group
X
Thus we find that the 30th harmonic number is 3.99, and the 31st
harmonic number is 4.02.
X
@node Programming Answer 8, Programming Answer 9, Programming Answer 7, Answers to Exercises
@subsection Programming Tutorial Exercise 8
X
@noindent
The first step is to compute the derivative @cite{f'(x)} and thus
the formula @c{$\displaystyle{x - {f(x) \over f'(x)}}$}
@cite{x - f(x)/f'(x)}.
X
(Because this definition is long, it will be repeated in concise form
below.  You can use @w{@kbd{M-# m}} to load it from there.  While you are
entering a @kbd{Z ` Z '} body in a macro, Calc simply collects
keystrokes without executing them.  In the following diagrams we'll
pretend Calc actually executed the keystrokes as you typed them,
just for purposes of illustration.)
X
@group
@smallexample
2:  sin(cos(x)) - 0.5            3:  4.5
1:  4.5                          2:  sin(cos(x)) - 0.5
X    .                            1:  -(sin(x) cos(cos(x)))
X                                     .
X
' sin(cos(x))-0.5 RET 4.5  m r  C-x ( Z `  TAB RET a d x RET
X
@end smallexample
@end group
@noindent
@group
@smallexample
2:  4.5
1:  x + (sin(cos(x)) - 0.5) / sin(x) cos(cos(x))
X    .
X
X    /  ' x RET TAB -   t 1
@end smallexample
@end group
X
Now, we enter the loop.  We'll use a repeat loop with a 20-repetition
limit just in case the method fails to converge for some reason.
(Normally, the @w{@kbd{Z /}} command will stop the loop before all 20
repetitions are done.)
X
@group
@smallexample
1:  4.5         3:  4.5                     2:  4.5
X    .           2:  x + (sin(cos(x)) ...    1:  5.24196456928
X                1:  4.5                         .
X                    .
X
X  20 Z <          RET r 1 TAB                 s l x RET
@end smallexample
@end group
X
This is the new guess for @cite{x}.  Now we compare it with the
old one to see if we've converged.
X
@group
@smallexample
3:  5.24196     2:  5.24196     1:  5.24196     1:  5.26345856348
2:  5.24196     1:  0               .               .
1:  4.5             .
X    .
X
X  RET M-TAB         a =             Z /             Z > Z ' C-x )
@end smallexample
@end group
X
The loop converges in just a few steps to this value.  To check
the result, we can simply substitute it back into the equation.
X
@group
@smallexample
2:  5.26345856348
1:  0.499999999997
X    .
X
X RET ' sin(cos($)) RET
@end smallexample
@end group
X
Let's test the new definition again:
X
@group
@smallexample
2:  x^2 - 9           1:  3.
1:  1                     .
X    .
X
X  ' x^2-9 RET 1           X
@end smallexample
@end group
X
Once again, here's the full Newton's Method definition:
X
@group
@example
C-x ( Z `  TAB RET a d x RET  /  ' x RET TAB -  t 1
X           20 Z <  RET r 1 TAB  s l x RET
X                   RET M-TAB  a =  Z /
X              Z >
X      Z '
C-x )
@end example
@end group
X
@c [fix-ref Nesting and Fixed Points]
It turns out that Calc has a built-in command for applying a formula
repeatedly until it converges to a number.  @xref{Nesting and Fixed Points},
to see how to use it.
X
@c [fix-ref Root Finding]
Also, of course, @kbd{a R} is a built-in command that uses Newton's
method (among others) to look for numerical solutions to any equation.
@xref{Root Finding}.
