Re: [math-fun] Theta function halfangles
From: rwg@sdf.lonestar.org [mailto:rwg@sdf.lonestar.org] Sent: Wed 7/9/2008 9:22 AM To: math-fun@mailman.xmission.com Cc: "rwg@sdf.lonestar.orgrwg"@sdf.lonestar.org Subject: Theta function halfangles
(I promised Cheny Xu, proprietor of Jade Palace Restaurant,
www.jadepalace.us
that I'd solve a problem with the help of some fish he fed me last night.)
It's not obvious how to extract a tractable halfangle formula from the many double angle formulae in W&W. After you eliminate the three undesired species, the resultant polynomial is too hairy. Actually, it would probably simplify if Macsyma or Mma only knew how, but I can't even get Mma 6.0 to EllipticTheta[1, Pi, q] -> 0 ! (Hey guys, need a programmer?)
Fortunately, I (re?)conjectured a family of theta determinant identities, http://www.tweedledum.com/rwg/thet.PNG (too tall for IE 7), arxiv.org/pdf/math/0703470, the first few (including the one used here) proved by Rich (and Hilarie?). These are nice when actually typeset:
x theta (-, q) = 1 2
3 3 3 - theta (0, q) theta (x, q) + theta (0, q) theta (x, q) - theta (0, q) theta (x, q) 4 4 3 3 2 2 1/4 (-----------------------------------------------------------------------------------) , 2
x theta (-, q) = 2 2
3 3 3 - theta (0, q) theta (x, q) + theta (0, q) theta (x, q) + theta (0, q) theta (x, q) 4 4 3 3 2 2 1/4 (-----------------------------------------------------------------------------------) , 2
x theta (-, q) = 3 2
3 3 3 theta (0, q) theta (x, q) + theta (0, q) theta (x, q) + theta (0, q) theta (x, q) 4 4 3 3 2 2 1/4 (---------------------------------------------------------------------------------) , 2
x theta (-, q) = 4 2
3 3 3 theta (0, q) theta (x, q) + theta (0, q) theta (x, q) - theta (0, q) theta (x, q) 4 4 3 3 2 2 1/4 (---------------------------------------------------------------------------------) . 2
E.g., pi pi theta (--, q) = theta (--, q) = 1 4 2 4
2 2 theta (0, q) theta (0, q) (theta (0, q) - theta (0, q)) 3 4 3 4 1/4 (-------------------------------------------------------) , 2
pi pi theta (--, q) = theta (--, q) = 3 4 4 4
2 2 theta (0, q) theta (0, q) (theta (0, q) + theta (0, q)) 3 4 4 3 1/4 (-------------------------------------------------------) . 2
But, duh, %pi %pi 4 theta (---, q) = theta (---, q) = theta (0, q ) 4 4 3 4 4
Finally, this must be in Tannery & Molk, but I don't recall seeing a Lambert type series come out as a pure theta instead of a log derivative theta:
inf 2 ==== j 2 j - 1 theta (0, q) - 1 \ (- 1) q 4 > --------------- = ----------------. / 2 j - 1 4 ==== q + 1 j = 1
There seems to be a lot of these. theta (0, q) 2 log(-------------------) inf 1/4 ==== n n 2 theta (0, q) q \ (- 1) q 3 > ---------- = ------------------------. / n 2 ==== n (q + 1) n = 1
--rwg
From: rwg@sdf.lonestar.org [mailto:rwg@sdf.lonestar.org] Sent: Wed 7/9/2008 9:22 AM To: math-fun@mailman.xmission.com [...] Finally, this must be in Tannery & Molk, but I don't recall seeing a Lambert type series come out as a pure theta instead of a log derivative theta:
inf 2 ==== j 2 j - 1 theta (0, q) - 1 \ (- 1) q 4 > --------------- = ----------------. / 2 j - 1 4 ==== q + 1 j = 1
There seems to be a lot of these. [...]
