# Pathological floating point problems

Pathological floating point problems
You are encouraged to solve this task according to the task description, using any language you may know.

Most programmers are familiar with the inexactness of floating point calculations in a binary processor.

The classic example being:

```0.1 + 0.2 =  0.30000000000000004
```

In many situations the amount of error in such calculations is very small and can be overlooked or eliminated with rounding.

There are pathological problems however, where seemingly simple, straight-forward calculations are extremely sensitive to even tiny amounts of imprecision.

This task's purpose is to show how your language deals with such classes of problems.

A sequence that seems to converge to a wrong limit.

Consider the sequence:

v1 = 2
v2 = -4
vn = 111   -   1130   /   vn-1   +   3000  /   (vn-1 * vn-2)

As   n   grows larger, the series should converge to   6   but small amounts of error will cause it to approach   100.

Display the values of the sequence where   n =   3, 4, 5, 6, 7, 8, 20, 30, 50 & 100   to at least 16 decimal places.

```    n = 3     18.5
n = 4      9.378378
n = 5      7.801153
n = 6      7.154414
n = 7      6.806785
n = 8      6.5926328
n = 20     6.0435521101892689
n = 30     6.006786093031205758530554
n = 50     6.0001758466271871889456140207471954695237
n = 100    6.000000019319477929104086803403585715024350675436952458072592750856521767230266
```

The Chaotic Bank Society   is offering a new investment account to their customers.

You first deposit   \$e - 1   where   e   is   2.7182818...   the base of natural logarithms.

After each year, your account balance will be multiplied by the number of years that have passed, and \$1 in service charges will be removed.

So ...

• after 1 year, your balance will be multiplied by 1 and \$1 will be removed for service charges.
• after 2 years your balance will be doubled and \$1 removed.
• after 3 years your balance will be tripled and \$1 removed.
• ...
• after 10 years, multiplied by 10 and \$1 removed, and so on.

What will your balance be after   25   years?

```   Starting balance: \$e-1
Balance = (Balance * year) - 1 for 25 years
Balance after 25 years: \$0.0399387296732302
```

Siegfried Rump's example.   Consider the following function, designed by Siegfried Rump in 1988.

f(a,b) = 333.75b6 + a2( 11a2b2 - b6 - 121b4 - 2 ) + 5.5b8 + a/(2b)
compute   f(a,b)   where   a=77617.0   and   b=33096.0
f(77617.0, 33096.0)   =   -0.827396059946821

Demonstrate how to solve at least one of the first two problems, or both, and the third if you're feeling particularly jaunty.

## 360 Assembly

The system/360 hexadecimal single precision floating point format is known to its weakness in precision. A lot of more precise formats have been added since.
A sequence that seems to converge to a wrong limit

`*        Pathological floating point problems  03/05/2016PATHOFP  CSECT         USING  PATHOFP,R13 SAVEAR   B      STM-SAVEAR(R15)         DC     17F'0'STM      STM    R14,R12,12(R13)         ST     R13,4(R15)         ST     R15,8(R13)         LR     R13,R15         LE     F0,=E'2'         STE    F0,U             u(1)=2         LE     F0,=E'-4'         STE    F0,U+4           u(2)=-4         LA     R6,3             n=3         LA     R7,U+4           @u(n-1)         LA     R8,U             @u(n-2)         LA     R9,U+8           @u(n)LOOPN    CH     R6,=H'100'       do n=3 to 100         BH     ELOOPN         LE     F4,0(R7)         u(n-1)         LE     F2,=E'1130'      1130         DER    F2,F4            1130/u(n-1)         LE     F0,=E'111'       111         SER    F0,F2            111-1130/u(n-1)         LE     F2,0(R7)         u(n-1)         LE     F4,0(R8)         u(n-2)         MER    F2,F4            u(n-1)*u(n-2)         LE     F6,=E'3000'      3000         DER    F6,F2            3000/(u(n-1)*u(n-2))         AER    F0,F6            111-1130/u(n-1)+3000/(u(n-1)*u(n-2))         STE    F0,0(R9)         store into u(n)         XDECO  R6,PG+0          n         LE     F0,0(R9)         u(n)         LA     R0,3             number of decimals         BAL    R14,FORMATF      format(u(n),'F13.3')         MVC    PG+12(13),0(R1)  put into buffer         XPRNT  PG,80            print buffer         LA     R6,1(R6)         n=n+1         LA     R7,4(R7)         @u(n-1)         LA     R8,4(R8)         @u(n-2)         LA     R9,4(R9)         @u(n)         B      LOOPNELOOPN   L      R13,4(0,R13)         LM     R14,R12,12(R13)         XR     R15,R15         BR     R14         COPY   FORMATF         LTORG  PG       DC     CL80' '          bufferU        DS     100E         YREGS         YFPREGS          END    PATHOFP`

The divergence comes very soon.

Output:
```           3       18.500
4        9.378
5        7.801
6        7.154
7        6.805
8        6.578
9        6.235
10        2.915
11     -111.573
12      111.905
13      100.661
14      100.040
15      100.002
16      100.000
17      100.000
18      100.000
...      100.000
```

`with Ada.Text_IO; procedure Converging_Sequence is    generic      type Num is digits <>;      After: Positive;   procedure Task_1;    procedure Task_1 is      package FIO is new Ada.Text_IO.Float_IO(Num);      package IIO is new Ada.Text_IO.Integer_IO(Integer);       procedure Output (I: Integer; N: Num) is      begin	 IIO.Put(Item => I, Width => 4);	 FIO.Put(Item => N, Fore => 4, Aft =>  After, Exp => 0);	 Ada.Text_IO.New_Line;      end Output;       Very_Old: Num :=  2.0;      Old:      Num := -4.0;      Now:        Num;   begin      Ada.Text_IO.Put_Line("Converging Sequence with" & Integer'Image(After) & 			     " digits");      for I in 3 .. 100 loop	 Now := 111.0  - 1130.0   /   Old   + 3000.0  /   (Old * Very_Old);	 Very_Old := Old;	 Old := Now;	 if (I < 9) or else (I=20 or I=30 or I=50 or I=100) then	    Output(I, Now);	 end if;      end loop;      Ada.Text_IO.New_Line;   end Task_1;    type Short is digits(8);   type Long  is digits(16);    procedure Task_With_Short is new Task_1(Short, 8);   procedure Task_With_Long  is new Task_1(Long, 16);begin   Task_With_Short;   Task_With_Long;end Converging_Sequence;`
Output:
```Converging Sequence with 8 digits
3  18.50000000
4   9.37837838
5   7.80115274
6   7.15441448
7   6.80678474
8   6.59263277
20  98.34950312
30 100.00000000
50 100.00000000
100 100.00000000

Converging Sequence with 16 digits
3  18.5000000000000000
4   9.3783783783783784
5   7.8011527377521614
6   7.1544144809752494
7   6.8067847369236337
8   6.5926327687044483
20   8.9530549789723472
30  99.9999999981565451
50 100.0000000000000000
100 100.0000000000000000```

### Task 2: Chaotic Bank Society

`with Ada.Text_IO, Ada.Numerics; procedure Chaotic_Bank is    generic     type Num is digits <>;     After: Positive;   procedure Task_2;    procedure Task_2 is      package IIO is new Ada.Text_IO.Integer_IO(Integer);      package FIO is new Ada.Text_IO.Float_IO(Num);      Balance: Num :=  Ada.Numerics.E - 1.0;   begin      Ada.Text_IO.Put_Line("Chaotic Bank Socienty with" & 			     Integer'Image(After) & " digits");      Ada.Text_IO.Put_Line("year        balance");      for year in 1 .. 25 loop	 Balance := (Balance * Num(year))- 1.0;	 IIO.Put(Item => Year, Width => 2);	 FIO.Put(Balance, Fore => 11, Aft => After, Exp => 0);	 Ada.Text_IO.New_Line;      end loop;      Ada.Text_IO.New_Line;   end Task_2;    type Short is digits(8);   type Long  is digits(16);    procedure Task_With_Short is new Task_2(Short, 8);   procedure Task_With_Long  is new Task_2(Long, 16); begin   Task_With_Short;   Task_With_Long;end Chaotic_Bank;`
Output:
```Chaotic Bank Socienty with 8 digits
year        balance
1          0.71828183
2          0.43656366
3          0.30969097
4          0.23876388
...         ...
16          0.06389363
17          0.08619166
18          0.55144980
19          9.47754622
20        188.55092437
21       3958.56941176
22      87087.52705873
23    2003012.12235075
24   48072289.93641794
25 1201807247.41044855

Chaotic Bank Socienty with 16 digits
year        balance
1          0.7182818284590452
2          0.4365636569180905
3          0.3096909707542714
4          0.2387638830170856
...         ...
17          0.0586186042274583
18          0.0551348760942503
19          0.0475626457907552
20         -0.0487470841848960
21         -2.0236887678828168
22        -45.5211528934219700
23      -1047.9865165487053100
24     -25152.6763971689275000
25    -628817.9099292231860000```

`with Ada.Text_IO; use Ada.Text_IO;                                                                         procedure Rumps_example is                                                                                    type Short is digits(8);                                                                                  type Long  is digits(16);                                                                                  A: constant := 77617.0;                                                                                   B: constant := 33096.0;                                                                                   C: constant := 333.75*B**6 + A**2*(11.0*A**2*B**2 - B**6 - 121.0*B**4 - 2.0) + 5.5*B**8 + A/(2.0*B);       package LIO is new Float_IO(Long);                                                                        package SIO is new Float_IO(Short);                                                                    begin                                                                                                        Put("Rump's Example, Short: ");                                                                           SIO.Put(C, Fore => 1, Aft => 8, Exp => 0);  New_Line;                                                     Put("Rump's Example, Long:  ");                                                                           LIO.Put(C, Fore => 1, Aft => 16, Exp => 0); New_Line;                                                  end Rumps_example;  `
Output:
```Rump's Example, Short: -0.82739606
Rump's Example, Long:  -0.827396059946821```

## ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

In Algol 68G, we can specify the precision of LONG LONG REAL values

`BEGIN    # task 1 #    BEGIN        PR precision 32 PR        print( ( " 32 digit REAL numbers", newline ) );        [ 1 : 100 ]LONG LONG REAL v;        v[ 1 ] := 2;        v[ 2 ] := -4;        FOR n FROM 3 TO UPB v DO v[ n ] := 111 - ( 1130 / v[ n - 1 ] ) + ( 3000 / ( v[ n - 1 ] * v[ n - 2 ] ) ) OD;        FOR n FROM  3 TO  8       DO print( ( "n = ", whole( n, 3 ), " ", fixed( v[ n ], -22, 16 ), newline ) ) OD;        FOR n FROM 20 BY 10 TO 50 DO print( ( "n = ", whole( n, 3 ), " ", fixed( v[ n ], -22, 16 ), newline ) ) OD;        print( ( "n = 100 ", fixed( v[ 100 ], -22, 16 ), newline ) )    END;    BEGIN        PR precision 120 PR        print( ( "120 digit REAL numbers", newline ) );        [ 1 : 100 ]LONG LONG REAL v;        v[ 1 ] := 2;        v[ 2 ] := -4;        FOR n FROM 3 TO UPB v DO v[ n ] := 111 - ( 1130 / v[ n - 1 ] ) + ( 3000 / ( v[ n - 1 ] * v[ n - 2 ] ) ) OD;        print( ( "n = 100 ", fixed( v[ 100 ], -22, 16 ), newline ) )    END;    print( ( newline ) );    # task 2 #    BEGIN        print( ( "single precision REAL numbers...", newline ) );        REAL chaotic balance := exp( 1 ) - 1;        print( ( "initial chaotic balance: ", fixed( chaotic balance, -22, 16 ), newline ) );        FOR i FROM 1 TO 25 DO ( chaotic balance *:= i ) -:= 1 OD;        print( ( "25 year chaotic balance: ", fixed( chaotic balance, -22, 16 ), newline ) )    END;    BEGIN        print( ( "double precision REAL numbers...", newline ) );        LONG REAL chaotic balance := long exp( 1 ) - 1;        print( ( "initial chaotic balance: ", fixed( chaotic balance, -22, 16 ), newline ) );        FOR i FROM 1 TO 25 DO ( chaotic balance *:= i ) -:= 1 OD;        print( ( "25 year chaotic balance: ", fixed( chaotic balance, -22, 16 ), newline ) )    END;    BEGIN        PR precision 32 PR        print( ( "        32 digit REAL numbers...", newline ) );        LONG LONG REAL chaotic balance := long long exp( 1 ) - 1;        print( ( "initial chaotic balance: ", fixed( chaotic balance, -22, 16 ), newline ) );        FOR i FROM 1 TO 25 DO ( chaotic balance *:= i ) -:= 1 OD;        print( ( "25 year chaotic balance: ", fixed( chaotic balance, -22, 16 ), newline ) )    ENDEND`
Output:
``` 32 digit REAL numbers
n =  +3    18.5000000000000000
n =  +4     9.3783783783783784
n =  +5     7.8011527377521614
n =  +6     7.1544144809752494
n =  +7     6.8067847369236330
n =  +8     6.5926327687044384
n = +20     6.0435521101892689
n = +30     6.0067860930262429
n = +40    -2.9367486132065552
n = +50   100.0000000006552004
n = 100   100.0000000000000000
120 digit REAL numbers
n = 100   100.0000000000000000

single precision REAL numbers...
initial chaotic balance:     1.7182818284590400
25 year chaotic balance: -2242373258.5701500000
double precision REAL numbers...
initial chaotic balance:     1.7182818284590452
25 year chaotic balance:     0.0406729916134442
32 digit REAL numbers...
initial chaotic balance:     1.7182818284590452
25 year chaotic balance:     0.0399387296732302
```

## AWK

 This example may be incorrect. Check if results are different for non-GNU awk Please verify it and remove this message. If the example does not match the requirements or does not work, replace this message with Template:incorrect or fix the code yourself.

GNU awk defaults to double-precision floating point numbers (not sure if this is true for other awk implementations?). GNU awk 4.1+ provides library support for arbitrary-precision floating point calculations, but not all available binaries have this compiled in.

awk code:

` BEGIN {    do_task1()    do_task2()    do_task3()    exit}  function do_task1(){    print "Task 1"    v[1] = 2    v[2] = -4    for (n=3; n<=100; n++) v[n] = 111 - 1130 / v[n-1] + 3000 / (v[n-1] * v[n-2])     for (i=3; i<=8; i++) print_results(i)    print_results(20)    print_results(30)    print_results(50)    print_results(100)} # This works because all awk variables are global, except when declared locallyfunction print_results(n){    printf("n = %d\t%20.16f\n", n, v[n])} # This function doesn't need any parameters; declaring balance and i in the function parameters makes them localfunction do_task2(      balance, i){    balance[0] = exp(1)-1    for (i=1; i<=25; i++) balance[i] = balance[i-1]*i-1    printf("\nTask 2\nBalance after 25 years: \$%12.10f", balance[25])} function do_task3(      a, b, f_ab){    a = 77617    b = 33096     f_ab = 333.75 * b^6 + a^2 * (11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b)     printf("\nTask 3\nf(%6.12f, %6.12f) = %10.24f", a, b, f_ab)}  `

This version doesn't include the arbitrary-precision libraries, so the program demonstrates the incorrect results:

