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Newsgroups: comp.sources.misc From: daveg@synaptics.com (David Gillespie) Subject: v24i083: gnucalc - GNU Emacs Calculator, v2.00, Part35/56 Message-ID: <1991Oct31.214552.2554@sparky.imd.sterling.com> X-Md4-Signature: 17d91c5c5faf3e2cf7d24ae52b02bd56 Date: Thu, 31 Oct 1991 21:45:52 GMT Approved: kent@sparky.imd.sterling.com Submitted-by: daveg@synaptics.com (David Gillespie) Posting-number: Volume 24, Issue 83 Archive-name: gnucalc/part35 Environment: Emacs Supersedes: gmcalc: Volume 13, Issue 27-45 ---- Cut Here and unpack ---- #!/bin/sh # do not concatenate these parts, unpack them in order with /bin/sh # file calc.texinfo continued # if test ! -r _shar_seq_.tmp; then echo 'Please unpack part 1 first!' exit 1 fi (read Scheck if test "$Scheck" != 35; then echo Please unpack part "$Scheck" next! exit 1 else exit 0 fi ) < _shar_seq_.tmp || exit 1 if test ! -f _shar_wnt_.tmp; then echo 'x - still skipping calc.texinfo' else echo 'x - continuing file calc.texinfo' sed 's/^X//' << 'SHAR_EOF' >> 'calc.texinfo' && * List Answer 14:: Random walk * Types Answer 1:: Square root of pi times rational * Types Answer 2:: Infinities * Types Answer 3:: What can "nan" be? * Types Answer 4:: Abbey Road * Types Answer 5:: Friday the 13th * Types Answer 6:: Leap years * Types Answer 7:: Erroneous donut * Types Answer 8:: Dividing intervals * Types Answer 9:: Squaring intervals * Types Answer 10:: Fermat's primality test * Types Answer 11:: pi * 10^7 seconds * Types Answer 12:: Abbey Road on CD * Types Answer 13:: Not quite pi * 10^7 seconds * Types Answer 14:: Supercomputers and c * Types Answer 15:: Sam the Slug * Algebra Answer 1:: Squares and square roots * Algebra Answer 2:: Solving a quartic equation * Algebra Answer 3:: Integral of x sin(pi x) * Algebra Answer 4:: Simpson's rule * Rewrites Answer 1:: Multiplying by conjugate * Rewrites Answer 2:: Alternative fib rule * Rewrites Answer 3:: Rewriting opt(a) + opt(b) x * Rewrites Answer 4:: Sequence of integers * Rewrites Answer 5:: Number of terms in sum * Rewrites Answer 6:: Truncated Taylor series * Programming Answer 1:: Fresnel's C(x) * Programming Answer 2:: Negate third stack element * Programming Answer 3:: Compute sin(x) / x, etc. * Programming Answer 4:: Average value of a list * Programming Answer 5:: Continued fraction phi * Programming Answer 6:: Matrix Fibonacci numbers * Programming Answer 7:: Harmonic number greater than 4 * Programming Answer 8:: Newton's method * Programming Answer 9:: Digamma function * Programming Answer 10:: Unpacking a polynomial * Programming Answer 11:: Recursive Stirling numbers * Programming Answer 12:: Stirling numbers with rewrites @end menu X @c The following kludgery prevents the individual answers from @c being entered on the table of contents. @tex \global\let\oldwrite=\write \global\def\skipwrite#1#2{\let\write=\oldwrite} \global\let\oldchapternofonts=\chapternofonts \global\def\chapternofonts{\let\write=\skipwrite\oldchapternofonts} @end tex X @node RPN Answer 1, RPN Answer 2, Answers to Exercises, Answers to Exercises @subsection RPN Tutorial Exercise 1 X @noindent @kbd{1 @key{RET} 2 @key{RET} 3 @key{RET} 4 + * -} X The result is @c{$1 - (2 \times (3 + 4)) = -13$} @cite{1 - (2 * (3 + 4)) = -13}. X @node RPN Answer 2, RPN Answer 3, RPN Answer 1, Answers to Exercises @subsection RPN Tutorial Exercise 2 X @noindent @c{$2\times4 + 7\times9.5 + {5\over4} = 75.75$} @cite{2*4 + 7*9.5 + 5/4 = 75.75} X After computing the intermediate term @c{$2\times4 = 8$} @cite{2*4 = 8}, you can leave that result on the stack while you compute the second term. With both of these results waiting on the stack you can then compute the final term, then press @kbd{+ +} to add everything up. X @group @smallexample 2: 2 1: 8 3: 8 2: 8 1: 4 . 2: 7 1: 66.5 X . 1: 9.5 . X . X X 2 RET 4 * 7 RET 9.5 * X @end smallexample @end group @noindent @group @smallexample 4: 8 3: 8 2: 8 1: 75.75 3: 66.5 2: 66.5 1: 67.75 . 2: 5 1: 1.25 . 1: 4 . X . X X 5 RET 4 / + + @end smallexample @end group X Alternatively, you could add the first two terms before going on with the third term. X @group @smallexample 2: 8 1: 74.5 3: 74.5 2: 74.5 1: 75.75 1: 66.5 . 2: 5 1: 1.25 . X . 1: 4 . X . X X ... + 5 RET 4 / + @end smallexample @end group X On an old-style RPN calculator this second method would have the advantage of using only three stack levels. But since Calc's stack can grow arbitrarily large this isn't really an issue. Which method you choose is purely a matter of taste. X @node RPN Answer 3, RPN Answer 4, RPN Answer 2, Answers to Exercises @subsection RPN Tutorial Exercise 3 X @noindent The @key{TAB} key provides a way to operate on the number in level 2. X @group @smallexample 3: 10 3: 10 4: 10 3: 10 3: 10 2: 20 2: 30 3: 30 2: 30 2: 21 1: 30 1: 20 2: 20 1: 21 1: 30 X . . 1: 1 . . X . X X TAB 1 + TAB @end smallexample @end group X Similarly, @key{M-TAB} gives you access to the number in level 3. X @group @smallexample 3: 10 3: 21 3: 21 3: 30 3: 11 2: 21 2: 30 2: 30 2: 11 2: 21 1: 30 1: 10 1: 11 1: 21 1: 30 X . . . . . X X M-TAB 1 + M-TAB M-TAB @end smallexample @end group X @node RPN Answer 4, Algebraic Answer 1, RPN Answer 3, Answers to Exercises @subsection RPN Tutorial Exercise 4 X @noindent Either @kbd{( 2 , 3 )} or @kbd{( 2 @key{SPC} 3 )} would have worked, but using both the comma and the space at once yields: X @group @smallexample 1: ( ... 2: ( ... 1: (2, ... 2: (2, ... 2: (2, ... X . 1: 2 . 1: (2, ... 1: (2, 3) X . . . X X ( 2 , SPC 3 ) @end smallexample @end group X Joe probably tried to type @kbd{@key{TAB} @key{DEL}} to swap the extra incomplete object to the top of the stack and delete it. But a feature of Calc is that @key{DEL} on an incomplete object deletes just one component out of that object, so he had to press @key{DEL} twice to finish the job. X @group @smallexample 2: (2, ... 2: (2, 3) 2: (2, 3) 1: (2, 3) 1: (2, 3) 1: (2, ... 