George Andrews kindly informs me that E. Grosswald ascribes this to Jacobi and his counting of representations as sums of two squares. It generalizes to inf 2 ==== n theta (0, q) theta (x, q) theta (x, q) \ q sin(2 n x) cot(x) 4 2 3
------------- = ------ - --------------------------------------. / n 4 4 theta (x, q) theta (x, q) ==== q + 1 1 4 n = 1
The reason for no thetaprimes is that it's the derivative of csc(x) theta (x, q) 1 log(-------------------) inf 1/4 ==== n 2 q theta (x, q) \ q cos(2 n x) 4 > ------------- = ------------------------, / n 2 ==== n (q + 1) n = 1 and the derivative of a quotient of two thetas comes out in thetas. Theta quotients come from subtracting variations on %pi 1/6 2 theta (---, q ) sin(z) 1 3 inf log(--------------------------) ==== n sqrt(3) theta (z, sqrt(q)) \ q cos(2 n z) 1 log(q) > ------------- = ------------------------------- + ------ / n 2 24 ==== n (1 - q ) n = 1 and %pi 1/3 theta (---, q ) 1 3 inf log(--------------------) ==== n sqrt(3) theta (z, q) \ q cos(2 n z) 4 log(q) > ------------- = ------------------------- - ------, / 2 n 2 24 ==== n (1 - q ) n = 1 which I dredged out of W&W. This gives us, e.g., inf ==== n \ q log(q) %pi 1/6 log(3) > ---------- = ------ - log(theta (---, q )) + ------, / n 24 1 3 2 ==== n (1 - q ) n = 1 but still nothing about plain old inf ==== n \ q > ------, / n ==== q - 1 n = 1 with which we could sum the reciprocal Fibonacci numbers. Here's another pure theta sum: inf ==== 2 n - 1 \ q sin((2 n - 1) z)
------------------------- = / 2 n - 1 ==== 1 - q n = 1
2 z z theta (0, sqrt(q)) theta (-, sqrt(q)) theta (-, sqrt(q)) 2 3 2 4 2 csc(z) -------------------------------------------------------- - ------ z z 4 8 theta (-, sqrt(q)) theta (-, sqrt(q)) 1 2 2 2 --rwg
* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Jul 12. 2008 16:01]:
[...]
but still nothing about plain old
inf ==== n \ q > ------, / n ==== q - 1 n = 1
with which we could sum the reciprocal Fibonacci numbers.
Have you checked Borwein/Borwein "Pi and the AGM"? pp.91-101 might make you happy.
* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Jul 12. 2008 16:01]:
[...]
but still nothing about plain old
inf ==== n \ q > ------, / n ==== q - 1 n = 1
with which we could sum the reciprocal Fibonacci numbers.
Have you checked Borwein/Borwein "Pi and the AGM"? pp.91-101 might make you happy.
Sadly not)-: They denote this Lambert series -L(q), but never give it in terms of Thetas or ThetaPrimes. Despite the section heading, they never deliver Sum 1/Fib(n) except in terms of L. It may well be inexpressible without some new special function. (They do give Sum 1/Fib(2*n+1) in Thetas, which is easy.) --rwg They say the brain is your largest sex organ. I don't believe it. I've gotten > 5000 pen!s enhancement spams (sic), and not one for enhancing my brain.
* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Jul 12. 2008 16:01]:
[...]
but still nothing about plain old
inf ==== n \ q > ------, / n ==== q - 1 n = 1
with which we could sum the reciprocal Fibonacci numbers.
Have you checked Borwein/Borwein "Pi and the AGM"? pp.91-101 might make you happy.
Sadly not)-: They denote this Lambert series -L(q), but never give it in terms of Thetas or ThetaPrimes. Despite the section heading, they never deliver Sum 1/Fib(n) except in terms of L. It may well be inexpressible without some new special function. (They do give Sum 1/Fib(2*n+1) in Thetas, which is easy.) --rwg
I wonder if the Borweins knew this generalization: inf %pi - x/3 ==== theta (---, %e ) \ cosh(n y) 1 3 x > ----------- = log(--------------------) + -- - log(sqrt(3)). / n sinh(n x) %i y - x 12 ==== theta (----, %e ) n = 1 4 2 Note d/dy gives a nice sum(sinh(ny)/sinh(nx)) in terms of theta'. Specializing x=2*y, and using the amazing Abel-Plana formula from a paper by Almkvist via Steve Finch: inf inf inf ==== / / \ [ h(%i y) - h(- %i y) [ h(0) > h(n) = %i I ------------------- dy + I h(x) dx + ----, / ] 2 %pi y ] 2 ==== / %e - 1 / n = 0 0 0 we get a peculiar theta' valued integral inf / [ sin(t y) I ---------------------------------------- dt = ] 2 %pi t / (%e - 1) (cos(2 t y) + cosh(2 y)) 0 %i y - 2 y y theta'(----, %e ) acot(%e ) 4 2 csch(2 y) - ----------- + ------------------------------- - ---------. 2 y sinh(y) %i y - 2 y 4 4 sinh(y) theta (----, %e ) 1 2 Note that by last week's halfangle formulae, 4 %i y - 2 y - 2 y 3 - 2 y theta (----, %e ) = (- %i theta (0, %e ) theta (0, %e ) 1 2 1 4 - 2 y 3 - 2 y + theta (0, %e ) theta (0, %e ) 2 3 3 - 2 y - 2 y y/2 - theta (0, %e ) theta (0, %e )) %e /2, 2 3 but I don't know any special values of theta' except theta'_1(0). --rwg
* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Jul 12. 2008 16:01]:
[...]
but still nothing about plain old
inf ==== n \ q > ------, / n ==== q - 1 n = 1
with which we could sum the reciprocal Fibonacci numbers.
Have you checked Borwein/Borwein "Pi and the AGM"? pp.91-101 might make you happy.