```Task 1
n = 3    18.5000000000000000
n = 4     9.3783783783783790
n = 5     7.8011527377521688
n = 6     7.1544144809753334
n = 7     6.8067847369248113
n = 8     6.5926327687217920
n = 20   98.3495031221653591
n = 30   99.9999999999989342
n = 50  100.0000000000000000
n = 100 100.0000000000000000

Balance after 25 years: \$-2242373258.570158004760742
f(77617.000000000000, 33096.000000000000) = -1180591620717411303424.000000000000000000000000
```

On versions with the libraries compiled in, the results depend on the level of precision specified. With 1024 bits, the results are as follows:

```Task 1
n = 3	 18.5000000000000000
n = 4	  9.3783783783783784
n = 5	  7.8011527377521614
n = 6	  7.1544144809752494
n = 7	  6.8067847369236330
n = 8	  6.5926327687044384
n = 20	  6.0435521101892689
n = 30	  6.0067860930312058
n = 50	  6.0001758466271872
n = 100	  6.0000000193194779

Balance after 25 years: \$0.0399387297
f(77617.000000000000, 33096.000000000000) = -0.827396059946821368141165
```

With 256 bits of precision, tasks 2 and 3 provide the same answer as above. Task 1 appears to be converging after 50 iterations, but by 100 iterations the answer has changed to 100.0

## C

Such exercises are very good examples that just because you have a nice library doesn't mean you won't get the wrong results. I was over-ambitious and the result is I have been wrangling with the Chaotic Bank task for a long time now, I will come back to it later but for now here are the first two cases, the "trivial" one of 0.1 + 0.2 and the Pathological series :

Library: GMP
` #include<stdio.h>#include<gmp.h> void firstCase(){	mpf_t a,b,c; 	mpf_inits(a,b,c,NULL); 	mpf_set_str(a,"0.1",10);	mpf_set_str(b,"0.2",10);	mpf_add(c,a,b); 	gmp_printf("\n0.1 + 0.2 = %.*Ff",20,c);} void pathologicalSeries(){	int n;	mpf_t v1, v2, vn, a1, a2, a3, t2, t3, prod; 	mpf_inits(v1,v2,vn, a1, a2, a3, t2, t3, prod,NULL); 	mpf_set_str(v1,"2",10);	mpf_set_str(v2,"-4",10);	mpf_set_str(a1,"111",10);	mpf_set_str(a2,"1130",10);	mpf_set_str(a3,"3000",10); 	for(n=3;n<=100;n++){		mpf_div(t2,a2,v2);		mpf_mul(prod,v1,v2);		mpf_div(t3,a3,prod);		mpf_add(vn,a1,t3);		mpf_sub(vn,vn,t2); 		if((n>=3&&n<=8) || n==20 || n==30 || n==50 || n==100){			gmp_printf("\nv_%d : %.*Ff",n,(n==3)?1:(n>=4&&n<=7)?6:(n==8)?7:(n==20)?16:(n==30)?24:(n==50)?40:78,vn);		} 		mpf_set(v1,v2);		mpf_set(v2,vn);	}} void healthySeries(){	int n; 	mpf_t num,denom,result;	mpq_t v1, v2, vn, a1, a2, a3, t2, t3, prod; 	mpf_inits(num,denom,result,NULL);	mpq_inits(v1,v2,vn, a1, a2, a3, t2, t3, prod,NULL); 	mpq_set_str(v1,"2",10);	mpq_set_str(v2,"-4",10);	mpq_set_str(a1,"111",10);	mpq_set_str(a2,"1130",10);	mpq_set_str(a3,"3000",10); 	for(n=3;n<=100;n++){		mpq_div(t2,a2,v2);		mpq_mul(prod,v1,v2);		mpq_div(t3,a3,prod);		mpq_add(vn,a1,t3);		mpq_sub(vn,vn,t2); 		if((n>=3&&n<=8) || n==20 || n==30 || n==50 || n==100){			mpf_set_z(num,mpq_numref(vn));			mpf_set_z(denom,mpq_denref(vn));			mpf_div(result,num,denom); 			gmp_printf("\nv_%d : %.*Ff",n,(n==3)?1:(n>=4&&n<=7)?6:(n==8)?7:(n==20)?16:(n==30)?24:(n==50)?40:78,result);		} 		mpq_set(v1,v2);		mpq_set(v2,vn);	}} int main(){		mpz_t rangeProd; 	firstCase(); 	printf("\n\nPathological Series : "); 	pathologicalSeries(); 	printf("\n\nNow a bit healthier : "); 	healthySeries(); 	return 0;} `

The reason I included the trivial case was the discovery that the value of 0.3 is stored inexactly even by GMP if 0.1 and 0.2 are set via the mpf_set_d function, a point observed also during the solution of the Currency task. Thus mpf_set_str has been used to set the values, for the 2nd task, a great learning was not to convert the values into floating points unless they have to be printed out. It's a polynomial with out a single floating point coefficient or exponent, a clear sign that the values must be treated as rationals for the highest accuracy. Thus there are two implementations for this task in the above code, one uses floating points, the other rationals. There is still a loss of accuracy even when floating points are used, probably during the conversion of a rational to a float.

```0.1 + 0.2 = 0.30000000000000000000

Pathological Series :
v_3 : 18.5
v_4 : 9.378378
v_5 : 7.801153
v_6 : 7.154414
v_7 : 6.806785
v_8 : 6.5926328
v_20 : 6.0751649921786439
v_30 : 99.999999824974113455900000
v_50 : 100.0000000000000000000000000000000000000000
v_100 : 100.000000000000000000000000000000000000000000000000000000000000000000000000000000

Now a bit healthier :
v_3 : 18.5
v_4 : 9.378378
v_5 : 7.801153
v_6 : 7.154414
v_7 : 6.806785
v_8 : 6.5926328
v_20 : 6.0435521101892689
v_30 : 6.006786093031205758530000
v_50 : 6.0001758466271871889100000000000000000000
v_100 : 6.000000019319477929060000000000000000000000000000000000000000000000000000000000
```

## Clojure

In Clojure, rational numbers are a first class data type! This allow us to avoid these floating point calculation problems. As long as the operations all involve integers, rational numbers will automatically be used behind the scenes. If for some twisted reason you really wanted to introduce the error, you could change a number to a floating point in the equation to force floating point calculations instead.

`(def converge-to-six ((fn task1 [a b] (lazy-seq (cons a (task1 b (+ (- 111 (/ 1130 b)) (/ 3000 (* b a))))))) 2 -4))(def values [3 4 5 6 7 8 20 30 50 100]); print decimal values:(pprint (sort (zipmap values (map double (map #(nth converge-to-six (dec %)) values))))); print rational values:(pprint (sort (zipmap values (map #(nth converge-to-six (dec %)) values))))`
Output:
```decimal representation:
([3 18.5]
[4 9.378378378378377]
[5 7.801152737752161]
[6 7.154414480975249]
[7 6.806784736923633]
[8 6.592632768704438]
[20 6.043552110189269]
[30 6.006786093031206]
[50 6.000175846627187]
[100 6.000000019319478])

rational representation:
([3 37/2]
[4 347/37]
[5 2707/347]
[6 19367/2707]
[7 131827/19367]
[8 869087/131827]
[20 2646751398406607/437946318679747]
[30 164874117215934539909207/27447975450168574034347]
[50
606122140256603032959120809808954954407/101017396114701505075745671950377583547]
[100
489988959736444127362399646251257712574995486122651802567073927074333549467407/81664826359787052820062672571300097001570504462706488724837986255629281356547])
```

### Task 2: Chaotic Bank Society

`(def e-ratio 106246577894593683/39085931702241241)(defn bank [n m] (- (* n m) 1))(double (reduce bank (- e-ratio 1) (range 1 26)))`
Output:
```Balance after 25 (decimal representation):
0.03993873004901714

Balance after 25 (rational representation):
1561042474970134/39085931702241241
```

`(defn rump [a b]  (+ (* (rationalize 333.75) (expt b 6))      (* (expt a 2)          (- (* 11 (expt a 2) (expt b 2)) (expt b 6) (* 121 (expt b 4)) 2))      (* (rationalize 5.5) (expt b 8))      (/ a (* 2 b)))); Using BigInt numeric literal style to avoid integer overflow(double (rump 77617 33096N))`
Output:
```-0.8273960599468214
```

## Excel

Works with: Excel 2003 version Excel 2015

A sequence that seems to converge to a wrong limit

`  A1: 2  A2: -4  A3: =111-1130/A2+3000/(A2*A1)  A4: =111-1130/A3+3000/(A3*A2)  ...`

The result converges to the wrong limit!

Output:
```       A
1      2
2     -4
3     18.5
4      9.378378378
5      7.801152738
6      7.154414481
7      6.806784737
8      6.592632769
9      6.449465934
10      6.348452061
11      6.274438663
12      6.218696769
13      6.175853856
14      6.142627170
15      6.120248705
16      6.166086560
17      7.235021166
18     22.06207846
19     78.57557489
20     98.34950312
21     99.89856927
22     99.99387099
23     99.99963039
24     99.99997773
25     99.99999866
26     99.99999992
27    100
...
30    100
...
40    100
...
50    100
...
100    100
```

## FreeBASIC

`' FB 1.05.0 Win64 ' As FB's native types have only 64 bit precision at most we need to use the' C library, GMP v6.1.0, for arbitrary precision arithmetic #Include Once "gmp.bi"mpf_set_default_prec(640) '' 640 bit precision, enough for this exercise Function v(n As UInteger, prev As __mpf_struct, prev2 As __mpf_struct) As __mpf_struct  Dim As __mpf_struct a, b, c  mpf_init(@a) : mpf_init(@b) : mpf_init(@c)  If n = 0 Then mpf_set_ui(@a, 0UL)  If n = 1 Then mpf_set_ui(@a, 2UL)  If n = 2 Then mpf_set_si(@a, -4L)  If n < 3 Then Return a   mpf_ui_div(@a, 1130UL, @prev)  mpf_mul(@b, @prev, @prev2)  mpf_ui_div(@c, 3000UL, @b)  mpf_ui_sub(@b, 111UL, @a)   mpf_add(@a, @b, @c)  mpf_clear(@b)  mpf_clear(@c)  Return aEnd Function Function f(a As Double, b As Double) As __mpf_Struct  Dim As __mpf_struct temp1, temp2, temp3, temp4, temp5, temp6, temp7, temp8  mpf_init(@temp1) : mpf_init(@temp2) : mpf_init(@temp3) : mpf_init(@temp4)  mpf_init(@temp5) : mpf_init(@temp6) : mpf_init(@temp7) : mpf_init(@temp8)  mpf_set_d(@temp1, a)               '' a  mpf_set_d(@temp2, b)               '' b   mpf_set_d(@temp3, 333.75)          '' 333.75  mpf_pow_ui(@temp4, @temp2, 6UL)    '' b ^ 6  mpf_mul(@temp3, @temp3, @temp4)    '' 333.75 * b^6  mpf_pow_ui(@temp5, @temp1, 2UL)    '' a^2  mpf_pow_ui(@temp6, @temp2, 2UL)    '' b^2  mpf_mul_ui(@temp7, @temp5, 11UL)   '' 11 * a^2  mpf_mul(@temp7, @temp7, @temp6)    '' 11 * a^2 * b^2  mpf_sub(@temp7, @temp7, @temp4)    '' 11 * a^2 * b^2 - b^6  mpf_pow_ui(@temp4, @temp2, 4UL)    '' b^4  mpf_mul_ui(@temp4, @temp4, 121UL)  '' 121 * b^4  mpf_sub(@temp7, @temp7, @temp4)    '' 11 * a^2 * b^2 - b^6 - 121 * b^4  mpf_sub_ui(@temp7, @temp7, 2UL)    '' 11 * a^2 * b^2 - b^6 - 121 * b^4 - 2  mpf_mul(@temp7, @temp7, @temp5)    '' (11 * a^2 * b^2 - b^6 - 121 * b^4 - 2) * a^2  mpf_add(@temp3, @temp3, @temp7)    '' 333.75 * b^6 + (11 * a^2 * b^2 - b^6 - 121 * b^4 - 2) * a^2  mpf_set_d(@temp4, 5.5)             '' 5.5  mpf_pow_ui(@temp5, @temp2, 8UL)    '' b^8    mpf_mul(@temp4, @temp4, @temp5)    '' 5.5 * b^8  mpf_add(@temp3, @temp3, @temp4)    '' 333.75 * b^6 + (11 * a^2 * b^2 - b^6 - 121 * b^4 - 2) * a^2 + 5.5 * b^8  mpf_mul_ui(@temp4, @temp2, 2UL)    '' 2 * b  mpf_div(@temp5, @temp1, @temp4)    '' a / (2 * b)  mpf_add(@temp3, @temp3, @temp5)    '' 333.75 * b^6 + (11 * a^2 * b^2 - b^6 - 121 * b^4 - 2) * a^2 + 5.5 * b^8 + a / (2 * b)  mpf_clear(@temp1) : mpf_clear(@temp2) : mpf_clear(@temp4) : mpf_clear(@temp5)  mpf_clear(@temp6) : mpf_clear(@temp7) : mpf_clear(@temp8)  Return temp3End Function Dim As Zstring * 60 zDim As __mpf_struct result, prev, prev2' We cache the two previous results to avoid recursive calls to vFor i As Integer = 1 To 100  result = v(i, prev, prev2)  If (i >= 3 AndAlso i <= 8) OrElse i = 20 OrElse i = 30 OrElse i = 50 OrElse i = 100 Then    gmp_sprintf(@z,"%53.50Ff",@result) '' express result to 50 decimal places    Print "n ="; i , z  End If   prev2 = prev  prev = result    Next mpf_clear(@prev) : mpf_clear(@prev2) '' note : prev = result Dim As __mpf_struct e, balance, ii, tempmpf_init(@e) : mpf_init(@balance) : mpf_init(@ii) : mpf_init(@temp)mpf_set_str(@e, "2.71828182845904523536028747135266249775724709369995", 10) '' e to 50 decimal placesmpf_sub_ui(@balance, @e, 1UL) For i As ULong = 1 To 25  mpf_set_ui(@ii, i)    mpf_mul(@temp, @balance, @ii)  mpf_sub_ui(@balance, @temp, 1UL) Next   PrintPrint "Chaotic B/S balance after 25 years : ";gmp_sprintf(@z,"%.16Ff",@balance) '' express balance to 16 decimal placesPrint zmpf_clear(@e) : mpf_clear(@balance) : mpf_clear(@ii) : mpf_clear(@temp)  PrintDim rump As __mpf_structrump = f(77617.0, 33096.0)gmp_sprintf(@z,"%.16Ff", @rump) '' express rump to 16 decimal placesPrint "f(77617.0, 33096.0) = "; z PrintPrint "Press any key to quit"Sleep`
Output:
```n = 3         18.50000000000000000000000000000000000000000000000000
n = 4          9.37837837837837837837837837837837837837837837837838
n = 5          7.80115273775216138328530259365994236311239193083573
n = 6          7.15441448097524935352789065386036202438123383819727
n = 7          6.80678473692363298394175659627200908762327670780193
n = 8          6.59263276870443839274200277636599482655298231773461
n = 20         6.04355211018926886777747736409754013318771500000612
n = 30         6.00678609303120575853055404795323970583307231443837
n = 50         6.00017584662718718894561402074719546952373517709933
n = 100        6.00000001931947792910408680340358571502435067543695