X
@node Programming Answer 9, Programming Answer 10, Programming Answer 8, Answers to Exercises
@subsection Programming Tutorial Exercise 9
X
@noindent
The first step is to adjust @cite{z} to be greater than 5.  A simple
``for'' loop will do the job here.  If @cite{z} is less than 5, we
reduce the problem using @c{$\psi(z) = \psi(z+1) - 1/z$}
@cite{psi(z) = psi(z+1) - 1/z}.  We go
on to compute @c{$\psi(z+1)$}
@cite{psi(z+1)}, and remember to add back a factor of
@cite{-1/z} when we're done.  This step is repeated until @cite{z > 5}.
X
(Because this definition is long, it will be repeated in concise form
below.  You can use @w{@kbd{M-# m}} to load it from there.  While you are
entering a @kbd{Z ` Z '} body in a macro, Calc simply collects
keystrokes without executing them.  In the following diagrams we'll
pretend Calc actually executed the keystrokes as you typed them,
just for purposes of illustration.)
X
@group
@smallexample
1:  1.             1:  1.
X    .                  .
X
X 1.0 RET       C-x ( Z `  s 1  0 t 2
@end smallexample
@end group
X
Here, variable 1 holds @cite{z} and variable 2 holds the adjustment
factor.  If @cite{z < 5}, we use a loop to increase it.
X
(By the way, we started with @samp{1.0} instead of the integer 1 because
otherwise the calculation below will try to do exact fractional arithmetic,
and will never converge because fractions compare equal only if they
are exactly equal, not just the same to within the current precision.)
X
@group
@smallexample
3:  1.      2:  1.       1:  6.
2:  1.      1:  1            .
1:  5           .
X    .
X
X  RET 5        a <    Z [  5 Z (  & s + 2  1 s + 1  1 Z ) r 1  Z ]
@end smallexample
@end group
X
Now we compute the initial part of the sum:  @c{$\ln z - {1 \over 2z}$}
@cite{ln(z) - 1/2z}
minus the adjustment factor.
X
@group
@smallexample
2:  1.79175946923      2:  1.7084261359      1:  -0.57490719743
1:  0.0833333333333    1:  2.28333333333         .
X    .                      .
X
X    L  r 1 2 * &           -  r 2                -
@end smallexample
@end group
X
Now we evaluate the series.  We'll use another ``for'' loop counting
up the value of @cite{2 n}.  (Calc does have a summation command,
@kbd{a +}, but we'll use loops just to get more practice with them.)
X
@group
@smallexample
3:  -0.5749       3:  -0.5749        4:  -0.5749      2:  -0.5749
2:  2             2:  1:6            3:  1:6          1:  2.3148e-3
1:  40            1:  2              2:  2                .
X    .                 .              1:  36.
X                                         .
X
X   2 RET 40        Z ( RET k b TAB     RET r 1 TAB ^      * /
X
@end smallexample
@end group
@noindent
@group
@smallexample
3:  -0.5749       3:  -0.5772      2:  -0.5772     1:  -0.577215664892
2:  -0.5749       2:  -0.5772      1:  0               .
1:  2.3148e-3     1:  -0.5749          .
X    .                 .
X
X  TAB RET M-TAB       - RET M-TAB      a =     Z /    2  Z )  Z ' C-x )
@end smallexample
@end group
X
This is the value of @c{$-\gamma$}
@cite{- gamma}, with a slight bit of roundoff error.
To get a full 12 digits, let's use a higher precision:
X
@group
@smallexample
2:  -0.577215664892      2:  -0.577215664892
1:  1.                   1:  -0.577215664901532
X
X    1. RET                   p 16 RET X
@end smallexample
@end group
X
Here's the complete sequence of keystrokes:
X
@group
@example
C-x ( Z `  s 1  0 t 2
X           RET 5 a <  Z [  5 Z (  & s + 2  1 s + 1  1 Z ) r 1  Z ]
X           L r 1 2 * & - r 2 -
X           2 RET 40  Z (  RET k b TAB RET r 1 TAB ^ * /
X                          TAB RET M-TAB - RET M-TAB a = Z /
X                  2  Z )
X      Z '
C-x )
@end example
@end group
X
@node Programming Answer 10, Programming Answer 11, Programming Answer 9, Answers to Exercises
@subsection Programming Tutorial Exercise 10
X
@noindent
Taking the derivative of a term of the form @cite{x^n} will produce
a term like @c{$n x^{n-1}$}
@cite{n x^(n-1)}.  Taking the derivative of a constant
produces zero.  From this it is easy to see that the @cite{n}th
derivative of a polynomial, evaluated at @cite{x = 0}, will equal the
coefficient on the @cite{x^n} term times @cite{n!}.