Sadly not)-: They denote this Lambert series -L(q), but never give it in terms of Thetas or ThetaPrimes. Despite the section heading, they never deliver Sum 1/Fib(n) except in terms of L. It may well be inexpressible without some new special function. (They do give Sum 1/Fib(2*n+1) in Thetas, which is easy.) --rwg
I wonder if the Borweins knew this generalization:
inf %pi - x/3 ==== theta (---, %e ) \ cosh(n y) 1 3 x > ----------- = log(--------------------) + -- - log(sqrt(3)). / n sinh(n x) %i y - x 12 ==== theta (----, %e ) n = 1 4 2
Note d/dy gives a nice sum(sinh(ny)/sinh(nx)) in terms of theta'. Specializing x=2*y, and using the amazing Abel-Plana formula from a paper by Almkvist via Steve Finch: inf inf inf ==== / / \ [ h(%i y) - h(- %i y) [ h(0) > h(n) = %i I ------------------- dy + I h(x) dx + ----, / ] 2 %pi y ] 2 ==== / %e - 1 / n = 0 0 0
we get a peculiar theta' valued integral
inf / [ sin(t y) I ---------------------------------------- dt = ] 2 %pi t / (%e - 1) (cos(2 t y) + cosh(2 y)) 0
%i y - 2 y y theta'(----, %e ) acot(%e ) 4 2 csch(2 y) - ----------- + ------------------------------- - ---------. 2 y sinh(y) %i y - 2 y 4 4 sinh(y) theta (----, %e ) 1 2
Note that by last week's halfangle formulae,
4 %i y - 2 y - 2 y 3 - 2 y theta (----, %e ) = (- %i theta (0, %e ) theta (0, %e ) 1 2 1 4
- 2 y 3 - 2 y + theta (0, %e ) theta (0, %e ) 2 3
3 - 2 y - 2 y y/2 - theta (0, %e ) theta (0, %e )) %e /2, 2 3
but I don't know any special values of theta' except theta'_1(0). --rwg
theta'(0, q) = theta (0, q) theta (0, q) theta (0, q) 1 2 3 4 and the conjugate points, e.g., 9/4 3 %i log(q) %pi = - %i q theta'(----------- - ---, q) 3 2 2 1/4 %i log(q) = %i q theta'(---------, q) = - q theta' (%i log(q), q) 4 2 1 %pi = - theta'(---, q) = . . . 2 2 But, duh, Macsyma already knew that theta_s'(0) = 0, s>1, (because those thetas are even). So now we can differentiate the halfangle formulae and get all the theta_s'(pi/2^n,q) as algebraic functions of the three theta_s(0,q). E.g., %pi %pi theta'(---, q) = - theta'(---, q) 1 4 2 4 4 1/4 1/4 theta (0, q) theta (0, q) theta (0, q) 2 3 4 = ------------------------------------------. 5/4 2 2 3/4 2 (theta (0, q) - theta (0, q)) 3 4 Tayloring these gives 1/4 9/4 25/4 49/4 sqrt(2) q + 3 sqrt(2) q - 5 sqrt(2) q - 7 sqrt(2) q 81/4 + 9 sqrt(2) q + . . ., so they are all just theta'(0, %i q) 1 = ---------------. 1/4 sqrt(2) %i I still need to look at the fractional pseudoperiods. --rwg
* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Jul 12. 2008 16:01]:
[...]
but still nothing about plain old
inf ==== n \ q > ------, / n ==== q - 1 n = 1
with which we could sum the reciprocal Fibonacci numbers.
Have you checked Borwein/Borwein "Pi and the AGM"? pp.91-101 might make you happy.
Sadly not)-: They denote this Lambert series -L(q), but never give it in terms of Thetas or ThetaPrimes. Despite the section heading, they never deliver Sum 1/Fib(n) except in terms of L. It may well be inexpressible without some new special function. (They do give Sum 1/Fib(2*n+1) in Thetas, which is easy.) --rwg
I wonder if the Borweins knew this generalization:
inf %pi - x/3 ==== theta (---, %e ) \ cosh(n y) 1 3 x > ----------- = log(--------------------) + -- - log(sqrt(3)). / n sinh(n x) %i y - x 12 ==== theta (----, %e ) n = 1 4 2
Note d/dy gives a nice sum(sinh(ny)/sinh(nx)) in terms of theta'. Specializing x=2*y, and using the amazing Abel-Plana formula from a paper by Almkvist via Steve Finch: inf inf inf ==== / / \ [ h(%i y) - h(- %i y) [ h(0) > h(n) = %i I ------------------- dy + I h(x) dx + ----, / ] 2 %pi y ] 2 ==== / %e - 1 / n = 0 0 0
we get a peculiar theta' valued integral
inf / [ sin(t y) I ---------------------------------------- dt = ] 2 %pi t / (%e - 1) (cos(2 t y) + cosh(2 y)) 0
%i y - 2 y y theta'(----, %e ) acot(%e ) 4 2 csch(2 y) - ----------- + ------------------------------- - ---------. 2 y sinh(y) %i y - 2 y 4 4 sinh(y) theta (----, %e ) 1 2
Note that by last week's halfangle formulae,
4 %i y - 2 y - 2 y 3 - 2 y theta (----, %e ) = (- %i theta (0, %e ) theta (0, %e ) 1 2 1 4
- 2 y 3 - 2 y + theta (0, %e ) theta (0, %e ) 2 3
3 - 2 y - 2 y y/2 - theta (0, %e ) theta (0, %e )) %e /2, 2 3