Chaotic B/S balance after 25 years : 0.0399387296732302

f(77617.0, 33096.0) = -0.8273960599468214
```

## Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.

The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.

## Fortran

### Compute from the hip

Problems arise because the floating-point arithmetic as performed by digital computers has only an oblique relationship to the arithmetic of Real numbers: many axia are violated, even if only by a little, and in certain situations. Most seriously, only a limited precision is available even if the floating-point variables are declared via such words as "REAL". Actions such as subtraction (of nearly-equal values) can in one step destroy many or all the digits of accuracy of the value being developed.

Fortran's only "built-in" assistance in this is provided via the ability to declare floating-point variables DOUBLE PRECISION, and on some systems, QUADRUPLE PRECISION is available. Earlier systems such as the IBM1620 supported decimal arithmetic of up to ninety-nine decimal digits (via hardware!), and the Fortran II compiler offered limited access to this via a control card at the start of the source file of the form `*ffkks` but the allowed range of ff was only 2 to 28, not 99. More modern compilers have abandoned this ability. Although the allowable syntax could admit something like `REAL*496`, the highest usually available is `REAL*8` for 64-bit floating-point numbers, and perhaps `REAL*10`, or `REAL*16` for QUADRUPLE PRECISION. Special "bignumber" arithmetic routines can be written supporting floating-point (or integer, or rational) arithmetic of hundreds or thousands or more words of storage per number, but this is not a standard arrangement.

Otherwise, a troublesome calculation might be recast into a different form that avoids a catastrophic loss of precision, probably after a lot of careful and difficult analysis and exploration and ingenuity.

Here, no such attempt is made. In the spirit of Formula Translation, this is a direct translation of the specified formulae into Fortran, with single and double precision results on display. There is no REAL*16 option, nor the REAL*10 that some systems allow to correspond to the eighty-bit floating-point format supported by the floating-point processor. The various integer constants cause no difficulty and I'm not bothering with writing them as <integer>.0 - the compiler can deal with this. The constants with fractional parts happen to be exactly represented in binary so there is no fuss over 333.75 and 333.75D0 whereas by contrast 0.1 and 0.1D0 are not equal. Similarly, there is no attempt to rearrange the formulae, for instance to have `A**2 * B**2` replaced by `(A*B)**2`, nor worry over `B**8` where 33096**8 = 1.439E36 and the largest possible single-precision number is 3.4028235E+38, in part because arithmetic within an expression can be conducted with a greater dynamic range. Most of all, no attention has been given to the subtractions...

This would be F77 style Fortran, except for certain conveniences offered by F90, especially the availability of generic functions such as EXP whose type is determined by the type of its parameter, rather than having to use EXP and DEXP for single and double precision respectively, or else... The END statement for subroutines and functions names the routine being ended, a useful matter to have checked.
`      SUBROUTINE MULLER       REAL*4 VN,VNL1,VNL2	!The exact precision and dynamic range       REAL*8 WN,WNL1,WNL2	!Depends on the format's precise usage of bits.       INTEGER I		!A stepper.        WRITE (6,1)		!A heading.    1   FORMAT ("Muller's sequence should converge to six...",/     1   "  N     Single      Double")        VNL1 = 2; VN = -4	!Initialise for N = 2.        WNL1 = 2; WN = -4	!No fractional parts yet.        DO I = 3,36			!No point going any further.          VNL2 = VNL1; VNL1 = VN		!Shuffle the values along one place.          WNL2 = WNL1; WNL1 = WN		!Ready for the next term's calculation.          VN = 111 - 1130/VNL1 + 3000/(VNL1*VNL2)	!Calculate the next term.          WN = 111 - 1130/WNL1 + 3000/(WNL1*WNL2)	!In double precision.          WRITE (6,2) I,VN,WN			!Show both.    2     FORMAT (I3,F12.7,F21.16)		!With too many fractional digits.        END DO				!On to the next term.      END SUBROUTINE MULLER	!That was easy. Too bad the results are wrong.       SUBROUTINE CBS		!The Chaotic Bank Society.       INTEGER YEAR	!A stepper.       REAL*4 V		!The balance.       REAL*8 W		!In double precision as well.        V = 1; W = 1		!Initial values, without dozy 1D0 stuff.        V = EXP(V) - 1		!Actual initial value desired is e - 1..,        W = EXP(W) - 1		!This relies on double-precision W selecting DEXP.        WRITE (6,1)		!Here we go.    1   FORMAT (///"The Chaotic Bank Society in action..."/"Year")        WRITE (6,2) 0,V,W	!Show the initial deposit.    2   FORMAT (I3,F16.7,F28.16)        DO YEAR = 1,25		!Step through some years.          V = V*YEAR - 1	!The specified procedure.          W = W*YEAR - 1	!The compiler handles type conversions.          WRITE (6,2) YEAR,V,W	!The current balance.        END DO			!On to the following year.      END SUBROUTINE CBS	!Madness!       REAL*4 FUNCTION SR4(A,B)	!Siegfried Rump's example function of 1988.       REAL*4 A,B        SR4 = 333.75*B**6     1      + A**2*(11*A**2*B**2 - B**6 - 121*B**4 - 2)     2      + 5.5*B**8 + A/(2*B)      END FUNCTION SR4      REAL*8 FUNCTION SR8(A,B)	!Siegfried Rump's example function.       REAL*8 A,B        SR8 = 333.75*B**6	!.75 is exactly represented in binary.     1      + A**2*(11*A**2*B**2 - B**6 - 121*B**4 - 2)     2      + 5.5*B**8 + A/(2*B)!.5 is exactly represented in binary.      END FUNCTION SR8       PROGRAM POKE      REAL*4 V	!Some example variables.      REAL*8 W	!Whose type goes to the inquiry function.      WRITE (6,1) RADIX(V),DIGITS(V),"single",DIGITS(W),"double"    1   FORMAT ("Floating-point arithmetic is conducted in base ",I0,/     1   2(I3," digits for ",A," precision",/))      WRITE (6,*) "Single precision limit",HUGE(V)      WRITE (6,*) "Double precision limit",HUGE(W)      WRITE (6,*)       CALL MULLER       CALL CBS       WRITE (6,10)   10 FORMAT (///"Evaluation of Siegfried Rump's function of 1988",     1 " where F(77617,33096) = -0.827396059946821")      WRITE (6,*) "Single precision:",SR4(77617.0,33096.0)      WRITE (6,*) "Double precision:",SR8(77617.0D0,33096.0D0)	!Must match the types.      END`

#### Output

Floating-point numbers in single and double precision use the "implicit leading one" binary format on this system: there have been many variations across different computers over the years. One can write strange routines that will test the workings of arithmetic (and other matters) so as to determine the situation on the computer of the moment, but F90 introduced special "inquiry" routines that reveal certain details as standard. This information could be used to make choices amongst calculation paths and ploys appropriate for different results, at of course a large expenditure in thought to produce a compound scheme that will (should?) work correctly on a variety of computers. No such effort has been made here!

Fifty-three binary digits corresponds to 15·95 decimal digits: there is no simple conversion so the usual ploy is to show additional decimal digits, knowing that the lower-order digits will be fuzz due to the binary/decimal conversion. The "Muller" sequence has for its fourth term 9.3783783783783790 - note that this is a recurring sequence, and its precision is less than the displayed sixteen decimal digits (seventeen digits of "precision" are on show) - when trying for maximum accuracy, converting a binary value to decimal adds confusion.

```Floating-point arithmetic is conducted in base 2
24 digits for single precision
53 digits for double precision

Single precision limit  3.4028235E+38
Double precision limit  1.797693134862316E+308

Muller's sequence should converge to six...
N     Single      Double
3  18.5000000  18.5000000000000000
4   9.3783779   9.3783783783783790
5   7.8011475   7.8011527377521688
6   7.1543465   7.1544144809753334
7   6.8058305   6.8067847369248113
8   6.5785794   6.5926327687217920
9   6.2355156   6.4494659340539329
10   2.9135900   6.3484520607466237
11-111.7097931   6.2744386627281159
12 111.8982391   6.2186967685821628
13 100.6615448   6.1758538558153901
14 100.0406036   6.1426271704810063
15 100.0024948   6.1202487045701588
16 100.0001526   6.1660865595980994
17 100.0000076   7.2350211655349312
18 100.0000000  22.0620784635257934
19 100.0000000  78.5755748878722358
20 100.0000000  98.3495031221653591
21 100.0000000  99.8985692661829034
22 100.0000000  99.9938709889027848
23 100.0000000  99.9996303872863450
24 100.0000000  99.9999777306794897
25 100.0000000  99.9999986592166863
26 100.0000000  99.9999999193218088
27 100.0000000  99.9999999951477605
28 100.0000000  99.9999999997082796
29 100.0000000  99.9999999999824638
30 100.0000000  99.9999999999989342
31 100.0000000  99.9999999999999289
32 100.0000000  99.9999999999999858
33 100.0000000 100.0000000000000000
34 100.0000000 100.0000000000000000
35 100.0000000 100.0000000000000000
36 100.0000000 100.0000000000000000

The Chaotic Bank Society in action...
Year
0       1.7182819          1.7182818284590453
1       0.7182819          0.7182818284590453
2       0.4365637          0.4365636569180906
3       0.3096912          0.3096909707542719
4       0.2387648          0.2387638830170875
5       0.1938238          0.1938194150854375
6       0.1629429          0.1629164905126252
7       0.1406002          0.1404154335883767
8       0.1248016          0.1233234687070137
9       0.1232147          0.1099112183631235
10       0.2321472          0.0991121836312345
11       1.5536194          0.0902340199435798
12      17.6434326          0.0828082393229579
13     228.3646240          0.0765071111984525
14    3196.1047363          0.0710995567783357
15   47940.5703125          0.0664933516750352
16  767048.1250000          0.0638936268005637
1713039817.0000000          0.0861916556095821
18****************          0.5514498009724775
19****************          9.4775462184770731
20****************        188.5509243695414625
21****************       3958.5694117603707127
22****************      87087.5270587281556800
23****************    2003012.1223507476970553
24****************   48072289.9364179447293282
25**************** 1201807247.4104485511779785