X
(Because this definition is long, it will be repeated in concise form
below.  You can use @w{@kbd{M-# m}} to load it from there.  While you are
entering a @kbd{Z ` Z '} body in a macro, Calc simply collects
keystrokes without executing them.  In the following diagrams we'll
pretend Calc actually executed the keystrokes as you typed them,
just for purposes of illustration.)
X
@group
@smallexample
2:  5 x^4 + (x + 1)^2          3:  5 x^4 + (x + 1)^2
1:  6                          2:  0
X    .                          1:  6
X                                   .
X
X  ' 5 x^4 + (x+1)^2 RET 6        C-x ( Z `  [ ] t 1  0 TAB
@end smallexample
@end group
X
@noindent
Variable 1 will accumulate the vector of coefficients.
X
@group
@smallexample
2:  0              3:  0                  2:  5 x^4 + ...
1:  5 x^4 + ...    2:  5 x^4 + ...        1:  1
X    .              1:  1                      .
X                       .
X
X   Z ( TAB         RET 0 s l x RET            M-TAB ! /  s | 1
@end smallexample
@end group
X
@noindent
Note that @kbd{s | 1} appends the top-of-stack value to the vector
in a variable; it is completely analogous to @kbd{s + 1}.  We could
have written instead, @kbd{r 1 TAB | t 1}.
X
@group
@smallexample
1:  20 x^3 + 2 x + 2      1:  0         1:  [1, 2, 1, 0, 5, 0, 0]
X    .                         .             .
X
X    a d x RET                 1 Z )         DEL r 1  Z ' C-x )
@end smallexample
@end group
X
To convert back, a simple method is just to map the coefficients
against a table of powers of @cite{x}.
X
@group
@smallexample
2:  [1, 2, 1, 0, 5, 0, 0]    2:  [1, 2, 1, 0, 5, 0, 0]
1:  6                        1:  [0, 1, 2, 3, 4, 5, 6]
X    .                            .
X
X    6 RET                        1 + 0 RET 1 C-u v x
X
@end smallexample
@end group
@noindent
@group
@smallexample
2:  [1, 2, 1, 0, 5, 0, 0]    2:  1 + 2 x + x^2 + 5 x^4
1:  [1, x, x^2, x^3, ... ]       .
X    .
X
X    ' x RET TAB V M ^            *
@end smallexample
@end group
X
Once again, here are the whole polynomial to/from vector programs:
X
@group
@example
C-x ( Z `  [ ] t 1  0 TAB
X           Z (  TAB RET 0 s l x RET M-TAB ! /  s | 1
X                a d x RET
X         1 Z ) r 1
X      Z '
C-x )
X
C-x (  1 + 0 RET 1 C-u v x ' x RET TAB V M ^ *  C-x )
@end example
@end group
X
@node Programming Answer 11, Programming Answer 12, Programming Answer 10, Answers to Exercises
@subsection Programming Tutorial Exercise 11
X
@noindent
First we define a dummy program to go on the @kbd{z s} key.  The true
@w{@kbd{z s}} key is supposed to take two numbers from the stack and
return one number, so @kbd{DEL} as a dummy definition will make
sure the stack comes out right.
X
@group
@smallexample
2:  4          1:  4                         2:  4
1:  2              .                         1:  2
X    .                                            .
