Degree 1 BBP-Type Zero Relations
                          Jaume Oliver Lafont
                            January 27, 2011


1    Introduction
Zero relations are BBP-type formulas that evaluate to zero. Although several
such formulas have been formally proved, others have remained unproved for
some time. This report outlines the proofs for eight previously unproved
binary zero relations.


2    Data
The compendium by D.H.Bailey includes eight zero relations found by PSLQ
but without any formal proof available. Five of them are length 24 base 212
relations, while the remaining three have length 40 and base 220 .


3    Methods
The series in the results section have been obtained through Fourier analysis
(written on PARI) of the normalized vector of coefficients from the P form
of the unproved zero relations. Vector coefficients are normalized in order to
cancel the 2n/2 factor. After identifying the real coefficients (using Plouffe’s
Inverter) simpler components have been obtained from the series, containing
only cosinus or sinus terms each.
    WolframAlpha is able to check the correctness of the first five equations.
The remaining two are also easy to prove, for example, by splitting into sev-
eral terms and plugging each term into WolframAlpha. This step guarantees
that the identification of floating point coefficients is correct. After that, ob-
taining the integer coefficients from the series is easily done using two lines
of PARI. Equations 103, 105, 106, 107 and 108 in the compendium (version
January 21, 2011) can be written as linear combinations of equations 1, 2,


                                      1
3, 4 and 5 in this report. Similarly, equations 109, 110 and 111 in the com-
pendium can be written as linear combinations of equations 6 and 7 and 8
below. This completes the proofs.


4     Results
4.1    Length 24
                            ∞                 πn
                                 1 − 2 cos     4 +    cos (πn)
                      0=
                           n=1
                                             n2n/2
                       =P (1, 24 , 8, (−8, 8, 4, 8, 2, 2, −1, 0))
                       =P (1, 212 , 24,                                                 (1)
                         (−2048, 2048, 1024, 2048, 512, 512, −256, 0,
                           − 128, 128, 64, 128, 32, 32, −16, 0,
                           − 8, 8, 4, 8, 2, 2, −1, 0))

                  ∞                  πn
                       1 − 2 cos      3     + cos (πn)
           0=
                                      n2n
                 n=1
             =P (1, 26 , 6, (−16, 24, 8, 6, −1, 0))
                  ∞                  πn               πn             5πn
                       1 − 2 cos      6     + 2 cos    2   − 2 cos    6    + cos (πn)
             =                                                                          (2)
                 n=1
                                                      n2n/2
             =P (1, 212 , 24,
                 (0, −1024, 0, 1536, 0, 512, 0, 384, 0, −64, 0, 0,
                 0, −16, 0, 24, 0, 8, 0, 6, 0, −1, 0, 0))

           ∞           πn             7πn
                 sin   12    − sin     12
      0=
           n=1
                            n2n/2
        =P (1, 212 , 24,                                                                (3)
          (2048, −2048, −2048, 0, −512, −1024, −256, 0, −256, −128, 64, 0,
           − 32, 32, 32, 0, 8, 16, 4, 0, 4, 2, −1, 0))

                  ∞             πn
                       cos      12   − 2 cos πn + cos
                                              4
                                                              7πn
                                                               12
           0=
                 n=1
                                          n2n/2
             =P (1, 212 , 24,                                                           (4)
                 (2048, 0, −4096, −3072, −512, 0, 256, 768, 512, 0, −64, 0,
                  − 32, 0, 64, 48, 8, 0, −4, −12, −8, 0, 1, 0))

                                                  2
∞            πn              πn             7πn              5πn
                  1      sin              12    − 2 sin    6    + sin     12    + 2 sin    6
              0 =√
                   3 n=1                                       n2n/2
                   =P (1, 212 , 24,                                                                         (5)
                      (2048, −4096, 0, −1024, 512, 0, 256, 768, 0, 256, 64, 0,
                      − 32, −64, 0, −48, −8, 0, −4, 4, 0, 4, −1, 0))

4.2     Length 40
                  ∞                   πn
                         1 − 2 cos     4 +      cos (πn)
           0=
                n=1
                                     n2n/2
            =P (1, 24 , 8, (−8, 8, 4, 8, 2, 2, −1, 0))
            =P (1, 220 , 40,
              (−524288, 524288, 262144, 524288, 131072, 131072, −65536, 0,                                  (6)
                  − 32768, 32768, 16384, 32768, 8192, 8192, −4096, 0,
                  − 2048, 2048, 1024, 2048, 512, 512, −256, 0,
                  − 128, 128, 64, 128, 32, 32, −16, 0,
                  − 8, 8, 4, 8, 2, 2, −1, 0))
       ∞            πn                πn               7πn              9πn            3πn           17πn
            cos     20     − 3 cos     4       + cos    20     + cos     20    − cos    4    + cos    20
 0=
      n=1
                                                              n2n/2
  =P (1, 220 , 40,
      (524288, 0, −262144, −1310720, −786432, 0, 65536, 327680,
      32768, 0, −16384, −8192, 0, −8192, 0, 24576, 20480,
      2048, 0, −1024, 0, −512, 0, 256, 1280,
      768, 0, −64, −320, −32, 0, 16, 80,
      8, 0, −24, −20, −2, 0, 1, 0))
                                                                                                            (7)
       ∞            πn               πn            7πn             9πn             3πn            17πn
            sin     20    − sin       4   − sin     20    + sin     20    + sin     4     + sin    20
 0=
      n=1
                                                          n2n/2
  =P (1, 220 , 40,
      (524288, −1572864, 262144, 0, 524288, 393216, −65536, 0,
                                                                                                            (8)
      32768, 65536, 16384, 0, −8192, 24576, 16384, 0,
      2048, −6144, 1024, 0, −512, 1536, −256, 0,
      − 512, −384, 64, 0, −32, −64, −16, 0,
      8, −24, −16, 0, −2, 6, −1, 0))


                                                          3
5    Conclusions
All degree-1 binary formulas currently listed in the Compendium are now proved.