Evaluation of Siegfried Rump's function of 1988 where F(77617,33096) = -0.827396059946821
Single precision: -1.1805916E+21
Double precision: -1.1805916E+21
```

None of the results are remotely correct! In the absence of a Fortran compiler supporting still higher precision (such as quadruple, or REAL*16) only two options remain: either devise multi-word high-precision arithmetic routines and try again with even more brute-force, or, analyse the calculation with a view to finding a way to avoid the loss of accuracy with calculations conducted in the available precision.

Alternatively, do not present various intermediate results such as might give rise to doubts, nor yet entertain any doubts, just declare the answer to be what appears, and move on. In a letter from F.S. Acton, "A former student of mine now hands out millions of dollars for computation ... and he dismally estimates that 70% of the "answers" are worthless because of poor analysis and poor programming."

### On putting some thought to the matter

#### The Chaotic Bank Society

From whom but an emissary of the Dark One could come a deposit of a transcendental sum of money? Following that lead, retreat from the swamp of finite-precision arithmetic to Real arithmetic, and consider the deposit's progress in a mathematical manner:

```Year        Deposit        =     Deposit.
0        e - 1                  e - 1      Initial deposit.
1       (e - 1).1 - 1           e - 2      At the end of the first year.
2       (e - 2).2 - 1          2e - 5
3      (2e - 5).3 - 1          6e - 16
4      (6e - 16).4 - 1        24e - 65
5     (24e - 65).5 - 1       120e - 326
6    (120e - 326).6 - 1      720e - 1957
7    (720e - 1957).7 - 1    5040e - 13700
```

Clearly, two numbers that are nearly equal are being subtracted, since the value of e is a little below three. For year n, the first term is e.n! (and here a pause to gloat over the arithmetic statement evaluator written in Turbo Pascal decades back whose precedence table had specially-crafted entries for factorial, so that e*n! was not evaluated as (e*n)!) The series expression for e is straightforward: e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + ... so, for the deposit at the end of year six for example,

e.6! = 6! + 6!/1! + 6!/2! + 6!/3! + 6!/4! + 6!/5! + 6!/6! + 6!/7! + 6!/8! + 6!/9! ...

e.6! = 720 + 720 + 360 + 120 + 30 + 6 + 1 + 6!/7! + 6!/8! + 6!/9! ...

e.6! = 1957 + 6!(1/7! + 1/8! + 1/9! + ...

And obviously, the 1957 exactly cancels: so this is the difference between e and the series for e that has been truncated. Further, the remnant need not be calculated as (large number) times (small number) because the factorial terms cancel as well, so the result is

Deposit = 1/7 + 1/7.8 + 1/7.8.9 + ...

Unlike a recurrence formula whereby a new result is calculated from previous results (thereby incurring the possibility of rapid amplification of any errors), each year's value is produced ab initio via a series that is easily calculated and which converges rapidly without instability, ever more rapidly for larger n. Indeed, a one-term approximation would suffice for approximate results and in decimal the values for 9 and 99 and 999, etc. can be achieved at a glance with mental arithmetic: just over 1/10, or 1/100, or 1/1000, etc. Adding an approximate adjustment from the second term is not much more effort. No need for a thousand-digit value for e, nor any slogging through multi-precision arithmetic...

A simple function CBSERIES handles the special case deposit. The only question is how many terms of the series are required to produce a value accurate to the full precision in use. Thanks to the enquiry function EPSILON(x) offered by F90, the smallest number such that 1 + eps differs from 1 for the precision of x is available without the need for cunning programming; this is a constant. An alternative form might be that EPSILON(X) returned the smallest number that, added to X, produced a result different from X in floating-point arithmetic of the precision of X - but this would not be a constant. Since the terms of the series are rapidly diminishing (and all are positive) a new term may be too small to affect the sum; this happens when S + T = S, or 1 + T/S = 1 + eps, thus the test in CBSERIES of T/S >= EPSILON(S) checks that the term affected the sum so that the loop stops for the first term that does not.

A misthimk had TINY(S) instead of EPSILON(S), and this demonstrates again the importance of providing output that shows the actual behaviour of a scheme and comparing it to expectations, since it showed that over a hundred terms were being calculated and the last term was tiny. Routine TINY(S) reports the smallest possible floating-point number in the precision of its parameter, which is not what is wanted! EPSILON(S) is tiny, but not so tiny as TINY(S). 2·220446049250313E-016 instead of 2·225073858507201E-308.
`      SUBROUTINE CBS	!The Chaotic Bank Society.       INTEGER YEAR	!A stepper.       REAL*4 V		!The balance.       REAL*8 W		!In double precision as well.       INTEGER NTERM	!Share information with CBSERIES.       REAL*8 T		!So as to show workings.        V = 1; W = 1		!Initial values, without dozy 1D0 stuff.        V = EXP(V) - 1		!Actual initial value desired is e - 1..,        W = EXP(W) - 1		!This relies on double-precision W selecting DEXP.        WRITE (6,1)		!Here we go.    1   FORMAT (///"The Chaotic Bank Society in action...",/,     *   "Year",9X,"Real*4",22X,"Real*8",12X,"Series summation",     *   9X,"Last term",2X,"Terms.")        WRITE (6,2) 0,V,W,CBSERIES(0),T,NTERM	!Show the initial deposit.    2   FORMAT (I3,F16.7,2F28.16,D18.8,I7)	!Not quite 16-digit precision for REAL*8.        DO YEAR = 1,25		!Step through some years.          V = V*YEAR - 1	!The specified procedure.          W = W*YEAR - 1	!The compiler handles type conversions.          WRITE (6,2) YEAR,V,W,CBSERIES(YEAR),T,NTERM	!The current balance.        END DO			!On to the following year.        CONTAINS		!An alternative.        REAL*8 FUNCTION CBSERIES(N)	!Calculates for the special deposit of e - 1.         INTEGER N	!Desire the balance after year N for the deposit of e - 1.         REAL*8 S	!Via a series summation.          S = 0			!Start the summation.          T = 1			!First term is 1/(N + 1)          I = N			!Second is 1/[(N + 1)*(N + 2)], etc.          NTERM = 0		!No terms so far.    3       NTERM = NTERM + 1	!Here we go.            I = I + 1		!Thus advance to the next divisor, and not divide by zero.            T = T/I		!Thus not compute the products from scratch each time.            S = S + T		!Add the term.            IF (T/S .GE. EPSILON(S)) GO TO 3	!If they're still making a difference, another.          CBSERIES = S		!Convergence is ever-faster as N increases.        END FUNCTION CBSERIES	!So this is easy.      END SUBROUTINE CBS	!Madness! `

And the output is (slightly decorated to show correct digits in bold):

```The Chaotic Bank Society in action...
Year         Real*4                      Real*8            Series summation         Last term  Terms.
0       1.7182819          1.7182818284590453          1.7182818284590455    0.15619207D-15     18
1       0.7182819          0.7182818284590453          0.7182818284590450    0.15619207D-15     17
2       0.4365637          0.4365636569180906          0.4365636569180904    0.16441270D-16     17
3       0.3096912          0.3096909707542719          0.3096909707542714    0.49323811D-16     16
4       0.2387648          0.2387638830170875          0.2387638830170856    0.98647623D-17     16
5       0.1938238          0.1938194150854375          0.1938194150854282    0.23487529D-17     16
6       0.1629429          0.1629164905126252          0.1629164905125695    0.14092518D-16     15
7       0.1406002          0.1404154335883767          0.1404154335879862    0.44839829D-17     15
8       0.1248016          0.1233234687070137          0.1233234687038897    0.15596462D-17     15
9       0.1232147          0.1099112183631235          0.1099112183350076    0.14036816D-16     14
10       0.2321472          0.0991121836312345          0.0991121833500754    0.58486733D-17     14
11       1.5536194          0.0902340199435798          0.0902340168508295    0.25734163D-17     14
12      17.6434326          0.0828082393229579          0.0828082022099543    0.11877306D-17     14
13     228.3646240          0.0765071111984525          0.0765066287294056    0.15440498D-16     13
14    3196.1047363          0.0710995567783357          0.0710928022116781    0.80061839D-17     13
15   47940.5703125          0.0664933516750352          0.0663920331751714    0.42890271D-17     13
16  767048.1250000          0.0638936268005637          0.0622725308027424    0.23663598D-17     13
1713039817.0000000          0.0861916556095821          0.0586330236466206    0.13409372D-17     13
18****************          0.5514498009724775          0.0553944256391715    0.77860870D-18     13
19****************          9.4775462184770731          0.0524940871442588    0.46229891D-18     13
20****************        188.5509243695414625          0.0498817428851763    0.92459783D-17     12
21****************       3958.5694117603707127          0.0475166005887012    0.58838044D-17     12
22****************      87087.5270587281556800          0.0453652129514256    0.38071675D-17     12
23****************    2003012.1223507476970553          0.0433998978827887    0.25018530D-17     12
24****************   48072289.9364179447293282          0.0415975491869292    0.16679020D-17     12
25**************** 1201807247.4104485511779785          0.0399387296732302    0.11269608D-17     12
```

## Go

`package main import (    "fmt"    "math/big") func main() {    sequence()    bank()    rump()} func sequence() {    // exact computations using big.Rat    var v, v1 big.Rat    v1.SetInt64(2)    v.SetInt64(-4)    n := 2    c111 := big.NewRat(111, 1)    c1130 := big.NewRat(1130, 1)    c3000 := big.NewRat(3000, 1)    var t2, t3 big.Rat    r := func() (vn big.Rat) {        vn.Add(vn.Sub(c111, t2.Quo(c1130, &v)), t3.Quo(c3000, t3.Mul(&v, &v1)))        return    }    fmt.Println("  n  sequence value")    for _, x := range []int{3, 4, 5, 6, 7, 8, 20, 30, 50, 100} {        for ; n < x; n++ {            v1, v = v, r()        }        f, _ := v.Float64()        fmt.Printf("%3d %19.16f\n", n, f)    }} func bank() {    // balance as integer multiples of e and whole dollars using big.Int    var balance struct{ e, d big.Int }    // initial balance    balance.e.SetInt64(1)    balance.d.SetInt64(-1)    // compute balance over 25 years    var m, one big.Int    one.SetInt64(1)    for y := 1; y <= 25; y++ {        m.SetInt64(int64(y))        balance.e.Mul(&m, &balance.e)        balance.d.Mul(&m, &balance.d)        balance.d.Sub(&balance.d, &one)    }    // sum account components using big.Float    var e, ef, df, b big.Float    e.SetPrec(100).SetString("2.71828182845904523536028747135")    ef.SetInt(&balance.e)    df.SetInt(&balance.d)    b.Add(b.Mul(&e, &ef), &df)    fmt.Printf("Bank balance after 25 years: \$%.2f\n", &b)} func rump() {    a, b := 77617., 33096.    fmt.Printf("Rump f(%g, %g): %g\n", a, b, f(a, b))} func f(a, b float64) float64 {    // computations done with big.Float with enough precision to give    // a correct answer.    fp := func(x float64) *big.Float { return big.NewFloat(x).SetPrec(128) }    a1 := fp(a)    b1 := fp(b)    a2 := new(big.Float).Mul(a1, a1)    b2 := new(big.Float).Mul(b1, b1)    b4 := new(big.Float).Mul(b2, b2)    b6 := new(big.Float).Mul(b2, b4)    b8 := new(big.Float).Mul(b4, b4)    two := fp(2)    t1 := fp(333.75)    t1.Mul(t1, b6)    t21 := fp(11)    t21.Mul(t21.Mul(t21, a2), b2)    t23 := fp(121)    t23.Mul(t23, b4)    t2 := new(big.Float).Sub(t21, b6)    t2.Mul(a2, t2.Sub(t2.Sub(t2, t23), two))    t3 := fp(5.5)    t3.Mul(t3, b8)    t4 := new(big.Float).Mul(two, b1)    t4.Quo(a1, t4)    s := new(big.Float).Add(t1, t2)    f64, _ := s.Add(s.Add(s, t3), t4).Float64()    return f64}`
Output:
```  n  sequence value
3 18.5000000000000000
4  9.3783783783783790
5  7.8011527377521617
6  7.1544144809752490
7  6.8067847369236327
8  6.5926327687044388
20  6.0435521101892693
30  6.0067860930312058
50  6.0001758466271875
100  6.0000000193194776
Bank balance after 25 years: \$0.04
Rump f(77617, 33096): -0.8273960599468214
```

## Icon and Unicon

Icon and Unicon support large integers. Used for scaling the intermediates. Task 1 includes an extra step, 200 iterations, to demonstrate a closer convergence.

`## Pathological floating point problems#procedure main()    sequence()    chaotic()end ## First task, sequence convergence#link printfprocedure sequence()     local l := [2, -4]     local iters := [3, 4, 5, 6, 7, 8, 20, 30, 50, 100, 200]     local i, j, k       local n := 1      write("Sequence convergence")     # Demonstrate the convergence problem with various precision values     every k := (100 | 300) do {         n := 10^k         write("\n", k, " digits of intermediate precision")          # numbers are scaled up using large integer powers of 10         every i := !iters do {             l := [2 * n, -4 * n]             printf("i: %3d", i)              every j := 3 to i do {                 # build out a list of intermediate passes                 # order of scaling operations matters                 put(l, 111 * n - (1130 * n * n / l[j - 1]) +                        (3000 * n * n * n / (l[j - 1] * l[j - 2])))             }             # down scale the result to a real             # some precision may be lost in the final display             printf(" %20.16r\n", l[i] * 1.0 / n)         }     }end ## Task 2, chaotic bank of Euler#procedure chaotic()    local euler, e, scale, show, y, d     write("\nChaotic Banking Society of Euler")    # format the number for listing, string form, way overboard on digits    euler :="2718281828459045235360287471352662497757247093699959574966967627724076630353_  547594571382178525166427427466391932003059921817413596629043572900334295260_  595630738132328627943490763233829880753195251019011573834187930702154089149_  934884167509244761460668082264800168477411853742345442437107539077744992069_  551702761838606261331384583000752044933826560297606737113200709328709127443_  747047230696977209310141692836819025515108657463772111252389784425056953696_  770785449969967946864454905987931636889230098793127736178215424999229576351_  482208269895193668033182528869398496465105820939239829488793320362509443117_  301238197068416140397019837679320683282376464804295311802328782509819455815_  301756717361332069811250996181881593041690351598888519345807273866738589422_  879228499892086805825749279610484198444363463244968487560233624827041978623_  209002160990235304369941849146314093431738143640546253152096183690888707016_  768396424378140592714563549061303107208510383750510115747704171898610687396_  9655212671546889570350354"     # precise math with long integers, string form just for pretty listing    e := integer(euler)     # 1000 digits after the decimal for scaling intermediates and service fee    scale := 10^1000     # initial deposit - \$1    d := e - scale     # show balance with 16 digits    show := 10^16    write("Starting balance:       \$", d * show / scale * 1.0 / show, "...")     # wait 25 years, with only a trivial \$1 service fee    every y := 1 to 25 do {        d := d * y - scale    }     # show final balance with 4 digits after the decimal (truncation)    show := 10^4    write("Balance after ", y, " years: \$", d * show / scale * 1.0 / show)end`
Output:
```prompt\$ time unicon -s patho.icn -x
Sequence convergence

