[math-fun] refractory telescopy
I like to present convergent telescoping identities as infinite sums, e.g., inf ==== \ x x tan(x) = > (2 csc(------) - csc(--)), / n - 1 n ==== 2 2 n = 0 with a parameter (here x), two instances of which can be subtracted to give the sum over any range. But this example immediately suggests k - 1 ==== x \ x cot(--) - cot(x) = > csc(--), k / n 2 ==== 2 n = 0 which blows up in one direction and goes nuts in the other. Integrating, inf ==== | n | \ log(|cot(2 x)|) > ---------------- = - 2 (log(abs(sin(x))) + log(2)), / n ==== 2 n = 0 sort of reenabling the infinite series convention, but the lhs is undefined for x = dyadic rational * pi, a dense set! There may also be x for which it doesn't converge. Further massaging finally gives the very convergent product, 2 x %e coth(x) sqrt(coth(2 x)) sqrt(sqrt(coth(4 x))) . . . = ---------- . 2 4 sinh (x) --rwg
Fourier series for Snowflake curve. E.g., www.tweedledum.com/rwg/cog.htm
Without the |abs|, the identity I gave yesterday is inf ==== n \ log(cot(2 x)) > -------------- = 2 %i x - 2 log(2 sin(x)), 0 < x < 2 pi, / n ==== 2 n = 0 or more generally, inf ==== n \ log(cot(2 x)) > -------------- = %i (%pi - 2 atan(cot(x))) / n ==== 2 n = 0 - 2 log(2 abs(sin(x))) . I'm surprised at the simplicity of the imagpart, whose source is the "random" subset of the cots which go negative as 2^n x galumphs off to +-infinity. rwg>Further massaging finally gives the very quadratically
convergent product, E.g.,
1 inf -- /===\ n | | 2 n - 2 x 2 (d5) | | tanh (2 x) = (1 - %e ) | | n = 0 (c6) apply_nouns(subst(3,inf,%)) 1/4 1/8 (d6) tanh(x) sqrt(tanh(2 x)) tanh (4 x) tanh (8 x) = - 2 x 2 (1 - %e ) (c10) taylor(trigexpand(subst(log(y),x,d6)),y,inf,33) 2 1 1 2 1 (d10)/T/ 1 - -- + -- + ----- + . . . = 1 - -- + -- + . . . 2 4 32 2 4 y y 8 y y y --rwg PYROCHEMICAL MICROCEPHALY
* R. William Gosper <rwg@osots.com> [Apr 14. 2005 22:18]:
Fourier series for Snowflake curve. E.g., www.tweedledum.com/rwg/cog.htm [...]
1/4 1/8 (d6) tanh(x) sqrt(tanh(2 x)) tanh (4 x) tanh (8 x) =
- 2 x 2 (1 - %e )
setting x=arctanh(y) gives prod( f@@n(y)^(1/2^n), n=0..infinity) == (y+1)^2/4 f@@0= y f@@1= 2 y + 1 ------ 2 y f@@2= 4 2 y + 6 y + 1 ------------- 3 4 y + 4 y f@@3= 8 6 4 2 y + 28 y + 70 y + 28 y + 1 ------------------------------ 7 5 3 8 y + 56 y + 56 y + 8 y f@@n = the iteration for sqrt(1) of order 2^n Using f((1+e)/(1-e))=(1+e^2)/(1-e^2) one finds prod( ((1+e^(2^k))/(1-e^(2^k)))^(1/2^k) ) == 1/(1-e)^2 (see page 350 of today's fxtbook draft). The first nontrivial sum/prod expressions with iterations I have seen so far. Nice!
[...]
All the best, jj -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
rwg>I like to present convergent telescoping identities as infinite sums, e.g.,
... But this example immediately suggests
k - 1 ==== x \ x cot(--) - cot(x) = > csc(--), k / n 2 ==== 2 n = 0
which blows up in one direction and goes nuts in the other. Duh, inf ==== 2 \ n ------ = > csch(2 x), x / e - 1 ==== n = 0
pretty much the log derivative of Joerg's observation. Test: apply_nouns(subst([x=.69105d0,inf=5],%)) 2.00841519141928d0 = 2.00841519141928d0 (With quadratic convergence, infinity = 5.) jj>[...] f@@n = the iteration for sqrt(1) of order 2^n Note that if we follow each Newton step with a true sqrt, 1 a + ------ n - 1 a n - 1 a = sqrt(---------------), n 2 with a a = sqrt(-), we get 0 b agm(a, b) a a a . . . = ---------. 0 1 2 b --rwg GRABBIEST BAGBITERS (Peter Samson's ancient PDP-1 music compiler used the error code agm to signify: Argument Greedily Masticates.)
* R. William Gosper <rwg@osots.com> [Apr 20. 2005 20:46]:
[...]
Note that if we follow each Newton step with a true sqrt,
1 a + ------ n - 1 a n - 1 a = sqrt(---------------), n 2
with a a = sqrt(-), we get 0 b
agm(a, b) a a a . . . = ---------. 0 1 2 b
= agm(a/b, 1) =: agm(z, 1) = agm((1+z)/2, sqrt(z)) = = sqrt(z)*agm((1+sqrt(z)^2)/(2*sqrt(z), 1) which leads to the slightly more obvious a_0 = a/b, a_{n+1}=(a_n+1)/(2*sqrt(a_n)) sqrt( a_0 * a_1 * ...) = agm(a/b, 1) I tried a_{n+1}=(a_n+1)/(2*a_n) with a_0 = (1+x)/(1-x) and got the generating function for sequence 123: G.F.: prod(k=0,inf,(1+x^(2^k))/(1-x^(2^k)) G.F.: prod(k=0,inf,(1+x^(2^k))^(k+1))/(1-x) G.F.: prod(k=0,inf,(1+x^(2^k))^(k+2)) (Submitted but not yet in). -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
participants (2)
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Joerg Arndt -
R. William Gosper