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[math-fun] Another weird leap year rule
by Keith F. Lynch 28 Apr '18

28 Apr '18

Some years ago, I asked about a calendar whose leap year rule was that you add a day for every odd number the year number is divisible by and subtract a day for every even number the year number is divisible by. The average year length was an interesting and unexpected number. I've finally come up with another weird leap year rule that gives an interesting and unexpected average year length. The rule is to add a day for every way in which the year number is the sum of two squares. (Squares include 0 and 1.) For instance 2018 is the sum of two squares in just one way, 13^2 + 43^2, so it would have one leap day. 2019 is not the sum of two squares, so it would have no leap days. 2005 would have two leap days because 2005 = 18^2 + 41^2 = 22^2 + 39^2. 1850 would have been the latest year with exactly 3, and 1885 would have been the most recent year with exactly 4. So, what is the average year length under this rule? As before, I already know the answer.

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Re: [math-fun] radix r numbers, |r|<1
by Keith F. Lynch 28 Apr '18

28 Apr '18

Henry Baker <hbaker1(a)pipeline.com> wrote: > What I was really trying to get at was irrational/transcendental > bases r, |r| close to 1, and what "numerals" to use. Radix r numbers and radix 1/r numbers are identical, except written in reverse order. For instance in radix 1/10, this is the year .8102 and one third is ...33333.0 Of course real numbers (unlike p-adics) that go on forever to the *left* look a little weird. You can't use fewer than two numerals. You might as well call them 0 and 1. There will in general be many ways to represent the same number. Of special interest are those radices (1<r<2) in which all positive integers have terminating representations. These include radix sqrt(2) and radix phi, but do not include any rational radices such as 1.1. I'm pretty sure they don't include any transcendental radices either. In radix phi, if you have the constraint that no two "1" digits can be adjacent, there's a unique representation for every number. That could be useful for encoding data on long-chain hydrocarbon "tapes," using single bonds for 0s and double bonds for 1s, given that it's unstable to have two adjacent double bonds. Another application is where the cost of 0 bits and 1 bits is unequal. For instance the SF author Charles Stross has proposed using C12 and C13 atoms for 0s and 1s to store data in diamond form. (Both isotopes are stable.) It would be extremely compact and extremely durable. Since there are 90 C12 atoms for every C13 atom, if the goal isn't to store data as compactly as possible, but to store as much data as possible in the total available carbon, it would make sense to use a radix in which 0 bits were 90 times more common than 1 bits. The old five-bit ITA2 teletype code, often wrongly called Baudot, treated 1-bits as more expensive than 0-bits (since the former cause more wear on mechanical teletypes), hence gave the most common letters the codes with the fewest 1-bits. This had the interesting side effect of making ITA2 a rich source of bitwise palindromes, like Morse code but unlike ASCII.

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Re: [math-fun] radix r numbers, |r|<1
by Henry Baker 28 Apr '18

28 Apr '18

Oops! Posted too soon. What I was really trying to get at was irrational/transcendental bases r, |r| close to 1, and what "numerals" to use. It appears that when 0.5<|r|<1.5, we have to leave digit-0 "holes". E.g., suppose for the moment that r=1.1. It will take (1.1)^8 = ~2.14 before we will need another "1" digit. So we get representations with subsequences of 0's in them. Ditto for r=0.9, where it will take 0.9^(-7) = ~2.09. At 09:16 AM 4/28/2018, Henry Baker wrote: >Positional number systems that can express *every* number >rely on the fact that the geometric series > > 2 3 4 5 6 7 8 9 10 >(%o7)/T/ 1 + r + r + r + r + r + r + r + r + r + r + . . . > >*diverges* for |r|>=1. > >But what if we don't care about representing *every* number? > >What if we only care about the interval |x|<2 ? > >In this case, r=1/2 should work, right? > >Then 1.5 = 1+1/2 = "1.1"; 1.25 = 1+1/4 = "1.01"; etc. > >A greedy algorithm should produce an acceptable representation.