100 digits of intermediate precision
i:   3  18.5000000000000000
i:   4   9.3783783783783790
i:   5   7.8011527377521620
i:   6   7.1544144809752490
i:   7   6.8067847369236330
i:   8   6.5926327687044380
i:  20   6.0435521101892680
i:  30   6.0067860930312060
i:  50   6.0001758466271870
i: 100  99.9999999999998400
i: 200 100.0000000000000000

300 digits of intermediate precision
i:   3  18.5000000000000000
i:   4   9.3783783783783790
i:   5   7.8011527377521610
i:   6   7.1544144809752490
i:   7   6.8067847369236320
i:   8   6.5926327687044380
i:  20   6.0435521101892680
i:  30   6.0067860930312060
i:  50   6.0001758466271870
i: 100   6.0000000193194780
i: 200   6.0000000000000000

Chaotic Banking Society of Euler
Starting balance:       \$1.718281828459045...
Balance after 25 years: \$0.0399

real    0m0.075s
user    0m0.044s
sys     0m0.020s```

## J

A sequence that seems to converge to a wrong limit.

Implementation of `vn`:

`   vn=: 111 +(_1130 % _1&{) + (3000 % _1&{ * _2&{)`

Example using IEEE-754 floating point:

`   3 21j16 ":"1] 3 4 5 6 7 8 20 30 50 100 ([,.{) (,vn)^:100(2 _4)  3   9.3783783783783861  4   7.8011527377522611  5   7.1544144809765555  6   6.8067847369419638  7   6.5926327689743687  8   6.4494659378910058 20  99.9934721906444960 30 100.0000000000000000 50 100.0000000000000000100 100.0000000000000000`

Example using exact arithmetic:

`   3 21j16 ":"1] 3 4 5 6 7 8 20 30 50 100 ([,.{) (,vn)^:100(2 _4x)  3   9.3783783783783784  4   7.8011527377521614  5   7.1544144809752494  6   6.8067847369236330  7   6.5926327687044384  8   6.4494659337902880 20   6.0360318810818568 30   6.0056486887714203 50   6.0001465345613879100   6.0000000160995649`

The Chaotic Bank Society

Let's start this example by using exact arithmetic, to make sure we have the right algorithm. We'll go a bit overboard, in representing e, so we don't have to worry too much about that.

`   e=: +/%1,*/\1+i.100x   81j76":e   2.7182818284590452353602874713526624977572470936999595749669676277240766303535   21j16":+`*/,_1,.(1+i.-25),e   0.0399387296732302`

(Aside: here, we are used the same mechanism for adding -1 to e that we are using to add -1 to the product of the year number and the running balance.)

Next, we will use ${\displaystyle 6157974361033\div 2265392166685}$ for e, to represent the limit of what can be expressed using 64 bit IEEE 754 floating point.

`   31j16":+`*/,_1,.(1+i.-25),6157974361033%2265392166685x_2053975868590.1852178761057505`

That's clearly way too low, so let's try instead using ${\displaystyle (1+6157974361033)\div 2265392166685}$ for e

`   31j16":+`*/,_1,.(1+i.-25),6157974361034%2265392166685x 4793054977300.3491517765983265`

So, our problem seems to be that there's no way we can express enough bits of e, using 64 bit IEEE-754 floating point arithmetic. Just to confirm:

`   1x12.71828   +`*/,_1,.(1+i.-25),1x1_2.24237e9`

Now let's take a closer look using our rational approximation for e:

`   21j16":+`*/,_1,.(1+i.-25),+/%1,*/\1+i.40x   0.0399387296732302   21j16":+`*/,_1,.(1+i.-25),+/%1,*/\1+i.30x   0.0399387277260840   21j16":+`*/,_1,.(1+i.-25),+/%1,*/\1+i.26x   0.0384615384615385   21j16":+`*/,_1,.(1+i.-25),+/%1,*/\1+i.25x   0.0000000000000000   21j16":+`*/,_1,.(1+i.-25),+/%1,*/\1+i.24x  _1.0000000000000000`

Things go haywire when our approximation for e uses the same number of terms as our bank's term. So, what does that look like, in terms of precision?

`   41j36":+/%1,*/\1+i.26x   2.718281828459045235360287471257428715   41j36":+/%1,*/\1+i.25x   2.718281828459045235360287468777832452   41j36":+/%1,*/\1+i.24x   2.718281828459045235360287404308329608`

In other words, we go astray when our approximation for e is inaccurate in the 26th position after the decimal point. But IEEE-754 floating point arithmetic can only represent approximately 16 decimal digits of precision.

Siegfried Rump's example.

Again, we use exact arithmetic to see if we have the algorithm right. That said, we'll also do this in small steps, to make sure we're being exact every step of the way, and to keep from building overly long lines:

`rump=:4 :0  NB. enforce exact arithmetic  add=. +&x:  sub=. -&x:  mul=. *&x:  div=. %&x:   a=. x  a2=. a mul a   b=. y  b2=. b mul b  b4=. b2 mul b2  b6=. b2 mul b4  b8=. b4 mul b4   c333_75=. 1335 div 4 NB. 333.75  term1=. c333_75 mul b6   t11a2b2=. 11 mul a2 mul b2  tnb6=. 0 sub b6  tn121b4=. 0 sub 121 mul b4  term2=. a2*(t11a2b2 + tnb6 + tn121b4 sub 2)   c5_5=. 11 div 2 NB. 5.5  term3=. c5_5 mul b8   term4=. a div 2 mul b   term1 add term2 add term3 add term4)`

Example use:

`   21j16": 77617 rump 33096  _0.8273960599468214`

Note that replacing the definitions of `add`, `sub`, `div`, `mul` with implementations which promote to floating point gives a very different result:

`   77617 rump 33096_1.39061e21`

But given that b8 is

`   33096^81.43947e36`

we're exceeding the limits of our representation here, if we're using 64 bit IEEE-754 floating point arithmetic.

## Java

Uses BigRational class: Arithmetic/Rational/Java. For comparison purposes, each task is also implemented with standard 64-bit floating point numbers.

`import java.math.BigDecimal;import java.math.RoundingMode; public class FPProblems {    public static void wrongConvergence() {        int[] INDEXES = new int[] { 3, 4, 5, 6, 7, 8, 20, 30, 50, 100 };         // Standard 64-bit floating point        double[] fpValues = new double[100];        fpValues[0] = 2.0;        fpValues[1] = -4.0;        for (int i = 2; i < fpValues.length; i++) {            fpValues[i] = 111.0 - 1130.0 / fpValues[i - 1] + 3000.0 / (fpValues[i - 1] * fpValues[i - 2]);        }         // Using rational representation        BigRational[] brValues = new BigRational[100];        brValues[0] = BigRational.valueOf(2);        brValues[1] = BigRational.valueOf(-4);        for (int i = 2; i < brValues.length; i++) {            // Using intermediate values for better readability            BigRational clause2 = BigRational.valueOf(1130).divide(brValues[i - 1]);            BigRational clause3 = BigRational.valueOf(3000).divide(brValues[i - 1].multiply(brValues[i - 2]));            brValues[i] = BigRational.valueOf(111).subtract(clause2).add(clause3);        }         System.out.println("Wrong Convergence Sequence");        for (int n : INDEXES) {            BigDecimal value = brValues[n - 1].toBigDecimal(16, RoundingMode.HALF_UP);            System.out.println("  For index " + n + ", FP value is " + fpValues[n - 1] + ", and rounded BigRational value is " + value.toPlainString());        }         return;    }     public static void chaoticBankSociety() {        System.out.println("Chaotic Bank Society");        double balance = Math.E - 1.0;         // Calculate e using first 1000 terms of the reciprocal of factorials formula        BigRational e = BigRational.ONE;        BigRational d = BigRational.ONE;        for (int i = 1; i < 1000; i++) {            d = d.multiply(BigRational.valueOf(i));            e = e.add(d.reciprocal());        }        System.out.println("DEBUG: e=" + e.toBigDecimal(100, RoundingMode.HALF_UP).toPlainString());         // Alternatively,        // BigRational e = BigRational.valueOf(Math.E);         BigRational brBalance = e.subtract(BigRational.ONE);        for (int year = 1; year <= 25; year++) {            balance = (balance * year) - 1.0;            brBalance = brBalance.multiply(BigRational.valueOf(year)).subtract(BigRational.ONE);            BigDecimal bdValue = brBalance.toBigDecimal(16, RoundingMode.HALF_UP);            System.out.println("  Year=" + year + ", FP balance=" + balance + ", BigRational balance=" + bdValue.toPlainString());        }    }     public static void siegfriedRump() {        System.out.println("Siegfried Rump formula");        double fpValue;        {            double a = 77617.0;            double b = 33096.0;            fpValue = 333.75 * Math.pow(b, 6) + a * a * (11.0 * a * a * b * b - Math.pow(b, 6) - 121.0 * Math.pow(b, 4) - 2.0) + 5.5 * Math.pow(b, 8) + a / (2.0 * b);        }         BigRational brValue;        {            BigRational a = BigRational.valueOf(77617);            BigRational b = BigRational.valueOf(33096);            BigRational clause1 = BigRational.valueOf(333.75).multiply(b.pow(6));            BigRational clause2a = BigRational.valueOf(11).multiply(a).multiply(a).multiply(b).multiply(b);            BigRational clause2b = b.pow(6).add(BigRational.valueOf(121).multiply(b.pow(4))).add(BigRational.valueOf(2));            BigRational clause2 = a.multiply(a).multiply(clause2a.subtract(clause2b));            BigRational clause3 = BigRational.valueOf(5.5).multiply(b.pow(8));            BigRational clause4 = a.divide(b.multiply(BigRational.valueOf(2)));            brValue = clause1.add(clause2).add(clause3).add(clause4);        }         System.out.println("  FP value is " + fpValue);        System.out.println("  BigRational rounded value is " + brValue.toBigDecimal(64, RoundingMode.HALF_UP).toPlainString());        System.out.println("  BigRational full value is " + brValue.toString());    }     public static void main(String... args) {        wrongConvergence();         System.out.println();        chaoticBankSociety();         System.out.println();        siegfriedRump();    }}`
Output:
```Wrong Convergence Sequence
For index 3, FP value is 18.5, and rounded BigRational value is 18.50000000000000
For index 4, FP value is 9.378378378378379, and rounded BigRational value is 9.378378378378378
For index 5, FP value is 7.801152737752169, and rounded BigRational value is 7.801152737752161
For index 6, FP value is 7.154414480975333, and rounded BigRational value is 7.154414480975249
For index 7, FP value is 6.806784736924811, and rounded BigRational value is 6.806784736923633
For index 8, FP value is 6.592632768721792, and rounded BigRational value is 6.592632768704438
For index 20, FP value is 98.34950312216536, and rounded BigRational value is 6.043552110189269
For index 30, FP value is 99.99999999999893, and rounded BigRational value is 6.006786093031206
For index 50, FP value is 100.0, and rounded BigRational value is 6.000175846627187
For index 100, FP value is 100.0, and rounded BigRational value is 6.000000019319478

Chaotic Bank Society
DEBUG: e=2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427
Year=1, FP balance=0.7182818284590451, BigRational balance=0.7182818284590452
Year=2, FP balance=0.4365636569180902, BigRational balance=0.4365636569180905
Year=3, FP balance=0.30969097075427054, BigRational balance=0.3096909707542714
Year=4, FP balance=0.23876388301708218, BigRational balance=0.2387638830170857
Year=5, FP balance=0.1938194150854109, BigRational balance=0.1938194150854282
Year=6, FP balance=0.16291649051246537, BigRational balance=0.1629164905125695
Year=7, FP balance=0.1404154335872576, BigRational balance=0.1404154335879862
Year=8, FP balance=0.12332346869806088, BigRational balance=0.1233234687038897
Year=9, FP balance=0.1099112182825479, BigRational balance=0.1099112183350075
Year=10, FP balance=0.09911218282547907, BigRational balance=0.09911218335007541
Year=11, FP balance=0.09023401108026974, BigRational balance=0.09023401685082952
Year=12, FP balance=0.08280813296323686, BigRational balance=0.08280820220995428
Year=13, FP balance=0.07650572852207915, BigRational balance=0.07650662872940558
Year=14, FP balance=0.07108019930910814, BigRational balance=0.07109280221167809
Year=15, FP balance=0.06620298963662208, BigRational balance=0.06639203317517140
Year=16, FP balance=0.05924783418595325, BigRational balance=0.06227253080274239
Year=17, FP balance=0.007213181161205284, BigRational balance=0.05863302364662064
Year=18, FP balance=-0.8701627390983049, BigRational balance=0.05539442563917152
Year=19, FP balance=-17.533092042867793, BigRational balance=0.05249408714425881
Year=20, FP balance=-351.66184085735586, BigRational balance=0.04988174288517625
Year=21, FP balance=-7385.898658004473, BigRational balance=0.04751660058870116
Year=22, FP balance=-162490.7704760984, BigRational balance=0.04536521295142560
Year=23, FP balance=-3737288.7209502636, BigRational balance=0.04339989788278872
Year=24, FP balance=-8.969493030280632E7, BigRational balance=0.04159754918692921
Year=25, FP balance=-2.242373258570158E9, BigRational balance=0.03993872967323021

Siegfried Rump formula
FP value is -1.1805916207174113E21
BigRational rounded value is -0.8273960599468213681411650954798162919990331157843848199178148417
BigRational full value is -54767/66192```

## jq

jq uses IEEE 754 64-bit numbers so it is easy to illustrate the pathological nature of the series. This is shown in the following implementation of the first problem. The second problem is solved using symbolic arithmetic.

### v series

The following implementation illustrates how a cache can be used in jq to avoid redundant computations. A JSON object is used as the cache.

`# Input: the value at which to compute vdef v:  # Input: cache  # Output: updated cache  def v_(n):    (n|tostring) as \$s    | . as \$cache    | if (\$cache | has(\$s)) then .      else if n == 1 then \$cache["1"] = 2           elif n == 2 then \$cache["2"] = -4           else (\$cache | v_(n-1) | v_(n - 2)) as \$new           | \$new[(n-1)|tostring] as \$x 	   | \$new[(n-2)|tostring] as \$y           | \$new + {(\$s):  ((111 - (1130 / \$x) + (3000 / (\$x * \$y)))) }	   end       end;    . as \$m | {} | v_(\$m) | .[(\$m|tostring)] ; `
Example:
`(3,4,5,6,7,8,20,30,50,100) | v`
Output:
```18.5
9.378378378378379
7.801152737752169
7.154414480975333
6.806784736924811
6.592632768721792
98.34950312216536
99.99999999999893
100
100```