1: ( ... . X . . . X X TAB DEL DEL @end smallexample @end group X (As it turns out, deleting the second-to-top stack entry happens often enough that Calc provides a special key, @kbd{M-DEL}, to do just that. @kbd{M-DEL} is just like @kbd{TAB DEL}, except that it doesn't exhibit the ``feature'' that tripped poor Joe.) X @node Algebraic Answer 1, Algebraic Answer 2, RPN Answer 4, Answers to Exercises @subsection Algebraic Entry Tutorial Exercise 1 X @noindent Type @kbd{' sqrt($) @key{RET}}. X If the @kbd{Q} key is broken, you could use @kbd{' $^0.5 @key{RET}}. Or, RPN style, @kbd{0.5 ^}. X (Actually, @samp{$^1:2}, using the fraction one-half as the power, is a closer equivalent, since @samp{9^0.5} yields @cite{3.0} whereas @samp{sqrt(9)} and @samp{9^1:2} yield the exact integer @cite{3}.) X @node Algebraic Answer 2, Algebraic Answer 3, Algebraic Answer 1, Answers to Exercises @subsection Algebraic Entry Tutorial Exercise 2 X @noindent In the formula @samp{2 x (1+y)}, @samp{x} was interpreted as a function name with @samp{1+y} as its argument. Assigning a value to a variable has no relation to a function by the same name. Joe needed to use an explicit @samp{*} symbol here: @samp{2 x*(1+y)}. X @node Algebraic Answer 3, Modes Answer 1, Algebraic Answer 2, Answers to Exercises @subsection Algebraic Entry Tutorial Exercise 3 X @noindent The result from @kbd{1 @key{RET} 0 /} will be the formula @cite{1 / 0}. The ``function'' @samp{/} cannot be evaluated when its second argument is zero, so it is left in symbolic form. When you now type @kbd{0 *}, the result will be zero because Calc uses the general rule that ``zero times anything is zero.'' X @c [fix-ref Infinities] The @kbd{m i} command enables an @dfn{infinite mode} in which @cite{1 / 0} results in a special symbol that represents ``infinity.'' If you multiply infinity by zero, Calc uses another special new symbol to show that the answer is ``indeterminate.'' @xref{Infinities}, for further discussion of infinite and indeterminate values. X @node Modes Answer 1, Modes Answer 2, Algebraic Answer 3, Answers to Exercises @subsection Modes Tutorial Exercise 1 X @noindent Calc always stores its numbers in decimal, so even though one-third has an exact base-3 representation (@samp{3#0.1}), it is still stored as 0.3333333 (chopped off after 12 or however many decimal digits) inside the calculator's memory. When this inexact number is converted back to base 3 for display, it may still be slightly inexact. When we multiply this number by 3, we get 0.999999, also an inexact value. X When Calc displays a number in base 3, it has to decide how many digits to show. If the current precision is 12 (decimal) digits, that corresponds to @samp{12 / log10(3) = 25.15} base-3 digits. Because 25.15 is not an exact integer, Calc shows only 25 digits, with the result that stored numbers carry a little bit of extra information that may not show up on the screen. When Joe entered @samp{3#0.2}, the stored number 0.666666 happened to round to a pleasing value when it lost that last 0.15 of a digit, but it was still inexact in Calc's memory. When he divided by 2, he still got the dreaded inexact value 0.333333. (Actually, he divided 0.666667 by 2 to get 0.333334, which is why he got something a little higher than 3#0.1 instead of a little lower.) X If Joe didn't want to be bothered with all this, he could have typed @kbd{M-24 d n} to display with one less digit than the default. (If you give @kbd{d n} a negative argument, it uses default-minus-that, so @kbd{M-- d n} would be an easier way to get the same effect.) Those inexact results would still be lurking there, but they would now be rounded to nice, natural-looking values for display purposes. (Remember, @samp{0.022222} in base 3 is like @samp{0.099999} in base 10; rounding off one digit will round the number up to @samp{0.1}.) Depending on the nature of your work, this hiding of the inexactness may be a benefit or a danger. With the @kbd{d n} command, Calc gives you the choice. X Incidentally, another consequence of all this is that if you type @kbd{M-30 d n} to display more digits than are ``really there,'' you'll see garbage digits at the end of the number. (In decimal display mode, with decimally-stored numbers, these garbage digits are always zero so they vanish and you don't notice them.) Because Calc rounds off that 0.15 digit, there is the danger that two numbers could be slightly different internally but still look the same. If you feel uneasy about this, set the @kbd{d n} precision to be a little higher than normal; you'll get ugly garbage digits, but you'll always be able to tell two distinct numbers apart. X An interesting side note is that most computers store their floating-point numbers in binary, and convert to decimal for display. Thus everyday programs have the same problem: Decimal 0.1 cannot be represented exactly in binary (try it: @kbd{0.1 d 2}), so @samp{0.1 * 10} comes out as an inexact approximation to 1 on some machines (though they generally arrange to hide it from you by rounding off one digit as we did above). Because Calc works in decimal instead of binary, you can be sure that numbers that look exact @emph{are} exact as long as you stay in decimal display mode. X It's not hard to show that any number that can be represented exactly in binary, octal, or hexadecimal is also exact in decimal, so the kinds of problems we saw in this exercise are likely to be severe only when you use a relatively unusual radix like 3. X @node Modes Answer 2, Modes Answer 3, Modes Answer 1, Answers to Exercises @subsection Modes Tutorial Exercise 2 X If the radix is 15 or higher, we can't use the letter @samp{e} to mark the exponent because @samp{e} is interpreted as a digit. When Calc needs to display scientific notation in a high radix, it writes @samp{16#F.E8F*16.^15}. You can enter a number like this as an algebraic entry. Also, pressing @kbd{e} without any digits before it normally types @kbd{1e}, but in a high radix it types @kbd{16.^} and puts you in algebraic entry: @kbd{16#f.e8f RET e 15 RET *} is another way to enter this number. X The reason Calc puts a decimal point in the @samp{16.