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[math-fun] radix r numbers, |r|<1
by Henry Baker 28 Apr '18

28 Apr '18

Positional number systems that can express *every* number rely on the fact that the geometric series 2 3 4 5 6 7 8 9 10 (%o7)/T/ 1 + r + r + r + r + r + r + r + r + r + r + . . . *diverges* for |r|>=1. But what if we don't care about representing *every* number? What if we only care about the interval |x|<2 ? In this case, r=1/2 should work, right? Then 1.5 = 1+1/2 = "1.1"; 1.25 = 1+1/4 = "1.01"; etc. A greedy algorithm should produce an acceptable representation.

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[math-fun] Verification of the result and sigma estimation
by françois mendzina essomba2 28 Apr '18

28 Apr '18

>From what I have previously posted: J[4*n]=((4*n-3)/(4*n-2))* J[4*n-4] ; J[4*n+1]=((4*n-2)/(4*n-1))* J[4*n-3] ; J[4n+2]=((4*n-1)/(4*n))* J[4*n-2] ; J[4*n+3]=((4*n)/(4*n+1))* J[4*n-1] ; Verification of the result and sigma estimation J[4*n+3]=((4*n)/(4*n+1))* J[4*n-1] ; J[4*n+4]=((4*n+1)/(4*n+2))* J[4*n] ; By expressing the two preceding expressions according to J [3] and J [4], we will have: J[4*m+3]=J[3]*product(((4*n)/(4*n+1)),n=1..m) ; J[4*m+4]=J[4]*product(((4*n+1)/(4*n+2)),n=1..m) ; We have, on the same principles as for the sinus function: Limit(J[4*m+3],m=infinity) = Limit( J[4*m+4],m=infinity) ; Which corresponds to : J[3]*product(((4*n)/(4*n+1)),n=1..infinity)= J[4]*product(((4*n+1)/(4*n+2)),n=1..infinity) ; we obtain : J[4]/J[3]= product(((4*n)*(4*n+2)/(4*n+1)^2),n=1..infinity) ; we have : J[3]=1 et J[4]=(1/2)*J[0]=(1/4)*sigma Finally i find (1/4)*sigma = product(((4*n)*(4*n+2)/(4*n+1)^2),n=1..infinity) ; FME...

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[math-fun] model of wallis's integrals...
by françois mendzina essomba2 28 Apr '18

28 Apr '18

Hello J[0]=integrate((qin(x))^(0), x= 0..sigma/2); J[0]=sigma/2 J[1]=integrate(qin(x),x=0..sigma/2) ; J[3]=integrate((qin(x))^(3), x= 0.. sigma/2)=1; On the model of the wallis's integrals. We will have: J[n]=integrate((qin(x))^(n), x= 0..sigma/2); By integration by part: J[n]=integrate((qin(x))^(n-3)*(qin(x))^(3), x, 0, sigma/2); Let : u = (qin(x))^(n-3) ; u’=(n-3)*(qos(x))^(3)* (qin(x))^(n-4) ; v’=(qin(x))^3 ; v=-qos(x) ; We are getting : J[n]=[-(qos(sigma/2))* (qin(sigma/2))^(n-3)+(qos(0))* (qin(0))^(n-3)] + (n-3)* integrate((qin(x))^(n-4)*(qos(x))^(4), x=0..sigma/2); [-(qos(sigma/2))* (qin(sigma/2))^(n-3)+(qos(0))* (qin(0))^(n-3)]=0 because qos(sigma/2)=0 and qin(0)=0 ; >From where : J[n]= (n-3)* integrate((qin(x))^(n-4)*(qos(x))^(4), x=0..sigma/2); Which gives us : J[n]= (n-3)* integrate((qin(x))^(n-4)*(1-(qin(x))^(4)), x=0..sigma/2) = (n-3)* integrate((qin(x))^(n-4)-(qin(x))^(n),x=0.. sigma/2); Which leads us to: J[n]=(n-3)* J[n-4]-(n-3)* J[n] And finally to: J[n]=((n-3)/(n-2))* J[n-4] Since the function is level 4, we will write: J[4n]=((4n-3)/(4n-2))* J[4n-4] ; J[4n+1]=((4n-2)/(4n-1))* J[4n-3] ; J[4n+2]=((4n-1)/(4n))* J[4n-2] ; J[4n+3]=((4n)/(4n+1))* J[4n-1] ; FME...