### Chaotic Bank Society

To avoid the pathological issues, the following uses symbolic arithmetic, with {"e": m, "c": n} representing (e*m + n).

`# Given the balance in the prior year, compute the new balance in year n.# Input: { e: m, c: n } representing m*e + ndef new_balance(n):  if n == 0 then {e: 1, c: -1}  else {e: (.e * n), c: (.c * n - 1) }  end; def balance(n):  def e: 1|exp;  reduce range(0;n) as \$i ({}; new_balance(\$i) )  | (.e * e) + .c;`
Example:
`balance(25)`
Output:
```   0
```

Well, at least that's a reasonable approximation to the value represented by {"e":620448401733239400000000,"c":-1686553615927922300000000}

## Julia

Works with: Julia version 0.6

The task could be completed even using ```Rational{BigFloat}``` type.

`# arbitrary precisionsetprecision(2000) # Task 1function seq(n)    len = maximum(n)    r = Vector{BigFloat}(len)    r[1] = 2    if len > 1 r[2] = -4 end     for i in 3:len        r[i] = 111 - 1130 / r[i-1] + 3000 / (r[i-1] * r[i-2])    end     return r[n]end n = [1, 2, 3, 5, 10, 100]v = seq(n)println("Task 1 - Sequence convergence:\n", join((@sprintf("v%-3i = %23.20f", i, s) for (i, s) in zip(n, v)), '\n')) # Task 2: solution with big float (precision can be set with setprecision function)function chaoticbankfund(years::Integer)    balance = big(e) - 1    for y in 1:years        balance = (balance * y) - 1    end     return balanceend println("\nTask 2 - Chaotic Bank fund after 25 years:\n", @sprintf "%.20f" chaoticbankfund(25)) # Task 3: solution with big floatf(a::Union{BigInt,BigFloat}, b::Union{BigInt,BigFloat}) =    333.75b ^ 6 + a ^ 2 * ( 11a ^ 2 * b ^ 2 - b ^ 6 - 121b ^ 4 - 2 ) + 5.5b ^ 8 + a / 2b println("\nTask 3 - Siegfried Rump's example:\nf(77617.0, 33096.0) = ", @sprintf "%.20f" f(big(77617.0), big(33096.0)))`
Output:
```Task 1 - Sequence convergence:
v1   =  2.00000000000000000000
v2   = -4.00000000000000000000
v3   = 18.50000000000000000000
v5   =  7.80115273775216138329
v10  =  6.34845205665435714710
v100 =  6.00000001931947792910

Task 2 - Chaotic Bank fund after 25 years:
0.03993872967323020890

Task 3 - Siegfried Rump's example:
f(77617.0, 33096.0) = -0.82739605994682136814```

## Kotlin

`// version 1.0.6 import java.math.* const val LIMIT = 100 val con480  = MathContext(480)val bigTwo =  BigDecimal(2)val bigE    = BigDecimal("2.71828182845904523536028747135266249775724709369995") // precise enough! fun main(args: Array<String>) {    // v(n) sequence task    val c1 = BigDecimal(111)    val c2 = BigDecimal(1130)    val c3 = BigDecimal(3000)    var v1 = bigTwo    var v2 = BigDecimal(-4)    var v3:  BigDecimal    for (i in 3 .. LIMIT) {        v3 = c1 - c2.divide(v2, con480) + c3.divide(v2 * v1, con480)        println("\${"%3d".format(i)} : \${"%19.16f".format(v3)}")        v1 = v2        v2 = v3    }     // Chaotic Building Society task    var balance = bigE - BigDecimal.ONE    for (year in 1..25) balance = balance.multiply(BigDecimal(year), con480) - BigDecimal.ONE    println("\nBalance after 25 years is \${"%18.16f".format(balance)}")     // Siegfried Rump task    val a  = BigDecimal(77617)    val b  = BigDecimal(33096)    val c4 = BigDecimal("333.75")    val c5 = BigDecimal(11)    val c6 = BigDecimal(121)    val c7 = BigDecimal("5.5")    var f  = c4 * b.pow(6, con480) + c7 * b.pow(8, con480) + a.divide(bigTwo * b, con480)    val c8 = c5 * a.pow(2, con480) * b.pow(2, con480) - b.pow(6, con480) - c6 * b.pow(4, con480) - bigTwo    f += c8 * a.pow(2, con480)    println("\nf(77617.0, 33096.0) is \${"%18.16f".format(f)}") }`
Output:
```  3 : 18.5000000000000000
4 :  9.3783783783783784
5 :  7.8011527377521614
6 :  7.1544144809752494
7 :  6.8067847369236330
8 :  6.5926327687044384
9 :  6.4494659337902880
10 :  6.3484520566543571
11 :  6.2744385982163279
12 :  6.2186957398023978
13 :  6.1758373049212301
14 :  6.1423590812383559
15 :  6.1158830665510808
16 :  6.0947394393336811
17 :  6.0777223048472427
18 :  6.0639403224998088
19 :  6.0527217610161522
20 :  6.0435521101892689
21 :  6.0360318810818568
22 :  6.0298473250239019
23 :  6.0247496523668479
24 :  6.0205399840615161
25 :  6.0170582573289876
26 :  6.0141749145508190
27 :  6.0117845878713337
28 :  6.0098012392984846
29 :  6.0081543789122289
30 :  6.0067860930312058
31 :  6.0056486887714203
32 :  6.0047028131881752
33 :  6.0039159416664605
34 :  6.0032611563057406
35 :  6.0027161539543513
36 :  6.0022624374405593
37 :  6.0018846538818819
38 :  6.0015700517342190
39 :  6.0013080341649643
40 :  6.0010897908901841
41 :  6.0009079941545271
42 :  6.0007565473053508
43 :  6.0006303766028389
44 :  6.0005252586505718
45 :  6.0004376772265183
46 :  6.0003647044182955
47 :  6.0003039018761868
48 :  6.0002532387368678
49 :  6.0002110233741743
50 :  6.0001758466271872
51 :  6.0001465345613879
52 :  6.0001221091522881
53 :  6.0001017555560260
54 :  6.0000847948586303
55 :  6.0000706613835716
56 :  6.0000588837928413
57 :  6.0000490693458029
58 :  6.0000408907870884
59 :  6.0000340754236785
60 :  6.0000283960251310
61 :  6.0000236632422855
62 :  6.0000197192908008
63 :  6.0000164326883272
64 :  6.0000136938694348
65 :  6.0000114115318177
66 :  6.0000095095917616
67 :  6.0000079246472413
68 :  6.0000066038639788
69 :  6.0000055032139253
70 :  6.0000045860073981
71 :  6.0000038216699107
72 :  6.0000031847228971
73 :  6.0000026539343389
74 :  6.0000022116109709
75 :  6.0000018430084630
76 :  6.0000015358399141
77 :  6.0000012798662675
78 :  6.0000010665549954
79 :  6.0000008887956715
80 :  6.0000007406629499
81 :  6.0000006172190487
82 :  6.0000005143491543
83 :  6.0000004286242585
84 :  6.0000003571868566
85 :  6.0000002976556961
86 :  6.0000002480464011
87 :  6.0000002067053257
88 :  6.0000001722544322
89 :  6.0000001435453560
90 :  6.0000001196211272
91 :  6.0000000996842706
92 :  6.0000000830702242
93 :  6.0000000692251858
94 :  6.0000000576876542
95 :  6.0000000480730447
96 :  6.0000000400608703
97 :  6.0000000333840583
98 :  6.0000000278200485
99 :  6.0000000231833736
100 :  6.0000000193194779

Balance after 25 years is 0.0399387296732302

f(77617.0, 33096.0) is -0.8273960599468214
```

## Mathematica

`v[1] = 2;v[2] = -4;v[n_] := Once[111 - 1130/v[n - 1] + 3000/(v[n - 1]*v[n - 2])]N[Map[v, {3, 4, 5, 6, 7, 8, 20, 30, 50, 100}], 80]`
Output:
```{18.500000000000000000000000000000000000000000000000000000000000000000000000000000,
9.3783783783783783783783783783783783783783783783783783783783783783783783783783784,
7.8011527377521613832853025936599423631123919308357348703170028818443804034582133,
7.1544144809752493535278906538603620243812338381972663465090506095308459549316587,
6.8067847369236329839417565962720090876232767078019311199463004079103629885888367,
6.5926327687044383927420027763659948265529823177346067194125634354115621230855591,
6.0435521101892688677774773640975401331877150000061201379728002521382151385271029,
6.0067860930312057585305540479532397058330723144383667645482877308904928243847153,
6.0001758466271871889456140207471954695237351770993318409845704023663691405177107,
6.0000000193194779291040868034035857150243506754369524580725927508565217672302663}```

`year = 1; N[Nest[# year++ - 1 &, E - 1, 25], 30]`
Output:
`0.0399387296732302089036714552104`

`f[a_, b_] := 333.75`100 b^6 + a^2 (11 a^2 b^2 - b^6 - 121 b^4 - 2) + 5.5`100 b^8 + a/(2 b)f[77617, 33096]`
Output:
`-0.827396059946821368141165095479816291999033115784384819917814842`

## PARI/GP

Task 1: Define recursive function V(n):

`V(n,a=2,v=-4.)=if(n < 3,return(v));V(n--,v,111-1130/v+3000/(v*a))`

In order to work set precision to at least 200 digits:

```\p 200: realprecision = 211 significant digits (200 digits displayed)

V(50):  6.000175846627187188945614020747195469523735177...
V(100): 6.0000000193194779291040868034035857150243506754369524580725927508565217672302663412282...
```

`balance(d,y)=d--;for(n=1,y,d=d*n-1);d`

Output balance(exp(1), 25):

`0.039938729673230208903...`

`f(a,b)=333.75*b^6+a*a*(11*a*a*b*b-b^6-121*b^4-2)+5.5*b^8+a/(2*b)`
Output:
`f(77617.0,33096.0): -0.827396059946821368141165...`

## Perl

All three tasks can be solved by using either the `bigrat` or `bignum` core modules. Both approaches are used, as demonstration.

#### Convergent series

The constants in the equation must be represented with a decimal point (even just `111.`) so that they are treated as rationals, not integers.

`use bigrat; @s = qw(2, -4);for my \$n (2..99) {    \$s[\$n]= 111.0 - 1130.0/\$s[-1] + 3000.0/(\$s[-1]*\$s[-2]);} for \$n (3..8, 20, 30, 35, 50, 100) {    (\$nu,\$de) = \$s[\$n-1] =~ m#^(\d+)/(\d+)#;;    printf "n = %3d %18.15f\n", \$n, \$nu/\$de;}`
Output:
```n =   3 18.500000000000000
n =   4  9.378378378378379
n =   5  7.801152737752162
n =   6  7.154414480975249
n =   7  6.806784736923633
n =   8  6.592632768704439
n =  20  6.043552110189269
n =  30  6.006786093031206
n =  35  6.002716153954351
n =  50  6.000175846627188
n = 100  6.000000019319478```

#### Chaotic bank society

The value of 𝑒 is imported from a module, but could also be calculated, as is done in Calculating the value of e

`use bignum qw(bexp);\$e       = bexp(1,43);\$years   = 25;\$balance = \$e - 1; print "Starting balance, \\$(e-1): \\$\$balance\n";for \$i (1..\$years) { \$balance = \$i * \$balance - 1 }printf "After year %d, you will have \\$%1.15g in your account.\n", \$years, \$balance;`
Output:
```Starting balance, \$(e-1): \$1.718281828459045235360287471352662497757247
After year 25, you will have \$0.0399387296732302 in your account.```

#### Rump's example

`use bignum; \$a = 77617;\$b = 33096;printf "%0.16f\n", 333.75*\$b**6 + \$a**2 * ( 11*\$a**2 * \$b**2 - \$b**6 - 121*\$b**4 - 2) + 5.5*\$b**8 + \$a/(2*\$b);`
Output:
`-0.8273960599468214`

## Perl 6

Works with: Rakudo version 2016-01

The simple solution to doing calculations where floating point numbers exhibit pathological behavior is: don't do floating point calculations. :-) Perl 6 is just as susceptible to floating point error as any other C based language, however, it offers built-in rational Types; where numbers are represented as a ratio of two integers. For normal precision it uses Rats - accurate to 1/2^64, and for arbitrary precision, FatRats, whose denominators can grow as large as available memory. Rats don't require any special setup to use. Any decimal number within its limits of precision is automatically stored as a Rat. FatRats require explicit coercion and are "sticky". Any FatRat operand in a calculation will cause all further results to be stored as FatRats.