but I don't know any special values of theta' except theta'_1(0). --rwg
theta'(0, q) = theta (0, q) theta (0, q) theta (0, q) 1 2 3 4
and the conjugate points, e.g.,
9/4 3 %i log(q) %pi = - %i q theta'(----------- - ---, q) 3 2 2
1/4 %i log(q) = %i q theta'(---------, q) = - q theta' (%i log(q), q) 4 2 1
%pi = - theta'(---, q) = . . . 2 2
But, duh, Macsyma already knew that theta_s'(0) = 0, s>1, (because those thetas are even). So now we can differentiate the halfangle formulae and get all the theta_s'(pi/2^n,q) as algebraic functions of the three theta_s(0,q). E.g.,
%pi %pi theta'(---, q) = - theta'(---, q) 1 4 2 4
4 1/4 1/4 theta (0, q) theta (0, q) theta (0, q) 2 3 4 = ------------------------------------------. 5/4 2 2 3/4 2 (theta (0, q) - theta (0, q)) 3 4 Tayloring these gives
1/4 9/4 25/4 49/4 sqrt(2) q + 3 sqrt(2) q - 5 sqrt(2) q - 7 sqrt(2) q
81/4 + 9 sqrt(2) q + . . ., so they are all just
theta'(0, %i q) 1 = ---------------. 1/4 sqrt(2) %i
I still need to look at the fractional pseudoperiods. --rwg
It basically works. Theta_4'(%i log(q)/2,q) came out 0/0, but that's congruent to theta_1'(0). Also, the four theta' quarter quasiperiod answers came out reversed! I assume it's just bad luck with the 4th root branches. Anyway, here's that weird Abel-Plana integral in terms of theta(0) only: inf / [ sin(y) (d188) I ------------------------------------- dy = ] 2 pi y / ------- 0 x (e - 1) (cos(2 y) + cosh(2 x)) 2 2 pi pi - --- - ---- x 2 x 2 x - 4 acot(e ) - x (sech(x) + 1) + pi theta (0, i e ) theta (0, i e ) 3 4 -----------------------------------------------------------------------------. 8 sinh(x) Testing for x=pi: (c189) dfloat(opsubst(nounify(integrate)=lambda([[l]],apply(quad_inf,l)),subst(%pi,x,%))) (d189) 0.00132907964796d0 = 0.00132907964796d0 I don't recall seeing elementary integrals for thetas. --rwg
Anyway, here's that weird Abel-Plana integral in terms of theta(0) only:
inf / [ sin(y) (d188) I ------------------------------------- dy = ] 2 pi y / ------- 0 x (e - 1) (cos(2 y) + cosh(2 x))
2 2 pi pi - --- - ---- x 2 x 2 x - 4 acot(e ) - x (sech(x) + 1) + pi theta (0, i e ) theta (0, i e ) 3 4 -----------------------------------------------------------------------------. 8 sinh(x)
Testing for x=pi:
(c189) dfloat(opsubst(nounify(integrate)=lambda([[l]],apply(quad_inf,l)),subst(%pi,x,%)))
(d189) 0.00132907964796d0 = 0.00132907964796d0
I don't recall seeing elementary integrals for thetas. --rwg
Duh, http://en.wikipedia.org/wiki/Theta_function#Integral_representations Looks like they just construct the powerseries as an infinite sum of residues. I overestimated the comprehensiveness of W&W. New business: Can someone tell me if Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124, or some other source answers whether A050788 is known infinite, i.e., does c^3+1=a^3+b^3 infinitely often? And does ceiling((a^n+b^n)^(1/n))^n - a^n - b^n < k have only finitely many solutions for n>3? --rwg
At 07:08 AM 7/22/2008, rwg@sdf.lonestar.org wrote:
New business: Can someone tell me if Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124,
or some other source answers whether A050788 is known infinite, i.e., does c^3+1=a^3+b^3 infinitely often?
There is an infinite family of solutions given by (a,b,c) = (9n^3 + 1, 9n^4, 9n^4 + 3n). A050788 actually asks about x^3+y^3=z^3-1 with x < y < z; for that we can take (x,y,z) = (9n^3 - 1, 9n^4 - 3n, 9n^4) for n > 1. I extracted these solutions from Theorem 235 in Hardy & Wright; the result shown there is that all nontrivial rational solutions of x^3 + y^3 = u^3 + v^3 are given by x = r(1 - (a - 3b)(a^2 + 3b^2)) y = r((a + 3b)(a^2 + 3b^2) - 1) u = r((a + 3b) - (a^2 + 3b^2)^2) v = r((a^2 + 3b^2)^2 - (a - 3b)) where r,a,b are rational and r is not zero. Specializing to r = 1, b = n/2 and a = 3n/2 gives x = 1 y = 9n^3 - 1 u = 3n - 9n^4 v = 9n^4. The solutions given above are obtained by changing signs and moving cubes from one side of the equation to the other as necessary. Unfortunately, not all integral solutions are found so easily: the third value in A050788 corresponds to 135^3 + 138^3 = 172^3 - 1; this is not produced by such simple choices of r,a,b.
