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• File r35/xlog/rounded.log — part of check-in [f2fda60abd] at 2011-09-02 18:13:33 on branch master — Some historical releases purely for archival purposes

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Codemist Standard Lisp 3.54 for DEC Alpha: May 23 1994
Dump file created: Mon May 23 10:39:11 1994
REDUCE 3.5, 15-Oct-93 ...
Memory allocation: 6023424 bytes

+++ About to read file ndotest.red


% Tests in the exact mode.

x := 1/2;


      1
x := ---
      2


y := x + 0.7;


      6
y := ---
      5


% Tests in approximate mode.

on rounded;



y;


1.2
   % as expected not converted to approximate form.

z := y+1.2;


z := 2.4


z/3;


0.8


% Let's raise this to a high power.

ws^24;


0.00472236648287


% Now a high exponent value.

% 10.2^821;

% Elementary function evaluation.

cos(pi);


 - 1


symbolic ws;


(!*sq ((!:rd!: . -1.0) . 1) t)


z := sin(pi);


z := 1.22464679915e-16


symbolic ws;


(!*sq ((!:rd!: . 1.2246467991474e-16) . 1) t)


% Handling very small quantities.

% With normal defaults, underflows are converted to 0.

exp(-100000.1**2);


0


% However, if you really want that small number, roundbf can be used.

on roundbf;



exp(-100000.1**2);


1.18441281937e-4342953505


off roundbf;



% Now let us evaluate pi.

pi;


3.14159265359



% Let us try a higher precision.

precision 50;


12


pi;


3.1415926535897932384626433832795028841971693993751


% Now find the cosine of pi/6.

cos(ws/6);


0.86602540378443864676372317075293618347140262690519


% This should be the sqrt(3)/2.

ws**2;


0.75



%Here are some well known examples which show the power of this system.

precision 10;


50


% This should give the usual default again.

let xx=e**(pi*sqrt(163));



let yy=1-2*cos((6*log(2)+log(10005))/sqrt(163));



% First notice that xx looks like an integer.

xx;


2.625374126e+17


% and that yy looks like zero.

yy;


0


% but of course it's an illusion.

precision 50;


10


xx;


2.6253741264076874399999999999925007259719818568888e+17


yy;


 - 1.2815256559456092775159749532170513334408547400481e-16


%now let's look at an unusual way of finding an old friend;

precision 50;


50


procedure agm;
  <<a := 1$ b := 1/sqrt 2$ u:= 1/4$ x := 1$ pn := 4$ repeat
   <<p := pn;
     y := a; a := (a+b)/2; b := sqrt(y*b); % Arith-geom mean.
     u := u-x*(a-y)**2; x := 2*x; pn := a**2/u;
     write "pn=",pn>> until pn>=p; p>>;


agm


let ag=agm();



ag;


pn=3.1876726427121086272019299705253692326510535718594

pn=3.1416802932976532939180704245600093827957194388154

pn=3.1415926538954464960029147588180434861088792372613

pn=3.1415926535897932384663606027066313217577024113424

pn=3.1415926535897932384626433832795028841971699491647

pn=3.1415926535897932384626433832795028841971693993751

pn=3.1415926535897932384626433832795028841971693993751

3.1415926535897932384626433832795028841971693993751


% The limit is obviously.

pi;


3.1415926535897932384626433832795028841971693993751


end;
(TIME:  rounded 683 683)


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