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[math-fun] "long tail" number notations?
by Henry Baker 27 Apr '18

27 Apr '18

Are there any number notations that aren't *based*, in the sense that one digit doesn't represent O(1/|b|) of the next higher-order digit, where b is the "base" ? I'm particularly looking for notations which *don't* converge quickly. This may sound silly, but why bother with a notation that converges more quickly than series you're trying to represent?

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[math-fun] limited development of some special functions
by françois mendzina essomba2 26 Apr '18

26 Apr '18

I noticed an error in my last post correction of derivations diff(qanh(x),x)=1/(qosh(x))^2;diff(cqnh(x),x)=-1/(qinh(x))^2; limited development of functions to order 16; I found qosh(x)=1+(1/4)*x^4+(9/160)*x^8+(149/9600)*x^(12)+(15147/3328000)*x^16; qinh(x)=x+(3/20)*x^5+(19/480)*x^9+(469/41600)*x^13+(189611/56576000)*x^17; I deduce without verifying that: qos(x)=1-(1/4)*x^4+(9/160)*x^8-(149/9600)*x^(12)+(15147/3328000)*x^16; qin(x)=x-(3/20)*x^5+(19/480)*x^9-(469/41600)*x^13+(189611/56576000)*x^17;

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[math-fun] Special functions
by françois mendzina essomba2 26 Apr '18

26 Apr '18

hello, referring me to sinusoidal functions i write: (qos(x))^4+(qin(x))^4=1;(qosh(x))^4-(qinh(x))^4=1; compound functions: qan(x)=qin(x)/qos(x); cqn(x)=qos(x)/qin(x); qanh(x)=qinh(x)/qosh(x); cqnh(x)=qosh(x)/qinh(x); derivates diff(qos(x),x)=-(qin(x))^3; diff(qin(x),x)=(qos(x))^3; diff(qosh(x),x)=(qinh(x))^3; diff(qinh(x),x)=(qosh(x))^3; diff(qan(x),x)=1/(qos(x))^2; diff(cqn(x),x)=-1/(qin(x))^2; diff(qanh(x),x)=1/(qos(x))^2; diff(cqnh(x),x)=-1/(qin(x))^2; reciprocal functions: aqos(x)=-integrate(1/(1-x^4)^(3/4), x); aqin(x)=integrate(1/(1-x^4)^(3/4), x); aqosh(x)=integrate(1/(x^4-1)^(3/4), x); aqinh(x)=integrate(1/(1+x^4)^(3/4), x); aqan(x)=integrate(1/(x^4+1)^(1/2), x); acqn(x)=-integrate(1/(1+x^4)^(1/2), x); a constancy noted sigma sigma=(Gamma(1/4))^2/(2*sqrt(Pi)); some special points. qos(0)=1 ; qin(0)=0 ; qos(sigma/2)=0; qin(sigma/2)=1; qos(sigma)=-1; qin(sigma)=0; qos(sigma/4)=1/2^(1/4); qin(sigma/4)=1/2^(1/4); qan(sigma/4)=1; some integrals integrate(1/sqrt(x^4+1), x= 0.. 1)=sigma/4; integrate(1/(1-x^4)^(3/4),x=0..1/2^(1/4))=sigma/4; integrate(1/(1-x^4)^(3/4),x=0..1)=sigma/2; integrate(1/(1-x^4)^(3/4),x=-1..1)=sigma; FME...

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[math-fun] Eisenstein representation
by Fred Lunnon 23 Apr '18

23 Apr '18

There exists an algorithm for finding integers m,n such that x = m^2 + n^2 , in amortised time proportional to (log x)^3 , where x is prime with x mod 4 = 1 . See Richard E. Crandall, Carl Pomerance "Prime Numbers: A Computational Perspective" Springer (2001), sect 2.3.2 pp 93--96 . Is there an analogous efficient algorithm for x = m^2 + m n + n^2 ? See https://en.wikipedia.org/wiki/Eisenstein_integer https://oeis.org/A034017 , https://oeis.org/A000086 What about more general binary quadratic forms? Fred Lunnon

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