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Find these constants I (Posted on 2005-11-10) Difficulty: 3 of 5
Given that
 k=∞          1           A*B-C*D
Σ    ------------------ = -------
 k=0 (3k+1)(3k+2)(3k+3)      E   
where A, B, C, D and E are, in some order, the number π (pi), the square root of an integer, the natural logarithm of an integer, and two integers, find their values.

See The Solution Submitted by pcbouhid    
Rating: 2.8000 (5 votes)

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Solution computer soln--spoiler--don't let this inhibit you from seeking analytic soln | Comment 1 of 10

 Doing the sum until the numbers didn't change resulted in
 
 .1787967688907531
 
 So we want a combination that comes out to about this.  The following program assumes the integer whose square root is used is under 23 as is the integer whose natural log is used.  The two integers that are used as-is are assumed to be under 45. The program tries all possibilities (of course not bothering to put, for example, B = pi, as A = pi already covers this sort of possibility).
 
 
 DEFDBL A-Z
 k = 0
 denom = 6
 DO
  term = 1 / denom
  ptot = tot
  tot = tot + term
  k = k + 1
  denom = (3 * k + 1) * (3 * k + 2) * (3 * k + 3)
 LOOP UNTIL ptot = tot
 PRINT tot
 
 pi = ATN(1) * 4
 FOR int1 = 1 TO 22
 sr = SQR(int1)
 FOR int2 = 1 TO 22
 lg = LOG(int2)
 FOR int3 = 1 TO 44
 FOR int4 = 1 TO 44
   a = pi
     b = sr
       c = lg
         d = int3: e = int4: GOSUB test
       e = lg
         d = int3: c = int4: GOSUB test
     c = sr
       b = lg
         d = int3: e = int4: GOSUB test
       d = lg
         b = int3: e = int4: GOSUB test
       e = lg
         b = int3: d = int4: GOSUB test
     e = sr
       b = lg
         d = int3: c = int4: GOSUB test
       d = lg
         b = int3: c = int4: GOSUB test
        
   c = pi
     b = sr
       a = lg
         d = int3: e = int4: GOSUB test
       e = lg
         d = int3: a = int4: GOSUB test
     a = sr
       b = lg
         d = int3: e = int4: GOSUB test
       d = lg
         b = int3: e = int4: GOSUB test
       e = lg
         b = int3: d = int4: GOSUB test
     e = sr
       b = lg
         d = int3: a = int4: GOSUB test
       d = lg
         b = int3: a = int4: GOSUB test
 
   e = pi
     a = sr: b = lg: c = int3: d = int4: GOSUB test
     b = int3: c = lg: GOSUB test
     a = int4: d = sr: GOSUB test
 NEXT
 NEXT
 NEXT
 NEXT
 END
 
 test:
  IF e <> 0 THEN
   v = (a * b - c * d) / e
   tst = ABS(tot / v)
   IF ABS(tst - 1) < 1E-08 THEN
    PRINT v; a; b; c; d; e
   END IF
  END IF
 RETURN
 
The following data result:


 
 .178796768891527  3.141592653589793  1.732050807568877  1.09861228866811  3 12
 .178796768891527  3.141592653589793  3.464101615137754  1.09861228866811  6 24
 .178796768891527  3.141592653589793  3.464101615137754  2.19722457733622  3 24

The value of the 5-number formula is shown; we presume the discrepancy from the total is due to rounding errors in the total and in not counting the total infinitely far out. The values of A, B, C, D and E are shown on the rest of each line.

The first line shows A=pi, B=sqrt(3), C=ln(3), D = 3 and E = 12, for

(pi*sqrt(3) - 3*ln(3))/12

The second line shows A=pi, B=sqrt(12), C=ln(3), D=6 and E=24, this being the same as the previous with numerator and denominator multiplied by 2.

The third line shows A=pi, B=sqrt(12), C=ln(9), D=3 and E=24, thereby doubling C and halving D from the previous solution.

In each case, of course, A and B can be interchanged as can C and D.


  Posted by Charlie on 2005-11-10 16:09:58
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