mini challenge: find the smallest cosine of an integer

10192021, 10:23 AM
Post: #1




mini challenge: find the smallest cosine of an integer
.
Naturally, we should use radians for this, and we're interested in the absolute value of the cosine. Some calculators will have a maximum integer that they are happy to take the cosine of, which will of course affect your result. But there are no prizes, so start your search  whether manually, programatically, or mathematically. Show your working! Spoiler padding follows... . .. . ... . .. . .... . .. . ... . .. . ..... . .. . ... . .. . .... . .. . ... . .. . cos(0) is the first, and the worst... cos(1) is better, and could hardly be otherwise... cos(2) is better yet cos(5) is a new record cos(8) is another winner cos(11) is pretty good! (And surely no coincidence) 

10192021, 11:33 AM
Post: #2




RE: mini challenge: find the smallest cosine of an integer
For getting smallest cos(x), we are only using anglereduction code of cosine function.
Tiny cos(x) implied x ≈ (2k+1)*(pi/2) cos(x) = sin(x  (2k+1)*(pi/2)) ≈ (x  (2k+1)*(pi/2)) Example, cos(11) ≈ 11  7*(pi/2) ≈ 0.0044257 So, it depends how good anglereduction code is. Most calculators are not good. Free42Decimal happened to have very good anglereduction code (at the cost of *huge* table) This is the smallest cos(x), for x < 1E50 9.341730789500356812974471132146251e46 COS ABS → 1.028415848209791685669808767152452e35 Above x (47 digits integer) is derived from convergents of y = (pi/2) / 10^(47  34) see https://www.hpmuseum.org/forum/thread17...#pid153032 

10192021, 11:58 AM
Post: #3




RE: mini challenge: find the smallest cosine of an integer
Brute force, HP 41
Code:
0.01 XEQ 'SC' yields 10.994 hence the integer 11 0.001 XEQ 'SC' yields 15,253.9997 hence the integer 15,254 0.0001 XEQ 'SC' yields 24,461.99999 hence the integer 24,462 Upon which the accuracy limit of the HP 41 is reached. Cheers PeterP Cheers, PeterP 

10192021, 01:30 PM
Post: #4




RE: mini challenge: find the smallest cosine of an integer
Limiting x to integer ≈ (2k+1)*(pi/2)
pi/2 ≈ x / (2k+1) Pick pi/2 convergents with odd denominator, then numerator = x c:\> spigot d22 C pi/2 1/1 2/1 3/2 11/7 344/219 355/226 51819/32989 52174/33215 260515/165849 573204/364913 4846147/3085153 5419351/3450066 37362253/23785549 42781604/27235615 122925461/78256779 411557987/262005952 534483448/340262731 2549491779/1623056876 3083975227/1963319607 17969367914/11439654911 21053343141/13402974518 881156436695/560961610149 902209779836/574364584667 HP50g: 902209779836 COS → 5.28216132855E13 (red part should be 3683) 

10192021, 02:44 PM
Post: #5




RE: mini challenge: find the smallest cosine of an integer
If calculator had 12digits precision, and we would like to search x of 14 digits:
pi/2/100 ≈ (x/100) / (2k+1) Pick (pi/2/100) convergents with odd denominator, then numerator = x/100 Note: To get x of 14 digits, numerator must be 12 digits. c:\> spigot d20 C pi/2/100  tail 5 1324880093/84344486322 8539351431/543631996417 9864231524/627976482739 205823981911/13103161651197 833160159168/53040623087527 HP50g: 833160159168E2 COS → 2.94461252684E14 (red part should be 2017582) 

10212021, 09:44 AM
(This post was last modified: 10212021 09:45 AM by EdS2.)
Post: #6