`say '1st: Convergent series';my @series = 2.FatRat, -4, { 111 - 1130 / \$^v + 3000 / ( \$^v * \$^u ) } ... *;for flat 3..8, 20, 30, 50, 100 -> \$n {say "n = {\$n.fmt("%3d")} @series[\$n-1]"}; say "\n2nd: Chaotic bank society";sub postfix:<!> (Int \$n) { [*] 2..\$n } # factorial operatormy \$years = 25;my \$balance = sum map { 1 / FatRat.new(\$_!) }, 1 .. \$years + 15; # Generate e-1  to sufficient precision with a Taylor seriesput "Starting balance, \\$(e-1): \\$\$balance";for 1..\$years -> \$i { \$balance = \$i * \$balance - 1 }printf("After year %d, you will have \\$%1.16g in your account.\n", \$years, \$balance); print "\n3rd: Rump's example: f(77617.0, 33096.0) = ";sub f (\a, \b) { 333.75*b⁶ + a²*( 11*a²*b² - b⁶ - 121*b⁴ - 2 ) + 5.5*b⁸ + a/(2*b) }say f(77617.0, 33096.0).fmt("%0.16g");`
Output:
```1st: Convergent series
n =   3 18.5
n =   4 9.378378
n =   5 7.801153
n =   6 7.154414
n =   7 6.806785
n =   8 6.5926328
n =  20 6.0435521101892689
n =  30 6.006786093031205758530554
n =  50 6.0001758466271871889456140207471954695237
n = 100 6.000000019319477929104086803403585715024350675436952458072592750856521767230266

2nd: Chaotic bank society
Starting balance, \$(e-1): \$1.7182818284590452353602874713526624977572470936999
After year 25, you will have \$0.0399387296732302 in your account.

3rd: Rump's example: f(77617.0, 33096.0) = -0.827396059946821
```

## Phix

Library: bigatom

Standard maths with the IEEE-754 hardware floats fails predictably (that code left in as comments), so roll out the bigatoms.
Task1: needs at least 120 decimal places to avoid serious divergance, and 196 to be accurate to 78 digits
Task2: needs at least 41 decimal places (and 42 passed as the first argument to ba_euler)
Task3: apparently only needs just 15 decimal places, then again ba_scale() defines the minimum, and it may (as in is permitted to) use more.

`include builtins\bigatom.eputs(1,"Task 1\n")constant {fns,fmts} = columnize({{3,1},{4,6},{5,6},{6,6},{7,6},{8,7},{20,16},{30,24},{50,40},{100,78}}){} = ba_scale(196)sequence v = {2,-4}for n=3 to 100 do--  v = append(v,111 - 1130/v[n-1] + 3000/(v[n-1]*v[n-2]))    v = append(v,ba_add(ba_sub(111,ba_divide(1130,v[n-1])),ba_divide(3000,ba_multiply(v[n-1],v[n-2]))))    if n<9 or find(n,{20,30,50,100}) then--      printf(1,"n = %-3d %20.16f\n", {n, v[n]})        string fmt = sprintf("%%.%dB",fmts[find(n,fns)])        printf(1,"n = %-3d %s\n", {n, ba_sprintf(fmt,v[n])})    end ifend for puts(1,"\nTask 2\n")--atom balance = exp(1)-1--for i=1 to 25 do balance = balance*i-1 end for--printf(1,"\nTask 2\nBalance after 25 years: \$%12.10f", balance){} = ba_scale(41)bigatom balance = ba_sub(ba_euler(42,true),1)for i=1 to 25 do balance = ba_sub(ba_multiply(balance,i),1) end forba_printf(1,"Balance after 25 years: \$%.16B\n\n", balance) puts(1,"Task 3\n"){} = ba_scale(15)   -- fine!integer a = 77617,        b = 33096--atom pa2 = power(a,2),--   pb2a211 = 11*pa2*power(b,2),--   pb4121 = 121*power(b,4),--   pb6 = power(b,6),--   pb855 = 5.5*power(b,8),--   f_ab = 333.75 * pb6 + pa2 * (pb2a211 - pb6 - pb4121 - 2) + pb855 + a/(2*b) --printf(1,"f(%d, %d) = %.15f\n\n", {a, b, f_ab})bigatom pa2 = ba_power(a,2),        pb2a211 = ba_multiply(11,ba_multiply(pa2,ba_power(b,2))),        pb4121 = ba_multiply(121,ba_power(b,4)),        pb6 = ba_power(b,6),        pa2mid = ba_multiply(pa2,ba_sub(ba_sub(ba_sub(pb2a211,pb6),pb4121),2)),        pb633375 = ba_multiply(333.75,pb6),        pb855 = ba_multiply(5.5,ba_power(b,8)),        f_ab = ba_add(ba_add(ba_add(pb633375,pa2mid),pb855),ba_divide(a,ba_multiply(2,b)))printf(1,"f(%d, %d) = %s\n", {a, b, ba_sprintf("%.15B",f_ab)})`

Aside: obviously you don't have to use lots of temps like that, but I find that style often makes things much easier to debug.

Output:
```Task 1
n = 3   18.5
n = 4   9.378378
n = 5   7.801153
n = 6   7.154414
n = 7   6.806785
n = 8   6.5926328
n = 20  6.0435521101892689
n = 30  6.006786093031205758530554
n = 50  6.0001758466271871889456140207471954695237
n = 100 6.000000019319477929104086803403585715024350675436952458072592750856521767230266

Balance after 25 years: \$0.0399387296732302

f(77617, 33096) = -0.827396059946821
```

## Python

Using rational numbers via standard library `fractions`

`from fractions import Fraction def muller_seq(n:int) -> float:    seq = [Fraction(0), Fraction(2), Fraction(-4)]    for i in range(3, n+1):        next_value = (111 - 1130/seq[i-1]            + 3000/(seq[i-1]*seq[i-2]))        seq.append(next_value)    return float(seq[n]) for n in [3, 4, 5, 6, 7, 8, 20, 30, 50, 100]:    print("{:4d} -> {}".format(n, muller_seq(n)))`
Output:
```   3 -> 18.5
4 -> 9.378378378378379
5 -> 7.801152737752162
6 -> 7.154414480975249
7 -> 6.806784736923633
8 -> 6.592632768704439
20 -> 6.043552110189269
30 -> 6.006786093031206
50 -> 6.0001758466271875
100 -> 6.000000019319478
```

### Task 2: The Chaotic Bank Society

Using `decimal` numbers with a high precision

`from decimal import Decimal, getcontext def bank(years:int) -> float:    """    Warning: still will diverge and return incorrect results after 250 years    the higher the precision, the more years will cover    """    getcontext().prec = 500    # standard math.e has not enough precision    e = Decimal('2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019011573834187930702154089149934884167509244761460668082264800168477411853742345442437107539077744992069551702761838606261331384583000752044933826560297606737113200709328709127443747047230696977209310141692836819025515108657463772111252389784425056953696770785449969967946864454905987931636889230098793127736178215424999229576351')    decimal_balance = e - 1    for year in range(1, years+1):        decimal_balance = decimal_balance * year - 1    return(float(decimal_balance)) print("Bank balance after 25 years = ", bank(25))`
Output:
```Bank balance after 25 years =  0.03993872967323021
```

but, still incorrectly diverging after some time, aprox. 250 years

`for year in range(200, 256, 5):    print(year, '->', bank(year)) `
Output:
```200 -> 0.004999875631110097
205 -> 0.004877933277184028
210 -> 0.004761797301186607
215 -> 0.0046510626428896236
220 -> 0.004545361061789591
225 -> 0.0044443570465329246
230 -> 0.004347744257820075
235 -> 0.004255242425346535
240 -> 0.004166594632576723
245 -> 0.004081564933953891
250 -> 0.003999846590933889
255 -> -92939.78784907148
```

### Task 3: Siegfried Rump's example

Using rational numbers via standard library `fractions`

`from fractions import Fraction def rump(generic_a, generic_b) -> float:    a = Fraction('{}'.format(generic_a))    b = Fraction('{}'.format(generic_b))    fractional_result = Fraction('333.75') * b**6 \        + a**2 * ( 11 * a**2 * b**2 - b**6 - 121 * b**4 - 2 ) \        + Fraction('5.5') * b**8 + a / (2 * b)    return(float(fractional_result))  print("rump(77617, 33096) = ", rump(77617.0, 33096.0)) `
Output:
```rump(77617, 33096) =  -0.8273960599468214
```

## Racket

Racket has the concept of exact (rational) and inexact (floating point) numbers, both real and complex. See: http://docs.racket-lang.org/guide/numbers.html for more details.

The examples below use real numbers, and the `x` function is used to transform them to floats, if desired, with the function `exact->inexact`.