And does ceiling((a^n+b^n)^(1/n))^n - a^n - b^n < k have only finitely many solutions for n>3?
With a,b positive that would be my guess, but I don't expect a proof any time soon. -- Fred W. Helenius fredh@ix.netcom.com
this infinite family of solutions puts the kibosh on the way i like to think about equations like x^3 + y^3 = z^3: namely, are the cubes dense enough that (set of cubes) + (set of cubes) intersects (set of cubes) ? in a sense, flt says no. but, by adding 1 to each element on the right-hand side, you get an infinite set of examples where (set of cubes) + (set of cubes) intersects (1+set of cubes). weird. bob baillie --- Fred W. Helenius wrote:
At 07:08 AM 7/22/2008, rwg@sdf.lonestar.org wrote:
New business: Can someone tell me if Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124,
or some other source answers whether A050788 is known infinite, i.e., does c^3+1=a^3+b^3 infinitely often?
There is an infinite family of solutions given by (a,b,c) = (9n^3 + 1, 9n^4, 9n^4 + 3n). A050788 actually asks about x^3+y^3=z^3-1 with x < y < z; for that we can take (x,y,z) = (9n^3 - 1, 9n^4 - 3n, 9n^4) for n > 1.
I extracted these solutions from Theorem 235 in Hardy & Wright; the result shown there is that all nontrivial rational solutions of x^3 + y^3 = u^3 + v^3 are given by
x = r(1 - (a - 3b)(a^2 + 3b^2)) y = r((a + 3b)(a^2 + 3b^2) - 1) u = r((a + 3b) - (a^2 + 3b^2)^2) v = r((a^2 + 3b^2)^2 - (a - 3b))
where r,a,b are rational and r is not zero.
Specializing to r = 1, b = n/2 and a = 3n/2 gives
x = 1 y = 9n^3 - 1 u = 3n - 9n^4 v = 9n^4.
The solutions given above are obtained by changing signs and moving cubes from one side of the equation to the other as necessary. Unfortunately, not all integral solutions are found so easily: the third value in A050788 corresponds to 135^3 + 138^3 = 172^3 - 1; this is not produced by such simple choices of r,a,b.
And does ceiling((a^n+b^n)^(1/n))^n - a^n - b^n < k have only finitely many solutions for n>3?
With a,b positive that would be my guess, but I don't expect a proof any time soon.
How do you get inf ==== 2 i + 1 \ (sqrt(2) - 1) li (sqrt(2) - 1) - li (1 - sqrt(2)) = 2 > -------------------- = 2 2 / 2 ==== (2 i + 1) i = 0 sqrt(2) - 1 / 2 2 [ atanh(x) %pi log (sqrt(2) + 1) 2 I -------- dx = ---- - ----------------- ] x 8 2 / 0 with the usual dilog relations? Way(?) harder: "Shew that" 2 (d592) 4 li (- sqrt(1 - x ) + %i x + 1) 2 2 2 - li (2 %i x sqrt(1 - x ) + 2 x ) 2 is pure imaginary for -1<=x<=1.
Yes they've been done to death, but here are two minor remarks that came up in a discussion with a (talented) youngster: [...] if a sequence obeys a (homogeneous) linear recurrence of order <= n, then the determinant of an n+1 by n+1 band matrix formed from any 2n+1 consecutive elements = 0. E.g., the fibonaccis are a 2nd order recurrence, so c393) genmatrix(lambda([i,j],fib(integer+i+j-2)),3) [ fib(integer) fib(integer + 1) fib(integer + 2) ] [ ] (d393) [ fib(integer + 1) fib(integer + 2) fib(integer + 3) ] [ ] [ fib(integer + 2) fib(integer + 3) fib(integer + 4) ] (c394) ratsimp(fibtophi(det(%))) (d394) 0 Now suppose you know that the square triangular numbers are a 3rd order recurrence, and you only know the first four: 0, 1, 36, and 1225; and you want more. The matrix needs 2n+1=7 values--what to do? Think about what the recurrence could possibly do for negative values. Are the squares of -1, -6, and -35 equal to the triangulars of index -2, -9, and -50? Yes! So we actually know seven sequence elements, and we only need six: c387) genmatrix(lambda([i,j],[36,1,0,1,36,1225,x][i+j-1]),4) [ 36 1 0 1 ] [ ] [ 1 0 1 36 ] (d387) [ ] [ 0 1 36 1225 ] [ ] [ 1 36 1225 x ] (c388) sqrt(solve(det(%))) (d388) [sqrt(x) = 204] so 41616 is the next one. The formula for the nth, showing which square equals which triangular: 2 n 2 n (sqrt(2) + 1) (1 - sqrt(2)) 2 (---------------- - ----------------) = 4 sqrt(2) 4 sqrt(2) n n n n (sqrt(2) + 1) (sqrt(2) - 1) 2 (sqrt(2) + 1) (sqrt(2) - 1) 2 ((-------------- - --------------) + 1) (-------------- - --------------) 2 2 2 2 ---------------------------------------------------------------------------. 