RE: mini challenge: find the smallest cosine of an integer
Thanks both, that's two very different approaches!
Peter, I like the simplicity and directness of your program. It's a good trick, I think, to look for n close to odd multiples of pi/2, and leave the cosines out of the loop. But there's something curious about your result. You get 24462 as the first n to have a cos less than 0.0001. And yet cos(24462) ≈ 0.0112. I suppose that adding pi some twenty thousand times might have caused a precision problem, but it seems a bit early for that to have happened. (In fact, it's about right: five digits to the left of the point and looking for small values in five digits to the right.) I think your program might do somewhat better by adding (pi3) instead of pi. The Online Encylopedia of Integer Sequences is quite handy. For this investigation, these two entries are of some use: A004112 Numbers k where cos(k) (or cosec(k) or cot(k)) decreases monotonically to 0; also tan(k), sec(k), sin(k) increases. A096456 Numerators of convergents to Pi/2. From the first, we find a link to this list of recordholders for small cos, and from that we see: cos(24129) ≈ 0.002375894172 cos(24484) ≈ 0.002345749902 which I think must only be coincidentally close to your 24462. But perhaps more notably we see cos(40459) ≈ 0.000989256 which ought to be the first one less than 0.001 Albert, I am not at all surprised that continued fractions and convergents would come into a mathematical solution! But your method for dealing with trailing zeros to find answers for calculators with limited precision is very nice. I wonder if the OEIS would like to add a series for the numerators of pi/2 convergents where the denominators are odd. I note also that Free 42 shows its strengths again, quite wonderfully. (I note also that wolfram alpha online needs to be given integers with lots of trailing zeros, to compute cosine with adequate precision: giving it those large inputs in floating form returns zero for cosine.) 

10212021, 12:01 PM
Post: #7




RE: mini challenge: find the smallest cosine of an integer
(10212021 09:44 AM)EdS2 Wrote: A004112 Numbers k where cos(k) (or cosec(k) or cot(k)) decreases monotonically to 0 Since it is impossible to input x to infinite precision, cos(x) does not decrease to zero. With limited precision, cos(x) is not much smaller than machineepsilon. Quote:cos(24129) ≈ 0.002375894172 FYI, all 3 x's comes from semiconvergents of pi/2: 344/219 355/226 51819/32989 24129 = 344 + 355*67 24484 = 344 + 355*68 40459 = 344 + 355*113 Quote:I wonder if the OEIS would like to add a series for the numerators of pi/2 convergents where the denominators are odd. No need. Just use convergents of pi instead pi/2 ≈ x / (2k+1) pi ≈ (2x) / (2k+1) With convergents fully reduced, even numerator guaranteed odd denominator. So, just pick convergents/semiconvergents of pi, with even numerator = 2x Note: convergents of pi/2 may have shifted to semiconvergents of pi. Example, convergents of pi: 333/106, 355/113 (333+355) / (106+113) = 2*(344/219) // (344/219) is a convergent of pi/2 

10212021, 01:09 PM
Post: #8




RE: mini challenge: find the smallest cosine of an integer
(10212021 12:01 PM)Albert Chan Wrote: With convergents fully reduced, even numerator guaranteed odd denominator. 2 neighboring convergents, p1/q1, p2/q2, we have p1*q2  p2*q1 = ± 1 // or, gap = p1/q1  p2/q2 = 1/q1*q2 If g = gcd(p1,q1) ≠ 1, RHS must be multiples of g, contradict above identity. Thus, convergents always fully reduced. 

10212021, 03:04 PM
Post: #9




RE: mini challenge: find the smallest cosine of an integer
Peter, I notice something else about your program: it seeks numbers which are just smaller than an integer, but in our case numbers which are just larger are also of interest. I think it should only take a line or two to address this.


10232021, 06:47 AM
Post: #10




RE: mini challenge: find the smallest cosine of an integer
Thinking a little more about Peter's program, if we separately accumulate the 3 and the pi3, and if we carry a 0.5 from the accumulation of fractions whenever it exceeds 0.5, into the accumulation of integers (now an accumulation of halfintegers), we'll find that we lose very little accuracy as proceed, which might extend the useful range of the search.
Indeed, I think we can do a little better, by also accumulating the even smaller fraction which holds the difference between the limited precision value of pi we're using and the real value of pi. We need to add this accumulation of tiny pieces to the accumulation of fractions when comparing to see how close we are to an integer. I don't think we'll need to split this tiny accumulation into big and small parts. Before I decided to trim the fractional accumulation by 0.5, I was going to trim it whenever it exceeded 1.0. With this approach, whenever we go over 1.0, we can as a next step accumulate 6 times our fraction, because we know the fraction is less than 1/7. So that could be a speedup. If that works, it could I think be modified for the more accurate approach of trimming at 0.5. The more mathematically inclined reader might be interested in this related paper I found: Rosenholtz, I. (1999). Tangent Sequences, World Records, π, and the Meaning of Life: Some Applications of Number Theory to Calculus. Mathematics Magazine, 72(5), 367. doi:10.2307/2690793 Quote:the author knows of no other instance in which knowing several million digits of pi is actually useful Quote:To summarize: If we desire to find tan(n)/n records, we need only look at numerators of those convergents of the continued fraction expansion of pi/2 having odd denominators 

