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Comments:
<0> Right then.
<1> % Sum[Log[1/n^2]], {n, 0, Infinity}
<2> matteo: $Failed
<1> damn
<1> shouldn't it converge?
<1> ah, maybe starting from 1 :P
<1> % Sum[Log[1/n^2]], {n, 1, Infinity}
<2> matteo: $Failed
<3> % Sum[Log[1/n^2]], {n, 0, Infinity}]
<2> Cale: $Failed
<3> % Sum[Log[1/n^2], {n, 1, Infinity}]
<2> Cale: Sum[Log[n^(-2)], {n, 1, Infinity}]
<3> diverges
<1> why $failed?
<3> $Failed usually means that you had a syntax error
<3> you didn't close the sum bracket, or closed it in the wrong place rather
<1> % Sum[Log[1/n^2]], {n, 1, Infinity}
<2> matteo: $Failed
<3> that's a syntax error
<1> where?
<3> your ] is in the wrong place
<3> it should be at the end of the expression
<3> % Sum[Log[1/n^2], {n, 1, Infinity}]
<1> % Sum[Log[1/n^2], {n, 1, Infinity}]
<2> Cale: Sum[Log[n^(-2)], {n, 1, Infinity}]
<2> matteo: Sum[Log[n^(-2)], {n, 1, Infinity}]
<3> it refuses to evaluate that because it diverges
<1> but 1/n converges
<3> what?
<3> sum over n >= 1 of 1/n doesn't converge
<4> % Limit[ 1/(x^2) - CoTan(x)^2, x->0 ]
<2> FBKK: Infinity
<3> % Limit[ 1/(x^2) - Cot[x]^2, x->0 ]
<2> Cale: 2/3
<4> cale , thanks
<3> np
<5> % Limit[Log[1/n^2], n->Infinity]
<2> Olathe: -Infinity
<1> % Sum[1/n^2, {1,Infinity}]
<2> matteo: Sum[1/n^2, {1, Infinity}]
<1> % Sum[1/n^2, {n, 1,Infinity}]
<2> matteo: Pi^2/6
<1> ok, and this converges
<1> now add Log[] and it diverges
<5> % Limit[1/n^2, n->Infinity]
<2> Olathe: 0
<1> mm
<3> For a series to converge, it's a necessary condition for the limit of the terms to be 0.
<3> It's not sufficient, though.
<1> because Inf is so big that even adding some small value gives Inf?
<3> well, roughly, yeah
<1> ok
<3> it's because of the Archimedean property of the integers
<3> For any real number x, there is an integer n such that n > x.
<1> numbers are infinite?
<1> i was thinking that 4294967295 is the biggest
<3> so, suppose you have sum over n >= 0 of a, where a is some constant
<3> that's defined as the limit as k -> infinity of the sum over n = 0 to k of a
<3> which is the limit as k -> infinity of (k+1) a
<3> which supposing that a > 0, diverges, since for any value M, there will be some k for which k * a > M
<6> % Sum[Log[1+1/n^2], {n, 1, Infinity}]
<2> int-e: -LogGamma[1 - I] - LogGamma[1 + I]
<3> % Sum[Log[1+1/n^2], {n, 1, Infinity}] //N
<2> Cale: 1.3018463986037148 + 0.*I
<7> 0.*I?
<1> Cale: but, 1/x^2 is always bigger than Log[1/x^2]
<3> _llll_: The imaginary component is numerically zero.
<6> Log[1/x^2] is negative for x>1
<3> % Table[Log[1/x^2],{x,1,20}]
<2> Cale: {0, -Log[4], -Log[9], -Log[16], -Log[25], -Log[36], -Log[49], -Log[64], -Log[81], -Log[100], -Log[121], -Log[144], -Log[169], -Log[196], -Log[225], -Log[256], -Log[289], -Log[324], -Log[361], -
<2> Log[400]}
<1> but my series starts from 1
<6> :P
<3> % Table[Log[1/x^2],{x,1,20}] //N
<2> Cale: {0., -1.3862943611198906, -2.1972245773362196, -2.772588722239781, -3.2188758248682006, -3.58351893845611, -3.8918202981106265, -4.1588830833596715, -4.394449154672439, -4.605170185988092, -4.