2 (Interesting computer algebra challenge: verify this identity noninductively.) I don't recall seeing the simple observation that this sequence is trivially constructed from the successive approximants to sqrt(2): Given a rational approximation a/b, the fraction 2 b/a errs on the opposite side, and their average would be the next Newton approximation. (Starting with a=b=1.) These form an exponentially sparse subsequence of the continued fraction (1+/2+/2+/2+/...) approximants resulting from taking the mediant, (a + 2 b)/(b + a), instead of the average. E.g., a 1 3 7 17 41 ... b 1 2 5 12 29 ... Then the square triangular numbers are just a^2 b^2 = binom(a^2+(n mod 2),2): (c607) define(inversetriang(n),rhs(solve(binomial(x+1,2) = n,x)[2])) sqrt(8 n + 1) - 1 (d607) inversetriang(n) := ----------------- 2 (c608) block([m : matrix([1,1],[1,0]),fancy_display : false], thru 9 do (print(m,(m[1,1]/m[2,1])^2,inversetriang((m[1,1]*m[2,1])^2)),m:m.matrix([2,1],[1,0])))$ [a a'] -1 [ ] a^2/b^2 triang (a^2 b^2) = a^2-(n+1 mod 2) [b b'] -------------------------------------------------------- [ 1 1 ] 1 [ ] - 1 [ 1 0 ] 1 [ 3 1 ] 9 [ ] - 8 [ 2 1 ] 4 [ 7 3 ] 49 [ ] -- 49 [ 5 2 ] 25 [ 17 7 ] 289 [ ] --- 288 [ 12 5 ] 144 [ 41 17 ] 1681 [ ] ---- 1681 [ 29 12 ] 841 [ 99 41 ] 9801 [ ] ---- 9800 [ 70 29 ] 4900 [ 239 99 ] 57121 [ ] ----- 57121 [ 169 70 ] 28561 [ 577 239 ] 332929 [ ] ------ 332928 [ 408 169 ] 166464 [ 1393 577 ] 1940449 [ ] ------- 1940449 [ 985 408 ] 970225 where [a a'] 1 1 2 1 n [ ] = ( )( ) and a'/b' is the previous a/b. [b b'] 1 0 1 0 --rwg
I don't recall seeing the simple observation that this sequence is trivially constructed from the successive approximants to sqrt(2): Given a rational approximation a/b, the fraction 2 b/a errs on the opposite side, and their average would be the next Newton approximation. (Starting with a=b=1.) These form an exponentially sparse subsequence of the continued fraction (1+/2+/2+/2+/...) approximants resulting from taking the mediant, (a + 2 b)/(b + a), instead of the average. E.g.,
a 1 3 7 17 41 ... b 1 2 5 12 29 ...
Then the square triangular numbers are just a^2 b^2 = binom(a^2+(n mod 2),2):
= (b[2n]/2)^2, where the nth a and b are a[n] and b[n]. I.e., b[2n] = 2 a[n] b[n].
(c607) define(inversetriang(n),rhs(solve(binomial(x+1,2) = n,x)[2]))
sqrt(8 n + 1) - 1 (d607) inversetriang(n) := ----------------- 2
(c608) block([m : matrix([1,1],[1,0]),fancy_display : false], thru 9 do (print(m,(m[1,1]/m[2,1])^2,inversetriang((m[1,1]*m[2,1])^2)),m:m.matrix([2,1],[1,0])))$
[a a'] -1 [ ] a^2/b^2 triang (a^2 b^2) = a^2-(n+1 mod 2) [b b'] --------------------------------------------------------
[ 1 1 ] 1 [ ] - 1 [ 1 0 ] 1
[ 3 1 ] 9 [ ] - 8 [ 2 1 ] 4
[ 7 3 ] 49 [ ] -- 49 [ 5 2 ] 25
[ 17 7 ] 289 [ ] --- 288 [ 12 5 ] 144
[ 41 17 ] 1681 [ ] ---- 1681 [ 29 12 ] 841
[ 99 41 ] 9801 [ ] ---- 9800 [ 70 29 ] 4900
[ 239 99 ] 57121 [ ] ----- 57121 [ 169 70 ] 28561
[ 577 239 ] 332929 [ ] ------ 332928 [ 408 169 ] 166464
[ 1393 577 ] 1940449 [ ] ------- 1940449 [ 985 408 ] 970225
where
[a a'] 1 1 2 1 n [ ] = ( )( ) and a'/b' is the previous a/b. [b b'] 1 0 1 0
--rwg
I don't recall seeing the simple observation that this sequence is trivially constructed from the successive approximants to sqrt(2): Given a rational approximation a/b, the fraction 2 b/a errs on the opposite side, and their average would be the next Newton approximation. (Starting with a=b=1.) These form an exponentially sparse subsequence of the continued fraction (1+/2+/2+/2+/...) approximants resulting from taking the mediant, (a + 2 b)/(b + a), instead of the average. E.g.,
a 1 3 7 17 41 ... b 1 2 5 12 29 ...