X
X  4 RET 2       C-x ( DEL C-x )  Z K s RET       2
@end smallexample
@end group
X
The last step replaces the 2 that was eaten during the creation
of the dummy @kbd{z s} command.  Now we move on to the real
definition.  The recurrence needs to be rewritten slightly,
to the form @cite{s(n,m) = s(n-1,m-1) - (n-1) s(n-1,m)}.
X
(Because this definition is long, it will be repeated in concise form
below.  You can use @kbd{M-# m} to load it from there.)
X
@group
@smallexample
2:  4        4:  4       3:  4       2:  4
1:  2        3:  2       2:  2       1:  2
X    .        2:  4       1:  0           .
X             1:  2           .
X                 .
X
X  C-x (       M-2 RET        a =         Z [  DEL DEL 1  Z :
X
@end smallexample
@end group
@noindent
@group
@smallexample
4:  4       2:  4                     2:  3      4:  3    4:  3    3:  3
3:  2       1:  2                     1:  2      3:  2    3:  2    2:  2
2:  2           .                         .      2:  3    2:  3    1:  3
1:  0                                            1:  2    1:  1        .
X    .                                                .        .
X
X  RET 0   a = Z [  DEL DEL 0  Z :  TAB 1 - TAB   M-2 RET     1 -      z s
@end smallexample
@end group
X
@noindent
(Note that the value 3 that our dummy @kbd{z s} produces is not correct;
it is merely a placeholder that will do just as well for now.)
X
@group
@smallexample
3:  3               4:  3           3:  3       2:  3      1:  -6
2:  3               3:  3           2:  3       1:  9          .
1:  2               2:  3           1:  3           .
X    .               1:  2               .
X                        .
X
X M-TAB M-TAB     TAB RET M-TAB         z s          *          -
X
@end smallexample
@end group
@noindent
@group
@smallexample
1:  -6                          2:  4          1:  11      2:  11
X    .                           1:  2              .       1:  11
X                                    .                          .
X
X  Z ] Z ] C-x )   Z K s RET      DEL 4 RET 2       z s      M-RET k s
@end smallexample
@end group
X
Even though the result that we got during the definition was highly
bogus, once the definition is complete the @kbd{z s} command gets
the right answers.
X
Here's the full program once again:
X
@group
@example
C-x (  M-2 RET a =
X       Z [  DEL DEL 1
X       Z :  RET 0 a =
X            Z [  DEL DEL 0
X            Z :  TAB 1 - TAB M-2 RET 1 - z s
X                 M-TAB M-TAB TAB RET M-TAB z s * -
X            Z ]
X       Z ]
C-x )
@end example
@end group
X
You can read this definition using @kbd{M-# m} (@code{read-kbd-macro})
followed by @kbd{Z K s}, without having to make a dummy definition
first, because @code{read-kbd-macro} doesn't need to execute the
definition as it reads it in.  For this reason, @code{M-# m} is often
the easiest way to create recursive programs in Calc.
X
@node Programming Answer 12, , Programming Answer 11, Answers to Exercises
@subsection Programming Tutorial Exercise 12
X
@noindent
This turns out to be a much easier way to solve the problem.  Let's
denote Stirling numbers as calls of the function @samp{s}.
X
First, we store the rewrite rules corresponding to the definition of
Stirling numbers in a convenient variable:
X
@smallexample
s e StirlingRules RET
[ s(n,n) := 1  :: n >= 0,
X  s(n,0) := 0  :: n > 0,
X  s(n,m) := s(n-1,m-1) - (n-1) s(n-1,m) :: n >= m :: m >= 1 ]
C-c C-c
@end smallexample
X
Now, it's just a matter of applying the rules:
X
@group
@smallexample
2:  4          1:  s(4, 2)              1:  11
1:  2              .                        .
X    .