<2> 795790545596741, -4.969813299576001, -5.1298987149230735, -5.278114659230517, -5.41610040220442, -5.545177444479562, -5.666426688112432, -5.780743515792329, -5.8888779583328805, -5.991464547107982}
<3> seems unlikely that the sum of those will converge :)
<1> to -Inf
<1> understood now
<1> % Sum[1/Log[n^2], {n, 1, Inf}]
<2> matteo: Sum[Log[n^2]^(-1), {n, 1, Inf}]
<1> % Sum[1/Log[n^2], {n, 1, Infinity}]
<2> matteo: Sum[Log[n^2]^(-1), {n, 1, Infinity}]
<3> the first term of that isn't defined anyway
<3> but it's not going to converge even without that
<3> % Sum[1/n^a, {n,1,Infinity}]
<2> Cale: Zeta[a]
<1> since Log[1] is 0 and 1/0 isn't defined?
<3> % Zeta[-1]
<2> Cale: -1/12
<3> hehe, 1 + 2 + 3 + ... = -1/12
<3> matteo: yeah
<1> % Sum[1/Log[n^2], {n, 2, Infinity}]
<2> matteo: Sum[Log[n^2]^(-1), {n, 2, Infinity}]
<1> again..
<3> still doesn't converge
<1> % Table[1/Log[n^2], {n, 2, 10}] //N
<2> matteo: {0.7213475204444817, 0.45511961331341866, 0.36067376022224085, 0.31066746727980593, 0.2790553132756236, 0.25694917118487537, 0.24044917348149392, 0.22755980665670933, 0.21714724095162588}
<3> 1/Log[n^2] > 1/n for large enough n
<6> % Sum[1/(n*Log[n^2]), {n, 2, Infinity}]
<6> even that doesn't converge
<2> int-e: Sum[1/(n*Log[n^2]), {n, 2, Infinity}]
<1> mm, Log[x^2] decreases too slowly
<3> yeah
<1> got it
<6> Log[n^2] = 2*Log[n] btw.
<1> i know
<6> so the exponent buys you nothing for convergence
<3> {1/n} is kind of right on the boundary of converging and diverging, in the sense that 1/n^a converges for every a > 1
<3> (the sum, of course)
<1> a >= 1
<3> no
<3> strictly greater than
<1> 1/n diverges too
<3> it diverges for a = 1
<6> yes, that's what Cale said
<3> and converges for larger a
<1> the x does matter?
<5> The x ?
<1> eg (1/(2*x^2)) converges too?
<8> what is the prefered way (fast and stable) to invert a 4x4 matrix ? I currently use SVD decomposition - is Cramer's rule faster ?
<5> Well, is the power on x greater than 1 ?
<1> y
<5> y ?
<1> Yes
<3> MatthiasM: I'd be very surprised if you could evaluate 8 4x4 determinants in less time :)
<3> er, 5 rather
<3> er, and that's per solution
<8> Cale: that means that SVD is faster then Cramer's ?
<3> I'd probably use Gauss-Jordan
<1> Det[] is O(n^3)?
<3> matteo: basically, it's really n^2.3 or n^2.7 or something like that if you use crazy algorithms, but the ordinary algorithms are n^3
<8> and n is 4 in case of 4x4 ?
<1> O(n^2.85)
<3> yeah
<6> http://www.intel.com/design/pentiumiii/sml/24504301.pdf ... intel said there that Cramer's rule is good, but they compared it to Gaussian elimination
<9> the record so far is n^2.37
<1> i p***es the algorithms and data structures test
<1> yes, 2^37
<3> 2.37
<1> mm, maybe that is for multiplications
<3> n^(2^37) would not be impressive :)
<6> I'll take n<=1.
<8> int-e: I'm aware of this document - but it uses ***embler which is not an option for me (java)
<1> Ln[7] (base 2)
<6> it has C code
<10> a nice trivia fact that way is that the permanent of a matrix is really hard to calculate
<3> int-e: That must be some usage of the term "Cramer's rule" of which I was previously unaware
<11> Hello.
<6> http://geosoft.no/software/matrix4x4/Matrix4x4.java.html has the same code in java
<1> % help::Log
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