Accurate Bernoulli numbers on the 41C, or "how close can you get"?

03092014, 07:12 PM
Post: #1




Accurate Bernoulli numbers on the 41C, or "how close can you get"?
In the last weeks there have been some discussions regarding various ways of determining Bernoulli numbers on the 41series and other calculators. The usual formulas included powers with exponents greater than 100, leading to reduced accuracy since an exact result would require at least twelve or thirteen digits for the base as opposed to the ten we have. Another problem is the available working range, so that the used algorithm has to make sure no intermediate result exceeds the limit at 9,999...E99.
So I wondered if there might be a way of evaluating all possible Bernoulli numbers within the working range sufficiently fast and, more important, as accurately as possible. Which leads to the question: how close can you get in the face of accumulating roundoff errors? Even a simple multiplication can be surprisingly inaccurate, the result may be off by up to 5 units in the last place. Try a simple \(\pi·\pi\) or \(e·e\), and the result on a correctly working 10digit calculator is 3 ULP high or low. So far, so bad. Here is the approach used in the following program. As usual, the lower Bernoulli numbers B_{0} to B_{8} are given directly. For n = 2...8 a simple quadratic equation can do the trick. For n = 10 to 116 (largest value within the 41's working range) the following formula was used: \(\large B_n = 4 \pi · \zeta(n) · e^{(0,5 + \frac{1}{12n}  \frac{1}{360n^3} + \frac{1}{1260n^5} + ...)} · (\frac{n}{2 \pi e})^{n+0,5} \) For n ≥ 10 and 10digit accuracy three terms of the series in the exponent of the efunction are sufficient. The expression \(\frac{1}{12n}  \frac{1}{360n^3} + \frac{1}{1260n^5}\) can be evaluated as \(\frac{210n^4  7n^2 + 2}{2520n^5}\). A literal implementation of the complete formula would yield results with substantial errors. At least the last two digits would be off. So a different way to handle this formula had to be found. Within the relevant domain, the factors at the left (\(4\pi=10·0,4\pi, \zeta\) and the exponential function) all start with 1. They do not vary much for n = 10...116: \(B_n = 10 · 1,256... · 1,000... · 1,65... · (\frac{n}{2 \pi e})^{n+0,5} \) The basic idea now is to evaluate all three factors minus one so that one additional digit is gained. Obtaining \(\zeta  1\) is trivial, and for the exponential function there is a dedicated \(e^x1\) command. The multiplication of three values close to 1 can be done in a way that preserves one additional digit of working precision. Since the product of the three factors is something between 2,07 and 2,09, the program even tries to calculate half of this minus 1 (and finally multiplies this +1 with twice the power), so that again a precious digit is saved. The program uses a 9digit approximation of \(0,4\pi  1 = rad(72°)1\) which is slightly low, so a correction term is applied. Its exact value should be near 7,2E10, but tests showed that in this case even better accuracy is obtained with a slighty lower value close to 6E10 (cf. line 89). Now let's look at the power at the right. For a correct 10digit result, the base would have to carry at least 12 or 13 digits. Here is how this is accomplished in the program: \((\frac{n}{2 \pi e})^{n+0,5}\) \( = (n · 0,05854983152432)^{n+0,5}\) \(\approx (n · 0,05854983)^{n+0,5} + 1,52432·10^{9}·n·(n+0,5)·(n · 0,05854983)^{n0,5}\) For n = 10...116 the base of the first power carries at most 9 digits, so both the base and the exponent are exact. However, the 41's power function sometimes truncates its result instead of rounding it, so the constant 1,52432E9 is better rounded up to 1,5244E9. Here is the 41C code: Code: 01 LBL"BN" One may now ask if the result is worth all the effort. I think it is. In total there are 60 possible nonzero results within the 41's working range (n = 0, 1, 2, 4, 6, 8, ..., 114, 116). The program returns 45 of these correctly rounded or truncated after 10 digits. The rest is 1 ULP high or low. I did not find any larger errors. In other words: the results are close to machine accuracy. BTW, while the largest possible result is B_{116}, the program can also provide B_{118}. The expected OUT OF RANGE error appears in the last calculation step when the program tries to multiply X by 10. At this point, pressing [X<>Y] reveals B_{118} as 6,116052000E+100. ;) Of course suggestions for improvements are always welcome. Dieter 

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