`#lang racket (define current-do-exact-calculations? (make-parameter exact->inexact)) (define (x n) (if (current-do-exact-calculations?) n (exact->inexact n))) (define (decimal.18 n)  (regexp-replace #px"0+\$" (real->decimal-string n 18) "")) (define (task-1 n)  (let ((c_1 (x 111)) (c_2 (x -1130)) (c_3 (x 3000)))    (let loop ((v_n-2 (x 2)) (v_n-1 (x -4)) (n (- n 2)))      (if (= n 0) v_n-1 (loop v_n-1 (+ c_1 (/ c_2 v_n-1) (/ c_3 (* v_n-1 v_n-2))) (- n 1)))))) (define (task-2) ; chaotic bank   (define e (if (current-do-exact-calculations?)                 #e2.71828182845904523536028747135266249775724709369995                 (exp 1)))  (for/fold ((b (- e 1))) ((y (in-range 1 26))) (- (* b y) 1))) (define (task-3 a b)    (+ (* (x #e333.75) (expt b 6))       (* (expt a 2) (- (* 11 (expt a 2) (expt b 2)) (expt b 6) (* 121 (expt b 4)) 2))       (* (x #e5.5) (expt b 8))       (/ a (* b 2)))) (define (all-tests)  (let ((classic-sum (+ (x #e0.2) (x #e0.1))))    (printf "Classic example: ~a = ~a~%" classic-sum (decimal.18 classic-sum)))   (displayln "TASK 1")  (for ((n (in-list '(3 4 5 6 7 8 20 30 50 100))))    (printf "n=~a\t~a~%" n (decimal.18 (task-1 n))))   (printf "TASK 2: balance after 25 years = ~a~%" (decimal.18 (task-2)))   (let ((t3 (task-3 77617 33096)))    (printf "TASK 3: f(77617, 33096) = ~a = ~a~%" t3 (decimal.18 t3)))) (module+ main  (displayln "INEXACT (Floating Point) NUMBERS")  (parameterize ([current-do-exact-calculations? #f])    (all-tests))  (newline)   (displayln "EXACT (Rational) NUMBERS")  (parameterize ([current-do-exact-calculations? #t])    (all-tests)))`
Output:
```INEXACT (Floating Point) NUMBERS
Classic example: 0.30000000000000004 = 0.300000000000000044
n=3	18.5
n=4	9.378378378378378954
n=5	7.801152737752168775
n=6	7.15441448097533339
n=7	6.806784736924811341
n=8	6.592632768721792047
n=20	98.349503122165359059
n=30	99.999999999998934186
n=50	100.
n=100	100.
TASK 2: balance after 25 years = -2242373258.570158004760742188
TASK 3: f(77617, 33096) = 1.1805916207174113e+021 = 1180591620717411303424.

EXACT (Rational) NUMBERS
Classic example: 3/10 = 0.3
n=3	18.5
n=4	9.378378378378378378
n=5	7.801152737752161383
n=6	7.154414480975249354
n=7	6.806784736923632984
n=8	6.592632768704438393
n=20	6.043552110189268868
n=30	6.006786093031205759
n=50	6.000175846627187189
n=100	6.000000019319477929
TASK 2: balance after 25 years = 0.039938729673230209
TASK 3: f(77617, 33096) = -54767/66192 = -0.827396059946821368```

## REXX

The REXX language uses character-based arithmetic.   So effectively, it looks, feels, and tastes like decimal floating point
(implemented in software).

So, the only (minor) problem is how many decimal digits should be used to solve these pathological floating point problems.

A little extra boilerplate code was added to support the specification of how many decimal digits that should be used for the
calculations,   as well how many decimal digits   (past the decimal point)   should be displayed.

### A sequence that seems to converge to a wrong limit

`/*REXX pgm (pathological FP problem):  a sequence that might converge to a wrong limit. */parse arg digs show .                            /*obtain optional arguments from the CL*/if digs=='' | digs==","  then digs= 150          /*Not specified?  Then use the default.*/if show=='' | show==","  then show=  20          /* "      "         "   "   "     "    */numeric digits digs                              /*have REXX use "digs" decimal digits. */#= 2 4 5 6 7 8 9 20 30 50 100                    /*the indices to display value of  V.n */fin= word(#, words(#) )                          /*find the last (largest) index number.*/w= length(fin)                                   /*  "   "  length (in dec digs) of FIN.*/v.1= 2                                           /*the value of the first   V  element. */v.2=-4                                           /* "    "    "  "  second  "     "     */      do n=3  to fin;   nm1= n-1;     nm2= n-2   /*compute some values of the V elements*/      v.n= 111 - 1130/v.nm1 + 3000/(v.nm1*v.nm2) /*   "      a  value  of  a  " element.*/                                                 /*display digs past the dec. point───┐ */      if wordpos(n, #)\==0  then say   'v.'left(n, w)       "="        format(v.n, , show)      end   /*n*/                                /*stick a fork in it,  we're all done. */`
output   when using the default inputs:
```v.4   = 9.37837837837837837838
v.5   = 7.80115273775216138329
v.6   = 7.15441448097524935353
v.7   = 6.80678473692363298394
v.8   = 6.59263276870443839274
v.9   = 6.44946593379028796887
v.20  = 6.04355211018926886778
v.30  = 6.00678609303120575853
v.50  = 6.00017584662718718895
v.100 = 6.00000001931947792910
```

### The Chaotic Bank Society

To be truly accurate, the number of decimal digits for   e   should have as many decimal digits as given on the
command line (if specified),   but what's currently coded will suffice for the (default) number of years.

However, it makes a difference computing the balance   (with 341 decimal digits)   after   183   years   (when at
that point,   the balance becomes negative and grows increasing negative fast).

With 150 decimal digits, the balance becomes negative after   96   years.

`/*REXX pgm (pathological FP problem): the chaotic bank society offering a new investment*/e=2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713,  ||8217852516642742746639193200305992181741359662904357290033429526059563073813232862794,  ||3490763233829880753195251019011573834187930702154089149934884167509244761460668082264,  ||8001684774118537423454424371075390777449920695517027618386062613313845830007520449338d = length(e)  -  1                              /*subtract unity for the decimal point.*/parse arg digs show y .                          /*obtain optional arguments from the CL*/if digs==''  |  digs==","  then digs=  d         /*Not specified?  Then use the default.*/if show==''  |  show==","  then show= 20         /* "      "         "   "   "     "    */if    y==''  |     y==","  then    y= 25         /* "      "         "   "   "     "    */numeric digits digs                              /*have REXX use "digs" decimal digits. */\$= e - 1                                         /*subtract \$1 from e, that's da deposit*/                                                 /* [↑]  value of newly opened account. */                           do n=1  for y         /*compute the value of the account/year*/                           \$= \$*n  -  1          /*   "     "    "    "  "  account now.*/                           end   /*n*/@@@= 'With '     d      " decimal digits, the balance after "      y      ' years is: 'say @@@    '\$'format(\$, , show) / 1              /*stick a fork in it,  we're all done. */`
output   when using the default inputs:
```With  341  decimal digits, the balance after  25  years is:  \$0.0399387296732302
```

### Siegfried Rump's example

`/*REXX pgm (pathological FP problem): the Siegfried Rump's example (problem dated 1988).*/parse arg digs show .                            /*obtain optional arguments from the CL*/if digs=='' | digs==","  then digs=150           /*Not specified?  Then use the default.*/if show=='' | show==","  then show= 20           /* "      "         "   "   "     "    */numeric digits digs                              /*have REXX use "digs" decimal digits. */a= 77617.0                                       /*initialize  A  to it's defined value.*/b= 33096.0                                       /*     "      B   "   "     "      "   */                                                 /*display SHOW digits past the dec. pt.*/say 'f(a,b)='    format(   f(a,b), , show)       /*display result from the  F  function.*/exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/f:  procedure; parse arg a,b;  return  333.75* b**6   +   a**2 * (11* a**2* b**2  -  b**6,                                     - 121*b**4  - 2)     +     5.5*b**8    +    a / (2*b)`
output   when using the default inputs:
```f(a,b)= -0.82739605994682136814
```

## Ring

` # Project : Pathological floating point problems decimals(8)v = list(100)v[1] = 2v[2] = -4for n = 3 to 100      v[n] = (111 - 1130 / v[n-1]) + 3000 / (v[n-1] * v[n-2])      if n < 9 or n = 20 or n = 30 or n = 50 or n = 100         see "n = " + n + "   " + v[n] + nl      oknext `

Output:

```n = 3   18.50000000
n = 4   9.37837838
n = 5   7.80115274
n = 6   7.15441448
n = 7   6.80678474
n = 8   6.59263277
n = 20  98.34950312
n = 30  100.00000000
n = 50  100
n = 100 100
```

## Ruby

Ruby numbers have a "quo" division method, which returns a rational (a fraction) when possible, avoiding Float inaccuracy.

`ar = [0, 2, -4]100.times{ar << (111 - 1130.quo(ar[-1])+ 3000.quo(ar[-1]*ar[-2])) } [3, 4, 5, 6, 7, 8, 20, 30, 50, 100].each do |n|  puts "%3d -> %0.16f" % [n, ar[n]]end `
Output:
```  3 -> 18.5000000000000000
4 -> 9.3783783783783784
5 -> 7.8011527377521614
6 -> 7.1544144809752494
7 -> 6.8067847369236330
8 -> 6.5926327687044384
20 -> 6.0435521101892689
30 -> 6.0067860930312058
50 -> 6.0001758466271872
100 -> 6.0000000193194779```

### Task 2: The Chaotic Bank Society

Using BigDecimal provides a way to specify the number of digits for E. 50 seems to be sufficient.

`require 'bigdecimal/math'balance = BigMath.E(50) - 11.upto(25){|y| balance = balance * y - 1}puts "Bank balance after 25 years = #{balance.to_f}"`
Output:
```Bank balance after 25 years = 0.03993872967323021
```

Rationals again.

`def rump(a,b)  a, b = a.to_r, b.to_r  333.75r * b**6 + a**2 * ( 11 * a**2 * b**2 - b**6 - 121 * b**4 - 2 )  + 5.5r *   b**8 + a / (2 * b)end puts "rump(77617, 33096) = #{rump(77617, 33096).to_f}"`
Output:
```rump(77617, 33096) = -0.8273960599468214
```

## Sidef

Muller's sequence

`func series (n) {    var (u, v) = (2, -4)    (n-2).times { (u, v) = (v, 111 - 1130/v + 3000/(v * u)) }    return v} [(3..8)..., 20, 30, 50, 100].each {|n|    printf("n = %3d -> %s\n", n, series(n))}`
Output:
```n =   3 -> 18.5
n =   4 -> 9.3783783783783783783783783783783783783783783783784
n =   5 -> 7.801152737752161383285302593659942363112391930836
n =   6 -> 7.154414480975249353527890653860362024381233838197
n =   7 -> 6.806784736923632983941756596272009087623276707802
n =   8 -> 6.592632768704438392742002776365994826552982317735
n =  20 -> 6.043552110189268867777477364097540133187715000006
n =  30 -> 6.006786093031205758530554047953239705833072314438
n =  50 -> 6.000175846627187188945614020747195469523735177099
n = 100 -> 6.000000019319477929104086803403585715024350675437
```

The Chaotic Bank Society

`var years = 25var balance = (1 .. years+15 -> sum_by {|n| 1 / n! })say "Starting balance, \$(e-1): \$#{balance}"for i in (1..years) { balance = (i*balance - 1) }printf("After year %d, you will have \$%1.16g in your account.\n", years, balance)`
Output:
```Starting balance, \$(e-1): \$1.7182818284590452353602874713526624977572470937
After year 25, you will have \$0.03993872967323021 in your account.
```

Siegfried Rump's example

`func f (a, b) {    (333.75 * b**6) + (a**2 * ((11 * a**2 * b**2) -      b**6 - (121 * b**4) - 2)) + (5.5 * b**8) + a/(2*b)} say f(77617.0, 33096.0)`
Output:
```-0.8273960599468213681411650954798162919990331157844
```

## Stata

`clearset obs 100gen n=_ngen v=.replace v=2 in 1replace v=-4 in 2replace v=111-1130/v[_n-1]+3000/(v[_n-1]*v[_n-2]) in 3/lformat %20.16f vlist if inlist(n,3,4,5,6,7,8,20,30,50,100), noobs`

Output

```  +----------------------------+
|   n                      v |
|----------------------------|
|   3    18.5000000000000000 |
|   4     9.3783783783783790 |
|   5     7.8011527377521688 |
|   6     7.1544144809753334 |
|   7     6.8067847369248113 |
|----------------------------|
|   8     6.5926327687217920 |
|  20    98.3495031221653590 |
|  30    99.9999999999989340 |
|  50   100.0000000000000000 |
| 100   100.0000000000000000 |
+----------------------------+
```

`clearset obs 26gen year=_n-1gen balance=exp(1)-1 in 1replace balance=year*balance[_n-1]-1 in 2/llist balance if year==25, noobs`

Output

```  +------------+
|    balance |
|------------|
| -2.242e+09 |
+------------+
```

If we replace exp(1)-1 by its value computed in higher precision by other means, the result is different, owing to the extreme sensibility of the computation.

`clearset obs 26gen year=_n-1gen balance=1.71828182845904523 in 1replace balance=year*balance[_n-1]-1 in 2/llist balance if year==25, noobs`

Output

```  +-----------+
|   balance |
|-----------|
| 1.202e+09 |
+-----------+
```

We can check the hexadecimal representation of both numbers. Note they differ only in the last bit:

`di %21x exp(1)-1di %21x 1.71828182845904523`

Output

```+1.b7e151628aed2X+000
+1.b7e151628aed3X+000
```

## TI-83 BASIC

A sequence that seems to converge to a wrong limit
Use the SEQ mode to enter the arithmetic progression. Note the way to set

``` u(1)=2
u(2)=-4
```
`   nMin=1   u(n)=111-1130/u(n-1) + 3000/(u(n-1)*u(n-2))   u(nMin)={-4;2}`

The result converges to the wrong limit!

Output:
```u(20)  : 100.055202
u(30)  : 100
u(50)  : 100
u(100) : 100
```

## zkl

zkl doesn't have a big rational or big float library (as of this writing) but does have big ints (via GNU GMP). It does have 64 bit doubles.

`Series:=Walker(fcn(vs){  // just keep appending new values to a list   vs.append(111.0 - 1130.0/vs[-1] + 3000.0/(vs[-1]*vs[-2])) }.fp(List(2,-4)));series:=Series.drop(100).value;`

We'll use the convenient formula given in the referenced paper to create a fraction with big ints

`var BN=Import("zklBigNum"), ten2n=BN(10).pow(64); fcn u(n){  // use formula to create a fraction of big ints   const B=-3, Y=4;   N:=BN(6).pow(n+1)*B + BN(5).pow(n+1)*Y;   D:=BN(6).pow(n)*B   + BN(5).pow(n)*Y;   tostr(N*ten2n/D,64,32)} fcn tostr(bn,m,r){ // convert big int (*10^m) to float string with len r remainder, flakey   str,d:=bn.toString(), str.len()-m;   if(d<0) String(".","0"*-d,str[0,r]);   else    String(str[0,d],".",str[d,r]);} println("1st: Convergent series");foreach n in (T(3,4,5,6,7,8,20,30,50,100)){    "n =%3d; %3.20F  %s".fmt(n,series[n-1],u(n-1)).println();}`
Output:

Note that, at n=100, we still have diverged (at the 15th place) from the Perl6 solution and 12th place from the J solution.

```1st: Convergent series
n =  3;  18.50000000000000000000  18.50000000000000000000000000000000
n =  4;   9.37837837837837895449  9.37837837837837837837837837837837
n =  5;   7.80115273775216877539  7.80115273775216138328530259365994
n =  6;   7.15441448097533339023  7.15441448097524935352789065386036
n =  7;   6.80678473692481134094  6.80678473692363298394175659627200
n =  8;   6.59263276872179204702  6.59263276870443839274200277636599
n = 20;  98.34950312216535905918  6.04355211018926886777747736409754
n = 30;  99.99999999999893418590  6.00678609303120575853055404795323
n = 50; 100.00000000000000000000  6.00017584662718718894561402074719
n =100; 100.00000000000000000000  6.00000001931947792910408680340358
```

Chaotic banking society is just nasty so we use a five hundred digit e (the e:= text is one long line).

`println("\n2nd: Chaotic banking society");e:="271828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742746639193200305992181741359662904357290033429526059563073813232862794349076323382988075319525101901157383418793070215408914993488416750924476146066808226480016847741185374234544243710753907774499206955170276183860626133138458300075204493382656029760673711320070932870912744374704723069697720931014169283681902551510865746377211125238978442505695369677078544996996794686445490598793163688923009879312";var en=(e.len()-1), tenEN=BN(10).pow(en);years,balance:=25, BN(e).sub(tenEN);  // in place mathbalance=[1..years].reduce(fcn(balance,i){ balance*i - tenEN },balance);balance=tostr(balance,en,2);println("After year %d, you will have \$%s in your account.".fmt(years,balance));`
Output:
```2nd: Chaotic banking society
After year 25, you will have \$.039 in your account.
```

For Rump's example, multiple the formula by 10ⁿ so we can use integer math.

`fcn rump(a,b){ b=BN(b);   b2,b4,b6,b8:=b.pow(2),b.pow(4),b.pow(6),b.pow(8);   a2:=BN(a).pow(2);   r:=( b6*33375 + a2*(a2*b2*11 - b6 - b4*121 - 2)*100 + b8*550 )*ten2n;   r+=BN(a)*ten2n*100/(2*b);   tostr(r,66,32)}println("\n3rd: Rump's example: f(77617.0, 33096.0) = ",rump(77617,33096));`
Output:
```3rd: Rump's example: f(77617.0, 33096.0) = -.82739605994682136814116509547981
```