10232021, 06:53 AM
Post: #11




RE: mini challenge: find the smallest cosine of an integer
Albert, thanks for your further thoughts. On this comment
Quote:Since it is impossible to input x to infinite precision, cos(x) does not decrease to zeroI think we are in danger of mixing mathematical results with the results of calculation. The statement Quote:A004112 Numbers k where cos(k) (or cosec(k) or cot(k)) decreases monotonically to 0surely stands, in mathematics, even if it fails in calculation. (It fails because our calculations have finite resources.) Is it certain, though, that there is no calculator input of a positive integer which causes the calculator to return a cos of zero? It's not clear to me, presently. 

10232021, 05:10 PM
(This post was last modified: 10242021 04:39 AM by Albert Chan.)
Post: #12




RE: mini challenge: find the smallest cosine of an integer
(10232021 06:53 AM)EdS2 Wrote: Is it certain, though, that there is no calculator input of a positive integer which causes the calculator to Let's explore what condition needed to have calculator return cos(x) of 0.0 Assumed correct anglereduction code, and no smalldifferencesincorrectlysettozero “feature" Again, using 12digits calculator, and integer x of 14 digits as example. Instead of convergents of pi/200, we do 200/pi Valid convergent/semiconvergent must have odd numerator, and denominator of 12 digits. c:> spigot d20 c l 200/pi 63;1,1,1,23,36,1,1,1,8,2,1,52,13,4,2,6,1,20,4,63 These are convergents of last 3 CF coefs: 20: p0/q0 = 13103161651197/205823981911 04: p1/q1 = 53040623087527/833160159168 63: p2/q2 = 3354662416165398/52694914009495 gap = p1/q1  p2/q2 = p1*q2p2*q1 / (q1*q2) = 1/(q1*q2) We expected next convergent much better than current convergent. Rough estimate, 200/pi  p1/q1 < gap = 1/(q1*q2) But, gap error is really shared between 2 convergents. Perhaps we increase q2 a bit: q2 = q0 + q1*CF < q1 * (CF+1) 200/pi  53040623087527/833160159168 ≈ 2.250E26 1/(833160159168^2 * (63+1)) ≈ 2.251E26 cos(x) // x = 100*q1, with odd p1 ≈ x  p1*(pi/2) = (100*q1) * 1  (p1/q1)/(200/pi) ≈ (100*q1) * 1/(q1^2*(CF+1)) / (200/pi) = (pi/2) / (q1*(CF+1)) ≈ 2.946E14 // cos(833160159168E2) ≈ 2.945E14, estimate not too bad Note that factor 100 get cancelled. We can generalize, for big x, with valid convergent p1/q1 (*) cos(x) ≈ (pi/2) / (q1*(CF+1)) ≈ (pi/2) / q2 To have cos(x) underflow to 0.0, q1*(CF+1) overflowed For a 12digits calculator that overflow if decimal exponent reached 500 (CF+1) > 10^(50012) = 10^488  12digits calculator, x all the way to 500 digits, x of 395 digits has smallest cos(x) Free42 Decimal: 0.700035629422e395 COS ABS → 2.056399907114786850992546180602331e15 For this number, CF = 1090, estimate formula is pretty close (pi/2) / (700035629422 * (1090+1)) ≈ 2.057e15 (*) if p1/q1, p2/q2 are not neighboring convergents, we need another estimate formula. Let pm/qm and p2/q2 be neighboring convergents. p1*q2  p2*q1 = (p2k*pm)*q2  p2*(q2k*qm) = k*(p2*qm  pm*q2) = ±k gap = p1/q1  p2/q2 = k / (q1*q2) → cos(x) ≈ (pi/2) / (q2/k) Example, x of 137 digits had a higher CF = 2573, yet cos(x) is not small Note: we reduced numerator of (2e125/pi), without disturbing its parity. c:\> spigot d28 C "2e125/pi mod 2"  tail 3 17206041685/23300343711 202744600696/274555819883 521679063632493/706455424902670 Convergent with 12digits denominator has even numerator. Best we could use is semiconvergent, with k = 2570 706455424902670  274555819883 * 2570 = 846967803360 Estimated cos(x) = (pi/2) / (706455424902670/2570) ≈ 5.714e12 Free42Decimal: 846967803360e125 COS ABS → 5.714370521060907484280466201567459e12 