^} is to prevent huge integers from being generated if the exponent is large (consider @samp{16#1.23*16^1000}, where we compute @samp{16^1000} as a giant exact integer and then throw away most of the digits when we multiply it by the floating-point @samp{16#1.23}). While this wouldn't normally matter for display purposes, it could give you a nasty surprise if you copied that number into a file and later yanked it back into Calc. X @node Modes Answer 3, Modes Answer 4, Modes Answer 2, Answers to Exercises @subsection Modes Tutorial Exercise 3 X @noindent The answer he got was @cite{0.5000000000006399}. X The problem is not that the square operation is inexact, but that the sine of 45 that was already on the stack was accurate to only 12 places. Arbitrary-precision calculations still only give answers as good as their inputs. X The real problem is that there is no 12-digit number which, when squared, comes out to 0.5 exactly. The @kbd{f [} and @kbd{f ]} commands decrease or increase a number by one unit in the last place (according to the current precision). They are useful for determining facts like this. X @group @smallexample 1: 0.707106781187 1: 0.500000000001 X . . X X 45 S 2 ^ X @end smallexample @end group @noindent @group @smallexample 1: 0.707106781187 1: 0.707106781186 1: 0.499999999999 X . . . X X U DEL f [ 2 ^ @end smallexample @end group X A high-precision calculation must be carried out in high precision all the way. The only number in the original problem which was known exactly was the quantity 45 degrees, so the precision must be raised before anything is done after the number 45 has been entered in order for the higher precision to be meaningful. X @node Modes Answer 4, Arithmetic Answer 1, Modes Answer 3, Answers to Exercises @subsection Modes Tutorial Exercise 4 X @noindent Many calculations involve real-world quantities, like the width and height of a piece of wood or the volume of a jar. Such quantities can't be measured exactly anyway, and if the data that is input to a calculation is inexact, doing exact arithmetic on it is a waste of time. X Fractions become unwieldy after too many calculations have been done with them. For example, the sum of the reciprocals of the integers from 1 to 10 is 7381:2520. The sum from 1 to 30 is 9304682830147:2329089562800. After a point it will take a long time to add even one more term to this sum, but a floating-point calculation of the sum will not have this problem. X Also, rational numbers cannot express the results of all calculations. There is no fractional form for the square root of two, so if you type @w{@kbd{2 Q}}, Calc has no choice but to give you a floating-point answer. X @node Arithmetic Answer 1, Arithmetic Answer 2, Modes Answer 4, Answers to Exercises @subsection Arithmetic Tutorial Exercise 1 X @noindent Dividing two integers that are larger than the current precision may give a floating-point result that is inaccurate even when rounded down to an integer. Consider @cite{123456789 / 2} when the current precision is 6 digits. The true answer is @cite{61728394.5}, but with a precision of 6 this will be rounded to @c{$12345700.0/2.0 = 61728500.0$} @cite{12345700.@: / 2.@: = 61728500.}. The result, when converted to an integer, will be off by 106. X Here are two solutions: Raise the precision enough that the floating-point round-off error is strictly to the right of the decimal point. Or, convert to fraction mode so that @cite{123456789 / 2} produces the exact fraction @cite{123456789:2}, which can be rounded down by the @kbd{F} command without ever switching to floating-point format. X @node Arithmetic Answer 2, Vector Answer 1, Arithmetic Answer 1, Answers to Exercises @subsection Arithmetic Tutorial Exercise 2 X @noindent @kbd{27 @key{RET} 9 B} could give the exact result @cite{3:2}, but it does a floating-point calculation instead and produces @cite{1.5}. X Calc will find an exact result for a logarithm if the result is an integer or the reciprocal of an integer. But there is no efficient way to search the space of all possible rational numbers for an exact answer, so Calc doesn't try. X @node Vector Answer 1, Vector Answer 2, Arithmetic Answer 2, Answers to Exercises @subsection Vector Tutorial Exercise 1 X @noindent Duplicate the vector, compute its length, then divide the vector by its length: @kbd{@key{RET} A /}. X @group @smallexample 1: [1, 2, 3] 2: [1, 2, 3] 1: [0.27, 0.53, 0.80] 1: 1. X . 1: 3.74165738677 . . X . X X r 1 RET A / A @end smallexample @end group X The final @kbd{A} command shows that the normalized vector does indeed have unit length. X @node Vector Answer 2, Matrix Answer 1, Vector Answer 1, Answers to Exercises @subsection Vector Tutorial Exercise 2 X @noindent The average position is equal to the sum of the products of the positions times their corresponding probabilities. This is the definition of the dot product operation. So all you need to do is to put the two vectors on the stack and press @kbd{*}. X @node Matrix Answer 1, Matrix Answer 2, Vector Answer 2, Answers to Exercises @subsection Matrix Tutorial Exercise 1 X @noindent The trick is to multiply by a vector of ones. Use @kbd{r 4 [1 1 1] *} to get the row sum. Similarly, use @kbd{[1 1] r 4 *} to get the column sum. X @node Matrix Answer 2, Matrix Answer 3, Matrix Answer 1, Answers to Exercises @subsection Matrix Tutorial Exercise 2 X @ifinfo @group @example X x + a y = 6 X x + b y = 10 @end example @end group @end ifinfo @tex \turnoffactive $$ \eqalign{ x &+ a y = 6 \cr X x &+ b y = 10} $$ @end tex X Just enter the righthand side vector, then divide by the lefthand side matrix as usual. X @group @smallexample 1: [6, 10] 2: [6, 10] 1: [6 - 4 a / (b - a), 4 / (b - a) ] X . 1: [ [ 1, a ] . X [ 1, b ] ] X . X ' [6 10] RET ' [1 a; 1 b] RET / @end smallexample @end group X This can be made more readable using @kbd{d B} to enable ``big'' display mode: X @group @smallexample X 4 a 4 1: [6 - -----, -----] X b - a b - a @end smallexample @end group X Type @kbd{d N} to return to ``normal'' display mode afterwards. X @node Matrix Answer 3, List Answer 1, Matrix Answer 2, Answers to Exercises @subsection Matrix Tutorial Exercise 3 X @noindent To solve @c{$A^T A \, X = A^T B$} @cite{trn(A) * A * X = trn(A) * B}, first we compute @c{$A' = A^T A$} @cite{A2 = trn(A) * A} and @c{$B' = A^T B$} @cite{B2 = trn(A) * B}; now, we have a system @c{$A' X = B'$} @cite{A2 * X = B2} which we can solve using Calc's @samp{/} command. X @ifinfo @group @example X a + 2b + 3c = 6 X 4a + 5b + 6c = 2 X 7a + 6b = 3 X 2a + 4b + 6c = 11 @end example @end group @end ifinfo @tex \turnoffactive $$ \openup1\jot \tabskip=0pt plus1fil \halign to\displaywidth{\tabskip=0pt X $\hfil#$&$\hfil{}#{}$& X $\hfil#$&$\hfil{}#{}$& X $\hfil#$&${}#\hfil$\tabskip=0pt