                                      4

Zero relations

  • 1.
    Degree 1 BBP-TypeZero Relations Jaume Oliver Lafont January 27, 2011 1 Introduction Zero relations are BBP-type formulas that evaluate to zero. Although several such formulas have been formally proved, others have remained unproved for some time. This report outlines the proofs for eight previously unproved binary zero relations. 2 Data The compendium by D.H.Bailey includes eight zero relations found by PSLQ but without any formal proof available. Five of them are length 24 base 212 relations, while the remaining three have length 40 and base 220 . 3 Methods The series in the results section have been obtained through Fourier analysis (written on PARI) of the normalized vector of coefficients from the P form of the unproved zero relations. Vector coefficients are normalized in order to cancel the 2n/2 factor. After identifying the real coefficients (using Plouffe’s Inverter) simpler components have been obtained from the series, containing only cosinus or sinus terms each. WolframAlpha is able to check the correctness of the first five equations. The remaining two are also easy to prove, for example, by splitting into sev- eral terms and plugging each term into WolframAlpha. This step guarantees that the identification of floating point coefficients is correct. After that, ob- taining the integer coefficients from the series is easily done using two lines of PARI. Equations 103, 105, 106, 107 and 108 in the compendium (version January 21, 2011) can be written as linear combinations of equations 1, 2, 1
  • 2.
    3, 4 and5 in this report. Similarly, equations 109, 110 and 111 in the com- pendium can be written as linear combinations of equations 6 and 7 and 8 below. This completes the proofs. 4 Results 4.1 Length 24 ∞ πn 1 − 2 cos 4 + cos (πn) 0= n=1 n2n/2 =P (1, 24 , 8, (−8, 8, 4, 8, 2, 2, −1, 0)) =P (1, 212 , 24, (1) (−2048, 2048, 1024, 2048, 512, 512, −256, 0, − 128, 128, 64, 128, 32, 32, −16, 0, − 8, 8, 4, 8, 2, 2, −1, 0)) ∞ πn 1 − 2 cos 3 + cos (πn) 0= n2n n=1 =P (1, 26 , 6, (−16, 24, 8, 6, −1, 0)) ∞ πn πn 5πn 1 − 2 cos 6 + 2 cos 2 − 2 cos 6 + cos (πn) = (2) n=1 n2n/2 =P (1, 212 , 24, (0, −1024, 0, 1536, 0, 512, 0, 384, 0, −64, 0, 0, 0, −16, 0, 24, 0, 8, 0, 6, 0, −1, 0, 0)) ∞ πn 7πn sin 12 − sin 12 0= n=1 n2n/2 =P (1, 212 , 24, (3) (2048, −2048, −2048, 0, −512, −1024, −256, 0, −256, −128, 64, 0, − 32, 32, 32, 0, 8, 16, 4, 0, 4, 2, −1, 0)) ∞ πn cos 12 − 2 cos πn + cos 4 7πn 12 0= n=1 n2n/2 =P (1, 212 , 24, (4) (2048, 0, −4096, −3072, −512, 0, 256, 768, 512, 0, −64, 0, − 32, 0, 64, 48, 8, 0, −4, −12, −8, 0, 1, 0)) 2
  • 3.
    πn πn 7πn 5πn 1 sin 12 − 2 sin 6 + sin 12 + 2 sin 6 0 =√ 3 n=1 n2n/2 =P (1, 212 , 24, (5) (2048, −4096, 0, −1024, 512, 0, 256, 768, 0, 256, 64, 0, − 32, −64, 0, −48, −8, 0, −4, 4, 0, 4, −1, 0)) 4.2 Length 40 ∞ πn 1 − 2 cos 4 + cos (πn) 0= n=1 n2n/2 =P (1, 24 , 8, (−8, 8, 4, 8, 2, 2, −1, 0)) =P (1, 220 , 40, (−524288, 524288, 262144, 524288, 131072, 131072, −65536, 0, (6) − 32768, 32768, 16384, 32768, 8192, 8192, −4096, 0, − 2048, 2048, 1024, 2048, 512, 512, −256, 0, − 128, 128, 64, 128, 32, 32, −16, 0, − 8, 8, 4, 8, 2, 2, −1, 0)) ∞ πn πn 7πn 9πn 3πn 17πn cos 20 − 3 cos 4 + cos 20 + cos 20 − cos 4 + cos 20 0= n=1 n2n/2 =P (1, 220 , 40, (524288, 0, −262144, −1310720, −786432, 0, 65536, 327680, 32768, 0, −16384, −8192, 0, −8192, 0, 24576, 20480, 2048, 0, −1024, 0, −512, 0, 256, 1280, 768, 0, −64, −320, −32, 0, 16, 80, 8, 0, −24, −20, −2, 0, 1, 0)) (7) ∞ πn πn 7πn 9πn 3πn 17πn sin 20 − sin 4 − sin 20 + sin 20 + sin 4 + sin 20 0= n=1 n2n/2 =P (1, 220 , 40, (524288, −1572864, 262144, 0, 524288, 393216, −65536, 0, (8) 32768, 65536, 16384, 0, −8192, 24576, 16384, 0, 2048, −6144, 1024, 0, −512, 1536, −256, 0, − 512, −384, 64, 0, −32, −64, −16, 0, 8, −24, −16, 0, −2, 6, −1, 0)) 3
  • 4.
    5 Conclusions All degree-1 binary formulas currently listed in the Compendium are now proved. 4