10242021, 06:56 AM
Post: #13




RE: mini challenge: find the smallest cosine of an integer
Ever since your first very insightful post I had been playing around with both of these ideas as indeed, the error accumulates much quicker than I had originally thought.
Alas, things get really slow when you add the test for 0.5. And just doing Pi3 actually only gives you about 12 extra digits and the error still accumulates quickly. I’m on my phone now but can post the improved code thanks to your insights later but would not advise using it  even in iOS i41CX it runs pretty slowly…. Thank you for the nice minichallenge. Thank you Albert for super impressive and instructive math, I love reading your posts and learn from you. Might have been nice to get someone from the 71b genius group share a code as well. Thank you again Ed for posting it! Cheers, PeterP 

10242021, 08:27 AM
Post: #14




RE: mini challenge: find the smallest cosine of an integer
Thanks Peter, that's very good to hear!
I would be interested to see your code. I have my 15C, 35S, and 30b ready, and all of those in principle could be programmed up for a search. (But instead I find myself playing around with Basic. And not even on a 71B.) And thanks Albert, again: I now have a glimmering of understanding that we can't get an actual zero. 

10242021, 10:28 AM
Post: #15




RE: mini challenge: find the smallest cosine of an integer
(10242021 08:27 AM)EdS2 Wrote: And thanks Albert, again: I now have a glimmering of understanding that we can't get an actual zero. I think it's partly a quality of implementation issue. I wouldn't be surprised if there is some calculator out there that would return 0 as the cosine of some integral number of radians, particularly when operating outside the domain of exact, consecutive integer representations. For example, when a decimal, floating point number represents a continuous span of at least 10, some points in that continuous span would have a cosine of 0, so a poor quality calculator might give up and return 0 (or some other value in the range [1,1]) for all such numbers! — Ian Abbott 

10242021, 12:46 PM
Post: #16




RE: mini challenge: find the smallest cosine of an integer
eps.bat Wrote:@spigot C "2e%1/pi mod 2"  awk vFS="/" ve="e%1" "length($2)>34 {print 1.5708/$2, q1 e; exit} {q1=$2}" Running this 1 liner overnight, x all the way to 6145 digits, we have a brute force proof Free42Decimal: cos(x) > 10^37, any real x if p1/q1 is valid convergent (odd p1, neighboring p2/q2), then cos(x) < ε = (pi/2)/q2 The requirement of q1 being 34 digits is not necessary. Example, both below suggested the same x (note that 2nd mantissa has 33 digits, not 34) c:\> eps 5981 → 1.60573e37 4272221537462525975023616239425960e5981 c:\> eps 5982 → 1.60509e37 427222153746252597502361623942596e5982 427222153746252597502361623942596e5982 COS ABS → 1.605082938363676519340189975883263e37 min(cos(x)) for integer x (also, for real x) c:\> eps 1381 → 1.03065e37 2344813655066356855719930664718056e1381 2344813655066356855719930664718056e1381 COS ABS → 1.030557387629248882465543827418861e37 eps.bat even work with negative exponent (below is min(cos(x)) for noninteger x) c:\> eps 25 → 1.55097e35 2589477332551582209163675623249625e25 2589477332551582209163675623249625e25 COS ABS → 1.532138639232766081410107492878833e35  eps.bat only get right a little over half the time. Neighboring convergents cannot both have p1,p2 even → p1 is more likely odd. see https://www.hpmuseum.org/forum/thread17...#pid153072 We need to check actual p1/q1, for odd p1 (or, just try cos(x)) For even p1, we may have to settle for semiconvergents, or p0/q0 (p0 guaranteed odd) (it is highly unlikely to produce small cos(x) though, see my previous post) pq.bat Wrote:@spigot C "2e%1/pi mod 2"  awk vS="/" "lengthindex($0,S)>34 {exit} {print}"  tail 2 Example: this get wrong eps (being conservative is OK; we don't miss anything) c:\> eps 4639 → 2.06902e38 9308532438209917461067659354862169e4639 c:\> pq 4639 1528624876453580931164937665986407/8775804805425001620222943906879558 // p0/q0 1621418726123738114106180500221048/9308532438209917461067659354862169 // p1/q1 p1 is even, we have to settle for p0/q0: cos(x = q0*1E4639) ≈ 1.68729e34 However, this suggested sin(x = q1*1E4639) is very close to 0. 9308532438209917461067659354862169e4639 SIN → 2.069013989544476107762446187843978e38 

