jlaporte wrote:I think you found a new case.
Congratulations.
Merci beaucoup Jacques.
It was not until you put the expression in the form
jlaporte wrote:first category
1.01 ^ k {1 .. 9}
second category
2.02 * (1.01 ^ k) {1 .. 9}
Did I realise the use of ln(2.02) in the Errata is indeed very strange!
The "constants" used in the HP35 that are greater than Ln(1.01) are
Ln(1.1), Ln(2) and Ln(10). Ln(2.02) is NOT a constant used in the HP35.
I decided to hunt again for errors, but this time start with
LN(1.01) with k = 1 to 9, in addition to Ln(1.1), Ln(2) and Ln(10).
As you can see below, we have found some "new" buggy numbers.
-------------------------------
ln(1.01) x 1 = 10 errors
exp( 9.950330850E-03 ) = 1.000000 , 0.990 % error
exp( 9.950330851E-03 ) = 1.000000 , 0.990 % error
exp( 9.950330852E-03 ) = 1.000000 , 0.990 % error
exp( 9.950330853E-03 ) = 1.000000 , 0.990 % error
exp( 9.950330854E-03 ) = 1.000000 , 0.990 % error
exp( 9.950330855E-03 ) = 1.000000 , 0.990 % error
exp( 9.950330856E-03 ) = 1.000000 , 0.990 % error
exp( 9.950330857E-03 ) = 1.000000 , 0.990 % error
exp( 9.950330858E-03 ) = 1.000000 , 0.990 % error
exp( 9.950330859E-03 ) = 1.000000 , 0.990 % error
ln(1.01) x 2 = 1 error
exp( 1.990066170E-02 ) = 1.010000 , 0.990 % error
ln(1.01) x 3 = 1 error
exp( 2.985099250E-02 ) = 1.020100 , 0.990 % error
ln(1.01) x 4 = 1 error
exp( 3.980132340E-02 ) = 1.030301 , 0.990 % error
ln(1.01) x 5 = 1 error
exp( 4.975165420E-02 ) = 1.040604 , 0.990 % error
ln(1.01) x 6 = 1 error
exp( 5.970198510E-02 ) = 1.051010 , 0.990 % error
ln(1.01) x 7 = 1 error
exp( 6.965231590E-02 ) = 1.061520 , 0.990 % error
ln(1.01) x 8 = 1 error
exp( 7.960264680E-02 ) = 1.072135 , 0.990 % error
ln(1.01) x 9 = 1 error
exp( 8.955297760E-02 ) = 1.082857 , 0.990 % error
-------------------------------
ln(1.1) + ( ln(1.01) x 1 ) = No error
ln(1.1) + ( ln(1.01) x 2 ) = 1 error
exp( 1.152108415E-01 ) = 1.111000 , 0.990 % error
ln(1.1) + ( ln(1.01) x 3 ) = No error
ln(1.1) + ( ln(1.01) x 4 ) = No error
ln(1.1) + ( ln(1.01) x 5 ) = No error
ln(1.1) + ( ln(1.01) x 6 ) = No error
ln(1.1) + ( ln(1.01) x 7 ) = No error
ln(1.1) + ( ln(1.01) x 8 ) = No error
ln(1.1) + ( ln(1.01) x 9 ) = No error
-------------------------------
ln(2) + ( ln(1.01) x 1 ) = 1 error
exp( 7.030975114E-01 ) = 2.000000 , 0.990 % error
ln(2) + ( ln(1.01) x 2 ) = No error
ln(2) + ( ln(1.01) x 3 ) = 1 error
exp( 7.229981731E-01 ) = 2.040200 , 0.990 % error
ln(2) + ( ln(1.01) x 4 ) = No error
ln(2) + ( ln(1.01) x 5 ) = 1 error
exp( 7.428988348E-01 ) = 2.081208 , 0.990 % error
ln(2) + ( ln(1.01) x 6 ) = No error
ln(2) + ( ln(1.01) x 7 ) = No error
ln(2) + ( ln(1.01) x 8 ) = No error
ln(2) + ( ln(1.01) x 9 ) = No error
-------------------------------
ln(10) + ( ln(1.01) x 1 ) = No error
ln(10) + ( ln(1.01) x 2 ) = No error
ln(10) + ( ln(1.01) x 3 ) = No error
ln(10) + ( ln(1.01) x 4 ) = No error
ln(10) + ( ln(1.01) x 5 ) = No error
ln(10) + ( ln(1.01) x 6 ) = No error
ln(10) + ( ln(1.01) x 7 ) = No error
ln(10) + ( ln(1.01) x 8 ) = No error
ln(10) + ( ln(1.01) x 9 ) = No error
-------------------------------
Although Ln(2.02) is not a constant used in the HP35,
let us look for bugs anyway.
ln(2.02) = 1 error
exp( 7.030975114E-01 ) = 2.000000 , 0.990 % error
ln(2.02) + ( ln(1.01) x 1 ) = No error
ln(2.02) + ( ln(1.01) x 2 ) = 1 error
exp( 7.229981731E-01 ) = 2.040200 , 0.990 % error
ln(2.02) + ( ln(1.01) x 3 ) = No error
ln(2.02) + ( ln(1.01) x 4 ) = 1 error
exp( 7.428988348E-01 ) = 2.081208 , 0.990 % error
ln(2.02) + ( ln(1.01) x 5 ) = No error
ln(2.02) + ( ln(1.01) x 6 ) = No error
ln(2.02) + ( ln(1.01) x 7 ) = No error
ln(2.02) + ( ln(1.01) x 8 ) = No error
ln(2.02) + ( ln(1.01) x 9 ) = No error
-------------------------------
If we take the Ln(2.02) errors,
we see they are symmetrical with the Ln(2) errors.
ln(2.02)
exp( 7.030975114E-01 ) = 2.000000 , 0.990 % error
is equivalent to:
ln(2) + ( ln(1.01) x 1 )
exp( 7.030975114E-01 ) = 2.000000 , 0.990 % error
ln(2.02) + ( ln(1.01) x 2 )
exp( 7.229981731E-01 ) = 2.040200 , 0.990 % error
is equivalent to:
ln(2) + ( ln(1.01) x 3 )
exp( 7.229981731E-01 ) = 2.040200 , 0.990 % error
ln(2.02) + ( ln(1.01) x 4 )
exp( 7.428988348E-01 ) = 2.081208 , 0.990 % error
is equivalent to:
ln(2) + ( ln(1.01) x 5 )
exp( 7.428988348E-01 ) = 2.081208 , 0.990 % error
I think we can conclude, that HP should have expressed their
HP35 Errata in terms of Ln(1.01) , Ln(1.1) and LN(2).
We can also see from above, that not ALL values of
LN(1.01) k = 1 to 9 produce errors with LN(1.1) and LN(2).
Although I am reasonably sure the above list of exp(X) errors
is complete, not all are symmetrical with the expression exp(ln(X))
I think the full list in this form of original HP35 errors would be:
exp(ln(1.0201))
exp(ln(1.04060401))
exp(ln(1.12211))
exp(ln(2.02))
exp(ln(2.060602))
Regards, Geoff Hitchcox (Christchurch, New Zealand).