plus1fil\cr X a&+&2b&+&3c&=6 \cr X 4a&+&5b&+&6c&=2 \cr X 7a&+&6b& & &=3 \cr X 2a&+&4b&+&6c&=11 \cr} $$ @end tex X The first step is to enter the coefficient matrix. We'll store it in quick variable number 7 for later reference. Next, we compute the @c{$B'$} @cite{B2} vector. X @group @smallexample 1: [ [ 1, 2, 3 ] 2: [ [ 1, 4, 7, 2 ] 1: [57, 84, 96] X [ 4, 5, 6 ] [ 2, 5, 6, 4 ] . X [ 7, 6, 0 ] [ 3, 6, 0, 6 ] ] X [ 2, 4, 6 ] ] 1: [6, 2, 3, 11] X . . X ' [1 2 3; 4 5 6; 7 6 0; 2 4 6] RET s 7 v t [6 2 3 11] * @end smallexample @end group X @noindent Now we compute the matrix @c{$A'$} @cite{A2} and divide. X @group @smallexample 2: [57, 84, 96] 1: [-11.64, 14.08, -3.64] 1: [ [ 70, 72, 39 ] . X [ 72, 81, 60 ] X [ 39, 60, 81 ] ] X . X X r 7 v t r 7 * / @end smallexample @end group X @noindent (The actual computed answer will be slightly inexact due to round-off error.) X Notice that the answers are similar to those for the @c{$3\times3$} @asis{3x3} system solved in the text. That's because the fourth equation that was added to the system is almost identical to the first one multiplied by two. (If it were identical, we would have gotten the exact same answer since the @c{$4\times3$} @asis{4x3} system would be equivalent to the original @c{$3\times3$} @asis{3x3} system.) X Since the first and fourth equations aren't quite equivalent, they can't both be satisfied at once. Let's plug our answers back into the original system of equations to see how well they match. X @group @smallexample 2: [-11.64, 14.08, -3.64] 1: [5.6, 2., 3., 11.2] 1: [ [ 1, 2, 3 ] . X [ 4, 5, 6 ] X [ 7, 6, 0 ] X [ 2, 4, 6 ] ] X . X X r 7 TAB * @end smallexample @end group X @noindent This is reasonably close to our original @cite{B} vector, @cite{[6, 2, 3, 11]}. X @node List Answer 1, List Answer 2, Matrix Answer 3, Answers to Exercises @subsection List Tutorial Exercise 1 X @noindent We can use @kbd{v x} to build a vector of integers. This needs to be adjusted to get the range of integers we desire. Mapping @samp{-} across the vector will accomplish this, although it turns out the plain @samp{-} key will work just as well. X @group @smallexample 2: 2 2: 2 1: [1, 2, 3, 4, 5, 6, 7, 8, 9] 1: [-4, -3, -2, -1, 0, 1, 2, 3, 4] X . . X X 2 v x 9 RET 5 V M - or 5 - @end smallexample @end group X @noindent Now we use @kbd{V M ^} to map the exponentiation operator across the vector. X @group @smallexample 1: [0.0625, 0.125, 0.25, 0.5, 1, 2, 4, 8, 16] X . X X V M ^ @end smallexample @end group X @node List Answer 2, List Answer 3, List Answer 1, Answers to Exercises @subsection List Tutorial Exercise 2 X @noindent Given @cite{x} and @cite{y} vectors in quick variables 1 and 2 as before, the first job is to form the matrix that describes the problem. X @ifinfo @example X m*x + b*1 = y @end example @end ifinfo @tex \turnoffactive $$ m \times x + b \times 1 = y $$ @end tex X Thus we want a @c{$19\times2$} @asis{19x2} matrix with our @cite{x} vector as one column and ones as the other column. So, first we build the column of ones, then we combine the two columns to form our @cite{A} matrix. X @group @smallexample 2: [1.34, 1.41, 1.49, ... ] 1: [ [ 1.34, 1 ] 1: [1, 1, 1, ...] [ 1.41, 1 ] X . [ 1.49, 1 ] X @dots{} X X r 1 1 v b 19 RET M-2 v p v t s 3 @end smallexample @end group X @noindent Now we compute @c{$A^T y$} @cite{trn(A) * y} and @c{$A^T A$} @cite{trn(A) * A} and divide. X @group @smallexample 1: [33.36554, 13.613] 2: [33.36554, 13.613] X . 1: [ [ 98.0003, 41.63 ] X [ 41.63, 19 ] ] X . X X v t r 2 * r 3 v t r 3 * @end smallexample @end group X @noindent (Hey, those numbers look familiar!) X @group @smallexample 1: [0.52141679, -0.425978] X . X X / @end smallexample @end group X Since we were solving equations of the form @c{$m \times x + b \times 1 = y$} @cite{m*x + b*1 = y}, these numbers should be @cite{m} and @cite{b}, respectively. Sure enough, they agree exactly with the result computed using @kbd{V M} and @kbd{V R}! X The moral of this story: @kbd{V M} and @kbd{V R} will probably solve your problem, but there is often an easier way using the higher-level arithmetic functions! X @c [fix-ref Curve Fitting] In fact, there is a built-in @kbd{a F} command that does least-squares fits. @xref{Curve Fitting}. X @node List Answer 3, List Answer 4, List Answer 2, Answers to Exercises @subsection List Tutorial Exercise 3 X @noindent Move to one end of the list and press @kbd{C-@@} (or @kbd{C-SPC} or whatever) to set the mark, then move to the other end of the list and type @w{@kbd{M-# g}}. X @group @smallexample 1: [2.3, 6, 22, 15.1, 7, 15, 14, 7.5, 2.5] X . @end smallexample @end group X To make things interesting, let's assume we don't know at a glance how many numbers are in this list. Then we could type: X @group @smallexample 2: [2.3, 6, 22, ... ] 2: [2.3, 6, 22, ... ] 1: [2.3, 6, 22, ... ] 1: 126356422.5 X . . X X RET V R * X @end smallexample @end group @noindent @group @smallexample 2: 126356422.5 2: 126356422.5 1: 7.94652913734 1: [2.3, 6, 22, ... ] 1: 9 . X . . X X TAB v l I ^ @end smallexample @end group X @noindent (The @kbd{I ^} command computes the @var{n}th root of a number. You could also type @kbd{& ^} to take the reciprocal of 9 and then raise the number to that power.) X @node List Answer 4, List Answer 5, List Answer 3, Answers to Exercises @subsection List Tutorial Exercise 4 X @noindent A number @cite{j} is a divisor of @cite{n} if @c{$n \mathbin{\hbox{\code{\%}}} j = 0$} @samp{n % j = 0}. The first step is to get a vector that identifies the divisors. X @group @smallexample 2: 30 2: [0, 0, 0, 2, ...] 1: [1, 1, 1, 0, ...] 1: [1, 2, 3, 4, ...] 1: 0 . X . . X X 30 RET v x 30 RET s 1 V M % 0 V M a = s 2 @end smallexample @end group X @noindent This vector has 1's marking divisors of 30 and 0's marking non-divisors. X The zeroth divisor function is just the total number of divisors. The first divisor function is the sum of the divisors. X @group @smallexample 1: 8 3: 8 2: 8 2: 8 X 2: [1, 2, 3, 4, ...] 1: [1, 2, 3, 0, ...] 