10242021, 05:36 PM
Post: #17




RE: mini challenge: find the smallest cosine of an integer
Ever since your first very insightful post I had been playing around with both of these ideas as indeed, the error accumulates much quicker than I had originally thought.
Alas, things get really slow when you add the test for 0.5. And just doing Pi3 actually only gives you about 12 extra digits and the error still accumulates quickly. I’m on my phone now but can post the improved code thanks to your insights later but would not advise using it  even in iOS i41CX it runs pretty slowly…. Thank you for the nice minichallenge. Thank you Albert for super impressive and instructive math, I love reading your posts and learn from you. Might have been nice to get someone from the 71b genius group share a code as well. Thank you again Ed for posting it! Cheers, PeterP 

10242021, 06:17 PM
Post: #18




RE: mini challenge: find the smallest cosine of an integer
(10242021 06:56 AM)PeterP Wrote: Alas, things get really slow when you add the test for 0.5. And just doing Pi3 actually only gives you about 12 extra digits and the error still accumulates quickly. A novel way is use SIN(X) >PI, SIN(PI) @ REM sum = 3.141592653589793238462643 (25 digits of pi) 3.14159265359 2.06761537357E13 This program get numerator of convergents of pi With internal 25 digits of stored pi, all convergents correct Building of convergents is fast, using formula, P = P0 + k*P1 (P's are all integers, no error is introduced building P's) 10 P0=0 @ P1=1 @ P=P0 @ S=2 20 P=P+P1 @ S2=ABS(SIN(P)) 30 IF S>S2 THEN S=S2 @ DISP P @ P0=P1 @ P1=P @ P=P0 40 IF P<1.E+12 THEN 20 >RUN 1 3 22 333 355 103993 104348 208341 312689 833719 1146408 4272943 5419351 80143857 165707065 245850922 411557987 1068966896 2549491779 6167950454 14885392687 21053343141 sin(2x) = 2*sin(x)*cos(x). If cos(x) is small, so does sin(2x) We pick convergents/semiconvergents with even numerator. 2x = 1+3, 1+3*3, 1+3*5, 22, 333+355, 333+355*3, 333+355*5 ... 333+355*291, 104348, ... x = 2, 5, 8, 11, 344, 699, 1054 ... 51819, 52174, ... 

10242021, 06:56 PM
Post: #19




RE: mini challenge: find the smallest cosine of an integer
Great stuff Albert! Thank you for sharing.
Cheers, PeterP 

10242021, 09:53 PM
(This post was last modified: 10112023 05:37 PM by Albert Chan.)
Post: #20




RE: mini challenge: find the smallest cosine of an integer
To complete symmetry, this is min(cos(x)) for Free42 binary, if angle reduction code is correct.
eps.bat Wrote:@spigot C "2*2^%1/pi mod 2"  awk vFS="/" ve="*0x1p%1" "$2>=2^53 {print 1.5708/$2, q1 e; exit} {q1=$2}" Just a change of constants needed for eps.bat This take less than 1 minute, x upto 971+53 = 1024 bits integer. c:\> (for /L %I in (52,1,971) do @(eps %I))  gsort g  head 4.69203e19 6381956970095103*0x1p797 6.19292e19 3205513981387887*0x1p46 6.20075e19 6411027962775774*0x1p47 9.38406e19 6381956970095103*0x1p798 1.23858e18 3205513981387887*0x1p45 ... The list is suggestion only, confirmation necessary. First does give minimum cos(x) see Accurate trigonometric functions for large arguments >>> from gmpy2 import * >>> x = mul_2exp(6381956970095103,797) # = 0x1.6ac5b262ca1ffp+849 >>> x mpfr('5.3193726483265414e+255') >>> cos(x) mpfr('4.6871659242546277e19') 

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