Then the square triangular numbers are just a^2 b^2 = binom(a^2+(n mod 2),2):
= (b[2n]/2)^2, where the nth a and b are a[n] and b[n]. I.e., b[2n] = 2 a[n] b[n].
I should've summarized this as A001110 = A001333^2 * A000129^2 = A000129[2n]^2/4 = binom(A001108,2). Mike Hirschhorn apparently found most or all of this in 1996, and refers me to ". . .Tom Beldon and Tony Gardiner, ``Triangular numbers and perfect squares'', The Mathematical Gazette, 2002, pp423--431, esp pp424--426. {This is a British publication for senior high--school students and their teachers.]" I still can't prove the polylog puzzles--I'm just detecting them with Rich's old lattice reducer. Ramanujan apparently found a similar relation between Li_2(1/3) and Li_2(-1/3). --rwg
Re: [math-fun] Presentation on floating point implementation Monday, July 28, 2008 9:16 PM From: "Mike Speciner" <ms@alum.mit.edu>
Those were nice. One of the slides reminded me that for solving quadratics (-b +- sqrt(b^2-4ac))/2a = 2c/(-b -+ sqrt(b^2-4ac)), so better precision for the two roots can be obtained by picking one root from each side so that both can have the sign of the sqrt same as -b. (And then it's easy to see the RHS root approaching -c/b as a->0.)
I don't remember where I first saw this trick, but I've certainly made use of it many times.
--ms
I think I first heard it from F J Corbato's 6.251 in '66 or '67. --rwg PS: Here are some sums I'm contributing to Steve Finch's Central Binomial Coeffs page http://algo.inria.fr/csolve/cbc.pdf, which needed rhss . inf ==== n 2 n + 1 \ (- 1) x > --------------------------- = / 2 ==== (2 n + 1) binomial(2 n, n) n = 0 2 2 2 x x x x 4 li (- sqrt(-- + 1) + - + 1) - li (x sqrt(-- + 1) - --). 2 4 2 2 4 2 Unfortunately, Mma can't integrate this/x dx without enormous coaching, resulting in a formula for inf ==== n 2 n + 1 \ (- 1) x > --------------------------- / 3 ==== (2 n + 1) binomial(2 n, n) n = 0 stretching many pages. [I've just got the non-alternating case down to a page.] The case x=2 is 3 2 log (sqrt(2) + 1) log(2) log (sqrt(2) + 1) - ----------------- - ------------------------ 3 2 2 1 3 log (2) log(sqrt(2) + 1) - 4 li (-------) log(sqrt(2) + 1) - -------------------------- 2 sqrt(2) 4 2 3 %pi log(sqrt(2) + 1) 1 + ----------------------- + 2 li (2 - sqrt(2)) + 2 li (1 - -------) 4 3 3 sqrt(2) 3 2 7 zeta(3) log (2) 7 %pi log(2) - --------- - ------- - ------------- ~ 1.88217914070106, 16 24 24 and I can't get it any simpler. (E.g., the ten terms satisfy no integer relation.) The x=1 case is similar, inf ==== n \ (- 1)
--------------------------- / 3 ==== (2 n + 1) binomial(2 n, n) n = 0
3 19 log (%phi) 2 %phi = ------------- - 2 log(2) log (%phi) - 4 li (----) log(%phi) 3 2 2 2 2 7 %pi log(%phi) 1 %phi - 4 log (2) log(%phi) + ---------------- - li (- - ----) 10 3 2 4 %phi 1 1 3 zeta(3) + 4 li (1 - ----) - 3 li (----) + 4 li (------) + --------- 3 2 3 %phi 3 2 %phi 10 ~ 0.98268607670276 There's no way to reduce the number of Li terms, but it's conceivable that transforming them into other Li terms could kill one or two of the Li-free terms. As you might also gather from math-fun, I simplified an enormous mess by non-rigorous means. FullSimplify's initial excretion included pages of terms like 2*I*Pi*ArcCsch[2]*(9*ArcCsch[2] - ArcSinh[3940598]) and (2/5)*Pi^2*(14*Log[2] + 17*Log[3 - Sqrt[5]] - Log[12360848946698171 + 5527939700884757*Sqrt[5]]). The latter comes from phi^77. inf ==== \ 1
--------------------------- / 3 ==== (2 n + 1) binomial(2 n, n) n = 0
= 4 Im(2 Li ((sqrt(3) - 2) %i + 1) + Li ((2 - sqrt(3)) %i)) 3 3 %pi li (4 sqrt(3) - 7) 2 - ---------------------- 6 3 %pi log(128 - 64 sqrt(3)) log(8 - 4 sqrt(3)) 7 %pi + -------------------------------------------- + ------ 6 72 ~ 1.0200208006525428 . This one's a pleasant surprise: inf ==== 2 n + 1 \ 2
--------------------------- / 3 ==== (2 n + 1) binomial(2 n, n) n = 0 2 3 1 - %i %i + 1 %pi log (2) %pi = 4 %i (li (------) - li (------)) - ----------- - ---- = 3 2 3 2 4 16
2 3 %pi log (2) 3 %pi = 4 %i (li (%i + 1) - li (1 - %i)) + ----------- + ------ 3 3 2 8 ~ 2.245380049461289, even shorter with IM or chi or Ti_3 notation.