X
X  4 RET 2       C-x (  ' s($$,$) RET     a r StirlingRules RET  C-x )
@end smallexample
@end group
X
As in the case of the @code{fib} rules, it would be useful to put these
rules in @code{EvalRules} and to add a @samp{:: remember} condition to
the last rule.
X
@c This ends the table-of-contents kludge from above:
@tex
\global\let\chapternofonts=\oldchapternofonts
@end tex
X
@c [reference]
X
@node Introduction, Data Types, Tutorial, Top
@chapter Introduction
X
@noindent
This chapter is the beginning of the Calc reference manual.
It covers basic concepts such as the stack, algebraic and
numeric entry, undo, numeric prefix arguments, etc.
X
@c [when-split]
@c (Chapter 2, the Tutorial, has been printed in a separate volume.)
X
@menu
* Basic Commands::
* Help Commands::
* Stack Basics::
* Numeric Entry::
* Algebraic Entry::
* Quick Calculator::
* Keypad Mode::
* Prefix Arguments::
* Undo::
* Error Messages::
* Multiple Calculators::
* Troubleshooting Commands::
@end menu
X
@node Basic Commands, Help Commands, Introduction, Introduction
@section Basic Commands
X
@noindent
@pindex calc
@pindex calc-mode
@cindex Starting the Calculator
@cindex Running the Calculator
To start the Calculator in its standard interface, type @kbd{M-x calc}.
By default this creates a pair of small windows, @samp{*Calculator*}
and @samp{*Calc Trail*}.  The former displays the contents of the
Calculator stack and is manipulated exclusively through Calc commands.
It is possible (though not usually necessary) to create several Calc
Mode buffers each of which has an independent stack, undo list, and
mode settings.  There is exactly one Calc Trail buffer; it records a
list of the results of all calculations that have been done.  The
Calc Trail buffer uses a variant of Calc Mode, so Calculator commands
still work when the trail buffer's window is selected.  It is possible
to turn the trail window off, but the @samp{*Calc Trail*} buffer still
exists and is updated silently.  @xref{Trail Commands}.@refill
X
@kindex M-# M-#
@kindex M-# #
@kindex M-# c
In most installations, the @kbd{M-#} (Meta-Shift-3) keystroke is a more
convenient way to start the Calculator.  If you don't have a Meta key,
press @key{ESC}, then @kbd{#}.  Follow @kbd{M-#} with a letter key
that says which Calc interface or function you desire.  @kbd{M-# c}
is equivalent to @kbd{M-x calc}.  @kbd{M-# M-#} or @kbd{M-# #} is
also generally equivalent to @kbd{M-x calc}.  @xref{Other M-# Commands}.
X
@kindex x
@kindex M-x
@pindex calc-execute-extended-command
Most Calc commands use one or two keystrokes.  Lower- and upper-case
letters are distinct.  Commands may also be entered in @kbd{M-x} form;
for some commands this is the only form.  As a convenience, the @kbd{x}
key (@code{calc-execute-extended-command})
is like @kbd{M-x} except that it enters the initial string @samp{calc-}
for you.  For example, the following key sequences are equivalent:
@kbd{S}, @kbd{M-x calc-sin @key{RET}}, @kbd{x sin @key{RET}}.@refill
X
@cindex Extensions module
@cindex @file{calc-ext} module
The Calculator exists in many parts.  When you type @kbd{M-x calc}, the
Emacs ``auto-load'' mechanism will bring in only the first part, which
contains the basic arithmetic functions.  The other parts will be
auto-loaded the first time you use the more advanced commands like trig
functions or matrix operations.  This is done to improve the response time
of the Calculator in the common case when all you need to do is a
little arithmetic.  If for some reason the Calculator fails to load an
extension module automatically, you can force it to load all the
extensions by using the @kbd{m X} or @kbd{M-# L}
(@code{calc-load-everything}) command.
@xref{Mode Settings}.@refill
X
If you type @kbd{M-x calc} or @kbd{M-# c} with any numeric prefix argument,
the Calculator is loaded if necessary, but it is not actually started.