1: 72 X 1: [1, 1, 1, 0, ...] . . X . X X V R + r 1 r 2 V M * V R + @end smallexample @end group X @noindent Once again, the last two steps just compute a dot product for which a simple @kbd{*} would have worked equally well. X @node List Answer 5, List Answer 6, List Answer 4, Answers to Exercises @subsection List Tutorial Exercise 5 X @noindent The obvious first step is to obtain the list of factors with @kbd{k f}. This list will always be in sorted order, so if there are duplicates they will be right next to each other. A suitable method is to compare the list with a copy of itself shifted over by one. X @group @smallexample 1: [3, 7, 7, 7, 19] 2: [3, 7, 7, 7, 19] 2: [3, 7, 7, 7, 19, 0] X . 1: [3, 7, 7, 7, 19, 0] 1: [0, 3, 7, 7, 7, 19] X . . X X 19551 k f RET 0 | TAB 0 TAB | X @end smallexample @end group @noindent @group @smallexample 1: [0, 0, 1, 1, 0, 0] 1: 2 1: 0 X . . . X X V M a = V R + 0 a = @end smallexample @end group X @noindent Note that we have to arrange for both vectors to have the same length so that the mapping operation works; no prime factor will ever be zero, so adding zeros on the left and right is safe. From then on the job is pretty straightforward. X Incidentally, Calc provides the @c{\dfn{M\"obius} $\mu$} @dfn{Moebius mu} function which is zero if and only if its argument is square-free. It would be a much more convenient way to do the above test in practice. X @node List Answer 6, List Answer 7, List Answer 5, Answers to Exercises @subsection List Tutorial Exercise 6 X @noindent First use @kbd{v x 6 RET} to get a list of integers, then @kbd{V M v x} to get a list of lists of integers! X @node List Answer 7, List Answer 8, List Answer 6, Answers to Exercises @subsection List Tutorial Exercise 7 X @noindent Here's one solution. First, compute the triangular list from the previous exercise and type @kbd{1 -} to subtract one from all the elements. X @group @smallexample 1: [ [0], X [0, 1], X [0, 1, 2], X @dots{} X X 1 - @end smallexample @end group X The numbers down the lefthand edge of the list we desire are called the ``triangular numbers'' (now you know why!). The @cite{n}th triangular number is the sum of the integers from 1 to @cite{n}, and can be computed directly by the formula @c{$n (n+1) \over 2$} @cite{n * (n+1) / 2}. X @group @smallexample 2: [ [0], [0, 1], ... ] 2: [ [0], [0, 1], ... ] 1: [0, 1, 2, 3, 4, 5] 1: [0, 1, 3, 6, 10, 15] X . . X X v x 6 RET 1 - V M ' $ ($+1)/2 RET @end smallexample @end group X @noindent Adding this list to the above list of lists produces the desired result: X @group @smallexample 1: [ [0], X [1, 2], X [3, 4, 5], X [6, 7, 8, 9], X [10, 11, 12, 13, 14], X [15, 16, 17, 18, 19, 20] ] X . X X V M + @end smallexample @end group X If we did not know the formula for triangular numbers, we could have computed them using a @kbd{V U +} command. We could also have gotten them the hard way by mapping a reduction across the original triangular list. X @group @smallexample 2: [ [0], [0, 1], ... ] 2: [ [0], [0, 1], ... ] 1: [ [0], [0, 1], ... ] 1: [0, 1, 3, 6, 10, 15] X . . X X RET V M V R + @end smallexample @end group X @noindent (This means ``map a @kbd{V R +} command across the vector,'' and since each element of the main vector is itself a small vector, @kbd{V R +} computes the sum of its elements.) X @node List Answer 8, List Answer 9, List Answer 7, Answers to Exercises @subsection List Tutorial Exercise 8 X @noindent The first step is to build a list of values of @cite{x}. X @group @smallexample 1: [1, 2, 3, ..., 21] 1: [0, 1, 2, ..., 20] 1: [0, 0.25, 0.5, ..., 5] X . . . X X v x 21 RET 1 - 4 / s 1 @end smallexample @end group X Next, we compute the Bessel function values. X @group @smallexample 1: [0., 0.124, 0.242, ..., -0.328] X . X X V M ' besJ(1,$) RET @end smallexample @end group X @noindent (Another way to do this would be @kbd{1 TAB V M f j}.) X A way to isolate the maximum value is to compute the maximum using @kbd{V R X}, then compare all the Bessel values with that maximum. X @group @smallexample 2: [0., 0.124, 0.242, ... ] 1: [0, 0, 0, ... ] 2: [0, 0, 0, ... ] 1: 0.5801562 . 1: 1 X . . X X RET V R X V M a = RET V R + DEL @end smallexample @end group X @noindent 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{$\sin x$} @cite{sin(x)} might have many points all equal to the maximum value, 1.) X The vector we have now has a single 1 in the position that indicates the maximum value of @cite{x}. Now it is a simple matter to convert this back into the corresponding value itself. X @group @smallexample 2: [0, 0, 0, ... ] 1: [0, 0., 0., ... ] 1: 1.75 1: [0, 0.25, 0.5, ... ] . . X . X X r 1 V M * V R + @end smallexample @end group X If @kbd{a =} had produced more than one @cite{1} value, this method would have given the sum of all maximum @cite{x} values; not very useful! In this case we could have used @kbd{v m} (@code{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 @cite{x} values. X The built-in @kbd{a X} command maximizes a function using more efficient methods. Just for illustration, let's use @kbd{a X} to maximize @samp{besJ(1,x)} over this same interval. X @group @smallexample 2: besJ(1, x) 1: [1.84115, 0.581865] 1: [0 .. 5] . X . X ' besJ(1,x), [0..5] RET a X x RET @end smallexample @end group X @noindent The output from @kbd{a X} is a vector containing the value of @cite{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. X @node List Answer 9, List Answer 10, List Answer 8, Answers to Exercises @subsection List Tutorial Exercise 9 X @noindent Step one is to convert our integer into vector notation. X @group @smallexample 1: 25129925999 3: 25129925999 X . 2: 10 X 1: [11, 10, 9, ..., 1, 0] X . X X 25129925999 RET 10 RET 12 RET v x 12 RET - X @end smallexample @end group @noindent @group @smallexample 1: 25129925999 1: [0, 2, 25, 251, 2512, ... ] 2: [100000000000, ... ] . X . X X V M ^ s 1 V M \ @end smallexample @end group X @noindent (Recall, the @kbd{\} command computes an integer quotient.) X @group @smallexample 1: [0, 2, 5, 1, 2, 9, 9, 2, 5, 9, 9, 9] X . X X 10 V M % s 2 @end smallexample @end group X Next we must increment this number. This involves adding one to the last digit, plus handling carries. There is a carry out of a digit to the left if that digit is a nine and all the digits to the right of it are nines. X @group @smallexample 1: [0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1] 1: [1, 1, 1, 0, 0, 1, ... ] X . . X X 9 V M a = v v X @end smallexample @end group @noindent @group @smallexample 1: [1, 1, 1, 0, 0, 0, ... ] 1: [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1] X . . X X V U * v v 1 | @end smallexample @end group X @noindent Accumulating @kbd{*} across a vector of ones and zeros will preserve only the initial run of ones. These are the carries into all digits except the rightmost digit. Concatenating a one on the right takes care of aligning the carries properly, and also adding one to the rightmost digit. X @group @smallexample 2: [0, 0, 0, 0, ... ] 1: [0, 0, 2, 5, 1, 2, 9, 9, 2, 6, 0, 0, 0] 1: [0, 0, 2, 5, ... ] . X . X X 0 r 2 | V M + 10 V M % @end smallexample @end group X @noindent Here we have concatenated 0 to the @emph{left} of the original number; this takes care of shifting the carries by one with respect to the digits that generated them. X Finally, we must convert this list back into an integer. X @group @smallexample 3: [0, 0, 2, 5, ... ] 2: [0, 0, 2, 5, ... ] 2: 1000000000000 1: [1000000000000, 100000000000, ... ] 1: [100000000000, ... ] . X . X X 10 RET 12 ^ r 1 | X @end smallexample @end group @noindent @group @smallexample 1: [0, 0, 20000000000, 5000000000, ... ] 1: 25129926000 X . . X X V M * V R + @end smallexample @end group X @noindent Another way to do this final step would be to reduce the formula @w{@cite{10 $$ + $}} across the vector of digits. X @group @smallexample 1: [0, 0, 2, 5, ... ] 1: 25129926000 X . . X X V R ' 10 $$ + $ RET @end smallexample @end group X @node List Answer 10, List Answer 11, List Answer 9, Answers to Exercises @subsection List Tutorial Exercise 10 X @noindent For the list @cite{[a, b, c, d]}, the result is @cite{((a = b) = c) = d}, which will compare @cite{a} and @cite{b} to produce a 1 or 0, which is then compared with @cite{c} to produce another 1 or 0, which is then compared with @cite{d}. This is not at all what Joe wanted. X Here's a more correct method: X @group @smallexample 1: [7, 7, 7, 8, 7] 2: [7, 7, 7, 8, 7] X . 1: 7 X . X X ' [7,7,7,8,7] RET RET v r 1 RET X @end smallexample @end group @noindent @group @smallexample 1: [1, 1, 1, 0, 1] 1: 0 X . . X X V M a = V R * @end smallexample @end group X @node List Answer 11, List Answer 12, List Answer 10, Answers to Exercises @subsection List Tutorial Exercise 11 X @noindent The circle of unit radius consists of those points @cite{(x,y)} for which @cite{x^2 + y^2 < 1}. We start by generating a vector of @cite{x^2} and a vector of @cite{y^2}. X We can make this go a bit faster by using the @kbd{v .} and @kbd{t .} commands. X @group @smallexample 2: [2., 2., ..., 2.] 2: [2., 2., ..., 2.] 1: [2., 2., ..., 2.] 1: [1.16, 1.98, ..., 0.81] X . . X X v . t . 2. v b 100 RET RET V M k r X @end smallexample @end group @noindent @group @smallexample 2: [2., 2., ..., 2.] 1: [0.026, 0.96, ..., 0.036] 1: [0.026, 0.96, ..., 0.036] 2: [0.53, 0.81, ..., 0.094] X . . X X 1 - 2 V M ^ TAB V M k r 1 - 2 V M ^ @end smallexample @end group X Now we sum the @cite{x^2} and @cite{y^2} values, compare with 1 to get a vector of 1/0 truth values, then sum the truth values. X @group @smallexample 1: [0.56, 1.78, ..., 0.13] 1: [1, 0, ..., 1] 1: 84 X . . . X X + 1 V M a < V R + @end smallexample @end group X @noindent The ratio @cite{84/100} should approximate the ratio @c{$\pi/4$} @cite{pi/4}. X @group @smallexample 1: 0.84 1: 3.36 2: 3.36 1: 1.0695 X . . 1: 3.14159 . X X 100 / 4 * P / @end smallexample @end group X @noindent Our estimate, 3.36, is off by about 7%. We could get a better estimate by taking more points (say, 1000), but it's clear that this method is not very efficient! X (Naturally, since this example uses random numbers your own answer will be slightly different than the one shown here!) X If you typed @kbd{v .} and @kbd{t .} before, type them again to return to full-sized display of vectors. X @node List Answer 12, List Answer 13, List Answer 11, Answers to Exercises @subsection List Tutorial Exercise 12 X @noindent This problem can be made a lot easier by taking advantage of some symmetries. First of all, after some thought it's clear that the @cite{y} axis can be ignored altogether. Just pick a random @cite{x} component for one end of the match, pick a random direction @c{$\theta$} @cite{theta}, and see if @cite{x} and @c{$x + \cos \theta$} @cite{x + cos(theta)} (which is the @cite{x} coordinate of the other endpoint) cross a line. The lines are at integer coordinates, so this happens when the two numbers surround an integer. X Since the two endpoints are equivalent, we may as well choose the leftmost of the two endpoints as @cite{x}. Then @cite{theta} is an angle pointing to the right, in the range -90 to 90 degrees. (We could use radians, but it would feel like cheating to refer to @c{$\pi/2$} @cite{pi/2} radians while trying to estimate @c{$\pi$} @cite{pi}!) X In fact, since the field of lines is infinite we can choose the coordinates 0 and 1 for the lines on either side of the leftmost endpoint. The rightmost endpoint will be between 0 and 1 if the match does not cross a line, or between 1 and 2 if it does. So: Pick random @cite{x} and @c{$\theta$} @cite{theta}, compute @c{$x + \cos \theta$} @cite{x + cos(theta)}, and count how many of the results are greater than one. Simple! X We can make this go a bit faster by using the @kbd{v .} and @kbd{t .} commands. X @group @smallexample 1: [0.52, 0.71, ..., 0.72] 2: [0.52, 0.71, ..., 0.72] X . 1: [78.4, 64.5, ..., -42.9] X . X v . t . 1. v b 100 RET V M k r 180. v b 100 RET V M k r 90 - @end smallexample @end group X @noindent (The next step may be slow, depending on the speed of your computer.) X @group @smallexample 2: [0.52, 0.71, ..., 0.72] 1: [0.72, 1.14, ..., 1.45] 1: [0.20, 0.43, ..., 0.73] . X . X X m d V M C + X @end smallexample @end group @noindent @group @smallexample 1: [0, 1, ..., 1] 1: 0.64 1: 3.125 X . . . X X 1 V M a > V R + 100 / 2 TAB / @end smallexample @end group X Let's try the third method, too. We'll use random integers up to one million. The @kbd{k r} command with an integer argument picks a random integer. X @group @smallexample 2: [1000000, 1000000, ..., 1000000] 2: [78489, 527587, ..., 814975] 1: [1000000, 1000000, ..., 1000000] 1: [324014, 358783, ..., 955450] X . . X X 1000000 v b 100 RET RET V M k r TAB V M k r X @end smallexample @end group @noindent @group @smallexample 1: [1, 1, ..., 25] 1: [1, 1, ..., 0] 1: 0.56 X . . . X X V M k g 1 V M a = V R + 100 / X @end smallexample @end group @noindent @group @smallexample 1: 10.714 1: 3.273 X . . X X 6 TAB / Q @end smallexample @end group X For a proof of this property of the GCD function, see section 4.5.2, exercise 10, of Knuth's @emph{Art of Computer Programming}, volume II. X If you typed @kbd{v .