By trivial bisection of the powerseries, t 4 4 theta (-, q) = theta (t, q ) + theta (t, q ) , 3 2 3 2 t 4 4 theta (-, q) = theta (t, q ) - theta (t, q ) 4 2 3 2 (The other two are perhaps not so simple.) Gene, Dick, George: How could you let me blather on like that? --rwg
Repeated integration by parts gives u k - 2 u / log(1 - t) log (-) [ t I ---------------------- dt ] t / 0 li (u) = - ----------------------------. k (k - 2)! Recklessly differentiating dk produces an integral that Mma expands into the termwise differentiated series (i.e., with a factor of -log n in the nth term). This series gives you (http://www.tweedledum.com/rwg/filds.dvi) t inf / / 2 2 [ [ s log(t + s ) I log(Gamma(x)) dx = I -------------- ds + t log(Gamma(t)) ] ] 2 %pi s / / %e - 1 1 0 2 2 t log(t) - t t 5 log(2 %pi) - %gamma + 1 zeta'(2) - ------------- + -- - ------------------------- - --------, 2 4 12 2 2 %pi which you can almost get from integrating Binet's second expansion of log Gamma(z) dz, but instead get mysterious integrals for some of the constants. Binet may have had this, but had no notation (nor popularity) for zeta'. For valuations of ilg(n/4 and n/6) (and hence the integral on the right), see www.tweedledum.com/rwg/idents.htm, (d1021) et seq. (near the end). The rhs integral (and Binet's) look suspiciously Abel-Plana, and might yield interesting series (or products!). With exp(ilg) (= "prodigal"(Gamma)), you can interpolate 1^1 2^2 ... n^n and 1! 2! ... n!, and derive "superStirling's" formulae. The reflection formula gives Catalan's constant. --rwg
rwg>This series gives you (http://gosper.org/filds.dvi) [...] Much nicer: Integrate from 0 and use Zeta'(-1): t inf / / 2 2 [ [ s log(t + s ) I log(Gamma(x)) dx = I -------------- ds + t log(Gamma(t)) ] ] 2 %pi s / / %e - 1 0 0 2 2 t log(t) t t - --------- + -- + - + Zeta'(-1). 2 4 2 rwg>For valuations of ilg(n/4 and n/6) (and hence the integral on the
right), see www.tweedledum.com/rwg/idents.htm, (d1021) et seq. (near the end).
Mma 6.0 can now do these (but not the infinite integral) in terms of (the superfluous) Log[Glaisher] instead of Zeta'[-1]. Use FunctionExpand if you get a PolyLog[-2,...] or a Derivative[1,0][Zeta][-1,...].
The rhs integral (and Binet's) look suspiciously Abel-Plana, and might yield interesting series (or product!).
Despite many stratagems, I can't get the sum to converge. --rwg ADROITLY DILATORY IDOLATRY
How do you get inf ==== 2 i + 1 \ (sqrt(2) - 1) li (sqrt(2) - 1) - li (1 - sqrt(2)) = 2 > -------------------- = 2 2 / 2 ==== (2 i + 1) i = 0
sqrt(2) - 1 / 2 2 [ atanh(x) %pi log (sqrt(2) + 1) 2 I -------- dx = ---- - ----------------- ] x 8 2 / 0
with the usual dilog relations?
Here's another: 2 2 1 1 %pi 3 log (%phi) li (-----) - li (- -----) = ---- - ------------. (%phi := golden ratio.) 2 3 2 3 12 2 %phi %phi (Yes, Henry, ln phi = asinh 1/2 .-) It's equivalent to 2 %phi 1 %phi 2 2 %pi li (----) + li (- - ----) = 2 log (%phi) - log (2) + ----. 2 2 2 2 2 12 How do you prove these?? They're obscuring drastic simplifications of huge expressions. GACK, look at this one! %phi 1 1 8 li (----) + 2 li (- 8 %phi - 4) - 12 li (-----) - li (-----) = 2 2 2 2 3 2 6 %phi %phi 2 2 2 %pi 35 log (%phi) - 40 log(2) log(%phi) - 4 log (2) + ---- 6 We're obviously missing a piece of technology to uncreate these things. Just now I got a twelve term relation with six different trilogs! Help. --rwg
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