If the argument is positive, the extensions are also loaded if necessary.
User-written Lisp code that wishes to make use of Calc's arithmetic
routines can use @samp{(calc 0)} or @samp{(calc 1)} to auto-load the
Calculator.@refill
X
@kindex M-# b
@pindex full-calc
If you type @kbd{M-# b}, then next time you use @kbd{M-# c} you
will get a Calculator that uses the full height of the Emacs screen.
When full-screen mode is on, @kbd{M-# c} runs the @code{full-calc}
command instead of @code{calc}.  From the Unix shell you can type
@samp{emacs -f full-calc} to start a new Emacs specifically for use
as a calculator.  When Calc is started from the Emacs command line
like this, Calc's normal ``quit'' commands actually quit Emacs itself.
X
@kindex M-# o
@pindex calc-other-window
The @kbd{M-# o} command is like @kbd{M-# c} except that the Calc
window is not actually selected.  If you are already in the Calc
window, @kbd{M-# o} switches you out of it.  (The regular Emacs
@kbd{C-x o} command would also work for this, but it has a
tendency to drop you into the Calc Trail window instead, which
@kbd{M-# o} takes care not to do.)
X
For one quick calculation, you can type @kbd{M-# q} (@code{quick-calc})
which prompts you for a formula (like @samp{2+3/4}).  The result is
displayed at the bottom of the Emacs screen without ever creating
any special Calculator windows.  @xref{Quick Calculator}.
X
Finally, if you are using the X window system you may want to try
@kbd{M-# k} (@code{calc-keypad}) which runs Calc with a
``calculator keypad'' picture as well as a stack display.  Click on
the keys with the mouse to operate the calculator.  @xref{Keypad Mode}.
X
@kindex q
@pindex calc-quit
@cindex Quitting the Calculator
@cindex Exiting the Calculator
The @kbd{q} key (@code{calc-quit}) exits Calc Mode and closes the
Calculator's window(s).  It does not delete the Calculator buffers.
If you type @kbd{M-x calc} again, the Calculator will reappear with the
contents of the stack intact.  Typing @kbd{M-# c} or @kbd{M-# M-#}
again from inside the Calculator buffer is equivalent to executing
@code{calc-quit}; you can think of @kbd{M-# M-#} as toggling the
Calculator on and off.@refill
X
@kindex M-# x
The @kbd{M-# x} key also turns the Calculator off, no matter which
user interface (standard, Keypad, or Embedded) is current active.
It also cancels @code{calc-edit} mode if used from there.
X
@kindex d SPC
@kindex d ~
@pindex calc-refresh
@cindex Refreshing a garbled display
@cindex Garbled displays, refreshing
The @kbd{d SPC} key sequence (@code{calc-refresh}) redraws the contents
of the Calculator buffer from memory.  Use this if the contents of the
buffer have been damaged somehow.  The @kbd{d ~} key is also bound to
@code{calc-refresh}.
X
The @kbd{o} key (@code{calc-realign}) moves the cursor back to its
``home'' position at the bottom of the Calculator buffer.
X
@kindex <
@kindex >
@pindex calc-scroll-left
@pindex calc-scroll-right
@cindex Horizontal scrolling
@cindex Scrolling
@cindex Wide text, scrolling
The @kbd{<} and @kbd{>} keys are bound to @code{calc-scroll-left} and
@code{calc-scroll-right}.  These are just like the normal horizontal
scrolling commands except that they scroll one half-screen at a time by
default.  (Calc formats its output to fit within the bounds of the
window whenever it can.)@refill
X
@kindex @{
@kindex @}
@pindex calc-scroll-down
@pindex calc-scroll-up
@cindex Vertical scrolling
The @kbd{@{} and @kbd{@}} keys are bound to @code{calc-scroll-down}
and @code{calc-scroll-up}.  They scroll up or down by one-half the
height of the Calc window.@refill
X
@pindex calc-version
The @code{calc-version} command displays the current version number
of Calc and the name of the person who installed it on your system.