} and @kbd{t .} before, type them again to return to full-sized display of vectors. X @node List Answer 13, List Answer 14, List Answer 12, Answers to Exercises @subsection List Tutorial Exercise 13 X @noindent First, we put the string on the stack as a vector of ASCII codes. X @group @smallexample 1: [84, 101, 115, ..., 51] X . X X "Testing, 1, 2, 3 RET @end smallexample @end group X @noindent Note that the @kbd{"} key, like @kbd{$}, initiates algebraic entry so there was no need to type an apostrophe. Also, Calc didn't mind that we omitted the closing @kbd{"}. (The same goes for all closing delimiters like @kbd{)} and @kbd{]} at the end of a formula. X We'll show two different approaches here. In the first, we note that if the input vector is @cite{[a, b, c, d]}, then the hash code is @cite{3 (3 (3a + b) + c) + d = 27a + 9b + 3c + d}. In other words, it's a sum of descending powers of three times the ASCII codes. X @group @smallexample 2: [84, 101, 115, ..., 51] 2: [84, 101, 115, ..., 51] 1: 16 1: [15, 14, 13, ..., 0] X . . X X RET v l v x 16 RET - X @end smallexample @end group @noindent @group @smallexample 2: [84, 101, 115, ..., 51] 1: 1960915098 1: 121 1: [14348907, ..., 1] . . X . X X 3 TAB V M ^ * 511 % @end smallexample @end group X @noindent Once again, @kbd{*} elegantly summarizes most of the computation. But there's an even more elegant approach: Reduce the formula @kbd{3 $$ + $} across the vector. Recall that this represents a function of two arguments that computes its first argument times three plus its second argument. X @group @smallexample 1: [84, 101, 115, ..., 51] 1: 1960915098 X . . X X "Testing, 1, 2, 3 RET V R ' 3$$+$ RET @end smallexample @end group X @noindent If you did the decimal arithmetic exercise, this will be familiar. Basically, we're turning a base-3 vector of digits into an integer, except that our ``digits'' are much larger than real digits. X Instead of typing @kbd{511 %} again to reduce the result, we can be cleverer still and notice that rather than computing a huge integer and taking the modulo at the end, we can take the modulo at each step without affecting the result. While this means there are more arithmetic operations, the numbers we operate on remain small so the operations are faster. X @group @smallexample 1: [84, 101, 115, ..., 51] 1: 121 X . . X X "Testing, 1, 2, 3 RET V R ' (3$$+$)%511 RET @end smallexample @end group X Why does this work? Think about a two-step computation: @w{@cite{3 (3a + b) + c}}. Taking a result modulo 511 basically means subtracting off enough 511's to put the result in the desired range. So the result when we take the modulo after every step is, X @ifinfo @example 3 (3 a + b - 511 m) + c - 511 n @end example @end ifinfo @tex \turnoffactive $$ 3 (3 a + b - 511 m) + c - 511 n $$ @end tex X @noindent for some suitable integers @cite{m} and @cite{n}. Expanding out by the distributive law yields X @ifinfo @example 9 a + 3 b + c - 511*3 m - 511 n @end example @end ifinfo @tex \turnoffactive $$ 9 a + 3 b + c - 511\times3 m - 511 n $$ @end tex X @noindent The @cite{m} term in the latter formula is redundant because any contribution it makes could just as easily be made by the @cite{n} term. So we can take it out to get an equivalent formula with @cite{n' = 3m + n}, X @ifinfo @example 9 a + 3 b + c - 511 n' @end example @end ifinfo @tex \turnoffactive $$ 9 a + 3 b + c - 511 n' $$ @end tex X @noindent which is just the formula for taking the modulo only at the end of the calculation. Therefore the two methods are essentially the same. X Later in the tutorial we will encounter @dfn{modulo forms}, which basically automate the idea of reducing every intermediate result modulo some value @i{M}. X @node List Answer 14, Types Answer 1, List Answer 13, Answers to Exercises @subsection List Tutorial Exercise 14 X We want to use @kbd{H V U} to nest a function which adds a random step to an @cite{(x,y)} coordinate. The function is a bit long, but otherwise the problem is quite straightforward. X @group @smallexample 2: [0, 0] 1: [ [ 0, 0 ] 1: 50 [ 0.4288, -0.1695 ] X . [ -0.4787, -0.9027 ] X ... X X [0,0] 50 H V U ' <# + [random(2.0)-1, random(2.0)-1]> RET @end smallexample @end group X Just as the text recommended, we used @samp{< >} nameless function notation to keep the two @code{random} calls from being evaluated before nesting even begins. X We now have a vector of @cite{[x, y]} sub-vectors, which by Calc's rules acts like a matrix. We can transpose this matrix and unpack to get a pair of vectors, @cite{x} and @cite{y}, suitable for graphing. X @group @smallexample 2: [ 0, 0.4288, -0.4787, ... ] 1: [ 0, -0.1696, -0.9027, ... ] X . X X v t v u g f @end smallexample @end group X Incidentally, because the @cite{x} and @cite{y} are completely independent in this case, we could have done two separate commands to create our @cite{x} and @cite{y} vectors of numbers directly. X To make a random walk of unit steps, we note that @code{sincos} of a random direction exactly gives us a @cite{[x, y]} step of unit length; in fact, the new nesting function is even briefer, though we might want to lower the precision a bit for it. X @group @smallexample 2: [0, 0] 1: [ [ 0, 0 ] 1: 50 [ 0.1318, 0.9912 ] X . [ -0.5965, 0.3061 ] X ... X X [0,0] 50 m d p 6 RET H V U ' <# + sincos(random(360.0))> RET @end smallexample @end group X Another @kbd{v t v u g f} sequence will graph this new random walk. X An interesting twist on these random walk functions would be to use complex numbers instead of 2-vectors to represent points on the plane. In the first example, we'd use something like @samp{random + random*(0,1)}, and in the second we could use polar complex numbers with random phase angles. (This exercise was first suggested in this form by Randal Schwartz.) X @node Types Answer 1, Types Answer 2, List Answer 14, Answers to Exercises @subsection Types Tutorial Exercise 1 X @noindent If the number is the square root of @c{$\pi$} @cite{pi} times a rational number, then its