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Bernoulli numbers and large factorials - Dieter - 02-09-2014 04:59 PM
In December Namir posted a program that evaluates the Bernoulli numbers on the HP-41. It can be found in the HP-41C Software Library. In January I suggested another version based on the same approach: \( B_n = 2 n! {(2\pi)^{-n}} \sum\limits_{i=1}^{\infty}i^{-n} \) For large n the sum rapidly approaches 1 so that at some point it may be omitted. On the other hand, for n as low as 10 only a few terms are required for the usual 10 or 12 digit precision. \(B_{0...8}\), which would require a substantial number of terms, may be given directly. So this is the easy part. Both Namir's and my solution suffer from a limitation due to overflow in the factorial function. For HP's 10-digit calculators with working range up to 9,999...E99 the limit is 69!, while most 12-digit devices with their upper limit at 9.999...E499 will work up to 253!. However, this is significantly less than the largest possible n that returns a Bernoulli number within the working range. Here, results up to \(B_{116}\) resp. \(B_{372}\) would be possible. There is an elegant way to overcome this problem if the calculator offers a permutation function (nPr). The essential idea is to split n! into two factors a and b which both fall within the calculator's working range. If factorials up to 253! are possible (the largest value below 9,999...E499) this can be done as follows: \( \begin{align*} n! &= \frac{n!}{253!} · 253!\\ &= \frac{n!}{(n - (n - 253))!} · 253!\\ &= Perm(n, n - 253) · 253! \end{align*} \) If n is less than 253, this constant is simply replaced with 0: \( Perm(n, n - 0) · 0! = n! · 1 = n!\) This method may also be written as follows: let r = max(0, n - 253) let a = Perm(n, r) let b = (n-r)! Then n! = a · b This way the factor \(2 n! {(2\pi)^{-n}}\) can be evaluated as \(2 a {(2\pi)^{-n} b}\) to avoid overflow. On calculators with a limit at 9,999...E99, for instance the 15C, simply use 69 instead of 253. Here is a complete program for the HP-35s. It evaluates all Bernoulli numbers within the working rage, i.e. from \(B_0\) to \(B_{372}\). Code: `B001 LBL B` The adjustment in line 051...056 tries to reduce the error in \((2\pi)^{-n}\). On the 35s I could not find errors larger than some units in the last place. Other calculators may require a different correction, or it may even be omitted completely if a somewhat larger error is acceptable. Usage: Enter n [XEQ] B [ENTER] => display shows n and Bn. Execution time: within approx. 3 seconds (at n = 10). Examples: 4 [XEQ] B [ENTER] => -0,333333333333 32 [XEQ] B [ENTER] => -15.116.315.767,1 100 [XEQ] B [ENTER] => 2,83322495708 E+78 372 [XEQ] B [ENTER] => -5,58475372908 E+499 Up to n = 26 the result may also be viewed in fraction mode. If no limitations are set (Flag 8 and 9 clear, /c ≥ 2730) the display shows the exact representation of the respective Bernoulli number: 22 [XEQ] B [ENTER] => 6.192,12318839 [FDISP] => 6192 17 / 138 ' exact result is \(6192 \frac{17}{138}\) or \(\frac{854513}{138}\) [RND] ' round to exact result [FDISP] => 6.192,12318841 Dieter RE: Bernoulli numbers and large factorials - Tugdual - 02-09-2014 07:47 PM
Nice job I checked the values on Wolfram Alpha and you are prettyu close (for the 10 first digits over 500 ;-) ). The Prime has Bernouilli numbers calculated internally but unfortunately the Bn function is not exposed in the library. So we can use the formula Bernoulli(x) := –x*Zeta(1–x) (Thanks Joe Horn) I tried it and couldn't calculate B372; the bigger number is B370. RE: Bernoulli numbers and large factorials - Marcus von Cube - 02-09-2014 08:01 PM
The WP 34S has Bn built-in. In double precision mode I can get B2122. Larger arguments return 0 (which can be considered a bug). RE: Bernoulli numbers and large factorials - Bunuel66 - 02-09-2014 08:42 PM
Another approach would be to compute iteratively the factorial using 1/2π factor at every step. As the factorial as the same number of factor as the term (2π)^−n this would limit the overflow. My two cents... RE: Bernoulli numbers and large factorials - Dieter - 02-09-2014 08:43 PM
(02-09-2014 07:47 PM)Tugdual Wrote: Nice job Thank you. :-) I just compared all possible 188 non-zero results with the correctly rounded 12-digit values. If I got it right, more than 80% are within ±1 ULP and more than 95% within ±2 ULP. The rest (<5%) is off by ±3 or 4 ULP. Dieter RE: Bernoulli numbers and large factorials - Dieter - 02-09-2014 09:26 PM
(02-09-2014 08:01 PM)Marcus von Cube Wrote: The WP 34S has Bn built-in. In double precision mode I can get B2122. Larger arguments return 0 (which can be considered a bug). On my 34s (v. 3.2 3405) both B2124 and Zeta (-2123) still return a result, while beyond that a "+∞ Error" is displayed. Do you really get a zero here? If 11 valid digits are sufficient, the 35s program does a good job and, compared to the 34s, it is really fast. I wonder how the 34s will perform with the same algorithm in user code. OK, 34 digits for n as low as 10 or 12 will take somewhat longer. ;-) EDIT: The reason for the 34s limit at B2122 probably is the same as the one mentioned in my original post: it's the factorial function. In DP mode the 34s still can handle 2122! but 2124! will cause an overflow. This could be overcome by using the permutation function. A quick-and-dirty test confirmed that B2776 = 1,01268...E+6140 can be done. The result is returned in about a second. Dieter RE: Bernoulli numbers and large factorials - Paul Dale - 02-09-2014 09:44 PM
The 34S is computing the Bernoulli numbers from the zeta function. A short series expansion like you've got here will be faster I expect. - Pauli RE: Bernoulli numbers and large factorials - Dieter - 02-12-2014 08:56 PM
(02-09-2014 09:44 PM)Paul Dale Wrote: The 34S is computing the Bernoulli numbers from the zeta function. A short series expansion like you've got here will be faster I expect. Well, this series expansion, i.e. the sum in the formula, actually is the Zeta function. ;-) \( \begin{align*} B_n &= 2 n! {(2\pi)^{-n}} \sum\limits_{i=1}^{\infty}i^{-n}\\ &= 2 n! {(2\pi)^{-n}} \zeta (n) \end{align*} \) The 35s program works so fast because the larger n gets, the less terms are required. The following table shows the number of terms needed for an error of at most 0,1 ULP in Zeta: Code: ` n 10 12 16 34 digits` This also explains why up to n = 8 the result is given directly. Otherwise the number of required terms would increase rapidly. I tried a program with the same algorithm on the 34s. In SP mode it is much faster than the internal Bernoulli function. For large n the result appears within a fraction of a second. As you will expect after a look at the above table, DP mode with 34 digit precision is a different story, at least for small n. ;-) Dieter |