(This information is also present in the @samp{*Calc Trail*} buffer,
and in the output of the @kbd{h h} command.)
X
@node Help Commands, Stack Basics, Basic Commands, Introduction
@section Help Commands
X
@noindent
@cindex Help commands
@kindex ?
@pindex calc-help
The @kbd{?} key (@code{calc-help}) displays a series of brief help messages.
Some keys (such as @kbd{b} and @kbd{d}) are prefix keys, like Emacs'
@key{ESC} and @kbd{C-x} prefixes.  You can type
@kbd{?} after a prefix to see a list of commands beginning with that
prefix.  (If the message includes @samp{[MORE]}, press @kbd{?} again
to see additional commands for that prefix.)
X
@kindex h h
@pindex calc-full-help
The @kbd{h h} (@code{calc-full-help}) command displays all the @kbd{?}
responses at once.  When printed, this makes a nice, compact (three pages)
summary of Calc keystrokes.
X
In general, the @kbd{h} key prefix introduces various commands that
provide help within Calc.  Many of the @kbd{h} key functions are
Calc-specific analogues to the @kbd{C-h} functions for Emacs help.
X
@kindex h i
@kindex M-# i
@kindex i
@pindex calc-info
The @kbd{h i} (@code{calc-info}) command runs the Emacs Info system
to read this manual on-line.  This is basically the same as typing
@kbd{C-h i} (the regular way to run the Info system), then, if Info
is not already in the Calc manual, selecting the beginning of the
manual.  The @kbd{M-# i} command is another way to read the Calc
manual; it is different from @kbd{h i} in that it works any time,
not just inside Calc.  The plain @kbd{i} key is also equivalent to
@kbd{h i}, though this key is obsolete and may be replaced with a
different command in a future version of Calc.
X
@kindex h t
@kindex M-# t
@pindex calc-tutorial
The @kbd{h t} (@code{calc-tutorial}) command runs the Info system on
the Tutorial section of the Calc manual.  It is like @kbd{h i},
except that it selects the starting node of the tutorial rather
than the beginning of the whole manual.  (It actually selects the
node @samp{Interactive Tutorial} which tells a few things about
using the Info system before going on to the actual tutorial.)
The @kbd{M-# t} key is equivalent to @kbd{h t} (but it works at
all times).
X
@kindex h s
@kindex M-# s
@pindex calc-info-summary
The @kbd{h s} (@code{calc-info-summary}) command runs the Info system
on the Summary node of the Calc manual.  @xref{Summary}.  The @kbd{M-# s}
key is equivalent to @kbd{h s}.
X
@kindex h k
@pindex calc-describe-key
The @kbd{h k} (@code{calc-describe-key}) command looks up a key
sequence in the Calc manual.  For example, @kbd{h k H a S} looks
up the documentation on the @kbd{H a S} (@code{calc-solve-for})
command.  This works by looking up the textual description of
the keys you press in the @samp{Key Index} of the manual, then
jumping to the node indicated by the index.
X
Most Calc commands do not have traditional Emacs documentation
strings, since the @kbd{h k} command is both more convenient and
more instructive.  This means the regular Emacs @kbd{C-h k}
(@code{describe-key}) command will not be useful for Calc keystrokes.
SHAR_EOF
true || echo 'restore of calc.texinfo failed'
fi
echo 'End of  part 36'
echo 'File calc.texinfo is continued in part 37'
echo 37 > _shar_seq_.tmp
exit 0
exit 0 # Just in case...
--
Kent Landfield                   INTERNET: [e-mail]
Sterling Software, IMD           UUCP:     uunet!sparky!kent
Phone:    [phone]         FAX:      [phone]
Please send comp.sources.misc-related mail to [e-mail].