square, divided by @c{$\pi$} @cite{pi}, should be a rational number. X @group @smallexample 1: 1.26508260337 1: 0.509433962268 1: 2486645810:4881193627 X . . . X X 2 ^ P / c F @end smallexample @end group X @noindent Technically speaking this is a rational number, but not one that is likely to have arisen in the original problem. More likely, it just happens to be the fraction which most closely represents some irrational number to within 12 digits. X But perhaps our result was not quite exact. Let's reduce the precision slightly and try again: X @group @smallexample 1: 0.509433962268 1: 27:53 X . . X X U p 10 RET c F @end smallexample @end group X @noindent Aha! It's unlikely that an irrational number would equal a fraction this simple to within ten digits, so our original number was probably @c{$\sqrt{27 \pi / 53}$} @cite{sqrt(27 pi / 53)}. X Notice that we didn't need to re-round the number when we reduced the precision. Remember, arithmetic operations always round their inputs to the current precision before starting. X @node Types Answer 2, Types Answer 3, Types Answer 1, Answers to Exercises @subsection Types Tutorial Exercise 2 X @noindent @samp{inf / inf = nan}. Perhaps @samp{1} is the ``obvious'' answer. But if @w{@samp{17 inf = inf}}, then @samp{17 inf / inf = inf / inf = 17}, too. X @samp{exp(inf) = inf}. It's tempting to say that the exponential of infinity must be ``bigger'' than ``regular'' infinity, but as far as Calc is concerned all infinities are as just as big. In other words, as @cite{x} goes to infinity, @cite{e^x} also goes to infinity, but the fact the @cite{e^x} grows much faster than @cite{x} is not relevant here. X @samp{exp(-inf) = 0}. Here we have a finite answer even though the input is infinite. X @samp{sqrt(-inf) = (0, 1) inf}. Remember that @cite{(0, 1)} represents the imaginary number @cite{i}. Here's a derivation: @samp{sqrt(-inf) = @w{sqrt((-1) * inf)} = sqrt(-1) * sqrt(inf)}. The first part is, by definition, @cite{i}; the second is @code{inf} because, once again, all infinities are the same size. X @samp{sqrt(uinf) = uinf}. In fact, we do know something about the direction because @code{sqrt} is defined to return a value in the right half of the complex plane. But Calc has no notation for this, so it settles for the conservative answer @code{uinf}. X @samp{abs(uinf) = inf}. No matter which direction @cite{x} points, @samp{abs(x)} always points along the positive real axis. X @samp{ln(0) = -inf}. Here we have an infinite answer to a finite input. As in the @cite{1 / 0} case, Calc will only use infinities here if you have turned on ``infinite'' mode. Otherwise, it will treat @samp{ln(0)} as an error. X @node Types Answer 3, Types Answer 4, Types Answer 2, Answers to Exercises @subsection Types Tutorial Exercise 3 X @noindent We can make @samp{inf - inf} be any real number we like, say, @cite{a}, just by claiming that we added @cite{a} to the first infinity but not to the second. This is just as true for complex values of @cite{a}, so @code{nan} can stand for a complex number. (And, similarly, @code{uinf} can stand for an infinity that points in any direction in the complex plane, such as @samp{(0, 1) inf}). X In fact, we can multiply the first @code{inf} by two. Surely @w{@samp{2 inf - inf = inf}}, but also @samp{2 inf - inf = inf - inf = nan}. So @code{nan} can even stand for infinity. Obviously it's just as easy to make it stand for minus infinity as for plus infinity. X The moral of this story is that ``infinity'' is a slippery fish indeed, and Calc tries to handle it by having a very simple model for infinities (only the direction counts, not the ``size''); but Calc is careful to write @code{nan} any time this simple model is unable to tell what the true answer is. X @node Types Answer 4, Types Answer 5, Types Answer 3, Answers to Exercises @subsection Types Tutorial Exercise 4 X @group @smallexample 2: 0@@ 47' 26" 1: 0@@ 2' 47.411765" 1: 17 . X . X X 0@@ 47' 26" RET 17 / @end smallexample @end group X @noindent The average song length is two minutes and 47.4 seconds. X @group @smallexample 2: 0@@ 2' 47.411765" 1: 0@@ 3' 7.411765" 1: 0@@ 53' 6.000005" 1: 0@@ 0' 20" . . X . X X 20" + 17 * @end smallexample @end group X @noindent The album would be 53 minutes and 6 seconds long. X @node Types Answer 5, Types Answer 6, Types Answer 4, Answers to Exercises @subsection Types Tutorial Exercise 5 X @noindent Let's suppose it's January 14, 1991. The easiest thing to do is to keep trying 13ths of months until Calc reports a Friday. We can do this by manually entering dates, or by using @kbd{t I}: X @group @smallexample 1: <Wed Feb 13, 1991> 1: <Wed Mar 13, 1991> 1: <Sat Apr 13, 1991> X . . . X X ' <2/13> RET DEL ' <3/13> RET t I @end smallexample @end group X @noindent (Calc assumes the current year if you don't say otherwise.) X This is getting tedious---we can keep advancing the date by typing @kbd{t I} over and over again, but let's automate the job by using vector mapping. The @kbd{t I} command actually takes a second ``how-many-months'' argument, which defaults to one. This argument is exactly what we want to map over: X @group @smallexample 2: <Sat Apr 13, 1991> 1: [<Mon May 13, 1991>, <Thu Jun 13, 1991>, 1: [1, 2, 3, 4, 5, 6] <Sat Jul 13, 1991>, <Tue Aug 13, 1991>, X . <Fri Sep 13, 1991>, <Sun Oct 13, 1991>] X . X X v x 6 RET V M t I @end smallexample @end group X @ifinfo @noindent Et voila, September 13, 1991 is a Friday. @end ifinfo @tex \noindent {\it Et voil{\accent"12 a}}, September 13, 1991 is a Friday. @end tex X @group @smallexample 1: 242 X . X ' <sep 13> - <jan 14> RET @end smallexample @end group X @noindent And the answer to our original question: 242 days to go. X @node Types Answer 6, Types Answer 7, Types Answer 5, Answers to Exercises @subsection Types Tutorial Exercise 6 X @noindent The full rule for leap years is that they occur in every year divisible SHAR_EOF true || echo 'restore of calc.texinfo failed' fi echo 'End of part 35' echo 'File calc.texinfo is continued in part 36' echo 36 > _shar_seq_.tmp exit 0 exit 0 # Just in case... -- Kent Landfield INTERNET: kent@sparky.IMD.Sterling.COM Sterling Software, IMD UUCP: uunet!sparky!kent Phone: (402) 291-8300 FAX: (402) 291-4362 Please send comp.sources.misc-related mail to kent@uunet.uu.net.